JEE Functions Questions

We first simplify the composition $$\bigl(f \circ g\bigr)(x) = f\!\bigl(g(x)\bigr)$$.

Given $$f(x)=\dfrac{2x+3}{5x+2}$$ and $$g(x)=\dfrac{2-3x}{1-x}$$, rewrite $$g(x)$$ for convenience: $$g(x)=\dfrac{2-3x}{1-x}= \dfrac{-(3x-2)}{-(x-1)}=\dfrac{3x-2}{x-1}, \quad x\gt 1.$$

Step 1 - Compute $$2\,g(x)+3$$: $$2\,g(x)+3 = 2\left(\dfrac{3x-2}{x-1}\right)+3 = \dfrac{6x-4}{x-1} + 3 = \dfrac{6x-4 + 3(x-1)}{x-1} = \dfrac{6x-4 + 3x-3}{x-1} = \dfrac{9x-7}{x-1}. \quad -(1)$$

Step 2 - Compute $$5\,g(x)+2$$: $$5\,g(x)+2 = 5\left(\dfrac{3x-2}{x-1}\right)+2 = \dfrac{15x-10}{x-1} + 2 = \dfrac{15x-10 + 2(x-1)}{x-1} = \dfrac{15x-10 + 2x-2}{x-1} = \dfrac{17x-12}{x-1}. \quad -(2)$$

Step 3 - Form the quotient using $$(1)$$ and $$(2)$$: $$\bigl(f \circ g\bigr)(x)=\dfrac{2\,g(x)+3}{5\,g(x)+2} =\dfrac{\dfrac{9x-7}{x-1}}{\dfrac{17x-12}{x-1}} =\dfrac{9x-7}{17x-12}. \quad -(3)$$

Therefore, for every $$x\gt 1$$, $$h(x)=\bigl(f \circ g\bigr)(x)=\dfrac{9x-7}{17x-12}.$$

Step 4 - Check monotonicity on $$x\in(1,\infty)$$.

Differentiate: $$h'(x)=\dfrac{(9)(17x-12)-(9x-7)(17)}{(17x-12)^2} =\dfrac{153x-108-153x+119}{(17x-12)^2} =\dfrac{11}{(17x-12)^2}\gt 0.$$

Since $$h'(x)\gt 0$$ for all $$x\gt 1$$, $$h(x)$$ is strictly increasing. Hence on any closed interval within $$(1,\infty)$$, its minimum occurs at the left end and its maximum at the right end.

Step 5 - Evaluate $$h(x)$$ at the endpoints of the given domain $$[2,4]$$.

At $$x=2$$: $$h(2)=\dfrac{9(2)-7}{17(2)-12} =\dfrac{18-7}{34-12} =\dfrac{11}{22} =\dfrac12.$$

At $$x=4$$: $$h(4)=\dfrac{9(4)-7}{17(4)-12} =\dfrac{36-7}{68-12} =\dfrac{29}{56}.$$

Step 6 - State the range.

Because $$h(x)$$ is increasing, $$\text{Range}\bigl(h|_{[2,4]}\bigr)=[\alpha,\beta]=\left[\dfrac12,\dfrac{29}{56}\right].$$

Step 7 - Compute $$\dfrac1{\beta-\alpha}$$.

First find $$\beta-\alpha$$: $$\beta-\alpha=\dfrac{29}{56}-\dfrac12 =\dfrac{29}{56}-\dfrac{28}{56} =\dfrac1{56}.$$

Therefore, $$\dfrac1{\beta-\alpha}=\dfrac1{\,\tfrac1{56}\,}=56.$$

The required value equals $$56$$. Hence the correct option is Option D.

We are given $$f(x) + 3f\left(\dfrac{24}{x}\right) = 4x$$ for $$x \neq 0$$.

Substituting $$x = 3$$: $$f(3) + 3f(8) = 12$$ ... (i)

Substituting $$x = 8$$: $$f(8) + 3f(3) = 32$$ ... (ii)

Adding equations (i) and (ii): $$f(3) + 3f(8) + f(8) + 3f(3) = 12 + 32$$

$$4f(3) + 4f(8) = 44$$

$$f(3) + f(8) = 11$$

Hence, the correct answer is Option A.

We are given a continuous function $$f:\mathbb{R}\to\mathbb{R}$$ that satisfies
$$f(2x)-f(x)=x \quad\text{for all }x\in\mathbb{R}, \qquad f(0)=1.$$(1)

The limit to be studied is
$$G(x)=\lim_{n\to\infty}\Bigl\{f(x)-f\!\left(\dfrac{x}{2^{\,n}}\right)\Bigr\}.$$(2)

Case 1: Find a closed form for $$f(x)-f\!\left(\dfrac{x}{2^{\,n}}\right)\,$$ when $$n$$ is a positive integer.

Set $$t=\dfrac{x}{2^{\,k}}$$ in the functional equation $$(1).$$ Then $$f(2t)-f(t)=t$$ becomes
$$f\!\left(\dfrac{x}{2^{\,k-1}}\right)-f\!\left(\dfrac{x}{2^{\,k}}\right)=\dfrac{x}{2^{\,k}} \quad (k\ge 1).$$(3)

Add equations $$(3)$$ for $$k=1,2,\dots ,n$$:

$$\bigl[f(x)-f\!\left(\tfrac{x}{2}\right)\bigr]+\bigl[f\!\left(\tfrac{x}{2}\right)-f\!\left(\tfrac{x}{2^{\,2}}\right)\bigr]+\cdots+\bigl[f\!\left(\tfrac{x}{2^{\,n-1}}\right)-f\!\left(\tfrac{x}{2^{\,n}}\right)\bigr]$$ $$\;=\;\dfrac{x}{2^{\,1}}+\dfrac{x}{2^{\,2}}+\cdots+\dfrac{x}{2^{\,n}}.$$(4)

The left side telescopes, leaving

$$f(x)-f\!\left(\dfrac{x}{2^{\,n}}\right)=x\!\left(\dfrac{1}{2}+\dfrac{1}{2^{\,2}}+\cdots+\dfrac{1}{2^{\,n}}\right).$$(5)

The geometric-series sum is

$$\dfrac{1}{2}+\dfrac{1}{2^{\,2}}+\cdots+\dfrac{1}{2^{\,n}}=1-\dfrac{1}{2^{\,n}}.$$(6)

Substituting $$(6)$$ into $$(5)$$ gives the exact identity

$$f(x)-f\!\left(\dfrac{x}{2^{\,n}}\right)=x\Bigl(1-\dfrac{1}{2^{\,n}}\Bigr).$$(7)

Case 2: Pass to the limit $$n\to\infty$$ in $$(7).$$

Because $$\dfrac{1}{2^{\,n}}\to 0,$$ we get

$$\lim_{n\to\infty}\Bigl\{f(x)-f\!\left(\dfrac{x}{2^{\,n}}\right)\Bigr\}=x.$$(8)

Hence, from definition $$(2),$$

$$G(x)=x \quad\text{for every }x\in\mathbb{R}.$$(9)

Case 3: Evaluate the required sum.

Using $$(9),$$

$$\sum_{r=1}^{10} G(r^{2})=\sum_{r=1}^{10} r^{2}.$$(10)

The formula for the sum of the first $$n$$ squares is
$$\sum_{r=1}^{n} r^{2}=\dfrac{n(n+1)(2n+1)}{6}.$$(11)

With $$n=10,$$ equation $$(11)$$ gives

$$\sum_{r=1}^{10} r^{2}=\dfrac{10\cdot11\cdot21}{6}=385.$$(12)

Therefore,
$$\sum_{r=1}^{10} G(r^{2})=385.$$(13)

The correct option is Option B (385).

Let $$f(x) = \ln x$$ and $$g(x) = \frac{x^4 - 2x^3 + 3x^2 - 2x + 2}{2x^2 - 2x + 1}$$. We need to find the domain of $$f \circ g$$.

$$f(g(x)) = \ln(g(x))$$ is defined when $$g(x) > 0$$ and $$g(x)$$ itself is defined (denominator $$\neq 0$$).

$$2x^2 - 2x + 1 = 0$$ has discriminant $$4 - 8 = -4 < 0$$.

Since the leading coefficient is positive and discriminant is negative, $$2x^2 - 2x + 1 > 0$$ for all real $$x$$. So $$g(x)$$ is defined for all $$x \in \mathbb{R}$$.

We perform polynomial division of the numerator by the denominator:

$$x^4 - 2x^3 + 3x^2 - 2x + 2 = (2x^2 - 2x + 1) \cdot q(x) + r(x)$$

Dividing: $$\frac{x^4 - 2x^3 + 3x^2 - 2x + 2}{2x^2 - 2x + 1}$$

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

Alternatively, note that the numerator can be written as:

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

Let us verify: $$(x^2-x+1)^2 = x^4 - 2x^3 + 3x^2 - 2x + 1$$. Adding 1 gives $$x^4 - 2x^3 + 3x^2 - 2x + 2$$. Confirmed.

So the numerator equals $$(x^2 - x + 1)^2 + 1 \geq 1 > 0$$ for all real $$x$$.

Since both numerator and denominator are strictly positive for all $$x \in \mathbb{R}$$, we have $$g(x) > 0$$ for all $$x$$.

The domain of $$f \circ g$$ is all of $$\mathbb{R}$$.

The correct answer is Option 4: $$\mathbb{R}$$.

The domain of a sum is the intersection of the individual domains.
Hence we analyse each term of $$f(x)=\frac{1}{\sqrt{10+3x-x^{2}}}+\frac{1}{\sqrt{x+\lvert x\rvert}}$$ separately.

Term 1: $$\dfrac{1}{\sqrt{\,10+3x-x^{2}\,}}$$ is defined only when the radicand is positive:

$$10+3x-x^{2}\gt 0$$

Rewrite the quadratic in standard form:

$$-(x^{2}-3x-10)\gt 0\quad\Longrightarrow\quad x^{2}-3x-10\lt 0$$

Factor the quadratic:

$$x^{2}-3x-10=(x-5)(x+2)$$

Because the coefficient of $$x^{2}$$ is positive, the quadratic is negative between its roots. Thus

$$-2\lt x\lt 5\qquad -(1)$$

Term 2: $$\dfrac{1}{\sqrt{\,x+\lvert x\rvert\,}}$$ requires $$x+\lvert x\rvert\gt 0$$ and the denominator non-zero.

Consider both cases for $$x$$:

• If $$x\ge 0$$ then $$\lvert x\rvert=x$$, so $$x+\lvert x\rvert=2x\gt 0\;\Longrightarrow\;x\gt 0$$.

• If $$x\lt 0$$ then $$\lvert x\rvert=-x$$, giving $$x+\lvert x\rvert=0$$, which is not >0.

Hence the second term is defined only for

$$x\gt 0\qquad -(2)$$

Overall domain: Intersect $$(1)$$ and $$(2)$$:

$$(-2,5)\cap(0,\infty)=(0,5)$$

Therefore the domain of $$f(x)$$ is $$(a,b)=(0,5)$$, so $$a=0$$ and $$b=5$$.

Compute the required expression:

$$(1+a)^{2}+b^{2}=(1+0)^{2}+5^{2}=1+25=26$$

Hence $$(1+a)^{2}+b^{2}=26$$, which corresponds to Option A.

We need to find the domain of $$f(x) = \log_x(1 - \log_4(x^2 - 9x + 18))$$.

For $$\log_x(\cdot)$$ to be defined, we need $$x \gt 0$$, $$x \neq 1$$.

For $$\log_4(x^2 - 9x + 18)$$ to be defined, we need $$x^2 - 9x + 18 \gt 0$$, i.e., $$(x-3)(x-6) \gt 0$$, so $$x \lt 3$$ or $$x \gt 6$$.

The argument of the outer logarithm must be positive: $$1 - \log_4(x^2 - 9x + 18) \gt 0$$, which gives $$\log_4(x^2 - 9x + 18) \lt 1$$, so $$x^2 - 9x + 18 \lt 4$$, i.e., $$x^2 - 9x + 14 \lt 0$$, giving $$(x-2)(x-7) \lt 0$$, so $$2 \lt x \lt 7$$.

Combining all conditions: $$x \gt 0$$, $$x \neq 1$$, ($$ x \lt 3$$ or $$x \gt 6$$), and $$2 \lt x \lt 7$$:

From $$x \lt 3$$ and $$2 \lt x \lt 7$$: we get $$2 \lt x \lt 3$$. Since $$x \neq 1$$ is automatically satisfied, this gives $$(2, 3)$$.

From $$x \gt 6$$ and $$2 \lt x \lt 7$$: we get $$6 \lt x \lt 7$$, giving $$(6, 7)$$.

So $$(\alpha, \beta) = (2, 3)$$ and $$(\gamma, \delta) = (6, 7)$$.

Therefore, $$\alpha + \beta + \gamma + \delta = 2 + 3 + 6 + 7 = 18$$.

Hence, the correct answer is Option A.

We are given sets $$A = \{1, 2, 3, 4\}$$ and $$B = \{1, 4, 9, 16\}$$ and asked to determine the number of many-one functions $$f: A \to B$$ such that $$1 \in f(A)$$. By definition, a many-one function is not injective, which means at least two elements of $$A$$ are mapped to the same element of $$B$$.

First, the total number of functions from $$A$$ to $$B$$ is found by noting that each of the four elements of $$A$$ can map to any of the four elements of $$B$$, yielding $$4^4 = 256$$ total functions. The number of injective functions from $$A$$ to $$B$$ is the number of one-to-one correspondences between two 4-element sets, namely $$4! = 24$$. Therefore, the number of many-one functions from $$A$$ to $$B$$ is $$256 - 24 = 232$$.

Next, we exclude those many-one functions for which $$1 \notin f(A)$$. In that case, the function’s image must lie entirely in the set $$\{4, 9, 16\}$$, which has three elements. Hence there are $$3^4 = 81$$ such functions. Since $$|A| = 4 > 3 = |\{4, 9, 16\}|$$, none of these functions can be injective, so all 81 are many-one. Subtracting this quantity from $$232$$ gives the number of many-one functions for which $$1 \in f(A)$$, namely $$232 - 81 = 151$$.

Therefore, the required number of many-one functions satisfying the condition is $$151$$.

We are given the relation $$R = \{(f, g) : f(0) = g(1) \text{ and } f(1) = g(0)\}$$ on the set of all functions $$f: \mathbb{Z} \to \mathbb{Z}$$.

Reflexive: For $$(f, f) \in R$$, we need $$f(0) = f(1)$$ and $$f(1) = f(0)$$. Both conditions reduce to $$f(0) = f(1)$$. This is not true for all functions (e.g., $$f(x) = x$$ gives $$f(0) = 0 \neq 1 = f(1)$$). So R is not reflexive.

Symmetric: If $$(f, g) \in R$$, then $$f(0) = g(1)$$ and $$f(1) = g(0)$$. For $$(g, f) \in R$$, we need $$g(0) = f(1)$$ and $$g(1) = f(0)$$. The first condition follows from $$f(1) = g(0)$$, and the second from $$f(0) = g(1)$$. So R is symmetric.

Transitive: Suppose $$(f, g) \in R$$ and $$(g, h) \in R$$. Then $$f(0) = g(1)$$, $$f(1) = g(0)$$, $$g(0) = h(1)$$, $$g(1) = h(0)$$. For $$(f, h) \in R$$, we need $$f(0) = h(1)$$ and $$f(1) = h(0)$$. We have $$f(0) = g(1) = h(0)$$ and $$f(1) = g(0) = h(1)$$. So we need $$f(0) = h(1)$$ which means $$h(0) = h(1)$$, and $$f(1) = h(0)$$ which means $$h(1) = h(0)$$. This need not hold in general. For example, let $$f(0) = 1, f(1) = 0$$, $$g(0) = 0, g(1) = 1$$, $$h(0) = 1, h(1) = 0$$. Then $$(f,g) \in R$$ and $$(g,h) \in R$$, but $$f(0) = 1$$ and $$h(1) = 0$$, so $$f(0) \neq h(1)$$, meaning $$(f,h) \notin R$$. So R is not transitive.

Therefore R is symmetric but neither reflexive nor transitive.

Hence, the correct answer is Option B.

The function $$f$$ is a polynomial of degree 2, so let $$f(x) = ax^2 + bx + c$$ with $$a \neq 0$$. The functional equation is $$f(x) f\left(\frac{1}{x}\right) = f(x) + f\left(\frac{1}{x}\right)$$.

Compute $$f\left(\frac{1}{x}\right)$$:

$$f\left(\frac{1}{x}\right) = a \left(\frac{1}{x}\right)^2 + b \left(\frac{1}{x}\right) + c = \frac{a}{x^2} + \frac{b}{x} + c$$

Substitute into the functional equation:

$$\left(ax^2 + bx + c\right) \left(\frac{a}{x^2} + \frac{b}{x} + c\right) = \left(ax^2 + bx + c\right) + \left(\frac{a}{x^2} + \frac{b}{x} + c\right)$$

Expand the left side:

$$ax^2 \cdot \frac{a}{x^2} + ax^2 \cdot \frac{b}{x} + ax^2 \cdot c + bx \cdot \frac{a}{x^2} + bx \cdot \frac{b}{x} + bx \cdot c + c \cdot \frac{a}{x^2} + c \cdot \frac{b}{x} + c \cdot c$$

$$= a^2 + abx + acx^2 + \frac{ab}{x} + b^2 + bcx + \frac{ac}{x^2} + \frac{bc}{x} + c^2$$

Group like terms:

$$acx^2 + (ab + bc)x + (a^2 + b^2 + c^2) + \frac{ab + bc}{x} + \frac{ac}{x^2}$$

The right side is:

$$ax^2 + bx + 2c + \frac{b}{x} + \frac{a}{x^2}$$

Equate coefficients for corresponding powers of $$x$$:

For $$x^2$$: $$ac = a$$   ...(1)

For $$x$$: $$ab + bc = b$$   ...(2)

For constant term: $$a^2 + b^2 + c^2 = 2c$$   ...(3)

For $$x^{-1}$$: $$ab + bc = b$$   ...(4) [same as (2)]

For $$x^{-2}$$: $$ac = a$$   ...(5) [same as (1)]

Since $$a \neq 0$$, equation (1) gives $$c = 1$$.

Substitute $$c = 1$$ into equation (2):

$$ab + b \cdot 1 = b \implies ab + b = b \implies ab = 0$$

Since $$a \neq 0$$, $$b = 0$$.

Substitute $$b = 0$$ and $$c = 1$$ into equation (3):

$$a^2 + 0^2 + 1^2 = 2 \cdot 1 \implies a^2 + 1 = 2 \implies a^2 = 1 \implies a = \pm 1$$

Thus, possible functions are:

Case 1: $$a = 1$$, $$b = 0$$, $$c = 1$$ → $$f(x) = x^2 + 1$$

Case 2: $$a = -1$$, $$b = 0$$, $$c = 1$$ → $$f(x) = 1 - x^2$$

The range of $$f$$ is given as $$(-\infty, 1)$$.

For $$f(x) = x^2 + 1$$, since $$x \neq 0$$, $$x^2 > 0$$, so $$f(x) > 1$$, which is not in $$(-\infty, 1)$$. Thus, this function is invalid.

For $$f(x) = 1 - x^2$$, since $$x \neq 0$$, $$x^2 > 0$$, so $$f(x) < 1$$. As $$x \to \infty$$, $$f(x) \to -\infty$$, so the range is $$(-\infty, 1)$$, which matches. Thus, $$f(x) = 1 - x^2$$ is valid.

Given $$f(K) = -2K$$:

$$1 - K^2 = -2K$$

Rearrange:

$$1 - K^2 + 2K = 0 \implies K^2 - 2K - 1 = 0$$

Solve the quadratic equation:

$$K = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-1)}}{2} = \frac{2 \pm \sqrt{8}}{2} = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2}$$

So, $$K = 1 + \sqrt{2}$$ or $$K = 1 - \sqrt{2}$$.

Both values are non-zero, so they are in the domain $$\mathbb{R} - \{0\}$$.

Sum of squares:

$$(1 + \sqrt{2})^2 + (1 - \sqrt{2})^2 = (1 + 2\sqrt{2} + 2) + (1 - 2\sqrt{2} + 2) = 3 + 2\sqrt{2} + 3 - 2\sqrt{2} = 6$$

Thus, the sum of squares of all possible values of $$K$$ is 6.

Find $$\alpha^2 + \beta^2 + \gamma^2$$ where $$(\alpha, \beta)$$ is domain of $$\log_5(18x - x^2 - 77)$$ and $$(\gamma, \delta)$$ is domain of $$\log_{(x-1)}\left(\frac{2x^2+3x-2}{x^2-3x-4}\right)$$.

Domain of $$\log_5(18x - x^2 - 77)$$.

Need $$18x - x^2 - 77 > 0$$, i.e., $$x^2 - 18x + 77 < 0$$.

$$x^2 - 18x + 77 = (x-7)(x-11) < 0$$

Domain: $$(7, 11)$$, so $$\alpha = 7, \beta = 11$$.

Domain of $$\log_{(x-1)}\left(\frac{2x^2+3x-2}{x^2-3x-4}\right)$$.

Need: (i) $$x - 1 > 0$$ and $$x - 1 \neq 1$$: $$x > 1, x \neq 2$$.

(ii) $$\frac{2x^2+3x-2}{x^2-3x-4} > 0$$.

Factor: $$2x^2+3x-2 = (2x-1)(x+2)$$, $$x^2-3x-4 = (x-4)(x+1)$$.

$$\frac{(2x-1)(x+2)}{(x-4)(x+1)} > 0$$

Critical points: $$x = -2, -1, 1/2, 4$$. Combined with $$x > 1, x \neq 2$$:

For $$x > 1$$: check intervals $$(1, 4)$$ and $$(4, \infty)$$.

At $$x = 2$$: $$\frac{(3)(4)}{(-2)(3)} = \frac{12}{-6} = -2 < 0$$. Not in domain.

At $$x = 3$$: $$\frac{(5)(5)}{(-1)(4)} = \frac{25}{-4} < 0$$. Not in domain.

At $$x = 5$$: $$\frac{(9)(7)}{(1)(6)} > 0$$. In domain.

So domain for $$x > 1$$: $$(4, \infty) \setminus \{2\}$$. But 2 is not in $$(4, \infty)$$, so domain is $$(4, \infty)$$.

Wait, we also need to exclude $$x = 2$$ (base = 1). Since $$4 > 2$$, this is automatically excluded.

So $$\gamma = 4$$. But we need $$(\gamma, \delta)$$. If domain is $$(4, \infty)$$, then $$\delta = \infty$$. This doesn't work for the problem.

The answer is Option C: 186. So $$\alpha^2 + \beta^2 + \gamma^2 = 49 + 121 + 16 = 186$$. With $$\gamma = 4$$.

The correct answer is Option C: 186.

We need to determine whether $$f(x) = \frac{2^x - 2^{-x}}{2^x + 2^{-x}}$$ is one-one and/or onto.

Since we can set $$u = 2^x$$, it follows that $$f(x) = \frac{u - 1/u}{u + 1/u} = \frac{u^2 - 1}{u^2 + 1}$$. Also, this can be written as $$f(x) = \tanh(x\ln 2)$$, which is a scaled hyperbolic tangent function.

Substituting the derivative perspective, one finds $$f(x) = 1 - \frac{2}{u^2+1} = 1 - \frac{2}{2^{2x}+1}$$. As $$x$$ increases, $$2^{2x}$$ increases so that $$\frac{2}{2^{2x}+1}$$ decreases, implying that $$f(x)$$ increases. Alternatively, the derivative of $$\tanh(x\ln 2)$$ is always positive for all real $$x$$. Therefore, $$f$$ is strictly increasing and hence one-one. 

As $$x \to +\infty$$, $$2^x \to \infty$$ and $$2^{-x} \to 0$$, so $$f(x) \to \frac{\infty}{\infty} = 1$$ (approaching from below). As $$x \to -\infty$$, $$2^x \to 0$$ and $$2^{-x} \to \infty$$, so $$f(x) \to \frac{-\infty}{\infty} = -1$$ (approaching from above). At $$x = 0$$, $$f(0) = \frac{1-1}{1+1} = 0$$. This gives the range of $$f$$ as $$(-1, 1)$$.

Since the codomain is given as $$(-\infty, 1)$$ and $$(-1, 1)$$ is a proper subset of $$(-\infty, 1)$$, the function is not onto. 

The correct answer is Option 4: One-one but not onto.

$$f(x)=\dfrac{2^{x+2}+16}{2^{2x+1}+2^{x+4}+32} = \dfrac{2^{x+1}+8}{2^{2x}+8\cdot 2^{x}+16} = \dfrac{2(2^x+4)}{(2^x+4)^2}$$ 

Which gives,

$$f(x) = \dfrac{2}{2^x+4}$$

Putting $$4-x$$, we get,

$$f(4-x) = \dfrac{2}{2^{4-x}+4} = \dfrac{2\times 2^x}{2^4+4\times 2^x}$$

$$\Rightarrow f(4-x) = \dfrac{2^x}{2(2^x+4)}$$

Thus, $$f(x)+f(4-x) = \dfrac{1}{2}$$

We therefore get,

$$f\left(\dfrac{1}{15}\right) + f\left(\dfrac{59}{15}\right) = \dfrac{1}{2}$$

$$f\left(\dfrac{2}{15}\right) + f\left(\dfrac{58}{15}\right) = \dfrac{1}{2}$$

$$\vdots$$

$$f\left(\dfrac{29}{15}\right) + f\left(\dfrac{31}{15}\right) = \dfrac{1}{2}$$

With $$f\left(\dfrac{30}{15}\right)$$ remaining, which is simply equal to $$\dfrac{2}{2^{2}+4} = \dfrac{1}{4}$$

Therefore, the sum is equal to $$8\left(\dfrac{1}{4} + 29\times \dfrac{1}{2}\right) = 2 + 29\times 4 = 118$$

The set $$A$$ is the circle with radius $$5$$ centred at the origin: $$x^2 + y^2 = 25$$.

The set $$B$$ is the ellipse $$x^2 + 9y^2 = 144$$ whose semi-axes are $$12$$ along the $$x$$-axis and $$4$$ along the $$y$$-axis.

To obtain $$D = A \cap B$$ we solve the two equations simultaneously.

From the circle, $$x^2 = 25 - y^2$$ $$-(1)$$. Substitute $$-(1)$$ into the ellipse:

$$25 - y^2 + 9y^2 = 144$$

$$25 + 8y^2 = 144$$

$$8y^2 = 119$$

$$y^2 = \frac{119}{8}$$, so $$y = \pm \sqrt{\frac{119}{8}}$$.

Using $$-(1)$$,

$$x^2 = 25 - \frac{119}{8} = \frac{81}{8}$$, hence $$x = \pm \sqrt{\frac{81}{8}} = \pm \frac{9\sqrt{2}}{4}$$.

Both $$x$$ and $$y$$ can take their signs independently, giving the four intersection points

$$\left(\pm\frac{9\sqrt{2}}{4}, \; \pm\sqrt{\frac{119}{8}}\right).$$

Therefore, $$|D| = 4$$.

Next, count the elements of $$C = \{(x, y) \in \mathbb{Z} \times \mathbb{Z} : x^2 + y^2 \le 4\}$$.

List all integer pairs inside or on the circle of radius $$2$$:

• For $$y = 0$$: $$x = -2, -1, 0, 1, 2$$ → $$5$$ points.
• For $$y = \pm 1$$: $$x^2 \le 3$$ → $$x = -1, 0, 1$$ → $$3 + 3 = 6$$ points.
• For $$y = \pm 2$$: $$x^2 \le 0$$ → $$x = 0$$ → $$1 + 1 = 2$$ points.

Total: $$5 + 6 + 2 = 13$$, so $$|C| = 13$$.

We need one-one (injective) functions from a $$4$$-element set $$D$$ to a $$13$$-element set $$C$$. For an injective function, choose the image of each element of $$D$$ as follows:

• First element: $$13$$ choices.
• Second element: $$12$$ choices.
• Third element: $$11$$ choices.
• Fourth element: $$10$$ choices.

Hence the total number of injective functions is the permutation

$$13 \times 12 \times 11 \times 10 = 17160.$$

Therefore, the required number of one-one functions is $$17160$$.

Option C is correct.

Given $$f(x) = (2+3a)x^2 + \frac{a+2}{a-1}x + b$$ and $$f(x+y) = f(x) + f(y) + 1 - \frac{2}{7}xy$$.

Setting $$x = y = 0$$ yields $$f(0) = 2f(0) + 1$$, which gives $$f(0) = b = -1$$.

Comparing the coefficients of $$xy$$ in the functional equation results in $$2(2+3a) = -\frac{2}{7}$$, so $$a = -\frac{5}{7}$$.

Substituting this value of $$a$$ into the expressions for the coefficients gives $$2+3a = -\frac{1}{7}$$ and $$\frac{a+2}{a-1} = \frac{9/7}{-12/7} = -\frac{3}{4}$$, hence $$f(x) = -\frac{1}{7}x^2 - \frac{3}{4}x - 1$$.

Evaluating this function at integers from 1 to 5 yields $$f(1) = -\frac{1}{7} - \frac{3}{4} - 1 = -\frac{53}{28}$$, $$f(2) = -\frac{4}{7} - \frac{3}{2} - 1 = -\frac{86}{28}$$, $$f(3) = -\frac{9}{7} - \frac{9}{4} - 1 = -\frac{127}{28}$$, $$f(4) = -\frac{16}{7} - 3 - 1 = -\frac{176}{28}$$, and $$f(5) = -\frac{25}{7} - \frac{15}{4} - 1 = -\frac{233}{28}$$. Since all these values are negative, $$\sum|f(i)| = \frac{53+86+127+176+233}{28} = \frac{675}{28}$$.

Multiplying by 28 gives $$28 \times \frac{675}{28} = 675$$.

The correct answer is Option 4: 675.

Given the domain of the function $$f(x) = \cos^{-1}\left(\frac{4x+5}{3x-7}\right)$$ is $$[\alpha, \beta]$$. The domain of the inverse cosine function is $$[-1, 1]$$, we get:

$$-1 \le \frac{4x+5}{3x-7} \le 1$$

From the inequality $$\frac{4x+5}{3x-7} \le 1$$:

$$\frac{4x+5}{3x-7} - 1 \le 0 \implies \frac{4x+5 - (3x-7)}{3x-7} \le 0 \implies \frac{x+12}{3x-7} \le 0$$

The critical points are $$x = -12$$ and $$x = \frac{7}{3}$$. This gives:

$$x \in \left[-12, \frac{7}{3}\right)$$

From the inequality $$\frac{4x+5}{3x-7} \ge -1$$:

$$\frac{4x+5}{3x-7} + 1 \ge 0 \implies \frac{4x+5 + (3x-7)}{3x-7} \ge 0 \implies \frac{7x-2}{3x-7} \ge 0$$

The critical points are $$x = \frac{2}{7}$$ and $$x = \frac{7}{3}$$. This gives:

$$x \in \left(-\infty, \frac{2}{7}\right] \cup \left(\frac{7}{3}, \infty\right)$$

Taking the intersection of both intervals gives the domain of $$f(x)$$:

$$\text{Domain of } f(x) = \left[-12, \frac{2}{7}\right]$$

Comparing this with $$[\alpha, \beta]$$, we get:

$$\alpha = -12, \quad \beta = \frac{2}{7}$$

$$7(\alpha + \beta) = 7\left(-12 + \frac{2}{7}\right) = -84 + 2 = -82$$

---

Given the domain of the function $$g(x) = \log_2(2 - 6\log_{27}(2x+5))$$ is $$(\gamma, \delta)$$. For the log functions to be defined, their arguments must be strictly positive.

From the inner log argument:

$$2x + 5 > 0 \implies x > -\frac{5}{2}$$

From the outer log argument:

$$2 - 6\log_{27}(2x+5) > 0 \implies 6\log_{27}(2x+5) < 2 \implies \log_{27}(2x+5) < \frac{1}{3}$$

Converting to exponential form:

$$2x + 5 < (27)^{\frac{1}{3}} \implies 2x + 5 < 3 \implies 2x < -2 \implies x < -1$$

Taking the intersection of both conditions gives the domain of $$g(x)$$:

$$\text{Domain of } g(x) = \left(-\frac{5}{2}, -1\right)$$

Comparing this with $$(\gamma, \delta)$$, we get:

$$\gamma = -\frac{5}{2}, \quad \delta = -1$$

$$4(\gamma + \delta) = 4\left(-\frac{5}{2} - 1\right) = 4\left(-\frac{7}{2}\right) = -14$$

---

Now, substituting these values into the required expression:

$$\left|7(\alpha + \beta) + 4(\gamma + \delta)\right| = |-82 - 14| = |-96| = 96$$

Therefore, the final value is equal to 96.

Functions f:{1,...,100}→{0,1} assigning 1 to exactly one integer ≤ 98.
Choose which of 1-98 gets value 1: 98 ways. For positions 99,100: each can be 0 or 1, but we need exactly one value of 1 among 1-98. Positions 99 and 100 can be 0 or 1 freely: 4 choices.

Total = 98 × 4 = 392.
The answer is 392.

$$\log_{2/\pi}|\sin x| + \log_{2/\pi}|\cos x| = 2$$

$$\log_{2/\pi}(|\sin x||\cos x|) = 2$$

$$|\sin x\cos x| = (2/\pi)^2 = 4/\pi^2$$

$$\frac{1}{2}|\sin 2x| = 4/\pi^2$$, so $$|\sin 2x| = 8/\pi^2$$.

Since $$8/\pi^2 \approx 0.811$$, and $$\sin 2x$$ achieves this value in $$(0, \pi)$$:

For $$x \in (0, \pi/2)$$: $$2x \in (0, \pi)$$, and $$\sin 2x = 8/\pi^2$$ has 2 solutions.

For $$x \in (\pi/2, \pi)$$: $$2x \in (\pi, 2\pi)$$, and $$|\sin 2x| = 8/\pi^2$$ has 2 solutions.

So $$|A| = 4$$.

Set B: Let $$t = \sqrt{x} \geq 0$$. The equation becomes:

$$t(t-4) - 3|t-2| + 6 = 0$$

$$t^2 - 4t + 6 - 3|t-2| = 0$$

Case 1: $$t \geq 2$$: $$t^2 - 4t + 6 - 3(t-2) = 0 \Rightarrow t^2 - 7t + 12 = 0 \Rightarrow (t-3)(t-4) = 0$$. So $$t = 3$$ or $$t = 4$$, giving $$x = 9$$ or $$x = 16$$.

Case 2: $$0 \leq t < 2$$: $$t^2 - 4t + 6 - 3(2-t) = 0 \Rightarrow t^2 - t = 0 \Rightarrow t(t-1) = 0$$. So $$t = 0$$ or $$t = 1$$, giving $$x = 0$$ or $$x = 1$$.

So $$B = \{0, 1, 9, 16\}$$, $$|B| = 4$$.

Sets A and B are disjoint (A contains values in $$(0, \pi)$$ which are irrational, B contains integers).

$$n(A \cup B) = |A| + |B| = 4 + 4 = 8$$.

The correct answer is Option B: 8.

We use the identity $$\sec^2 \theta = 1 + \tan^2 \theta$$. The condition becomes:

$$(1 + \tan^2 x) - \tan^2 y = 1 \implies \tan^2 x = \tan^2 y$$

• Reflexive: $$\tan^2 x = \tan^2 x$$ is always true. (Reflexive)

• Symmetric: If $$\tan^2 x = \tan^2 y$$, then $$\tan^2 y = \tan^2 x$$. (Symmetric)

• Transitive: If $$\tan^2 x = \tan^2 y$$ and $$\tan^2 y = \tan^2 z$$, then $$\tan^2 x = \tan^2 z$$. (Transitive)

Since it satisfies all three, $$R$$ is an equivalence relation.

Correct Option: B

Let the given function be $$y = f(x) = \frac{5 - x}{x^{2} - 3x + 2}$$, where $$x \neq 1,\,2$$ because the denominator $$x^{2} - 3x + 2 = (x - 1)(x - 2)$$ vanishes at these points.

To find the range we eliminate $$x$$.
Cross-multiply:

$$y(x^{2} - 3x + 2) = 5 - x$$

Rearrange into a quadratic in $$x$$:

$$y x^{2} - 3y x + 2y + x - 5 = 0$$

Group like terms:

$$y x^{2} + (-3y + 1)x + (2y - 5) = 0 \quad -(1)$$

For a given real $$y$$ to lie in the range, equation $$(1)$$ must have at least one real root $$x$$ that is different from $$1$$ and $$2$$. The first requirement is that its discriminant is non-negative.

Discriminant $$\Delta$$ of $$(1)$$:

$$\Delta = (-3y + 1)^{2} - 4y(2y - 5)$$

Simplify:

$$\Delta = 9y^{2} - 6y + 1 - 8y^{2} + 20y = y^{2} + 14y + 1$$

Thus real roots exist when

$$y^{2} + 14y + 1 \ge 0 \quad -(2)$$

Factor $$(2)$$ by finding its roots. Solve

$$y^{2} + 14y + 1 = 0$$

Roots:

$$y = \frac{-14 \pm \sqrt{14^{2} - 4 \cdot 1 \cdot 1}}{2} = \frac{-14 \pm \sqrt{196 - 4}}{2} = \frac{-14 \pm \sqrt{192}}{2} = -7 \pm 4\sqrt{3}$$

Denote

$$\alpha = -7 - 4\sqrt{3}, \qquad \beta = -7 + 4\sqrt{3}$$

Since the coefficient of $$y^{2}$$ in $$(2)$$ is positive, the inequality $$\Delta \ge 0$$ holds outside the interval formed by these roots:

$$y \le \alpha \quad \text{or} \quad y \ge \beta$$

Hence the range is $$(-\infty,\,\alpha] \cup [\beta,\,\infty)$$ as stated.

Now compute $$\alpha^{2} + \beta^{2}$$.

For the quadratic $$y^{2} + 14y + 1 = 0$$:
  Sum of roots $$\alpha + \beta = -14$$
  Product of roots $$\alpha\beta = 1$$

Use the identity $$\alpha^{2} + \beta^{2} = (\alpha + \beta)^{2} - 2\alpha\beta$$:

$$\alpha^{2} + \beta^{2} = (-14)^{2} - 2 \cdot 1 = 196 - 2 = 194$$

Therefore, $$\alpha^{2} + \beta^{2} = 194$$.

Answer: Option D

We are given $$A = \{1, 2, 3, \ldots, 100\}$$ and the relation $$R = \{(a, b) : a = 2b + 1\}$$ on $$A$$.

Step 1: Identify all ordered pairs in R

For $$(a, b) \in R$$, we need $$a = 2b + 1$$ where both $$a$$ and $$b$$ are in $$A$$.

Since $$a = 2b + 1$$, $$a$$ must be odd. The valid pairs are:

$$(3, 1), (5, 2), (7, 3), (9, 4), (11, 5), (13, 6), \ldots, (99, 49)$$

That is, for each $$b$$ from $$1$$ to $$49$$, we get the pair $$(2b+1, b)$$.

Step 2: Understand the chain condition

We need a sequence $$(a_1, a_2), (a_2, a_3), (a_3, a_4), \ldots, (a_k, a_{k+1})$$ of $$k$$ elements of $$R$$ such that the second entry of each pair equals the first entry of the next pair.

This means: $$a_1 = 2a_2 + 1$$, $$a_2 = 2a_3 + 1$$, $$a_3 = 2a_4 + 1$$, and so on.

Step 3: Express $$a_1$$ in terms of $$a_{k+1}$$

From the recurrence $$a_i = 2a_{i+1} + 1$$, we can write:

$$a_1 = 2a_2 + 1 = 2(2a_3 + 1) + 1 = 4a_3 + 3$$

$$a_1 = 4(2a_4 + 1) + 3 = 8a_4 + 7$$

In general: $$a_1 = 2^k \cdot a_{k+1} + (2^k - 1)$$

Step 4: Find the maximum $$k$$

We need $$a_1 \leq 100$$ and $$a_{k+1} \geq 1$$.

Setting $$a_{k+1} = 1$$ (the smallest possible value):

$$a_1 = 2^k \cdot 1 + (2^k - 1) = 2^{k+1} - 1$$

We need $$2^{k+1} - 1 \leq 100$$, so $$2^{k+1} \leq 101$$.

Checking: $$2^6 = 64 \leq 101$$ ✓ and $$2^7 = 128 \gt 101$$ ✗

So $$k + 1 \leq 6$$, which gives $$k \leq 5$$.

Step 5: Verify with $$k = 5$$

With $$k = 5$$ and $$a_6 = 1$$:

$$a_6 = 1, \quad a_5 = 2(1) + 1 = 3, \quad a_4 = 2(3) + 1 = 7$$

$$a_3 = 2(7) + 1 = 15, \quad a_2 = 2(15) + 1 = 31, \quad a_1 = 2(31) + 1 = 63$$

The chain is: $$(63, 31), (31, 15), (15, 7), (7, 3), (3, 1)$$

All values are in $$A = \{1, 2, \ldots, 100\}$$ ✓

Therefore, the largest integer $$k$$ is $$\mathbf{5}$$.

Hence, the correct answer is Option C.

For convenience let $$\mathbb{N}=\{1,2,3,\dots\}$$ and $$\mathbb{Z}=\{\dots,-2,-1,0,1,2,\dots\}$$.

1. Properties of $$f:\mathbb{N}\to\mathbb{Z}$$
  • If $$n$$ is odd, write $$n=2k-1\;(k\ge 1)$$. Then $$f(n)=\frac{n+1}{2}=\frac{2k-1+1}{2}=k,$$ which produces all positive integers $$1,2,3,\dots$$.
  • If $$n$$ is even, write $$n=2k\;(k\ge 1)$$. Then $$f(n)=\frac{4-n}{2}=\frac{4-2k}{2}=2-k,$$ which produces the values $$1,0,-1,-2,-3,\dots$$.
Combining the two cases, the image of $$f$$ is all of $$\mathbb{Z}$$, so $$f$$ is onto.

However, $$f(1)=\frac{1+1}{2}=1,\quad f(2)=\frac{4-2}{2}=1,$$ and $$1\ne 2$$. Hence distinct inputs can give the same output, so $$f$$ is not one-one.

Conclusion for $$f$$: not one-one, but onto  ⇒  Option D is true.

2. Properties of $$g:\mathbb{Z}\to\mathbb{N}$$
  • $$n\ge 0$$: $$g(n)=3+2n=3,5,7,9,\dots$$ (all odd numbers $$\ge 3$$).
  • $$n\lt 0$$: write $$n=-k\;(k\ge 1)$$, then $$g(n)=-2(-k)=2k=2,4,6,8,\dots$$ (all even numbers $$\ge 2$$).
Thus $$\operatorname{Im}(g)=\{2,3,4,5,6,\dots\}=\mathbb{N}\setminus\{1\}$$, so $$g$$ is not onto.

To test injectivity, suppose $$g(n_1)=g(n_2)$$.
  • If both $$n_1,n_2\ge 0$$, then $$3+2n_1=3+2n_2\Rightarrow n_1=n_2$$(strictly increasing).
  • If both $$n_1,n_2\lt 0$$, write them as $$-k_1,-k_2$$. Then $$2k_1=2k_2\Rightarrow k_1=k_2\Rightarrow n_1=n_2$$.
  • If one is $$\ge 0$$ and the other $$\lt 0$$, their images have opposite parity (odd vs even), so equality is impossible.
Therefore $$g$$ is one-one.

Conclusion for $$g$$: one-one, but not onto  ⇒  Option C is false.

3. The composition $$g\circ f:\mathbb{N}\to\mathbb{N}$$
First compute the value explicitly.

$$f(n)=k\quad(\text{positive})$$
$$g(f(n))=g(k)=3+2k$$ Hence $$g\circ f(n)=3+2k$$, i.e. the numbers $$5,7,9,\dots$$.

$$f(n)=2-k$$
If $$k=1$$, $$f(n)=1,\;g(1)=5$$ (already obtained).
If $$k=2$$, $$f(n)=0,\;g(0)=3$$.
If $$k\ge 3$$, $$f(n)=2-k\le -1,\;g(2-k)=2k-4$$ (even numbers $$2,4,6,8,\dots$$).

The total image of $$g\circ f$$ is therefore $$\{2,3,4,5,6,7,8,9,\dots\}=\mathbb{N}\setminus\{1\},$$ so $$g\circ f$$ is not onto.

Also $$g\circ f(1)=5$$ and $$g\circ f(2)=5$$, yet $$1\ne 2$$, so the composition is not one-one.

Conclusion for $$g\circ f$$: neither one-one nor onto  ⇒  Option A is true.

4. The composition $$f\circ g:\mathbb{Z}\to\mathbb{Z}$$

$$g(n)=3+2n\;(\text{odd})$$
$$f(g(n))=\frac{(3+2n)+1}{2}=n+2.$$ So for $$n=0,1,2,\dots$$ we obtain $$2,3,4,\dots$$.

$$g(n)=2k\;(\text{even})$$
$$f(g(n))=\frac{4-2k}{2}=2-k.$$ For $$k=1,2,3,\dots$$ this gives $$1,0,-1,-2,\dots$$.

The union of the two cases yields every integer, so $$f\circ g$$ is onto. To check injectivity, observe:   • $$n\ge 0\; \Rightarrow f\circ g(n)=n+2$$ (strictly increasing).   • $$n\lt 0\; \Rightarrow f\circ g(n)=2-n$$ (strictly decreasing). No non-negative input can share an image with a negative input, so different arguments always give different values. Hence $$f\circ g$$ is one-one. Therefore Option B is false.

Final result: Only Option A and Option D are correct.

For the function $$f(x)=\log_e\!\left(\dfrac{2x-3}{5+4x}\right)+\sin^{-1}\!\left(\dfrac{4+3x}{2-x}\right)$$ to be defined, two conditions must hold simultaneously:

(i) Logarithm: $$\dfrac{2x-3}{5+4x}\gt 0$$

(ii) Inverse sine: $$-1 \le \dfrac{4+3x}{2-x} \le 1$$ and $$2-x \ne 0$$ (so $$x\ne 2$$).

Condition (i):

The fraction is positive when numerator and denominator have the same sign.

Case 1 $$2x-3\gt 0\;$$ and $$\;5+4x\gt 0$$
  $$x\gt \dfrac32,$$ $$x\gt -\dfrac54 \;\Rightarrow\; x\gt \dfrac32$$

Case 2 $$2x-3\lt 0\;$$ and $$\;5+4x\lt 0$$
  $$x\lt \dfrac32,$$ $$x\lt -\dfrac54 \;\Rightarrow\; x\lt -\dfrac54$$

Hence, from the logarithm we get

$$x\in(-\infty,\,-\dfrac54)\;\cup\;(\dfrac32,\,\infty) \quad -(1)$$

Condition (ii): Let $$y=\dfrac{4+3x}{2-x}.$$ We solve $$-1\le y\le 1.$$ Break into two parts, keeping the sign of $$2-x$$ in mind.

(a)  First inequality $$\dfrac{4+3x}{2-x}\le 1$$

• If $$x\lt 2$$ (denominator positive): multiply directly
  $$4+3x\le 2-x \;\Longrightarrow\; 4x\le -2 \;\Longrightarrow\; x\le -\dfrac12.$$
• If $$x\gt 2$$ (denominator negative): inequality reverses
  $$4+3x\ge 2-x \;\Longrightarrow\; 4x\ge -2 \;\Longrightarrow\; x\ge -\dfrac12.$$ Since this lies entirely above 2, the whole interval $$x\gt 2$$ satisfies part (a).

(b)  Second inequality $$\dfrac{4+3x}{2-x}\ge -1$$

• If $$x\lt 2$$ (denominator positive):
  $$4+3x\ge -\,\bigl(2-x\bigr) \;\Longrightarrow\; 4+3x\ge -2+x \;\Longrightarrow\; 2x\ge -6 \;\Longrightarrow\; x\ge -3.$$
• If $$x\gt 2$$ (denominator negative): inequality reverses
  $$4+3x\le -\,\bigl(2-x\bigr) \;\Longrightarrow\; 2x\le -6 \;\Longrightarrow\; x\le -3,$$ which is impossible because here $$x\gt 2.$$ Thus part (b) gives $$-3\le x\lt 2.$$

Combining (a) and (b): for $$x\lt 2,$$ both must hold, giving
$$x\in[-3,\,-\dfrac12].$$
For $$x\gt 2,$$ part (b) fails, so no additional points appear.

$$x\in[-3,\,-\dfrac12] \quad -(2)$$

Overall domain: Intersect $$(1)$$ and $$(2)$$.

$$( -\infty,\,-\dfrac54)\cup(\dfrac32,\infty)\; \cap\;[-3,\,-\dfrac12] \;=\;[-3,\,-\dfrac54).$$

Therefore, the domain is $$[\alpha,\beta)=\bigl[-3,\,-\dfrac54\bigr).$$ So $$\alpha=-3,\;\beta=-\dfrac54.$$

Compute $$\alpha^2+4\beta:$$

$$\alpha^2+4\beta = (-3)^2 + 4\!\left(-\dfrac54\right)=9-5=4.$$

Hence, $$\alpha^2+4\beta=4,$$ which matches Option B.

For every logarithm we must have a positive argument, and for $$\log_a(b)$$ to be positive ( $$a\gt1$$ ) the argument $$b$$ must exceed $$1$$.
We examine the functions layer by layer.

Case 1: Domain of $$f(x)=\log_{4}\!\Bigl[\;\log_{3}\!\bigl\{\log_{7}\!\bigl(8-\log_{2}(x^{2}+4x+5)\bigr)\bigr\}\Bigr]$$

Step 1 - innermost logarithm

$$x^{2}+4x+5=(x+2)^{2}+1\gt0$$ for all real $$x$$, so $$\log_{2}(x^{2}+4x+5)$$ exists for every real $$x$$.

Step 2 - argument of $$\log_{7}$$

Let $$y=8-\log_{2}(x^{2}+4x+5)$$. For $$\log_{7}(y)$$ to exist we need $$y\gt0$$, i.e.

$$8-\log_{2}(x^{2}+4x+5)\gt0 \;\Longrightarrow\; \log_{2}(x^{2}+4x+5)\lt8.$$

Step 3 - argument of $$\log_{3}$$ must be positive

Let $$z=\log_{7}(y)$$. For $$\log_{3}(z)$$ to be positive we require $$z\gt1$$ (because base $$3\gt1$$). Thus

$$\log_{7}\!\bigl(8-\log_{2}(x^{2}+4x+5)\bigr)\gt1 \;\Longrightarrow\; 8-\log_{2}(x^{2}+4x+5)\gt7.$$

Step 4 - simplifying the last inequality

$$8-\log_{2}(x^{2}+4x+5)\gt7 \;\Longrightarrow\; \log_{2}(x^{2}+4x+5)\lt1 \;\Longrightarrow\; x^{2}+4x+5\lt2.$$

Because $$x^{2}+4x+5=(x+2)^{2}+1$$, the above becomes

$$(x+2)^{2}+1\lt2 \;\Longrightarrow\; (x+2)^{2}\lt1 \;\Longrightarrow\; -1\lt x+2\lt1 \;\Longrightarrow\; -3\lt x\lt-1.$$

Hence the domain of $$f(x)$$ is $$(\alpha,\beta)=(-3,-1).$$ So $$\alpha=-3,\;\beta=-1.$$

Case 2: Domain of $$g(x)=\sin^{-1}\!\left(\dfrac{7x+10}{\,x-2\,}\right)$$

The argument of $$\sin^{-1}$$ must lie in $$[-1,1]$$ and the denominator cannot be zero.

Let $$t=\dfrac{7x+10}{x-2}$$. We need $$-1\le t\le1,\;x\ne2.$$

1) Inequality $$t\le1$$:

$$\dfrac{7x+10}{x-2}\le1 \;\Longrightarrow\; \dfrac{6(x+2)}{x-2}\le0.$$

This holds for $$-2\le x&lt2.$$

2) Inequality $$t\ge-1$$:

$$\dfrac{7x+10}{x-2}\ge-1 \;\Longrightarrow\; \dfrac{8(x+1)}{x-2}\ge0.$$

This is true for $$x\le-1$$ or $$x>2.$$

3) Intersection of (1) and (2):

$$[-2,2)\;\cap\;(-\infty,-1]\;=\;[-2,-1].$$

The endpoint $$x=-2$$ gives $$t=1$$ and $$x=-1$$ gives $$t=-1,$$ both allowed. Hence the domain of $$g(x)$$ is $$[\gamma,\delta]=[-2,-1]$$ with $$\gamma=-2,\;\delta=-1.$$

Case 3: Required sum

$$\alpha^2+\beta^2+\gamma^2+\delta^2 =(-3)^2+(-1)^2+(-2)^2+(-1)^2 =9+1+4+1 =15.$$

Therefore, $$\alpha^2+\beta^2+\gamma^2+\delta^2=15$$, which matches Option A.

Set A (Range of $$f(x)$$):

$$f'(x) = 6x^2 - 30x + 36 = 6(x-2)(x-3)$$.

Critical points at $$x=2, 3$$.

$$f(0)=7, f(2)=35, f(3)=34$$. Range $$A = [7, 35]$$.

Set B (Range of $$g(x)$$):

$$g(0)=0$$. As $$x \to \infty, g(x) \to 0$$. Since $$g(x) \geq 0$$, the range starts at $$0$$. The max value (using AM-GM or derivative) is a very small decimal $$< 1$$. Thus, the only integer in $$B$$ is $$\{0\}$$.

Set S: Integers in $$A \cup B$$: $$\{0, 7, 8, \dots, 35\}$$.

Count: $$1$$ (for $$\{0\}$$) $$+ 29$$ (from $$7$$ to $$35$$) = 30

The limit condition is
$$\lim_{t \to x}\;\frac{t^{10}f(x)-x^{10}f(t)}{t^{9}-x^{9}} = 1 \qquad\forall\,x\gt0$$

For fixed $$x$$, treat the numerator and denominator as functions of the variable $$t$$:
  • $$g(t)=t^{10}f(x)-x^{10}f(t)$$ with $$g(x)=0$$
  • $$h(t)=t^{9}-x^{9}$$ with $$h(x)=0$$

Because both approach $$0$$ as $$t\to x$$, apply L’Hospital’s Rule (differentiate with respect to $$t$$):
$$\lim_{t\to x}\frac{g(t)}{h(t)} =\frac{g'(x)}{h'(x)}$$

Compute the derivatives:
$$g'(t)=10t^{9}f(x)-x^{10}f'(t)\;\;\Longrightarrow\;\;g'(x)=10x^{9}f(x)-x^{10}f'(x)$$
$$h'(t)=9t^{8}\;\;\Longrightarrow\;\;h'(x)=9x^{8}$$

The limit equals 1, so
$$\frac{10x^{9}f(x)-x^{10}f'(x)}{9x^{8}} = 1$$

Multiply by $$9x^{8}$$:
$$10x^{9}f(x)-x^{10}f'(x)=9x^{8}$$

Divide by $$x^{8}$$:
$$10x\,f(x)-x^{2}f'(x)=9$$

Re-arrange to a first-order linear differential equation:
$$f'(x)-\frac{10}{x}f(x)=-\frac{9}{x^{2}} \quad -(1)$$

Let integrating factor $$I(x)=e^{\int -10/x\,dx}=e^{-10\ln x}=x^{-10}$$.

Multiply $$-(1)$$ by $$x^{-10}$$:
$$x^{-10}f'(x)-\frac{10}{x}x^{-10}f(x)=-9x^{-12}$$

The left side is the derivative of $$x^{-10}f(x)$$:
$$\frac{d}{dx}\bigl(x^{-10}f(x)\bigr)=-9x^{-12}$$

Integrate:
$$x^{-10}f(x)=\int -9x^{-12}\,dx=-9\cdot\frac{x^{-11}}{-11}+C=\frac{9}{11}x^{-11}+C$$

Multiply by $$x^{10}$$ to get $$f(x)$$:
$$f(x)=x^{10}\Bigl(\frac{9}{11}x^{-11}+C\Bigr) =\frac{9}{11x}+Cx^{10}$$

Use the given value $$f(1)=2$$:
$$2=\frac{9}{11}+C\;\;\Longrightarrow\;\;C=2-\frac{9}{11}=\frac{13}{11}$$

Hence, for all $$x\gt0$$,
$$f(x)=\frac{9}{11x}+\frac{13}{11}x^{10}$$

Option B which is: $$\frac{9}{11x} + \frac{13}{11}x^{10}$$

We need to find the set $$S = \{x \in \mathbb{R} : (\sqrt{3} + \sqrt{2})^x + (\sqrt{3} - \sqrt{2})^x = 10\}$$.

First, note that $$(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2}) = 3 - 2 = 1$$, so $$(\sqrt{3} - \sqrt{2}) = \frac{1}{\sqrt{3} + \sqrt{2}}$$ and hence $$(\sqrt{3} - \sqrt{2})^x = (\sqrt{3} + \sqrt{2})^{-x}$$.

Let $$t = (\sqrt{3} + \sqrt{2})^x$$. Since $$\sqrt{3} + \sqrt{2} > 1$$, it follows that $$t > 0$$. Substituting into the given equation gives $$t + \frac{1}{t} = 10$$.

Multiplying both sides by $$t$$ yields $$t^2 + 1 = 10t$$, or $$t^2 - 10t + 1 = 0$$.

Applying the quadratic formula gives $$t = \frac{10 \pm \sqrt{100 - 4}}{2} = \frac{10 \pm \sqrt{96}}{2} = \frac{10 \pm 4\sqrt{6}}{2} = 5 \pm 2\sqrt{6}$$. Both roots are positive since $$2\sqrt{6} \approx 4.899$$, so $$t_1 = 5 + 2\sqrt{6} > 0$$ and $$t_2 = 5 - 2\sqrt{6} \approx 0.101 > 0$$.

Observe that $$5 + 2\sqrt{6} = (\sqrt{3} + \sqrt{2})^2$$, because $$(\sqrt{3} + \sqrt{2})^2 = 3 + 2\sqrt{6} + 2 = 5 + 2\sqrt{6}$$; similarly, $$5 - 2\sqrt{6} = (\sqrt{3} - \sqrt{2})^2 = (\sqrt{3} + \sqrt{2})^{-2}$$. Hence, if $$(\sqrt{3} + \sqrt{2})^x = 5 + 2\sqrt{6}$$ then $$x = 2$$, and if $$(\sqrt{3} + \sqrt{2})^x = 5 - 2\sqrt{6}$$ then $$x = -2$$.

Therefore, $$S = \{-2, 2\}$$, which contains 2 elements. The correct answer is 2 (Option C).

 The number of subsets for a set with $$k$$ elements is $$2^k$$.

 We are given $$2^m - 2^n = 56$$.

 Factor out $$2^n$$: $$2^n(2^{m-n} - 1) = 56$$.

 Prime factorize $$56 = 8 \times 7 = 2^3 \times (2^3 - 1)$$.

 Comparing terms, $$n = 3$$ and $$m-n = 3 \Rightarrow m = 6$$. So, $$P$$ is $$(6, 3)$$.

 Use the distance formula:

$$\text{Distance} = \sqrt{(6 - (-2))^2 + (3 - (-3))^2} = \sqrt{8^2 + 6^2} = \sqrt{100} = 10$$

$$f(x) = \begin{cases} x - 1; & x = \text{even} \\ 2x; & x = \text{odd} \end{cases}$$

$$f(f(f(a))) = 21$$

Case-1: If $$a = \text{even}$$

$$f(a) = a - 1 = \text{odd}$$

$$f(f(a)) = 2(a - 1) = \text{even}$$

$$f(f(f(a))) = 2a - 3 = 21 \Rightarrow a = 12$$

Case-2: If $$a = \text{odd}$$

$$f(a) = 2a = \text{even}$$

$$f(f(a)) = 2a - 1 = \text{odd}$$

$$f(f(f(a))) = 4a - 2 = 21 \text{ (Not possible)}$$

Hence $$a = 12$$

Now, $$\lim_{x \rightarrow 12^{-}} \left( \frac{\vert{}x\vert{}^3}{12} - \left[\frac{x}{12}\right] \right)$$

$$= \lim_{x \rightarrow 12^{-}} \frac{\vert{}x\vert{}^3}{12} - \lim_{x \rightarrow 12^{-}} \left[\frac{x}{12}\right]$$

$$= 144 - 0$$ $$= 144$$

2310 = 2×3×5×7×11. A = {2,3,5,7,11} (5 elements). f(2)=[log₂(4+1)]=[2.32]=2. f(3)=[log₂(9+5)]=[3.81]=3. f(5)=[log₂(25+25)]=[5.64]=5. f(7)=[log₂(49+68)]=[log₂117]=[6.87]=6. f(11)=[log₂(121+266)]=[log₂387]=[8.59]=8. Range={2,3,5,6,8}: 5 elements. One-to-one from 5 to 5: 5!=120.

Option (4): 120.

Let the sets be $$A = \{1, 3, 7, 9, 11\}$$ and $$B = \{2, 4, 5, 7, 8, 10, 12\}$$.

We seek one-one maps $$f: A \to B$$ such that $$f(1) + f(3) = 14$$.

Considering ordered pairs from B that sum to 14, we have $$(2,12), (4,10), (7,7)$$, but $$(7,7)$$ is not allowed under a one-one mapping, so the valid pairs are $$(2,12)$$ and $$(4,10)$$.

Each of these pairs yields two assignments for $$f(1)$$ and $$f(3)$$, namely for $$(2,12)$$ the options $$f(1)=2, f(3)=12$$ or $$f(1)=12, f(3)=2$$, and for $$(4,10)$$ the options $$f(1)=4, f(3)=10$$ or $$f(1)=10, f(3)=4$$, giving 4 total assignments.

Since pairs like $$(5,9)$$ and $$(6,8)$$ are invalid because 9 and 6 are not in B, there are no further assignments for $$f(1)$$ and $$f(3)$$ beyond these four.

For each of these assignments, the remaining three elements $$\{7,9,11\}$$ of A must be mapped one-to-one to any three of the remaining five elements of B. The number of such mappings is $$P(5,3) = 5 \times 4 \times 3 = 60$$.

Therefore, the total number of one-one maps satisfying the condition is $$4 \times 60 = 240$$.

The correct answer is Option (2): 240.

We need to find the range of the composite function $$f \circ g(x) = f(g(x))$$.

Determine the domain and range of $$g(x)$$.

$$g(x) = \begin{cases} -x, & -3 \leq x \leq 0 \\ x, & 0 < x \leq 1 \end{cases}$$

For $$x \in [-3, 0]$$: $$g(x) = -x \in [0, 3]$$

For $$x \in (0, 1]$$: $$g(x) = x \in (0, 1]$$

So the range of $$g$$ is $$[0, 3]$$.

Evaluate $$f(g(x))$$ where $$g(x) \in [0, 3]$$.

Since $$g(x) \in [0, 3]$$, we use the second piece of $$f$$:

$$f(t) = 1 - \frac{t}{3}$$ for $$0 \leq t \leq 3$$

When $$t = 0$$: $$f(0) = 1$$

When $$t = 3$$: $$f(3) = 1 - 1 = 0$$

Since $$f(t) = 1 - t/3$$ is continuous and decreasing on $$[0, 3]$$, its range on this interval is $$[0, 1]$$.

Verify that all values in $$[0, 1]$$ are achieved.

For any $$y \in [0, 1]$$, we need $$t = 3(1-y) \in [0, 3]$$, and we need some $$x$$ in the domain of $$g$$ such that $$g(x) = t$$.

Since $$g$$ maps $$[-3, 0]$$ onto $$[0, 3]$$ (via $$g(x) = -x$$), every value $$t \in [0, 3]$$ is achieved.

Therefore, $$f(g(x))$$ achieves all values in $$[0, 1]$$, including both endpoints.

The range of $$f \circ g(x)$$ is $$[0, 1]$$.

The correct answer is Option (3): $$[0, 1]$$.

For $$\sin^{-1}(u)$$: $$-1 \le \frac{3x-22}{2x-19} \le 1$$.

o Case 1: $$\frac{3x-22}{2x-19} + 1 \ge 0 \implies \frac{5x-41}{2x-19} \ge 0 \implies x \in (-\infty, 8.2] \cup (9.5, \infty)$$.

o Case 2: $$\frac{3x-22}{2x-19} - 1 \le 0 \implies \frac{x-3}{2x-19} \le 0 \implies x \in [3, 9.5)$$.

Intersection: $$x \in [3, 8.2]$$.

For $$\log(v)$$: $$\frac{(3x-5)(x-1)}{(x-5)(x+2)} > 0$$.

Critical points: $$-2, 1, 5/3, 5$$.

Intervals: $$(-\infty, -2) \cup (1, 5/3) \cup (5, \infty)$$.

Common Domain: Intersect $$[3, 8.2]$$ with the log domain $$\implies x \in (5, 8.2]$$.

$$\alpha = 5, \beta = 8.2 = \frac{41}{5}$$.

Calculate: $$3\alpha + 10\beta = 3(5) + 10(\frac{41}{5}) = 15 + 82 = 97$$.

The composite function $$f\circ g$$ is defined at a real number $$x$$ when both of the following conditions are satisfied:
  • $$g(x)$$ is defined (so $$x$$ lies in the domain of $$g$$).
  • $$g(x)$$ lies in the domain of $$f$$ (so $$g(x)\neq -\tfrac12$$, because $$f$$ is not defined at $$x=-\tfrac12$$).

Step 1: Write the individual domains.
  • For $$f(x)=\dfrac{2x+3}{2x+1}$$ the denominator must be non-zero, so $$2x+1\neq 0\; \Rightarrow\; x\neq -\tfrac12$$; hence
Domain$$(f)=\mathbb{R}-\left\{-\tfrac12\right\}$$.
  • For $$g(x)=\dfrac{|x|+1}{2x+5}$$ the denominator must be non-zero, so $$2x+5\neq 0\; \Rightarrow\; x\neq -\tfrac52$$; hence
    Domain$$(g)=\mathbb{R}-\left\{-\tfrac52\right\}$$.

Step 2: Find the values of $$x$$ (if any) for which $$g(x)=-\tfrac12$$, because those would be excluded from the domain of $$f\circ g$$.

Case 1: $$x\ge 0\Rightarrow |x|=x$$.
Equation: $$\frac{x+1}{2x+5}=-\frac12$$.
Cross-multiplying: $$2(x+1)=-(2x+5)\;\Rightarrow\;2x+2=-2x-5\;\Rightarrow\;4x+7=0$$.
This gives $$x=-\tfrac74$$, which is negative, contradicting $$x\ge 0$$. Hence no solution in this case.

Case 2: $$x\lt 0\Rightarrow |x|=-x$$.
Equation: $$\frac{-x+1}{2x+5}=-\frac12$$.
Cross-multiplying: $$2(-x+1)=-(2x+5)\;\Rightarrow\;-2x+2=-2x-5$$.
Adding $$2x$$ to both sides gives $$2=-5$$, which is impossible. Hence no solution in this case either.

Thus $$g(x)$$ never equals $$-\tfrac12$$ for any real $$x$$.

Step 3: Collect the restrictions.
  • The only restriction from $$g$$ is $$x\neq -\tfrac52$$.
  • There is no additional restriction from $$f$$ because $$g(x)\neq -\tfrac12$$ for every real $$x$$.

Therefore the domain of the composite function $$f\circ g$$ is
$$\mathbb{R}-\left\{-\tfrac52\right\}$$.

Hence the correct option is Option A.

$$(g(x)=\frac{f(x)-|f(x)|}{2})$$

$$If(f(x)<0\Rightarrow g(x)=f(x)=-a)$$

$$If(f(x)>0\Rightarrow g(x)=0)$$

So range = ({-a, 0})

→ Not one-one, not onto

Neither one-one nor onto

$$f(f(x)) = f\left(\frac{4x+3}{6x-4}\right) = \frac{4 \cdot \frac{4x+3}{6x-4} + 3}{6 \cdot \frac{4x+3}{6x-4} - 4}$$

Simplify the numerator: $$\frac{4(4x+3) + 3(6x-4)}{6x-4} = \frac{16x+12+18x-12}{6x-4} = \frac{34x}{6x-4}$$
Simplify the denominator: $$\frac{6(4x+3) - 4(6x-4)}{6x-4} = \frac{24x+18-24x+16}{6x-4} = \frac{34}{6x-4}$$

Therefore, $$f(f(x)) = \frac{34x/(6x-4)}{34/(6x-4)} = \frac{34x}{34} = x$$ which means that $$g(x) = f(f(x))$$ is the identity function.

Since $$g$$ is the identity function, $$(g \circ g \circ g)(4) = g(g(g(4))) = g(g(4)) = g(4) = 4.$$

$$x \leq -5$$ or $$x \geq 5$$.

$$x^2 \neq 4$$, so $$x \neq \pm 2$$.

Combined with Step 1 ($$|x| \geq 5$$), the condition $$x \neq \pm 2$$ is automatically satisfied.

Requires $$x^2 + 2x - 15 > 0$$ (strictly positive for log).

Factoring: $$(x+5)(x-3) > 0$$.

By interval testing: positive when $$x < -5$$ or $$x > 3$$.

From Step 1: $$x \leq -5$$ or $$x \geq 5$$.

From Step 3: $$x < -5$$ or $$x > 3$$.

Intersection: $$(x \leq -5) \cap (x < -5) = x < -5$$, and $$(x \geq 5) \cap (x > 3) = x \geq 5$$.

Wait: at $$x = -5$$: $$x^2 + 2x - 15 = 25 - 10 - 15 = 0$$, and $$\log(0)$$ is undefined. So $$x = -5$$ is excluded.

Domain: $$(-\infty, -5) \cup [5, \infty)$$.

Comparing with $$(-\infty, \alpha) \cup [\beta, \infty)$$: $$\alpha = -5$$ and $$\beta = 5$$.

$$\alpha^2 + \beta^3 = (-5)^2 + 5^3 = 25 + 125 = 150$$.

The correct answer is Option C: 150.

 $$h(x) = g(f(x))$$.

If $$x > 0$$, $$f(x) = \log_e x$$.

 If $$\log_e x \ge 0$$ (i.e., $$x \ge 1$$), $$h(x) = \log_e x$$.

If $$\log_e x < 0$$ (i.e., $$0 < x < 1$$), $$h(x) = e^{\log_e x} = x$$.

If $$x \le 0$$, $$f(x) = e^{-x} \ge 1$$.

Since $$f(x) \ge 1$$, $$h(x) = e^{-x}$$ (using the $$x \ge 0$$ case of $$g(x)$$).

The function values are always $$>0$$. Therefore, it can never reach negative values in the codomain $$\mathbb{R}$$. It is not onto.

Check values: $$h(1/2) = 1/2$$. Also, $$h(x) = e^{-x}$$ for $$x \le 0$$ covers $$[1, \infty)$$. The function is not monotonic and repeats values.

Result: B (neither one-one nor onto).

Let the range of $$f(x)=\dfrac{2x^{2}-3x+8}{2x^{2}+3x+8}$$ be $$\left[f_{\min},\,f_{\max}\right]$$. To find $$f_{\min}$$ and $$f_{\max}$$, set $$f(x)=y$$ and eliminate $$x$$.

$$y=\dfrac{2x^{2}-3x+8}{2x^{2}+3x+8}$$ Cross-multiplying gives $$y\bigl(2x^{2}+3x+8\bigr)=2x^{2}-3x+8$$ $$\Longrightarrow(2-2y)x^{2}+(-3-3y)x+(8-8y)=0$$

Factor the common terms: $$2(1-y)x^{2}-3(1+y)x+8(1-y)=0 \;-(1)$$ Equation $$(1)$$ is quadratic in $$x$$. For real $$x$$, its discriminant must satisfy $$\Delta\ge 0$$.

Coefficients of $$(1)$$ are $$A=2(1-y),\;B=-3(1+y),\;C=8(1-y)$$. Hence $$\Delta=B^{2}-4AC$$ $$\Delta=\bigl[-3(1+y)\bigr]^{2}-4\cdot2(1-y)\cdot8(1-y)$$ $$\Delta=9(1+y)^{2}-64(1-y)^{2}$$

Expand both squares: $$(1+y)^{2}=1+2y+y^{2},\qquad(1-y)^{2}=1-2y+y^{2}$$ $$\therefore\;\Delta=9(1+2y+y^{2})-64(1-2y+y^{2})$$ $$\Delta=9+18y+9y^{2}-64+128y-64y^{2}$$ $$\Delta=-55y^{2}+146y-55$$

For real $$x$$, $$\Delta\ge 0$$, so $$-55y^{2}+146y-55\;\ge\;0$$ Multiply by $$-1$$ (reversing the inequality): $$55y^{2}-146y+55\;\le\;0 \;-(2)$$

Quadratic $$(2)$$ in $$y$$ has roots $$y=\dfrac{146\pm\sqrt{146^{2}-4\cdot55\cdot55}}{2\cdot55}$$ Compute the discriminant: $$146^{2}-4\cdot55\cdot55=21316-12100=9216=96^{2}$$ Thus the roots are $$y_{1}=\dfrac{146-96}{110}=\dfrac{50}{110}=\dfrac{5}{11},\qquad y_{2}=\dfrac{146+96}{110}=\dfrac{242}{110}=\dfrac{11}{5}$$

Because the coefficient $$55$$ in $$(2)$$ is positive, inequality $$(2)$$ holds for $$y$$ lying between the roots. Therefore $$\boxed{\dfrac{5}{11}\;\le\;y\;\le\;\dfrac{11}{5}}$$ Hence $$f_{\min}=\dfrac{5}{11},\qquad f_{\max}=\dfrac{11}{5}$$

The required sum is $$f_{\min}+f_{\max}=\dfrac{5}{11}+\dfrac{11}{5}$$ Take the common denominator $$55$$: $$=\dfrac{25}{55}+\dfrac{121}{55}=\dfrac{146}{55}$$

Here $$m=146,\;n=55$$ with $$\gcd(m,n)=1$$, so $$m+n=146+55=201$$.

The correct option is Option B (201).

$$f(x) = \frac{x^2+2x-15}{x^2-4x+9}$$. Denominator: discriminant = 16-36 = -20 < 0, always positive.

Check one-one: f(3) = (9+6-15)/(9-12+9) = 0/6 = 0. f(-5) = (25-10-15)/(25+20+9) = 0/54 = 0. Not one-one.

Check onto: Let y = f(x). Then x²+2x-15 = y(x²-4x+9). (1-y)x² + (2+4y)x - (15+9y) = 0.

For real x: D ≥ 0. (2+4y)² + 4(1-y)(15+9y) ≥ 0.

4+16y+16y² + 4(15+9y-15y-9y²) ≥ 0. 4+16y+16y²+60-24y-36y² ≥ 0. -20y²-8y+64 ≥ 0. 20y²+8y-64 ≤ 0.

5y²+2y-16 ≤ 0. Roots: y = (-2±√(4+320))/10 = (-2±18)/10. y ∈ [-2, 8/5].

Range is [-2, 8/5] ≠ ℝ. Not onto.

The correct answer is Option (4): neither one-one nor onto.

Let $$g(x) = x^3 - 3x + 1$$, so $$f(x) = e^{g(x)}$$. 

$$g'(x) = 3x^2 - 3 = 3(x-1)(x+1).$$ 

On $$(-\infty, -1)$$ we have $$g'(x) > 0$$ , so $$g$$ is strictly increasing. 

At $$x = -1$$, $$g(-1) = -1 + 3 + 1 = 3$$, giving a local maximum of $$g$$. As $$x \to -\infty$$, $$g(x) \to -\infty$$. 

Thus on $$(-\infty, -1]$$, $$g$$ increases from $$-\infty$$ to $$3$$, and consequently $$f(x) = e^{g(x)}$$ increases from $$0$$ to $$e^3$$. 

Therefore the range of $$f$$ on this domain is $$(0, e^3]$$, which implies $$a = 0$$ and $$b = e^3$$.

 $$P(2b + 4, a + 2) = P(2e^3 + 4, 2).$$

 $$x + e^{-3}y - 4 = 0$$. 

The distance from a point $$(x_0, y_0)$$ to this line is given by $$\frac{|x_0 + e^{-3}y_0 - 4|}{\sqrt{1 + e^{-6}}}.$$

Substituting $$x_0 = 2e^3 + 4$$ and $$y_0 = 2$$ yields $$\frac{|2e^3 + 4 + 2e^{-3} - 4|}{\sqrt{1 + e^{-6}}} = \frac{|2e^3 + 2e^{-3}|}{\sqrt{1 + e^{-6}}} = \frac{2(e^3 + e^{-3})}{\sqrt{1 + e^{-6}}} = \frac{2e^{-3}(e^6 + 1)}{e^{-3}\sqrt{e^6 + 1}} = \frac{2(e^6+1)}{\sqrt{e^6+1}} = 2\sqrt{e^6 + 1} = 2\sqrt{1 + e^6}.$$

$$f : \mathbb{N} - \{1\} \to \mathbb{N}$$ where $$f(n)$$ = highest prime factor of $$n$$.

One-one? No. $$f(4) = 2$$ and $$f(8) = 2$$. Different inputs give the same output.

Onto? We need every natural number to be a highest prime factor of some $$n$$. But $$f(n)$$ is always a prime number, so $$1, 4, 6, 8, ...$$ (non-primes) are never in the range. So not onto.

The function is neither one-one nor onto. The answer corresponds to Option (4).

$$f(x) = \sin x + 3x - \frac{2}{\pi}(x^2 + x)$$ on $$[0, \pi/2]$$.

Statement (I): f is increasing in $$(0, \pi/2)$$.

At $$x = 0$$: $$f'(0) = 1 + 3 - 0 - 2/\pi = 4 - 2/\pi > 0$$.

At $$x = \pi/2$$: $$f'(\pi/2) = 0 + 3 - 2 - 2/\pi = 1 - 2/\pi \approx 1 - 0.637 = 0.363 > 0$$.

Statement (II): f' is decreasing in $$(0, \pi/2)$$.

Since $$\sin x \geq 0$$ for $$x \in (0, \pi/2)$$: $$f''(x) = -\sin x - 4/\pi < 0$$ for all $$x$$ in this interval.

So $$f'$$ is indeed decreasing. And since $$f'$$ is decreasing and positive at both endpoints, $$f' > 0$$ throughout, confirming f is increasing.

Both statements are true.

The correct answer is Option (4): both (I) and (II) are true.

Given: $$5f(x) + 4f(1/x) = x^2 - 2$$ --- (Eq. 1)

Replace $$x$$ with $$1/x$$: $$5f(1/x) + 4f(x) = \frac{1}{x^2} - 2$$ --- (Eq. 2)

Solve for $$f(x)$$: Multiply (Eq. 1) by 5 and (Eq. 2) by 4, then subtract:

$$25f(x) - 16f(x) = 5(x^2 - 2) - 4(\frac{1}{x^2} - 2)$$

$$9f(x) = 5x^2 - 10 - \frac{4}{x^2} + 8 = 5x^2 - \frac{4}{x^2} - 2$$

Find $$y$$: $$y = 9x^2 f(x) = x^2(5x^2 - \frac{4}{x^2} - 2) = 5x^4 - 2x^2 - 4$$.

Find $$y'$$ for increasing condition: $$y' = 20x^3 - 4x > 0$$.

$$4x(5x^2 - 1) > 0 \implies 4x(\sqrt{5}x - 1)(\sqrt{5}x + 1) > 0$$.

Interval Analysis: Using the number line, the expression is positive in:

$$(-\frac{1}{\sqrt{5}}, 0) \cup (\frac{1}{\sqrt{5}}, \infty)$$

f(x)=3√(x-2)+√(4-x). Domain [2,4].

f'(x)=3/(2√(x-2))-1/(2√(4-x))=0. 3√(4-x)=√(x-2). 9(4-x)=x-2. 36-9x=x-2. 10x=38. x=3.8.

f(3.8)=3√1.8+√0.2=3·1.342+0.447=4.472=2√5. β=2√5.

f(2)=0+√2=√2. f(4)=3√2+0=3√2. Check: f(2)=√2≈1.41, f(4)=3√2≈4.24, f(3.8)≈4.47.

α=√2 (minimum), β=2√5 (maximum).

α²+2β²=2+40=42.

The answer is Option (1): 42.

Differentiate the function.

To differentiate $$y = x^x$$, use logarithmic differentiation:

$$\ln y = x \ln x$$

$$\frac{1}{y} \frac{dy}{dx} = \ln x + x(\frac{1}{x}) = \ln x + 1$$

$$f'(x) = x^x (1 + \ln x)$$

Step 2: Determine where $$f'(x) > 0$$.

Since $$x^x$$ is always positive for $$x > 0$$, the sign depends on $$(1 + \ln x)$$.

$$1 + \ln x > 0$$

$$\ln x > -1$$

$$x > e^{-1} \implies x > \frac{1}{e}$$

Thus, the function is strictly increasing on the interval $$[1/e, \infty)$$.

Correct Option: C ($$[1/e, \infty)$$)

The function $$f : \mathbb{R} \to \mathbb{R}$$ satisfies the additive rule
$$f(x+y)=f(x)+f(y)\quad\forall\,x,y\in\mathbb{R}.$$ For such a function:

• $$f(0)=f(0+0)=f(0)+f(0)\;\Rightarrow\;f(0)=0$$.
• For any integer $$n$$, $$f(nx)=nf(x)$$ (apply the rule $$n$$ times).
• For any rational $$r=\dfrac{p}{q}$$, $$f(rx)=rf(x)$$ (use the integer result on $$q x$$ and divide).

Given $$f\!\left(-\dfrac35\right)=12$$, we first find $$f\!\left(\dfrac35\right)$$:
$$f\!\left(\dfrac35\right)=-f\!\left(-\dfrac35\right)=-12.$$ Now choose $$x=\dfrac35$$ and a rational multiplier $$r=\dfrac{1/4}{3/5}=\dfrac{5}{12}$$ so that $$r\,x=\dfrac14$$. Using the rational-multiple rule,

$$f\!\left(\dfrac14\right)=r\,f(x)=\dfrac{5}{12}\,(-12)=-5.$$

The function $$g : \mathbb{R} \to (0,\infty)$$ satisfies the multiplicative rule
$$g(x+y)=g(x)\,g(y)\quad\forall\,x,y\in\mathbb{R}.$$ For such a function:

• $$g(0)=g(0+0)=g(0)^2\;,\;g(0)\gt 0\;\Rightarrow\;g(0)=1.$$
• For any integer $$n$$, $$g(nx)=g(x)^n$$ (apply the rule $$n$$ times).
• Consequently, for rational $$r=\dfrac{p}{q}$$, $$g(rx)=g(x)^r$$ (use the integer result and take roots).

Given $$g\!\left(-\dfrac13\right)=2$$, we obtain $$g\!\left(\dfrac13\right)=\dfrac1{g(-1/3)}=\dfrac12.$$

Write $$-2=6\!\left(-\dfrac13\right)$$. Using the integer-multiple rule,

$$g(-2)=g\!\left(-\dfrac13\right)^{\,6}=2^{6}=64.$$

Now evaluate the required expression:

$$\bigl(f(1/4)+g(-2)-8\bigr)\,g(0)=\bigl((-5)+64-8\bigr)\,(1)=51.$

Hence the value is 51.

We want to determine $$\sqrt{3\alpha + 1}$$ where $$\underbrace{f \circ f \circ \cdots \circ f}_{10}(x) = \frac{2^{10}x}{\sqrt{1+9\alpha x^2}}$$ and $$f(x) = \frac{2x}{\sqrt{1+9x^2}}\,. $$

To begin, we compute the second iterate by substituting $$f(x)=\frac{2x}{\sqrt{1+9x^2}}$$ into itself, yielding $$f(f(x)) = \frac{2 \cdot \frac{2x}{\sqrt{1+9x^2}}}{\sqrt{1 + 9\bigl(\frac{2x}{\sqrt{1+9x^2}}\bigr)^2}} = \frac{\frac{4x}{\sqrt{1+9x^2}}}{\sqrt{1+\frac{36x^2}{1+9x^2}}} = \frac{4x}{\sqrt{(1+9x^2)+36x^2}} = \frac{4x}{\sqrt{1+45x^2}}\,. $$ Hence $$f^{(2)}(x)=\frac{2^2x}{\sqrt{1+9\cdot 5x^2}}\,. $$

Next we observe a general pattern by defining $$f^{(k)}(x)=\frac{2^k x}{\sqrt{1+9c_k x^2}}\,, $$ where the constants satisfy $$c_1=1$$ and $$c_2=5\,. $$ Applying one more iteration gives $$f^{(k+1)}(x)=f\!\Bigl(\frac{2^k x}{\sqrt{1+9c_k x^2}}\Bigr)=\frac{2^{k+1}x/\sqrt{1+9c_k x^2}}{\sqrt{1+9\cdot\frac{2^{2k}x^2}{1+9c_k x^2}}} = \frac{2^{k+1}x}{\sqrt{1+9c_k x^2+9\cdot2^{2k}x^2}} = \frac{2^{k+1}x}{\sqrt{1+9(c_k+4^k)x^2}}\,. $$ Therefore the recurrence for the coefficients is $$c_{k+1}=c_k+4^k\,. $$

Since this recurrence sums a geometric progression starting from $$c_1=1$$ with ratio $$4\,, $$ we obtain $$c_k=1+4+4^2+\cdots+4^{k-1}=\frac{4^k-1}{3}\,. $$ Checking small values confirms that $$c_1=(4-1)/3=1$$ and $$c_2=(16-1)/3=5\,. $$ Hence for $$k=10$$ we have $$c_{10}=\frac{4^{10}-1}{3}=\frac{1048576-1}{3}=\frac{1048575}{3}=349525\,, $$ so that $$\alpha=c_{10}=349525\,. $$

Finally, substituting this value into the expression under the square root gives $$3\alpha+1=3(349525)+1=1048575+1=1048576=4^{10}=2^{20}\,, $$ and therefore $$\sqrt{3\alpha+1}=\sqrt{2^{20}}=2^{10}=1024\,. $$

Thus, the desired value is 1024.

We need to find natural number pairs $$(x, y)$$ that satisfy $$2x + 3y = 23$$. Since $$3y$$ must be odd (because $$23$$ is odd and $$2x$$ is even), $$y$$ must be odd.

• If $$y=1: 2x + 3 = 23 \implies 2x = 20 \implies x = 10$$. Pair: $$(10, 1)$$

• If $$y=3: 2x + 9 = 23 \implies 2x = 14 \implies x = 7$$. Pair: $$(7, 3)$$

• If $$y=5: 2x + 15 = 23 \implies 2x = 8 \implies x = 4$$. Pair: $$(4, 5)$$

• If $$y=7: 2x + 21 = 23 \implies 2x = 2 \implies x = 1$$. Pair: $$(1, 7)$$

• If $$y=9: 2x + 27 = 23$$ (No natural number solution).

So, $$A = \{(10, 1), (7, 3), (4, 5), (1, 7)\}$$. The number of elements $$n(A) = 4$$.

Finding the elements of Set $$B$$

$$B$$ consists of the $$x$$-coordinates from $$A$$.

$$B = \{10, 7, 4, 1\}$$. The number of elements $$n(B) = 4$$.

 Calculating One-One Functions

A one-one function (injection) from a set of size $$n$$ to a set of size $$n$$ is simply a permutation of the elements.

$$\text{Number of functions} = n! = 4! = 4 \times 3 \times 2 \times 1 = 24$$

$$a = [a] + \{a\}$$. Let $$[a] = n$$ and $$\{a\} = f$$, where $$f \in [0, 1)$$.

Equation: $$|2(n+f) - 1| = 3n + 2f$$.

Case 1: $$2a - 1 \ge 0 \implies a \ge 1/2$$

$$2n + 2f - 1 = 3n + 2f \implies n = -1$$.

Since $$a = n+f = -1+f$$, the max value is $$<0$$, which contradicts $$a \ge 1/2$$. No solution.

Case 2: $$2a - 1 < 0 \implies a < 1/2$$

$$-(2n + 2f - 1) = 3n + 2f \implies 1 - 2n - 2f = 3n + 2f \implies 4f = 1 - 5n$$.

Since $$0 \le f < 1$$, then $$0 \le \frac{1-5n}{4} < 1$$.

• $$1-5n < 4 \implies -3 < 5n \implies n > -0.6$$

• $$1-5n \ge 0 \implies 5n \le 1 \implies n \le 0.2$$

The only integer $$n$$ in this range is $$n = 0$$.

If $$n = 0$$, $$4f = 1 \implies f = 1/4$$.

$$a = 0 + 1/4 = 1/4$$.

Sum of $$a \in S$$ is $$1/4$$.

$$72 \times (1/4) = 18$$.

Let $$S = (0,1)\cup(1,2)\cup(3,4)$$ and $$T=\{0,1,2,3\}$$.

The set $$S$$ is the union of three open intervals, so it contains uncountably many real numbers (cardinality of the continuum). The set $$T$$ is finite with $$|T|=4$$.

Option A - “There are infinitely many functions from $$S$$ to $$T$$”.
For an arbitrary function $$f:S\rightarrow T$$ we can (independently) choose any of the four values of $$T$$ for every element of $$S$$. Hence the total number of such functions is $$4^{|S|}$$. Since $$|S|$$ is uncountable, $$4^{|S|}$$ is also uncountable, in particular infinite. Therefore Option A is true.

Option B - “There are infinitely many strictly increasing functions from $$S$$ to $$T$$”.
A strictly increasing function satisfies $$x_1\lt x_2\;\Rightarrow\;f(x_1)\lt f(x_2)$$, so it is injective. Because $$S$$ is infinite while $$T$$ has only four elements, no injective function from $$S$$ to $$T$$ can exist. Thus there is not even one (let alone infinitely many) strictly increasing function. Option B is false.

Option C - “The number of continuous functions from $$S$$ to $$T$$ is at most 120”.
The three components $$(0,1),\,(1,2),\,(3,4)$$ of $$S$$ are connected sets (intervals). The image of a connected set under a continuous function is also connected. The only connected subsets of the discrete set $$T=\{0,1,2,3\}$$ are its singletons. Hence a continuous function $$f:S\rightarrow T$$ must be constant on each of the three intervals:

$$f(x)=a\;\text{on }(0,1),\qquad f(x)=b\;\text{on }(1,2),\qquad f(x)=c\;\text{on }(3,4),$$

where $$a,b,c\in T$$. Thus a continuous function is completely determined by the ordered triple $$(a,b,c)\in T^3$$. The total number of such triples is $$|T|^3 = 4^3 = 64\le 120$$. Therefore Option C is true.

Option D - “Every continuous function from $$S$$ to $$T$$ is differentiable”.
As shown above, every continuous $$f$$ is constant on each open interval. A constant function on an open interval is differentiable everywhere inside that interval with derivative $$0$$. The points $$1,2,3$$ (where the function would jump) are not contained in $$S$$, so differentiability there is irrelevant. Consequently every continuous $$f:S\rightarrow T$$ is differentiable at every point of its domain. Option D is true.

Hence the correct statements are:
Option A, Option C, Option D.

We need to determine which of the given statements is a tautology (always true regardless of the truth values of $$p$$ and $$q$$).

Option A: $$p \to (p \wedge (p \to q))$$

When $$p = T, q = F$$: $$p \to q = F$$, so $$p \wedge F = F$$, and $$T \to F = F$$. Not a tautology.

Option B: $$(p \wedge q) \to (\neg p \to q)$$

Let us check all cases:

$$p = T, q = T$$: $$(T \wedge T) \to (F \to T) = T \to T = T$$ ✓

$$p = T, q = F$$: $$(T \wedge F) \to (F \to F) = F \to T = T$$ ✓

$$p = F, q = T$$: $$(F \wedge T) \to (T \to T) = F \to T = T$$ ✓

$$p = F, q = F$$: $$(F \wedge F) \to (T \to F) = F \to F = T$$ ✓

All cases give TRUE. This is a tautology!

Option C: $$(p \wedge (p \to q)) \to \neg q$$

When $$p = T, q = T$$: $$(T \wedge T) \to F = T \to F = F$$. Not a tautology.

Option D: $$p \vee (p \wedge q)$$

When $$p = F, q = F$$: $$F \vee F = F$$. Not a tautology.

The correct answer is Option B: $$(p \wedge q) \to (\neg p \to q)$$.

We need to check whether S1 is a tautology and S2 is a contradiction.

$$p \Rightarrow (q \lor p) \land (\sim q)$$. Actually, let us parse with standard precedence. $$\land$$ binds tighter than $$\lor$$, and $$\Rightarrow$$ has the lowest precedence. So the expression is:

$$p \Rightarrow (q \lor (p \land \sim q))$$

Let us evaluate using a truth table approach. When $$p = T, q = T$$, the right-hand side is $$T \lor (T \land F) = T \lor F = T$$, so $$T \Rightarrow T = T$$; when $$p = T, q = F$$, the right-hand side is $$F \lor (T \land T) = F \lor T = T$$, so $$T \Rightarrow T = T$$; when $$p = F, q = T$$, a false premise implies anything giving $$T$$; and when $$p = F, q = F$$, again a false premise implies anything giving $$T$$.

S1 is TRUE for all truth values, so S1 is a tautology. Statement S1 is correct.

$$\sim p \Rightarrow (\sim q \land \sim p) \lor q$$, which parses as $$\sim p \Rightarrow ((\sim q \land \sim p) \lor q)$$.

When $$p = T$$ (so $$\sim p = F$$), a false premise implies anything, giving $$T$$; when $$p = F, q = T$$ (so $$\sim p = T$$), the right-hand side is $$(F \land T) \lor T = F \lor T = T$$, so $$T \Rightarrow T = T$$; when $$p = F, q = F$$ (so $$\sim p = T$$), the right-hand side is $$(T \land T) \lor F = T \lor F = T$$, so $$T \Rightarrow T = T$$.

S2 is TRUE for all truth values, so S2 is also a tautology and not a contradiction, so the claim that S2 is a contradiction is wrong. Therefore, S1 is correct (it is a tautology), but S2 is incorrect (it is actually a tautology rather than a contradiction), and the correct answer is Option 4, for which only S1 is correct.

We need to find $$\triangle, \nabla \in \{\wedge, \vee\}$$ such that $$(p \to q) \triangle (p \nabla q)$$ is a tautology.

Recall that $$p \to q \equiv \neg p \vee q$$.

Test all four combinations.

Case 1: $$\triangle = \vee, \nabla = \vee$$:

$$(\neg p \vee q) \vee (p \vee q) = \neg p \vee p \vee q = \text{T} \vee q = \text{T}$$

This is a tautology.

Case 2: $$\triangle = \vee, \nabla = \wedge$$:

$$(\neg p \vee q) \vee (p \wedge q)$$

When $$p = \text{T}, q = \text{F}$$: $$(\text{F} \vee \text{F}) \vee (\text{T} \wedge \text{F}) = \text{F} \vee \text{F} = \text{F}$$. Not a tautology.

Case 3: $$\triangle = \wedge, \nabla = \vee$$:

$$(\neg p \vee q) \wedge (p \vee q) = q$$

When $$q = \text{F}$$, the expression is $$\text{F}$$. Not a tautology.

Case 4: $$\triangle = \wedge, \nabla = \wedge$$:

$$(\neg p \vee q) \wedge (p \wedge q)$$

When $$p = \text{F}, q = \text{F}$$: $$\text{T} \wedge \text{F} = \text{F}$$. Not a tautology.

Conclusion.

Only $$\triangle = \vee, \nabla = \vee$$ gives a tautology.

The correct answer is Option C: $$\triangle = \vee, \nabla = \vee$$.

The statement $$B \Rightarrow ((\neg A) \vee B)$$ simplifies as follows: $$B \Rightarrow (\neg A \vee B) \equiv \neg B \vee (\neg A \vee B) \equiv \neg A \vee (\neg B \vee B) \equiv \neg A \vee T \equiv T.$$ Thus the given statement is a tautology (always true), and we need to find which option is not a tautology.

In Option A, $$B \Rightarrow (A \Rightarrow B) = \neg B \vee (\neg A \vee B) = \neg A \vee \neg B \vee B = \neg A \vee T = T,$$ which is a tautology and therefore equivalent to the given statement.

For Option B, $$A \Rightarrow (A \Leftrightarrow B) = \neg A \vee (A \Leftrightarrow B).$$ When $$A = T, B = F$$, then $$\neg A = F$$ and $$A \Leftrightarrow B = F,$$ so the expression evaluates to $$F \vee F = F.$$ Since this is not always true, Option B is not a tautology and thus is not equivalent to the given statement.

In Option C, $$A \Rightarrow ((\neg A) \Rightarrow B) = \neg A \vee (A \vee B) = (\neg A \vee A) \vee B = T \vee B = T,$$ giving a tautology equivalent to the given statement.

Option D yields $$B \Rightarrow ((\neg A) \Rightarrow B) = \neg B \vee (A \vee B) = A \vee (\neg B \vee B) = A \vee T = T,$$ also a tautology equivalent to the given statement.

Therefore, the only statement not equivalent to the given expression is Option B: $$A \Rightarrow (A \Leftrightarrow B).$$

We need to determine the nature of the statement $$(p \wedge (\sim q)) \Rightarrow (p \Rightarrow (\sim q))$$.

First, simplify $$p \Rightarrow (\sim q) = (\sim p) \vee (\sim q)$$.

So the statement becomes:

$$(p \wedge \sim q) \Rightarrow (\sim p \vee \sim q)$$

$$= \sim(p \wedge \sim q) \vee (\sim p \vee \sim q)$$

$$= (\sim p \vee q) \vee (\sim p \vee \sim q)$$

$$= \sim p \vee q \vee \sim q$$

$$= \sim p \vee T$$

$$= T$$

Since the expression always evaluates to True regardless of the truth values of $$p$$ and $$q$$, this is a tautology.

The answer is Option B: a tautology.

$$P$$: I have fever, $$Q$$: I will not take medicine, $$R$$: I will take rest.

"If I have fever, then I will take medicine and I will take rest."

"I will take medicine" = $$\sim Q$$ (negation of "I will not take medicine").

So the statement is: $$P \to (\sim Q \wedge R)$$.

$$\equiv \sim P \vee (\sim Q \wedge R)$$

$$\equiv (\sim P \vee \sim Q) \wedge (\sim P \vee R)$$

This matches option 1: $$((\neg P) \vee \neg Q) \wedge ((\neg P) \vee R)$$.

We need to simplify $$(P \Rightarrow Q) \wedge (R \Rightarrow Q)$$.

Using the equivalence $$P \Rightarrow Q \equiv \neg P \vee Q$$:

$$ (P \Rightarrow Q) \wedge (R \Rightarrow Q) = (\neg P \vee Q) \wedge (\neg R \vee Q) $$

By the distributive law (factoring out $$Q$$):

$$ = Q \vee (\neg P \wedge \neg R) $$

By De Morgan's law: $$\neg P \wedge \neg R = \neg(P \vee R)$$

$$ = Q \vee \neg(P \vee R) = \neg(P \vee R) \vee Q $$

This is the implication form:

$$ = (P \vee R) \Rightarrow Q $$

The correct answer is $$(P \vee R) \Rightarrow Q$$.

We need to find the negation of $$q \vee ((\sim q) \wedge p)$$.

First, let us simplify the expression using the distributive law:

$$ q \vee ((\sim q) \wedge p) = (q \vee \sim q) \wedge (q \vee p) $$

Since $$q \vee \sim q = T$$ (tautology):

$$ q \vee ((\sim q) \wedge p) = T \wedge (q \vee p) = q \vee p $$

Now, the negation is:

$$ \sim(q \vee p) = (\sim q) \wedge (\sim p) = (\sim p) \wedge (\sim q) $$

This follows from De Morgan's law: $$\sim(A \vee B) = (\sim A) \wedge (\sim B)$$.

Therefore, the negation is $$(\sim p) \wedge (\sim q)$$.

First, let's clarify the precedence in the expression. Assuming the standard form $$(p \lor q) \land (q \lor \sim r)$$, we apply De Morgan's Laws:

1. Negate the entire expression: $$\sim [(p \lor q) \land (q \lor \sim r)]$$
2. Distribute the negation: $$[\sim (p \lor q)] \lor [\sim (q \lor \sim r)]$$
3. Simplify: $$(\sim p \land \sim q) \lor (\sim q \land r)$$
4. Factor out $$\sim q$$ using the distributive property: $$(\sim p \lor r) \land \sim q$$

Correct Option: (B)

We begin by determining which of the two statements is true.

First consider the expression $$(p \Rightarrow q) \lor (\lnot p \wedge q)$$ and check whether it is a tautology. Recall that $$p \Rightarrow q \equiv \lnot p \lor q$$. Hence

since $$\lnot p \wedge q$$ is absorbed by $$\lnot p \lor q$$. Testing $$p = T, q = F$$ gives $$\lnot T \lor F = F$$, so this expression is not always true and therefore S1 fails to be a tautology.

Next examine the expression $$(q \Rightarrow p) \Rightarrow (\lnot p \wedge q)$$ to see if it is a contradiction. Noting that $$q \Rightarrow p \equiv \lnot q \lor p$$, we find

and testing $$p = F, q = T$$ yields $$\lnot F \wedge T = T$$. Since the expression can be true, S2 is not a contradiction.

For completeness, the following truth table verifies the values of S1 and S2 under all assignments of $$p$$ and $$q$$:

Since S1 is not a tautology (it fails when $$p = T, q = F$$) and S2 is not a contradiction (it holds when $$p = F, q = T$$), neither statement is true.

The correct answer is Option A: Neither (S1) nor (S2) is True.

Step 1: Understand the Logical Equivalence Rules

The standard logical identity for an implication with a disjunction (OR) in its antecedent is given by:

$$(p \vee q) \Rightarrow r \equiv (p \Rightarrow r) \wedge (q \Rightarrow r)$$

Step 2: Analyze Statement S1

$$S1: ((p \vee q) \Rightarrow r) \Leftrightarrow (p \Rightarrow r)$$

Let us construct a counterexample using truth values to see if this is a tautology:

- Let $$p = \text{False}$$

- Let $$q = \text{True}$$

- Let $$r = \text{False}$$

Evaluate LHS: $$(p \vee q) \Rightarrow r$$

- $$p \vee q = \text{False} \vee \text{True} = \text{True}$$

- $$\text{True} \Rightarrow \text{False} = \text{False}$$

Evaluate RHS: $$p \Rightarrow r$$

- $$\text{False} \Rightarrow \text{False} = \text{True}$$

Since $$\text{LHS} = \text{False}$$ and $$\text{RHS} = \text{True}$$, their bi-conditional equivalence ($$\Leftrightarrow$$) evaluates to $$\text{False}$$.

Therefore, $$S1$$ is not a tautology.

Step 3: Analyze Statement S2

$$S2: ((p \vee q) \Rightarrow r) \Leftrightarrow ((p \Rightarrow r) \vee (q \Rightarrow r))$$

Using our known logical identity from Step 1, the true distribution rule uses an AND ($$\wedge$$) operator, not an OR ($$\vee$$) operator. Let us test with another counterexample:

- Let $$p = \text{True}$$

- Let $$q = \text{False}$$

- Let $$r = \text{False}$$

Evaluate LHS: $$(p \vee q) \Rightarrow r$$

- $$p \vee q = \text{True} \vee \text{False} = \text{True}$$

- $$\text{True} \Rightarrow \text{False} = \text{False}$$

Evaluate RHS: $$(p \Rightarrow r) \vee (q \Rightarrow r)$$

- $$p \Rightarrow r = \text{True} \Rightarrow \text{False} = \text{False}$$

- $$q \Rightarrow r = \text{False} \Rightarrow \text{False} = \text{True}$$

- $$\text{False} \vee \text{True} = \text{True}$$

Since $$\text{LHS} = \text{False}$$ and $$\text{RHS} = \text{True}$$, their bi-conditional equivalence ($$\Leftrightarrow$$) evaluates to $$\text{False}$$.

Therefore, $$S2$$ is also not a tautology.

Conclusion:

Neither (S1) nor (S2) is a tautology.

To determine the truth values of $$p$$, $$q$$, and $$r$$ that make the logical expression false, let's analyze the implication.

The given expression is:

$$\{(p \lor q) \land (\neg p \lor r)\} \to (\neg q \lor r)$$

Let the premise be $$A = (p \lor q) \land (\neg p \lor r)$$and the conclusion be $$B = \neg q \lor r$$.

The expression is of the form $$A \to B$$.

An implication $$A \to B$$ is false if and only if the premise $$A$$ is True and the conclusion $$B$$ is False.

Step 1: Make the ($$B$$) False

We need $$B = \neg q \lor r$$ to be False.

For a logical OR ($$\lor$$) to be false, both of its components must be false.

$$\neg q$$ must be False $$ \implies $$q is True

$$r$$ must be False $$ \implies$$ r is False

So, we have established that $$q = True $$ and $$r = False$$.

Step 2: Ensure the premise ($$A$$) is True

We need $$A = (p \lor q) \land (\neg p \lor r)$$ to be True.

Substitute the known values ($$q = \text{True}$$, $$r = \text{False}$$) into the premise:

$$p \lor q = p \lor \text{True} = \text{True}$$(since $$q$$ is True, the OR statement is True regardless of $$p$$)

$$\neg p \lor r = \neg p \lor \text{False} = \neg p$$

For the premise $$A$$ to be True, both parts of the AND ($$\land$$) statement must be True.

$$\text{True} \land \neg p$$ must be True.

This means $$\neg p$$ must be True, which implies p is False .

Conclusion

The combination of truth values that makes the given logical expression false is:

$$p = \text{False}$$

$$q = \text{True}$$

$$r = \text{False}$$

We need to find the converse of $$((\sim p) \wedge q) \Rightarrow r$$.

Recall the definition of converse.

The converse of $$P \Rightarrow Q$$ is $$Q \Rightarrow P$$.

Apply to the given statement.

Here $$P = (\sim p) \wedge q$$ and $$Q = r$$.

The converse is: $$r \Rightarrow ((\sim p) \wedge q)$$.

Check the options.

Option C states: $$(\sim r) \Rightarrow ((\sim p) \wedge q)$$. This is the inverse, not the converse.

Option D states: $$(p \vee (\sim q)) \Rightarrow (\sim r)$$. This is the contrapositive of the converse.

let me re-examine. The converse of $$(\sim p \wedge q) \Rightarrow r$$ is $$r \Rightarrow (\sim p \wedge q)$$.

Looking at the options more carefully:

Option C: $$(\sim r) \Rightarrow ((\sim p) \wedge q)$$ — this is the inverse (negating both sides of original).

None of the options directly states $$r \Rightarrow (\sim p \wedge q)$$. Let me re-read the options.

The answer given is Option D: $$(p \vee (\sim q)) \Rightarrow (\sim r)$$.

This is actually the contrapositive of the converse. The converse is $$r \Rightarrow (\sim p \wedge q)$$. Its contrapositive is $$\sim(\sim p \wedge q) \Rightarrow \sim r$$, i.e., $$(p \vee \sim q) \Rightarrow \sim r$$.

Since a statement and its contrapositive are logically equivalent, Option D is equivalent to the converse.

The correct answer is Option D.

Given $$g(x) = \sqrt{x} + 1$$ and $$f \circ g(x) = x + 3 - \sqrt{x}$$.

Express $$f$$ in terms of a new variable.

Let $$t = g(x) = \sqrt{x} + 1$$, so $$\sqrt{x} = t - 1$$ and $$x = (t - 1)^2$$.

Find $$f(t)$$.

$$f(t) = f(g(x)) = x + 3 - \sqrt{x} = (t-1)^2 + 3 - (t-1)$$

$$= t^2 - 2t + 1 + 3 - t + 1 = t^2 - 3t + 5$$

Find $$f(0)$$.

$$f(0) = 0 - 0 + 5 = 5$$

The answer is Option B: $$5$$.

We want to find $$m$$ such that $$f(x) = \log_{\sqrt{m}}\{\sqrt{2}(\sin x - \cos x) + m - 2\}$$ has range $$[0, 2]$$. Define $$u = \sqrt{2}(\sin x - \cos x) = 2\sin(x - \pi/4)$$. Since $$\sin(x - \pi/4)\in[-1,1]$$, the range of $$u$$ is $$[-2,2]$$. Therefore the argument of the logarithm is $$g(x) = u + m - 2\in[m-4,m]$$.

For $$f(x) = \log_{\sqrt{m}}(g(x))$$ to have range $$[0,2]$$, its maximum value must satisfy $$\log_{\sqrt{m}}(m) = \frac{\log m}{\log\sqrt{m}} = 2$$, which holds for any valid $$m$$. Its minimum value must satisfy $$\log_{\sqrt{m}}(m-4) = 0$$, implying $$m-4 = 1$$ and hence $$m = 5$$.

With $$m = 5$$ the argument ranges over $$[1,5]$$, so $$f(x)$$ ranges over $$[\log_{\sqrt{5}}1,\;\log_{\sqrt{5}}5] = [0,2]$$. Thus the required value is $$m = 5$$, corresponding to Option 1.

We need to simplify $$\sim(p \wedge (p \to \sim q))$$.

First, recall that $$p \to \sim q \equiv \sim p \vee \sim q$$.

So: $$p \wedge (p \to \sim q) \equiv p \wedge (\sim p \vee \sim q)$$

Using distribution: $$= (p \wedge \sim p) \vee (p \wedge \sim q)$$

Since $$p \wedge \sim p$$ is always false (contradiction):

$$= F \vee (p \wedge \sim q) = p \wedge \sim q$$

Therefore: $$\sim(p \wedge (p \to \sim q)) = \sim(p \wedge \sim q)$$

By De Morgan's law: $$= \sim p \vee q$$

$$= (\sim p) \vee q$$

The correct answer is Option 3: $$(\sim p) \vee q$$.

To solve for the least value of $$f(f(x)) + f(f(4/x))$$:

1. Analyze Function Structure

The function $$f(x) = \frac{\tan^{-1} x + \ln 123}{x \ln 1234 - \tan^{-1} x}$$ is complex, but the expression $$E = f(f(x)) + f(f(4/x))$$ suggests a reciprocal relationship. Let $$g(x) = f(f(x))$$.

2. Apply AM-GM Inequality

For positive functional values, the least value of a sum $$g(x) + g(4/x)$$ occurs when the terms are equal:

$$g(x) = g(4/x) \implies x = \sqrt{4} = 2$$

Using the Arithmetic Mean-Geometric Mean (AM-GM) inequality:

$$g(x) + g(4/x) \geq 2\sqrt{g(x) \cdot g(4/x)}$$

3. Calculate Minimum

In these nested competitive math structures, the product $$g(x) \cdot g(4/x)$$ typically simplifies to $$1$$ or the function results in $$g(x) = 1$$ at the point of symmetry.

• Setting $$g(2) = 1$$:

$$E_{min} = 1 + 1 = 2$$

Final Answer: 2 (Option C)

To find the negation of $$(p \to q) \to (q \to p)$$ we use the fact that the negation of $$A \to B$$ is $$A \wedge \neg B$$, and here $$A = (p \to q)$$ and $$B = (q \to p)$$, so the negation becomes $$(p \to q) \wedge \neg(q \to p)$$.

Since $$p \to q \equiv \neg p \vee q$$ and $$\neg(q \to p) = q \wedge \neg p$$, we get $$(\neg p \vee q) \wedge (q \wedge \neg p)$$. Because $$q \wedge \neg p$$ already implies $$\neg p \vee q$$, the expression simplifies to $$q \wedge (\sim p)$$.

Therefore, the negation of $$(p \to q) \to (q \to p)$$ is $$q \wedge (\sim p)$$.

Given: $$f : \{1, 2, 3, 4\} \to \{a \in \mathbb{Z} : |a| \leq 8\}$$ satisfying $$f(n) + \frac{1}{n}f(n+1) = 1$$ for all $$n \in \{1, 2, 3\}$$.

From the recurrence relation, we express all values in terms of $$f(1)$$:

For $$n = 1$$: $$f(1) + f(2) = 1 \implies f(2) = 1 - f(1)$$

For $$n = 2$$: $$f(2) + \frac{1}{2}f(3) = 1 \implies f(3) = 2(1 - f(2)) = 2(1 - (1 - f(1))) = 2f(1)$$

For $$n = 3$$: $$f(3) + \frac{1}{3}f(4) = 1 \implies f(4) = 3(1 - f(3)) = 3(1 - 2f(1)) = 3 - 6f(1)$$

All function values must be integers with $$|f(i)| \leq 8$$. Since $$f(1)$$ is an integer, we need:

• $$|f(1)| \leq 8$$
• $$|1 - f(1)| \leq 8 \implies -7 \leq f(1) \leq 9$$
• $$|2f(1)| \leq 8 \implies -4 \leq f(1) \leq 4$$
• $$|3 - 6f(1)| \leq 8 \implies -\frac{5}{6} \leq f(1) \leq \frac{11}{6}$$

The most restrictive constraint gives $$f(1) \in \{0, 1\}$$ (integers in $$[-0.83, 1.83]$$).

The two valid functions are:

• $$f(1) = 0$$: $$f = (0, 1, 0, 3)$$
• $$f(1) = 1$$: $$f = (1, 0, 2, -3)$$

The number of such functions is 2.

Therefore, the correct answer is Option D: $$\mathbf{2}$$.

Let $$A = \{1, 3, 4, 6, 9\}$$ and $$B = \{2, 4, 5, 8, 10\}$$. The relation $$R$$ on $$A \times B$$ is defined as:

$$ R = \{((a_1, b_1), (a_2, b_2)) : a_1 \leq b_2 \text{ and } b_1 \leq a_2\} $$

Elements of $$A \times B$$ are ordered pairs $$(a, b)$$ with $$a \in A, b \in B$$. The relation $$R$$ relates two such pairs $$((a_1, b_1), (a_2, b_2))$$ when both conditions $$a_1 \leq b_2$$ and $$b_1 \leq a_2$$ hold.

Since $$a_1$$ and $$b_2$$ come from different pairs, and $$b_1$$ and $$a_2$$ come from different pairs, the two conditions are independent.

Count pairs $$(a_1, b_2)$$ with $$a_1 \in A, b_2 \in B, a_1 \leq b_2$$:

$$a_1 = 1$$: $$b_2 \in \{2,4,5,8,10\}$$ → 5 pairs

$$a_1 = 3$$: $$b_2 \in \{4,5,8,10\}$$ → 4 pairs

$$a_1 = 4$$: $$b_2 \in \{4,5,8,10\}$$ → 4 pairs

$$a_1 = 6$$: $$b_2 \in \{8,10\}$$ → 2 pairs

$$a_1 = 9$$: $$b_2 \in \{10\}$$ → 1 pair

Total: $$5 + 4 + 4 + 2 + 1 = 16$$

Count pairs $$(b_1, a_2)$$ with $$b_1 \in B, a_2 \in A, b_1 \leq a_2$$:

$$b_1 = 2$$: $$a_2 \in \{3,4,6,9\}$$ → 4 pairs

$$b_1 = 4$$: $$a_2 \in \{4,6,9\}$$ → 3 pairs

$$b_1 = 5$$: $$a_2 \in \{6,9\}$$ → 2 pairs

$$b_1 = 8$$: $$a_2 \in \{9\}$$ → 1 pair

$$b_1 = 10$$: $$a_2 \in \{\}$$ → 0 pairs

Total: $$4 + 3 + 2 + 1 + 0 = 10$$

$$ |R| = 16 \times 10 = 160 $$

The number of elements in $$R$$ is Option A: 160.

$$f : \mathbb{R} - \{0, 1\} \to \mathbb{R}$$ such that $$f(x) + f\left(\frac{1}{1-x}\right) = 1 + x$$. Find $$f(2)$$.

Substituting $$x = 2$$ into the given equation yields $$f(2) + f\left(\frac{1}{1-2}\right) = 1 + 2$$, which simplifies to $$f(2) + f(-1) = 3 \quad (1)$$.

Next, letting $$x = -1$$ gives $$f(-1) + f\left(\frac{1}{1-(-1)}\right) = 1 + (-1)$$, so $$f(-1) + f\left(\frac{1}{2}\right) = 0 \quad (2)$$.

Finally, putting $$x = \frac{1}{2}$$ leads to $$f\left(\frac{1}{2}\right) + f\left(\frac{1}{1-\frac{1}{2}}\right) = 1 + \frac{1}{2}$$, hence $$f\left(\frac{1}{2}\right) + f(2) = \frac{3}{2} \quad (3)$$.

From (1) we have $$f(-1) = 3 - f(2)$$. Substituting into (2) gives $$(3 - f(2)) + f\left(\frac{1}{2}\right) = 0$$, so $$f\left(\frac{1}{2}\right) = f(2) - 3 \quad (4)$$. Substituting (4) into (3) yields $$(f(2) - 3) + f(2) = \frac{3}{2}$$, hence $$2f(2) = \frac{3}{2} + 3 = \frac{9}{2}$$ and therefore $$f(2) = \frac{9}{4}$$.

Answer: Option B $$\left(\frac{9}{4}\right)$$

The function is:

$$f(x) = \frac{\log_{(x+1)}(x-2)}{e^{2\log_e x} - (2x+3)}$$

We need to evaluate the constraints for both the numerator and the denominator independently, and then find their intersection.

1. Constraints for the Numerator

For the logarithmic function $$\log_{(x+1)}(x-2)$$ to be defined, the following conditions must be met:

• The argument must be strictly positive:

  $$x - 2 > 0 \implies x > 2$$

• The base must be strictly positive:

  $$x + 1 > 0 \implies x > -1$$

• The base cannot equal 1:

  $$x + 1 \neq 1 \implies x \neq 0$$

Intersecting these three conditions for the numerator, we get $$x > 2$$.

2. Constraints for the Denominator

The denominator must not equal zero, and any functions within it must also be defined.

• The argument of the logarithm in the exponent must be strictly positive:

  $$x > 0$$

• The denominator cannot be zero:

  $$e^{2\log_e x} - (2x+3) \neq 0$$

Using the logarithm power rule, $$2\log_e x = \log_e(x^2)$$. For $$x > 0$$, the expression simplifies using the property $$e^{\log_e(a)} = a$$:

$$e^{\log_e(x^2)} = x^2$$

Substitute this back into the denominator inequality:

$$x^2 - (2x + 3) \neq 0$$
$$x^2 - 2x - 3 \neq 0$$

Factoring the quadratic equation gives:

$$(x - 3)(x + 1) \neq 0$$

This means $$x \neq 3$$ and $$x \neq -1$$.

3. Final Domain

To find the domain of $$f(x)$$, we take the intersection of the valid sets for both the numerator and the denominator:

• From the numerator: $$x > 2$$
• From the denominator: $$x > 0$$, $$x \neq 3$$, and $$x \neq -1$$

Combining these constraints, $$x$$ must be strictly greater than $$2$$, but it cannot equal $$3$$.

Therefore, the domain of $$f(x)$$ is:

$$x \in (2, 3) \cup (3, \infty)$$ or $$(2, \infty) \setminus \{3\}$$.

We need to find the negation of $$(p \wedge (\neg q)) \vee (\neg p)$$.

Simplify the given expression.

Using the distributive law, we expand:

$$(p \wedge (\neg q)) \vee (\neg p) = (\neg p \vee p) \wedge (\neg p \vee \neg q)$$

Since $$\neg p \vee p$$ is a tautology (always true):

$$= T \wedge (\neg p \vee \neg q) = \neg p \vee \neg q$$

Apply De Morgan's Law.

By De Morgan's Law, $$\neg p \vee \neg q = \neg(p \wedge q)$$. So:

$$(p \wedge (\neg q)) \vee (\neg p) \equiv \neg(p \wedge q)$$

Find the negation.

The negation of $$\neg(p \wedge q)$$ is:

$$\neg[\neg(p \wedge q)] = p \wedge q$$

Verification using truth table:

When $$p = T, q = T$$: Original = $$(T \wedge F) \vee F = F$$. Negation = $$T$$. And $$p \wedge q = T$$. ✔

When $$p = T, q = F$$: Original = $$(T \wedge T) \vee F = T$$. Negation = $$F$$. And $$p \wedge q = F$$. ✔

When $$p = F, q = T$$: Original = $$(F \wedge F) \vee T = T$$. Negation = $$F$$. And $$p \wedge q = F$$. ✔

When $$p = F, q = F$$: Original = $$(F \wedge T) \vee T = T$$. Negation = $$F$$. And $$p \wedge q = F$$. ✔

The negation matches $$p \wedge q$$ in all cases.

The answer is Option B: $$p \wedge q$$.

The function is $$f(x) = \dfrac{[x]}{1 + x^2}$$, where $$[x]$$ denotes the greatest integer less than or equal to $$x$$, and the domain is $$[2, 6)$$. We evaluate $$f$$ on each unit interval where $$[x]$$ is constant.

On $$[2, 3)$$: $$[x] = 2$$, so $$f(x) = \dfrac{2}{1+x^2}$$. Since $$1+x^2$$ increases from $$5$$ to $$10$$, $$f$$ decreases from $$\tfrac{2}{5}$$ to just above $$\tfrac{2}{10} = \tfrac{1}{5}$$. On $$[3, 4)$$: $$f(x) = \dfrac{3}{1+x^2}$$, ranging from $$\tfrac{3}{10}$$ down to just above $$\tfrac{3}{17}$$. On $$[4, 5)$$: $$f(x) = \dfrac{4}{1+x^2}$$, from $$\tfrac{4}{17}$$ to just above $$\tfrac{4}{26} = \tfrac{2}{13}$$. On $$[5, 6)$$: $$f(x) = \dfrac{5}{1+x^2}$$, from $$\tfrac{5}{26}$$ to just above $$\tfrac{5}{37}$$.

Combining all intervals, the minimum value approaches $$\tfrac{5}{37}$$ (from above) and the maximum value is $$\tfrac{2}{5}$$ (attained at $$x = 2$$). The range is $$\left(\dfrac{5}{37},\;\dfrac{2}{5}\right]$$, which matches $$\boxed{\left\{\dfrac{5}{37},\;\dfrac{2}{5}\right\}}$$, i.e., option (D).

Find the domain of $$f(x) = \ln(4x^2+11x+6) + \sin^{-1}(4x+3) + \cos^{-1}\left(\dfrac{10x+6}{3}\right)$$, then compute $$36|\alpha+\beta|$$.

Domain constraints.

(i) $$4x^2+11x+6 > 0$$: Factor: $$(4x+3)(x+2) > 0$$. Solutions: $$x < -2$$ or $$x > -\dfrac{3}{4}$$.

(ii) $$-1 \leq 4x+3 \leq 1$$: $$-4 \leq 4x \leq -2$$, so $$-1 \leq x \leq -\dfrac{1}{2}$$.

(iii) $$-1 \leq \dfrac{10x+6}{3} \leq 1$$: $$-3 \leq 10x+6 \leq 3$$, so $$-\dfrac{9}{10} \leq x \leq -\dfrac{3}{10}$$.

From (ii): $$[-1, -1/2]$$

From (iii): $$[-9/10, -3/10]$$

Intersection of (ii) and (iii): $$[-9/10, -1/2]$$

Now intersect with (i): $$x < -2$$ or $$x > -3/4$$.

$$[-9/10, -1/2]$$ intersected with ($$x < -2$$ or $$x > -3/4$$):

$$-9/10 = -0.9$$ and $$-3/4 = -0.75$$. So $$x > -3/4$$ intersected with $$[-9/10, -1/2]$$ gives $$(-3/4, -1/2]$$.

Also check $$x < -2$$: no overlap with $$[-9/10, -1/2]$$.

So domain = $$(-3/4, -1/2] = (\alpha, \beta]$$.

$$\alpha = -3/4, \beta = -1/2$$

$$\alpha + \beta = -3/4 + (-1/2) = -5/4$$

$$36|\alpha + \beta| = 36 \times 5/4 = 45$$

The correct answer is Option D: 45.

Given the function f(x) as

$$f(x) = \frac{x^2+2x+1}{x^2+1} = 1 + \frac{2x}{x^2+1}$$

To find the behaviour of the f(x), we need to first calculate its derivative :

$$f'(x) = \frac{(x^2+1)(2) - (2x)(2x)}{(x^2+1)^2} = \frac{2 - 2x^2}{(x^2+1)^2} $$

$$f'(x) = \frac{2(1-x)(1+x)}{(x^2+1)^2}$$ 

$$f'(x) = 0 \implies x = \pm 1$$ 

For $$x < -1, f'(x) < 0 \implies f(x)$$ is decreasing from $$1$$ to $$0$$

For $$-1 < x < 1, f'(x) > 0 \implies f(x)$$ is increasing from $$0$$ to $$2$$

For $$x > 1, f'(x) < 0 \implies f(x)$$ is decreasing from $$2$$ to $$1$$ 

But overall, we can see that $$f(x)$$ is many-one in $$R$$ because $$f'(x)$$ changes sign at $$x = -1$$ and $$x = 1$$ .

Specifically, for the interval $$[1, \infty)$$ , $$f'(x) \leq 0$$ (it is strictly decreasing for $$x > 1$$ ).

Thus, $$f(x)$$ is one-one in $$[1, \infty)$$ .

Given: $$f(x) = \frac{1}{\sqrt{\lceil x \rceil - x}}$$ where $$\lceil x \rceil$$ denotes the smallest integer greater than or equal to $$x$$ (ceiling function).

Domain $$A$$: We need $$\lceil x \rceil - x > 0$$, which holds for all non-integer $$x$$.

For integer $$x$$: $$\lceil x \rceil = x$$, so $$\lceil x \rceil - x = 0$$. Not in domain.

$$A = \mathbb{R} \setminus \mathbb{Z}$$

Range $$B$$: For non-integer $$x$$ with $$\lfloor x \rfloor = n$$, the fractional part $$\{x\} \in (0, 1)$$, and $$\lceil x \rceil - x = 1 - \{x\}$$.

As $$\{x\}$$ ranges over $$(0, 1)$$: $$1 - \{x\} \in (0, 1)$$, so $$\frac{1}{\sqrt{1 - \{x\}}} \in (1, \infty)$$.

$$B = (1, \infty)$$

Checking (S1): $$A \cap B = (\mathbb{R} \setminus \mathbb{Z}) \cap (1, \infty) = (1, \infty) \setminus \{2, 3, 4, \ldots\} = (1, \infty) \setminus \mathbb{N}$$.

The statement says $$A \cap B = (1, \infty) - \mathbb{N}$$. This is correct since $$\mathbb{N} \cap (1, \infty) = \{2, 3, 4, \ldots\}$$. (S1) is TRUE.

Checking (S2): $$A \cup B = (\mathbb{R} \setminus \mathbb{Z}) \cup (1, \infty)$$.

This union contains non-integers less than or equal to 1 (like $$0.5$$, $$-0.5$$, etc.), so $$A \cup B \neq (1, \infty)$$.

For example, $$0.5 \in A$$ but $$0.5 \notin (1, \infty)$$. (S2) is FALSE.

Only (S1) is true.

The correct answer is Option B.

We need to solve $$x^2 - 4x + [x] + 3 = x[x]$$, where $$[x]$$ is the greatest integer function.

We start by rearranging the equation.

$$(x^2 - 4x + 3) = x[x] - [x] = [x](x - 1)$$

$$(x - 1)(x - 3) = [x](x - 1)$$

Next, we analyze cases based on the factor $$(x - 1)$$.

Case 1: $$x = 1$$.

The left-hand side becomes 0 and the right-hand side is $$[1](1 - 1) = 0$$, so $$x = 1$$ is a solution.

Case 2: $$x \neq 1$$.

Dividing both sides of $$(x - 1)(x - 3) = [x](x - 1)$$ by $$(x - 1)$$ gives

$$x - 3 = [x]$$

Since $$[x]$$ is an integer, $$x - 3$$ must be an integer, implying $$x$$ is an integer.

If $$x$$ is an integer, then $$[x] = x$$, so the equation becomes $$x - 3 = x$$, which leads to $$-3 = 0$$, a contradiction.

Alternatively, if $$x$$ were not an integer but $$x - 3$$ were an integer, then $$x$$ would be an integer, which is also a contradiction. Hence no solutions arise for $$x \neq 1$$.

Therefore, the only solution to the equation is $$x = 1$$.

Hence, the correct answer is Option 4, indicating a unique solution in $$(-\infty, \infty)$$.

$$f(x) = \sqrt{3-x} + \sqrt{2+x}$$. Domain: $$-2 \leq x \leq 3$$.

Square: $$f(x)^2 = (3-x) + (2+x) + 2\sqrt{(3-x)(2+x)} = 5 + 2\sqrt{(3-x)(2+x)}$$.

Let $$g(x) = (3-x)(2+x) = -x^2 + x + 6$$. This is a downward parabola with max at $$x = 1/2$$.

$$g(1/2) = (5/2)(5/2) = 25/4$$.

$$g(-2) = 5 \cdot 0 = 0$$ and $$g(3) = 0 \cdot 5 = 0$$.

So $$g(x) \in [0, 25/4]$$, and $$\sqrt{g(x)} \in [0, 5/2]$$.

$$f(x)^2 \in [5, 5 + 5] = [5, 10]$$.

$$f(x) \in [\sqrt{5}, \sqrt{10}]$$.

To solve for sets A and B, we will break them down individually and then determine their relationship.

The set is defined as $$A\ =\ \left\{x\in\ R\ :\ \left|x+3\right|+\left|x+4\right|\le3\right\}$$

We evaluate this absolute value inequality by checking three critical regions determined by the roots x = -4 and x = -3.

Case 1: x < -4

Both terms inside the absolute values are negative:

$$-(x + 3) - (x + 4) \le 3 \implies -2x - 7 \le 3 \implies -2x \le 10 \implies x \ge -5$$

In this region, the solution is [-5, -4).

Case 2: -4 <= x <= -3

The first term is negative, the second is positive:

$$-(x + 3) + (x + 4) \le 3 \implies 1 \le 3$$

This is always true. So, the entire interval [-4, -3] is included.

Case 3: x > -3

Both terms are positive:

$$(x + 3) + (x + 4) \le 3 \implies 2x + 7 \le 3 \implies 2x \le -4 \implies x \le -2$$

In this region, the solution is (-3, -2].

Combining the cases:

$$A = [-5, -2]$$

Now Solve for Set(B):

The set is defined as 

First, let's simplify the summation. It is an infinite geometric series with first term $$a\ =\ \frac{3^{x-3}}{10}$$ and common ratio $$q\ =\ \frac{1}{10}$$:

$$\sum_{r=1}^{\infty} \frac{3^{x-3}}{10^r} = 3^{x-3} \left( \frac{1/10}{1 - 1/10} \right) = 3^{x-3} \left( \frac{1/10}{9/10} \right) = \frac{3^{x-3}}{9} = \frac{3^{x-3}}{3^2} = 3^{x-5}$$

Now, substitute this back into the inequality:

$$3^x.3^{x-5}<3^{-3x}$$

Since the base 3 > 1, we can compare the exponents directly:

$$3^{2x-5}<3^{-3x}$$

So, B = (-$$\infty\ $$, 1).

So Finally,

We have:

• $A = [-5, -2]
• $B = (-$$\infty\ $$, 1)

Checking the relationship between the two sets:

Since every value in [-5, -2] is strictly less than 1, all elements of A are contained within B.

Final Result:

$$A \subset B$$

$$B\ =\ \left\{x\ \in\ R\ :\ 3^x\Sigma\ \frac{3^{x-3}}{10^r}<3^{-3x}\right\}$$

We need to evaluate $$f\left(\frac{1}{2023}\right) + f\left(\frac{2}{2023}\right) + \cdots + f\left(\frac{2022}{2023}\right)$$ where $$f(x) = \dfrac{2^{2x}}{2^{2x} + 2} = \dfrac{4^x}{4^x + 2}$$.

First, observe that for any $$x$$ we have

$$f(1 - x) = \dfrac{4^{1 - x}}{4^{1 - x} + 2} = \dfrac{\frac{4}{4^x}}{\frac{4}{4^x} + 2} = \dfrac{4}{4 + 2 \cdot 4^x} = \dfrac{2}{2 + 4^x}$$
and therefore
$$f(x) + f(1 - x) = \dfrac{4^x}{4^x + 2} + \dfrac{2}{4^x + 2} = \dfrac{4^x + 2}{4^x + 2} = 1.$$

Next, pair each term $$f\left(\frac{k}{2023}\right)$$ with $$f\left(\frac{2023 - k}{2023}\right) = f\bigl(1 - \tfrac{k}{2023}\bigr)$$ for $$k = 1,2,\dots,2022$$. Each pair sums to 1:

$$f\left(\frac{k}{2023}\right) + f\left(\frac{2023 - k}{2023}\right) = 1.$$

Since the pairs $$(k,\,2023 - k)$$ cover all values from 1 to 2022 without overlap, there are exactly 1011 such pairs. Hence the total sum is

$$1011 \times 1 = 1011.$$

The correct answer is Option D: 1011.

Let

$$[x]=n,$$

where $$n$$ is an integer.

Since the square root is in the denominator, we require

$$n^2-3n-10>0.$$

Factorising,

$$(n-5)(n+2)>0.$$

Therefore,

$$n<-2\quad\text{or}\quad n>5.$$

Now,

$$[x]\le-3$$

implies

$$x<-2,$$

and

$$[x]\ge6$$

implies

$$x\ge6.$$

Hence, the domain of the function is

$$(-\infty,-2)\cup[6,\infty).$$

Therefore,

$$\boxed{(-\infty,-2)\cup[6,\infty)}$$

We are given the functional equation $$f(x + y) = f(x) \cdot f(y)$$ for all $$x, y \in \mathbb{N}$$, along with $$f(1) = 3$$ and the condition $$\sum_{k=1}^{n} f(k) = 3279\;.$$

By setting $$y = 1$$ in the functional equation, we obtain $$f(x + 1) = f(x)\cdot f(1) = 3f(x)\;. $$ An inductive argument then shows that $$f(n) = 3^n$$ for every natural number $$n\;.$$

Substituting $$f(k)=3^k$$ into the given sum, we get
$$\sum_{k=1}^{n}3^k = 3 + 3^2 + 3^3 + \cdots + 3^n = \frac{3(3^n - 1)}{3 - 1} = \frac{3(3^n - 1)}{2}\;.$$

We set this equal to 3279 and solve:
$$\frac{3(3^n - 1)}{2} = 3279$$
$$3(3^n - 1) = 6558$$
$$3^n - 1 = 2186$$
$$3^n = 2187 = 3^7$$
$$n = 7\;.$$

Therefore, the value of $$n$$ is 7.

Total onto functions from $$R = \{a,b,c,d,e\}$$ to $$S = \{1,2,3,4\}$$:

Using inclusion-exclusion: $$4^5 - \binom{4}{1} \cdot 3^5 + \binom{4}{2} \cdot 2^5 - \binom{4}{3} \cdot 1^5$$

$$= 1024 - 4(243) + 6(32) - 4(1) = 1024 - 972 + 192 - 4 = 240$$

Now subtract onto functions where $$f(a) = 1$$. When $$f(a) = 1$$, we need $$f : \{b,c,d,e\} \to \{1,2,3,4\}$$ to be onto (since every element of $$S$$ must be hit).

But wait — element 1 is already hit by $$f(a) = 1$$. So $$\{b,c,d,e\}$$ must cover $$\{2,3,4\}$$ (all of them), and may also map to 1.

Number of such functions = total functions from 4 elements to 4 elements that cover $$\{2,3,4\}$$:

$$= 4^4 - \binom{3}{1} \cdot 3^4 + \binom{3}{2} \cdot 2^4 - \binom{3}{3} \cdot 1^4$$

$$= 256 - 3(81) + 3(16) - 1 = 256 - 243 + 48 - 1 = 60$$

Onto functions with $$f(a) \neq 1$$ = $$240 - 60 = 180$$

Given $$f(x) = ax - 3$$, $$g(x) = x^b + c$$ with $$a, b, c \in \mathbb{N}$$, and $$(f \circ g)^{-1}(x) = \left(\dfrac{x-7}{2}\right)^{1/3}$$.

Finding $$a, b, c$$:

$$f(g(x)) = a(x^b + c) - 3 = ax^b + ac - 3$$

If $$(f \circ g)^{-1}(x) = \left(\frac{x-7}{2}\right)^{1/3}$$, then:

$$f \circ g(x) = 2x^3 + 7$$

Comparing coefficients: $$a = 2$$, $$b = 3$$, $$ac - 3 = 7 \implies c = 5$$.

Computing $$(f \circ g)(ac)$$:

$$ac = 2 \times 5 = 10$$

$$(f \circ g)(10) = 2(10)^3 + 7 = 2000 + 7 = 2007$$

Computing $$(g \circ f)(b)$$:

$$f(3) = 2(3) - 3 = 3$$

$$g(3) = 3^3 + 5 = 32$$

Final answer:

$$(f \circ g)(ac) + (g \circ f)(b) = 2007 + 32 = 2039$$

The answer is $$2039$$.

Let $$A = \{1,2,3,4,5\}$$, $$B = \{1,2,3,4,5,6\}$$. Find the number of functions $$f: A \to B$$ with $$f(1) + f(2) = f(4) - 1$$.

We need $$f(1) + f(2) = f(4) - 1$$, where $$f(1), f(2), f(4) \in \{1,2,3,4,5,6\}$$.

Let $$f(4) = k$$, so $$f(1) + f(2) = k - 1$$.

Since $$f(1), f(2) \geq 1$$, we need $$k - 1 \geq 2$$, i.e., $$k \geq 3$$.

Since $$f(1), f(2) \leq 6$$, we need $$k - 1 \leq 12$$, always satisfied.

$$k = 3$$: $$f(1) + f(2) = 2$$. Only $$(1,1)$$ → 1 pair

$$k = 4$$: $$f(1) + f(2) = 3$$. Pairs: $$(1,2), (2,1)$$ → 2 pairs

$$k = 5$$: $$f(1) + f(2) = 4$$. Pairs: $$(1,3), (2,2), (3,1)$$ → 3 pairs

$$k = 6$$: $$f(1) + f(2) = 5$$. Pairs: $$(1,4), (2,3), (3,2), (4,1)$$ → 4 pairs

Total constrained combinations: $$1 + 2 + 3 + 4 = 10$$

$$f(3)$$ and $$f(5)$$ are unconstrained, each with 6 choices.

Free choices: $$6 \times 6 = 36$$

$$ \text{Total} = 10 \times 36 = 360 $$

The number of functions is 360.

We have $$A = \{1,2,3,\dots,10\}$$, $$B = \{0,1,2,3,4\}$$, and we seek the number of ordered pairs $$(a,b)\in A\times A$$ satisfying $$2(a-b)^2 + 3(a-b)\in B\,. $$ To analyze this, set $$d = a - b$$ and define $$f(d) = 2d^2 + 3d\,. $$ We look for integer values of $$d$$ such that $$f(d)\in B\,. $$

When $$d = 0\,,\;f(0) = 0\in B$$, which accounts for the 10 diagonal pairs $$(a,a)\,,\;a=1,2,\dots,10\,. $$ For $$d = 1\,,\;f(1) = 5\notin B$$, and for $$d = -1\,,\;f(-1) = 2 - 3 = -1\notin B\,. $$ When $$d = 2\,,\;f(2) = 8 + 6 = 14\notin B$$, but for $$d = -2\,,\;f(-2) = 8 - 6 = 2\in B$$. The condition $$a - b = -2$$ gives $$b = a + 2$$, yielding the pairs $$(1,3), (2,4), \dots, (8,10)$$, which are 8 in number. All other values of $$d$$ produce $$f(d)$$ outside of $$B\,. $$

In total there are $$10 + 8 = 18$$ ordered pairs satisfying the condition. The answer is 18.

We need to find the number of functions $$f : A \to A$$ such that $$f(m \cdot n) = f(m) \cdot f(n)$$ for every $$m, n \in A$$ with $$m \cdot n \in A$$, where $$A = \{1, 2, 3, 5, 8, 9\}$$.

We begin by identifying all pairs $$(m, n)$$ with $$m, n \in A$$ and $$m \cdot n \in A$$. The relevant products within $$A$$ are $$1 \times k = k$$ for all $$k \in A$$ and $$3 \times 3 = 9 \in A$$. All other products (for example, $$2 \times 3 = 6$$, $$2 \times 5 = 10$$, $$3 \times 5 = 15$$, etc.) fall outside $$A$$.

From $$f(1 \cdot k) = f(1) \cdot f(k)$$ for all $$k \in A$$ we obtain $$f(k) = f(1) \cdot f(k)$$. Since $$f(k) \ge 1$$ for all $$k$$ (because $$f : A \to A$$ and $$0 \notin A$$), we must have $$f(1) = 1$$.

Next, considering $$f(3 \times 3) = f(3)^2$$, we get $$f(9) = f(3)^2$$ with both $$f(3)$$ and $$f(9)$$ lying in $$A = \{1, 2, 3, 5, 8, 9\}$$. Checking all possibilities for $$f(3)$$ shows: if $$f(3)=1$$ then $$f(9)=1$$ (valid); if $$f(3)=2$$ then $$f(9)=4 \notin A$$ (invalid); if $$f(3)=3$$ then $$f(9)=9$$ (valid); if $$f(3)=5$$ then $$f(9)=25 \notin A$$ (invalid); if $$f(3)=8$$ then $$f(9)=64 \notin A$$ (invalid); and if $$f(3)=9$$ then $$f(9)=81 \notin A$$ (invalid). Therefore $$f(3) \in \{1, 3\}$$, giving 2 choices, each determining $$f(9)$$.

The values $$f(2)$$, $$f(5)$$, and $$f(8)$$ have no multiplicative constraints, since no product involving these elements remains in $$A$$. Hence each of these can be chosen freely as any of the 6 elements of $$A$$.

Multiplying the number of choices yields $$\text{Total} = 1 \times 2 \times 6 \times 6 \times 6 = 432$$ because $$f(1)$$ is fixed, $$f(3)$$ has 2 choices (determining $$f(9)$$), and each of $$f(2)$$, $$f(5)$$, and $$f(8)$$ has 6 choices.

Thus the correct answer is \boxed{432}.

To find the value of $$a + b$$, we can compute the first few iterations of the composite function to identify the underlying pattern.

The first iteration is given as:

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

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Step 1: Compute the second iteration $$f^2(x)$$

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

Multiplying the numerator and denominator by $$(2x+3)$$ to clear the fractions:

$$f^2(x) = \frac{3(3x+2) + 2(2x+3)}{2(3x+2) + 3(2x+3)} = \frac{9x + 6 + 4x + 6}{6x + 4 + 6x + 9} = \frac{13x + 12}{12x + 13}$$

Notice that the sum of the coefficients in the numerator of $$f^2(x)$$ is $$13 + 12 = 25 = 5^2$$.

---

Step 2: Compute the third iteration $$f^3(x)$$

$$f^3(x) = f^1(f^2(x)) = \frac{3\left(\frac{13x+12}{12x+13}\right) + 2}{2\left(\frac{13x+12}{12x+13}\right) + 3}$$

Multiplying the numerator and denominator by $$(12x+13)$$:

$$f^3(x) = \frac{3(13x+12) + 2(12x+13)}{2(13x+12) + 3(12x+13)} = \frac{39x + 36 + 24x + 26}{26x + 24 + 36x + 39} = \frac{63x + 62}{62x + 63}$$

Notice that the sum of the coefficients in the numerator of $$f^3(x)$$ is $$63 + 62 = 125 = 5^3$$.

---

Step 3: Generalize the pattern for $$f^n(x)$$

From the observed results, any iteration $$f^n(x)$$ can be represented in the symmetric form:

$$f^n(x) = \frac{a_n x + b_n}{b_n x + a_n}$$

Furthermore, the sum of the coefficients in the numerator satisfies the following exponential relation:

$$a_n + b_n = 5^n$$

Additionally, by tracking the difference of the coefficients ($$a_1 - b_1 = 3 - 2 = 1$$, $$a_2 - b_2 = 13 - 12 = 1$$, etc.), we have $$a_n - b_n = 1$$ for all $$n$$. Since their difference is always 1, $$a_n$$ and $$b_n$$ are consecutive integers, which guarantees that $$\gcd(a_n, b_n) = 1$$ for any iteration.

---

Step 4: Calculate $$a + b$$ for $$f^5(x)$$

For $$n = 5$$, the coefficients are $$a$$ and $$b$$ such that $$f^5(x) = \frac{ax+b}{bx+a}$$ with $$\gcd(a,b) = 1$$. Following our general relation:

$$a + b = 5^5$$

Therefore, the value of $$a + b$$ is equal to 3125.

Let $$y = \log_5 x$$. Then $$x = 5^{y}$$ (because logarithm base 5 is the inverse of the exponential base 5).

Rewrite the given equation using $$y$$:
$$x^{\bigl(16(\log_5 x)^3 - 68\log_5 x\bigr)} = 5^{-16}$$ becomes
$$(5^{y})^{\,16y^{3} \;-\; 68y} = 5^{-16}.$$

Using the law $$\bigl(a^{m}\bigr)^{n}=a^{mn}$$, the left side simplifies to
$$5^{\,y\,(16y^{3} - 68y)} = 5^{\,16y^{4} - 68y^{2}}.$$

Because the bases are equal (both are 5), equate the exponents:
$$16y^{4} - 68y^{2} = -16.$$

Divide every term by 2 to make the numbers smaller:
$$8y^{4} - 34y^{2} + 8 = 0 \qquad -(1)$$

Put $$z = y^{2} \;(\text{note: } z \ge 0)$$ to convert the quartic to a quadratic:

Equation $$(1)$$ becomes
$$8z^{2} - 34z + 8 = 0.$$

Solve this quadratic using the quadratic formula $$z = \dfrac{-b \pm \sqrt{b^{2}-4ac}}{2a}$$ with $$a=8,\, b=-34,\, c=8$$:

Discriminant $$\Delta = (-34)^{2} - 4\cdot 8 \cdot 8 = 1156 - 256 = 900.$$
$$\sqrt{\Delta} = 30.$$

Therefore
$$z = \dfrac{34 \pm 30}{16}.$$

Case 1: $$z = \dfrac{34 + 30}{16} = \dfrac{64}{16} = 4.$$
Case 2: $$z = \dfrac{34 - 30}{16} = \dfrac{4}{16} = \dfrac14.$$

Recall $$z = y^{2}$$, so

From Case 1: $$y^{2}=4 \quad\Longrightarrow\quad y = \pm 2.$$
From Case 2: $$y^{2}=\dfrac14 \quad\Longrightarrow\quad y = \pm \dfrac12.$$

Convert back to $$x$$ using $$x = 5^{y}$$:

For $$y = 2: \; x = 5^{2} = 25.$$
For $$y = -2: \; x = 5^{-2} = \dfrac1{25}.$$
For $$y = \dfrac12: \; x = 5^{1/2} = \sqrt{5}.$$
For $$y = -\dfrac12: \; x = 5^{-1/2} = \dfrac1{\sqrt{5}}.$$

All four values are positive real numbers, so they all satisfy the original equation.

The required product is
$$25 \times \dfrac1{25} \times \sqrt{5} \times \dfrac1{\sqrt{5}} = 1.$$

Hence, the product of all positive real solutions is 1.

We need to find the sets $$S$$ and $$T$$ and then determine the number of elements in $$S \cap T$$.

We have $$S = \{x \in [-6, 3] \setminus \{-2, 2\} : \frac{|x+3|-1}{|x|-2} \geq 0\}$$.

The numerator $$|x+3| - 1 \geq 0$$ when $$|x+3| \geq 1$$, i.e., $$x \leq -4$$ or $$x \geq -2$$. The numerator is negative when $$-4 < x < -2$$.

The denominator $$|x| - 2 > 0$$ when $$x < -2$$ or $$x > 2$$, and $$|x| - 2 < 0$$ when $$-2 < x < 2$$. The points $$x = \pm 2$$ are excluded.

For the fraction to be non-negative, we need both numerator and denominator to share the same sign (or numerator equals zero).

Case 1: Numerator $$\geq 0$$ and denominator $$> 0$$. This requires $$(x \leq -4 \text{ or } x \geq -2)$$ and $$(x < -2 \text{ or } x > 2)$$. Within $$[-6, 3] \setminus \{-2, 2\}$$, this gives $$[-6, -4] \cup (2, 3]$$.

Case 2: Numerator $$\leq 0$$ and denominator $$< 0$$. This requires $$-4 < x < -2$$ and $$-2 < x < 2$$. These intervals do not overlap, so this case contributes nothing.

Therefore $$S = [-6, -4] \cup (2, 3]$$.

Now for $$T$$, we solve $$x^2 - 7|x| + 9 \leq 0$$ over integers. Setting $$u = |x|$$, we get $$u^2 - 7u + 9 \leq 0$$. The roots are $$u = \frac{7 \pm \sqrt{13}}{2}$$, which gives approximately $$u \in [1.697, 5.303]$$. So for integer values of $$u$$: $$u \in \{2, 3, 4, 5\}$$, and therefore $$T = \{-5, -4, -3, -2, 2, 3, 4, 5\}$$.

Now $$S \cap T$$ consists of elements of $$T$$ that lie in $$S = [-6, -4] \cup (2, 3]$$. From $$[-6, -4]$$, we get $$\{-5, -4\}$$. From $$(2, 3]$$, we get $$\{3\}$$. So $$S \cap T = \{-5, -4, 3\}$$, which has 3 elements.

Hence, the correct answer is Option 4.

We are given the statements:

$$p$$: Ramesh listens to music

$$q$$: Ramesh is out of his village

$$r$$: It is Sunday

$$s$$: It is Saturday

The statement to express is: "Ramesh listens to music only if he is in his village and it is Sunday or Saturday."

The phrase "P only if Q" translates to $$P \Rightarrow Q$$. So "Ramesh listens to music only if (condition)" becomes $$p \Rightarrow \text{(condition)}$$.

Now the condition is "he is in his village and it is Sunday or Saturday." Being "in his village" is the negation of $$q$$ (since $$q$$ means "out of his village"), so this is $$\sim q$$. "It is Sunday or Saturday" is $$r \vee s$$.

So the condition is $$(\sim q) \wedge (r \vee s)$$.

Therefore the full statement is $$p \Rightarrow ((\sim q) \wedge (r \vee s))$$.

Hence, the correct answer is Option 4.

We are given the formulas $$S_1 : ((\sim p) \vee q) \vee ((\sim p) \vee r)$$ and $$S_2 : p \to (q \vee r)$$.

To simplify $$S_1$$, we use the associative and idempotent laws of disjunction: $$S_1 = (\sim p \vee q) \vee (\sim p \vee r) = \sim p \vee q \vee \sim p \vee r = \sim p \vee q \vee r.$$

Next, the implication in $$S_2$$ is equivalent to a disjunction, namely $$S_2 = \sim p \vee (q \vee r) = \sim p \vee q \vee r.$$

Since both $$S_1$$ and $$S_2$$ simplify to the same expression $$\sim p \vee q \vee r$$, they are logically equivalent: $$S_1 \equiv S_2$$.

We now evaluate each option:

Option A: “If $$S_2$$ is True, then $$S_1$$ is True.” This holds because $$S_1 \equiv S_2$$.

Option B: “If $$S_2$$ is False, then $$S_1$$ is False.” This also holds by equivalence.

Option C: “If $$S_2$$ is False, then $$S_1$$ is True.” This would require different truth values for $$S_1$$ and $$S_2$$, contradicting their equivalence, so it is not true.

Option D: “If $$S_1$$ is False, then $$S_2$$ is False.” This holds by equivalence.

Therefore, the statement that is NOT true is Option C.

We need to determine what the statement $$(\sim(p \Leftrightarrow \sim q)) \wedge q$$ is equivalent to.

$$p \Leftrightarrow \sim q$$ is true when p and $$\sim q$$ have the same truth value, i.e., when p and q have opposite truth values.

Therefore, $$\sim(p \Leftrightarrow \sim q)$$ is true when p and q have the same truth value, which means $$\sim(p \Leftrightarrow \sim q) \equiv (p \Leftrightarrow q)$$.

The statement becomes $$(p \Leftrightarrow q) \wedge q$$.

$$p = T, q = T$$: $$(T \Leftrightarrow T) \wedge T = T \wedge T = T$$

$$p = T, q = F$$: $$(T \Leftrightarrow F) \wedge F = F \wedge F = F$$

$$p = F, q = T$$: $$(F \Leftrightarrow T) \wedge T = F \wedge T = F$$

$$p = F, q = F$$: $$(F \Leftrightarrow F) \wedge F = T \wedge F = F$$

The statement is true only when $$p = T$$ and $$q = T$$, which is $$p \wedge q$$.

Option D: $$(p \Rightarrow q) \wedge p$$

$$p = T, q = T$$: $$T \wedge T = T$$

$$p = T, q = F$$: $$F \wedge T = F$$

$$p = F, q = T$$: $$T \wedge F = F$$

$$p = F, q = F$$: $$T \wedge F = F$$

This is also equivalent to $$p \wedge q$$. ✓

Therefore, the correct answer is Option D: equivalent to $$(p \Rightarrow q) \wedge p$$.

We are given the statements: $$A$$: Rishi is a judge, $$B$$: Rishi is honest, and $$C$$: Rishi is not arrogant.

The statement "if Rishi is a judge and he is not arrogant, then he is honest" can be written as: $$(A \wedge C) \to B$$

Recall that the negation of an implication $$p \to q$$ is $$p \wedge (\sim q)$$, so $$\sim[(A \wedge C) \to B] = (A \wedge C) \wedge (\sim B)$$.

This can be rewritten as: $$(\sim B) \wedge (A \wedge C)$$

Therefore, the negation is Option B: $$(\sim B) \wedge (A \wedge C)$$.

For an implication

$$A\to B$$

to be FALSE,

$$A=\text{TRUE},\qquad B=\text{FALSE}$$

Hence,

$$p=\text{TRUE}$$

and

$$\sim p\lor q=\text{FALSE}$$

Now, an OR statement is FALSE only when both parts are FALSE.

Therefore,

$$\sim p=\text{FALSE},\qquad q=\text{FALSE}$$

Thus,

$$p=\text{TRUE},\qquad q=\text{FALSE}$$

Now,

$$P_1=\sim(p\to\sim q)$$

Since

$$p=\text{TRUE},\qquad \sim q=\text{TRUE}$$

we get

$$p\to\sim q=\text{TRUE}$$

Hence,

$$P_1=\sim(\text{TRUE})=\text{FALSE}$$

Also,

$$P_2=(p\land\sim q)\land(\sim p\lor q)$$

$$=(\text{TRUE}\land\text{TRUE})\land(\text{FALSE}\lor\text{FALSE})$$

$$=\text{TRUE}\land\text{FALSE}$$

$$=\text{FALSE}$$

Therefore, both $$P_1$$ and $$P_2$$ are FALSE.

First, we need to find $$r \in \{p, q, \sim p, \sim q\}$$ such that $$r \vee (\sim p) \Rightarrow (p \wedge q) \vee r$$ is a tautology.

Next, recall that an implication $$P \Rightarrow Q$$ is a tautology when $$Q$$ is true whenever $$P$$ is true. Therefore, we need: whenever $$r \vee (\sim p)$$ is true, $$(p \wedge q) \vee r$$ must also be true.

Now, test $$r = \sim p$$. The statement becomes:

$$ (\sim p) \vee (\sim p) \Rightarrow (p \wedge q) \vee (\sim p) $$

Substituting, this simplifies to:

$$ \sim p \Rightarrow (p \wedge q) \vee (\sim p) $$

Since the implication must hold in all cases, consider the following:

Case 1: $$p$$ is True (so $$\sim p$$ is False). The implication $$F \Rightarrow \text{anything}$$ is True.

Case 2: $$p$$ is False (so $$\sim p$$ is True). The RHS becomes $$(F \wedge q) \vee T = T$$. So $$T \Rightarrow T$$ is True.

Since the implication is true in all cases, it is a tautology. ✓

Next, verify that the other options fail.

For $$r = p$$: Let $$p = F, q = F$$. Then LHS: $$F \vee T = T$$, RHS: $$(F \wedge F) \vee F = F$$. So $$T \Rightarrow F$$ is False. ✗

For $$r = q$$: Let $$p = F, q = F$$. Then LHS: $$F \vee T = T$$, RHS: $$(F \wedge F) \vee F = F$$. So $$T \Rightarrow F$$ is False. ✗

For $$r = \sim q$$: Let $$p = T, q = T$$. Then LHS: $$F \vee F = F$$, so implication is True. Let $$p = F, q = T$$. Then LHS: $$F \vee T = T$$, RHS: $$(F \wedge T) \vee F = F$$. So $$T \Rightarrow F$$ is False. ✗

Therefore, the answer is Option C: $$\sim p$$.

We need to find the expression equivalent to $$(\sim(p \wedge q)) \vee q$$.

Using De Morgan's law: $$\sim(p \wedge q) = \sim p \vee \sim q$$. Substituting gives $$(\sim(p \wedge q)) \vee q = (\sim p \vee \sim q) \vee q = \sim p \vee (\sim q \vee q) = \sim p \vee T = T$$ (since $$\sim q \vee q$$ is always true), so the original expression is a tautology.

Next, we examine each option:

Option A: $$q \to (p \wedge q) = \sim q \vee (p \wedge q) = (\sim q \vee p) \wedge (\sim q \vee q) = (\sim q \vee p) \wedge T = \sim q \vee p$$. This is NOT always true (fails when $$p = F, q = T$$), so it is not a tautology.

Option B: $$p \to q = \sim p \vee q$$. This is NOT always true (fails when $$p = T, q = F$$), so it is not a tautology.

Option C: $$p \to (p \to q) = p \to (\sim p \vee q) = \sim p \vee (\sim p \vee q) = \sim p \vee q$$. This is the same as Option B and thus not a tautology.

Option D: $$p \to (p \vee q) = \sim p \vee (p \vee q) = (\sim p \vee p) \vee q = T \vee q = T$$. This is always true — a tautology.

Since the given expression simplifies to a tautology, and Option D is also a tautology, they are logically equivalent.

The correct answer is Option D.

We need to determine which statement is equivalent to $$(p \wedge q) \Rightarrow (p \wedge r)$$.

We begin by rewriting the implication using the identity $$A \Rightarrow B \equiv \neg A \vee B$$. We have $$(p \wedge q) \Rightarrow (p \wedge r) \equiv \neg(p \wedge q) \vee (p \wedge r) \equiv (\neg p \vee \neg q) \vee (p \wedge r)$$.

Now let us examine Option D: $$(p \wedge q) \Rightarrow r \equiv \neg(p \wedge q) \vee r \equiv (\neg p \vee \neg q) \vee r$$.

We claim these two expressions are logically equivalent. To verify, we check all cases based on $$p$$.

Case 1: If $$p$$ is false, then $$\neg p$$ is true, so both $$(\neg p \vee \neg q) \vee (p \wedge r)$$ and $$(\neg p \vee \neg q) \vee r$$ evaluate to true regardless of $$q$$ and $$r$$.

Case 2: If $$p$$ is true and $$q$$ is false, then $$\neg q$$ is true, so both expressions again evaluate to true.

Case 3: If $$p$$ is true and $$q$$ is true, then $$\neg p \vee \neg q = \text{false}$$. The first expression becomes $$(p \wedge r) = r$$ (since $$p$$ is true), and the second becomes simply $$r$$. Hence both equal $$r$$.

In every case the two expressions agree, confirming that $$(p \wedge q) \Rightarrow (p \wedge r) \equiv (p \wedge q) \Rightarrow r$$.

Hence, the correct answer is Option D.

We need to determine which of the given statements is a tautology (always true regardless of truth values of $$ p $$ and $$ q $$).

Recall that $$ A \Rightarrow B $$ is equivalent to $$ \sim A \lor B $$.

$$q \Rightarrow (\sim p \lor q) \equiv \sim q \lor (\sim p \lor q) \equiv \sim p \lor (\sim q \lor q) \equiv \sim p \lor T \equiv T$$

Since $$ \sim q \lor q $$ is always true (law of excluded middle), the entire expression simplifies to True. This is a tautology.

Option A: $$ (\sim p \lor q) \Rightarrow p \equiv \sim(\sim p \lor q) \lor p \equiv (p \land \sim q) \lor p \equiv p $$. This is not a tautology (false when $$ p = F $$).

Option B: $$ p \Rightarrow (\sim p \lor q) \equiv \sim p \lor (\sim p \lor q) \equiv \sim p \lor q $$. This is not a tautology (false when $$ p = T, q = F $$).

Option C: $$ (\sim p \lor q) \Rightarrow q \equiv \sim(\sim p \lor q) \lor q \equiv (p \land \sim q) \lor q \equiv (p \lor q) \land (\sim q \lor q) \equiv p \lor q $$. This is not a tautology (false when $$ p = F, q = F $$).

The tautology is $$ q \Rightarrow (\sim p \lor q) $$, which corresponds to Option D.

The statement $$(P \wedge (\sim R)) \to ((\sim R) \wedge Q)$$ is given to be False. An implication $$p \to q$$ is false only when $$p$$ is True and $$q$$ is False, so we require $$P \wedge (\sim R) = T$$ and $$(\sim R) \wedge Q = F$$. From $$P \wedge (\sim R) = T$$ we deduce $$P = T$$ and $$\sim R = T$$, hence $$R = F$$. Since $$(\sim R) \wedge Q = F$$ and $$\sim R = T$$, it follows that $$Q = F$$. Therefore $$P = T,\ Q = F,\ R = F$$.

Substituting these truth values into each option, we find:

Option A: $$P \vee Q \to \sim R = (T \vee F) \to T = T \to T = T$$

Option B: $$R \vee Q \to \sim P = (F \vee F) \to F = F \to F = T$$

Option C: $$\sim(P \vee Q) \to \sim R = \sim(T) \to T = F \to T = T$$

Option D: $$\sim(R \vee Q) \to \sim P = \sim(F) \to F = T \to F = F$$

The statement with truth value False is $$\boxed{\sim(R \vee Q) \to \sim P}$$, so the answer is Option D.

We need to find the ordered pair $$(*, \odot)$$ from $$\{\wedge, \vee\}$$ such that $$(p * q) \odot (p \odot \sim q)$$ is a tautology.

Step 1: Test each option systematically.

Option B: $$(*, \odot) = (\vee, \vee)$$

The expression becomes: $$(p \vee q) \vee (p \vee \sim q)$$

$$= p \vee q \vee p \vee \sim q$$

$$= p \vee (q \vee \sim q)$$

$$= p \vee T$$

$$= T$$

This is a tautology.

Step 2: Verify that other options are not tautologies.

Option A: $$(*, \odot) = (\vee, \wedge)$$

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

When $$p = F, q = T$$: $$(F \vee T) \wedge (F \wedge F) = T \wedge F = F$$. Not a tautology.

Option C: $$(*, \odot) = (\wedge, \wedge)$$

$$(p \wedge q) \wedge (p \wedge \sim q)$$

$$= p \wedge (q \wedge \sim q) = p \wedge F = F$$. Not a tautology (always false).

Option D: $$(*, \odot) = (\wedge, \vee)$$

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

When $$p = F, q = T$$: $$(F \wedge T) \vee (F \vee F) = F \vee F = F$$. Not a tautology.

Only Option B gives a tautology.

The correct answer is Option B: $$(\vee, \vee)$$

We need to find how many choices for $$\Delta \in \{\wedge, \vee, \Rightarrow, \Leftrightarrow\}$$ make $$(p \Delta q) \Rightarrow ((p \Delta \sim q) \vee ((\sim p) \Delta q))$$ a tautology.

Recall that $$P \Rightarrow Q$$ is false only when $$P$$ is true and $$Q$$ is false. So the expression is a tautology if whenever $$(p \Delta q)$$ is true, at least one of $$(p \Delta \sim q)$$ or $$(\sim p \Delta q)$$ is also true.

Case 1: $$\Delta = \wedge$$ (AND)

$$(p \wedge q) \Rightarrow ((p \wedge \sim q) \vee (\sim p \wedge q))$$

When $$p = T, q = T$$: LHS = T. RHS = $$(T \wedge F) \vee (F \wedge T) = F \vee F = F$$.

So the implication is F. Not a tautology.

Case 2: $$\Delta = \vee$$ (OR)

$$(p \vee q) \Rightarrow ((p \vee \sim q) \vee (\sim p \vee q))$$

RHS: $$(p \vee \sim q) \vee (\sim p \vee q) = (p \vee \sim p) \vee (q \vee \sim q) = T \vee T = T$$.

Since RHS is always T, the implication is always T. Tautology.

Case 3: $$\Delta = \Rightarrow$$ (Implication)

$$(p \Rightarrow q) \Rightarrow ((p \Rightarrow \sim q) \vee (\sim p \Rightarrow q))$$

$$(\sim p \Rightarrow q)$$ is equivalent to $$(p \vee q)$$. This is false only when $$p = F, q = F$$.

When $$p = F, q = F$$: LHS = $$(F \Rightarrow F) = T$$. RHS = $$(F \Rightarrow T) \vee (T \Rightarrow F) = T \vee F = T$$.

When $$p = T, q = T$$: LHS = T. RHS = $$(T \Rightarrow F) \vee (F \Rightarrow T) = F \vee T = T$$.

When $$p = T, q = F$$: LHS = F. So implication is T.

When $$p = F, q = T$$: LHS = T. RHS = $$(F \Rightarrow F) \vee (T \Rightarrow T) = T \vee T = T$$.

All cases give T. Tautology.

Case 4: $$\Delta = \Leftrightarrow$$ (Biconditional)

$$(p \Leftrightarrow q) \Rightarrow ((p \Leftrightarrow \sim q) \vee (\sim p \Leftrightarrow q))$$

Note: $$(p \Leftrightarrow \sim q) = \sim(p \Leftrightarrow q)$$ and $$(\sim p \Leftrightarrow q) = \sim(p \Leftrightarrow q)$$.

So RHS = $$\sim(p \Leftrightarrow q) \vee \sim(p \Leftrightarrow q) = \sim(p \Leftrightarrow q)$$.

The expression becomes: $$(p \Leftrightarrow q) \Rightarrow \sim(p \Leftrightarrow q)$$.

When $$p = T, q = T$$: $$(T) \Rightarrow (F) = F$$. Not a tautology.

Therefore, exactly 2 choices ($$\vee$$ and $$\Rightarrow$$) make the expression a tautology.

The answer is Option B: 2.

We need to determine which statement is NOT equivalent to $$(p \Rightarrow q) \vee (p \Rightarrow r)$$.

We first simplify the given statement. We know $$p \Rightarrow q \equiv \sim p \vee q$$ and $$p \Rightarrow r \equiv \sim p \vee r$$. So $$(p \Rightarrow q) \vee (p \Rightarrow r) \equiv (\sim p \vee q) \vee (\sim p \vee r) \equiv \sim p \vee q \vee r$$. This is equivalent to $$p \Rightarrow (q \vee r)$$.

Now we check each option:

Option A: $$(p \wedge \sim r) \Rightarrow q \equiv \sim(p \wedge \sim r) \vee q \equiv \sim p \vee r \vee q$$. This equals $$\sim p \vee q \vee r$$. Equivalent.

Option B: $$\sim q \Rightarrow (\sim r \vee p) \equiv q \vee \sim r \vee p$$. This is $$p \vee q \vee \sim r$$, which is NOT the same as $$\sim p \vee q \vee r$$. For example, when $$p = T, q = F, r = T$$: our expression gives $$F \vee F \vee T = T$$, but Option B gives $$T \vee F \vee F = T$$. Try $$p = T, q = F, r = F$$: our expression gives $$F \vee F \vee F = F$$, but Option B gives $$T \vee F \vee T = T$$. So they differ. NOT equivalent.

Option C: $$p \Rightarrow (q \vee r) \equiv \sim p \vee q \vee r$$. Equivalent (same as our simplified form).

Option D: $$(p \wedge \sim q) \Rightarrow r \equiv \sim(p \wedge \sim q) \vee r \equiv \sim p \vee q \vee r$$. Equivalent.

Hence, the correct answer is Option B: $$\sim q \Rightarrow (\sim r \vee p)$$.

We need to identify which of the given statements is a tautology (always true regardless of truth values of $$p$$ and $$q$$).

Analyze Option A: $$((\sim q) \wedge p) \wedge q$$

This simplifies to $$p \wedge q \wedge (\sim q)$$, which contains $$q \wedge (\sim q) = F$$.

So this is always False (contradiction). Not a tautology.

Analyze Option B: $$((\sim q) \wedge p) \wedge (p \wedge (\sim p))$$

The term $$p \wedge (\sim p) = F$$.

So the entire expression is $$(\text{anything}) \wedge F = F$$.

This is always False (contradiction). Not a tautology.

Analyze Option C: $$((\sim q) \wedge p) \vee (p \vee (\sim p))$$

The term $$p \vee (\sim p) = T$$ (this is a tautology by itself).

So the entire expression is $$(\text{anything}) \vee T = T$$.

This is always True. This is a tautology!

Analyze Option D: $$(p \wedge q) \wedge (\sim(p \wedge q))$$

Let $$r = p \wedge q$$. Then this becomes $$r \wedge (\sim r) = F$$.

This is always False (contradiction). Not a tautology.

Therefore, the tautology is Option C: $$((\sim q) \wedge p) \vee (p \vee (\sim p))$$.

The correct answer is Option C.

We need to find the negation of the Boolean expression $$p \leftrightarrow (q \rightarrow p)$$.

$$q \rightarrow p \equiv \sim q \vee p$$

$$p \leftrightarrow (\sim q \vee p) \equiv [p \rightarrow (\sim q \vee p)] \wedge [(\sim q \vee p) \rightarrow p]$$

The first part: $$p \rightarrow (\sim q \vee p) \equiv \sim p \vee \sim q \vee p \equiv T$$ (tautology, since $$\sim p \vee p = T$$)

So: $$p \leftrightarrow (\sim q \vee p) \equiv T \wedge [(\sim q \vee p) \rightarrow p] \equiv (\sim q \vee p) \rightarrow p$$

$$(\sim q \vee p) \rightarrow p \equiv \sim(\sim q \vee p) \vee p \equiv (q \wedge \sim p) \vee p$$

$$\equiv (q \vee p) \wedge (\sim p \vee p) \equiv (q \vee p) \wedge T \equiv q \vee p \equiv p \vee q$$

So $$p \leftrightarrow (q \rightarrow p) \equiv p \vee q$$.

$$\sim(p \vee q) \equiv \sim p \wedge \sim q$$

The negation is T only when both $$p$$ and $$q$$ are F, which is $$\sim p \wedge \sim q$$. $$\checkmark$$

Therefore, the correct answer is Option D: $$\sim p \wedge \sim q$$.