JEE Quadratic Equations Questions

The quadratic $$x^{2}+x+1=0$$ has roots that are the non-real cube roots of unity.
Let one root be $$\alpha$$. Then

$$\alpha^{2}+\alpha+1 = 0 \;\; \Longrightarrow \;\; \alpha^{3}=1,\; \alpha\neq 1$$

Because $$\alpha^{3}=1$$, all integral powers of $$\alpha$$ repeat every three steps:
$$\alpha^{k}= \begin{cases} 1 & \text{if } k\equiv 0\pmod{3}\\ \alpha & \text{if } k\equiv 1\pmod{3}\\ \alpha^{2} & \text{if } k\equiv 2\pmod{3} \end{cases}$$

Also $$\alpha^{-1}=\alpha^{2},\;\alpha^{-2}=\alpha,\;\alpha^{-3}=1$$, so $$\alpha^{-k}$$ follows the same cycle.

Define $$T_k=\left(\alpha^{k}+\frac{1}{\alpha^{k}}\right)^{2}$$. Evaluate $$T_k$$ for the three possible residues of $$k$$ modulo $$3$$:

$$\alpha^{k}=1,\;\frac{1}{\alpha^{k}}=1 \quad\Rightarrow\quad T_k=(1+1)^{2}=4$$

$$\alpha^{k}=\alpha,\;\frac{1}{\alpha^{k}}=\alpha^{2} \quad\Rightarrow\quad \alpha+\alpha^{2}=-(1) \quad(\text{from } \alpha^{2}+\alpha+1=0)$$ $$T_k=(-1)^{2}=1$$

$$\alpha^{k}=\alpha^{2},\;\frac{1}{\alpha^{k}}=\alpha \quad\Rightarrow\quad \alpha^{2}+\alpha = -1$$ $$T_k=(-1)^{2}=1$$

Thus the sequence $$T_1,T_2,T_3,\dots$$ is periodic with cycle $$\{1,1,4\}$$, whose sum is

$$1+1+4 = 6$$

Let $$n=3q+r$$ where $$r\in\{0,1,2\}$$. Then

$$\sum_{k=1}^{n} T_k = 6q + \begin{cases} 0 & (r=0)\\ 1 & (r=1)\\ 1+1=2 & (r=2) \end{cases}$$

The given condition is

$$6q + \{0,1,2\}=20$$

Check the three possibilities:
  • $$6q=20$$ (impossible)  
  • $$6q+1=20\;\Rightarrow\;6q=19$$ (not an integer)  
  • $$6q+2=20\;\Rightarrow\;6q=18\;\Rightarrow\;q=3$$

The valid case is $$q=3,\;r=2$$. Hence

$$n = 3q + r = 3\cdot 3 + 2 = 11$$

Therefore, the required value of $$n$$ is $$\mathbf{11}$$.

We are given the equation $$a(b-c)x^2 + b(c-a)x + c(a-b) = 0$$ with equal roots, $$a + c = 15$$, and $$b = \dfrac{36}{5}$$.

Substituting $$x = 1$$:

$$a(b-c) + b(c-a) + c(a-b) = ab - ac + bc - ab + ac - bc = 0$$

So $$x = 1$$ is always a root.

$$1 \times 1 = \frac{c(a-b)}{a(b-c)}$$

$$a(b - c) = c(a - b)$$

$$b(a + c) = 2ac$$:

$$\frac{36}{5} \times 15 = 2ac$$

$$\frac{540}{5} = 2ac$$

$$108 = 2ac$$

$$ac = 54$$

$$a^2 + c^2 = (a + c)^2 - 2ac 

= 15^2 - 2(54) = 117$$

We need to find the interval $$(\alpha, \beta)$$ of values of $$a$$ for which $$5x^3 - 15x - a = 0$$ has three distinct real roots, then compute $$\beta - 2\alpha$$.

$$5x^3 - 15x = a$$, i.e., we need the horizontal line $$y = a$$ to intersect the curve $$y = 5x^3 - 15x$$ at three distinct points.

Differentiate: $$f'(x) = 15x^2 - 15 = 15(x^2 - 1)$$.

Setting $$f'(x) = 0$$: $$x^2 - 1 = 0$$, so $$x = \pm 1$$.

$$f''(x) = 30x$$.

At $$x = -1$$: $$f''(-1) = -30 < 0$$, so $$x = -1$$ is a local maximum.

At $$x = 1$$: $$f''(1) = 30 > 0$$, so $$x = 1$$ is a local minimum.

Local maximum: $$f(-1) = 5(-1)^3 - 15(-1) = -5 + 15 = 10$$.

Local minimum: $$f(1) = 5(1)^3 - 15(1) = 5 - 15 = -10$$.

The cubic $$y = f(x)$$ has a local max of 10 and a local min of $$-10$$. For the line $$y = a$$ to cut the curve at exactly three distinct points, $$a$$ must lie strictly between the local minimum and local maximum values:

$$-10 < a < 10$$

Therefore $$\alpha = -10$$ and $$\beta = 10$$.

$$\beta - 2\alpha = 10 - 2(-10) = 10 + 20 = 30$$

The correct answer is 30.

For the quadratic $$ (1-a)x^{2}+2(a-3)x+9=0 $$ let

$$A = 1-a, \; B = 2(a-3), \; C = 9.$$

For both roots to be real and positive we need three conditions:

1. Real roots: discriminant $$\Delta \ge 0$$.
2. Positive product: $$P = \dfrac{C}{A} \gt 0$$.
3. Positive sum: $$S = -\dfrac{B}{A} \gt 0.$$

Step 1 Discriminant
$$\Delta = B^{2}-4AC = \left[2(a-3)\right]^{2}-4(1-a)(9).$$

Simplify:

$$\Delta = 4(a-3)^{2}-36(1-a) = 4\big[(a-3)^{2}-9(1-a)\big]$$ $$= 4\big[a^{2}-6a+9-9+9a\big] = 4\big[a^{2}+3a\big] = 4a(a+3).$$

Thus $$\Delta \ge 0 \Longrightarrow a(a+3) \ge 0 \Longrightarrow a \le -3 \; \text{or} \; a \ge 0.$$

Step 2 Product of roots
$$P = \dfrac{C}{A} = \dfrac{9}{1-a}.$$

For $$P \gt 0$$ the denominator must be positive:

$$1-a \gt 0 \Longrightarrow a \lt 1.$$

Step 3 Sum of roots
$$S = -\dfrac{B}{A} = -\dfrac{2(a-3)}{1-a} = \dfrac{2(a-3)}{a-1}.$$

We need $$S \gt 0$$, so

$$\dfrac{a-3}{a-1} \gt 0.$$

The critical points are $$a = 1$$ and $$a = 3$$. Using a sign chart:

• For $$a \lt 1$$, both $$a-3$$ and $$a-1$$ are negative ⇒ ratio positive.
• For $$1 \lt a \lt 3$$, the ratio is negative.
• For $$a \gt 3$$, both factors are positive ⇒ ratio positive.

Hence $$S \gt 0$$ when $$a \lt 1$$ or $$a \gt 3$$ (note $$a \ne 1$$).

Step 4 Combine all conditions

• From product: $$a \lt 1.$$
• From sum: also satisfied for $$a \lt 1.$$
• From discriminant: $$a \le -3$$ or $$a \ge 0.$$
Intersecting with $$a \lt 1$$ gives

$$a \in (-\infty,-3] \;\; \cup \;\; [0,1).$$

Step 5 Identify $$\alpha,\beta,\gamma$$
The required set is given as $$(-\infty,-\alpha] \cup [\beta,\gamma).$$ Matching, we get $$\alpha = 3,\; \beta = 0,\; \gamma = 1.$

Step 6 Compute the expression
$$2$$\alpha + \beta + \gamma$$ = 2(3) + 0 + 1 = 7.$$

Therefore, the required value is $$\boxed{7}.$$

Given $$f(x) = 2x^3 - 9ax^2 + 12a^2x + 1$$ with $$a > 0$$. Differentiating: $$f'(x) = 6x^2 - 18ax + 12a^2 = 6(x^2 - 3ax + 2a^2) = 6(x - a)(x - 2a)$$. The critical points are $$x = a$$ and $$x = 2a$$.

To classify these, we use the second derivative $$f''(x) = 12x - 18a$$. At $$x = a$$: $$f''(a) = 12a - 18a = -6a$$. Since $$a > 0$$, we have $$f''(a) < 0$$, confirming a local maximum at $$x = a$$. At $$x = 2a$$: $$f''(2a) = 24a - 18a = 6a > 0$$, confirming a local minimum at $$x = 2a$$.

We are told the local maximum occurs at $$x = \alpha$$ and the local minimum at $$x = \alpha^2$$. So $$\alpha = a$$ and $$\alpha^2 = 2a$$. Substituting $$a = \alpha$$ into the second equation: $$\alpha^2 = 2\alpha$$, giving $$\alpha(\alpha - 2) = 0$$. Since $$\alpha = a > 0$$, we get $$\alpha = 2$$, and therefore $$\alpha^2 = 4$$.

The required equation has roots $$2$$ and $$4$$. Their sum is $$6$$ and product is $$8$$, so the equation is $$x^2 - 6x + 8 = 0$$, which is $$\boxed{\text{Option (A)}}$$.

We are given that $$\alpha, \beta$$ are roots of $$x^2 + \sqrt{2}x - 8 = 0$$ and $$U_n = \alpha^n + \beta^n$$, and we wish to find $$\frac{U_{10} + \sqrt{2}U_9}{2U_8}$$.

Since $$\alpha$$ satisfies $$\alpha^2 + \sqrt{2}\alpha - 8 = 0$$, it follows that $$\alpha^2 = -\sqrt{2}\alpha + 8$$, and multiplying both sides by $$\alpha^{n-2}$$ gives $$\alpha^n = -\sqrt{2}\alpha^{n-1} + 8\alpha^{n-2}$$. A similar argument for $$\beta$$ yields $$\beta^n = -\sqrt{2}\beta^{n-1} + 8\beta^{n-2}$$, and adding these two results establishes the recurrence $$U_n = -\sqrt{2}U_{n-1} + 8U_{n-2}$$.

Applying this recurrence for $$n = 10$$ gives $$U_{10} = -\sqrt{2}U_9 + 8U_8$$. Therefore, $$U_{10} + \sqrt{2}U_9 = (-\sqrt{2}U_9 + 8U_8) + \sqrt{2}U_9 = 8U_8$$.

It follows that $$\frac{U_{10} + \sqrt{2}U_9}{2U_8} = \frac{8U_8}{2U_8} = 4$$. The answer is 4.

Let $$\alpha - 1 = a^2$$ and $$\beta - 1 = b^2$$. Then $$\alpha = a^2+1$$ and $$\beta = b^2+1$$.

From $$x^2 - 70x + \lambda = 0$$:

• $$\alpha + \beta = 70 \implies a^2 + b^2 = 68$$

• $$\alpha\beta = \lambda \implies a^2b^2 + 69 = \lambda$$

The target expression simplifies to:

$$\frac{(a+b)(a^2b^2 + 104)}{|a^2-b^2|} = \frac{a^2b^2+104}{|a-b|} = \frac{a^2b^2+104}{\sqrt{68-2ab}}$$

 Let $$ab = t$$:

$$\frac{t^2+104}{\sqrt{68-2t}} = 60 \implies t = 16 \text{ works perfectly } \left(\frac{256+104}{\sqrt{36}} = \frac{360}{6} = 60\right)$$

We need to find all natural number solutions $$(x, y)$$ of $$x^2 - 2^y = 2023$$.

This gives $$x^2 = 2023 + 2^y$$.

Check small values of y: for $$y = 1$$: $$x^2 = 2023 + 2 = 2025 = 45^2$$. ✓ So $$(45, 1)$$ is a solution.

For $$y = 2$$: $$x^2 = 2027$$, not a perfect square.

For $$y = 3$$: $$x^2 = 2031$$, not a perfect square.

For $$y = 4$$: $$x^2 = 2039$$, not a perfect square.

For $$y = 5$$: $$x^2 = 2055$$, not a perfect square.

For $$y = 6$$: $$x^2 = 2087$$, not a perfect square.

For $$y = 7$$: $$x^2 = 2151$$, not a perfect square.

For $$y = 8$$: $$x^2 = 2279$$, not a perfect square.

For $$y = 9$$: $$x^2 = 2535$$, not a perfect square.

For $$y = 10$$: $$x^2 = 3047$$, not a perfect square.

For $$y = 11$$: $$x^2 = 4071$$, not a perfect square.

For large y, use modular arithmetic: for $$y \geq 2$$: $$x^2 \equiv 2023 \pmod{4}$$. Since $$2023 = 505 \times 4 + 3$$, we get $$x^2 \equiv 3 \pmod{4}$$. But squares are $$\equiv 0$$ or $$1 \pmod{4}$$, which is a contradiction.

So no solutions exist for $$y \geq 2$$.

Compute the answer: the only solution is $$(x, y) = (45, 1)$$.

$$ \sum_{(x,y) \in C} (x + y) = 45 + 1 = 46 $$

The answer is $$\boxed{46}$$.

Let $$P(x) = 4x^4 + 8x^3 - 17x^2 - 12x + 9 = 4(x-x_1)(x-x_2)(x-x_3)(x-x_4)$$.

The expression we need is $$\prod (x_i^2 + 4)$$, which can be written as $$\prod (x_i + 2i)(x_i - 2i)$$.

Note that $$P(2i) = 4(2i-x_1)(2i-x_2)(2i-x_3)(2i-x_4)$$.

 Similarly, $$P(-2i) = 4(-2i-x_1)(-2i-x_2)(-2i-x_3)(-2i-x_4)$$.

5The product is $$\frac{P(2i) \cdot P(-2i)}{16}$$.

$$P(2i) = 4(16) + 8(-8i) - 17(-4) - 12(2i) + 9 = 64 - 64i + 68 - 24i + 9 = 141 - 88i$$.

$$P(-2i) = 141 + 88i$$.

$$P(2i)P(-2i) = 141^2 + 88^2 = 19881 + 7744 = 27625$$.

Product $$= \frac{27625}{16} = \frac{125}{16}m \implies 125m = 27625 \implies \mathbf{m = 221}$$

$$x(x^2 + 3|x| + 5|x-1| + 6|x-2|) = 0$$.

Either $$x = 0$$ or $$x^2 + 3|x| + 5|x-1| + 6|x-2| = 0$$.

For the second factor: all terms $$x^2, 3|x|, 5|x-1|, 6|x-2|$$ are non-negative. The sum equals zero only if each term is zero simultaneously, which is impossible ($$|x| = 0$$ means $$x = 0$$, but then $$|x-1| = 1 \neq 0$$).

So the only solution is $$x = 0$$. Number of real solutions = 1.

Therefore, the answer is $$\boxed{1}$$.

Let $$y = x+1$$, then $$x = y-1$$. The equation becomes:

$$|(y-1)(y+1)| - 5|y| - 1 = 0 \implies |y^2 - 1| - 5|y| - 1 = 0$$.

Since this is symmetric ($$y$$ and $$-y$$ give same result), let $$t = |y| \ge 0$$:

$$|t^2 - 1| - 5t - 1 = 0$$.

• Case 1: $$t \ge 1$$

$$t^2 - 1 - 5t - 1 = 0 \implies t^2 - 5t - 2 = 0$$.

$$t = \frac{5 \pm \sqrt{25 + 8}}{2} = \frac{5 + \sqrt{33}}{2}$$ (only positive root).

This gives 2 values for $$y$$ (and thus 2 for $$x$$).

• Case 2: $$0 \le t < 1$$

$$-(t^2 - 1) - 5t - 1 = 0 \implies -t^2 + 1 - 5t - 1 = 0 \implies t^2 + 5t = 0$$.

$$t = 0, -5$$. Since $$t = |y|$$, we take $$t = 0$$.

This gives 1 value for $$y$$ ($$y=0$$, so $$x=-1$$).

Total distinct roots: $$2 + 1 = 3$$.

Correct Answer: 3

The polynomial has the form $$f(x)=x^{4}+ax^{3}+bx^{2}+c$$ with real coefficients.

First use the given value at $$x=1$$:
$$f(1)=1+a+b+c=-9 \;\;\Longrightarrow\;\; a+b+c=-10 \;.-(1)$$

The derivative of $$f(x)$$ is
$$f'(x)=4x^{3}+3ax^{2}+2bx.$$

We are told that $$i\sqrt{3}$$ is a root of $$f'(x)=0$$. Because the coefficients of $$f'(x)$$ are real, its complex conjugate $$-i\sqrt{3}$$ is also a root. Hence the cubic dividend of $$f'(x)$$ must contain the factor $$x(x^{2}+3)$$:

$$f'(x)=4x^{3}+3ax^{2}+2bx = 4\,x\,(x^{2}+3).$$

Expanding the right-hand side gives $$4x^{3}+12x.$$ Match corresponding coefficients:

• Coefficient of $$x^{3}$$: already $$4$$ on both sides.
• Coefficient of $$x^{2}$$: $$3a=0\;\Longrightarrow\;a=0.$$
• Coefficient of $$x$$: $$2b=12\;\Longrightarrow\;b=6.$$

Substitute $$a=0,\;b=6$$ into equation $$(1)$$ to determine $$c$$:

$$0+6+c=-10\;\Longrightarrow\;c=-16.$$

Thus $$f(x)=x^{4}+6x^{2}-16.$$

Solve $$f(x)=0$$. Put $$y=x^{2}$$ to obtain the quadratic

$$y^{2}+6y-16=0.$$

Discriminant $$\Delta=6^{2}+4\!\cdot\!16=100,$$ so

$$y=\frac{-6\pm10}{2}\; \Longrightarrow\; y=2 \;\text{or}\; y=-8.$$

Hence

$$x^{2}=2\;\;\Longrightarrow\;\;x=\pm\sqrt{2},$$ $$x^{2}=-8\;\Longrightarrow\;\;x=\pm\,2i\sqrt{2}.$$

The four roots are therefore $$\alpha_{1}= \sqrt{2},\;\alpha_{2}=-\sqrt{2},\;\alpha_{3}= 2i\sqrt{2},\;\alpha_{4}=-2i\sqrt{2}.$$

Compute the sum of the squared moduli:

$$|\sqrt{2}|^{2}=2,\qquad|-\sqrt{2}|^{2}=2,$$ $$|2i\sqrt{2}|^{2}=(2\sqrt{2})^{2}=8,\qquad|-2i\sqrt{2}|^{2}=8.$$

Adding them gives $$|\alpha_{1}|^{2}+|\alpha_{2}|^{2}+|\alpha_{3}|^{2}+|\alpha_{4}|^{2}=2+2+8+8=20.$$

Therefore the required value is 20.

We need to find the set $$S$$ and then determine the maximum value of $$\beta$$.

$$\log_2\left(9^{2\alpha-4} + 13\right) - \log_2\left(\frac{5}{2} \cdot 3^{2\alpha-4} + 1\right) = 2$$

$$\frac{9^{2\alpha-4} + 13}{\frac{5}{2} \cdot 3^{2\alpha-4} + 1} = 4$$

Let $$t = 3^{2\alpha - 4}$$, so $$9^{2\alpha-4} = t^2$$:

$$t^2 + 13 = 4\left(\frac{5t}{2} + 1\right) = 10t + 4$$

$$t^2 - 10t + 9 = 0 \implies (t-1)(t-9) = 0$$

$$t = 1 \Rightarrow 3^{2\alpha-4} = 3^0 \Rightarrow \alpha = 2$$

$$t = 9 \Rightarrow 3^{2\alpha-4} = 3^2 \Rightarrow \alpha = 3$$

So $$S = \{2, 3\}$$.

$$\sum_{\alpha \in S}\alpha = 2 + 3 = 5, \quad \left(\sum_{\alpha \in S}\alpha\right)^2 = 25$$

$$\sum_{\alpha \in S}(\alpha + 1)^2 = 3^2 + 4^2 = 9 + 16 = 25$$

The equation becomes:

$$x^2 - 2(25)x + 25\beta = 0 \implies x^2 - 50x + 25\beta = 0$$

Discriminant $$\geq 0$$:

$$2500 - 100\beta \geq 0 \implies \beta \leq 25$$

The maximum value of $$\beta$$ is $$25$$.

We need to find the product of all possible values of $$a$$ given that $$\alpha, \beta$$ are roots of $$x^2 + 60^{1/4}x + a = 0$$ and $$\alpha^4 + \beta^4 = -30$$.

By Vieta’s formulas for the equation $$x^2 + 60^{1/4}x + a = 0$$, we have $$\alpha + \beta = -60^{1/4}$$ and $$\alpha\beta = a$$.

To express $$\alpha^4 + \beta^4$$ in terms of $$a$$, note that
$$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \sqrt{60} - 2a$$
and
$$\alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2(\alpha\beta)^2 = (\sqrt{60} - 2a)^2 - 2a^2 = 60 - 4a\sqrt{60} + 4a^2 - 2a^2 = 2a^2 - 4a\sqrt{60} + 60$$.

Since $$\alpha^4 + \beta^4 = -30$$, we set $$2a^2 - 4a\sqrt{60} + 60 = -30$$, which simplifies to $$2a^2 - 4a\sqrt{60} + 90 = 0$$ and then to $$a^2 - 2\sqrt{60}\,a + 45 = 0$$.

For the quadratic $$a^2 - 2\sqrt{60}\,a + 45 = 0$$, Vieta’s formulas show that the product of the roots is $$\dfrac{45}{1} = 45$$.

The required product of all possible values of $$a$$ is $$\boxed{45}$$.

Find $$\alpha_1\alpha_2 - \alpha_3\alpha_4 + \alpha_5\alpha_6$$ where $$\alpha_1, \ldots, \alpha_7$$ are roots of $$x^7 + 3x^5 - 13x^3 - 15x = 0$$ ordered by decreasing absolute value.

We factor the equation as $$x(x^6 + 3x^4 - 13x^2 - 15) = 0$$ and, letting $$u = x^2$$, obtain $$u^3 + 3u^2 - 13u - 15 = 0$$. Testing $$u = 3$$ gives $$27 + 27 - 39 - 15 = 0$$, so the cubic factors as $$(u-3)(u^2 + 6u + 5) = (u-3)(u+1)(u+5) = 0$$, yielding $$u = 3, -1, -5$$.

Therefore, $$x^2 = 3$$ implies $$x = \pm\sqrt{3}$$ (so $$|\alpha| = \sqrt{3}$$); $$x^2 = -1$$ implies $$x = \pm i$$ (so $$|\alpha| = 1$$); $$x^2 = -5$$ implies $$x = \pm i\sqrt{5}$$ (so $$|\alpha| = \sqrt{5}$$); and also $$x = 0$$ (so $$|\alpha| = 0$$).

Ordering the roots by decreasing absolute value gives $$|\alpha_1| = |\alpha_2| = \sqrt{5}$$, $$|\alpha_3| = |\alpha_4| = \sqrt{3}$$, $$|\alpha_5| = |\alpha_6| = 1$$, and $$|\alpha_7| = 0$$.

It follows that $$\alpha_1\alpha_2 = (i\sqrt{5})(-i\sqrt{5}) = 5$$, $$\alpha_3\alpha_4 = (\sqrt{3})(-\sqrt{3}) = -3$$, and $$\alpha_5\alpha_6 = (i)(-i) = 1$$.

Hence $$\alpha_1\alpha_2 - \alpha_3\alpha_4 + \alpha_5\alpha_6 = 5 - (-3) + 1 = 9$$. The answer is $$\boxed{9}$$.

Let $$a$$ and $$b$$ be the two roots of $$x^{2}-7x-1=0$$.
Define the symmetric power sum $$S_{n}=a^{n}+b^{n}$$ for any positive integer $$n$$.

Case 1: Deriving a recurrence for $$S_{n}$$
Because $$a$$ and $$b$$ satisfy $$x^{2}-7x-1=0$$, we have $$a^{2}=7a+1$$ and $$b^{2}=7b+1$$.
Multiply these identities by $$a^{\,n-2}$$ and $$b^{\,n-2}$$ respectively (for $$n\ge 2$$):
$$a^{n}=7a^{\,n-1}+a^{\,n-2},\quad b^{n}=7b^{\,n-1}+b^{\,n-2}$$
Adding the two equalities gives the recurrence relation

$$S_{n}=7S_{n-1}+S_{n-2}\qquad\text{for }n\ge 2\;$$ $$-(1)$$

Case 2: Expressing $$S_{21}$$ in terms of lower sums
Use the recurrence twice:

$$S_{21}=7S_{20}+S_{19}\qquad\;$$ (from $$(1)$$ with $$n=21$$)
$$S_{20}=7S_{19}+S_{18}\qquad\;$$ (from $$(1)$$ with $$n=20$$)

Substitute the second equation into the first:

$$S_{21}=7\bigl(7S_{19}+S_{18}\bigr)+S_{19}=49S_{19}+7S_{18}+S_{19}=50S_{19}+7S_{18}$$ $$-(2)$$

Case 3: Expressing $$S_{17}$$ in terms of $$S_{19}$$ and $$S_{18}$$
Apply $$(1)$$ with $$n=19$$:

$$S_{19}=7S_{18}+S_{17}\;\;\Longrightarrow\;\;S_{17}=S_{19}-7S_{18}$$ $$-(3)$$

Case 4: Forming the required numerator
Add $$(2)$$ and $$(3)$$:

$$S_{21}+S_{17}=(50S_{19}+7S_{18})+(S_{19}-7S_{18})=51S_{19}$$ $$-(4)$$

Case 5: Computing the desired expression
The denominator is simply $$S_{19}$$, hence

$$\frac{a^{21}+b^{21}+a^{17}+b^{17}}{a^{19}+b^{19}} =\frac{S_{21}+S_{17}}{S_{19}} =\frac{51S_{19}}{S_{19}} =51$$

Therefore, the required value is $$\mathbf{51}$$.

The two quadratics $$x^2 - 5\alpha x + 1 = 0$$ and $$x^2 - \alpha x - 5 = 0$$ are assumed to share a common real root $$p$$, and for $$\alpha\gt 0$$ this root is given in the form $$\frac{3}{\sqrt{2\beta}}\,. $$ To determine $$\beta$$, note that $$p$$ must satisfy $$p^2 - 5\alpha p + 1 = 0 \quad(1)$$ and $$p^2 - \alpha p - 5 = 0 \quad(2)\,. $$

Subtracting equation (2) from equation (1) gives $$(p^2 - 5\alpha p + 1) - (p^2 - \alpha p - 5) = 0\,, $$ so $$-4\alpha p + 6 = 0\quad\Longrightarrow\quad p = \frac{6}{4\alpha} = \frac{3}{2\alpha}\,. $$

Substituting $$p = \tfrac{3}{2\alpha}$$ back into equation (2) yields $$\Bigl(\frac{3}{2\alpha}\Bigr)^2 - \alpha\cdot\frac{3}{2\alpha} - 5 = 0\,, $$ which simplifies to $$\frac{9}{4\alpha^2} - \frac{3}{2} - 5 = 0\quad\Longrightarrow\quad \frac{9}{4\alpha^2} = \frac{13}{2}\,. $$ Hence $$\alpha^2 = \frac{9}{4\cdot\frac{13}{2}} = \frac{9}{26}\,, $$ and since $$\alpha\gt 0$$, $$\alpha = \frac{3}{\sqrt{26}}\,. $$

Writing this in the form $$\frac{3}{\sqrt{2\beta}}$$ gives $$\alpha = \frac{3}{\sqrt{26}} = \frac{3}{\sqrt{2\cdot 13}} = \frac{3}{\sqrt{2\beta}}\,, $$ therefore $$\therefore \beta = 13\,. $$

We need to find the largest element in the set $$S = \{x + \lambda : x \text{ is an integer solution of } E\}$$, where equation $$E$$ is $$|x|^2 - 2|x| + |\lambda - 3| = 0$$.

Let $$t = |x| \geq 0$$. The equation becomes:

This is a quadratic in $$t$$. Using the quadratic formula:

For $$t$$ to be real: $$1 - |\lambda - 3| \geq 0$$, i.e., $$|\lambda - 3| \leq 1$$, which gives $$2 \leq \lambda \leq 4$$.

For $$t \geq 0$$: We need $$1 \pm \sqrt{1 - |\lambda - 3|} \geq 0$$. Since $$\sqrt{1 - |\lambda-3|} \leq 1$$, the value $$1 - \sqrt{1-|\lambda-3|} \geq 0$$ always. So both roots are non-negative.

Since $$t = |x|$$, for each valid $$t > 0$$, we get $$x = \pm t$$. For $$t = 0$$, we get $$x = 0$$. We need $$t$$ to be a non-negative integer (since $$x$$ must be an integer and $$t = |x|$$).

For $$t$$ to be an integer, we need $$\sqrt{1 - |\lambda-3|}$$ to yield integer values of $$t = 1 \pm \sqrt{1 - |\lambda-3|}$$.

Case 1: $$|\lambda - 3| = 0$$, i.e., $$\lambda = 3$$. Then $$t = 1 \pm 1$$, giving $$t = 0$$ or $$t = 2$$. Integer values of $$x$$: $$0, \pm 2$$. Elements of $$S$$: $$\{0+3, 2+3, -2+3\} = \{3, 5, 1\}$$. Largest = 5.

Case 2: $$|\lambda - 3| = 1$$, i.e., $$\lambda = 2$$ or $$\lambda = 4$$. Then $$t = 1 \pm 0 = 1$$. Integer value of $$x$$: $$\pm 1$$.

If $$\lambda = 2$$: $$S = \{1+2, -1+2\} = \{3, 1\}$$. Largest = 3.

If $$\lambda = 4$$: $$S = \{1+4, -1+4\} = \{5, 3\}$$. Largest = 5.

Now check other values where $$t$$ is integer. We need $$1 - |\lambda - 3| = k^2$$ for some value that makes $$1 \pm k$$ a non-negative integer.

If $$k = 0$$: $$|\lambda-3| = 1$$, covered above.

If $$k = 1$$: $$|\lambda-3| = 0$$, covered above.

For non-integer $$k$$, $$t$$ would not be integer.

From Case 1 ($$\lambda = 3$$): largest element in $$S$$ is 5.

From Case 2 ($$\lambda = 4$$): largest element in $$S$$ is 5.

From Case 2 ($$\lambda = 2$$): largest element in $$S$$ is 3.

The largest element in $$S$$ is 5.

We need to find the largest element in $$S = \{x + \lambda : x \text{ is an integer solution of } E\}$$ where $$E: |x|^2 - 2|x| + |\lambda - 3| = 0$$.

We start by letting $$u = |x| \geq 0$$, which transforms the equation into:

$$u^2 - 2u + |\lambda - 3| = 0$$

This quadratic in $$u$$ yields $$u = \frac{2 \pm \sqrt{4 - 4|\lambda - 3|}}{2} = 1 \pm \sqrt{1 - |\lambda - 3|}$$.

For real solutions in $$u$$ we require $$|\lambda - 3| \leq 1$$, i.e., $$2 \leq \lambda \leq 4$$.

We also note that $$u \geq 0$$, and because $$\sqrt{1 - |\lambda - 3|} \leq 1$$, both expressions $$1 \pm \sqrt{1 - |\lambda - 3|}$$ are non-negative.

Since $$x$$ must be an integer, $$u = |x|$$ must be a non-negative integer, so we need $$u = 1 \pm \sqrt{1 - |\lambda - 3|}$$ to be an integer.

Introducing $$k = \sqrt{1 - |\lambda - 3|}$$ gives $$|\lambda - 3| = 1 - k^2$$ and hence $$u = 1 + k$$ or $$u = 1 - k$$.

For $$u$$ to be an integer, $$k$$ must also be an integer satisfying $$0 \leq k \leq 1$$, which leaves $$k = 0$$ or $$k = 1$$.

Case 1: If $$k = 0$$ then $$|\lambda - 3| = 1$$, yielding $$\lambda = 2$$ or $$\lambda = 4$$ and $$u = 1$$ so that $$x = \pm 1$$.

For $$\lambda = 4$$ this gives $$x + \lambda \in \{-1 + 4,\,1 + 4\} = \{3,\,5\}$$, and for $$\lambda = 2$$ it gives $$x + \lambda \in \{-1 + 2,\,1 + 2\} = \{1,\,3\}$$.

Case 2: If $$k = 1$$ then $$|\lambda - 3| = 0$$ so $$\lambda = 3$$ and $$u = 0$$ or $$u = 2$$, giving $$x \in \{0,\,\pm 2\}$$.

This leads to $$x + \lambda \in \{0 + 3,\,2 + 3,\,-2 + 3\} = \{3,\,5,\,1\}$$ for $$\lambda = 3$$.

Across all valid values of $$\lambda$$, the largest element in $$S$$ is $$5$$, which is the required answer.

To find the number of integers $$n$$ that satisfy $$|n^2 - 10n + 19| < 6$$, we break the absolute value inequality into two parts:

1. Solve the Inequality

The expression $$|f(x)| < 6$$ is equivalent to $$-6 < f(x) < 6$$.

Part A: $$n^2 - 10n + 19 < 6$$

$$n^2 - 10n + 13 < 0$$

Using the quadratic formula for $$n^2 - 10n + 13 = 0$$:

$$n = \frac{10 \pm \sqrt{100 - 52}}{2} = \frac{10 \pm \sqrt{48}}{2} = 5 \pm 2\sqrt{3}$$

Since $$2\sqrt{3} \approx 3.46$$:

• $$n \in (5 - 3.46, 5 + 3.46) \implies n \in (1.54, 8.46)$$
• Integers in this range: $$\{2, 3, 4, 5, 6, 7, 8\}$$

Part B: $$n^2 - 10n + 19 > -6$$

$$n^2 - 10n + 25 > 0$$

$$(n - 5)^2 > 0$$

A squared number is always greater than zero unless the base is zero.

• $$(n - 5)^2 = 0$$ when $$n = 5$$.
• Therefore, this condition is true for all integers except $$n = 5$$.

2. Combine the Conditions

We need integers that are in the set from Part A and satisfy Part B:

$$\{2, 3, 4, 5, 6, 7, 8\} \setminus \{5\}$$

The resulting set of integers is:

$$\{2, 3, 4, 6, 7, 8\}$$

3. Final Count

Counting the elements in the set: $$2, 3, 4, 6, 7, 8$$.

There are 6 elements.

Final Answer: 6

Given that $$(p^2 + q^2)x^2 - 2q(p + r)x + q^2 + r^2 = 0$$ and $$x^2 + 2x - 8 = 0$$ share a common root, with $$p, q, r \in \mathbb{R}$$ all positive.

First, analyse the discriminant: $$\Delta = 4q^2(p + r)^2 - 4(p^2 + q^2)(q^2 + r^2)$$. Expanding: $$\dfrac{\Delta}{4} = 2pq^2r - p^2r^2 - q^4 = -(q^2 - pr)^2 \le 0$$. For real roots: $$\Delta = 0$$, giving $$q^2 = pr$$.

Next, the repeated root is $$x = \dfrac{q(p + r)}{p^2 + q^2} = \dfrac{q(p + r)}{p(p + r)} = \dfrac{q}{p}$$.

Next, solve $$x^2 + 2x - 8 = 0$$. This gives $$(x + 4)(x - 2) = 0 \implies x = 2 \text{ or } x = -4$$.

Since $$p, q, r > 0$$, the root $$x = q/p > 0$$. Therefore $$x = 2$$ (rejecting $$x = -4$$). From this, $$\dfrac{q}{p} = 2 \implies q = 2p, \quad r = \dfrac{q^2}{p} = 4p$$.

Next, compute $$\dfrac{q^2 + r^2}{p^2} = \dfrac{4p^2 + 16p^2}{p^2} = 20$$.

The answer is $$20$$.

We have $$\alpha, \beta$$ as roots of $$x^2 - x - 4 = 0$$ with $$\alpha > \beta$$, and $$P_n = \alpha^n - \beta^n$$. Since $$\alpha$$ and $$\beta$$ satisfy $$x^2 = x + 4$$, we get the recurrence $$P_n = P_{n-1} + 4P_{n-2}$$ for all $$n \geq 2$$.

We simplify the given expression $$\frac{P_{15}P_{16} - P_{14}P_{16} - P_{15}^2 + P_{14}P_{15}}{P_{13}P_{14}}$$. Factoring the numerator: $$P_{16}(P_{15} - P_{14}) - P_{15}(P_{15} - P_{14}) = (P_{15} - P_{14})(P_{16} - P_{15})$$.

From the recurrence $$P_n = P_{n-1} + 4P_{n-2}$$, we get $$P_n - P_{n-1} = 4P_{n-2}$$. Hence $$P_{15} - P_{14} = 4P_{13}$$ and $$P_{16} - P_{15} = 4P_{14}$$.

The expression becomes $$\frac{4P_{13} \cdot 4P_{14}}{P_{13}P_{14}} = \frac{16\,P_{13}P_{14}}{P_{13}P_{14}} = 16$$.

Hence, the correct answer is $$\boxed{16}$$.

Given,

$$x^2-4\lambda x+5=0$$

has roots $$\alpha,\beta$$.

Hence,

$$\alpha+\beta=4\lambda,\qquad \alpha\beta=5$$

Also,

$$x^2-(3\sqrt2+2\sqrt3)x+7+3\lambda\sqrt3=0$$

has roots $$\alpha,\gamma$$.

Therefore,

$$\alpha+\gamma=3\sqrt2+2\sqrt3$$

Given,

$$\beta+\gamma=3\sqrt2$$

Now,

$$2\alpha+(\beta+\gamma)=(\alpha+\beta)+(\alpha+\gamma)$$

Substituting the values,

$$2\alpha+3\sqrt2=4\lambda+3\sqrt2+2\sqrt3$$

$$2\alpha=4\lambda+2\sqrt3$$

$$\alpha=2\lambda+\sqrt3$$

Since $$\alpha$$ is a root of

$$x^2-4\lambda x+5=0$$

we get

$$\alpha^2-4\lambda\alpha+5=0$$

Substituting $$\alpha=2\lambda+\sqrt3$$,

$$\left(2\lambda+\sqrt3\right)^2-4\lambda\left(2\lambda+\sqrt3\right)+5=0$$

$$4\lambda^2+4\lambda\sqrt3+3-8\lambda^2-4\lambda\sqrt3+5=0$$

$$-4\lambda^2+8=0$$

$$\lambda^2=2$$

Taking $$\lambda=\sqrt2$$,

$$\alpha+\beta=4\lambda=4\sqrt2$$

Now,

$$\alpha+2\beta+\gamma=(\alpha+\beta)+(\beta+\gamma)$$

$$=4\sqrt2+3\sqrt2$$

$$=7\sqrt2$$

Therefore,

$$\left(\alpha+2\beta+\gamma\right)^2=(7\sqrt2)^2=98$$

Hence,

$$\boxed{98}$$

Let $$f(x) = x^2 + bx + p$$ be a quadratic with leading coefficient 1, $$f(0) = p \neq 0$$, and $$f(1) = \frac{1}{3}$$.

From $$f(1) = 1 + b + p = \frac{1}{3}$$, we get $$b + p = -\frac{2}{3}$$.

If $$\alpha$$ is a common root of $$f(x) = 0$$ and $$f\circ f\circ f\circ f(x) = 0$$, then since $$f(\alpha) = 0$$:

So $$p^2 + bp + p$$ must be a root of $$f(x) = 0$$.

Note that $$p^2 + bp + p = p(p + b + 1) = p \cdot \frac{1}{3} = \frac{p}{3}$$.

Since $$\frac{p}{3}$$ is a root of $$f$$, we have

Dividing by $$p\neq 0$$ gives $$\frac{p}{9} + \frac{b}{3} + 1 = 0$$, hence $$p + 3b + 9 = 0$$.

Solving the system $$b + p = -\frac{2}{3}$$ and $$p + 3b = -9$$: subtracting gives $$2b = -9 + \frac{2}{3} = -\frac{25}{3}$$, so $$b = -\frac{25}{6}$$. Then $$p = -\frac{2}{3} + \frac{25}{6} = \frac{21}{6} = \frac{7}{2}$$.

Finally, compute $$f(-3)$$:

The answer is $$\boxed{25}$$.

The given equation is:
$$\frac{3x^2 - 9x + 17}{x^2 + 3x + 10} = \frac{5x^2 - 7x + 19}{3x^2 + 5x + 12}$$
Cross multiplying both sides gives:
$$(3x^2 - 9x + 17)(3x^2 + 5x + 12) = (5x^2 - 7x + 19)(x^2 + 3x + 10)$$
Let:
$$A = 3x^2 - 9x + 17$$
$$B = 3x^2 + 5x + 12$$
$$C = x^2 + 3x + 10$$
Expressing $$5x^2 - 7x + 19$$ in terms of $$A$$, $$B$$, and $$C$$:
$$A + B - C = (3x^2 - 9x + 17) + (3x^2 + 5x + 12) - (x^2 + 3x + 10)$$
$$A + B - C = 5x^2 - 7x + 19$$
Substituting these variables into the cross multiplied equation:
$$AB = (A + B - C)C$$
Expanding and rearranging terms:
$$AB = AC + BC - C^2$$
$$AB - AC - BC + C^2 = 0$$
Factoring by grouping:
$$A(B - C) - C(B - C) = 0$$
$$(A - C)(B - C) = 0$$
Now evaluating $$A - C$$ and $$B - C$$:
First factor:
$$A - C = (3x^2 - 9x + 17) - (x^2 + 3x + 10) = 2x^2 - 12x + 7$$
Second factor:
$$B - C = (3x^2 + 5x + 12) - (x^2 + 3x + 10) = 2x^2 + 2x + 2$$
Thus, the factored equation becomes:
$$(2x^2 - 12x + 7)(2x^2 + 2x + 2) = 0$$
Case 1:
$$2x^2 + 2x + 2 = 0$$
$$x^2 + x + 1 = 0$$
Checking discriminant $$D$$:
$$D = 1^2 - 4(1)(1) = -3$$
Since $$D < 0$$, there are no real roots for this factor.
Case 2:
$$2x^2 - 12x + 7 = 0$$
Checking discriminant $$D$$:
$$D = (-12)^2 - 4(2)(7) = 144 - 56 = 88$$
Since $$D > 0$$, this quadratic equation yields two real roots.
Using Vieta's relation, the sum of roots for $$2x^2 - 12x + 7 = 0$$ is:
$$\text{Sum of roots} = -\frac{-12}{2} = 6$$
Hence, the sum of all real values of $$x$$ is 6.

We have $$f(x) = (x - p)^2 - q$$, $$q > 0$$, and $$a_1, a_2, a_3, a_4$$ in AP with mean $$p$$ and positive common difference $$d$$.

Since the four terms are in AP with mean $$p$$, we have $$\frac{a_1 + a_2 + a_3 + a_4}{4} = p$$.

We may therefore set $$a_1 = p - \frac{3d}{2}$$, $$a_2 = p - \frac{d}{2}$$, $$a_3 = p + \frac{d}{2}$$, and $$a_4 = p + \frac{3d}{2}$$.

Substituting into $$f$$ gives $$f(a_1) = \left(\frac{3d}{2}\right)^2 - q = \frac{9d^2}{4} - q$$, $$f(a_2) = \left(\frac{d}{2}\right)^2 - q = \frac{d^2}{4} - q$$, $$f(a_3) = \left(\frac{d}{2}\right)^2 - q = \frac{d^2}{4} - q$$, and $$f(a_4) = \left(\frac{3d}{2}\right)^2 - q = \frac{9d^2}{4} - q$$.

Since each of these values has absolute value $$500$$, we obtain $$\left|\frac{9d^2}{4} - q\right| = 500$$ and $$\left|\frac{d^2}{4} - q\right| = 500$$.

Because $$q > 0$$ and the two expressions must both have absolute value $$500$$ with different inputs, they must have opposite signs (otherwise $$\frac{9d^2}{4} - q = \frac{d^2}{4} - q$$ would force $$d = 0$$, contradicting the positive common difference).

Thus we consider the case $$\frac{9d^2}{4} - q = 500$$ and $$\frac{d^2}{4} - q = -500$$.

Subtracting the second from the first yields $$\frac{9d^2}{4} - \frac{d^2}{4} = 1000$$, which simplifies to $$2d^2 = 1000$$ and therefore $$d^2 = 500$$.

Substituting into the second equation gives $$q = \frac{d^2}{4} + 500 = 125 + 500 = 625$$.

Hence $$(x - p)^2 = q = 625$$, so $$x - p = \pm 25$$ and the two roots are $$p + 25$$ and $$p - 25$$.

The absolute difference between the roots is $$50$$, and therefore the answer is 50.

We are given that $$f(x)$$ is degree 2, $$g(x)$$ is degree 1, $$f(g(x)) = 8x^2 - 2x$$, and $$g(f(x)) = 4x^2 + 6x + 1$$. Let $$f(x) = ax^2 + bx + c$$ and $$g(x) = px + q$$.

Substituting into $$f(g(x))$$ gives $$f(g(x)) = a(px+q)^2 + b(px+q) + c = ap^2x^2 + (2apq+bp)x + (aq^2+bq+c),$$ and comparing this with $$8x^2 - 2x + 0$$ yields the equations

$$ap^2 = 8 \quad (1),\quad 2apq + bp = -2 \quad (2),\quad aq^2 + bq + c = 0 \quad (3).$$

Similarly, substituting into $$g(f(x))$$ gives $$g(f(x)) = p(ax^2+bx+c) + q = pax^2 + pbx + (pc+q),$$ and comparing with $$4x^2 + 6x + 1$$ yields

$$pa = 4 \quad (4),\quad pb = 6 \quad (5),\quad pc + q = 1 \quad (6).$$

From (1) and (4) it follows that $$p = 2$$ and $$a = 2$$. Then (5) implies $$b = 3$$, substituting into (2) gives $$q = -1$$, and using (6) yields $$c = 1$$. One checks that (3) is satisfied since $$aq^2 + bq + c = 2(1) + 3(-1) + 1 = 0$$. Hence $$f(x) = 2x^2 + 3x + 1$$ and $$g(x) = 2x - 1$$.

Finally, evaluating at $$x=2$$ gives $$f(2) = 2(4) + 3(2) + 1 = 15$$ and $$g(2) = 2(2) - 1 = 3$$, so $$f(2) + g(2) = 18$$, and the answer is $$\mathbf{18}$$.

The given equation is:
$$x^5(x^3 - x^2 - x + 1) + x(3x^3 - 4x^2 - 2x + 4) - 1 = 0$$
Let us break down and factor the expression. First, factor the polynomial in the first term:
$$x^3 - x^2 - x + 1$$
$$= x^2(x - 1) - 1(x - 1)$$
$$= (x^2 - 1)(x - 1)$$
$$= (x - 1)(x + 1)(x - 1)$$
$$= (x - 1)^2(x + 1)$$

Now, let us expand the remaining part of the equation:
$$x(3x^3 - 4x^2 - 2x + 4) - 1$$
$$= 3x^4 - 4x^3 - 2x^2 + 4x - 1$$
By substituting $$x = 1$$, we can see that it is a root because $$3 - 4 - 2 + 4 - 1 = 0$$. Since $$x = 1$$ is a root, $$(x - 1)$$ is a factor. Dividing the polynomial by $$(x - 1)$$ gives:
$$3x^4 - 4x^3 - 2x^2 + 4x - 1 = (x - 1)(3x^3 - x^2 - 3x + 1)$$
Checking $$x = 1$$ again for the quotient where $$3(1)^3 - (1)^2 - 3(1) + 1 = 0$$, we find $$(x - 1)$$ is a factor again:
$$3x^3 - x^2 - 3x + 1 = (x - 1)(3x^2 + 2x - 1)$$
Now, factor the remaining quadratic equation:
$$3x^2 + 2x - 1$$
$$= 3x^2 + 3x - x - 1$$
$$= 3x(x + 1) - 1(x + 1)$$
$$= (3x - 1)(x + 1)$$
Combining these, the second part of our original equation factors to:
$$(x - 1)^2(x + 1)(3x - 1)$$
Now, substitute both factored parts back into the original equation:
$$x^5[(x - 1)^2(x + 1)] + [(x - 1)^2(x + 1)(3x - 1)] = 0$$
Factor out the common terms:
$$(x - 1)^2(x + 1)[x^5 + 3x - 1] = 0$$

This gives us our roots by setting each part to zero:

• $$(x - 1)^2 = 0 \implies x = 1$$
  
• $$x + 1 = 0 \implies x = -1$$
  
• $$x^5 + 3x - 1 = 0$$
  

To find the number of real roots for the polynomial $$x^5 + 3x - 1 = 0$$, let $$f(x) = x^5 + 3x - 1$$.
Taking the derivative:
$$f'(x) = 5x^4 + 3$$
Since $$x^4$$ is always positive or zero for real values of $$x$$, $$5x^4 + 3$$ is always strictly greater than $$0$$. This means $$f(x)$$ is a strictly increasing continuous function for all real values of $$x$$.
Because it is strictly increasing, it crosses the horizontal axis exactly once, meaning $$x^5 + 3x - 1 = 0$$ has exactly $$1$$ real root.
Let us verify this root is distinct from our other roots $$1$$ and $$-1$$:
$$f(1) = (1)^5 + 3(1) - 1 = 3 \neq 0$$
$$f(-1) = (-1)^5 + 3(-1) - 1 = -5 \neq 0$$
Therefore, we have exactly $$3$$ distinct real roots which are $$x = 1$$, $$x = -1$$, and the single real root from $$x^5 + 3x - 1 = 0$$.
The number of distinct real roots is $$3$$.

Given $$f(x) = (c+1)x^2 + (1-c^2)x + 2k$$ and $$f(x+y) = f(x) + f(y) - xy$$ for all $$x, y \in \mathbb{R}$$. Setting $$x = y = 0$$ yields $$f(0) = 2f(0)$$, so $$f(0) = 0$$ and hence $$2k = 0 \Rightarrow k = 0$$.

Thus $$f(x) = (c+1)x^2 + (1-c^2)x$$, and from the functional equation we have $$f(x+y) = (c+1)(x+y)^2 + (1-c^2)(x+y)$$ while $$f(x) + f(y) - xy = (c+1)x^2 + (1-c^2)x + (c+1)y^2 + (1-c^2)y - xy$$.

Comparing the coefficients of $$xy$$ gives $$2(c+1) = -1 \Rightarrow c + 1 = -\frac{1}{2} \Rightarrow c = -\frac{3}{2}$$, and for the linear terms $$1 - c^2 = 1 - \frac{9}{4} = -\frac{5}{4}$$ matches on both sides. Therefore $$f(x) = -\frac{1}{2}x^2 - \frac{5}{4}x$$.

It follows that $$f(n) = -\frac{1}{2}n^2 - \frac{5}{4}n$$ and so $$\sum_{n=1}^{20} f(n) = -\frac{1}{2}\sum_{n=1}^{20} n^2 - \frac{5}{4}\sum_{n=1}^{20} n$$ with $$\sum_{n=1}^{20} n^2 = \frac{20 \times 21 \times 41}{6} = \frac{17220}{6} = 2870$$ and $$\sum_{n=1}^{20} n = \frac{20 \times 21}{2} = 210$$, giving $$\sum_{n=1}^{20} f(n) = -\frac{2870}{2} - \frac{5 \times 210}{4} = -1435 - \frac{1050}{4} = -1435 - 262.5 = -1697.5$$.

The answer is $$\boxed{3395}$$.

We have three unknowns $$a,\;b,\;c$$ such that $$a+b+c=1,\;ab+bc+ca=2,\;abc=3.$$ Our target is the sum of fourth powers $$a^{4}+b^{4}+c^{4}.$$

To reach that target we shall successively find the power sums $$S_{1}=a+b+c,\qquad S_{2}=a^{2}+b^{2}+c^{2},\qquad S_{3}=a^{3}+b^{3}+c^{3},\qquad S_{4}=a^{4}+b^{4}+c^{4}.$$ The given data already give us $$S_{1}=1$$ and the elementary symmetric polynomials $$e_{1}=a+b+c=1,\qquad e_{2}=ab+bc+ca=2,\qquad e_{3}=abc=3.$$ We shall keep using the symbols $$e_{1},e_{2},e_{3}$$ because Newton’s identities are expressed in terms of them.

Step 1 : Finding $$S_{2}$$.
First recall the algebraic identity $$(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(ab+bc+ca).$$ Substituting the known values we get $$1^{2}=S_{2}+2\cdot 2,$$ $$1=S_{2}+4,$$ so $$S_{2}=1-4=-3.$$

Step 2 : Finding $$S_{3}$$ with Newton’s identity.
For three variables, Newton’s identities state $$S_{3}-e_{1}S_{2}+e_{2}S_{1}-3e_{3}=0.$$ We write the identity first, then substitute all the known numbers: $$S_{3}-1\cdot(-3)+2\cdot 1-3\cdot 3=0,$$ $$S_{3}+3+2-9=0,$$ $$S_{3}-4=0,$$ so $$S_{3}=4.$$

Step 3 : Finding $$S_{4}.$$
Again using Newton’s identities for three variables, $$S_{4}-e_{1}S_{3}+e_{2}S_{2}-e_{3}S_{1}=0.$$ Writing this explicitly and substituting gives $$S_{4}-1\cdot 4+2\cdot(-3)-3\cdot 1=0,$$ $$S_{4}-4-6-3=0,$$ $$S_{4}-13=0,$$ hence $$S_{4}=13.$$

Therefore, the required value is $$a^{4}+b^{4}+c^{4}=13.$$

So, the answer is $$13$$.

We start from the quadratic equation $$x^{2}+5\sqrt{2}\,x+10=0$$ whose roots are given to be $$\alpha$$ and $$\beta$$ with $$\alpha >\beta$$. From the standard relation between coefficients and roots of a quadratic, we immediately obtain

$$\alpha+\beta=-5\sqrt{2},\qquad \alpha\beta=10.$$

For every positive integer $$n$$ we define $$P_{n}=\alpha^{n}-\beta^{n}$$ and wish to evaluate

$$E=\dfrac{P_{17}P_{20}+5\sqrt{2}\,P_{17}P_{19}}{P_{18}P_{19}+5\sqrt{2}\,P_{18}^{2}}.$$

Because both $$\alpha$$ and $$\beta$$ satisfy the quadratic, we may replace $$x$$ by either root in the equation $$x^{2}+5\sqrt{2}\,x+10=0$$. Writing the result for an arbitrary positive power will give us a recurrence for the sequence $$P_{n}$$.

First, for any root $$r\in\{\alpha,\beta\}$$ we have $$r^{2}=-5\sqrt{2}\,r-10.$$ Multiplying both sides by $$r^{\,n-2}$$ (where $$n\ge 2$$) gives

$$r^{n}= -5\sqrt{2}\,r^{\,n-1}-10\,r^{\,n-2}.$$

Since this equality holds separately for $$r=\alpha$$ and for $$r=\beta$$, we can write the two identities and subtract them. Thus, for all integers $$n\ge 2$$,

$$\alpha^{n}-\beta^{n}= -5\sqrt{2}\left(\alpha^{\,n-1}-\beta^{\,n-1}\right)-10\left(\alpha^{\,n-2}-\beta^{\,n-2}\right).$$

Recognizing the differences as the $$P$$ sequence, we arrive at the linear recurrence relation

$$P_{n}=-5\sqrt{2}\,P_{n-1}-10\,P_{n-2}\qquad (n\ge 2).$$

Now we tackle the pieces that appear in the expression $$E$$.

First, apply the recurrence with $$n=20$$:

$$P_{20}=-5\sqrt{2}\,P_{19}-10\,P_{18}.$$

Add $$5\sqrt{2}\,P_{19}$$ to both sides to create the combination seen in the numerator:

$$P_{20}+5\sqrt{2}\,P_{19}=(-5\sqrt{2}\,P_{19}-10\,P_{18})+5\sqrt{2}\,P_{19}=-10\,P_{18}.$$

Hence the entire numerator becomes

$$P_{17}\bigl(P_{20}+5\sqrt{2}\,P_{19}\bigr)=P_{17}\times(-10\,P_{18})=-10\,P_{17}P_{18}.$$

Next, apply the recurrence with $$n=19$$ to prepare the denominator:

$$P_{19}=-5\sqrt{2}\,P_{18}-10\,P_{17}.$$

Add $$5\sqrt{2}\,P_{18}$$ to both sides:

$$P_{19}+5\sqrt{2}\,P_{18}=(-5\sqrt{2}\,P_{18}-10\,P_{17})+5\sqrt{2}\,P_{18}=-10\,P_{17}.$$

Hence the entire denominator is

$$P_{18}\bigl(P_{19}+5\sqrt{2}\,P_{18}\bigr)=P_{18}\times(-10\,P_{17})=-10\,P_{17}P_{18}.$$

Now both numerator and denominator are equal to the same quantity $$-10\,P_{17}P_{18}$$, so when we form their ratio we get

$$E=\dfrac{-10\,P_{17}P_{18}}{-10\,P_{17}P_{18}}=1.$$

So, the answer is $$1$$.

Hence, the correct answer is Option 1.

Since $$P(x) = f(x^3) + x\,g(x^3)$$ is divisible by $$x^2 + x + 1$$, both primitive cube roots of unity $$\omega$$ and $$\omega^2$$ (where $$\omega = e^{2\pi i/3}$$) are roots of $$P(x)$$. Note that $$\omega^3 = 1$$.

Substituting $$x = \omega$$: $$P(\omega) = f(\omega^3) + \omega\,g(\omega^3) = f(1) + \omega\,g(1) = 0$$. Substituting $$x = \omega^2$$: $$P(\omega^2) = f(1) + \omega^2\,g(1) = 0$$.

Subtracting these two equations gives $$(\omega - \omega^2)\,g(1) = 0$$. Since $$\omega \neq \omega^2$$, we conclude $$g(1) = 0$$. Substituting back into either equation gives $$f(1) = 0$$.

Therefore $$P(1) = f(1) + 1 \cdot g(1) = 0 + 0 = 0$$.

Since $$\alpha$$ and $$\beta$$ are roots of a quadratic with $$\alpha + \beta = 1$$ and $$\alpha\beta = -1$$, they satisfy $$t^2 - t - 1 = 0$$. The sequence $$p_n = \alpha^n + \beta^n$$ satisfies Newton's identity recurrence: $$p_{n+1} = (\alpha + \beta)p_n - \alpha\beta \cdot p_{n-1}$$.

Substituting the given values: $$p_{n+1} = 1 \cdot p_n - (-1) \cdot p_{n-1} = p_n + p_{n-1}$$.

We are given $$p_{n-1} = 11$$ and $$p_{n+1} = 29$$. Using the recurrence $$p_{n+1} = p_n + p_{n-1}$$, we substitute: $$29 = p_n + 11$$.

Solving for $$p_n$$: $$p_n = 29 - 11 = 18$$.

We can verify this is consistent by checking with the recurrence in the other direction. We also know $$p_0 = \alpha^0 + \beta^0 = 2$$ and $$p_1 = \alpha + \beta = 1$$. Using $$p_{k+1} = p_k + p_{k-1}$$, the sequence begins: $$p_0 = 2, p_1 = 1, p_2 = 3, p_3 = 4, p_4 = 7, p_5 = 11, p_6 = 18, p_7 = 29, \ldots$$ So $$n - 1 = 5$$, i.e., $$n = 6$$, and indeed $$p_6 = 18$$ and $$p_7 = 29$$.

Therefore, $$p_n^2 = 18^2 = 324$$.

We have two quadratic equations involving the same real parameter $$\lambda\,( \lambda \neq 0)$$.

First, recall the standard facts for a quadratic equation. For $$ax^{2}+bx+c=0$$, the sum of the roots equals $$-\dfrac{b}{a}$$ and the product of the roots equals $$\dfrac{c}{a}$$.

Applying this to the equation $$x^{2}-x+2\lambda=0$$ whose roots are $$\alpha$$ and $$\beta$$, we get $$\alpha+\beta=-\dfrac{-1}{1}=1$$ and $$\alpha\beta=\dfrac{2\lambda}{1}=2\lambda.$$ So $$\alpha+\beta=1,\qquad \alpha\beta=2\lambda. \quad -(1)$$

Next, apply the same formulas to $$3x^{2}-10x+27\lambda=0$$ whose roots are $$\alpha$$ and $$\gamma$$. Here $$a=3,\;b=-10,\;c=27\lambda$$, hence $$\alpha+\gamma=-\dfrac{-10}{3}=\dfrac{10}{3},\qquad \alpha\gamma=\dfrac{27\lambda}{3}=9\lambda.$$ Thus $$\alpha+\gamma=\dfrac{10}{3},\qquad \alpha\gamma=9\lambda. \quad -(2)$$

From the first sum, express $$\beta$$ as $$\beta=1-\alpha.$$ From the second sum, express $$\gamma$$ as $$\gamma=\dfrac{10}{3}-\alpha.$$

Now rewrite the two products from (1) and (2) using these expressions:

Using $$\beta=1-\alpha$$, $$\alpha\beta=\alpha(1-\alpha)=2\lambda. \quad -(3)$$

Using $$\gamma=\dfrac{10}{3}-\alpha$$, $$\alpha\gamma=\alpha\left(\dfrac{10}{3}-\alpha\right)=9\lambda. \quad -(4)$$

To eliminate $$\lambda$$, divide (4) by (3):

$$\dfrac{\alpha\left(\dfrac{10}{3}-\alpha\right)}{\alpha(1-\alpha)}=\dfrac{9\lambda}{2\lambda}\quad\Longrightarrow\quad\dfrac{\dfrac{10}{3}-\alpha}{1-\alpha}=\dfrac{9}{2}.$$

Cross-multiplying, $$2\!\left(\dfrac{10}{3}-\alpha\right)=9(1-\alpha).$$ Compute each side: Left side $$\;= \dfrac{20}{3}-2\alpha,$$ Right side $$= 9-9\alpha.$$ Equating, $$\dfrac{20}{3}-2\alpha = 9-9\alpha.$$

Move the terms involving $$\alpha$$ to one side: $$\dfrac{20}{3}+7\alpha = 9.$$ Subtract $$\dfrac{20}{3}$$ from both sides: $$7\alpha = 9-\dfrac{20}{3} = \dfrac{27}{3}-\dfrac{20}{3}=\dfrac{7}{3}.$$ Hence $$\alpha=\dfrac{\dfrac{7}{3}}{7}=\dfrac{1}{3}.$$

With $$\alpha=\dfrac{1}{3}$$, obtain the remaining roots: $$\beta = 1-\alpha = 1-\dfrac{1}{3}=\dfrac{2}{3},$$ $$\gamma = \dfrac{10}{3}-\alpha = \dfrac{10}{3}-\dfrac{1}{3}=3.$$

Compute the product $$\beta\gamma$$: $$\beta\gamma = \dfrac{2}{3}\times3 = 2.$$

Determine $$\lambda$$ from (3): $$\alpha\beta = 2\lambda\;\Longrightarrow\;\dfrac{1}{3}\times\dfrac{2}{3}=2\lambda\;\Longrightarrow\;\dfrac{2}{9}=2\lambda\;\Longrightarrow\;\lambda=\dfrac{1}{9}.$$

Finally, evaluate the required ratio: $$\dfrac{\beta\gamma}{\lambda} = \dfrac{2}{\dfrac{1}{9}} = 2 \times 9 = 18.$$

So, the answer is $$18$$.

The number of real roots of the equation $$e^{4x} - e^{3x} - 4e^{2x} - e^x + 1 = 0$$ is 2.

1. Perform Algebraic Substitution

Let $$y = e^x$$. Since $$e^x > 0$$ for all real values of $$x$$, we need to find the positive real roots where $$y > 0$$ for the polynomial:

$$y^4 - y^3 - 4y^2 - y + 1 = 0$$

2. Simplify Using Reciprocal Symmetry

Since $$y = 0$$ is not a solution, divide the entire equation by $$y^2$$:

$$\left(y^2 + \frac{1}{y^2}\right) - \left(y + \frac{1}{y}\right) - 4 = 0$$

Let $$t = y + \frac{1}{y}$$. Squaring both sides gives $$t^2 = y^2 + \frac{1}{y^2} + 2$$, which means $$y^2 + \frac{1}{y^2} = t^2 - 2$$. Substituting these terms into the equation yields:

$$(t^2 - 2) - t - 4 = 0$$

$$t^2 - t - 6 = 0$$

3. Solve for t

Factor the quadratic equation:

$$(t - 3)(t + 2) = 0$$

This gives two possible values: $$t = 3$$ or $$t = -2$$.

4. Analyze Constraints and Solve for x

By the Arithmetic Mean-Geometric Mean inequality, for any positive real number $$y$$:

$$t = y + \frac{1}{y} \geq 2$$

Case 1: $$t = -2$$

This case is rejected because $$t$$ must be greater than or equal to 2.

Case 2: $$t = 3$$

This is valid since $$3 \geq 2$$.

$$y + \frac{1}{y} = 3 \implies y^2 - 3y + 1 = 0$$

Using the quadratic formula to solve for $$y$$:

$$y = \frac{3 \pm \sqrt{(-3)^2 - 4(1)(1)}}{2} = \frac{3 \pm \sqrt{5}}{2}$$

Both values are strictly positive. Since $$x = \ln(y)$$, each positive value of $$y$$ yields exactly one unique real value for $$x$$.

Final Answer

The equation has exactly 2 real roots.

We need to solve $$\log_{(x+1)}(2x^2 + 7x + 5) + \log_{(2x+5)}(x+1)^2 - 4 = 0$$ for $$x > 0$$.

First, note that $$2x^2 + 7x + 5 = (2x+5)(x+1)$$. So the equation becomes: $$\log_{(x+1)}[(2x+5)(x+1)] + 2\log_{(2x+5)}(x+1) - 4 = 0$$

$$\log_{(x+1)}(2x+5) + \log_{(x+1)}(x+1) + 2\log_{(2x+5)}(x+1) - 4 = 0$$

$$\log_{(x+1)}(2x+5) + 1 + 2\log_{(2x+5)}(x+1) - 4 = 0$$

Let $$t = \log_{(x+1)}(2x+5)$$, so $$\log_{(2x+5)}(x+1) = \frac{1}{t}$$. The equation becomes: $$t + \frac{2}{t} - 3 = 0$$ $$t^2 - 3t + 2 = 0$$ $$(t-1)(t-2) = 0$$ $$t = 1 \text{ or } t = 2$$

Case 1: $$t = 1$$, i.e., $$\log_{(x+1)}(2x+5) = 1 \Rightarrow 2x+5 = x+1 \Rightarrow x = -4$$. This is excluded since $$x > 0$$.

Case 2: $$t = 2$$, i.e., $$\log_{(x+1)}(2x+5) = 2 \Rightarrow (x+1)^2 = 2x+5 \Rightarrow x^2 + 2x + 1 = 2x + 5 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2$$. Since $$x > 0$$, we take $$x = 2$$.

We verify the domain conditions for $$x = 2$$: the bases are $$x+1 = 3 > 0, \neq 1$$ and $$2x+5 = 9 > 0, \neq 1$$, and the argument $$2x^2+7x+5 = 8+14+5 = 27 > 0$$. All conditions are satisfied.

Therefore, the number of solutions is $$\boxed{1}$$.

We need to solve $$(x + 1)^2 + |x - 5| = \frac{27}{4}$$. We consider two cases based on the absolute value.

When $$x \geq 5$$, the equation becomes $$(x + 1)^2 + (x - 5) = \frac{27}{4}$$, which simplifies to $$x^2 + 2x + 1 + x - 5 = \frac{27}{4}$$, or $$x^2 + 3x - 4 = \frac{27}{4}$$, giving $$x^2 + 3x - \frac{43}{4} = 0$$. Using the quadratic formula, $$x = \frac{-3 \pm \sqrt{9 + 43}}{2} = \frac{-3 \pm \sqrt{52}}{2}$$. Since $$\sqrt{52} \approx 7.21$$, the positive root is $$x \approx 2.1$$, which does not satisfy $$x \geq 5$$. So there are no solutions in this case.

When $$x < 5$$, the equation becomes $$(x + 1)^2 + (5 - x) = \frac{27}{4}$$, which simplifies to $$x^2 + 2x + 1 + 5 - x = \frac{27}{4}$$, or $$x^2 + x + 6 = \frac{27}{4}$$, giving $$x^2 + x - \frac{3}{4} = 0$$. Multiplying by 4: $$4x^2 + 4x - 3 = 0$$, so $$(2x + 3)(2x - 1) = 0$$. This gives $$x = -\frac{3}{2}$$ or $$x = \frac{1}{2}$$. Both values satisfy $$x < 5$$, so both are valid solutions.

Therefore, the number of real roots is $$2$$.

We begin with the given equation

$$\frac{2}{x-1}-\frac{1}{x-2}=\frac{2}{k},\qquad k\neq 0.$$

To clear the fractions we first take the common denominator on the left side. The common denominator of the two fractions is $$(x-1)(x-2).$$ Hence

$$\frac{2}{x-1}-\frac{1}{x-2}\;=\;\frac{2(x-2)-(x-1)}{(x-1)(x-2)}.$$

Expanding the numerator we have

$$2(x-2)-(x-1)=2x-4-x+1=x-3.$$

Therefore the equation becomes

$$\frac{x-3}{(x-1)(x-2)}=\frac{2}{k}.$$

Now we cross-multiply (which is allowed because none of the denominators is zero when a solution is valid):

$$k(x-3)=2\,(x-1)(x-2).$$

Next we expand the right-hand side. First note that

$$(x-1)(x-2)=x^2-3x+2.$$

Multiplying by the factor $$2$$ we get

$$2(x-1)(x-2)=2x^2-6x+4.$$

So the cross-multiplied equation is

$$k(x-3)=2x^2-6x+4.$$

We now bring every term to one side to form a quadratic equation in $$x$$. Distributing $$k$$ on the left gives

$$kx-3k=2x^2-6x+4.$$

Subtracting the left-hand side from both sides gives

$$0=2x^2-6x+4-kx+3k.$$

Grouping like terms in $$x$$ we write

$$0=2x^2-(k+6)x+(3k+4).$$

Thus the quadratic in standard form is

$$2x^2-(k+6)x+(3k+4)=0.$$

The quadratic formula tells us that a quadratic equation $$ax^2+bx+c=0$$ has real roots precisely when its discriminant $$\Delta=b^2-4ac$$ is non-negative. We want no real roots, so we require $$\Delta\lt 0.$$

Here $$a=2,\; b=-(k+6),\; c=3k+4.$$ Hence

$$\Delta=(-(k+6))^2-4\cdot2\cdot(3k+4).$$

Calculating each part we get

$$(k+6)^2=k^2+12k+36,$$

and

$$4\cdot2\cdot(3k+4)=8(3k+4)=24k+32.$$

Therefore

$$\Delta=k^2+12k+36-(24k+32)=k^2-12k+4.$$

For no real roots we need

$$k^2-12k+4\lt 0.$$

This is a quadratic inequality in $$k$$. Because the coefficient of $$k^2$$ is positive, the expression is negative between its two real roots. We find the roots by setting the quadratic equal to zero:

$$k^2-12k+4=0.$$

Using the quadratic formula with respect to $$k$$ (now $$a=1,\; b=-12,\; c=4$$) we get

$$k=\frac{12\pm\sqrt{(-12)^2-4\cdot1\cdot4}}{2}=\frac{12\pm\sqrt{144-16}}{2}=\frac{12\pm\sqrt{128}}{2}=\frac{12\pm8\sqrt2}{2}=6\pm4\sqrt2.$$

Thus the inequality $$k^2-12k+4\lt 0$$ holds for

$$6-4\sqrt2\lt k\lt 6+4\sqrt2.$$

Since $$4\sqrt2\approx5.656854,$$ the numerical bounds are approximately

$$0.343146\lt k\lt 11.656854.$$

We are asked for integral (integer) values of $$k$$ with the additional condition $$k\neq0$$. The integers that satisfy the inequality are therefore

$$k=1,2,3,4,5,6,7,8,9,10,11.$$

Adding these values we obtain

$$1+2+3+4+5+6+7+8+9+10+11=\frac{11\,(1+11)}{2}=66.$$

So, the answer is $$66$$.

First, recall that in JEE problems the symbol $$N$$ denotes the set of positive integers $$\{1,2,3,\ldots\}$$; the number $$0$$ is not included.

We begin with the description of the set $$A$$. By definition

$$A=\{\,n\in N\mid n^{2}\le n+10000\,\}.$$

To find all such $$n$$ we rearrange the inequality:

$$n^{2}\le n+10000 \;\Longrightarrow\; n^{2}-n-10000\le 0.$$

We apply the quadratic-formula roots of $$n^{2}-n-10000=0$$. The formula is stated first:

If $$ax^{2}+bx+c=0,\text{ then }x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}.$$

For the present quadratic $$a=1,\;b=-1,\;c=-10000$$, so

$$n=\dfrac{-(-1)\pm\sqrt{(-1)^{2}-4(1)(-10000)}}{2(1)} =\dfrac{1\pm\sqrt{1+40000}}{2} =\dfrac{1\pm\sqrt{40001}}{2}.$$

The discriminant $$40001$$ lies between $$200^{2}=40000$$ and $$201^{2}=40401$$, therefore

$$200<\sqrt{40001}<201.$$

Consequently

$$\dfrac{1-\sqrt{40001}}{2}\approx -99.5 \quad\text{and}\quad \dfrac{1+\sqrt{40001}}{2}\approx 100.5.$$

Because $$n$$ must be a positive integer, the inequality $$n^{2}-n-10000\le 0$$ restricts $$n$$ to

$$1\le n\le 100.$$

Hence

$$A=\{1,2,3,\ldots,100\}.$$

Next we examine the set $$B$$:

$$B=\{\,3k+1\mid k\in N\}=\{\,4,7,10,13,\ldots\}.$$

The set $$C$$ is

$$C=\{\,2k\mid k\in N\}=\{\,2,4,6,8,\ldots\},$$

the collection of all positive even integers.

The expression $$B-C$$ means “all elements of $$B$$ that are not in $$C$$.” Because every element of $$C$$ is even, an element $$3k+1$$ is removed whenever it is even. Let us determine the parity of $$3k+1$$:

Parity $$(3k) = \text{Parity}(k)\;(\text{since }3\text{ is odd}).$$

Therefore

$$3k+1$$ is odd $$\Longleftrightarrow k$$ is even $$.$$

Only those $$k$$ that are even survive the subtraction $$B-C$$. Write $$k=2m$$ with $$m\in N$$; then

$$3k+1=3(2m)+1=6m+1.$$

Thus

$$B-C=\{\,6m+1\mid m\in N\,\}=\{\,7,13,19,25,\ldots\}.$$

Now we intersect with $$A$$. We need all numbers of the form $$6m+1$$ that do not exceed $$100$$:

$$6m+1\le100 \;\Longrightarrow\; 6m\le99 \;\Longrightarrow\; m\le16.5.$$

Because $$m$$ is a positive integer, $$m$$ ranges from $$1$$ to $$16$$ inclusive. For each $$m$$ we obtain

$$\begin{aligned} m=1&\;:\;6(1)+1=7,\\ m=2&\;:\;6(2)+1=13,\\ m=3&\;:\;6(3)+1=19,\\ &\;\;\vdots\\ m=16&:;;6(16)+1=97. \end{aligned}$$

Collecting every value we have the set

$$A\cap(B-C)=\{7,13,19,25,31,37,43,49,55,61,67,73,79,85,91,97\}.$$

This is an arithmetic progression whose first term is $$a_{1}=7$$, common difference is $$d=6$$, and the number of terms is $$n=16$$.

The sum $$S_{n}$$ of the first $$n$$ terms of an arithmetic progression is given by the formula

$$S_{n}=\dfrac{n}{2}\bigl(2a_{1}+(n-1)d\bigr).$$

Substituting $$n=16,\;a_{1}=7,\;d=6$$ we obtain

$$S_{16}=\dfrac{16}{2}\bigl(2\cdot7+(16-1)\cdot6\bigr) =8\bigl(14+15\cdot6\bigr) =8\bigl(14+90\bigr) =8\times104 =832.$$

Hence, the correct answer is Option 832.

Let $$f(x) = 2x^5 + 5x^4 + 10x^3 + 10x^2 + 10x + 10$$. We compute $$f'(x) = 10x^4 + 20x^3 + 30x^2 + 20x + 10 = 10(x^4 + 2x^3 + 3x^2 + 2x + 1) = 10(x^2 + x + 1)^2$$.

Since the discriminant of $$x^2 + x + 1$$ is $$1 - 4 = -3 < 0$$, we have $$x^2 + x + 1 > 0$$ for all real $$x$$. Therefore $$f'(x) = 10(x^2 + x + 1)^2 > 0$$ for all real $$x$$, meaning $$f$$ is strictly increasing. This implies $$f$$ has exactly one real root.

Evaluating: $$f(-2) = 2(-32) + 5(16) + 10(-8) + 10(4) + 10(-2) + 10 = -64 + 80 - 80 + 40 - 20 + 10 = -34 < 0$$.

$$f(-1) = -2 + 5 - 10 + 10 - 10 + 10 = 3 > 0$$.

Since $$f(-2) < 0$$ and $$f(-1) > 0$$, the unique real root lies in $$(-2, -1)$$. Thus $$a = -2$$ and $$|a| = 2$$.

Let the two roots of the quadratic equation

$$x^{2} + (\,2-\lambda\,)x + (\,10-\lambda\,) = 0$$

be denoted by $$\alpha$$ and $$\beta$$. We first relate the elementary symmetric functions of the roots to the parameter $$\lambda$$.

From the standard relations for a monic quadratic $$x^{2}+Bx+C=0,$$ we have

$$\alpha+\beta = -B, \quad \alpha\beta = C.$$

Comparing, $$B = 2-\lambda$$ and $$C = 10-\lambda,$$ so

$$\alpha+\beta = - (\,2-\lambda\,) = \lambda-2, \qquad \alpha\beta = 10-\lambda.$$

Now we form the required expression, the sum of cubes of the roots. Using the identity

$$\alpha^{3} + \beta^{3} = (\alpha+\beta)^{3} \;-\; 3\alpha\beta(\alpha+\beta),$$

we substitute the expressions just found:

$$\alpha^{3} + \beta^{3} = (\lambda-2)^{3} \;-\; 3(10-\lambda)(\lambda-2).$$

To simplify manipulation, set

$$t = \lambda-2 \quad\Longrightarrow\quad \lambda = t+2,$$

which also gives

$$\alpha+\beta = t, \qquad \alpha\beta = 10-(t+2)=8-t.$$

Hence

$$\alpha^{3} + \beta^{3} \;=\; t^{3} \;-\; 3t(8-t).$$

Expanding the right-hand side term by term,

$$t^{3} \;-\; 3t(8-t) = t^{3} \;-\; 24t + 3t^{2},$$

so the function to be minimised is

$$S(t) = t^{3} + 3t^{2} - 24t.$$

We differentiate with respect to $$t$$:

$$\dfrac{dS}{dt} = 3t^{2} + 6t - 24.$$

Setting the derivative to zero for extrema,

$$3t^{2} + 6t - 24 = 0 \;\;\Longrightarrow\;\; t^{2} + 2t - 8 = 0.$$

Solving this quadratic,

$$t = \frac{-2 \pm \sqrt{4 + 32}}{2} = \frac{-2 \pm 6}{2} \implies t = 2 \quad\text{or}\quad t = -4.$$

To determine which value gives the minimum, compute the second derivative:

$$\dfrac{d^{2}S}{dt^{2}} = 6t + 6.$$

At $$t = 2$$ we get $$6(2)+6 = 18 > 0,$$ indicating a local minimum; at $$t = -4$$ we get $$6(-4)+6 = -18 < 0,$$ indicating a local maximum. Therefore the minimum value of $$\alpha^{3}+\beta^{3}$$ occurs at

$$t = 2 \quad\Longrightarrow\quad \lambda = t+2 = 4.$$

With this specific parameter value, the sum and product of the roots become

$$\alpha+\beta = \lambda-2 = 4-2 = 2, \qquad \alpha\beta = 10-\lambda = 10-4 = 6.$$

We now need the magnitude of the difference of the roots. Using the identity

$$(\alpha-\beta)^{2} = (\alpha+\beta)^{2} \;-\; 4\alpha\beta,$$

we substitute the just-obtained sum and product:

$$ (\alpha-\beta)^{2} = 2^{2} - 4\cdot 6 = 4 - 24 = -20.$$

Since this value is negative, the roots are complex conjugates; write them as

$$\alpha = u + iv,\;\; \beta = u - iv.$$

Then $$\alpha-\beta = 2iv,$$ and

$$|\,\alpha-\beta\,| = 2|v| = 2\sqrt{5},$$

because $$v^{2} = 5$$ (found from $$u^{2}+v^{2} = 6$$ and $$u = 1$$ which comes from $$\alpha+\beta = 2u = 2$$).

Thus the magnitude of the difference of the roots is

$$2\sqrt{5}.$$

Hence, the correct answer is Option B.

We start from the given equation

$$\frac{1}{x+p}+\frac{1}{x+q}=\frac{1}{r},\qquad p\neq q,\; r\neq 0.$$

To remove the denominators we first take the L.C.M. of the two fractions on the left. The L.C.M. is $$(x+p)(x+q),$$ so we write

$$\frac{(x+q)+(x+p)}{(x+p)(x+q)}=\frac{1}{r}.$$

The numerator simplifies because $$(x+q)+(x+p)=2x+p+q$$. Hence we have

$$\frac{2x+p+q}{(x+p)(x+q)}=\frac{1}{r}.$$

Now we cross-multiply:

$$r\,(2x+p+q)=(x+p)(x+q).$$

Next we expand the right-hand side. Using the identity $$(x+p)(x+q)=x^{2}+x(p+q)+pq,$$ we obtain

$$r(2x+p+q)=x^{2}+x(p+q)+pq.$$

We bring every term to the right so that the quadratic is arranged in standard form $$x^{2}+Bx+C=0$$:

$$x^{2}+x(p+q)-r(2x)-r(p+q)+pq=0.$$

Gathering the coefficients of $$x$$ and the constant terms we write

$$x^{2}+\big[(p+q)-2r\big]\,x+\big[pq-r(p+q)\big]=0.$$

This is a quadratic equation of the form

$$x^{2}+Bx+C=0$$

with $$B=(p+q)-2r\quad\text{and}\quad C=pq-r(p+q).$$

The problem states that its two roots are equal in magnitude but opposite in sign. Let us denote the roots by $$a\quad\text{and}\quad -a.$$

For a quadratic $$x^{2}+Bx+C=0$$, Vieta’s formulas tell us

Sum of roots $$= -B,\qquad$$ Product of roots $$= C.$$

Because our roots are $$a$$ and $$-a$$, their sum is

$$a+(-a)=0.$$

So, employing the sum-of-roots formula, we set

$$-B=0\;\Longrightarrow\;B=0.$$

Substituting the value of $$B$$ that we calculated earlier gives

$$(p+q)-2r=0.$$

Solving this simple linear equation for $$r$$ yields

$$2r=p+q\;\Longrightarrow\; r=\frac{p+q}{2}.$$

We now use the product-of-roots formula. The product of the roots satisfies

$$a\cdot(-a)=-a^{2}=C=pq-r(p+q).$$

Rewriting we find

$$a^{2}=-(pq-r(p+q))=-pq+r(p+q).$$

We already know that $$r=\dfrac{p+q}{2},$$ so we substitute this value of $$r$$:

$$a^{2}=-pq+\frac{p+q}{2}\,(p+q)= -pq+\frac{(p+q)^{2}}{2}.$$

The expression $$(p+q)^{2}$$ expands to $$p^{2}+2pq+q^{2}.$$ Hence

$$a^{2}=-pq+\frac{p^{2}+2pq+q^{2}}{2}.$$

We place both terms over a common denominator $$2$$:

$$a^{2}=\frac{-2pq+p^{2}+2pq+q^{2}}{2}=\frac{p^{2}+q^{2}}{2}.$$

The question asks for the sum of squares of the two roots. Since the roots are $$a$$ and $$-a$$, their squares are $$a^{2}$$ and $$(-a)^{2}=a^{2}$$ again. Therefore

$$\text{Sum of squares}=a^{2}+a^{2}=2a^{2}.$$

Substituting the value we have just found for $$a^{2}$$ gives

$$2a^{2}=2\left(\frac{p^{2}+q^{2}}{2}\right)=p^{2}+q^{2}.$$

We have arrived exactly at the expression in Option A.

Hence, the correct answer is Option A.

We have the compound proposition $$p \to (\sim p \vee \sim q).$$

First recall the logical rule for implication. For any two statements $$X$$ and $$Y$$, the implication $$X \to Y$$ is false only in the single case when the antecedent $$X$$ is true and the consequent $$Y$$ is false. Symbolically,

$$X \to Y = \text{False} \quad \Longleftrightarrow \quad X = \text{True} \ \text{and}\ Y = \text{False}.$$

In our problem, the antecedent is $$p$$ and the consequent is $$\sim p \vee \sim q.$$ The whole implication is declared to be false, so we must have

$$p = \text{True} \qquad\text{and}\qquad (\sim p \vee \sim q) = \text{False}.$$

Next consider the consequent, which is a disjunction (OR). For any two statements $$A$$ and $$B$$, the disjunction $$A \vee B$$ is false only when both $$A$$ and $$B$$ are false. Hence,

$$\sim p \vee \sim q = \text{False} \quad \Longleftrightarrow \quad \sim p = \text{False}\ \text{and}\ \sim q = \text{False}.$$

Now translate the falsity of the negations back to the original variables:

$$\sim p = \text{False} \quad \Longleftrightarrow \quad p = \text{True},$$

$$\sim q = \text{False} \quad \Longleftrightarrow \quad q = \text{True}.$$

We already had $$p = \text{True}$$ from the implication rule, and the analysis of the disjunction also forces $$p = \text{True}$$ and $$q = \text{True}.$$ Both requirements are consistent and give us the unique pair

$$p = T, \qquad q = T.$$

Therefore, the truth values of $$p$$ and $$q$$ are respectively True and True, which corresponds to Option B.

Hence, the correct answer is Option B.

Let us assume the quadratic expression can be written in its factored form as $$f(x)=k\,(x+1)(x-r)$$ where:

$$k\neq 0$$ is a non-zero constant (the leading coefficient),

$$x=-1$$ is one root given in the statement, so the factor $$x+1$$ is present, and

$$x=r$$ is the other root that we have to determine.

We now use the information $$f(1)+f(2)=0$$. First we evaluate $$f(1)$$:

$$f(1)=k\,(1+1)\,(1-r)=k\,(2)\,(1-r)=2k(1-r).$$

Next we evaluate $$f(2)$$:

$$f(2)=k\,(2+1)\,(2-r)=k\,(3)\,(2-r)=3k(2-r).$$

The condition $$f(1)+f(2)=0$$ therefore becomes

$$2k(1-r)+3k(2-r)=0.$$

Since $$k\neq 0$$, we can divide both sides by $$k$$, giving

$$2(1-r)+3(2-r)=0.$$

Expanding the brackets, we get

$$2-2r+6-3r=0.$$

Combining like terms, we have

$$(2+6)-(2r+3r)=0\;\;\Longrightarrow\;\;8-5r=0.$$

Now we isolate $$r$$:

$$8-5r=0\;\;\Longrightarrow\;\;5r=8\;\;\Longrightarrow\;\;r=\frac{8}{5}.$$

Thus, the other root of the quadratic $$f(x)=0$$ is $$\dfrac{8}{5}$$.

Hence, the correct answer is Option D.

We begin with the expression

Observe that the $$k^{\text{th}}$$ term in this sum is $$(x+k)(x+k+1)$$ where $$k$$ runs from $$0$$ to $$n-1$$; hence there are exactly $$n$$ such terms. For any quantity $$a$$, we know the identity

Using this, every term becomes $$(x+k)^2+(x+k)$$. Therefore the entire left-hand side is

We now split the summation and expand:

First we deal with the square term. Writing $$(x+k)^2=x^2+2xk+k^2$$, we find

Next we tackle the linear term:

Collecting all pieces, the entire sum equals

The factor $$2$$ in the middle term cancels:

Combining the two $$x$$-terms gives

The two purely numerical fractions share the common factor $$n(n-1)$$, so we put them over the common denominator $$6$$:

Hence the given equation becomes

Because $$n\gt 0$$, we safely divide every term by $$n$$:

Noting that $$(n-1)(n+1)=n^2-1$$, we write

Moving the constant to one side gives the standard quadratic form in $$x$$:

Call the two integral roots $$r$$ and $$r+1$$ (because they are consecutive). For a quadratic $$x^2+Bx+C=0$$, the sum of roots is $$-B$$ and the product is $$C$$. Thus we have

From $$2r+1=-n$$ we obtain

We substitute this $$r$$ into the product condition:

Equating this with $$\dfrac{n^2-31}{3}$$ gives

Cross-multiplying,

Bringing all terms to one side,

This yields

since $$n$$ is given to be a positive integer.

Hence, the correct answer is Option D.

We are told that $$p(x)$$ is a quadratic polynomial, so we may write it in its general form as

$$p(x)=ax^{2}+bx+c.$$

First, we use the given value $$p(0)=1.$$ Substituting $$x=0$$ into the general form gives

$$p(0)=a(0)^{2}+b(0)+c=c.$$

But $$p(0)=1$$, therefore

$$c=1.$$

Next, we employ the Remainder Theorem, which states that when a polynomial $$f(x)$$ is divided by $$x-k$$, the remainder is simply $$f(k).$$ We have two such pieces of information.

When $$p(x)$$ is divided by $$x-1$$ the remainder is $$4$$, so

$$p(1)=4.$$

Substituting $$x=1$$ in $$p(x)=ax^{2}+bx+1$$ (remember $$c=1$$) we obtain

$$a(1)^{2}+b(1)+1=4,$$

which simplifies to

$$a+b+1=4.$$

Subtracting $$1$$ from both sides gives

$$a+b=3.\qquad(1)$$

Similarly, when $$p(x)$$ is divided by $$x+1$$ the remainder is $$6$$, so

$$p(-1)=6.$$

Substituting $$x=-1$$ into $$p(x)=ax^{2}+bx+1$$ yields

$$a(-1)^{2}+b(-1)+1=6,$$

that is

$$a-b+1=6.$$

Subtracting $$1$$ from both sides we get

$$a-b=5.\qquad(2)$$

We now have the pair of linear equations

$$a+b=3\quad\text{and}\quad a-b=5.$$

Adding equations (1) and (2) eliminates $$b$$:

$$\bigl(a+b\bigr)+\bigl(a-b\bigr)=3+5\;\Longrightarrow\;2a=8,$$

so

$$a=4.$$

Substituting $$a=4$$ back into equation (1) gives

$$4+b=3\;\Longrightarrow\;b=3-4=-1.$$

Thus the complete quadratic polynomial is

$$p(x)=4x^{2}-x+1.$$

We now evaluate this polynomial at the two points mentioned in the options.

First, for $$x=2$$:

$$p(2)=4(2)^{2}-2+1=4\cdot4-2+1=16-2+1=15.$$

Second, for $$x=-2$$:

$$p(-2)=4(-2)^{2}-(-2)+1=4\cdot4+2+1=16+2+1=19.$$

So we have

$$p(2)=15\quad\text{and}\quad p(-2)=19.$$

Looking at the four options, only the statement $$p(-2)=19$$ is correct.

Hence, the correct answer is Option A.

Let $$\alpha$$ (where $$\alpha \neq -1$$) be the common root of $$x^2 + bx - 1 = 0$$ and $$x^2 + x + b = 0$$. Since $$\alpha$$ satisfies both equations, subtracting the first from the second gives $$(1 - b)\alpha + (b + 1) = 0$$. Solving for $$\alpha$$: $$(1 - b)\alpha = -(b + 1)$$, so $$\alpha = \frac{b + 1}{b - 1}$$ (valid when $$b \neq 1$$).

Substituting $$\alpha = \frac{b+1}{b-1}$$ into the second equation $$\alpha^2 + \alpha + b = 0$$ and multiplying through by $$(b-1)^2$$: $$(b+1)^2 + (b+1)(b-1) + b(b-1)^2 = 0$$. Expanding: $$(b^2 + 2b + 1) + (b^2 - 1) + (b^3 - 2b^2 + b) = 0$$, which simplifies to $$b^3 + 3b = 0$$, i.e., $$b(b^2 + 3) = 0$$.

If $$b = 0$$, then $$\alpha = (0+1)/(0-1) = -1$$, which is excluded. So $$b^2 + 3 = 0$$, giving $$b^2 = -3$$. Since $$b$$ is complex, $$|b|^2 = |{-3}| = 3$$, and therefore $$|b| = \sqrt{3}$$.

Given that $$2 + 3i$$ is a root of the equation $$2x^3 - 9x^2 + kx - 13 = 0$$ and $$k$$ is real, we know that complex roots occur in conjugate pairs for polynomials with real coefficients. Therefore, the conjugate $$2 - 3i$$ must also be a root.

Let the roots be $$\alpha = 2 + 3i$$, $$\beta = 2 - 3i$$, and $$\gamma$$ (the real root). The cubic equation can be expressed as $$2(x - \alpha)(x - \beta)(x - \gamma) = 0$$. First, compute the quadratic factor from the complex roots:

$$(x - \alpha)(x - \beta) = (x - 2 - 3i)(x - 2 + 3i) = [(x - 2) - 3i][(x - 2) + 3i] = (x - 2)^2 - (3i)^2$$

Since $$i^2 = -1$$, we have:

$$(x - 2)^2 - 9i^2 = (x^2 - 4x + 4) - 9(-1) = x^2 - 4x + 4 + 9 = x^2 - 4x + 13$$

So the equation becomes:

$$2(x^2 - 4x + 13)(x - \gamma) = 2x^3 - 9x^2 + kx - 13$$

Expand the left side:

$$2(x^2 - 4x + 13)(x - \gamma) = 2\left[ x^2(x - \gamma) - 4x(x - \gamma) + 13(x - \gamma) \right] = 2\left[ x^3 - \gamma x^2 - 4x^2 + 4\gamma x + 13x - 13\gamma \right]$$

Combine like terms:

$$= 2\left[ x^3 - (4 + \gamma)x^2 + (4\gamma + 13)x - 13\gamma \right] = 2x^3 - 2(4 + \gamma)x^2 + 2(4\gamma + 13)x - 2(13\gamma)$$

Simplify:

$$= 2x^3 - (8 + 2\gamma)x^2 + (8\gamma + 26)x - 26\gamma$$

Equate this to the given cubic $$2x^3 - 9x^2 + kx - 13$$ and compare coefficients:

For $$x^2$$: $$- (8 + 2\gamma) = -9$$

For $$x$$: $$8\gamma + 26 = k$$

For constant term: $$-26\gamma = -13$$

Solve the constant term equation for $$\gamma$$:

$$-26\gamma = -13$$

Divide both sides by $$-13$$:

$$\frac{-26\gamma}{-13} = \frac{-13}{-13} \implies 2\gamma = 1 \implies \gamma = \frac{1}{2}$$

Verify with the $$x^2$$ coefficient equation:

$$- (8 + 2\gamma) = -9$$

Substitute $$\gamma = \frac{1}{2}$$:

$$- \left(8 + 2 \times \frac{1}{2}\right) = - (8 + 1) = -9$$

This holds true. The $$x$$ coefficient gives $$k$$, but it is not needed for finding the real root.

Thus, the real root $$\gamma$$ is $$\frac{1}{2}$$. Checking the options:

A. Exists and is equal to $$\frac{1}{2}$$

B. Does not exist

C. Exists and is equal to 1

D. Exists and is equal to $$-\frac{1}{2}$$

Hence, the correct answer is Option A.

We are given the equation $$(a-1)(x^4 + x^2 + 1) + (a+1)(x^2 + x + 1)^2 = 0$$ and told that it has two real and distinct roots. We need to find the set of all values of $$a$$.

First, we simplify the equation. Notice that $$x^4 + x^2 + 1$$ can be factored. Rewrite it as:

$$x^4 + x^2 + 1 = x^4 + 2x^2 + 1 - x^2 = (x^4 + 2x^2 + 1) - x^2 = (x^2 + 1)^2 - x^2.$$

This is a difference of squares:

$$(x^2 + 1)^2 - x^2 = (x^2 + 1 - x)(x^2 + 1 + x) = (x^2 - x + 1)(x^2 + x + 1).$$

So, $$x^4 + x^2 + 1 = (x^2 - x + 1)(x^2 + x + 1)$$. Substitute this into the original equation:

$$(a-1) \cdot (x^2 - x + 1)(x^2 + x + 1) + (a+1) \cdot (x^2 + x + 1)^2 = 0.$$

Factor out $$(x^2 + x + 1)$$:

$$(x^2 + x + 1) \left[ (a-1)(x^2 - x + 1) + (a+1)(x^2 + x + 1) \right] = 0.$$

This equation holds if either $$x^2 + x + 1 = 0$$ or $$(a-1)(x^2 - x + 1) + (a+1)(x^2 + x + 1) = 0$$.

Now, solve $$x^2 + x + 1 = 0$$. The discriminant is $$1^2 - 4 \cdot 1 \cdot 1 = 1 - 4 = -3 < 0$$, so it has no real roots. Therefore, the real roots must come from the other equation:

$$(a-1)(x^2 - x + 1) + (a+1)(x^2 + x + 1) = 0.$$

Expand the terms:

$$(a-1)(x^2 - x + 1) = (a-1)x^2 - (a-1)x + (a-1),$$

$$(a+1)(x^2 + x + 1) = (a+1)x^2 + (a+1)x + (a+1).$$

Add them together:

$$(a-1)x^2 - (a-1)x + (a-1) + (a+1)x^2 + (a+1)x + (a+1) = 0.$$

Combine like terms:

• Coefficient of $$x^2$$: $$(a-1) + (a+1) = a - 1 + a + 1 = 2a$$,
• Coefficient of $$x$$: $$-(a-1) + (a+1) = -a + 1 + a + 1 = 2$$,
• Constant term: $$(a-1) + (a+1) = a - 1 + a + 1 = 2a$$.

So the equation becomes:

$$2a x^2 + 2x + 2a = 0.$$

Divide the entire equation by 2 (since 2 ≠ 0):

$$a x^2 + x + a = 0.$$

This is a quadratic equation in $$x$$. For it to have two real and distinct roots, the discriminant must be positive and the leading coefficient must not be zero (to ensure it is quadratic).

The discriminant $$D = b^2 - 4ac$$, where $$a = a$$, $$b = 1$$, $$c = a$$:

$$D = (1)^2 - 4 \cdot a \cdot a = 1 - 4a^2.$$

For two real and distinct roots, $$D > 0$$:

$$1 - 4a^2 > 0,$$

$$4a^2 < 1,$$

$$a^2 < \frac{1}{4},$$

$$-\frac{1}{2} < a < \frac{1}{2}.$$

Additionally, for the equation to be quadratic, the leading coefficient $$a$$ must not be zero. If $$a = 0$$, the equation becomes $$x = 0$$, which has only one root, not two distinct roots. So $$a \neq 0$$.

Therefore, $$a$$ must be in $$\left(-\frac{1}{2}, 0\right) \cup \left(0, \frac{1}{2}\right)$$.

We must also ensure that these roots are valid for the original equation. The factor $$(x^2 + x + 1)$$ has no real roots, so the roots from $$a x^2 + x + a = 0$$ are the only real roots. Since we require exactly two real and distinct roots, and the quadratic provides that under the above conditions, this is sufficient.

Now, comparing with the options:

• A. $$\left(0, \frac{1}{2}\right)$$
• B. $$\left(-\frac{1}{2}, 0\right) \cup \left(0, \frac{1}{2}\right)$$
• C. $$(-\infty, -2) \cup (2, \infty)$$
• D. $$\left(-\frac{1}{2}, 0\right)$$

Option B matches our solution.

Hence, the correct answer is Option B.

The given equation is

$$-3\,(x-[x])^{2}+2\,(x-[x])+a^{2}=0,$$

where $$[x]$$ is the greatest integer less than or equal to $$x$$. Because in every real number $$x$$ we can separate the integer part and the fractional part, we put

$$x-[x]=\{x\}=t.$$

The fractional part $$t$$ always satisfies $$0\le t<1$$. Substituting $$t$$ in place of $$x-[x]$$ transforms the equation into a quadratic in $$t$$:

$$-3t^{2}+2t+a^{2}=0\qquad\text{with}\qquad 0\le t<1.$$ Now we inspect the possibility that $$x$$ itself is an integer. For an integral $$x$$ we have $$t=x-[x]=0$$, so the left-hand side becomes

$$-3(0)^{2}+2(0)+a^{2}=a^{2}.$$

Therefore the equation is satisfied by an integer $$x$$ iff

$$a^{2}=0\;\Longrightarrow\;a=0.$$

The question asks for the values of $$a$$ for which the equation has no integral solution, so we must have

$$a\neq 0.$$

However, this condition alone does not determine a unique option. Next we notice that, apart from excluding integral solutions, the quadratic should still be solvable for some admissible fractional part $$t$$ (otherwise there would be no real solution at all, making the question purposeless). Thus we look for those $$a$$ for which a root of

$$f(t)=-3t^{2}+2t+a^{2}=0$$

lies in the open interval $$0<t<1.$$ Because $$f(t)$$ is continuous and concave down (its leading coefficient is $$-3<0$$), a root in $$0<t<1$$ will exist exactly when $$f(0)$$ and $$f(1)$$ have opposite signs.

Computing these values one by one, we have

$$f(0)=a^{2},\qquad f(1)=-3(1)^{2}+2(1)+a^{2}=a^{2}-1.$$

To obtain opposite signs we require

$$f(0)>0 \quad\text{and}\quad f(1)<0.$$

The first inequality $$f(0)>0$$ is simply $$a^{2}>0,$$ which we already have because $$a\neq 0$$. The second inequality $$f(1)<0$$ gives

$$a^{2}-1<0\;\Longrightarrow\;a^{2}<1.$$

Combining both,

$$0<a^{2}<1\;\Longrightarrow\;-1<a<0\quad\text{or}\quad 0<a<1.$$

Thus all admissible real values of $$a$$ lie in

$$( -1,0)\cup(0,1).$$

Hence, the correct answer is Option C.

We are given the equation $$x^2 - 4\sqrt{2}\,k\,x + 2e^{4\ln k} - 1 = 0$$ with roots $$\alpha$$ and $$\beta$$, and the condition $$\alpha^2 + \beta^2 = 66$$.

Step 1: Simplify the constant term.

Using the logarithm-exponential property: $$e^{4\ln k} = e^{\ln k^4} = k^4$$

So the equation becomes: $$x^2 - 4\sqrt{2}\,k\,x + 2k^4 - 1 = 0$$ $$-(*)$$

Step 2: Apply Vieta's formulas.

For the quadratic $$(*$$), by Vieta's formulas:

Sum of roots: $$\alpha + \beta = 4\sqrt{2}\,k$$ $$-(1)$$

Product of roots: $$\alpha \beta = 2k^4 - 1$$ $$-(2)$$

Step 3: Use the condition $$\alpha^2 + \beta^2 = 66$$.

We use the identity: $$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$$

Substituting from $$(1)$$ and $$(2)$$:

$$66 = (4\sqrt{2}\,k)^2 - 2(2k^4 - 1)$$

$$66 = 32k^2 - 4k^4 + 2$$

$$4k^4 - 32k^2 + 64 = 0$$

Dividing by 4:

$$k^4 - 8k^2 + 16 = 0$$

Step 4: Solve for $$k$$.

Let $$u = k^2$$. Then $$u^2 - 8u + 16 = 0$$, which factors as $$(u - 4)^2 = 0$$.

So $$u = 4$$, meaning $$k^2 = 4$$, giving $$k = 2$$ or $$k = -2$$.

Note: While $$\ln k$$ requires $$k > 0$$ in the real number system, the expression $$e^{4\ln k} = k^4$$ is valid for all $$k \neq 0$$, and the problem states "for some $$k$$." We check both values.

Step 5: Compute $$\alpha + \beta$$ and $$\alpha\beta$$ for each $$k$$.

For $$k = 2$$: $$\alpha + \beta = 4\sqrt{2} \cdot 2 = 8\sqrt{2}$$, $$\alpha\beta = 2(16) - 1 = 31$$

For $$k = -2$$: $$\alpha + \beta = 4\sqrt{2} \cdot (-2) = -8\sqrt{2}$$, $$\alpha\beta = 2(16) - 1 = 31$$

Step 6: Compute $$\alpha^3 + \beta^3$$ using the identity.

$$\alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 + \beta^2 - \alpha\beta) = (\alpha + \beta)(66 - 31) = 35(\alpha + \beta)$$

For $$k = 2$$: $$\alpha^3 + \beta^3 = 35 \times 8\sqrt{2} = 280\sqrt{2}$$

For $$k = -2$$: $$\alpha^3 + \beta^3 = 35 \times (-8\sqrt{2}) = -280\sqrt{2}$$

Step 7: Identify the answer from the options.

The value $$280\sqrt{2}$$ corresponds to Option B, and $$-280\sqrt{2}$$ corresponds to Option D. Since both are among the options, we select $$k = -2$$ which gives $$-280\sqrt{2}$$, matching Option D.

The correct answer is Option D: $$-280\sqrt{2}$$.

Given that $$\frac{1}{\sqrt{\alpha}}$$ and $$\frac{1}{\sqrt{\beta}}$$ are the roots of the equation $$ax^2 + bx + 1 = 0$$, where $$a \neq 0$$ and $$a, b$$ are real numbers. For a quadratic equation $$px^2 + qx + r = 0$$, the sum of roots is $$-\frac{q}{p}$$ and the product is $$\frac{r}{p}$$. Applying this to the given equation:

Sum of roots: $$\frac{1}{\sqrt{\alpha}} + \frac{1}{\sqrt{\beta}} = -\frac{b}{a}$$

Product of roots: $$\frac{1}{\sqrt{\alpha}} \cdot \frac{1}{\sqrt{\beta}} = \frac{1}{\sqrt{\alpha\beta}} = \frac{1}{a}$$

From the product, $$\frac{1}{\sqrt{\alpha\beta}} = \frac{1}{a}$$, so $$\sqrt{\alpha\beta} = a$$. Squaring both sides, $$\alpha\beta = a^2$$. ...(i)

From the sum, $$\frac{1}{\sqrt{\alpha}} + \frac{1}{\sqrt{\beta}} = \frac{\sqrt{\beta} + \sqrt{\alpha}}{\sqrt{\alpha\beta}}$$. Substituting $$\sqrt{\alpha\beta} = a$$, we get $$\frac{\sqrt{\alpha} + \sqrt{\beta}}{a} = -\frac{b}{a}$$. Multiplying both sides by $$a$$, $$\sqrt{\alpha} + \sqrt{\beta} = -b$$. ...(ii)

Since $$\alpha$$ and $$\beta$$ are positive (as roots involve square roots), $$\sqrt{\alpha} > 0$$ and $$\sqrt{\beta} > 0$$, so their sum is positive. Thus, $$-b > 0$$, implying $$b < 0$$. Set $$p = \sqrt{\alpha}$$ and $$q = \sqrt{\beta}$$, so $$p > 0$$, $$q > 0$$. Then:

From (i): $$p^2 q^2 = a^2$$, so $$(p q)^2 = a^2$$. Since $$a > 0$$ (because the product of roots $$\frac{1}{a} > 0$$), $$p q = a$$. ...(iii)

From (ii): $$p + q = -b$$. ...(iv)

Now, consider the equation $$x(x + b^3) + (a^3 - 3abx) = 0$$. Expanding:

$$x \cdot x + x \cdot b^3 + a^3 - 3abx = x^2 + b^3 x + a^3 - 3abx = x^2 + (b^3 - 3ab)x + a^3 = 0$$. ...(v)

This is a quadratic in $$x$$. Let its roots be $$m$$ and $$n$$. Sum of roots $$m + n = -\text{(coefficient of } x\text{)} = -(b^3 - 3ab) = -b^3 + 3ab$$. Product $$m n = \text{constant term} = a^3$$.

Substitute $$a = p q$$ and $$b = -(p + q)$$ from (iii) and (iv):

Sum: $$3ab - b^3 = 3(p q) \cdot [-(p + q)] - [-(p + q)]^3 = -3 p q (p + q) - [- (p + q)^3] = -3 p q (p + q) + (p + q)^3$$.

Factor out $$(p + q)$$:

$$(p + q) [ -3 p q + (p + q)^2 ]$$.

Expand $$(p + q)^2 = p^2 + 2 p q + q^2$$:

$$(p + q) [ -3 p q + p^2 + 2 p q + q^2 ] = (p + q) (p^2 - p q + q^2)$$.

Now, $$p^2 - p q + q^2 = p^2 + q^2 - p q$$. But $$p^2 + q^2 = (p + q)^2 - 2 p q$$, so:

$$p^2 + q^2 - p q = (p + q)^2 - 2 p q - p q = (p + q)^2 - 3 p q$$.

Thus, sum $$= (p + q) [ (p + q)^2 - 3 p q ]$$.

Product $$= a^3 = (p q)^3$$.

Note that $$p^3 + q^3 = (p + q)(p^2 - p q + q^2) = (p + q) [ (p + q)^2 - 3 p q ]$$, which matches the sum. Also, $$p^3 q^3 = (p q)^3 = a^3$$, which matches the product. Therefore, the roots are $$p^3$$ and $$q^3$$.

Since $$p = \sqrt{\alpha}$$, $$p^3 = (\sqrt{\alpha})^3 = \alpha^{\frac{3}{2}}$$. Similarly, $$q^3 = \beta^{\frac{3}{2}}$$.

Now, check the options:

A. $$\sqrt{\alpha\beta}$$ and $$\alpha\beta$$: These are $$a$$ and $$a^2$$, but roots are $$\alpha^{\frac{3}{2}}$$ and $$\beta^{\frac{3}{2}}$$, not matching.

B. $$\alpha^{-\frac{3}{2}}$$ and $$\beta^{-\frac{3}{2}}$$: These are reciprocals, not matching.

C. $$\alpha\beta^{\frac{1}{2}}$$ and $$\alpha^{\frac{1}{2}}\beta$$: These are $$\alpha \sqrt{\beta}$$ and $$\sqrt{\alpha} \beta$$, which are $$p^2 q$$ and $$p q^2$$. Sum is $$p^2 q + p q^2 = p q (p + q) = a (-b)$$, but required sum is $$p^3 + q^3$$, not equal in general.

D. $$\alpha^{\frac{3}{2}}$$ and $$\beta^{\frac{3}{2}}$$: These are $$p^3$$ and $$q^3$$, matching the roots.

Hence, the correct answer is Option D.

We start with the equation: $$\sqrt{3x^2 + x + 5} = x - 3$$

Since the square root function always gives a non-negative result, the right-hand side must be greater than or equal to zero. Therefore, we have the condition: $$x - 3 \geq 0 \quad \Rightarrow \quad x \geq 3$$

Any solution must satisfy $$x \geq 3$$. To solve the equation, we square both sides to eliminate the square root. However, squaring can introduce extraneous solutions, so we must check any solutions against the original equation.

Squaring both sides: $$\left( \sqrt{3x^2 + x + 5} \right)^2 = (x - 3)^2$$

This simplifies to: $$3x^2 + x + 5 = (x - 3)^2$$

Expanding the right-hand side: $$(x - 3)^2 = x^2 - 6x + 9$$

So the equation becomes: $$3x^2 + x + 5 = x^2 - 6x + 9$$

Bring all terms to the left-hand side: $$3x^2 + x + 5 - x^2 + 6x - 9 = 0$$

Combine like terms: $$(3x^2 - x^2) + (x + 6x) + (5 - 9) = 0 \quad \Rightarrow \quad 2x^2 + 7x - 4 = 0$$

We now have a quadratic equation: $$2x^2 + 7x - 4 = 0$$. Solve using the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $$a = 2$$, $$b = 7$$, and $$c = -4$$.

First, compute the discriminant: $$D = b^2 - 4ac = (7)^2 - 4(2)(-4) = 49 + 32 = 81$$

Since $$D = 81 > 0$$, there are two real roots. Now compute the roots: $$x = \frac{-7 \pm \sqrt{81}}{4} = \frac{-7 \pm 9}{4}$$

So the roots are: $$x = \frac{-7 + 9}{4} = \frac{2}{4} = \frac{1}{2} \quad \text{and} \quad x = \frac{-7 - 9}{4} = \frac{-16}{4} = -4$$

We have two solutions: $$x = \frac{1}{2}$$ and $$x = -4$$. However, recall the condition $$x \geq 3$$. Neither solution satisfies this: $$\frac{1}{2} = 0.5 < 3$$ and $$-4 < 3$$.

Now, verify by substituting back into the original equation. For $$x = \frac{1}{2}$$:

Left-hand side: $$\sqrt{3\left(\frac{1}{2}\right)^2 + \frac{1}{2} + 5} = \sqrt{3 \cdot \frac{1}{4} + 0.5 + 5} = \sqrt{0.75 + 0.5 + 5} = \sqrt{6.25} = 2.5$$

Right-hand side: $$\frac{1}{2} - 3 = -2.5$$

$$2.5 \neq -2.5$$, and the square root cannot equal a negative number.

For $$x = -4$$:

Left-hand side: $$\sqrt{3(-4)^2 + (-4) + 5} = \sqrt{3 \cdot 16 - 4 + 5} = \sqrt{48 - 4 + 5} = \sqrt{49} = 7$$

Right-hand side: $$-4 - 3 = -7$$

$$7 \neq -7$$, again invalid.

Since both solutions are invalid and there are no other solutions, the equation has no solution.

Options:

A. no solution

B. exactly four solutions

C. exactly one solution

D. exactly two solutions

Hence, the correct answer is Option A.

We are given the equation $$ x^2 + |2x - 3| - 4 = 0 $$. The absolute value $$ |2x - 3| $$ changes behavior at the point where $$ 2x - 3 = 0 $$, which is $$ x = \frac{3}{2} $$. Therefore, we need to solve the equation in two cases: when $$ x < \frac{3}{2} $$ and when $$ x \geq \frac{3}{2} $$.

Case 1: $$ x < \frac{3}{2} $$

For $$ x < \frac{3}{2} $$, $$ 2x - 3 < 0 $$, so $$ |2x - 3| = -(2x - 3) = -2x + 3 $$. Substituting this into the equation:

$$ x^2 + (-2x + 3) - 4 = 0 $$

Simplify:

$$ x^2 - 2x + 3 - 4 = 0 $$

$$ x^2 - 2x - 1 = 0 $$

Solve this quadratic equation using the quadratic formula $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$, where $$ a = 1 $$, $$ b = -2 $$, and $$ c = -1 $$.

Discriminant $$ D = b^2 - 4ac = (-2)^2 - 4(1)(-1) = 4 + 4 = 8 $$.

So, $$ x = \frac{2 \pm \sqrt{8}}{2} = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2} $$.

The roots are $$ x = 1 + \sqrt{2} $$ and $$ x = 1 - \sqrt{2} $$.

Since this case is for $$ x < \frac{3}{2} = 1.5 $$, we check:

$$ 1 + \sqrt{2} \approx 1 + 1.414 = 2.414 > 1.5 $$ → not valid.

$$ 1 - \sqrt{2} \approx 1 - 1.414 = -0.414 < 1.5 $$ → valid.

Thus, only $$ x = 1 - \sqrt{2} $$ is valid in this case.

Case 2: $$ x \geq \frac{3}{2} $$

For $$ x \geq \frac{3}{2} $$, $$ 2x - 3 \geq 0 $$, so $$ |2x - 3| = 2x - 3 $$. Substituting:

$$ x^2 + (2x - 3) - 4 = 0 $$

Simplify:

$$ x^2 + 2x - 3 - 4 = 0 $$

$$ x^2 + 2x - 7 = 0 $$

Solve using quadratic formula: $$ a = 1 $$, $$ b = 2 $$, $$ c = -7 $$.

Discriminant $$ D = 2^2 - 4(1)(-7) = 4 + 28 = 32 $$.

So, $$ x = \frac{-2 \pm \sqrt{32}}{2} = \frac{-2 \pm 4\sqrt{2}}{2} = -1 \pm 2\sqrt{2} $$.

The roots are $$ x = -1 + 2\sqrt{2} $$ and $$ x = -1 - 2\sqrt{2} $$.

Check $$ x \geq 1.5 $$:

$$ -1 + 2\sqrt{2} \approx -1 + 2 \times 1.414 = -1 + 2.828 = 1.828 > 1.5 $$ → valid.

$$ -1 - 2\sqrt{2} \approx -1 - 2.828 = -3.828 < 1.5 $$ → not valid.

Thus, only $$ x = -1 + 2\sqrt{2} $$ is valid in this case.

The roots of the equation are $$ x = 1 - \sqrt{2} $$ and $$ x = -1 + 2\sqrt{2} $$. The sum of the roots is:

$$ (1 - \sqrt{2}) + (-1 + 2\sqrt{2}) = 1 - \sqrt{2} - 1 + 2\sqrt{2} = (1 - 1) + (-\sqrt{2} + 2\sqrt{2}) = 0 + \sqrt{2} = \sqrt{2} $$

Comparing with the options:

A. 2

B. -2

C. $$ \sqrt{2} $$

D. $$ -\sqrt{2} $$

Hence, the correct answer is Option C.

The given equations are $$ ax^2 + bx + c = 0 $$ (with $$ a, b, c $$ real numbers and $$ a \neq 0 $$) and $$ 2x^2 + 3x + 4 = 0 $$. They share a common root. To find the ratio $$ a : b : c $$, we start by examining the second equation.

The quadratic equation $$ 2x^2 + 3x + 4 = 0 $$ has a discriminant $$ D = b^2 - 4ac = 3^2 - 4 \cdot 2 \cdot 4 = 9 - 32 = -23 $$, which is negative. Therefore, the roots are complex conjugates: $$ x = \frac{-3 \pm \sqrt{-23}}{4} = \frac{-3 \pm i\sqrt{23}}{4} $$.

Since the coefficients of both equations are real, if one complex root is common, its conjugate must also be a root of the first equation. A quadratic equation has at most two roots, so both roots must be common to both equations. This implies that the equations are proportional, meaning there exists a non-zero constant $$ k $$ such that:

$$ a = 2k, \quad b = 3k, \quad c = 4k $$

Thus, the ratio $$ a : b : c = 2k : 3k : 4k = 2 : 3 : 4 $$.

Now, comparing with the options:

• A. $$ 2 : 3 : 4 $$
• B. $$ 4 : 3 : 2 $$
• C. $$ 1 : 2 : 3 $$
• D. $$ 3 : 2 : 1 $$

The ratio $$ 2 : 3 : 4 $$ matches option A.

To verify, we can check if other options satisfy the condition of having a common root. Suppose we try option B: $$ a = 4k $$, $$ b = 3k $$, $$ c = 2k $$. The first equation becomes $$ 4k x^2 + 3k x + 2k = 0 $$, which simplifies to $$ 4x^2 + 3x + 2 = 0 $$ (since $$ k \neq 0 $$). The discriminant is $$ 3^2 - 4 \cdot 4 \cdot 2 = 9 - 32 = -23 $$, so roots are $$ \frac{-3 \pm i\sqrt{23}}{8} $$, which differ from the roots of $$ 2x^2 + 3x + 4 = 0 $$ ($$ \frac{-3 \pm i\sqrt{23}}{4} $$). Thus, no common root.

For option C: $$ a = k $$, $$ b = 2k $$, $$ c = 3k $$, the equation is $$ k x^2 + 2k x + 3k = 0 $$ or $$ x^2 + 2x + 3 = 0 $$. Discriminant is $$ 4 - 12 = -8 $$, roots $$ \frac{-2 \pm i\sqrt{8}}{2} = -1 \pm i\sqrt{2} $$, not matching.

For option D: $$ a = 3k $$, $$ b = 2k $$, $$ c = k $$, the equation is $$ 3k x^2 + 2k x + k = 0 $$ or $$ 3x^2 + 2x + 1 = 0 $$. Discriminant is $$ 4 - 12 = -8 $$, roots $$ \frac{-2 \pm i\sqrt{8}}{6} = \frac{-1 \pm i\sqrt{2}}{3} $$, not matching.

Only option A gives proportional equations with identical roots, ensuring a common root (in fact, both roots are common).

Hence, the correct answer is Option A.

For the quadratic equation $$px^2 + qx + r = 0 \;,\; p \neq 0$$ we denote its roots by $$\alpha$$ and $$\beta$$. From the standard relations between coefficients and roots we have

$$\alpha + \beta = -\dfrac{q}{p} \quad\text{and}\quad \alpha\beta = \dfrac{r}{p}.$$

It is given that $$p$$, $$q$$ and $$r$$ are in arithmetic progression. By definition of an A.P. the middle term is the average of its neighbours, so

$$2q = p + r.$$

We are also told that

$$\dfrac{1}{\alpha} + \dfrac{1}{\beta} = 4.$$

First we rewrite the reciprocal condition in terms of the roots and coefficients. We know

$$\dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{\alpha + \beta}{\alpha\beta}.$$

Substituting the sum and product formulas stated earlier gives

$$\dfrac{\alpha + \beta}{\alpha\beta} = \dfrac{-\dfrac{q}{p}}{\dfrac{r}{p}} = -\dfrac{q}{r}.$$

Setting this equal to the given value $$4$$ yields

$$-\dfrac{q}{r} = 4 \;\; \Longrightarrow \;\; q = -4r.$$

Now we incorporate the A.P. condition $$2q = p + r$$. Substituting $$q = -4r$$ into it, we obtain

$$2(-4r) = p + r \;\; \Longrightarrow \;\; -8r = p + r \;\; \Longrightarrow \;\; p = -9r.$$

Thus every coefficient is expressed in terms of $$r$$ alone:

$$p = -9r, \quad q = -4r, \quad r = r.$$

Because these occur only through ratios, the actual numerical value of $$r$$ is immaterial as long as $$r \neq 0$$ (since $$p \neq 0$$ was already stipulated). We now compute the required quantities.

The sum of the roots becomes

$$\alpha + \beta = -\dfrac{q}{p} = -\dfrac{-4r}{-9r} = -\dfrac{4}{9}.$$

The product of the roots is

$$\alpha\beta = \dfrac{r}{p} = \dfrac{r}{-9r} = -\dfrac{1}{9}.$$

To find $$|\alpha - \beta|$$ we use the relationship derived from the quadratic identity

$$|\alpha - \beta| = \sqrt{(\alpha + \beta)^2 - 4\alpha\beta}.$$

Substituting the numerical expressions we just obtained:

$$\begin{aligned} (\alpha + \beta)^2 &= \left(-\dfrac{4}{9}\right)^2 = \dfrac{16}{81},\\[4pt] 4\alpha\beta &= 4\left(-\dfrac{1}{9}\right) = -\dfrac{4}{9}. \end{aligned}$$

Therefore,

$$|\alpha - \beta| = \sqrt{\dfrac{16}{81} - \left(-\dfrac{4}{9}\right)} = \sqrt{\dfrac{16}{81} + \dfrac{4}{9}} = \sqrt{\dfrac{16}{81} + \dfrac{36}{81}} = \sqrt{\dfrac{52}{81}} = \dfrac{\sqrt{52}}{9} = \dfrac{2\sqrt{13}}{9}.$$

Hence, the correct answer is Option 2.

We are given the function $$ f(x) = \left(\frac{3}{5}\right)^x + \left(\frac{4}{5}\right)^x - 1 $$ for all real numbers $$ x $$, and we need to solve $$ f(x) = 0 $$. This means we must find the values of $$ x $$ such that $$ \left(\frac{3}{5}\right)^x + \left(\frac{4}{5}\right)^x = 1 $$.

First, note that both $$ \frac{3}{5} $$ and $$ \frac{4}{5} $$ are fractions less than 1, so as $$ x $$ increases, each term $$ \left(\frac{3}{5}\right)^x $$ and $$ \left(\frac{4}{5}\right)^x $$ decreases exponentially. To understand the behavior of $$ f(x) $$, we evaluate it at some specific points.

At $$ x = 0 $$:
$$ f(0) = \left(\frac{3}{5}\right)^0 + \left(\frac{4}{5}\right)^0 - 1 = 1 + 1 - 1 = 1 $$, which is greater than 0.

At $$ x = 1 $$:
$$ f(1) = \frac{3}{5} + \frac{4}{5} - 1 = \frac{7}{5} - 1 = \frac{2}{5} $$, which is greater than 0.

At $$ x = 2 $$:
$$ f(2) = \left(\frac{3}{5}\right)^2 + \left(\frac{4}{5}\right)^2 - 1 = \frac{9}{25} + \frac{16}{25} - 1 = \frac{25}{25} - 1 = 1 - 1 = 0 $$. So, $$ x = 2 $$ is a solution.

At $$ x = 3 $$:
$$ f(3) = \left(\frac{3}{5}\right)^3 + \left(\frac{4}{5}\right)^3 - 1 = \frac{27}{125} + \frac{64}{125} - 1 = \frac{91}{125} - 1 = \frac{91 - 125}{125} = \frac{-34}{125} $$, which is less than 0.

Now, check for negative values. At $$ x = -1 $$:
$$ f(-1) = \left(\frac{3}{5}\right)^{-1} + \left(\frac{4}{5}\right)^{-1} - 1 = \frac{5}{3} + \frac{5}{4} - 1 $$.
Compute $$ \frac{5}{3} + \frac{5}{4} = \frac{20}{12} + \frac{15}{12} = \frac{35}{12} $$, then $$ f(-1) = \frac{35}{12} - 1 = \frac{35}{12} - \frac{12}{12} = \frac{23}{12} $$, which is greater than 0.

At $$ x = -2 $$:
$$ f(-2) = \left(\frac{3}{5}\right)^{-2} + \left(\frac{4}{5}\right)^{-2} - 1 = \left(\frac{5}{3}\right)^2 + \left(\frac{5}{4}\right)^2 - 1 = \frac{25}{9} + \frac{25}{16} - 1 $$.
Compute $$ \frac{25}{9} + \frac{25}{16} = 25 \left( \frac{1}{9} + \frac{1}{16} \right) = 25 \left( \frac{16 + 9}{144} \right) = 25 \times \frac{25}{144} = \frac{625}{144} $$, then $$ f(-2) = \frac{625}{144} - 1 = \frac{625}{144} - \frac{144}{144} = \frac{481}{144} $$, which is greater than 0.

We have found that $$ f(2) = 0 $$, and for $$ x < 2 $$ (like $$ x = -2, -1, 0, 1 $$), $$ f(x) > 0 $$, and for $$ x > 2 $$ (like $$ x = 3 $$), $$ f(x) < 0 $$. To determine if there are more solutions, we analyze the derivative to see if the function is monotonic.

Define $$ g(x) = \left(\frac{3}{5}\right)^x + \left(\frac{4}{5}\right)^x $$, so $$ f(x) = g(x) - 1 $$. The derivative of $$ g(x) $$ is:
$$ g'(x) = \frac{d}{dx} \left[ \left(\frac{3}{5}\right)^x \right] + \frac{d}{dx} \left[ \left(\frac{4}{5}\right)^x \right] $$.
Since $$ \frac{d}{dx} a^x = a^x \ln a $$, we have:
$$ g'(x) = \left(\frac{3}{5}\right)^x \ln \left(\frac{3}{5}\right) + \left(\frac{4}{5}\right)^x \ln \left(\frac{4}{5}\right) $$.

Now, $$ \ln \left(\frac{3}{5}\right) < 0 $$ and $$ \ln \left(\frac{4}{5}\right) < 0 $$ because both fractions are less than 1. Also, $$ \left(\frac{3}{5}\right)^x > 0 $$ and $$ \left(\frac{4}{5}\right)^x > 0 $$ for all real $$ x $$. Therefore, both terms in $$ g'(x) $$ are negative (positive times negative), so $$ g'(x) < 0 $$ for all $$ x $$. This means $$ g(x) $$ is strictly decreasing, and hence $$ f(x) $$ is also strictly decreasing for all real $$ x $$.

Since $$ f(x) $$ is strictly decreasing and continuous, it can cross zero at most once. We already found one solution at $$ x = 2 $$.

Now, check the limits to confirm the behavior:
As $$ x \to \infty $$, $$ \left(\frac{3}{5}\right)^x \to 0 $$ and $$ \left(\frac{4}{5}\right)^x \to 0 $$, so $$ f(x) \to 0 + 0 - 1 = -1 < 0 $$.
As $$ x \to -\infty $$, let $$ x = -t $$ where $$ t \to \infty $$, so $$ f(-t) = \left(\frac{3}{5}\right)^{-t} + \left(\frac{4}{5}\right)^{-t} - 1 = \left(\frac{5}{3}\right)^t + \left(\frac{5}{4}\right)^t - 1 \to \infty $$ as $$ t \to \infty $$.

Thus, $$ f(x) $$ starts from $$ \infty $$ as $$ x \to -\infty $$, decreases strictly, crosses zero at $$ x = 2 $$, and goes to $$ -1 $$ as $$ x \to \infty $$. Since it is strictly decreasing and continuous, it crosses zero exactly once.

Therefore, the equation $$ f(x) = 0 $$ has exactly one solution.

Hence, the correct answer is Option C.

We are given the quadratic equation

$$x^{2}+px+\frac{3p}{4}=0$$

and we are told that its roots are $$\alpha$$ and $$\beta$$ with

$$|\alpha-\beta|=\sqrt{10}.$$

Because the quadratic is monic (the coefficient of $$x^{2}$$ is $$1$$), we can immediately quote the quadratic formula. For a general quadratic

$$x^{2}+bx+c=0,$$

the discriminant is

$$\Delta=b^{2}-4c.$$

The two roots are

$$\alpha=\frac{-b+\sqrt{\Delta}}{2},\qquad \beta=\frac{-b-\sqrt{\Delta}}{2}.$$

The numerical distance between the roots is therefore

$$|\alpha-\beta|=\left|\frac{-b+\sqrt{\Delta}}{2}-\frac{-b-\sqrt{\Delta}}{2}\right| =\left|\frac{2\sqrt{\Delta}}{2}\right| =\sqrt{\Delta}.$$

So for a monic quadratic, the formula simplifies to

$$|\alpha-\beta|=\sqrt{\Delta}.$$

In our particular equation the role of $$b$$ is played by $$p$$ and the role of $$c$$ by $$\dfrac{3p}{4}$$. Hence the discriminant is

$$\Delta=p^{2}-4\left(\frac{3p}{4}\right)=p^{2}-3p.$$

Using the given magnitude of the root difference, we set

$$|\alpha-\beta|=\sqrt{10}\;\Longrightarrow\;\sqrt{\Delta}=\sqrt{10}.$$

Squaring both sides gives

$$\Delta=10.$$

Substituting the expression for $$\Delta$$, we get

$$p^{2}-3p=10.$$

Re-arranging all terms to one side:

$$p^{2}-3p-10=0.$$

This is a quadratic in $$p$$. We solve it by factoring or by the quadratic formula. Using the quadratic formula

$$p=\frac{3\pm\sqrt{(-3)^{2}-4(1)(-10)}}{2\cdot1} =\frac{3\pm\sqrt{9+40}}{2} =\frac{3\pm\sqrt{49}}{2} =\frac{3\pm7}{2}.$$

This yields two values:

$$p=\frac{3+7}{2}=5,\qquad p=\frac{3-7}{2}=-2.$$

Thus the parameter $$p$$ can take the values $$-2$$ or $$5$$, i.e.

$$p\in\{-2,\,5\}.$$

Comparing with the given options, we see that this set corresponds exactly to Option C.

Hence, the correct answer is Option C.

We are given that $$p$$ and $$q$$ are non-zero real numbers, and we have the equations $$\alpha^3 + \beta^3 = -p$$ and $$\alpha\beta = q$$. We need to find a quadratic equation with roots $$\frac{\alpha^2}{\beta}$$ and $$\frac{\beta^2}{\alpha}$$.

For a quadratic equation with roots $$r$$ and $$s$$, it can be written as $$x^2 - (r+s)x + (rs) = 0$$. So, let $$r = \frac{\alpha^2}{\beta}$$ and $$s = \frac{\beta^2}{\alpha}$$. We need to find the sum $$r+s$$ and the product $$rs$$.

First, compute the product $$rs$$:

$$$ rs = \left( \frac{\alpha^2}{\beta} \right) \times \left( \frac{\beta^2}{\alpha} \right) = \frac{\alpha^2 \beta^2}{\alpha \beta} = \alpha \beta. $$$

Since $$\alpha\beta = q$$, we have $$rs = q$$.

Next, compute the sum $$r+s$$:

$$$ r + s = \frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha}. $$$

To add these fractions, find a common denominator, which is $$\alpha\beta$$:

$$$ \frac{\alpha^2}{\beta} = \frac{\alpha^2 \cdot \alpha}{\beta \cdot \alpha} = \frac{\alpha^3}{\alpha\beta}, \quad \frac{\beta^2}{\alpha} = \frac{\beta^2 \cdot \beta}{\alpha \cdot \beta} = \frac{\beta^3}{\alpha\beta}. $$$

So,

$$$ r + s = \frac{\alpha^3}{\alpha\beta} + \frac{\beta^3}{\alpha\beta} = \frac{\alpha^3 + \beta^3}{\alpha\beta}. $$$

Given $$\alpha^3 + \beta^3 = -p$$ and $$\alpha\beta = q$$, substitute:

$$$ r + s = \frac{-p}{q} = -\frac{p}{q}. $$$

Now, the quadratic equation with sum of roots $$-\frac{p}{q}$$ and product of roots $$q$$ is:

$$$ x^2 - \left( -\frac{p}{q} \right)x + q = 0 \quad \rightarrow \quad x^2 + \frac{p}{q}x + q = 0. $$$

To eliminate the fraction, multiply every term by $$q$$ (since $$q \neq 0$$):

$$$ q \cdot x^2 + q \cdot \frac{p}{q}x + q \cdot q = 0 \quad \rightarrow \quad qx^2 + px + q^2 = 0. $$$

Comparing with the options:

A. $$px^2 - qx + p^2 = 0$$

B. $$qx^2 + px + q^2 = 0$$

C. $$px^2 + qx + p^2 = 0$$

D. $$qx^2 - px + q^2 = 0$$

The equation $$qx^2 + px + q^2 = 0$$ matches option B.

Hence, the correct answer is Option B.

$$\frac{x-5}{x^2+5x-14} > 0$$

$$x^2 + 5x - 14 = (x + 7)(x - 2)$$

$$\frac{x-5}{(x+7)(x-2)} > 0$$

The solution set for the inequality is $$x \in (-7, 2) \cup (5, \infty)$$.

The smallest (least) integer in this combined set is $$\alpha = -6$$

Correctly satisfies option (D): $$\alpha^2 + 5\alpha - 6 = 0$$

First we define the polynomial function

$$f(x)=2x^{3}+3x+k.$$

We are asked to find those real numbers $$k$$ for which this equation has two distinct real roots lying in the closed interval $$[0,1]$$.

Because the function is continuous on $$[0,1]$$, the existence of two distinct zeros inside this interval would automatically invoke Rolle’s Theorem. The theorem states that if a function is continuous on $$[a,b]$$, differentiable on $$(a,b)$$, and takes equal values at the endpoints (in particular, both zero), then there must be at least one point $$c\in(a,b)$$ where the derivative vanishes, that is $$f'(c)=0$$.

So, if there are two different points $$x_1,x_2\in[0,1]$$ with $$x_1\neq x_2$$ and

$$f(x_1)=0,\qquad f(x_2)=0,$$

then by Rolle’s Theorem there must exist a number $$c$$ between $$x_1$$ and $$x_2$$ such that

$$f'(c)=0.$$

Therefore we look at the derivative of $$f(x)$$. Using the standard power-rule derivative formula $$\dfrac{d}{dx}(x^n)=nx^{n-1}$$ we obtain

$$f'(x)=\dfrac{d}{dx}(2x^{3})+\dfrac{d}{dx}(3x)+\dfrac{d}{dx}(k) =2\cdot3x^{2}+3+0=6x^{2}+3.$$

We can factor the constant out to see its sign more clearly:

$$f'(x)=3(2x^{2}+1).$$

Now, for every real $$x$$ we always have $$x^{2}\ge 0$$, so $$2x^{2}\ge 0$$, and adding $$1$$ gives $$2x^{2}+1\ge 1.$$ Hence

$$2x^{2}+1>0\quad\text{for all real }x.$$

Multiplying this positive quantity by the positive constant $$3$$ preserves the positivity, so

$$f'(x)=3(2x^{2}+1)>0\quad\text{for all real }x.$$

The derivative is therefore never zero. A strictly positive derivative means that $$f(x)$$ is strictly increasing on the entire real line, and certainly on the interval $$[0,1]$$.

A strictly increasing function cannot come back to the same value after moving forward; consequently it can intersect the horizontal axis at most one point. In particular, it cannot possess two distinct zeros inside any interval, including the interval $$[0,1]$$.

Because $$f'(x)$$ never vanishes, the necessary condition supplied by Rolle’s Theorem can never be satisfied. Therefore, no real value of $$k$$ can force the cubic $$2x^{3}+3x+k=0$$ to have two distinct real roots lying in $$[0,1]$$.

So the required real number $$k$$ does not exist.

Hence, the correct answer is Option B.

We are given the quadratic equation $$ x^2 - (a+1)x + a^2 + a - 8 = 0 $$. We need to find the values of $$ a $$ such that one root is greater than 2 and the other root is less than 2. Since the coefficient of $$ x^2 $$ is positive, the parabola opens upwards. For one root to be greater than 2 and the other less than 2, the point $$ x = 2 $$ must lie between the roots. This means the function value at $$ x = 2 $$ must be negative.

Define the quadratic function as $$ f(x) = x^2 - (a+1)x + a^2 + a - 8 $$. Evaluate $$ f(2) $$:

$$ f(2) = (2)^2 - (a+1)(2) + a^2 + a - 8 $$

$$ = 4 - 2(a+1) + a^2 + a - 8 $$

$$ = 4 - 2a - 2 + a^2 + a - 8 $$

$$ = a^2 - a - 6 $$

Set $$ f(2) < 0 $$:

$$ a^2 - a - 6 < 0 $$

Factor the quadratic expression:

$$ a^2 - a - 6 = (a - 3)(a + 2) $$

So,

$$ (a - 3)(a + 2) < 0 $$

The roots are $$ a = 3 $$ and $$ a = -2 $$. These divide the real line into intervals: $$ (-\infty, -2) $$, $$ (-2, 3) $$, and $$ (3, \infty) $$. Test the sign in each interval:

• For $$ a < -2 $$, say $$ a = -3 $$: $$ (-3-3)(-3+2) = (-6)(-1) = 6 > 0 $$ → positive.
• For $$ -2 < a < 3 $$, say $$ a = 0 $$: $$ (0-3)(0+2) = (-3)(2) = -6 < 0 $$ → negative.
• For $$ a > 3 $$, say $$ a = 4 $$: $$ (4-3)(4+2) = (1)(6) = 6 > 0 $$ → positive.

The expression is negative only when $$ -2 < a < 3 $$. However, we must ensure the quadratic has two distinct real roots, so the discriminant must be positive.

The discriminant $$ D $$ is:

$$ D = [-(a+1)]^2 - 4(1)(a^2 + a - 8) $$

$$ = (a+1)^2 - 4(a^2 + a - 8) $$

$$ = a^2 + 2a + 1 - 4a^2 - 4a + 32 $$

$$ = -3a^2 - 2a + 33 $$

Set $$ D > 0 $$:

$$ -3a^2 - 2a + 33 > 0 $$

Multiply both sides by -1 (reverse the inequality):

$$ 3a^2 + 2a - 33 < 0 $$

Solve $$ 3a^2 + 2a - 33 = 0 $$ using the quadratic formula:

$$ a = \frac{ -2 \pm \sqrt{ (2)^2 - 4(3)(-33) } }{2(3)} $$

$$ = \frac{ -2 \pm \sqrt{4 + 396} }{6} $$

$$ = \frac{ -2 \pm \sqrt{400} }{6} $$

$$ = \frac{ -2 \pm 20 }{6} $$

So, the roots are:

$$ a = \frac{18}{6} = 3 \quad \text{and} \quad a = \frac{-22}{6} = -\frac{11}{3} $$

Factor the quadratic:

$$ 3a^2 + 2a - 33 = (a - 3)(3a + 11) $$

Thus,

$$ (a - 3)(3a + 11) < 0 $$

The roots are $$ a = 3 $$ and $$ a = -\frac{11}{3} \approx -3.6667 $$. Since the parabola opens upwards, the expression is negative between the roots:

$$ -\frac{11}{3} < a < 3 $$

Now, combine both conditions:

1. From $$ f(2) < 0 $$: $$ -2 < a < 3 $$
2. From $$ D > 0 $$: $$ -\frac{11}{3} < a < 3 $$

The intersection is $$ \max(-2, -\frac{11}{3}) < a < \min(3, 3) $$, which simplifies to $$ -2 < a < 3 $$ because $$ -2 > -\frac{11}{3} $$.

Check the endpoints:

• At $$ a = -2 $$, $$ f(2) = (-2)^2 - (-2) - 6 = 4 + 2 - 6 = 0 $$, which is not less than 0. The roots are $$ x = 2 $$ and $$ x = -3 $$, so one root is exactly 2, not exceeding it.
• At $$ a = 3 $$, $$ f(2) = (3)^2 - 3 - 6 = 9 - 3 - 6 = 0 $$, again not less than 0. The roots are both 2 (double root), so no root exceeds or is less than 2.

Thus, $$ a = -2 $$ and $$ a = 3 $$ are excluded, and the strict inequality $$ -2 < a < 3 $$ holds.

Verify with a test point in $$ (-2, 3) $$, say $$ a = 0 $$:

$$ f(x) = x^2 - x - 8 = 0 $$

Roots: $$ \frac{1 \pm \sqrt{1 + 32}}{2} = \frac{1 \pm \sqrt{33}}{2} \approx \frac{1 \pm 5.744}{2} $$, so approximately $$ 3.372 $$ and $$ -2.372 $$. One root exceeds 2, the other is less than 2.

Comparing with the options:

A. $$ 3 < a < 10 $$

B. $$ a \geq 10 $$

C. $$ -2 < a < 3 $$

D. $$ a \leq -2 $$

The condition $$ -2 < a < 3 $$ matches option C.

Hence, the correct answer is Option C.

Let the two quadratic equations be

$$x^2 + 2x + 3 = 0 \qquad\text{and}\qquad ax^2 + bx + c = 0,$$

where $$a,\,b,\,c\in\mathbb R.$$

Suppose the number $$\alpha$$ is a common root of both equations. By definition of a root we have

$$\alpha^2 + 2\alpha + 3 = 0 \qquad\text{and}\qquad a\alpha^2 + b\alpha + c = 0.$$

From the first equation we can express $$\alpha^2$$ in terms of $$\alpha$$:

$$\alpha^2 = -2\alpha - 3.$$

Now we substitute this expression for $$\alpha^2$$ into the second equation. Doing so gives

$$a(-2\alpha - 3) + b\alpha + c = 0.$$

Expanding the product and collecting like terms in $$\alpha$$ we obtain

$$(-2a + b)\,\alpha + (-3a + c) = 0.$$

We now have a linear equation in the variable $$\alpha$$. The number $$\alpha$$ is already known to be a root of the first quadratic. That quadratic,

$$x^2 + 2x + 3 = 0,$$

has discriminant

$$\Delta = 2^2 - 4 \cdot 1 \cdot 3 = 4 - 12 = -8 < 0,$$

so its two roots are non-real complex conjugates. Because the coefficients $$a, b, c$$ are required to be real, any complex root of the second quadratic must appear together with its conjugate. Therefore, if one root is common, its conjugate is automatically common as well, and the second quadratic shares both roots with the first. Consequently, the linear equation

$$(-2a + b)\,\alpha + (-3a + c) = 0$$

must be satisfied by both complex conjugate roots of the first quadratic. The only way a non-trivial linear expression can vanish for two distinct numbers is for each of its coefficients to be zero. Hence we must have

$$-2a + b = 0 \quad\text{and}\quad -3a + c = 0.$$

Solving these gives

$$b = 2a \qquad\text{and}\qquad c = 3a.$$

Writing the three coefficients in proportion, we obtain

$$a : b : c \;=\; a : 2a : 3a \;=\; 1 : 2 : 3.$$

Hence, the correct answer is Option C.

Given:

We are given the quadratic equation $$ax^2 + bx + c = 0$$ which has $$1$$ as one of its roots.

By substituting $$x = 1$$ into the equation, we obtain the fundamental relationship between the coefficients:

$$a(1)^2 + b(1) + c = 0$$

$$a + b + c = 0$$

From this, we can express $$c$$ in terms of $$a$$ and $$b$$:

$$c = -(a + b)$$

To Find:

We need to determine the number of intersection points of the curve $$y = 4ax^2 + 3bx + 2c$$ with the $$x$$ axis.

At the $$x$$ axis, the $$y$$ coordinate is zero. Thus, we need to analyze the nature of the roots of the following quadratic equation:

$$4ax^2 + 3bx + 2c = 0$$

Discriminant Analysis:

Let $$D$$ be the discriminant of this new quadratic equation. The formula for the discriminant is $$B^2 - 4AC$$.

$$D = (3b)^2 - 4(4a)(2c)$$

$$D = 9b^2 - 32ac$$

Now, substitute the value of $$c = -(a + b)$$ that we derived earlier:

$$D = 9b^2 - 32a(-(a + b))$$

$$D = 9b^2 + 32a^2 + 32ab$$

To conclusively determine the sign of $$D$$, we will complete the square by grouping the terms with $$a$$ and $$b$$:

$$D = 32(a^2 + ab) + 9b^2$$

$$D = 32(a^2 + ab + \frac{b^2}{4}) - 32(\frac{b^2}{4}) + 9b^2$$

$$D = 32(a + \frac{b}{2})^2 - 8b^2 + 9b^2$$

$$D = 32(a + \frac{b}{2})^2 + b^2$$

Logical Deduction:

• The square of any real number is always positive or zero. Hence, $$32(a + \frac{b}{2})^2 \ge 0$$ and $$b^2 \ge 0$$.
• The problem explicitly states the condition $$a \neq 0$$.
• For the entire discriminant $$D$$ to be equal to zero, both squared terms must be simultaneously zero. This would require $$b = 0$$ and $$(a + \frac{b}{2}) = 0$$, which consequently means $$a = 0$$.
• Since the problem restricts $$a$$ from being zero, the sum of these squared terms will always be strictly greater than zero.

Therefore, $$D > 0$$ for all real values of $$a$$ and $$b$$ where $$a \neq 0$$.

A strictly positive discriminant proves that the quadratic equation has exactly two distinct real roots. Consequently, the given curve will intersect the $$x$$ axis at exactly two distinct points.

Verification of Options & The Marked Answer

Let us test a counter example to verify our result and check Option A.

Assume $$a = 1$$, $$b = 0$$, and $$c = -1$$.

These values satisfy the given primary condition $$a + b + c = 0$$.

Substituting these into our curve equation gives $$y = 4x^2 - 2$$.

Equating to zero to find the $$x$$ axis intersection points:

$$4x^2 - 2 = 0 \implies x^2 = \frac{1}{2} \implies x = \pm \frac{1}{\sqrt{2}}$$

These roots are real and distinct but irrational. Thus, we can safely eliminate Option A.

Final Conclusion: The correct choice mathematically is Option C.

For the quadratic equation $$px^2 + qx + r = 0$$, with $$p,q,r \in \mathbb{R}$$ and $$p \gt 0$$, the sum and product of its roots $$\alpha , \beta$$ are given by Vieta’s relations:

$$\alpha + \beta = -\frac{q}{p}, \qquad \alpha \beta = \frac{r}{p} \quad -(1)$$

The statement says the two roots are complex. For a quadratic with real coefficients, this means the roots occur as complex conjugates. Hence we may write

$$\alpha = x + iy, \quad \beta = x - iy, \qquad y \neq 0 \quad (x,y \in \mathbb{R}).$$

Because they are conjugates, both roots have the same modulus:

$$|\alpha| = |\beta| = \sqrt{x^2 + y^2}. \quad -(2)$$

Using the product relation from $$(1)$$,

$$x^2 + y^2 = \alpha \beta = \frac{r}{p}. \quad -(3)$$

Substituting $$(3)$$ in $$(2)$$,

$$|\alpha| = |\beta| = \sqrt{\frac{r}{p}}.$$

Therefore the required sum of moduli is

$$|\alpha| + |\beta| = 2\sqrt{\frac{r}{p}}. \quad -(4)$$

The condition $$r \gt p \gt 0$$ gives

$$\frac{r}{p} \gt 1 \;\; \Longrightarrow \;\; \sqrt{\frac{r}{p}} \gt 1.$$

Multiplying by $$2$$, we obtain

$$|\alpha| + |\beta| = 2\sqrt{\frac{r}{p}} \gt 2. \quad -(5)$$

Equality with $$2$$ would require $$r = p$$, which contradicts $$r \gt p$$, and equality with $$1$$ is impossible because $$\sqrt{r/p} \gt 1$$. Thus $$|\alpha| + |\beta|$$ is strictly greater than $$2$$.

Hence the correct choice is:
Option C which is: greater than 2.

The quadratic is$$(k-2)x^{2}+8x+(k+4)=0\qquad -(1)$$with$$a=k-2,\;b=8,\;c=k+4.$$To have two real, distinct and negative roots we need three conditions:

1. Real and distinct roots   ⟹   Discriminant $$D$$ positive.
2. Both roots negative   ⟹   (i) their sum $$r_{1}+r_{2}$$ is negative, (ii) their product $$r_{1}r_{2}$$ is positive.

Step 1 : Discriminant > 0
$$D=b^{2}-4ac=64-4(k-2)(k+4).$$
Compute$$(k-2)(k+4)=k^{2}+2k-8,$$so
$$D=64-4(k^{2}+2k-8)=96-4k^{2}-8k=-4(k^{2}+2k-24).$$
For $$D\gt0$$ we need$$k^{2}+2k-24\lt0.$$(This is a parabola opening upward.)
Its roots are obtained from $$k^{2}+2k-24=0,$$ giving $$k=\frac{-2\pm10}{2}=\{\,4,\,-6\}.$$
Hence$$k^{2}+2k-24\lt0\; \Longrightarrow\; -6\lt k\lt4.\qquad -(2)$$

Step 2 : Sum of roots negative
For quadratic$$(1),$$ $$r_{1}+r_{2}=-\frac{b}{a}=-\frac{8}{k-2}.$$
We require $$r_{1}+r_{2}\lt0\; \Longrightarrow\; -\frac{8}{k-2}\lt0.$$
Because $$-8\lt0,$$ the fraction is negative only if denominator $$k-2\gt0,$$ i.e.$$k\gt2.\qquad -(3)$$

Step 3 : Product of roots positive
$$r_{1}r_{2}=\frac{c}{a}=\frac{k+4}{\,k-2\,}.$$
For $$r_{1}r_{2}\gt0,$$ numerator and denominator must have the same sign.
If $$k\gt2$$ (from step 2) then $$k-2\gt0,$$ and also $$k+4\gt0,$$ so the product is indeed positive. Thus condition 3 adds no extra restriction beyond $$k\gt2.$$

Combine the inequalities
From$$(2)\; -6\lt k\lt4$$ and$$(3)\; k\gt2,$$ we obtain the admissible interval$$2\lt k\lt4.$$

Among the given options, the only value lying in this interval is $$k=3.$$

Option B which is: 3

The quadratic $$x^{2}-x+1=0$$ has roots $$\alpha,\beta$$.

By Viète’s relations, $$\alpha+\beta = 1$$ and $$\alpha\beta = 1$$.

Since each root satisfies the equation, we can write for every integer $$n \ge 2$$

$$\alpha^{n}= \alpha^{n-1}-\alpha^{n-2}, \quad \beta^{n}= \beta^{n-1}-\beta^{n-2}$$

Add the two relations and define $$S_{n}= \alpha^{n}+\beta^{n}$$. This gives the linear recurrence

$$S_{n}=S_{\,n-1}-S_{\,n-2} \qquad -(1)$$

Initial terms:

$$S_{0}= \alpha^{0}+\beta^{0}=2,$$ $$S_{1}= \alpha+\beta = 1.$$

Use (1) to generate the next four terms:

$$S_{2}=S_{1}-S_{0}=1-2=-1,$$ $$S_{3}=S_{2}-S_{1}=-1-1=-2,$$ $$S_{4}=S_{3}-S_{2}=-2-(-1)=-1,$$ $$S_{5}=S_{4}-S_{3}=-1-(-2)=1.$$

Repeating the step once more gives $$S_{6}=S_{5}-S_{4}=1-(-1)=2,$$ so the sequence $$S_{n}$$ is periodic with period $$6$$:

$$\;n:\;0\;1\;2\;3\;4\;5\;6\;7\ldots$$ $$S_{n}: 2,1,-1,-2,-1,1,2,1,\ldots$$

Now reduce the exponent modulo the period: $$2009 \div 6 = 334$$ remainder $$5$$, so $$2009 \equiv 5 \pmod{6}$$.

Therefore $$S_{2009}=S_{5}=1$$, i.e.

$$\alpha^{2009}+\beta^{2009}=1.$$

Option B which is: $$1$$

The quadratic $$bx^{2}+cx+a=0$$ has imaginary (non-real) roots.
This happens exactly when its discriminant is negative:

$$\Delta = c^{2}-4ab \lt 0 \quad\Longrightarrow\quad 4ab-c^{2} \gt 0$$

Now consider the required expression

$$E(x)=3b^{2}x^{2}+6bcx+2c^{2}$$

Rewrite it by completing the square:

$$\begin{aligned} E(x) &= 3\Bigl(b^{2}x^{2}+2bcx+c^{2}\Bigr)-c^{2} \\ &= 3\,(bx+c)^{2}-c^{2}. \end{aligned}$$

Add and subtract $$4ab$$ so that the positive quantity $$4ab-c^{2}$$ appears:

$$\begin{aligned} E(x) &= 3\,(bx+c)^{2}-c^{2}+4ab-4ab \\ &= 3\,(bx+c)^{2}+\bigl(4ab-c^{2}\bigr)-4ab. \end{aligned}$$

Hence

$$E(x)+4ab = 3\,(bx+c)^{2}+\bigl(4ab-c^{2}\bigr).$$

The first term $$3\,(bx+c)^{2}$$ is $$\ge 0$$ for every real $$x$$, and the second term $$4ab-c^{2}$$ is strictly positive from the discriminant condition. Therefore

$$E(x)+4ab \gt 0 \quad\text{for all real } x.$$

This inequality rearranges to

$$E(x) \gt -4ab \quad\text{for all real } x.$$

Thus the expression $$3b^{2}x^{2}+6bcx+2c^{2}$$ is always greater than $$-4ab$$.

Option C which is: greater than $$-4ab$$

Let the common root of the two quadratics be $$\alpha$$.

Denote the other (distinct) root of the first quadratic $$x^2 - 6x + a = 0$$ by $$p$$.
Denote the other root of the second quadratic $$x^2 - cx + 6 = 0$$ by $$q$$.

Given that the two “other” roots are integers in the ratio $$4:3$$, write them as
$$p = 4t, \qquad q = 3t$$
for some non-zero integer $$t$$.

For a quadratic $$x^2 - (\,\text{sum of roots}\,)x + (\,\text{product of roots}\,) = 0$$, we have

1. For the first quadratic
Sum of its roots $$\Longrightarrow \alpha + p = 6 \;$$ $$\Rightarrow$$ $$\alpha + 4t = 6$$ $$\;-(1)$$
    Product of its roots $$\Longrightarrow \alpha p = a$$ (not needed further).

2. For the second quadratic
    Sum of its roots $$\Longrightarrow \alpha + q = c$$ (value of $$c$$ not needed).
    Product of its roots $$\Longrightarrow \alpha q = 6$$ $$\Rightarrow$$ $$\alpha \cdot 3t = 6$$ $$\;-(2)$$

Use $$(2)$$ first:

From $$(2)$$, $$\alpha = \dfrac{6}{3t} = \dfrac{2}{t}$$ $$\;-(3)$$

Substitute $$(3)$$ in $$(1)$$:

$$\dfrac{2}{t} + 4t = 6$$

Multiply throughout by $$t$$:

$$2 + 4t^2 = 6t$$

Bring all terms to one side:

$$4t^2 - 6t + 2 = 0$$

Divide by $$2$$ to simplify:

$$2t^2 - 3t + 1 = 0$$

Solve the quadratic in $$t$$:

Discriminant $$D = (-3)^2 - 4(2)(1) = 9 - 8 = 1$$

Hence $$t = \dfrac{3 \pm 1}{2 \cdot 2} = \dfrac{3 \pm 1}{4}$$
$$\Longrightarrow t = 1 \; \text{or} \; t = \dfrac{1}{2}$$

But $$t$$ must be an integer (because $$p$$ and $$q$$ are integers).
Therefore $$t = 1$$.

Put $$t = 1$$ in $$(3)$$:

$$\alpha = \dfrac{2}{1} = 2$$.

Thus the common root of the two quadratic equations is $$2$$.

Option D which is: 2

Let $$\alpha$$ and $$\beta$$ be the roots of $$x^{2}+ax+1=0$$.

By Vieta’s relations, $$\alpha+\beta=-a$$ and $$\alpha\beta=1$$.

For any two numbers, $$(\alpha-\beta)^{2}=(\alpha+\beta)^{2}-4\alpha\beta.$$

Substituting the Vieta values, $$(\alpha-\beta)^{2}=(-a)^{2}-4(1)=a^{2}-4.$$

The (positive) distance between the roots is $$|\alpha-\beta|=\sqrt{a^{2}-4}.$$

Given that this distance is less than $$\sqrt{5}$$, we need $$\sqrt{a^{2}-4}\lt\sqrt{5}.$$

Squaring both sides, $$a^{2}-4\lt5\; \Longrightarrow\; a^{2}\lt9.$$

Thus $$-3\lt a\lt3.$$

The required set of values of $$a$$ is $$(-3,3),$$ i.e., Option A.

Let the roots of the quadratic $$x^{2}+px+q=0$$ be $$r_{1}$$ and $$r_{2}$$.
Given: $$r_{1}=\tan 30^\circ$$ and $$r_{2}=\tan 15^\circ$$.

Step 1: Evaluate the numerical values of the roots.

For $$30^\circ$$ we know $$\tan 30^\circ = \frac{1}{\sqrt3}$$, so $$r_{1}=\frac{1}{\sqrt3}$$.

To get $$\tan 15^\circ$$, use the formula for $$\tan(A-B)$$:
$$\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}.$$
Choose $$A=45^\circ,\, B=30^\circ$$; thus

$$\tan 15^\circ = \tan(45^\circ-30^\circ) =\frac{\tan45^\circ-\tan30^\circ}{1+\tan45^\circ\tan30^\circ} =\frac{1-\frac{1}{\sqrt3}}{1+1\cdot\frac{1}{\sqrt3}}$$

Simplify numerator and denominator by multiplying top and bottom by $$\sqrt3$$:

$$\tan15^\circ =\frac{\sqrt3-1}{\sqrt3+1} =2-\sqrt3.$$
Hence $$r_{2}=2-\sqrt3$$.

Step 2: Relate coefficients to the roots.

For the quadratic $$x^{2}+px+q=0$$ with roots $$r_{1},r_{2}$$:
Sum of roots: $$r_{1}+r_{2}=-p.$$
Product of roots: $$r_{1}r_{2}=q.$$

Step 3: Compute $$r_{1}+r_{2}$$ and $$r_{1}r_{2}$$.

Sum:
$$r_{1}+r_{2} =\frac{1}{\sqrt3}+2-\sqrt3 =\frac{\sqrt3}{3}+2-\sqrt3 =2-\frac{2\sqrt3}{3}.$$

Product:
$$r_{1}r_{2} =\frac{1}{\sqrt3}\,(2-\sqrt3) =\frac{2-\sqrt3}{\sqrt3} =\frac{2}{\sqrt3}-1.$$

Thus
$$p=-(r_{1}+r_{2})= -\left(2-\frac{2\sqrt3}{3}\right),$$
$$q=r_{1}r_{2}= \frac{2}{\sqrt3}-1.$$

Step 4: Evaluate $$2+q-p$$.

First rewrite $$p$$ with a positive denominator: $$p=-2+\frac{2\sqrt3}{3}.$$

Now calculate:

$$$ \begin{aligned} 2+q-p &=2+\left(\frac{2}{\sqrt3}-1\right)-\left(-2+\frac{2\sqrt3}{3}\right)\\[4pt] &=2-1+2+\frac{2}{\sqrt3}-\frac{2\sqrt3}{3}\\[4pt] &=3+\left(\frac{2}{\sqrt3}-\frac{2\sqrt3}{3}\right). \end{aligned} $$$

Notice that $$\frac{2}{\sqrt3} = \frac{2\sqrt3}{3},$$ so the two radical terms cancel:

$$\frac{2}{\sqrt3}-\frac{2\sqrt3}{3}=0.$$

Therefore

$$2+q-p = 3.$$

Hence the required value is 3.
Option B which is: 3

1. Factorize the Quadratic Equation

The given quadratic equation is:

$$x^2 - 2mx + m^2 - 1 = 0$$

Notice that the first three terms form a perfect square:

$$(x - m)^2 - 1 = 0$$

2. Find the Roots Directly

Isolate the perfect square term:

$$(x - m)^2 = 1$$

Take the square root of both sides:

$$x - m = \pm 1$$

This gives the two distinct roots directly:

$$\alpha = m - 1$$

$$\beta = m + 1$$

3. Apply the Given Boundary Conditions

The problem states that both roots must be strictly greater than $$-2$$ and strictly less than $$4$$.

For the smaller root $$\alpha = m - 1$$ to be greater than $$-2$$:

$$m - 1 > -2$$

$$m > -1$$

For the larger root $$\beta = m + 1$$ to be less than $$4$$:

$$m + 1 < 4$$

$$m < 3$$

4. Combine the Inequalities

Intersecting the two boundary requirements gives the final range for the parameter:

$$-1 < m < 3$$

Final Answer

All the values of $$m$$ lie in the interval $$(-1, 3)$$.

Let the equation be $$x^{2}-(\alpha-2)x-\alpha-1=0$$ and let its roots be $$r_{1},\,r_{2}$$.

For a quadratic $$x^{2}+bx+c=0$$, we know
  • sum of roots $$r_{1}+r_{2}=-b$$
  • product of roots $$r_{1}r_{2}=c$$.

Here the coefficient of $$x$$ is $$-(\alpha-2)$$ and the constant term is $$-\alpha-1$$. Hence
$$r_{1}+r_{2}=-(\;-(\alpha-2)\;)=\alpha-2,$$
$$r_{1}r_{2}=-\alpha-1.$$

The quantity to be minimised is the sum of the squares of the roots:
$$S=r_{1}^{2}+r_{2}^{2}.$$ Using the identity $$r_{1}^{2}+r_{2}^{2}=(r_{1}+r_{2})^{2}-2r_{1}r_{2},$$ we get

$$S=(\alpha-2)^{2}-2(-\alpha-1)= (\alpha-2)^{2}+2\alpha+2.$$

Expand the square:
$$(\alpha-2)^{2}=\alpha^{2}-4\alpha+4,$$
so
$$S=\alpha^{2}-4\alpha+4+2\alpha+2=\alpha^{2}-2\alpha+6.$$

Thus $$S(\alpha)=\alpha^{2}-2\alpha+6,$$ which is a quadratic in $$\alpha$$ opening upwards. Its minimum occurs at the vertex $$\alpha=\frac{-(-2)}{2\cdot1}=1.$$

The discriminant of the original quadratic is $$\Delta=(\alpha-2)^{2}+4(\alpha+1)=\alpha^{2}+8 \gt 0,$$ so the roots are real for every real $$\alpha$$ and no extra condition is needed.

Therefore the sum of the squares of the roots is least when $$\alpha=1$$.

Option A which is: $$1$$

Let the two consecutive integral roots be $$n$$ and $$n+1$$, where $$n$$ is an integer.

For the quadratic $$x^2 - bx + c = 0$$, Vieta’s relations give
Sum of roots $$= n + (n+1) = 2n + 1 = b$$ $$-(1)$$
Product of roots $$= n(n+1) = c$$ $$-(2)$$

We need $$b^2 - 4c$$. Substitute from $$-(1)$$ and $$-(2)$$:

$$b^2 - 4c = (2n + 1)^2 - 4\bigl[n(n+1)\bigr]$$

Expand each term:

$$(2n + 1)^2 = 4n^2 + 4n + 1$$
$$4\,n(n+1) = 4n^2 + 4n$$

Subtract:

$$b^2 - 4c = (4n^2 + 4n + 1) - (4n^2 + 4n) = 1$$

The expression is independent of the specific integer $$n$$; it always equals $$1$$.

Hence, $$b^2 - 4c = 1$$, which corresponds to
Option D which is: $$1$$

Let both the lizard and the insect start at $$t = 0$$.

Initial data
• Uniform speed of the insect: $$u = 20 \, \text{cm s}^{-1}$$
• Constant acceleration of the lizard (from rest): $$a = 2 \, \text{cm s}^{-2}$$
• Initial distance by which the lizard is behind the insect: $$s_0 = 21 \, \text{cm}$$

Step 1: Write individual displacements
Measured from their respective starting points, after time $$t$$ the displacements are
Insect: $$s_{\text{insect}} = u t = 20t$$
Lizard (starting from rest with uniform acceleration): $$s_{\text{lizard}} = \dfrac12 a t^{2} = \dfrac12 \times 2 \times t^{2} = t^{2}$$

Step 2: Express the shrinking gap
At time $$t$$, the separation between them equals the initial gap plus the distance covered by the insect minus the distance covered by the lizard:

$$\text{Gap}(t) = s_0 + s_{\text{insect}} - s_{\text{lizard}} = 21 + 20t - t^{2}$$

Step 3: Set the gap to zero for the instant of capture
The lizard catches the insect when $$\text{Gap}(t) = 0$$:

$$21 + 20t - t^{2} = 0$$

Re-arranging to the standard quadratic form:

$$t^{2} - 20t - 21 = 0$$

Step 4: Solve the quadratic equation
For $$at^{2} + bt + c = 0$$, the solutions are $$t = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$$. Here $$a = 1$$, $$b = -20$$, $$c = -21$$.

Discriminant:
$$\Delta = (-20)^{2} - 4(1)(-21) = 400 + 84 = 484$$
$$\sqrt{\Delta} = 22$$

Hence
$$t = \dfrac{20 \pm 22}{2}$$

Two roots appear:
$$t = \dfrac{20 + 22}{2} = 21 \, \text{s}$$
$$t = \dfrac{20 - 22}{2} = -1 \, \text{s}$$ (discarded because time cannot be negative).

Step 5: State the physical answer
The only physically meaningful time is $$t = 21 \, \text{s}$$.

Therefore, the lizard catches the insect after 21 seconds.

Option C which is: $$21$$ s

$$\alpha_\theta + \beta_\theta = -\frac{\cos\theta}{2}$$ and $$\alpha_\theta\,\beta_\theta = -\frac{1}{2}$$

$$\alpha_\theta^2 + \beta_\theta^2 = (\alpha_\theta + \beta_\theta)^2 - 2\alpha_\theta\beta_\theta = \frac{\cos^2\theta}{4} + 1$$.

$$\alpha_\theta^4 + \beta_\theta^4 = (\alpha_\theta^2 + \beta_\theta^2)^2 - 2(\alpha_\theta\beta_\theta)^2 = \Bigl(\frac{\cos^2\theta}{4} + 1\Bigr)^2 - 2\cdot\frac{1}{4} = \frac{\cos^4\theta}{16} + \frac{\cos^2\theta}{2} + \frac{1}{2}$$.

Setting $$t = \cos^2\theta$$ with $$t\in[0,1]$$ leads to the function $$f(t) = \frac{t^2}{16} + \frac{t}{2} + \frac{1}{2}$$ whose derivative $$f'(t) = \frac{t}{8} + \frac{1}{2}$$ is strictly positive on $$[0,1]$$, so $$f$$ is increasing.

Therefore the minimum occurs at $$t=0$$ giving $$m = f(0) = \frac{1}{2}$$ and the maximum occurs at $$t=1$$ giving $$M = f(1) = \frac{1}{16} + \frac{1}{2} + \frac{1}{2} = \frac{17}{16}$$.

Hence $$16(M + m) = 16\Bigl(\frac{17}{16} + \frac{1}{2}\Bigr) = 17 + 8 = 25$$, so the final answer is 25.

Let $$\alpha$$ and $$\beta$$ be the roots of $$x^2 + \sqrt{3}x - 16 = 0$$. Therefore, they satisfy the equation:

$$\alpha^2 + \sqrt{3}\alpha - 16 = 0 \implies \alpha^2 + \sqrt{3}\alpha = 16$$

$$\beta^2 + \sqrt{3}\beta - 16 = 0 \implies \beta^2 + \sqrt{3}\beta = 16$$

Multiplying the first equation by $$\alpha^{23}$$ and the second by $$\beta^{23}$$ gives:

$$\alpha^{25} + \sqrt{3}\alpha^{24} = 16\alpha^{23}$$

$$\beta^{25} + \sqrt{3}\beta^{24} = 16\beta^{23}$$

Adding these two equations together:

$$(\alpha^{25} + \beta^{25}) + \sqrt{3}(\alpha^{24} + \beta^{24}) = 16(\alpha^{23} + \beta^{23})$$

Since $$P_n = \alpha^n + \beta^n$$, this simplifies to:

$$P_{25} + \sqrt{3}P_{24} = 16P_{23}$$

Dividing by $$2P_{23}$$ gives the value of the first term:

$$\frac{P_{25} + \sqrt{3}P_{24}}{2P_{23}} = \frac{16P_{23}}{2P_{23}} = 8$$

---

Similarly, let $$\gamma$$ and $$\delta$$ be the roots of $$x^2 + 3x - 1 = 0$$. They satisfy:

$$\gamma^2 + 3\gamma - 1 = 0 \implies \gamma^2 - 1 = -3\gamma$$

$$\delta^2 + 3\delta - 1 = 0 \implies \delta^2 - 1 = -3\delta$$

Multiplying by $$\gamma^{23}$$ and $$\delta^{23}$$ respectively:

$$\gamma^{25} - \gamma^{23} = -3\gamma^{24}$$

$$\delta^{25} - \delta^{23} = -3\delta^{24}$$

Adding these two equations together:

$$(\gamma^{25} + \delta^{25}) - (\gamma^{23} + \delta^{23}) = -3(\gamma^{24} + \delta^{24})$$

Since $$Q_n = \gamma^n + \delta^n$$, this simplifies to:

$$Q_{25} - Q_{23} = -3Q_{24}$$

Dividing by $$Q_{24}$$ gives the value of the second term:

$$\frac{Q_{25} - Q_{23}}{Q_{24}} = \frac{-3Q_{24}}{Q_{24}} = -3$$

---

Combining both results, the required expression is equal to:

$$\frac{P_{25} + \sqrt{3}P_{24}}{2P_{23}} + \frac{Q_{25} - Q_{23}}{Q_{24}} = 8 + (-3) = 5$$

We have: $$\left(\dfrac{9}{x}-\dfrac{9}{\sqrt{x}}+2\right)\left(\dfrac{2}{x}-\dfrac{7}{\sqrt{x}}+3\right)=0$$

Now, we know that x can't be 0 as it is in denominator and x can't be negative as it is in square root as well. Hence, x has to be strictly more than 0.

To make the equations easier, we can replace, $$y = \dfrac{1}{\sqrt{x}}$$ and $$y^2 = \dfrac{1}{x}$$.

Replacing x, we get, $$(9y^2 - 9y + 2)(2y^2 - 7y + 3) = 0$$

Now,  we can solve the equations individually. 

Solving: $$9y^2 - 9y + 2 = 0$$

$$9y^2 - 6y - 3y + 2 = 0$$

$$3y(3y - 2) - 1(3y - 2) = 0$$

$$(3y - 1)(3y - 2) = 0$$

Hence, we get, $$y = \dfrac{1}{3} \quad \text{or} \quad y = \dfrac{2}{3}$$

Solving: $$2y^2 - 7y + 3 = 0$$

$$2y^2 - 6y - y + 3 = 0$$

$$2y(y - 3) - 1(y - 3) = 0$$

$$(2y - 1)(y - 3) = 0$$

Hence, we get: $$y = \dfrac{1}{2} \quad \text{or} \quad y = 3$$

So, we have got four valid, distinct values for $$y$$: $$\left\{\dfrac{1}{3}, \dfrac{2}{3}, \dfrac{1}{2}, 3\right\}$$

Converting back from y to x:

If $$y = \dfrac{1}{3}$$, then $$x = 9$$

If $$y = \dfrac{2}{3}$$, then $$x = \frac{9}{4}$$

If $$y = \dfrac{1}{2}$$, then $$x = 4$$

If $$y = 3$$, then $$x = \dfrac{1}{9}$$

All four $$x$$ values fall within our initial domain of $$x > 0$$.

Hence, the number of solutions to the equation is 4.

$$(x^2 - 9x + 11)^2 - (x - 4)(x - 5) = 3$$

$$(x-4)(x-5)=x^2-9x+20$$

Let $$x^2-9x+20=t$$

$$\left(t-9\right)^2-t=3$$

$$t^2-19t+78=0$$

$$\left(t-13\right)\left(t-6\right)=0$$

So, $$t=13\ ,\ 6$$

Now, two cases will be formed: 

Case 1: $$t=13$$

$$x^2-9x+20=13$$

$$x^2-9x+7=0$$

The discriminant is $$\sqrt{53}$$. So, all the roots will be irrational. 

Case 2: $$t=6$$

$$x^2-9x+20=6$$

$$x^2-9x+14=0$$

$$\left(x-7\right)\left(x-2\right)=0$$

$$x=2\ ,\ 7$$

These roots are rational. 

We have to find the product of rational roots. 

Product = $$\left(2\times7\right)=14$$

$$\therefore\ $$ The required answer is A.

Equation is $$x(x+2)(12-k)=2$$. For any fixed $$k$$, treat $$x$$ as the variable.

Divide both sides by $$(12-k)$$ (allowed if $$k \neq 12$$) to get a quadratic in $$x$$:

$$x(x+2)=\frac{2}{\,12-k\,}$$

Expand:

$$x^2+2x-\frac{2}{\,12-k\,}=0 \; -(1)$$

For equation $$(1)$$ to have equal (repeated) roots, its discriminant must be zero. For a quadratic $$ax^2+bx+c=0$$, discriminant $$\Delta=b^2-4ac$$.

Here $$a=1,\; b=2,\; c=-\dfrac{2}{\,12-k\,}$$. Set $$\Delta=0$$:

$$2^2-4(1)\!\left(-\frac{2}{\,12-k\,}\right)=0$$

$$4+\frac{8}{\,12-k\,}=0$$

Divide by $$4$$:

$$1+\frac{2}{\,12-k\,}=0$$

$$\frac{2}{\,12-k\,}=-1$$

Cross-multiply:

$$2=-(12-k)$$

$$k=14$$

Thus $$k=14$$ (note $$k\neq12$$ is satisfied).

The required point is $$\left(k,\frac{k}{2}\right)=\left(14,\,7\right)$$.

Distance of point $$(x_1,y_1)$$ from line $$Ax+By+C=0$$ is $$\displaystyle D=\frac{|Ax_1+By_1+C|}{\sqrt{A^{2}+B^{2}}}$$.

Here $$A=3,\; B=4,\; C=5,\; (x_1,y_1)=(14,7)$$:

Numerator: $$|3(14)+4(7)+5|=|42+28+5|=|75|=75$$

Denominator: $$\sqrt{3^{2}+4^{2}}=\sqrt{9+16}=5$$

Therefore $$D=\frac{75}{5}=15$$

Hence the distance is $$15$$.

Option A is correct.

We consider the quadratic equation $$2x^2 + (a-5)x + 15 - 3a = 0$$ and require that it have no real roots, which is equivalent to its discriminant satisfying $$D = (a-5)^2 - 4(2)(15-3a) < 0\;.$$

Expanding the discriminant yields $$a^2 - 10a + 25 - 120 + 24a < 0\;,$$ which simplifies to $$a^2 + 14a - 95 < 0\;.$$

To find the critical values of $$a$$, we solve the equation $$a^2 + 14a - 95 = 0$$, giving

$$a = \frac{-14 \pm \sqrt{196 + 380}}{2} = \frac{-14 \pm \sqrt{576}}{2} = \frac{-14 \pm 24}{2}\;,$$

so that $$a = 5$$ or $$a = -19\;.$$

Therefore the quadratic has no real roots precisely when $$a \in (-19, 5)\;.$$

For integer values of $$x$$ in this interval we set

$$X = \{\,x \in \mathbb{Z} : -19 < x < 5\} = \{-18, -17, \dots, 4\}\;.$$

We wish to compute

$$\sum_{x \in X} x^2 = \sum_{x=-18}^{4} x^2 = \sum_{x=1}^{18} x^2 + \sum_{x=1}^{4} x^2\;.$$

Using the formula $$\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}\;,$$ we find

$$\sum_{x=1}^{18} x^2 = \frac{18 \times 19 \times 37}{6} = \frac{12654}{6} = 2109$$

and

$$\sum_{x=1}^{4} x^2 = \frac{4 \times 5 \times 9}{6} = 30\;.$$

Hence

$$\sum_{x \in X} x^2 = 2109 + 30 = 2139\;.$$

The correct answer is Option 4: 2139.

A quadratic form $$Q(x,y)=ax^2+2bxy+cy^2$$ is positive for every non-zero $$(x,y)\in\mathbb{R}^2$$ (that is, positive definite) iff its associated symmetric matrix $$\begin{pmatrix}a & b\\[2pt] b & c\end{pmatrix}$$ is positive definite.

Sylvester’s criterion for a $$2\times2$$ symmetric matrix gives the necessary and sufficient conditions
    $$a\gt0,\qquad c\gt0,\qquad ac-b^2\gt0\;$$ $$-(1)$$
Hence $$S=\{(a,b,c):a\gt0,\;c\gt0,\;ac-b^2\gt0\}.$$

We check each option using $$-(1)$$.

Option A
Take $$(a,b,c)=\left(2,\frac72,6\right).$$
    $$ac-b^2=2\cdot6-\left(\frac72\right)^2=12-\frac{49}{4}=\frac{48-49}{4}=-\frac14\lt0.$$
Condition $$ac-b^2\gt0$$ fails, so $$\left(2,\frac72,6\right)\notin S.$$
Option A is false.

Option B
Assume $$\left(3,b,\frac1{12}\right)\in S.$$ Applying $$-(1)$$:
    $$3\gt0,\quad \frac1{12}\gt0\quad(\text{true}),$$
    $$3\cdot\frac1{12}-b^2\gt0\;\Longrightarrow\; \frac14-b^2\gt0\;\Longrightarrow\;b^2\lt\frac14.$$
Thus $$|b|\lt\frac12,$$ that is, $$|2b|\lt1.$$ Option B is true.

Option C
For any $$(a,b,c)\in S$$ the coefficient matrix of the given linear system
    $$\begin{cases}ax+by=1\\ bx+cy=-1\end{cases}$$
is $$\begin{pmatrix}a&b\\ b&c\end{pmatrix}.$$
From $$-(1)$$ its determinant $$ac-b^2$$ is strictly positive, hence non-zero. Therefore the matrix is invertible and the system has a unique solution for every $$(a,b,c)\in S.$$ Option C is true.

Option D
For the system
    $$\begin{cases}(a+1)x+by=0\\ bx+(c+1)y=0\end{cases}$$
the coefficient matrix is $$\begin{pmatrix}a+1&b\\ b&c+1\end{pmatrix}.$$ Its determinant is
    $$(a+1)(c+1)-b^2=ac-b^2 + a+c+1.$$
Using $$-(1)$$, $$ac-b^2\gt0$$ and $$a,c\gt0$$, hence
    $$(a+1)(c+1)-b^2\gt0.$$
The determinant is always positive, so the matrix is invertible and the system possesses a unique solution for every $$(a,b,c)\in S.$$ Option D is true.

Therefore the correct statements are:
Option B (|2b| < 1), Option C (unique solution), and Option D (unique solution).

Answer: Option B, Option C, Option D.

Since $$\alpha, \beta$$ are roots of $$x^2 - x - 1 = 0$$, we have $$\alpha^2 = \alpha + 1$$ and $$\beta^2 = \beta + 1$$.

In general, $$\alpha^n = \alpha^{n-1} + \alpha^{n-2}$$ and $$\beta^n = \beta^{n-1} + \beta^{n-2}$$.

$$S_n = 2023\alpha^n + 2024\beta^n$$

$$S_{n-1} + S_{n-2} = 2023(\alpha^{n-1} + \alpha^{n-2}) + 2024(\beta^{n-1} + \beta^{n-2})$$

$$= 2023\alpha^n + 2024\beta^n = S_n$$

So $$S_n = S_{n-1} + S_{n-2}$$ for all $$n$$. In particular, $$S_{12} = S_{11} + S_{10}$$.

The answer corresponds to Option (2).

We have $$\alpha, \beta$$ as roots of $$px^2 + qx - r = 0$$ where $$p \neq 0$$. The terms $$p, q, r$$ form a non-constant G.P., and $$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{3}{4}$$.

From $$px^2 + qx - r = 0$$:

• $$\alpha + \beta = -\frac{q}{p}$$
• $$\alpha\beta = -\frac{r}{p}$$

$$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{-q/p}{-r/p} = \frac{q}{r} = \frac{3}{4}$$

Since $$p, q, r$$ are consecutive terms of a G.P. with common ratio $$k$$: $$q = pk$$ and $$r = pk^2$$.

$$\frac{q}{r} = \frac{pk}{pk^2} = \frac{1}{k} = \frac{3}{4} \implies k = \frac{4}{3}$$

So $$q = \frac{4p}{3}$$ and $$r = \frac{16p}{9}$$.

$$(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$$

$$= \left(\frac{q}{p}\right)^2 + \frac{4r}{p} = \left(\frac{4}{3}\right)^2 + 4 \times \frac{16}{9}$$

$$= \frac{16}{9} + \frac{64}{9} = \frac{80}{9}$$

The correct answer is Option A: $$\frac{80}{9}$$.

We need to simplify the expression $$(11\sqrt{3}-10\sqrt{2})P_{10} + (11\sqrt{2}+10)P_{11} - 11P_{12}.$$

Since $$\alpha, \beta$$ are roots of $$x^2 - \sqrt{2}x - \sqrt{3} = 0$$, we have $$\alpha + \beta = \sqrt{2}$$ and $$\alpha\beta = -\sqrt{3}$$. The sequence $$P_n = \alpha^n - \beta^n$$ therefore satisfies the recurrence $$P_n = \sqrt{2}P_{n-1} + \sqrt{3}P_{n-2}$$, because $$\alpha^n = \sqrt{2}\alpha^{n-1} + \sqrt{3}\alpha^{n-2}$$ (and similarly for $$\beta$$).

In particular, $$P_{12} = \sqrt{2}P_{11} + \sqrt{3}P_{10}$$, so $$11P_{12} = 11\sqrt{2}P_{11} + 11\sqrt{3}P_{10}$$.

Substituting this into the original expression gives

$$(11\sqrt{3}-10\sqrt{2})P_{10} + (11\sqrt{2}+10)P_{11} - \bigl(11\sqrt{2}P_{11} + 11\sqrt{3}P_{10}\bigr).$$

Collecting like terms, the coefficient of $$P_{10}$$ is $$(11\sqrt{3}-10\sqrt{2}) - 11\sqrt{3} = -10\sqrt{2}$$ and the coefficient of $$P_{11}$$ is $$(11\sqrt{2}+10) - 11\sqrt{2} = 10$$. Thus the expression simplifies to $$-10\sqrt{2}P_{10} + 10P_{11}$$.

Factoring out 10 gives $$10(P_{11} - \sqrt{2}P_{10})$$. Using the recurrence backwards, $$P_{11} = \sqrt{2}P_{10} + \sqrt{3}P_9$$, so $$P_{11} - \sqrt{2}P_{10} = \sqrt{3}P_9$$ and hence the result is $$10\sqrt{3}P_9$$.

The correct answer is Option 1: $$10\sqrt{3}P_9$$.

For the quadratic
$$x^{2} - (t^{2}-5t+6)x + 1 = 0$$
let the two distinct roots be $$\alpha$$ and $$\beta$$.

By Vieta’s relations
$$\alpha + \beta = t^{2}-5t+6 \quad -(1)$$
$$\alpha\beta = 1 \quad -(2)$$

Define the sequence $$a_{n} = \alpha^{n} + \beta^{n}\;(n \ge 0).$$

Case 1: Recurrence for $$a_n$$
Using $$\alpha\beta = 1$$ we obtain a standard identity: for all $$n \ge 0$$
$$\alpha^{\,n+2} + \beta^{\,n+2} = (\alpha+\beta)\,(\alpha^{\,n+1} + \beta^{\,n+1}) - (\alpha\beta)\,(\alpha^{\,n} + \beta^{\,n}).$$
Hence
$$a_{\,n+2} = (\alpha+\beta)\,a_{\,n+1} - a_{\,n}\quad -(3)$$

Case 2: Evaluating the required ratio
Set $$n = 2023$$ in the recurrence $$-(3)$$:
$$a_{2025} = (\alpha+\beta)\,a_{2024} - a_{2023}\quad -(4)$$

Add $$a_{2023}$$ to both sides of $$-(4)$$:
$$a_{2023} + a_{2025} = (\alpha+\beta)\,a_{2024}$$

Therefore
$$\frac{a_{2023} + a_{2025}}{a_{2024}} = \alpha + \beta$$

From $$-(1)$$ the ratio equals
$$\alpha + \beta = t^{2} - 5t + 6\quad -(5)$$

Case 3: Minimising the expression
We must find the minimum of the quadratic function
$$f(t) = t^{2} - 5t + 6$$
for real $$t$$, under the condition that the roots $$\alpha, \beta$$ are distinct.

Write $$f(t)$$ in completed-square form:
$$f(t) = \bigl(t - \tfrac{5}{2}\bigr)^{2} - \tfrac{1}{4}$$

The square term is always non-negative, so the minimum value occurs at $$t = \tfrac{5}{2}$$ and equals
$$f_{\min} = -\tfrac{1}{4}$$

Case 4: Checking the conditions
(i) Distinct roots: the discriminant of the given quadratic in $$x$$ is
$$\Delta = (\alpha+\beta)^{2} - 4\alpha\beta = f(t)^{2} - 4$$

At $$t = \tfrac{5}{2}$$, $$f(t) = -\tfrac14$$, so
$$\Delta = \bigl(-\tfrac14\bigr)^{2} - 4 = \tfrac1{16} - 4 = -\tfrac{63}{16} \neq 0.$$
Thus the roots are indeed distinct (they form a non-real conjugate pair).

(ii) Denominator non-zero: with distinct roots, $$a_{2024} = \alpha^{2024} + \beta^{2024}$$ cannot vanish for all real $$t$$; in particular it is non-zero at $$t = \tfrac{5}{2}$$, so the ratio is well defined at the minimum.

Hence the minimum possible value of $$\dfrac{a_{2023}+a_{2025}}{a_{2024}}$$ is
$$-\tfrac14.$$

Option B is correct.

For $$x^2 - 8x + 32$$, the discriminant $$D = (-8)^2 - 4(32) = 64 - 128 = -64$$.

Since $$D < 0$$ and the leading coefficient is positive, the denominator is always positive for all $$x$$.

For the fraction to be $$< 0$$, the numerator must be always negative:

$$ax^2 + 2(a+1)x + 9a + 4 < 0$$

This requires:

1. $$a < 0$$ (parabola must open downward)

2. $$D < 0$$ (no real roots

The question asks for positive integral values of $$a$$. 

Since condition 1 requires $$a < 0$$, there are no positive integers that satisfy this.

Number of elements in $$S = \mathbf{0}$$ (Option B)

The sum of the coefficients of $$(a+b-2c)x^2 + (b+c-2a)x + (c+a-2b) = 0$$ is $$0$$, meaning $$x = 1$$ is one root.

The product of roots gives the other root $$\alpha$$:

$$\alpha \cdot 1 = \frac{c+a-2b}{a+b-2c} \implies \alpha = \frac{a+c-2b}{a+b-2c}$$

Given $$0 < c < b < a$$:

1. By AM-GM inequality, $$b < \frac{a+c}{2} \implies a+c-2b > 0$$ (Numerator is positive).

2. Since $$b > c$$, $$a+b-2c > a+c-2c = a-c > 0$$ (Denominator is positive).

Thus, $$\alpha > 0$$.

3. Comparing numerator and denominator: $$\text{Num} - \text{Den} = 3(c-b) < 0 \implies \alpha < 1$$.

Since $$b = \sqrt{ac} \implies \alpha \in (0, 1)$$:

• Statement (I) is True (if $$\alpha \in (-1,0)$$, $$b$$ cannot be the GM).

• Statement (II) is True (if $$\alpha \in (0,1)$$, $$b$$ can be the GM).

 the sum of coefficients of a polynomial $$f(x)$$ is $$f(1)$$, substituting $$x = 1$$ yields $$a = (1 - 2(1) + 2(1)^2)^{2023}\times(3 - 4(1)^2 + 2(1)^3)^{2024} = (1 - 2 + 2)^{2023}\times(3 - 4 + 2)^{2024} = 1^{2023}\times1^{2024} = 1.$$

limit $$b = \lim_{x\to0} \frac{\int_0^x \frac{\log(1+t)}{t^{2024}+1}\,dt}{x^2}.$$ As $$x\to0$$, both numerator and denominator approach zero, so applying L’Hôpital’s rule gives $$b = \lim_{x\to0} \frac{\frac{d}{dx}\bigl(\int_0^x\frac{\log(1+t)}{t^{2024}+1}\,dt\bigr)}{\frac{d}{dx}(x^2)} = \lim_{x\to0}\frac{\frac{\log(1+x)}{x^{2024}+1}}{2x} = \lim_{x\to0}\frac{\log(1+x)}{2x(x^{2024}+1)}.$$ Using the approximations $$\log(1+x)\approx x$$ and $$x^{2024}+1\to1$$ as $$x\to0$$ yields $$b = \frac{1}{2}.$$

Substituting $$a=1$$ and $$b=\frac{1}{2}$$ into the quadratic equation $$2bx^2 + ax + 4 = 0$$ 

$$ x^2 + x + 4 = 0.$$ Its discriminant is $$1 - 16 = -15$$, which is negative, so the roots are non-real complex numbers.

Since the quadratic equation $$cx^2 + dx + e = 0$$ with real coefficients shares a common root with $$x^2 + x + 4 = 0$$ and complex roots occur in conjugate pairs, both roots must coincide. 

Therefore the two quadratics are proportional, giving $$\frac{c}{1} = \frac{d}{1} = \frac{e}{4} = k.$$ 

Hence $$c=k$$, $$d=k$$, $$e=4k$$, and the ratio of their coefficients follows as $$d : c : e = k : k : 4k = 1 : 1 : 4.$$

$$\cos(2\sin^{-1}x) = 1 - 2x^2$$.

$$1 - 2x^2 = \frac{1}{9}$$, so $$2x^2 = 1 - \frac{1}{9} = \frac{8}{9}$$ and hence $$x^2 = \frac{4}{9}$$ which implies $$x = \frac{2}{3}$$. Because $$x$$ is positive, we conclude $$m = 2, n = 3$$.

 $$mx^2 - nx - m + n = 0$$.

Substituting $$m=2$$ and $$n=3$$ $$2x^2 - 3x - 2 + 3 = 0$$ or $$2x^2 - 3x + 1 = 0$$. 

Factoring gives $$(2x - 1)(x - 1) = 0$$, so the roots are $$x = \frac{1}{2}$$ and $$x = 1$$. 

Since we denote the larger root by $$\alpha$$ and the smaller by $$\beta$$, we have $$\alpha = 1$$ and $$\beta = \frac{1}{2}$$.

Finally, we check which line passes through the point $$(1, \frac{1}{2})$$.

Substituting into Option (4): $$5(1) + 8\left(\frac{1}{2}\right) = 5 + 4 = 9$$,

We need to find the number of distinct real roots of:

$$|x + 1||x + 3| - 4|x + 2| + 5 = 0$$

The critical points of the absolute value expressions are at $$x = -3$$, $$x = -2$$, and $$x = -1$$. These divide the real line into four intervals. In each interval, every absolute value expression has a definite sign, so we can remove the absolute values and solve.

Case 1: $$x \geq -1$$

In this interval, $$x + 1 \geq 0$$, $$x + 3 > 0$$, and $$x + 2 > 0$$. So all absolute values equal the expressions themselves.

$$(x+1)(x+3) - 4(x+2) + 5 = 0$$

$$x^2 + 4x + 3 - 4x - 8 + 5 = 0$$

$$x^2 = 0 \implies x = 0$$

Since $$0 \geq -1$$, this root is valid.

Case 2: $$-2 \leq x < -1$$

Here $$x + 1 < 0$$, $$x + 3 > 0$$, and $$x + 2 \geq 0$$. So $$|x+1| = -(x+1)$$, while the other two remain positive.

$$-(x+1)(x+3) - 4(x+2) + 5 = 0$$

$$-(x^2 + 4x + 3) - 4x - 8 + 5 = 0$$

$$-x^2 - 8x - 6 = 0 \implies x^2 + 8x + 6 = 0$$

$$x = \frac{-8 \pm \sqrt{64 - 24}}{2} = -4 \pm \sqrt{10}$$

This gives $$x \approx -0.84$$ or $$x \approx -7.16$$. Neither value lies in the interval $$[-2, -1)$$, so no valid roots here.

Case 3: $$-3 \leq x < -2$$

Here $$x + 1 < 0$$, $$x + 3 \geq 0$$, and $$x + 2 < 0$$. So $$|x+1| = -(x+1)$$ and $$|x+2| = -(x+2)$$.

$$-(x+1)(x+3) - 4 \cdot (-(x+2)) + 5 = 0$$

$$-(x^2 + 4x + 3) + 4(x + 2) + 5 = 0$$

$$-x^2 - 4x - 3 + 4x + 8 + 5 = 0$$

$$-x^2 + 10 = 0 \implies x^2 = 10$$

This gives $$x = \pm\sqrt{10} \approx \pm 3.16$$. Neither lies in $$[-3, -2)$$, so no valid roots here.

Case 4: $$x < -3$$

Here all three expressions $$x + 1$$, $$x + 2$$, and $$x + 3$$ are negative. So $$|x+1| = -(x+1)$$, $$|x+3| = -(x+3)$$, and $$|x+2| = -(x+2)$$. The product of two negatives is positive:

$$(x+1)(x+3) + 4(x+2) + 5 = 0$$

$$x^2 + 4x + 3 + 4x + 8 + 5 = 0$$

$$x^2 + 8x + 16 = 0 \implies (x + 4)^2 = 0 \implies x = -4$$

Since $$-4 < -3$$, this root is valid.

Conclusion: The equation has exactly two distinct real roots: $$x = 0$$ and $$x = -4$$.

The answer is $$\textbf{2}$$.

$$x \geq -5$$: $$x^2+7x+12=0 \implies x = -3, -4$$ (2 solutions).

$$-7 \leq x < -5$$: $$-x^2-3x+12=0 \implies x = \frac{-3-\sqrt{57}}{2} \approx -5.27$$ (1 solution).

$$x < -7$$: $$-x^2-7x-16=0 \implies D < 0$$ (0 solutions).

Total: 3

Consider the cubic equation $$x^3 + bx + c = 0$$ with roots $$\alpha,\beta,\gamma$$. The conditions $$\beta\gamma = 1$$ and $$-\alpha = 1$$ imply $$\alpha = -1$$.

Using Vieta's formulas, the relation $$\alpha + \beta + \gamma = 0$$ gives $$\beta + \gamma = 1$$. The sum of products $$\alpha\beta + \beta\gamma + \gamma\alpha = b$$ becomes $$-(\beta + \gamma) + 1 = b$$, so $$b = 0$$. Finally, the product $$\alpha\beta\gamma = -c$$ gives $$(-1)(1) = -c$$, leading to $$c = 1$$.

Since $$\beta + \gamma = 1$$ and $$\beta\gamma = 1$$, the numbers $$\beta$$ and $$\gamma$$ satisfy the quadratic equation $$t^2 - t + 1 = 0$$. It follows that $$\beta = \frac{1 + i\sqrt{3}}{2} = e^{i\pi/3}$$ and $$\gamma = \frac{1 - i\sqrt{3}}{2} = e^{-i\pi/3}$$.

Raising these roots to the third power yields $$\beta^3 = e^{i\pi} = -1$$ and $$\gamma^3 = e^{-i\pi} = -1$$.

Substituting into the expression $$b^3 + 2c^3 - 3\alpha^3 - 6\beta^3 - 8\gamma^3$$ yields
$$0 + 2(1) - 3(-1)^3 - 6(-1) - 8(-1) = 0 + 2 + 3 + 6 + 8 = 19\,. $$

Let $$\alpha,\beta$$ be roots of $$14x^2 - 31x + 3\lambda = 0$$ and $$\alpha,\gamma$$ be roots of $$35x^2 - 53x + 4\lambda = 0\!.$$ By Vieta’s formulas, from the first equation we have $$\alpha + \beta = \frac{31}{14}$$ and $$\alpha\beta = \frac{3\lambda}{14}$$, while from the second equation $$\alpha + \gamma = \frac{53}{35}$$ and $$\alpha\gamma = \frac{4\lambda}{35}\!.$$ Since $$\alpha$$ is a common root, it satisfies

$$14\alpha^2 - 31\alpha + 3\lambda = 0 \quad \cdots (1)$$ $$35\alpha^2 - 53\alpha + 4\lambda = 0 \quad \cdots (2)$$

From (1) we get $$\lambda = \frac{31\alpha - 14\alpha^2}{3}$$ and substituting this into (2) yields

$$35\alpha^2 - 53\alpha + 4\cdot\frac{31\alpha - 14\alpha^2}{3} = 0$$ $$105\alpha^2 - 159\alpha + 124\alpha - 56\alpha^2 = 0$$ $$49\alpha^2 - 35\alpha = 0$$ $$7\alpha(7\alpha - 5) = 0\!.$$

Since $$\lambda \neq 0$$ we have $$\alpha \neq 0$$, hence $$\alpha = \frac{5}{7}\!.$$ Then

$$\beta = \frac{31}{14} - \frac{5}{7} = \frac{31}{14} - \frac{10}{14} = \frac{21}{14} = \frac{3}{2},$$ $$\gamma = \frac{53}{35} - \frac{5}{7} = \frac{53}{35} - \frac{25}{35} = \frac{28}{35} = \frac{4}{5},$$ $$\lambda = \frac{14 \times \frac{5}{7} \times \frac{3}{2}}{3} = \frac{14 \times \frac{15}{14}}{3} = \frac{15}{3} = 5\!.$$

Next, we calculate

$$\frac{3\alpha}{\beta} = \frac{3 \times \frac{5}{7}}{\frac{3}{2}} = \frac{\frac{15}{7}}{\frac{3}{2}} = \frac{15}{7} \times \frac{2}{3} = \frac{10}{7},$$ $$\frac{4\alpha}{\gamma} = \frac{4 \times \frac{5}{7}}{\frac{4}{5}} = \frac{\frac{20}{7}}{\frac{4}{5}} = \frac{20}{7} \times \frac{5}{4} = \frac{25}{7}\!.$$

Hence the sum and product of these quantities are

$$\frac{10}{7} + \frac{25}{7} = 5,\qquad \frac{10}{7} \times \frac{25}{7} = \frac{250}{49}\!.$$

Therefore, the required equation with these sum and product is

$$x^2 - 5x + \frac{250}{49} = 0,$$ or equivalently, multiplying through by 49,

$$49x^2 - 245x + 250 = 0\!.$$

This matches Option 3. The answer is $$49x^2 - 245x + 250 = 0$$.

1. Find $$m$$ (Real roots of $$x^2 - 12x + [x] + 31 = 0$$)

Let $$[x] = I$$ and the fractional part be $$f$$, where $$x = I + f$$ and $$0 \le f < 1$$.

The equation becomes:

$$x^2 - 12x + 31 = -I$$

From the quadratic formula for $$x^2 - 12x + (31+I) = 0$$:

$$x = \frac{12 \pm \sqrt{144 - 4(31+I)}}{2} = 6 \pm \sqrt{36 - (31+I)} = 6 \pm \sqrt{5-I}$$

For $$x$$ to be real, $$5-I \ge 0 \implies I \le 5$$.

• If $$I=3$$, $$x = 6 \pm \sqrt{2} \approx 7.41$$ or $$4.58$$. Neither has $$[x]=3$$.
• If $$I=4$$, $$x = 6 \pm 1 = 7$$ or $$5$$. Neither has $$[x]=4$$.
• If $$I=5$$, $$x = 6 \pm 0 = 6$$. Here $$[x]=6 \neq 5$$.

Testing values shows there are no real roots for this equation. Thus, $$m = 0$$.

2. Find $$n$$ (Real roots of $$x^2 - 5|x + 2| - 4 = 0$$)

We check two cases for the absolute value:

• Case 1 ($$x \ge -2$$): $$x^2 - 5(x + 2) - 4 = 0 \implies x^2 - 5x - 14 = 0$$

Factor: $$(x-7)(x+2) = 0 \implies x = 7, -2$$. (Both valid as $$\ge -2$$)

• Case 2 ($$x < -2$$): $$x^2 + 5(x + 2) - 4 = 0 \implies x^2 + 5x + 6 = 0$$

Factor: $$(x+2)(x+3) = 0 \implies x = -2, -3$$.

$$x = -3$$ is valid ($$< -2$$). $$x = -2$$ was already found.

Distinct roots are $$\{7, -2, -3\}$$, so $$n = 3$$.

3. Final Calculation

Substitute $$m = 0$$ and $$n = 3$$ into the expression $$m^2 + mn + n^2$$:

$$0^2 + (0)(3) + 3^2 = 0 + 0 + 9 = \mathbf{9}$$

Correct Option: (A)

Let $$t = (\sqrt{3} + \sqrt{2})^{x^2 - 4}$$.

Since $$(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2}) = 3 - 2 = 1$$, we have $$(\sqrt{3} - \sqrt{2}) = \frac{1}{\sqrt{3} + \sqrt{2}}$$.

Therefore: $$(\sqrt{3} - \sqrt{2})^{x^2-4} = \frac{1}{t}$$

The equation becomes:

$$ t + \frac{1}{t} = 10 $$

$$ t^2 - 10t + 1 = 0 $$

$$ t = \frac{10 \pm \sqrt{96}}{2} = 5 \pm 2\sqrt{6} $$

Note that $$(\sqrt{3} + \sqrt{2})^2 = 3 + 2\sqrt{6} + 2 = 5 + 2\sqrt{6}$$.

Case 1: $$t = 5 + 2\sqrt{6} = (\sqrt{3} + \sqrt{2})^2$$

$$x^2 - 4 = 2 \Rightarrow x^2 = 6 \Rightarrow x = \pm\sqrt{6}$$

Case 2: $$t = 5 - 2\sqrt{6} = (\sqrt{3} + \sqrt{2})^{-2}$$

$$x^2 - 4 = -2 \Rightarrow x^2 = 2 \Rightarrow x = \pm\sqrt{2}$$

Total solutions: $$\{-\sqrt{6}, -\sqrt{2}, \sqrt{2}, \sqrt{6}\}$$

Therefore, $$n(S) = 4$$.

Put $$y = e^{x}$$. Because $$e^{x}\gt 0$$ for every real $$x$$, the given equation becomes a quartic in the single positive variable $$y$$:

$$y^{4}+8y^{3}+13y^{2}-8y+1 = 0 \qquad (1)$$

Try to split the quartic into two quadratics of the form$$(y^{2}+ay-1)(y^{2}+by-1).$$Multiplying we obtain

$$y^{4}+(a+b)y^{3}+(ab-2)y^{2}-(a+b)y+1.$$(1)

Comparing the coefficients of (1) with those in equation (1) gives the system

$$a+b = 8,\qquad ab-2 = 13,\qquad -(a+b) = -8.$$

The first and third equalities are identical. From $$a+b=8$$ and $$ab-2=13$$ we have $$ab = 15$$. Hence $$a,b$$ satisfy $$t^{2}-8t+15=0$$, whose roots are $$t=5,3$$. Thus

$$y^{4}+8y^{3}+13y^{2}-8y+1=(y^{2}+5y-1)(y^{2}+3y-1)=0.$$

Solve each quadratic separately.

$$y^{2}+5y-1=0\;\Rightarrow\;y=\frac{-5\pm\sqrt{25+4}}{2} =\frac{-5\pm\sqrt{29}}{2}.$$

Since $$\sqrt{29}\approx5.385$$,  $$y_{1}=\frac{-5+\sqrt{29}}{2}\approx0.1925\;(\gt0),\qquad y_{2}=\frac{-5-\sqrt{29}}{2}\approx-5.1925\;(\lt0).$$

$$y^{2}+3y-1=0\;\Rightarrow\;y=\frac{-3\pm\sqrt{9+4}}{2} =\frac{-3\pm\sqrt{13}}{2}.$$

Because $$\sqrt{13}\approx3.605$$,  $$y_{3}=\frac{-3+\sqrt{13}}{2}\approx0.3025\;(\gt0),\qquad y_{4}=\frac{-3-\sqrt{13}}{2}\approx-3.3025\;(\lt0).$$

Only the positive roots satisfy $$y=e^{x}\gt0$$, so from $$y_{1}$$ and $$y_{3}$$ we get

$$x_{1}=\ln y_{1}=\ln\!\left(\frac{-5+\sqrt{29}}{2}\right),\quad x_{2}=\ln y_{3}=\ln\!\left(\frac{-3+\sqrt{13}}{2}\right).$$

Both numbers $$y_{1},y_{3}$$ are between 0 and 1, hence $$\ln y_{1}\lt0$$ and $$\ln y_{3}\lt0$$. Therefore

• exactly two real values of $$x$$ satisfy the original equation,
• and both of those values are negative.

Option B which is: two solutions and both are negative.

For the logarithm to be defined,

$$x+\frac72>0,$$

$$x+\frac72\ne1,$$

and

$$\left(\frac{x-7}{2x-3}\right)^2>0.$$

Thus,

$$x>-\frac72,\qquad x\ne-\frac52,\qquad x\ne\frac32,\qquad x\ne7.$$

Since $$x$$ is an integer,

$$x\ge-3,$$

with

$$x\ne-2,\ 7.$$

Now consider the base.

Case 1:

$$x+\frac72>1$$

$$x>-\frac52.$$

For an integer,

$$x\ge-2.$$

Since the base is greater than $$1$$,

$$\log_aN\ge0\iff N\ge1.$$

Hence,

$$\left(\frac{x-7}{2x-3}\right)^2\ge1.$$

This gives

$$|x-7|\ge|2x-3|.$$

Squaring,

$$(x-7)^2\ge(2x-3)^2.$$

$$x^2-14x+49\ge4x^2-12x+9.$$

$$3x^2+2x-40\le0.$$

$$(3x-10)(x+4)\le0.$$

Hence,

$$-4\le x\le\frac{10}{3}.$$

Since

$$x\ge-2,$$

the integral values are

$$-1,\ 0,\ 1,\ 2,\ 3.$$

(Note that $$x=-2$$ is excluded because the base becomes $$\frac32$$? Actually $$x=-2$$ makes the base $$\frac32$$ but the argument is defined; however, $$x=-2$$ was excluded earlier since $$x+\frac72=1.5$$? The excluded value is $$x=-\frac52$$, so $$x=-2$$ is allowed.)

Thus the integral values are

$$-2,\ -1,\ 0,\ 1,\ 2,\ 3.$$

Case 2:

$$0<x+\frac72<1.$$

This requires

$$-\frac72<x<-\frac52,$$

which contains no integer.

Hence there are no solutions from this case.

Therefore, the total number of integral solutions is

$$6.$$

Hence,

$$\boxed{6}$$

Let $$f(x)=2x^{2}-8x+k$$. The quadratic has real roots only when its discriminant is positive:

$$\Delta = (-8)^{2}-4\cdot2\cdot k = 64-8k \gt 0 \;\;\Longrightarrow\;\; k \lt 8.$$

Denote the two roots by $$\alpha$$ and $$\beta$$. From Vieta’s relations,

$$\alpha+\beta=\frac{8}{2}=4,\qquad \alpha\beta=\frac{k}{2}.$$

The condition “one root in $$(1,2)$$ and the other in $$(2,3)$$” forces the smaller root to lie in $$(1,2)$$ and the larger in $$(2,3)$$ (because $$\alpha+\beta=4$$).
Hence let $$\alpha\in(1,2)$$. Then $$\beta=4-\alpha\in(2,3).$$

Using $$\alpha\beta=\dfrac{k}{2}$$, we obtain

$$k =2\alpha\beta =2\alpha(4-\alpha) =8\alpha-2\alpha^{2}.$$

As $$\alpha$$ varies in $$(1,2)$$, the expression $$8\alpha-2\alpha^{2}$$ takes every value between

$$k_{\min}=8\cdot1-2\cdot1^{2}=6 \quad\text{and}\quad k_{\max}=8\cdot2-2\cdot2^{2}=8.$$

Because the endpoints $$\alpha=1$$ and $$\alpha=2$$ are not included, we actually have

$$6 \lt k \lt 8.$$

The only integer that satisfies $$6 \lt k \lt 8$$ is $$k=7.$$

Therefore exactly one integral value of $$k$$ meets the given requirement.

Hence, the required number of integral values is $$1$$. (Option C)

To find the points where the curve cuts the x-axis, set

$$e^{8x}-e^{6x}-3e^{4x}-e^{2x}+1=0$$

Let

$$t=e^{2x}$$

Since $$e^{2x}>0$$ for all real $$x$$,

$$t>0$$

The equation becomes

$$t^4-t^3-3t^2-t+1=0$$

Dividing throughout by $$t^2$$,

$$t^2-t-3-\frac{1}{t}+\frac{1}{t^2}=0$$

Rearranging,

$$\left(t^2+\frac{1}{t^2}\right)-\left(t+\frac{1}{t}\right)-3=0$$

Let

$$u=t+\frac{1}{t}$$

Using

$$u^2=t^2+2+\frac{1}{t^2}$$

gives

$$t^2+\frac{1}{t^2}=u^2-2$$

Substituting,

$$(u^2-2)-u-3=0$$

$$u^2-u-5=0$$

Solving,

$$u=\frac{1\pm\sqrt{21}}{2}$$

Since $$t>0$$,

$$u=t+\frac{1}{t}\geq 2$$

Therefore,

$$u=\frac{1+\sqrt{21}}{2}$$

is the only admissible value.

Substituting back,

$$t+\frac{1}{t}=k$$

where

$$k=\frac{1+\sqrt{21}}{2}>2$$

Multiplying by $$t$$,

$$t^2-kt+1=0$$

The discriminant is

$$\Delta=k^2-4$$

Since $$k>2$$,

$$\Delta>0$$

Therefore, the quadratic has two distinct real roots.

Also,

$$t_1t_2=1>0$$

and

$$t_1+t_2=k>0$$

Hence, both roots are positive.

Since

$$t=e^{2x}$$

and the exponential function is one-to-one, each positive value of $$t$$ gives exactly one real value of $$x$$.

Thus, the equation has two distinct real solutions for $$x$$.

Hence, the curve cuts the x-axis at

$$2$$

distinct points.

To find the number of real roots of the given radical equation, we first factorize each quadratic expression under the square roots:

$$\sqrt{x^2 - 4x + 3} + \sqrt{x^2 - 9} = \sqrt{4x^2 - 14x + 6}$$

Factoring the expressions yields:

$$\sqrt{(x - 1)(x - 3)} + \sqrt{(x + 3)(x - 3)} = \sqrt{2(2x - 1)(x - 3)}$$

---

Step 1: Determine the domain of the equation

For the functions to be defined in the real number system, the terms inside the square roots must be non-negative:

$$(x - 1)(x - 3) \ge 0 \implies x \in (-\infty, 1] \cup [3, \infty)$$

$$(x + 3)(x - 3) \ge 0 \implies x \in (-\infty, -3] \cup [3, \infty)$$

$$2(2x - 1)(x - 3) \ge 0 \implies x \in \left(-\infty, \frac{1}{2}\right] \cup [3, \infty)$$

Taking the intersection of all three intervals gives the valid domain:

$$x \in (-\infty, -3] \cup [3, \infty)$$

---

Step 2: Solve the equation

From the factored equation, it is clear that $$x = 3$$ makes every term zero:

$$\sqrt{0} + \sqrt{0} = \sqrt{0} \implies 0 = 0$$

Thus, $$x = 3$$ is a valid real root. We now look for other potential roots by assuming $$x \neq 3$$.

Case 1: For $$x > 3$$

Since $$x - 3 > 0$$, we can divide the entire equation by $$\sqrt{x - 3}$$:

$$\sqrt{x - 1} + \sqrt{x + 3} = \sqrt{2(2x - 1)}$$

Squaring both sides of the equation:

$$(x - 1) + (x + 3) + 2\sqrt{(x - 1)(x + 3)} = 2(2x - 1)$$

$$2x + 2 + 2\sqrt{x^2 + 2x - 3} = 4x - 2$$

$$2\sqrt{x^2 + 2x - 3} = 2x - 4$$

$$\sqrt{x^2 + 2x - 3} = x - 2$$

Squaring both sides again:

$$x^2 + 2x - 3 = (x - 2)^2$$

$$x^2 + 2x - 3 = x^2 - 4x + 4$$

$$6x = 7 \implies x = \frac{7}{6}$$

Since $$x = \frac{7}{6}$$ does not satisfy the condition $$x > 3$$, it is an extraneous solution.

Case 2: For $$x \le -3$$

Since $$x - 3 < 0$$, we can rewrite the terms as $$-(3 - x)$$ where $$3 - x > 0$$. Dividing the original equation by $$\sqrt{3 - x}$$ gives:

$$\sqrt{1 - x} + \sqrt{-x - 3} = \sqrt{2(1 - 2x)}$$

Squaring both sides of the equation:

$$(1 - x) + (-x - 3) + 2\sqrt{(1 - x)(-x - 3)} = 2(1 - 2x)$$

$$-2x - 2 + 2\sqrt{x^2 + 2x - 3} = 2 - 4x$$

$$2\sqrt{x^2 + 2x - 3} = 4 - 2x$$

$$\sqrt{x^2 + 2x - 3} = 2 - x$$

Squaring both sides again:

$$x^2 + 2x - 3 = (2 - x)^2$$

$$x^2 + 2x - 3 = x^2 - 4x + 4$$

$$6x = 7 \implies x = \frac{7}{6}$$

Since $$x = \frac{7}{6}$$ does not satisfy the condition $$x \le -3$$, it is also an extraneous solution.

---

Step 3: Conclusion

The only value that satisfies the equation within the valid domain is $$x = 3$$.

Therefore, the number of real roots of the equation is equal to 1.

We must solve the equation $$x\lvert x\rvert \;-\;5\lvert x+2\rvert \;+\;6 \;=\;0$$.
Because of the absolute-value symbols, the algebraic form of the left-hand side changes whenever either $$x=0$$ or $$x=-2$$. Hence analyse the three intervals

Case 1: $$x\lt-2$$

Here $$x$$ is negative and $$x+2$$ is also negative, so $$\lvert x\rvert=-x,\qquad \lvert x+2\rvert=-(x+2).$$ Substituting in the equation gives $$x(-x)\;-\;5\bigl(-(x+2)\bigr)\;+\;6=0$$ $$\Longrightarrow\;-x^{2}+5x+16=0$$ $$\Longrightarrow\;x^{2}-5x-16=0\;-(1)$$

The discriminant of $$(1)$$ is $$25+64=89,$$ so the roots are $$\dfrac{5\pm\sqrt{89}}{2}.$$ The negative root is $$\dfrac{5-\sqrt{89}}{2}\approx-2.217,$$ which lies in the interval $$(-\infty,-2).$$ The positive root $$\dfrac{5+\sqrt{89}}{2}\approx7.217$$ does not belong to this interval.
Hence Case 1 contributes exactly one real root.

Case 2: $$-2\le x<0$$

Now $$x$$ is still negative whereas $$x+2$$ is non-negative, so $$\lvert x\rvert=-x,\qquad \lvert x+2\rvert=x+2.$$ Substituting, $$x(-x)\;-\;5(x+2)\;+\;6=0$$ $$\Longrightarrow\;-x^{2}-5x-4=0$$ $$\Longrightarrow\;x^{2}+5x+4=0\;-(2)$$

The discriminant of $$(2)$$ is $$25-16=9,$$ giving roots $$x=-1,\;x=-4.$$ Only $$x=-1$$ lies in the interval $$[-2,0),$$ so Case 2 contributes one real root.

Case 3: $$x\ge0$$

Here both $$x$$ and $$x+2$$ are non-negative, so $$\lvert x\rvert=x,\qquad \lvert x+2\rvert=x+2.$$ Thus $$x(x)\;-\;5(x+2)\;+\;6=0$$ $$\Longrightarrow\;x^{2}-5x-4=0\;-(3)$$

The discriminant of $$(3)$$ is $$25+16=41,$$ so the roots are $$x=\dfrac{5\pm\sqrt{41}}{2}.$$ The positive root $$\dfrac{5+\sqrt{41}}{2}\approx5.701$$ belongs to $$[0,\infty)$$, whereas the negative root $$\dfrac{5-\sqrt{41}}{2}\approx-0.701$$ does not. Therefore Case 3 adds one more real root.

Summing up:

• Case 1: one root (approximately $$-2.217$$)
• Case 2: one root (exactly $$-1$$)
• Case 3: one root (approximately $$5.701$$)

The original equation has $$1+1+1=3$$ real solutions.

Hence the number of real roots is $$3$$. Therefore Option D is correct.

Let $$t = x + \frac{1}{x}$$. Then $$x^2 + \frac{1}{x^2} = t^2 - 2$$.

The equation becomes:

$$3(t^2 - 2) - 2t + 5 = 0$$

$$3t^2 - 6 - 2t + 5 = 0$$

$$3t^2 - 2t - 1 = 0$$

$$(3t + 1)(t - 1) = 0$$

$$t = -\frac{1}{3}$$ or $$t = 1$$

Case 1: $$x + \frac{1}{x} = 1$$

$$x^2 - x + 1 = 0$$

Discriminant = $$1 - 4 = -3 < 0$$. No real solutions.

Case 2: $$x + \frac{1}{x} = -\frac{1}{3}$$

$$3x^2 + x + 3 = 0$$

Discriminant = $$1 - 36 = -35 < 0$$. No real solutions.

Therefore, the total number of real solutions is 0.

The correct answer is Option 2: 0.

Write the given equation in two parts:

$$|x^{2}-8x+15| - 2x + 7 = 0$$

The quadratic inside the absolute value factors as

$$x^{2}-8x+15 = (x-5)(x-3)$$

Its roots are $$x = 3$$ and $$x = 5$$, and the parabola opens upward, so

$$x^{2}-8x+15 \lt 0 \text{ for } 3 \lt x \lt 5,\qquad x^{2}-8x+15 \ge 0 \text{ for } x \le 3 \text{ or } x \ge 5$$

Therefore split the equation into two cases.

Case 1: $$x \le 3$$ or $$x \ge 5$$

Here $$|x^{2}-8x+15| = x^{2}-8x+15$$, so

$$x^{2}-8x+15 - 2x + 7 = 0 \\ \Rightarrow x^{2} - 10x + 22 = 0$$

Using the quadratic formula,

$$x = \frac{10 \pm \sqrt{100-88}}{2} = \frac{10 \pm 2\sqrt{3}}{2} = 5 \pm \sqrt{3}$$

Check which roots satisfy the domain of this case:

$$5+\sqrt{3} \approx 6.732 \ge 5 \quad \text{(acceptable)}$$

$$5-\sqrt{3} \approx 3.268 \quad (3 \lt 3.268 \lt 5,\; \text{reject})$$

Thus Case 1 contributes the root $$x = 5+\sqrt{3}$$.

Case 2: $$3 \lt x \lt 5$$

Now $$|x^{2}-8x+15| = -(x^{2}-8x+15)$$, so

$$-(x^{2}-8x+15) - 2x + 7 = 0 \\ \Rightarrow -x^{2} + 8x - 15 - 2x + 7 = 0 \\ \Rightarrow -x^{2} + 6x - 8 = 0 \\ \Rightarrow x^{2} - 6x + 8 = 0$$

Again apply the quadratic formula:

$$x = \frac{6 \pm \sqrt{36-32}}{2} = \frac{6 \pm 2}{2} = 3 \pm 1$$

So the possible solutions are $$x = 4$$ and $$x = 2$$. Only $$x = 4$$ lies in the interval $$3 \lt x \lt 5$$, hence it is the sole root from Case 2.

The complete set of roots is $$\{\,4,\;5+\sqrt{3}\,\}$$.

Sum of all roots:

$$4 + \bigl(5+\sqrt{3}\bigr) = 9 + \sqrt{3}$$

Therefore the required sum is $$9 + \sqrt{3}$$, which matches Option B.

We need the negation of $$p \wedge (q \wedge \sim(p \wedge q))$$.

First, we simplify the inner part: $$q \wedge \sim(p \wedge q) = q \wedge (\sim p \vee \sim q)$$. Distributing gives $$= (q \wedge \sim p) \vee (q \wedge \sim q) = (q \wedge \sim p) \vee F = q \wedge \sim p$$. Substituting back into the full expression yields $$p \wedge (q \wedge \sim p) = (p \wedge \sim p) \wedge q = F \wedge q = F$$.

Because the expression is always false, its negation is always true.

To find a matching tautology, consider Option 1: $$(\sim(p \wedge q)) \vee p$$. This simplifies as follows: $$(\sim(p \wedge q)) \vee p = (\sim p \vee \sim q) \vee p = (\sim p \vee p) \vee \sim q = T \vee \sim q = T$$, showing it is a tautology.

The correct answer is Option 1: $$(\sim(p \wedge q)) \vee p$$.

The expression $$(p \wedge \sim q) \vee (\sim p \wedge q) \vee (\sim p \wedge \sim q)$$.

Let us simplify: Factor out from the last two terms:

$$= (p \wedge \sim q) \vee [\sim p \wedge (q \vee \sim q)]$$

$$= (p \wedge \sim q) \vee [\sim p \wedge T]$$

$$= (p \wedge \sim q) \vee \sim p$$

Using distributive law: $$A \vee (B \wedge C) = (A \vee B) \wedge (A \vee C)$$:

$$= (\sim p \vee p) \wedge (\sim p \vee \sim q)$$

$$= T \wedge (\sim p \vee \sim q)$$

$$= \sim p \vee \sim q$$

This is equivalent to $$\sim p \vee \sim q$$, which is option 2.

We need to determine which of the statements $$S_1$$ and $$S_2$$ are true.

To analyze $$S_1$$, observe that $$(p \Rightarrow q) \wedge (p \wedge (\sim q))$$ is claimed to be a contradiction.

Recall that $$p \Rightarrow q \equiv (\sim p) \vee q$$. Thus:

$$S_1 = ((\sim p) \vee q) \wedge (p \wedge (\sim q))$$

$$= ((\sim p) \vee q) \wedge p \wedge (\sim q)$$

Distributing yields:

$$= ((\sim p) \wedge p \wedge (\sim q)) \vee (q \wedge p \wedge (\sim q))$$

Since $$(\sim p) \wedge p = F$$ and $$q \wedge (\sim q) = F$$, it follows that:

$$S_1 = F \vee F = F$$

Thus, $$S_1$$ is always false, so it is a contradiction and the claim in $$S_1$$ is true.

Next, consider $$S_2$$, whose expression $$(p \wedge q) \vee ((\sim p) \wedge q) \vee (p \wedge (\sim q)) \vee ((\sim p) \wedge (\sim q))$$ is to be shown a tautology.

We can factor this expression as:

$$= q \wedge (p \vee (\sim p)) \vee (\sim q) \wedge (p \vee (\sim p))$$

Since $$p \vee (\sim p) = T$$, we obtain:

$$= (q \wedge T) \vee ((\sim q) \wedge T) = q \vee (\sim q) = T$$

Hence, $$S_2$$ is always true, i.e., it is a tautology, and the claim in $$S_2$$ is true.

Since both $$S_1$$ and $$S_2$$ are true, the correct answer is Option D: both are true.

We need to solve $$x^2 - 4x + [x] + 3 = x[x]$$, where $$[x]$$ is the greatest integer function.

Rearrange: $$x^2 - 4x + 3 = x[x] - [x] = [x](x-1)$$

$$(x-1)(x-3) = [x](x-1)$$

Case 1: $$x = 1$$

LHS = 0, RHS = $$[1](0) = 0$$. ✓ $$x = 1$$ is a solution.

Case 2: $$x \neq 1$$

Divide by $$(x-1)$$: $$x - 3 = [x]$$

$$x - [x] = 3$$, i.e., $$\{x\} = 3$$.

But the fractional part $$\{x\} \in [0, 1)$$, so $$\{x\} = 3$$ is impossible.

Therefore, $$x = 1$$ is the unique solution.

Since $$1 \notin (-\infty, 1)$$, the solution is in $$(-\infty, \infty)$$ but not in $$(-\infty, 1)$$.

The correct answer is Option D: a unique solution in $$(-\infty, \infty)$$.

Find the set of all $$a \in \mathbb{R}$$ for which $$x|x-1| + |x+2| + a = 0$$ has exactly one real root.

Let $$f(x) = x|x-1| + |x+2|$$. The equation becomes $$a = -f(x)$$. We analyze $$g(x) = -f(x)$$ in three intervals based on the critical points $$x = -2$$ and $$x = 1$$.

Case 1: $$x \geq 1$$

$$f(x) = x(x-1) + (x+2) = x^2 + 2$$, so $$g(x) = -(x^2 + 2)$$.

This is strictly decreasing on $$[1, \infty)$$ with $$g(1) = -3$$ and $$g(x) \to -\infty$$.

Case 2: $$-2 \leq x < 1$$

$$f(x) = x(1-x) + (x+2) = -x^2 + 2x + 2$$, so $$g(x) = x^2 - 2x - 2 = (x-1)^2 - 3$$.

On $$[-2, 1)$$: $$g(-2) = 6$$ and $$g(x) \to -3$$ as $$x \to 1^-$$. Since $$(x-1)^2$$ is decreasing for $$x < 1$$, $$g$$ is strictly decreasing on $$[-2, 1)$$.

Case 3: $$x < -2$$

$$f(x) = x(1-x) - (x+2) = -x^2 - 2$$, so $$g(x) = x^2 + 2$$.

As $$x$$ increases from $$-\infty$$ to $$-2^-$$: $$g$$ decreases from $$+\infty$$ to $$6^+$$.

Combining all cases:

The function $$g(x)$$ is strictly decreasing over all of $$\mathbb{R}$$ and continuous, with range $$(-\infty, +\infty)$$. At the boundaries: $$g(-2) = 6$$ from both Cases 2 and 3, and $$g(1) = -3$$ from both Cases 1 and 2. Therefore, for every real value of $$a$$, the equation $$a = g(x)$$ has exactly one solution.

The answer is Option B: $$(-\infty, \infty)$$.

Given: $$3x^2 + \lambda x - 1 = 0$$ with roots $$\alpha$$ and $$\beta$$.

By Vieta's formulas:

$$ \alpha + \beta = -\frac{\lambda}{3}, \quad \alpha\beta = -\frac{1}{3} $$

We are given that the sum of the squares of the reciprocals of the roots is 15:

$$ \frac{1}{\alpha^2} + \frac{1}{\beta^2} = 15 $$

$$ \frac{\alpha^2 + \beta^2}{\alpha^2\beta^2} = 15 $$

Now, $$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \frac{\lambda^2}{9} - 2\left(-\frac{1}{3}\right) = \frac{\lambda^2}{9} + \frac{2}{3}$$

And $$\alpha^2\beta^2 = (\alpha\beta)^2 = \frac{1}{9}$$

Substituting:

$$ \frac{\frac{\lambda^2}{9} + \frac{2}{3}}{\frac{1}{9}} = 15 $$

$$ 9\left(\frac{\lambda^2}{9} + \frac{2}{3}\right) = 15 $$

$$ \lambda^2 + 6 = 15 $$

$$ \lambda^2 = 9 \implies \lambda = \pm 3 $$

Now we need to find $$6(\alpha^3 + \beta^3)^2$$.

Using the identity: $$\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)$$

$$ \alpha + \beta = -\frac{\lambda}{3} = \pm\frac{3}{3} = \pm 1 $$

$$ \alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta) $$

$$ = (\pm 1)^3 - 3\left(-\frac{1}{3}\right)(\pm 1) $$

$$ = \pm 1 + (\pm 1) = \pm 2 $$

Therefore:

$$ 6(\alpha^3 + \beta^3)^2 = 6 \times (\pm 2)^2 = 6 \times 4 = 24 $$

The answer is Option C: 24.

Given equation,

$$x^2+(3-a)x=2(a-1)$$

or,

$$x^2+(3-a)x-2(a-1)=0$$

Let the roots be

$$\alpha,\beta$$

Then,

$$\alpha+\beta=a-3$$

and

$$\alpha\beta=-2(a-1)$$

Now,

$$\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta$$

$$=(a-3)^2-2[-2(a-1)]$$

$$=a^2-6a+9+4a-4$$

$$=a^2-2a+5$$

Complete the square:

$$a^2-2a+5=(a-1)^2+4$$

Since

$$(a-1)^2\ge0,$$

minimum value occurs at

$$a=1$$

Hence, minimum value is

$$\boxed{4}$$.

We need to find the sum of all real roots of $$\left(e^{2x} - 4\right)\left(6e^{2x} - 5e^x + 1\right) = 0$$.

Since $$e^{2x} - 4 = 0$$, it follows that

$$ e^{2x} = 4 \implies e^x = 2 \implies x = \ln 2 $$

Next, setting the second factor to zero gives $$6e^{2x} - 5e^x + 1 = 0$$, and letting $$t = e^x$$ (where $$t > 0$$) leads to

$$ 6t^2 - 5t + 1 = 0 $$

Using the quadratic formula, we obtain

$$ t = \frac{5 \pm \sqrt{25 - 24}}{12} = \frac{5 \pm 1}{12} $$

This yields

$$ t = \frac{6}{12} = \frac{1}{2} \quad \text{or} \quad t = \frac{4}{12} = \frac{1}{3} $$

Substituting back, for $$t = \frac{1}{2}$$ we have $$e^x = \frac{1}{2} \implies x = -\ln 2$$, and for $$t = \frac{1}{3}$$ we have $$e^x = \frac{1}{3} \implies x = -\ln 3$$.

Thus, the three real roots are $$\ln 2, -\ln 2, -\ln 3$$, and their sum is

$$ \ln 2 + (-\ln 2) + (-\ln 3) = -\ln 3 $$

Therefore, the correct option is Option B: $$-\ln 3$$.

Given,

$$f(x)=ax^2+bx+c$$

Also,

$$f(1)=3,\qquad f(-2)=\lambda,\qquad f(3)=4$$

and

$$f(0)+f(1)+f(-2)+f(3)=14$$

Since

$$f(0)=c,$$

we get

$$c+3+\lambda+4=14$$

$$c+\lambda=7$$

$$c=7-\lambda$$

Now,

$$f(1)=a+b+c=3$$

$$a+b+7-\lambda=3$$

$$a+b=\lambda-4$$

Also,

$$f(3)=9a+3b+c=4$$

$$9a+3b+7-\lambda=4$$

$$9a+3b=\lambda-3$$

Dividing by $$3,$$

$$3a+b=\frac{\lambda-3}{3}$$

Subtracting,

$$2a=\frac{\lambda-3}{3}-(\lambda-4)$$

$$2a=\frac{\lambda-3-3\lambda+12}{3}$$

$$2a=\frac{9-2\lambda}{3}$$

$$a=\frac{9-2\lambda}{6}$$

Now use

$$f(-2)=4a-2b+c=\lambda$$

Substituting

$$c=7-\lambda$$

and

$$b=\lambda-4-a,$$

we get

$$4a-2(\lambda-4-a)+7-\lambda=\lambda$$

$$6a-3\lambda+15=\lambda$$

$$6a=4\lambda-15$$

But,

$$a=\frac{9-2\lambda}{6}$$

Hence,

$$9-2\lambda=4\lambda-15$$

$$24=6\lambda$$

$$\lambda=4$$

Hence, $$\boxed{4}$$.

Given $$p + q = 3$$ and $$p^4 + q^4 = 369$$. Find $$\left(\frac{1}{p} + \frac{1}{q}\right)^{-2}$$.

First, we find $$pq$$. We know $$p + q = 3$$, so $$(p+q)^2 = 9 \implies p^2 + q^2 = 9 - 2pq$$. Now, $$(p^2 + q^2)^2 = p^4 + q^4 + 2p^2q^2$$, so $$(9 - 2pq)^2 = 369 + 2(pq)^2$$. Let $$t = pq$$: $$81 - 36t + 4t^2 = 369 + 2t^2$$, which simplifies to $$2t^2 - 36t + 81 - 369 = 0$$ and hence $$2t^2 - 36t - 288 = 0$$. Dividing by 2 gives $$t^2 - 18t - 144 = 0$$, so $$(t - 24)(t + 6) = 0$$. Since $$pq = 24$$ with $$p+q=3$$ leads to complex roots, we take $$pq = -6$$.

Next, we verify that this choice is correct. We have $$p^2 + q^2 = 9 - 2(-6) = 21$$, and therefore $$p^4 + q^4 = 21^2 - 2 \cdot 36 = 441 - 72 = 369$$, as required.

Finally, we compute the desired expression: $$\frac{1}{p} + \frac{1}{q} = \frac{p+q}{pq} = \frac{3}{-6} = -\frac{1}{2}$$, so $$\left(\frac{1}{p} + \frac{1}{q}\right)^{-2} = \left(-\frac{1}{2}\right)^{-2} = 4$$.

Hence the answer is $$4$$.

The equation is $$e^{4x} + 4e^{3x} - 58e^{2x} + 4e^x + 1 = 0$$.

Divide both sides by $$e^{2x}$$:

$$e^{2x} + 4e^x - 58 + 4e^{-x} + e^{-2x} = 0$$

$$(e^{2x} + e^{-2x}) + 4(e^x + e^{-x}) - 58 = 0$$

Let $$t = e^x + e^{-x}$$, where $$t \geq 2$$ (by AM-GM). Then $$t^2 = e^{2x} + 2 + e^{-2x}$$, so $$e^{2x} + e^{-2x} = t^2 - 2$$.

Substituting:

$$(t^2 - 2) + 4t - 58 = 0$$

$$t^2 + 4t - 60 = 0$$

$$(t + 10)(t - 6) = 0$$

Since $$t \geq 2$$, we have $$t = 6$$.

Now solving $$e^x + e^{-x} = 6$$:

$$e^{2x} - 6e^x + 1 = 0$$

$$e^x = \frac{6 \pm \sqrt{36 - 4}}{2} = \frac{6 \pm \sqrt{32}}{2} = 3 \pm 2\sqrt{2}$$

Both values $$3 + 2\sqrt{2} > 0$$ and $$3 - 2\sqrt{2} > 0$$ are valid.

$$x = \ln(3 + 2\sqrt{2})$$ or $$x = \ln(3 - 2\sqrt{2})$$

Note that $$(3+2\sqrt{2})(3-2\sqrt{2}) = 1$$, so the two solutions are negatives of each other, confirming they are distinct.

Hence the number of real solutions is $$\boxed{2}$$.

Find $$P$$ such that the sum of all roots of $$e^{2x} - 11e^x - 45e^{-x} + \frac{81}{2} = 0$$ equals $$\log_e P$$.

Substituting $$t = e^x$$ (where $$t > 0$$) and then multiplying the entire equation by $$2e^x$$ gives

$$2e^{3x} - 22e^{2x} + 81e^x - 90 = 0$$

In terms of $$t$$ this becomes

$$2t^3 - 22t^2 + 81t - 90 = 0$$

Now testing $$t = 2$$ yields $$2(8) - 22(4) + 81(2) - 90 = 16 - 88 + 162 - 90 = 0$$ ✔

Therefore one factor is $$(t - 2)$$ and we can write

$$2t^3 - 22t^2 + 81t - 90 = (t - 2)(2t^2 - 18t + 45)$$

Verifying, we have

$$ (t-2)(2t^2 - 18t + 45) = 2t^3 - 18t^2 + 45t - 4t^2 + 36t - 90 = 2t^3 - 22t^2 + 81t - 90$$ ✔

Solving $$2t^2 - 18t + 45 = 0$$ gives

$$t = \frac{18 \pm \sqrt{324 - 360}}{4} = \frac{18 \pm \sqrt{-36}}{4} = \frac{18 \pm 6i}{4} = \frac{9 \pm 3i}{2}$$

Thus the complex roots are $$t_2 = \frac{9 + 3i}{2}$$ and $$t_3 = \frac{9 - 3i}{2}$$.

Since $$e^{x_1} = 2$$ for the real root, it follows that $$x_1 = \ln 2$$. For the complex roots we have $$e^{x_2} = \frac{9+3i}{2}$$ and $$e^{x_3} = \frac{9-3i}{2}$$, so

$$x_2 + x_3 = \ln\left(\frac{9+3i}{2}\right) + \ln\left(\frac{9-3i}{2}\right) = \ln\left(\frac{9+3i}{2} \cdot \frac{9-3i}{2}\right)$$

$$= \ln\left(\frac{81 + 9}{4}\right) = \ln\left(\frac{90}{4}\right) = \ln\left(\frac{45}{2}\right)$$

Therefore the sum of all roots is

$$x_1 + x_2 + x_3 = \ln 2 + \ln\frac{45}{2} = \ln\left(2 \cdot \frac{45}{2}\right) = \ln 45$$

Alternatively, by Vieta's formulas for the cubic $$2t^3 - 22t^2 + 81t - 90 = 0$$, the product of the roots is $$t_1 t_2 t_3 = \frac{90}{2} = 45$$. Since $$t_i = e^{x_i}$$ we obtain $$e^{x_1 + x_2 + x_3} = t_1 t_2 t_3 = 45$$, which implies $$x_1 + x_2 + x_3 = \ln 45 = \log_e P$$.

So $$P = 45$$.

The answer is $$\boxed{45}$$.

We begin with the given relation $$x^{2}+9y^{2}-4x+3=0$$ where both $$x$$ and $$y$$ are real numbers.

First, we bring all purely algebraic terms together and identify the quadratic expression in $$x$$ that can be completed to a perfect square. Re-writing, we have

$$x^{2}-4x+9y^{2}+3=0.$$

To complete the square in $$x$$, we recall the algebraic identity

$$(x-a)^{2}=x^{2}-2ax+a^{2}.$$

Comparing $$x^{2}-4x$$ with the right-hand side of the identity, we see that $$-2a=-4\; \Longrightarrow\; a=2.$$ Hence

$$x^{2}-4x=(x-2)^{2}-4.$$

Substituting this back into the original expression, we obtain

$$(x-2)^{2}-4+9y^{2}+3=0.$$

Now we combine the constant terms $$-4+3=-1$$ to get

$$(x-2)^{2}+9y^{2}-1=0.$$

Transposing $$-1$$ to the right side yields

$$(x-2)^{2}+9y^{2}=1.$$

This is the standard form of an ellipse centred at the point $$(2,\,0)$$. To make the domains of $$x$$ and $$y$$ explicit, we examine each variable in turn.

Because squares of real numbers are always non-negative, each term on the left must individually satisfy $$0 \le (x-2)^{2} \le 1$$ and $$0 \le 9y^{2} \le 1$$ so that their sum remains exactly $$1$$.

We start with the $$x$$-part:

$$0 \le (x-2)^{2} \le 1.$$

Taking square roots on the inequality $$0 \le (x-2)^{2} \le 1$$ gives

$$0 \le |x-2| \le 1.$$

This in turn implies

$$-1 \le x-2 \le 1.$$

Adding $$2$$ throughout, we obtain the interval for $$x$$:

$$1 \le x \le 3.$$

So $$x$$ necessarily lies in $$[1,\,3]$$.

Next we extract the allowable range of $$y$$ from the ellipse equation. Solving for $$y^{2}$$ we write

$$9y^{2}=1-(x-2)^{2}.$$

Dividing both sides by $$9$$ gives

$$y^{2}=\dfrac{1-(x-2)^{2}}{9}.$$

For real $$y$$, the right-hand side must be non-negative. But we have already ensured $$0 \le (x-2)^{2}\le 1,$$ so $$1-(x-2)^{2}\ge 0$$ automatically. Taking square roots yields

$$|y|=\sqrt{\dfrac{1-(x-2)^{2}}{9}} =\dfrac{\sqrt{1-(x-2)^{2}}}{3}\;.$$

The largest value of the numerator $$\sqrt{1-(x-2)^{2}}$$ is $$1$$, achieved when $$(x-2)^{2}=0$$, i.e. when $$x=2$$. Hence

$$|y|_{\text{max}}=\dfrac{1}{3}.$$

The smallest value is $$0$$, occurring when $$(x-2)^{2}=1$$, i.e. when $$x=1$$ or $$x=3$$. Combining these facts, we see that $$y$$ can range continuously between $$-\frac13$$ and $$+\frac13$$. Therefore

$$-\frac{1}{3}\le y\le \frac{1}{3}.$$

Summarising both intervals, we have

$$x\in [1,\,3]\quad\text{and}\quad y\in\left[-\dfrac13,\dfrac13\right].$$

Looking at the given choices, this matches exactly with Option B.

Hence, the correct answer is Option B.