JEE Quadratic Equations Questions

To solve this systematically, let's break down the Right Hand Side (RHS) first.

The function $$\min\{|x - 3|, |x + 2|\}$$ gives the smaller of the two distances from $$x$$ to $$3$$ and from $$x$$ to $$-2$$. The point where these distances are exactly equal is the midpoint:

$$x = \frac{3 + (-2)}{2} = 0.5$$

This allows us to redefine the RHS piecewise:

• For $$x \le 0.5$$, the distance to $$-2$$ is smaller, so $$\text{RHS} = |x + 2|$$
• For $$x > 0.5$$, the distance to $$3$$ is smaller, so $$\text{RHS} = |x - 3|$$

Now, split the original equation into these two main regions and solve.

Region 1: For $$x \le 0.5$$

The equation becomes: $$x^2 + 3x + 2 = |x + 2|$$

Since the absolute value changes behavior at $$x = -2$$, we check two sub-cases:

• Case 1a ($$x \le -2$$):

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

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

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

  (Since $$-2 \le -2$$, this is a valid solution).

• Case 1b ($$-2 < x \le 0.5$$):

  $$x^2 + 3x + 2 = +(x + 2)$$

  $$x^2 + 2x = 0$$

  $$x(x + 2) = 0 \implies x = 0 \text{ or } x = -2$$

  (Since $$x = 0$$ falls within $$-2 < x \le 0.5$$, this is a valid solution).

Region 2: For $$x > 0.5$$

The equation becomes: $$x^2 + 3x + 2 = |x - 3|$$

Again, split at the critical point $$x = 3$$:

• Case 2a ($$0.5 < x < 3$$):

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

  $$x^2 + 4x - 1 = 0$$

  Using the quadratic formula: $$x = \frac{-4 \pm \sqrt{16 - 4(1)(-1)}}{2} = -2 \pm \sqrt{5}$$

  Since $$\sqrt{5} \approx 2.23$$, the roots are roughly $$0.23$$ and $$-4.23$$. Neither of these falls in the interval $$(0.5, 3)$$.

  (No solution here).

• Case 2b ($$x \ge 3$$):

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

  $$x^2 + 2x + 5 = 0$$

  Checking the discriminant: $$D = 2^2 - 4(1)(5) = -16$$

  Since $$D < 0$$, there are no real roots.

  (No solution here).

Conclusion:

The only valid real solutions are $$x = -2$$ and $$x = 0$$.

Therefore, the number of real solutions is 2.

We have |2x-3| so the sign changes at x = 3/2

Let's take case 1: $$x\ge\dfrac{3}{2}$$

In this case: $$|2x-3| = 2x-3$$.

Hence, the equation will be $$x^{2} + (2x - 3) - 4 = 0$$

Or, $$x^{2} + 2x - 7 = 0$$

Solving using the discriminant formula i.e. $$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

$$x = \dfrac{-2 \pm \sqrt{2^2 - 4(1)(-7)}}{2(1)}$$

$$x = \dfrac{-2 \pm \sqrt{4 + 28}}{2}$$

$$x = \dfrac{-2 \pm \sqrt{32}}{2}$$

$$x = \dfrac{-2 \pm 4\sqrt{2}}{2}$$

$$x = -1 \pm 2\sqrt{2}$$

Now, let's check if the roots satisfy the initial assumption or not. 

$$x_1 = -1 + 2\sqrt{2} \approx -1 + 2(1.414) = 1.828$$. This is greater than 1.5, so it is a valid root.

$$x_2 = -1 - 2\sqrt{2} \approx -1 - 2.828 = -3.828$$. This is less than 1.5, so we reject it.

Let's take case 1:$$x<\dfrac{3}{2}$$

Here: $$|2x - 3| = -(2x - 3) = -2x + 3$$

Hence, the quadratic equation will be $$x^{2} + (-2x + 3) - 4 = 0$$

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

Again, use the quadratic formula:

$$x = \dfrac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-1)}}{2(1)}$$

$$x = \dfrac{2 \pm \sqrt{4 + 4}}{2}$$

$$x = \dfrac{2 \pm \sqrt{8}}{2}$$

$$x = \dfrac{2 \pm 2\sqrt{2}}{2}$$

$$x = 1 \pm \sqrt{2}$$

Again, we need to check if the roots satisfy the range or not. 

$$x_3 = 1 + \sqrt{2} \approx 1 + 1.414 = 2.414$$. This is greater than $$1.5$$, so we reject it.

$$x_4 = 1 - \sqrt{2} \approx 1 - 1.414 = -0.414$$ This is less than $$1.5$$, so it is a valid root.

Now, calculating the sum of the squares:

$$\alpha = -1 + 2\sqrt{2}$$ and $$\beta = 1 - \sqrt{2}$$

$$\alpha^{2} = (-1 + 2\sqrt{2})^{2} = 1 - 4\sqrt{2} + 4(2) = 1 - 4\sqrt{2} + 8 = 9 - 4\sqrt{2}$$

$$\beta^{2} = (1 - \sqrt{2})^{2} = 1 - 2\sqrt{2} + 2 = 3 - 2\sqrt{2}$$

Adding them: $$\alpha^{2} + \beta^{2} = (9 - 4\sqrt{2}) + (3 - 2\sqrt{2})$$

$$\alpha^2+\beta^2=12-6\sqrt{2}\ =\ 6\left(2-\sqrt{2}\right)$$

We are given $$P_n = \alpha^n + \beta^n$$ with $$P_{10} = 123$$, $$P_9 = 76$$, $$P_8 = 47$$, and $$P_1 = 1$$.

If $$\alpha$$ and $$\beta$$ are roots of $$x^2 - sx + p = 0$$ where $$s = \alpha + \beta$$ and $$p = \alpha\beta$$, then by Newton's identity: $$P_n = s \cdot P_{n-1} - p \cdot P_{n-2}$$.

Using $$n = 10$$: $$P_{10} = s \cdot P_9 - p \cdot P_8$$, so $$123 = 76s - 47p$$ ...(1)

We also know $$P_1 = \alpha + \beta = s = 1$$.

Substituting $$s = 1$$ in (1): $$123 = 76 - 47p$$, so $$47p = 76 - 123 = -47$$, giving $$p = -1$$.

Verification: $$P_2 = s \cdot P_1 - 2p = 1 - (-2) = 3$$. $$P_3 = s \cdot P_2 - p \cdot P_1 = 3 + 1 = 4$$. Continuing this recurrence should yield $$P_8 = 47$$, $$P_9 = 76$$, $$P_{10} = 123$$. Let us verify: $$P_4 = P_3 + P_2 = 7$$, $$P_5 = 11$$, $$P_6 = 18$$, $$P_7 = 29$$, $$P_8 = 47$$, $$P_9 = 76$$, $$P_{10} = 123$$. Confirmed.

The quadratic with roots $$\alpha$$ and $$\beta$$ is $$x^2 - x - 1 = 0$$ (since $$s = 1$$ and $$p = -1$$).

We need the quadratic with roots $$\frac{1}{\alpha}$$ and $$\frac{1}{\beta}$$. The sum is $$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{1}{-1} = -1$$. The product is $$\frac{1}{\alpha\beta} = \frac{1}{-1} = -1$$.

So the quadratic is $$x^2 - (-1)x + (-1) = 0$$, which is $$x^2 + x - 1 = 0$$.

Hence, the correct answer is Option B.

$$\alpha_\theta + \beta_\theta = -\frac{\cos\theta}{2}$$ 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.

First split the real line at the points where the absolute-value expressions change their form, i.e. at $$x = 2$$ and $$x = 3$$. This gives three disjoint intervals:

$$x \lt 2$$ Here $$|x-2| = 2-x$$ and $$|x-3| = 3-x$$.

The equation becomes $$x(2-x) + 3(3-x) + 1 = 0$$ $$\Rightarrow 2x - x^{2} + 9 - 3x + 1 = 0$$ $$\Rightarrow -x^{2} - x + 10 = 0$$ $$\Rightarrow x^{2} + x - 10 = 0$$ $$-(1)$$

Using the quadratic formula on $$(1)$$: $$x = \frac{-1 \pm \sqrt{1 + 40}}{2} = \frac{-1 \pm \sqrt{41}}{2}$$

Numerically, $$\dfrac{-1 - \sqrt{41}}{2} \approx -3.70$$ and $$\dfrac{-1 + \sqrt{41}}{2} \approx 2.70$$. Only the first value satisfies $$x \lt 2$$. Hence Case 1 contributes exactly one real root: $$x = \dfrac{-1 - \sqrt{41}}{2}.$$

$$2 \le x \lt 3$$ Here $$|x-2| = x-2$$ and $$|x-3| = 3-x$$.

The equation becomes $$x(x-2) + 3(3-x) + 1 = 0$$ $$\Rightarrow x^{2} - 2x + 9 - 3x + 1 = 0$$ $$\Rightarrow x^{2} - 5x + 10 = 0$$ $$-(2)$$

For $$(2)$$, the discriminant is $$\Delta = (-5)^{2} - 4(1)(10) = 25 - 40 = -15 \lt 0,$$ so there are no real roots in Case 2.

$$x \ge 3$$ Here $$|x-2| = x-2$$ and $$|x-3| = x-3$$.

The equation becomes $$x(x-2) + 3(x-3) + 1 = 0$$ $$\Rightarrow x^{2} - 2x + 3x - 9 + 1 = 0$$ $$\Rightarrow x^{2} + x - 8 = 0$$ $$-(3)$$

For $$(3)$$, the roots are $$x = \frac{-1 \pm \sqrt{1 + 32}}{2} = \frac{-1 \pm \sqrt{33}}{2}.$$ The positive root is $$\dfrac{-1 + \sqrt{33}}{2} \approx 2.37,$$ which is less than $$3,$$ and the negative root is even smaller. Hence neither root lies in the interval $$x \ge 3,$$ so Case 3 contributes no real roots.

Combining all three cases, only one value of $$x$$ satisfies the given equation.

Therefore the total number of real roots is $$1$$, corresponding to Option C.

Let $$\alpha$$ and $$\beta$$ be the roots of the equation.

$$\Rightarrow$$ $$\alpha + \beta = -4$$ and $$\alpha\beta = -n$$

The roots of the equation should be integral.

We want pairs such that the sum of the roots is $$-4$$ and the product of the roots is $$-n$$, where $$n$$ lies in $$[20, 100]$$. Hence, the following possible pairs exist. We can enumerate the cases and find out the number of distinct values of $$n$$

$$(3, -7) \Rightarrow n = 21$$

$$(4, -8) \Rightarrow n = 32$$

$$(5, -9) \Rightarrow n = 45$$

$$(6, -10) \Rightarrow n = 60$$

$$(7, -11) \Rightarrow n = 77$$

$$(8, -12) \Rightarrow n = 96$$

This gives 6 possible values of $$n$$

Thus, option C is the correct choice.

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 $$\boxed{7}.$$

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.

Let $$y=\ln\ x$$

Since $$x>0$$, this substitution is valid.

$$\therefore$$ $$x=e^y$$

and $$x^8=\left(e^y\right)^8=e^{8y}$$

Substitute into the equation 

$$e^{5\left(\ln\ x\right)^2-3}=e^{8y}$$

Since $$\left(\ln\ x\right)=y$$

$$\therefore$$ $$e^{5\left(y\right)^2-3}=e^{8y}$$

Equate the exponents as the bases are same (e)

$$5y^2-3=8y$$

$$\therefore$$ $$5y^2-8y-3=0$$

Sum of the quadratic equation =  $$y_1+y_2$$ which is equal to $$-\dfrac{b}{a}$$

$$\therefore$$ $$y_1+y_2=8/5$$------(i)

To find: $$x_1\times\ x_2$$

We know  $$x=e^y$$ 

$$\therefore$$ $$x_1\times\ x_2=e^{y_1+y_2}$$

Substitute the value of  $$y_1+y_2$$ from---(i)

$$\therefore$$  $$x_1\times\ x_2=e^\dfrac{8}{5}$$.

Hence, option A is the correct answer

$$(x^2 - 9x + 11)^2 - (x - 4)(x - 5) = 3$$

$$(x-4)(x-5)=x^2-9x+20$$

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.

We need to find the sum of the squares of the roots of two equations.

Equation 1: $$|x - 2|^2 + |x - 2| - 2 = 0$$

Let $$t = |x - 2|$$ where $$t \geq 0$$. The equation becomes:

$$$t^2 + t - 2 = 0$$$

Factoring:

$$$(t + 2)(t - 1) = 0$$$

So $$t = -2$$ or $$t = 1$$. Since $$t = |x - 2| \geq 0$$, we reject $$t = -2$$.

Therefore $$|x - 2| = 1$$, which gives:

$$x - 2 = 1 \implies x = 3$$ or $$x - 2 = -1 \implies x = 1$$

Sum of squares of roots from Equation 1: $$3^2 + 1^2 = 9 + 1 = 10$$

Equation 2: $$x^2 - 2|x - 3| - 5 = 0$$

Case (a): $$x \geq 3$$

$$|x - 3| = x - 3$$, so the equation becomes:

$$$x^2 - 2(x - 3) - 5 = 0$$$

$$$x^2 - 2x + 6 - 5 = 0$$$

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

$$$(x - 1)^2 = 0 \implies x = 1$$$

But $$x = 1 \lt 3$$, so this does not satisfy our assumption $$x \geq 3$$. No valid root in this case.

Case (b): $$x \lt 3$$

$$|x - 3| = -(x - 3) = 3 - x$$, so the equation becomes:

$$$x^2 - 2(3 - x) - 5 = 0$$$

$$$x^2 + 2x - 6 - 5 = 0$$$

$$$x^2 + 2x - 11 = 0$$$

Using the quadratic formula:

$$$x = \frac{-2 \pm \sqrt{4 + 44}}{2} = \frac{-2 \pm \sqrt{48}}{2} = \frac{-2 \pm 4\sqrt{3}}{2} = -1 \pm 2\sqrt{3}$$$

Checking: $$x_1 = -1 + 2\sqrt{3} = -1 + 3.464 = 2.464 \lt 3$$ ✓

$$x_2 = -1 - 2\sqrt{3} = -1 - 3.464 = -4.464 \lt 3$$ ✓

Both roots are valid.

Sum of squares of roots from Equation 2:

$$x_1^2 + x_2^2 = (-1 + 2\sqrt{3})^2 + (-1 - 2\sqrt{3})^2$$

Expanding each:

$$(-1 + 2\sqrt{3})^2 = 1 - 4\sqrt{3} + 12 = 13 - 4\sqrt{3}$$

$$(-1 - 2\sqrt{3})^2 = 1 + 4\sqrt{3} + 12 = 13 + 4\sqrt{3}$$

Adding: $$x_1^2 + x_2^2 = (13 - 4\sqrt{3}) + (13 + 4\sqrt{3}) = 26$$

Total sum of squares:

$$$10 + 26 = 36$$$

Hence, the correct answer is Option B.

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.

Let the given quadratic be $$ax^2 + bx + 1 = 0$$.

Because 2 and 6 are its roots, each satisfies the equation.

Substituting $$x = 2$$ gives $$4a + 2b + 1 = 0$$ $$-(1)$$

Substituting $$x = 6$$ gives $$36a + 6b + 1 = 0$$ $$-(2)$$

Subtract $$(1)$$ from $$(2)$$:

$$32a + 4b = 0 \;\;\Longrightarrow\;\; 8a + b = 0$$ $$-(3)$$

From $$(3)$$, $$b = -8a$$. Insert this value of $$b$$ into $$(1)$$:

$$4a + 2(-8a) + 1 = 0 \;\;\Longrightarrow\;\; 4a - 16a + 1 = 0$$

$$-12a + 1 = 0 \;\;\Longrightarrow\;\; a = \frac{1}{12}$$

Using $$b = -8a$$, we get $$b = -\frac{8}{12} = -\frac{2}{3}$$.

Next, find $$2a + b$$ and $$6a + b$$:

$$2a + b = 2\left(\frac{1}{12}\right) - \frac{2}{3} = \frac{1}{6} - \frac{2}{3} = -\frac{1}{2}$$

$$6a + b = 6\left(\frac{1}{12}\right) - \frac{2}{3} = \frac{1}{2} - \frac{2}{3} = -\frac{1}{6}$$

The required new roots are $$\frac{1}{2a+b}$$ and $$\frac{1}{6a+b}$$:

$$\frac{1}{2a+b} = \frac{1}{-\frac{1}{2}} = -2$$

$$\frac{1}{6a+b} = \frac{1}{-\frac{1}{6}} = -6$$

Hence the new quadratic, with roots $$-2$$ and $$-6$$, is obtained by

$$(x + 2)(x + 6) = 0$$

Multiplying out gives $$x^2 + 8x + 12 = 0$$.

Therefore, the required quadratic equation is $$x^2 + 8x + 12 = 0$$, which corresponds to Option B.

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).

Given $$\left|\frac{z_1 - 2z_2}{\frac{1}{2} - z_1\bar{z}_2}\right| = 2$$.

Multiply numerator and denominator by 2:

$$\left|\frac{2(z_1 - 2z_2)}{1 - 2z_1\bar{z}_2}\right| = 2$$

$$\left|\frac{z_1 - 2z_2}{\frac{1}{2} - z_1\bar{z}_2}\right| = 2$$

Squaring both sides:

$$|z_1 - 2z_2|^2 = 4\left|\frac{1}{2} - z_1\bar{z}_2\right|^2$$

Expanding the left side:

$$|z_1|^2 - 2z_1\overline{2z_2} - 2z_2\bar{z}_1 + 4|z_2|^2 = |z_1|^2 - 2(z_1\bar{z}_2 + \bar{z}_1 z_2) + 4|z_2|^2$$

Wait, let me be more careful. $$|z_1 - 2z_2|^2 = (z_1 - 2z_2)(\bar{z}_1 - 2\bar{z}_2) = |z_1|^2 - 2z_1\bar{z}_2 - 2\bar{z}_1 z_2 + 4|z_2|^2$$.

Right side: $$4\left|\frac{1}{2} - z_1\bar{z}_2\right|^2 = 4\left(\frac{1}{2} - z_1\bar{z}_2\right)\left(\frac{1}{2} - \bar{z}_1 z_2\right)$$

$$= 4\left(\frac{1}{4} - \frac{1}{2}\bar{z}_1 z_2 - \frac{1}{2}z_1\bar{z}_2 + |z_1|^2|z_2|^2\right)$$

$$= 1 - 2\bar{z}_1 z_2 - 2z_1\bar{z}_2 + 4|z_1|^2|z_2|^2$$

Setting LHS = RHS:

$$|z_1|^2 - 2z_1\bar{z}_2 - 2\bar{z}_1 z_2 + 4|z_2|^2 = 1 - 2\bar{z}_1 z_2 - 2z_1\bar{z}_2 + 4|z_1|^2|z_2|^2$$

The $$-2z_1\bar{z}_2 - 2\bar{z}_1 z_2$$ terms cancel from both sides:

$$|z_1|^2 + 4|z_2|^2 = 1 + 4|z_1|^2|z_2|^2$$

Rearranging: $$|z_1|^2 - 1 + 4|z_2|^2 - 4|z_1|^2|z_2|^2 = 0$$

$$(|z_1|^2 - 1) - 4|z_2|^2(|z_1|^2 - 1) = 0$$

$$(|z_1|^2 - 1)(1 - 4|z_2|^2) = 0$$

So either $$|z_1|^2 = 1$$ or $$|z_2|^2 = \frac{1}{4}$$.

This means either $$|z_1| = 1$$ ($$z_1$$ lies on a circle of radius 1) or $$|z_2| = \frac{1}{2}$$ ($$z_2$$ lies on a circle of radius $$\frac{1}{2}$$).

The answer is Option D.

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.

Given $$x^2 + 2\sqrt{2}x - 1 = 0$$ with roots $$\alpha, \beta$$.

By Vieta's: $$\alpha + \beta = -2\sqrt{2}$$, $$\alpha\beta = -1$$.

$$\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = 8 + 2 = 10$$

$$\alpha^4 + \beta^4 = (\alpha^2+\beta^2)^2 - 2(\alpha\beta)^2 = 100 - 2 = 98$$

$$\alpha^6 + \beta^6 = (\alpha^2+\beta^2)^3 - 3(\alpha\beta)^2(\alpha^2+\beta^2) = 1000 - 3(1)(10) = 970$$

So the two roots of the required equation are: $$\alpha^4 + \beta^4 = 98$$ and $$\frac{1}{10}(\alpha^6+\beta^6) = \frac{970}{10} = 97$$.

Sum of roots = $$98 + 97 = 195$$.

Product of roots = $$98 \times 97 = 9506$$.

The equation is $$x^2 - 195x + 9506 = 0$$.

The correct answer is Option 3.

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)

Let $$t=e^{\sin x}$$. 

Then $$t-2/t=2$$, so $$t^2-2t-2=0$$, $$t=1+\sqrt{3}$$ or $$t=1-\sqrt{3}<0$$ (rejected).

$$e^{\sin x}=1+\sqrt{3}\approx 2.73$$, $$\sin x=\ln(1+\sqrt{3})\approx 1.005>1$$.

Since $$|\sin x|\leq 1$$, no solution.

0 solution

Given: $$(8)^{2x} - 16 \cdot (8)^x + 48 = 0$$

Let $$y=8^x$$---(i); 

Therefore, the equation can also be written as $$y^2-16y+48$$

 Factorize the equation: $$\left(y-12\right)\left(y-4\right)$$

Hence, replace values of y in the equation, (i) we get $$4=8^{x\ }and\ 12=8^x$$

Now take the log with base 8 on both sides. 

$$\therefore$$ $$x_1=\log_84$$ and $$x_2=\log_812$$

 Sum of all solutions $$\Rightarrow$$ $$x_1+x_2=\log_8\left(12\right)\left(4\right)$$

It can be also written as $$\log_88+\log_{_8}6$$ $$\Rightarrow$$ $$1+\log_{_8}6$$

Hence, the correct answer is option 1.

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): $$5(1) + 8\left(\frac{1}{2}\right) = 5 + 4 = 9$$,

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

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

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

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.

Let $$\alpha, \beta$$ be roots of $$x^2 + \sqrt{6}x + 3 = 0$$. Then $$\alpha + \beta = -\sqrt{6}$$ and $$\alpha\beta = 3$$. Define $$S_n = \alpha^n + \beta^n$$, which satisfies the recurrence $$S_n = (\alpha+\beta)S_{n-1} - \alpha\beta S_{n-2} = -\sqrt{6}S_{n-1} - 3S_{n-2}$$.

We have $$S_0 = 2$$
$$S_1 = -\sqrt{6}$$
$$S_2 = (\alpha+\beta)^2 - 2\alpha\beta = 6 - 6 = 0$$
$$S_3 = -\sqrt{6}(0) - 3(-\sqrt{6}) = 3\sqrt{6}$$
$$S_4 = -\sqrt{6}(3\sqrt{6}) - 3(0) = -18$$
$$S_5 = -\sqrt{6}(-18) - 3(3\sqrt{6}) = 18\sqrt{6} - 9\sqrt{6} = 9\sqrt{6}$$
$$S_6 = -\sqrt{6}(9\sqrt{6}) - 3(-18) = -54 + 54 = 0$$

Since $$S_2 = 0$$ and $$S_6 = 0$$ and in general $$S_{4k+2} = 0$$ for all $$k \geq 0$$, it follows that $$S_{10} = 0$$ and $$S_{14} = 0$$.

Therefore, the given expression simplifies as $$\frac{S_{23} + S_{14}}{S_{15} + S_{10}} = \frac{S_{23} + 0}{S_{15} + 0} = \frac{S_{23}}{S_{15}}$$, and from the recurrence one finds $$S_{n+8} = 81 S_n$$ so that $$\frac{S_{23}}{S_{15}} = \frac{S_{15+8}}{S_{15}} = 81$$.

Alternatively, one may note that $$S_{23} = -\sqrt{6} S_{22} - 3 S_{21}$$ and continue the pattern with $$S_{15}$$ as reference to obtain the same ratio.

81

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, $$n(S) = 4$$.

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.

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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: $$(\sim(p \wedge q)) \vee p$$.

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 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 $$\mathbf{51}$$.

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

Solution :

Given quadratic equation :

$$x^2-\left(5+3^{\sqrt{\log_3 5}}-5^{\sqrt{\log_5 3}}\right)x + 3\left(3^{(\log_3 5)^{1/3}}-5^{(\log_5 3)^{2/3}}-1\right)=0$$

First simplify the coefficients.

Using identity :

$$a^{\log_a b}=b$$

Also,

$$3^{\sqrt{\log_3 5}} = 5^{\sqrt{\log_5 3}}$$

Therefore,

coefficient of $$x$$ becomes :

$$5+3^{\sqrt{\log_3 5}}-5^{\sqrt{\log_5 3}}=5$$

Hence,

$$\alpha+\beta = 5$$

Now consider constant term :

$$3^{(\log_3 5)^{1/3}} = 5^{(\log_5 3)^{2/3}}$$

Thus,

$$3\left(3^{(\log_3 5)^{1/3}}-5^{(\log_5 3)^{2/3}}-1\right)$$

$$=3(0-1)$$

$$=-3$$

Therefore,

$$\alpha\beta=-3$$

Now required roots are :

$$\alpha+\frac1\beta$$

and

$$\beta+\frac1\alpha$$

Sum of roots :$$S = \alpha+\beta+\frac1\alpha+\frac1\beta$$

$$= (\alpha+\beta)+\frac{\alpha+\beta}{\alpha\beta}$$

$$= 5+\frac{5}{-3}$$

$$= \frac{10}{3}$$

Product of roots :

$$P = \left(\alpha+\frac1\beta\right) \left(\beta+\frac1\alpha\right)$$

$$=\alpha\beta+1+1+\frac1{\alpha\beta}$$

$$=-3+2-\frac13$$

$$= -\frac43$$

Hence required equation :

$$x^2-Sx+P=0$$

$$x^2-\frac{10}{3}x-\frac43=0$$

Multiplying by 3 :

$$3x^2-10x-4=0$$

Final Answer :

$$3x^2-10x-4=0$$

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.

Let $$f(x)$$ be a quadratic polynomial with one root equal to $$-1$$.

Write the general form:

Since $$-1$$ is a root, we can write:

$$f(x) = a(x + 1)(x - r)$$

where $$r$$ is the other root and $$a \neq 0$$.

Apply the condition $$f(-2) + f(3) = 0$$:

$$f(-2) = a(-2 + 1)(-2 - r) = a(-1)(-2 - r) = a(2 + r)$$

$$f(3) = a(3 + 1)(3 - r) = 4a(3 - r)$$

Setting their sum to zero:

$$a(2 + r) + 4a(3 - r) = 0$$

$$a[(2 + r) + 4(3 - r)] = 0$$

Since $$a \neq 0$$:

$$2 + r + 12 - 4r = 0$$

$$14 - 3r = 0$$

$$r = \frac{14}{3}$$

Find the sum of the roots:

$$\text{Sum of roots} = -1 + \frac{14}{3} = \frac{-3 + 14}{3} = \frac{11}{3}$$

Hence, the sum of the roots is $$\dfrac{11}{3}$$.

The correct answer is Option A.

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,

$$ax^2-2bx+15=0$$

has repeated root $$\alpha$$.

Hence, discriminant is zero:

$$(-2b)^2-4(a)(15)=0$$

$$4b^2-60a=0$$

$$b^2=15a$$

Now, for repeated root,

$$\alpha=\frac{2b}{2a}=\frac ba$$

Using

$$b^2=15a$$

we get

$$\alpha=\frac{15}{b}$$

Also, $$\alpha$$ and $$\beta$$ are roots of

$$x^2-2bx+21=0$$

Hence,

$$\alpha+\beta=2b,\qquad \alpha\beta=21$$

Since $$\alpha$$ is a root,

$$\alpha^2-2b\alpha+21=0$$

Substituting

$$\alpha=\frac{15}{b}$$

we get

$$\left(\frac{15}{b}\right)^2-2b\left(\frac{15}{b}\right)+21=0$$

$$\frac{225}{b^2}-30+21=0$$

$$\frac{225}{b^2}=9$$

$$b^2=25$$

Now,

$$\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta$$

$$=(2b)^2-2(21)$$

$$=4(25)-42$$

$$=58$$

Hence, $$\boxed{58}$$.

We are given that $$\alpha, \beta$$ are roots of $$x^2 - \sqrt{2}x + \sqrt{6} = 0$$. By Vieta's formulas, $$\alpha + \beta = \sqrt{2}$$ and $$\alpha\beta = \sqrt{6}$$.

Now $$\frac{1}{\alpha^2+1}$$ and $$\frac{1}{\beta^2+1}$$ are the roots of $$x^2 + ax + b = 0$$. We compute their sum and product.

First, $$\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = 2 - 2\sqrt{6}$$.

The product of the denominators is $$(\alpha^2+1)(\beta^2+1) = (\alpha\beta)^2 + (\alpha^2 + \beta^2) + 1 = 6 + (2 - 2\sqrt{6}) + 1 = 9 - 2\sqrt{6}$$.

So the sum of the new roots is:

And the product is:

We need $$a + b = \frac{2\sqrt{6} - 4 + 1}{9 - 2\sqrt{6}} = \frac{2\sqrt{6} - 3}{9 - 2\sqrt{6}}$$. Rationalizing:

The final equation is $$x^2 - (a+b-2)x + (a+b+2) = 0$$. With $$s = a + b = \frac{4\sqrt{6}-1}{19}$$:

The sum of roots $$= s - 2 = \frac{4\sqrt{6} - 39}{19}$$. Since $$4\sqrt{6} \approx 9.8$$, this is approximately $$\frac{-29.2}{19} < 0$$.

The product of roots $$= s + 2 = \frac{4\sqrt{6} + 37}{19} > 0$$.

When the sum of two roots is negative and their product is positive, both roots must be negative (since two numbers with a positive product have the same sign, and their negative sum forces both to be negative).

Hence, the correct answer is Option 2.

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 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 $$20$$.

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 $$4$$.

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 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 $$\boxed{2}$$.

We need to solve $$\frac{3x^2 - 9x + 17}{x^2 + 3x + 10} = \frac{5x^2 - 7x + 19}{3x^2 + 5x + 12}$$.

First, cross-multiplying yields $$(3x^2 - 9x + 17)(3x^2 + 5x + 12) = (5x^2 - 7x + 19)(x^2 + 3x + 10)$$.

Next, we look for a common structure by introducing $$A = x^2 + 3x + 10$$ and $$B = 3x^2 + 5x + 12$$.

Computing the first numerator minus the first denominator, we find $$(3x^2 - 9x + 17) - (x^2 + 3x + 10) = 2x^2 - 12x + 7$$.

Similarly, computing the second numerator minus the second denominator gives $$(5x^2 - 7x + 19) - (3x^2 + 5x + 12) = 2x^2 - 12x + 7$$.

Since both differences equal the same expression, we let $$k = 2x^2 - 12x + 7$$.

Therefore the first numerator can be written as $$A + k$$ and the second numerator as $$B + k$$.

Substituting these expressions into the equation yields $$\frac{A + k}{A} = \frac{B + k}{B}$$.

From this we obtain $$1 + \frac{k}{A} = 1 + \frac{k}{B}$$, which simplifies to $$\frac{k}{A} = \frac{k}{B}$$ and hence $$k\left(\frac{1}{A} - \frac{1}{B}\right) = 0$$.

This gives us two possibilities: either $$k = 0$$ or $$A = B$$.

If $$k = 0$$ then $$2x^2 - 12x + 7 = 0$$. By the quadratic formula we have $$x = \frac{12 \pm \sqrt{144 - 56}}{4} = \frac{12 \pm \sqrt{88}}{4}$$, and both roots are real. By Vieta's formulas, the sum of the roots is $$\frac{12}{2} = 6$$. We must also verify these roots do not make the denominators zero: since $$A = x^2 + 3x + 10$$ has discriminant $$9 - 40 = -31 < 0$$, it is always positive, and similarly $$B = 3x^2 + 5x + 12$$ has discriminant $$25 - 144 = -119 < 0$$ and is always positive, so neither denominator vanishes. Hence both roots are valid.

On the other hand, if $$A = B$$ then $$x^2 + 3x + 10 = 3x^2 + 5x + 12$$, which simplifies to $$2x^2 + 2x + 2 = 0$$ or $$x^2 + x + 1 = 0$$. Since the discriminant $$1 - 4 = -3 < 0$$, there are no real roots in this case.

Therefore, the only real solutions come from $$k = 0$$, and their sum is $$6$$.

The answer 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.

Let $$f(x) = ax^2 + bx + c$$. We are given $$f(1) = 3$$, $$f(-2) = \lambda$$, $$f(3) = 4$$, and $$f(0) + f(1) + f(-2) + f(3) = 14$$.

From $$f(0) = c$$, the sum condition gives $$c + 3 + \lambda + 4 = 14$$, so $$c + \lambda = 7$$ $$-(1)$$.

From $$f(1) = a + b + c = 3$$ $$-(2)$$.

From $$f(3) = 9a + 3b + c = 4$$ $$-(3)$$.

Subtracting $$(2)$$ from $$(3)$$: $$8a + 2b = 1$$ $$-(4)$$.

From $$f(-2) = 4a - 2b + c = \lambda$$ $$-(5)$$.

From $$(1)$$: $$c = 7 - \lambda$$. Substituting into $$(2)$$: $$a + b = 3 - (7 - \lambda) = \lambda - 4$$ $$-(6)$$.

Substituting $$c = 7 - \lambda$$ into $$(5)$$: $$4a - 2b + 7 - \lambda = \lambda$$, so $$4a - 2b = 2\lambda - 7$$ $$-(7)$$.

From $$(6)$$: $$b = \lambda - 4 - a$$. Substituting into $$(4)$$: $$8a + 2(\lambda - 4 - a) = 1$$, so $$6a + 2\lambda - 8 = 1$$, giving $$a = \dfrac{9 - 2\lambda}{6}$$ $$-(8)$$.

From $$(6)$$: $$b = \lambda - 4 - \dfrac{9 - 2\lambda}{6} = \dfrac{6\lambda - 24 - 9 + 2\lambda}{6} = \dfrac{8\lambda - 33}{6}$$.

Substituting into $$(7)$$: $$4 \cdot \dfrac{9-2\lambda}{6} - 2 \cdot \dfrac{8\lambda-33}{6} = 2\lambda - 7$$.

$$\dfrac{36 - 8\lambda - 16\lambda + 66}{6} = 2\lambda - 7$$

$$\dfrac{102 - 24\lambda}{6} = 2\lambda - 7$$

$$102 - 24\lambda = 12\lambda - 42$$

$$144 = 36\lambda$$, so $$\lambda = 4$$.

The answer is $$4$$.

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 $$\mathbf{18}$$.

Given,

$$x^5(x^3-x^2-x+1)+x(3x^3-4x^2-2x+4)-1=0$$

First factor each bracket.

For

$$x^3-x^2-x+1,$$

group terms:

$$x^2(x-1)-1(x-1)$$

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

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

Now factor

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

Grouping,

$$x^2(3x-4)-2(3x-4)$$

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

Hence equation becomes

$$x^5(x-1)^2(x+1)+x(3x-4)(x^2-2)-1=0$$

Now expand completely:

$$x^8-x^7-x^6+x^5+3x^4-4x^3-2x^2+4x-1=0$$

Now check rational roots.

Substituting

$$x=1,$$

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

Hence,

$$(x-1)$$

is a factor.

Using synthetic division:

$$x^8-x^7-x^6+x^5+3x^4-4x^3-2x^2+4x-1$$

$$=(x-1)(x^7-x^5+3x^3-x^2-3x+1)$$

Now check

$$x=-1$$

in the quotient:

$$-1+1+3-1+3+1=0$$

Hence,

$$(x+1)$$

is also a factor.

Dividing again,

$$=(x-1)(x+1)(x^6-x^5+x^4-x^3+2x^2-3x+1)$$

Now test

$$x=1$$

again:

$$1-1+1-1+2-3+1=0$$

Hence another factor

$$(x-1)$$

appears.

Therefore,

$$=(x-1)^2(x+1)(x^5+x^3+2x-1)$$

Now consider

$$x^5+x^3+2x-1$$

Its derivative is

$$5x^4+3x^2+2>0$$

Hence this function is strictly increasing and has exactly one real root.

Thus real roots are:

- $$x=1$$

- $$x=-1$$

- one real root from $$x^5+x^3+2x-1=0$$

Therefore, number of distinct real roots is

$$\boxed{3}$$.

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 $$\boxed{45}$$.

The equation is $$x^2 + 3^{1/4}x + 3^{1/2} = 0$$. Let $$\omega = 3^{1/4}$$, so the equation becomes $$x^2 + \omega x + \omega^2 = 0$$.

The roots are: $$x = \frac{-\omega \pm \sqrt{\omega^2 - 4\omega^2}}{2} = \frac{-\omega \pm \sqrt{-3\omega^2}}{2} = \frac{-\omega \pm i\omega\sqrt{3}}{2} = \omega \cdot \frac{-1 \pm i\sqrt{3}}{2}$$

Recognizing that $$\frac{-1 \pm i\sqrt{3}}{2}$$ are the complex cube roots of unity, we have $$\alpha = \omega \cdot \zeta$$ and $$\beta = \omega \cdot \zeta^2$$, where $$\zeta = e^{2\pi i/3}$$ satisfies $$\zeta^3 = 1$$ and $$1 + \zeta + \zeta^2 = 0$$.

Now compute $$\alpha^{96}(\alpha^{12} - 1) + \beta^{96}(\beta^{12} - 1)$$.

First, $$\alpha^{12} = \omega^{12} \cdot \zeta^{12} = (3^{1/4})^{12} \cdot (\zeta^3)^4 = 3^3 \cdot 1 = 27$$. Similarly, $$\beta^{12} = \omega^{12} \cdot \zeta^{24} = 27 \cdot (\zeta^3)^8 = 27$$.

So $$\alpha^{12} - 1 = 26$$ and $$\beta^{12} - 1 = 26$$.

Next, $$\alpha^{96} = \omega^{96} \cdot \zeta^{96} = (3^{1/4})^{96} \cdot (\zeta^3)^{32} = 3^{24} \cdot 1 = 3^{24}$$. Similarly, $$\beta^{96} = 3^{24}$$.

Therefore: $$\alpha^{96}(\alpha^{12}-1) + \beta^{96}(\beta^{12}-1) = 3^{24} \cdot 26 + 3^{24} \cdot 26 = 2 \times 26 \times 3^{24} = 52 \times 3^{24}$$

This corresponds to Option 3: $$52 \times 3^{24}$$.

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.

Given that $$\alpha$$ and $$\beta$$ are the roots of $$x^2 - 6x - 2 = 0$$, by Vieta's formulas we have $$\alpha + \beta = 6$$ and $$\alpha\beta = -2$$. Also, since both $$\alpha$$ and $$\beta$$ satisfy the equation, we have $$\alpha^2 = 6\alpha + 2$$ and $$\beta^2 = 6\beta + 2$$.

We need to find $$\frac{a_{10} - 2a_8}{3a_9}$$ where $$a_n = \alpha^n - \beta^n$$.

Consider $$a_{10} - 2a_8 = (\alpha^{10} - \beta^{10}) - 2(\alpha^8 - \beta^8) = \alpha^8(\alpha^2 - 2) - \beta^8(\beta^2 - 2)$$.

Using $$\alpha^2 = 6\alpha + 2$$, we get $$\alpha^2 - 2 = 6\alpha$$. Similarly, $$\beta^2 - 2 = 6\beta$$.

Substituting: $$a_{10} - 2a_8 = \alpha^8(6\alpha) - \beta^8(6\beta) = 6(\alpha^9 - \beta^9) = 6a_9$$.

Therefore, $$\frac{a_{10} - 2a_8}{3a_9} = \frac{6a_9}{3a_9} = 2$$.

The answer is 2, which corresponds to option (3).

We start with the quadratic equation $$x^{2} + (20)^{1/4}\,x + (5)^{1/2}=0$$ whose two roots are denoted by $$\alpha$$ and $$\beta$$.

For any quadratic $$x^{2}+px+q=0$$ with roots $$\alpha,\,\beta$$ we always have the Vieta relations

$$\alpha+\beta=-p,\qquad \alpha\beta=q.$$

Matching the given equation with the standard form we identify

$$p=(20)^{1/4},\qquad q=(5)^{1/2}.$$

Hence

$$\alpha+\beta=-\,(20)^{1/4},\qquad \alpha\beta=(5)^{1/2}.$$

We wish to evaluate $$\alpha^{8}+\beta^{8}.$$ To avoid expanding high powers directly, we use the fact that every power of a root can be reduced with the help of the quadratic itself. Because each root satisfies $$x^{2}+px+q=0,$$ we can write

$$x^{2}=-px-q.$$

If we set $$R_{k}=\alpha^{k}+\beta^{k}$$ then, multiplying the above identity by $$x^{k-2}$$ and adding for $$x=\alpha$$ and $$x=\beta,$$ we obtain the recurrence relation

$$R_{k}=-p\,R_{k-1}-q\,R_{k-2}\qquad(k\ge 2).$$

First we list the initial values:

$$R_{0}=\alpha^{0}+\beta^{0}=1+1=2,$$

$$R_{1}=\alpha+\beta=-p=-(20)^{1/4}.$$

For convenience let us denote $$r=(20)^{1/4}$$ so that $$p=r$$ and $$q=\sqrt5.$$ It will be useful to note

$$r^{2}=(20)^{1/2}=\sqrt{20}=2\sqrt5.$$

Now we compute successive $$R_k$$ using the recurrence.

Second power

$$R_{2}=-p\,R_{1}-q\,R_{0}=-(r)(-r)-(\sqrt5)(2)=r^{2}-2\sqrt5.$$

Substituting $$r^{2}=2\sqrt5$$ gives

$$R_{2}=2\sqrt5-2\sqrt5=0.$$

Third power

$$R_{3}=-p\,R_{2}-q\,R_{1}=-(r)(0)-(\sqrt5)(-r)=\sqrt5\,r.$$

Fourth power

$$R_{4}=-p\,R_{3}-q\,R_{2}=-(r)(\sqrt5\,r)-(\sqrt5)(0)=-\sqrt5\,r^{2}.$$

Because $$r^{2}=2\sqrt5,$$ we obtain

$$R_{4}=-\sqrt5\,(2\sqrt5)=-2\cdot5=-10.$$

Fifth power

$$R_{5}=-p\,R_{4}-q\,R_{3}=-(r)(-10)-(\sqrt5)(\sqrt5\,r)=10r-5r=5r.$$

Sixth power

$$R_{6}=-p\,R_{5}-q\,R_{4}=-(r)(5r)-(\sqrt5)(-10)=-5r^{2}+10\sqrt5.$$

Again inserting $$r^{2}=2\sqrt5$$ yields

$$R_{6}=-5(2\sqrt5)+10\sqrt5=-10\sqrt5+10\sqrt5=0.$$

Seventh power

$$R_{7}=-p\,R_{6}-q\,R_{5}=-(r)(0)-(\sqrt5)(5r)=-5\sqrt5\,r.$$

Eighth power (desired term)

$$R_{8}=-p\,R_{7}-q\,R_{6}=-(r)(-5\sqrt5\,r)-(\sqrt5)(0)=5\sqrt5\,r^{2}.$$

Using $$r^{2}=2\sqrt5$$ one final time gives

$$R_{8}=5\sqrt5\,(2\sqrt5)=5\cdot2\cdot5=50.$$

Therefore we have found

$$\alpha^{8}+\beta^{8}=50.$$

Hence, the correct answer is Option C.

First, let us rewrite the given expression in a single base so that the maximisation and minimisation become easier.

We have $$8^{2\sin 3x}\cdot 4^{4\cos 3x}.$$

Recall that $$8 = 2^3 \quad\text{and}\quad 4 = 2^2.$$ Substituting these powers of two, we get

$$\bigl(2^3\bigr)^{2\sin 3x}\cdot\bigl(2^2\bigr)^{4\cos 3x} = 2^{3\cdot 2\sin 3x}\cdot 2^{2\cdot 4\cos 3x} = 2^{6\sin 3x}\cdot 2^{8\cos 3x}.$$

Whenever we multiply two numbers with the same base, we add the exponents, so

$$2^{6\sin 3x}\cdot 2^{8\cos 3x}=2^{\,6\sin 3x+8\cos 3x}.$$ Hence, maximising or minimising the original expression is the same as maximising or minimising the exponent

$$f(t)=6\sin t+8\cos t,$$

where we have set $$t = 3x,$$ because $$\sin 3x$$ and $$\cos 3x$$ both depend only on $$3x.$$

Now, for any expression of the form $$A\sin t + B\cos t,$$ its maximum value is $$\sqrt{A^{2}+B^{2}}$$ and its minimum value is $$-\sqrt{A^{2}+B^{2}}.$$ This comes directly from interpreting the linear combination as a shifted sine wave with amplitude $$\sqrt{A^{2}+B^{2}}.$$ Stating the result clearly,

$$\max(A\sin t+B\cos t)=\sqrt{A^2+B^2},\qquad \min(A\sin t+B\cos t)=-\sqrt{A^2+B^2}.$$

Here, $$A=6$$ and $$B=8,$$ so

$$\sqrt{A^{2}+B^{2}}=\sqrt{6^{2}+8^{2}} =\sqrt{36+64} =\sqrt{100}=10.$$

Therefore

$$\max_{t\in\mathbb R} f(t)=10,\qquad \min_{t\in\mathbb R} f(t)=-10.$$

Remembering that the original quantity is $$2^{f(t)},$$ we get

$$\alpha = 2^{10}=1024,\qquad \beta = 2^{-10}=\dfrac1{1024}.$$

The problem says that $$\alpha^{1/5}$$ and $$\beta^{1/5}$$ are the roots of the quadratic $$8x^{2}+bx+c=0.$$ Let us denote the two roots by

$$r_{1}=\alpha^{1/5},\qquad r_{2}=\beta^{1/5}.$$

We now compute these roots explicitly. Since $$\alpha=1024=2^{10},$$

$$r_{1}=\bigl(2^{10}\bigr)^{1/5}=2^{10/5}=2^{2}=4.$$

Similarly, $$\beta=2^{-10},$$ so

$$r_{2}=\bigl(2^{-10}\bigr)^{1/5}=2^{-10/5}=2^{-2}=\dfrac14.$$

Next, recall the standard relations between the coefficients of a quadratic and its roots. For a quadratic $$ax^{2}+bx+c=0$$ with roots $$r_{1},\,r_{2},$$ we have

$$r_{1}+r_{2}=-\dfrac{b}{a},\qquad r_{1}r_{2}=\dfrac{c}{a}.$$

In our particular equation, $$a=8,$$ so

$$r_{1}+r_{2}=-\dfrac{b}{8},\qquad r_{1}r_{2}=\dfrac{c}{8}.$$

Substituting the numerical values we have found:

Sum of roots: $$r_{1}+r_{2}=4+\dfrac14=\dfrac{16}{4}+\dfrac14=\dfrac{17}{4}.$$

Thus $$-\dfrac{b}{8}=\dfrac{17}{4}\quad\Longrightarrow\quad b=-8\cdot\dfrac{17}{4}=-34.$$

Product of roots: $$r_{1}r_{2}=4\cdot\dfrac14=1.$$

Thus $$\dfrac{c}{8}=1\quad\Longrightarrow\quad c=8.$$

The question finally asks for $$c-b.$$ We compute

$$c-b = 8 - (-34)=8+34=42.$$

Hence, the correct answer is Option A.

We are given $$p + q = 2$$ and $$p^4 + q^4 = 272$$, and we need to find the quadratic equation whose roots are $$p$$ and $$q$$.

We know that $$p^2 + q^2 = (p + q)^2 - 2pq = 4 - 2pq$$.

Now, $$p^4 + q^4 = (p^2 + q^2)^2 - 2p^2q^2$$. Substituting, we get $$(4 - 2pq)^2 - 2(pq)^2 = 272$$.

Let $$s = pq$$. Then $$16 - 16s + 4s^2 - 2s^2 = 272$$, which gives $$2s^2 - 16s + 16 = 272$$.

So $$2s^2 - 16s - 256 = 0$$, or $$s^2 - 8s - 128 = 0$$.

Using the quadratic formula, $$s = \frac{8 \pm \sqrt{64 + 512}}{2} = \frac{8 \pm \sqrt{576}}{2} = \frac{8 \pm 24}{2}$$.

This gives $$s = 16$$ or $$s = -8$$.

The quadratic equation with sum of roots $$p + q = 2$$ and product $$pq = 16$$ is $$x^2 - 2x + 16 = 0$$, and with $$pq = -8$$ it is $$x^2 - 2x - 8 = 0$$. Among the given options, only $$x^2 - 2x + 16 = 0$$ is present.

Hence, the correct answer is Option D.

For the quadratic expression $$x^2 - 2(3k - 1)x + 8k^2 - 7$$ to be strictly positive for every real $$x$$, the coefficient of $$x^2$$ must be positive (which it is, since it equals 1) and the discriminant must be strictly negative.

The discriminant is: $$D = [2(3k-1)]^2 - 4(1)(8k^2 - 7) = 4(3k-1)^2 - 4(8k^2 - 7)$$

Dividing by 4: $$\frac{D}{4} = (3k-1)^2 - (8k^2 - 7) = 9k^2 - 6k + 1 - 8k^2 + 7 = k^2 - 6k + 8$$

We need $$k^2 - 6k + 8 < 0$$. Factoring: $$(k - 2)(k - 4) < 0$$

This inequality holds when $$2 < k < 4$$. The only integer value of $$k$$ in this open interval is $$k = 3$$.

Therefore, the correct answer is Option 3: $$k = 3$$.

Given quadratic equation,

$$x^2+ax+b=0$$

Let its roots be

$$x_1,x_2$$

Condition given:

if $$\alpha$$ is a root, then

$$\alpha^2-2$$

is also a root of the same equation.

Thus the set of roots remains invariant under the mapping

$$f(x)=x^2-2$$

Now find all possible cases.

Case 1: Fixed points

Suppose

$$\alpha^2-2=\alpha$$

Then,

$$\alpha^2-\alpha-2=0$$

$$(\alpha-2)(\alpha+1)=0$$

Hence,

$$\alpha=2\quad \text{or}\quad \alpha=-1$$

Subcase 1(a): Roots $$\{-1,-1\}$$

Equation becomes

$$x^2+2x+1=0$$

giving

$$(a,b)=(2,1)$$

Subcase 1(b): Roots $$\{2,2\}$$

Equation becomes

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

giving

$$(a,b)=(-4,4)$$

Subcase 1(c): Roots $$\{2,-1\}$$

Equation becomes

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

giving

$$(a,b)=(-1,-2)$$

Case 2: Two-cycle

Suppose

$$x_1^2-2=x_2$$

and

$$x_2^2-2=x_1$$

Then,

$$f(f(x))=x$$

So,

$$((x^2-2)^2-2)=x$$

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

Factorizing,

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

The roots $$2$$ and $$-1$$ are already counted.

Hence remaining roots satisfy

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

Thus equation is

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

giving

$$(a,b)=(1,-1)$$

Case 3: One root maps to a fixed point

Subcase 3(a):

Suppose one root maps to $$2.$$

Then,

$$x^2-2=2$$

$$x^2=4$$

$$x=\pm2$$

The pair $$\{2,2\}$$ is already counted, so new roots are

$$\{2,-2\}$$

Equation becomes

$$x^2-4=0$$

giving

$$(a,b)=(0,-4)$$

Subcase 3(b):

Suppose one root maps to $$-1.$$

Then,

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

$$x^2=1$$

$$x=\pm1$$

The pair $$\{-1,-1\}$$ is already counted, so new roots are

$$\{-1,1\}$$

Equation becomes

$$x^2-1=0$$

giving

$$(a,b)=(0,-1)$$

Thus all possible pairs are:

$$(2,1),\ (-4,4),\ (-1,-2),\ (1,-1),\ (0,-4),\ (0,-1)$$

Therefore, total number of pairs $$(a,b)$$ is

$$\boxed{6}$$

We have to solve the equation $$x^{2}-|x|-12=0$$ and count how many real values of $$x$$ satisfy it.

The expression contains the absolute value $$|x|$$, so we split the work into two cases depending on whether $$x$$ is non-negative or non-positive. Recall the definition $$|x|=\begin{cases}x,&x\ge 0\\-x,&x\le 0\end{cases}.$$

Case 1 : $$x\ge 0$$. Here $$|x|=x$$, so substituting this into the equation gives

$$x^{2}-x-12=0.$$

This is a quadratic equation of the standard form $$ax^{2}+bx+c=0$$ with $$a=1,\;b=-1,\;c=-12$$. The quadratic formula states $$x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}.$$ Applying it,

$$x=\dfrac{-(-1)\pm\sqrt{(-1)^{2}-4(1)(-12)}}{2(1)} =\dfrac{1\pm\sqrt{1+48}}{2} =\dfrac{1\pm 7}{2}.$$

So the two roots are

$$x=\dfrac{1+7}{2}=4\quad\text{and}\quad x=\dfrac{1-7}{2}=-3.$$

Because we are inside the case $$x\ge 0$$, only $$x=4$$ is admissible; $$x=-3$$ must be discarded.

Case 2 : $$x\le 0$$. Now $$|x|=-x$$, and the equation becomes

$$x^{2}+x-12=0.$$

Again using the quadratic formula with $$a=1,\;b=1,\;c=-12$$, we obtain

$$x=\dfrac{-1\pm\sqrt{1^{2}-4(1)(-12)}}{2(1)} =\dfrac{-1\pm\sqrt{1+48}}{2} =\dfrac{-1\pm 7}{2}.$$

Thus the roots are

$$x=\dfrac{-1+7}{2}=3\quad\text{and}\quad x=\dfrac{-1-7}{2}=-4.$$

But we must respect the condition $$x\le 0$$ for this case, so only $$x=-4$$ is acceptable; $$x=3$$ is rejected.

Gathering results from both cases, the values that satisfy the original equation are

$$x=4\quad\text{and}\quad x=-4.$$

There are exactly two real solutions.

Hence, the correct answer is Option A.

We start with the given equation

$$ (3x^2+4x+3)^2-(k+1)(3x^2+4x+3)(3x^2+4x+2)+k(3x^2+4x+2)^2=0,\qquad k\gt -1. $$

Notice that the two quadratic expressions which appear repeatedly differ by exactly $$1$$:

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

So we introduce the substitution

$$ A=3x^2+4x+2, \qquad \text{hence } 3x^2+4x+3=A+1. $$

Re-writing every occurrence in terms of $$A$$ gives

$$ (A+1)^2-(k+1)(A+1)A+kA^2=0. $$

Now we expand every product:

$$ (A+1)^2=A^2+2A+1, $$

$$ (A+1)A=A^2+A. $$

Substituting these expansions, the whole equation becomes

$$ (A^2+2A+1)-(k+1)(A^2+A)+kA^2=0. $$

We now collect like terms. First the $$A^2$$ terms:

$$ A^2 - (k+1)A^2 + kA^2 \;=\; \bigl[1-(k+1)+k\bigr]A^2 \;=\; (1-k-1+k)A^2 \;=\; 0\cdot A^2 \;=\;0. $$

Thus the entire $$A^2$$ part cancels out completely.

Next the $$A$$ terms:

$$ 2A-(k+1)A \;=\; (2-k-1)A \;=\; (1-k)A. $$

Finally, the constant term is simply $$+1$$.

So the original rather complicated equation has reduced to the linear equation in $$A$$

$$ (1-k)A+1=0. $$

If $$k\neq1$$, we can solve for $$A$$ straight away:

$$ A=\frac{-1}{1-k}=\frac{1}{k-1}. $$

However, if $$k=1$$ the coefficient in front of $$A$$ becomes zero and we are left with the contradictory statement $$1=0$$. Therefore

$$ k=1 \quad\text{gives no root at all and must be excluded}. $$

For every other value of $$k$$ (still keeping $$k\gt -1$$ in mind) the problem now boils down to asking when the quadratic in $$x$$

$$ 3x^2+4x+2=\frac{1}{k-1} $$

has real solutions. Rearranging this gives a standard quadratic equation:

$$ 3x^2+4x+\Bigl(2-\frac{1}{k-1}\Bigr)=0. $$

For such a quadratic $$ax^2+bx+c=0$$ to have real roots, its discriminant must satisfy $$\Delta=b^2-4ac\ge 0.$$ Here

$$ a=3,\qquad b=4,\qquad c=2-\frac{1}{k-1}. $$

Therefore

$$ \Delta =4^2-4\cdot3\Bigl(2-\frac{1}{k-1}\Bigr) =16-12\Bigl(2-\frac{1}{k-1}\Bigr). $$

Expanding the right-hand side, we get

$$ \Delta =16-24+12\cdot\frac{1}{k-1} =-8+\frac{12}{k-1}. $$

Requiring $$\Delta\ge0$$ gives

$$ -8+\frac{12}{k-1}\ge0 \;\;\Longrightarrow\;\; \frac{12}{k-1}\ge8 \;\;\Longrightarrow\;\; \frac{1}{k-1}\ge\frac{2}{3}. $$

Now we solve the inequality $$\dfrac{1}{k-1}\ge\dfrac{2}{3}$$ carefully, taking into account the sign of the denominator.

Case I: $$k-1\gt 0$$ (that is, $$k\gt 1$$).
Multiplying by the positive quantity $$(k-1)$$ preserves the inequality:

$$ 1\;\ge\;\frac{2}{3}(k-1) \;\;\Longrightarrow\;\; k-1\;\le\;\frac{3}{2} \;\;\Longrightarrow\;\; k\;\le\;\frac{5}{2}. $$

Together with $$k\gt 1$$, this yields the interval

$$ 1\lt k\le\frac{5}{2}. $$

Case II: $$k-1\lt 0$$ (that is, $$k\lt 1$$).
Here the denominator is negative, so multiplying reverses the inequality:

$$ 1\;\le\;\frac{2}{3}(k-1). $$

The right side is negative because $$(k-1)\lt 0$$, while the left side is positive, which is impossible. Thus no solutions arise from this case.

Combining everything, remembering also that $$k=1$$ is excluded and the original condition was $$k\gt -1$$, we obtain the final set of all permissible values of $$k$$:

$$ \,1\lt k\le\frac{5}{2}\,. $$

Looking at the options, this interval is exactly Option B.

Hence, the correct answer is Option B.

The given equation is:

$$x + 1 - 2 \log_2(3 + 2^x) + 2 \log_4(10 - 2^{-x}) = 0$$

First, we simplify the term $$2 \log_4(10 - 2^{-x})$$ using the base change property $$\log_{a^k} b = \frac{1}{k} \log_a b$$:

$$2 \log_{2^2}(10 - 2^{-x}) = 2 \cdot \frac{1}{2} \log_2(10 - 2^{-x}) = \log_2(10 - 2^{-x})$$

Now, substitute this back into the equation:

$$x + 1 - 2 \log_2(3 + 2^x) + \log_2(10 - 2^{-x}) = 0$$

Move the logarithmic terms to one side and the linear terms to the other:

$$x + 1 = 2 \log_2(3 + 2^x) - \log_2(10 - 2^{-x})$$

$$x + 1 = \log_2((3 + 2^x)^2) - \log_2(10 - 2^{-x})$$

Using the quotient rule $$\log_a M - \log_a N = \log_a \frac{M}{N}$$:

$$x + 1 = \log_2 \left( \frac{(3 + 2^x)^2}{10 - 2^{-x}} \right)$$

Convert from logarithmic form to exponential form ($$2^{x+1}$$):

$$2^{x+1} = \frac{(3 + 2^x)^2}{10 - 2^{-x}}$$ 

$$2 \cdot 2^x = \frac{(3 + 2^x)^2}{10 - \frac{1}{2^x}}$$

Let $$2^x = t$$. The equation becomes:

$$2t = \frac{(3 + t)^2}{10 - \frac{1}{t}}$$ $$2t = \frac{(3 + t)^2}{\frac{10t - 1}{t}}$$

$$2t = \frac{t(3 + t)^2}{10t - 1}$$

Since $$t = 2^x$$, $$t$$ cannot be $$0$$. We can divide both sides by $$t$$:

$$2 = \frac{(3 + t)^2}{10t - 1}$$

$$2(10t - 1) = (3 + t)^2$$

$$20t - 2 = 9 + 6t + t^2$$

$$t^2 - 14t + 11 = 0$$

Let the roots of the quadratic equation in $$t$$ be $$t_1$$ and $$t_2$$. These correspond to $$2^{x_1}$$ and $$2^{x_2}$$ where $$x_1, x_2$$ are the roots of the original equation.

The product of the roots $$t_1 t_2$$ is:

$$t_1 t_2 = \frac{c}{a} = 11$$

Since $$t_1 = 2^{x_1}$$ and $$t_2 = 2^{x_2}$$:

$$2^{x_1} \cdot 2^{x_2} = 11$$

$$2^{x_1 + x_2} = 11$$

Taking $$\log_2$$ on both sides:

$$x_1 + x_2 = \log_2 11$$

We have to examine two logical sentences.

First consider $$S_1 : (p \rightarrow q)\;\vee\;(\sim q \rightarrow p).$$

Before simplifying, we recall the basic implication equivalence:
For any propositions $$a$$ and $$b,$$ the implication $$a \rightarrow b$$ is equivalent to $$\sim a \;\vee\; b.$$

Applying this rule to each implication inside $$S_1,$$ we get

$$ (p \rightarrow q)=\; \sim p \;\vee\; q,\\ (\sim q \rightarrow p)=\; \sim (\sim q) \;\vee\; p=\; q \;\vee\; p. $$

So $$S_1$$ becomes

$$ (\sim p \;\vee\; q)\;\vee\;(q \;\vee\; p). $$

Using the associative and commutative laws of $$\vee,$$ we may regroup and list all literals together:

$$ (\sim p \;\vee\; p)\;\vee\;(q \;\vee\; q). $$

Now, by the law of excluded middle, $$\sim p \;\vee\; p = \text{T}$$ (always true).
Also, $$q \;\vee\; q = q.$$

Hence

$$ (\sim p \;\vee\; p)\;\vee\;(q \;\vee\; q)=\text{T}\;\vee\; q=\text{T}. $$

Because the disjunction reduces to the constant true value $$\text{T},$$ the formula is true for every possible truth assignment of $$p$$ and $$q.$$ Therefore $$S_1$$ is indeed a tautology.

Now we examine $$S_2 : (p \wedge \sim q)\;\wedge\;(\sim p \vee q).$$

For this conjunction to be true, both parts must be true simultaneously.

Start with the left conjunct $$p \wedge \sim q.$$ For it to be true we must have

$$p = \text{T} \quad\text{and}\quad q = \text{F}.$$

With these same truth values, evaluate the right conjunct $$\sim p \vee q.$$ Since $$p=\text{T},$$ we have $$\sim p=\text{F};$$ and with $$q=\text{F},$$ we obtain

$$\sim p \vee q = \text{F} \vee \text{F} = \text{F}.$$

Thus, whenever the first conjunct is true, the second conjunct is forced to be false, making the whole conjunction false. No other truth assignment can make the first conjunct true, so there is no assignment that makes the entire expression true.

Therefore the statement $$(p \wedge \sim q)\;\wedge\;(\sim p \vee q)$$ is always false; it is a fallacy. Hence the meta-statement $$S_2$$ saying “this formula is a fallacy” is true.

We have established that $$S_1$$ is a tautology (true) and $$S_2$$ correctly asserts a fallacy (so $$S_2$$ is also true). Therefore, both statements are true.

Hence, the correct answer is Option D.

We begin with the compound implication $$\big((p \vee q) \wedge (q \rightarrow r) \wedge (\sim r)\big) \rightarrow (p \wedge q).$$

The standard truth table rule is: an implication $$A \rightarrow B$$ is false only when $$A$$ is true and $$B$$ is false. In every other case it is true.

Because the whole expression is stated to be false, we must therefore have

$$\big((p \vee q) \wedge (q \rightarrow r) \wedge (\sim r)\big)=\text{True}$$

and simultaneously

$$ (p \wedge q)=\text{False}. $$

Let us analyse each part in turn.

First, the consequent $$p \wedge q$$ is false. A conjunction is false whenever at least one of its components is false, so at least one of $$p$$ or $$q$$ must be false.

Next, the antecedent is a conjunction of three statements, and we are told the conjunction is true. Therefore each of its three factors must be true:

1. $$(p \vee q)=\text{True}.$$

2. $$(q \rightarrow r)=\text{True}.$$

3. $$(\sim r)=\text{True}.$$

From the third condition $$\sim r=\text{True}$$ we obtain immediately

$$ r=\text{False}. $$

Now consider the second condition $$q \rightarrow r=\text{True}.$$ Recall the logical equivalence $$q \rightarrow r \equiv (\sim q) \vee r.$$

Substituting the already known value $$r=\text{False},$$ we get

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

The only way this disjunction can be true is if $$\sim q=\text{True},$$ hence

$$ q=\text{False}. $$

Finally, look at the first factor $$p \vee q=\text{True}.$$ We have just found $$q=\text{False},$$ so for the disjunction to be true we must have

$$ p=\text{True}. $$

Collecting our results, we have

$$ p=\text{True}, \quad q=\text{False}, \quad r=\text{False}. $$

Written in the order $$p,\,q,\,r$$ this is $$T F F,$$ which matches Option B.

Hence, the correct answer is Option 2.

We start with the given statement:

$$\text{(I)}\;:\;\forall\,M \gt 0,\;\exists\,x\in S\text{ such that }x\ge M.$$

The words “for all” correspond to the universal quantifier $$\forall$$ and the words “there exists” correspond to the existential quantifier $$\exists$$. To find the negation of a quantified statement, we use the standard logical rules:

1. The negation of $$\forall$$ (for all) is $$\exists$$ (there exists).
2. The negation of $$\exists$$ (there exists) is $$\forall$$ (for all).
3. Inside, we must also negate the inner predicate.

Applying rule 1 to statement (I) we replace the leading $$\forall$$ by $$\exists$$. So we write:

$$\neg\text{(I)}:\;\exists\,M \gt 0\ \bigl[\neg\,(\exists\,x\in S\text{ such that }x\ge M)\bigr].$$

Now we focus on the inner part $$\exists\,x\in S\text{ such that }x\ge M$$ and apply rule 2: its negation changes $$\exists$$ to $$\forall$$ and, according to rule 3, the condition $$x\ge M$$ switches to its opposite $$x\lt M$$. Thus,

$$\neg\bigl(\exists\,x\in S\text{ such that }x\ge M\bigr)\;=\;\forall\,x\in S,\;x\lt M.$$

Substituting this negation back, we get

$$\neg\text{(I)}:\;\exists\,M \gt 0\text{ such that }\forall\,x\in S,\;x\lt M.$$

Reading this sentence in plain words: “There exists a positive number $$M$$ such that every element $$x$$ of the set $$S$$ is strictly less than $$M$$.”

Now we compare this with the options given:

A. There exists $$M \gt 0$$ such that $$x \lt M$$ for all $$x \in S$$.
B. There exists $$M \gt 0$$, there exists $$x \in S$$ such that $$x \ge M$$.
C. There exists $$M \gt 0$$, there exists $$x \in S$$ such that $$x \lt M$$.
D. There exists $$M \gt 0$$ such that $$x \ge M$$ for all $$x \in S$$.

Option A reproduces exactly the negated statement we have derived. The other options either keep both quantifiers as existential (B, C) or invert the inequality without changing both quantifiers appropriately (D). Therefore Option A is the only correct negation.

Hence, the correct answer is Option A.

Let us translate every part of the given English sentence into symbolic logic so that each subsequent algebraic-style step is absolutely clear.

First, we introduce three simple statements:

$$P : \text{"The match will be played"}$$

$$Q : \text{"The weather is good"}$$

$$R : \text{"The ground is not wet"}$$

The original statement reads: “The match will be played only if the weather is good and the ground is not wet.” In propositional logic, the phrase “only if” is expressed by the implication arrow $$\rightarrow$$, with the condition coming after the arrow. Hence we write

$$P \;\rightarrow\; (Q \land R).$$

Our task is to find the negation of this entire implication. We start by recalling the standard logical equivalence for the negation of an implication:

Formula to be used: $$\neg(A \rightarrow B) \equiv A \land \neg B.$$

Here $$A$$ is $$P$$ and $$B$$ is $$(Q \land R)$$. Applying the formula gives

$$\neg\bigl(P \rightarrow (Q \land R)\bigr) \;=\; P \land \neg(Q \land R).$$

Now we still have the negation of a conjunction inside. We therefore invoke De Morgan’s law, which states

Formula to be used: $$\neg(X \land Y) \equiv \neg X \;\lor\; \neg Y.$$

Taking $$X = Q$$ and $$Y = R$$, De Morgan’s law yields

$$\neg(Q \land R) \;=\; \neg Q \;\lor\; \neg R.$$

Substituting this back into our earlier result, we obtain

$$\neg\bigl(P \rightarrow (Q \land R)\bigr) \;=\; P \land (\neg Q \lor \neg R).$$

Let us now translate this final symbolic form back into ordinary language:

• $$P$$ is true  ⇒  “The match will be played.”
• $$\neg Q$$ is true  ⇒  “The weather is not good.”
• $$\neg R$$ is true  ⇒  “The ground is wet.”

So the entire expression $$P \land (\neg Q \lor \neg R)$$ becomes

“The match will be played and (the weather is not good or the ground is wet).”

Scanning the given options, we see that Option C states exactly this sentence:

“The match will be played and weather is not good or ground is wet.”

Hence, the correct answer is Option C.

We have the compound statement

$$\bigl((P \vee Q) \wedge (\sim P)\bigr)\;\Rightarrow\;Q.$$

First we recall the basic logical equivalence for an implication:

$$A \Rightarrow B \;\equiv\; \sim A \;\vee\; B.$$

Here,

$$A = (P \vee Q)\wedge(\sim P) \quad\text{and}\quad B = Q.$$

Applying the rule we obtain

$$\bigl((P \vee Q)\wedge(\sim P)\bigr)\Rightarrow Q \;\equiv\; \sim\bigl((P \vee Q)\wedge(\sim P)\bigr)\;\vee\;Q.$$

Now we remove the negation inside by using De Morgan’s law:

$$\sim(X\wedge Y)\;\equiv\; \sim X\;\vee\;\sim Y.$$

So, with

$$X = (P \vee Q)\quad\text{and}\quad Y = (\sim P),$$

we have

$$\sim\bigl((P \vee Q)\wedge(\sim P)\bigr) = \bigl(\sim(P \vee Q)\bigr)\;\vee\;\bigl(\sim(\sim P)\bigr) = (\sim P\wedge\sim Q)\;\vee\;P.$$

The expression therefore becomes

$$\bigl((\sim P\wedge\sim Q)\;\vee\;P\bigr)\;\vee\;Q.$$

The associative and commutative laws of $$\vee$$ allow us to regroup and reorder the disjunction:

$$ \bigl((\sim P\wedge\sim Q)\;\vee\;P\bigr)\;\vee\;Q \;=\; P\;\vee\;Q\;\vee\;(\sim P\wedge\sim Q). $$

Next we distribute $$\vee$$ over $$\wedge$$ to simplify the mixed term:

$$ P\;\vee\;Q\;\vee\;(\sim P\wedge\sim Q) = (P\;\vee\;Q)\;\vee\;(\sim P\wedge\sim Q) = \bigl(P\;\vee\;Q\;\vee\;\sim P\bigr)\;\wedge\;\bigl(P\;\vee\;Q\;\vee\;\sim Q\bigr). $$

Each of the two factors contains a full pair of complementary literals:

$$P\;\vee\;\sim P = \text{T},\quad Q\;\vee\;\sim Q = \text{T}.$$

Hence both brackets reduce to T (truth), and we are left with

$$\text{T}\;\wedge\;\text{T} = \text{T}.$$

Thus the given compound statement is a tautology; it is always true for every possible truth-value assignment to $$P$$ and $$Q$$.

Among the provided options, Option D is

$$\sim(P \Rightarrow Q)\;\Leftrightarrow\;P\wedge\sim Q,$$

which is itself a biconditional connecting two logically equivalent expressions (since $$\sim(P \Rightarrow Q)\equiv P\wedge\sim Q$$). A biconditional between two identical propositions is also a tautology, i.e. it is always true. Therefore the original statement and Option D are equivalent.

Hence, the correct answer is Option 4.

We solve $$(|x| - 3)|x + 4| = 6$$ by considering cases based on the sign of $$x$$ and $$x + 4$$.

Case 1: $$x \geq 0$$. Then $$|x| = x$$ and $$|x + 4| = x + 4$$, so $$(x - 3)(x + 4) = 6$$. Expanding: $$x^2 + x - 12 = 6$$, i.e., $$x^2 + x - 18 = 0$$. Using the quadratic formula: $$x = \frac{-1 \pm \sqrt{1 + 72}}{2} = \frac{-1 \pm \sqrt{73}}{2}$$. Since $$x \geq 0$$, only $$x = \frac{-1 + \sqrt{73}}{2} \approx 3.77$$ is valid. This gives one solution.

Case 2: $$-4 \leq x < 0$$. Then $$|x| = -x$$ and $$|x + 4| = x + 4$$, so $$(-x - 3)(x + 4) = 6$$. This gives $$-(x + 3)(x + 4) = 6$$, or $$x^2 + 7x + 12 = -6$$, i.e., $$x^2 + 7x + 18 = 0$$. The discriminant is $$49 - 72 = -23 < 0$$, so there are no real solutions in this case.

Case 3: $$x < -4$$. Then $$|x| = -x$$ and $$|x + 4| = -(x + 4)$$, so $$(-x - 3)(-(x + 4)) = 6$$, which simplifies to $$(x + 3)(x + 4) = 6$$. Expanding: $$x^2 + 7x + 12 = 6$$, i.e., $$x^2 + 7x + 6 = 0$$, giving $$(x + 1)(x + 6) = 0$$. Since $$x < -4$$, only $$x = -6$$ is valid. This gives one solution.

In total, there are 2 elements in the set.

We first notice that the unknown appears only through the exponential of $$x$$. A very standard trick in such cases is to declare a new variable equal to the common base of all exponentials.

Let us write  $$y=e^{x}\,.$$

Because the exponential function is always positive, we immediately have

$$y=e^{x}>0.$$

Now we rewrite every exponential term in the equation in powers of $$y$$:

$$$\begin{aligned} e^{6x}-e^{4x}-2e^{3x}-12e^{2x}+e^{x}+1 &= y^{6}-y^{4}-2y^{3}-12y^{2}+y+1. \end{aligned}$$$

So the original transcendental equation is equivalent to the algebraic one

$$$y^{6}-y^{4}-2y^{3}-12y^{2}+y+1=0,\qquad\text{with }y>0.$$$

We now tackle this sixth-degree polynomial. The coefficients suggest that it might break into lower-degree factors. To discover a factorisation we try to rearrange the terms in pairs and look for a common pattern.

Group the first three and the last three terms:

$$y^{6}-y^{4}-2y^{3}\ \;-\;\ 12y^{2}+y+1.$$

Within the first group  $$y^{6}-y^{4}-2y^{3}$$  we can factor out $$y^{3}$$:

$$y^{6}-y^{4}-2y^{3}=y^{3}\bigl(y^{3}-y-2\bigr).$$

Within the second group  $$-12y^{2}+y+1$$  we can factor out $$-1$$ to make the inner bracket resemble the previous bracket:

$$-12y^{2}+y+1 = -\bigl(12y^{2}-y-1\bigr).$$

We next notice that $$12y^{2}-y-1$$ happens to be a scalar multiple of $$y^{3}-y-2$$ when we evaluate at suitable $$y$$. To test this idea we divide $$y^{3}-y-2$$ by $$y-2$$:

Using the factor theorem,
$$$\bigl(y^{3}-y-2\bigr)\big|_{y=2}=8-2-2=4\neq0,\qquad \bigl(y^{3}-y-2\bigr)\big|_{y=1}=1-1-2=-2\neq0.$$$

Thus $$y-1$$ and $$y-2$$ are not factors. We abandon that line and instead try a direct polynomial long division of the full sixth-degree expression by the quadratic $$y^{2}+1$$ which is often helpful with symmetric-looking ends (the constant term is $$1$$ matching the leading coefficient).

Divide $$y^{6}-y^{4}-2y^{3}-12y^{2}+y+1$$ by $$y^{2}+1$$:

1. The leading term $$y^{6}$$ divided by $$y^{2}$$ gives $$y^{4}$$. Multiply back: $$y^{4}\cdot(y^{2}+1)=y^{6}+y^{4}.$$

Subtract this from the dividend:

$$\bigl(y^{6}-y^{4}\bigr)-\bigl(y^{6}+y^{4}\bigr)=-2y^{4}.$$

Bring down the rest to make a new dividend:

$$-2y^{4}-2y^{3}-12y^{2}+y+1.$$

2. Now divide the new leading term $$-2y^{4}$$ by $$y^{2}$$ to get $$-2y^{2}$$. Multiply back: $$-2y^{2}\cdot(y^{2}+1)=-2y^{4}-2y^{2}.$$

Subtract again:

$$$\bigl(-2y^{4}-2y^{3}-12y^{2}\bigr)-\bigl(-2y^{4}-2y^{2}\bigr)=-2y^{3}-10y^{2}.$$$

Bring down the remaining $$+y+1$$:

$$-2y^{3}-10y^{2}+y+1.$$

3. Divide the new leading term $$-2y^{3}$$ by $$y^{2}$$ to get $$-2y$$. Multiply back: $$-2y(y^{2}+1)=-2y^{3}-2y.$$

Subtract:

$$$\bigl(-2y^{3}-10y^{2}+y\bigr)-\bigl(-2y^{3}-2y\bigr)=-10y^{2}+3y.$$$

Again bring down the constant $$+1$$:

$$-10y^{2}+3y+1.$$

4. Divide $$-10y^{2}$$ by $$y^{2}$$ to get $$-10$$. Multiply back: $$-10(y^{2}+1)=-10y^{2}-10.$$

Subtract once more:

$$$\bigl(-10y^{2}+3y+1\bigr)-\bigl(-10y^{2}-10\bigr)=3y+11.$$$

The remainder is $$3y+11,$$ not zero. Hence $$y^{2}+1$$ is not a factor. However we have gleaned useful intermediate coefficients: the (synthetic) quotient so far is $$y^{4}-2y^{2}-2y-10.$$

At this point systematic long division has not exposed a factor that eliminates the remainder, so we switch to the Rational Root Theorem for the polynomial in $$y$$. Possible rational roots are of the form $$\pm1,\pm\frac1{1},\pm\frac1{\,1}$$ - in short only $$\pm1$$. We already evaluated $$P(1)=-12\neq0$$ and $$P(-1)=-10\neq0$$, so our polynomial has no rational roots.

Because $$y>0,$$ we are concerned only with positive real zeros. Let us therefore apply Descartes’ Rule of Signs to count the number of positive roots of

$$P(y)=y^{6}-y^{4}-2y^{3}-12y^{2}+y+1.$$

Write down the sequence of coefficients with their signs:

$$+\;-\;-\;-\;+\;+\;.$$

Count the sign changes:
$$$+ \to - \;(1),\; - \to -\;(0),\; - \to -\;(0),\; - \to +\;(1),\; + \to +\;(0).$$$

There are exactly $$2$$ sign changes. Descartes’ Rule tells us that the number of positive real roots is either exactly $$2$$ or less than $$2$$ by an even integer. Thus the only possibilities are $$2$$ or $$0.$$

To decide which of the two occurs, we study the behaviour of $$P(y)$$ for $$y>0.$$ We compute its value at two convenient positive points to see whether the polynomial changes sign.

At $$y=1$$: $$P(1)=1-1-2-12+1+1=-12<0.$$

At $$y=0^{+}$$ (that is, for $$y$$ tending to $$0$$ from the right) the dominant terms are the constant $$+1$$ and the linear term $$+y$$; every higher power vanishes, so

$$P(0^{+})\approx1>0.$$

Thus the sign goes from positive near $$0$$ to negative at $$y=1$$, guaranteeing at least one root in the interval $$(0,1).$$ So we cannot have the “zero roots” scenario. Consequently there must be exactly $$2$$ positive real roots.

However the question asks for the number of real roots of the original equation in $$x$$. Because $$y=e^{x}$$ is a one-to-one, strictly increasing correspondence from the entire real line onto the interval $$\bigl(0,\infty\bigr),$$ each positive root $$y_{0}$$ of the polynomial produces exactly one real root $$x_{0}=\ln y_{0}$$ of the exponential equation, and no other roots occur.

Hence the exponential equation possesses precisely $$2$$ real roots.

Hence, the correct answer is Option A.

We have to solve the exponential equation

$$e^{4x}+2e^{3x}-e^{x}-6=0.$$

Because every exponential term contains the same base $$\,e\,$$ raised to a multiple of $$x$$, it is natural to set

$$t=e^{x}\qquad\bigl(\text{so }t\gt 0\bigr).$$

Substituting $$\,e^{x}=t\,$$ into each term gives

$$\begin{aligned} e^{4x}&=(e^{x})^{4}=t^{4},\\[2mm] 2e^{3x}&=2(e^{x})^{3}=2t^{3},\\[2mm] -e^{x}&=-t. \end{aligned}$$

Hence the original equation becomes

$$t^{4}+2t^{3}-t-6=0.$$

Now we try to factor this quartic. A convenient way is to look for a product of two quadratic factors

$$t^{4}+2t^{3}-t-6 =(t^{2}+at+b)(t^{2}+ct+d).$$

Expanding the right-hand side, we obtain

$$t^{4}+(a+c)t^{3}+(ac+b+d)t^{2}+(ad+bc)t+bd.$$

Comparing coefficients with $$\,t^{4}+2t^{3}-t-6\,$$ we get the system

$$\begin{cases} a+c=2,\\ ac+b+d=0,\\ ad+bc=-1,\\ bd=-6. \end{cases}$$

A little trial shows that choosing

$$b=2,\;d=-3,\;a=1,\;c=1$$

satisfies all four conditions, because

$$\begin{aligned} a+c&=1+1=2,\\[2mm] ac+b+d&=(1)(1)+2+(-3)=0,\\[2mm] ad+bc&=(1)(-3)+(2)(1)=-3+2=-1,\\[2mm] bd&=(2)(-3)=-6. \end{aligned}$$

Therefore

$$t^{4}+2t^{3}-t-6=(t^{2}+t+2)(t^{2}+t-3)=0.$$

So we must solve

$$t^{2}+t+2=0\quad\text{or}\quad t^{2}+t-3=0.$$

First consider $$\,t^{2}+t+2=0.$$ Its discriminant is

$$\Delta_1=1^{2}-4(1)(2)=1-8=-7\lt 0,$$

so this quadratic has no real solutions. Because $$\,t=e^{x}\,,$$ which is always positive and real, we discard this factor entirely.

Next solve $$\,t^{2}+t-3=0.$$ The discriminant here is

$$\Delta_2=1^{2}-4(1)(-3)=1+12=13\gt 0,$$

giving two real roots

$$t=\frac{-1\pm\sqrt{13}}{2}.$$

Numerically,

$$\frac{-1+\sqrt{13}}{2}\approx\frac{-1+3.605}{2}\approx1.3027\gt 0,$$

while

$$\frac{-1-\sqrt{13}}{2}\approx\frac{-1-3.605}{2}\approx-2.3027\lt 0.$$

The second value is negative and therefore impossible, because $$\,t=e^{x}\gt 0\,$$ for every real $$x$$. Only the positive root survives:

$$t=\frac{-1+\sqrt{13}}{2}.$$

Finally revert to $$\,t=e^{x}.$$ We obtain

$$e^{x}=\frac{-1+\sqrt{13}}{2}\;\;\Longrightarrow\;\;x=\ln\!\left(\frac{-1+\sqrt{13}}{2}\right).$$

This is a single real value of $$x$$. Therefore the given equation possesses exactly one real root.

So, the answer is $$1$$.

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