JEE Sets, Relations & 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 a seven-digit number $$x$$ contain $$a_0$$ zeros, $$a_1$$ ones and $$a_2$$ twos, so that

$$a_0 + a_1 + a_2 = 7$$ and $$a_0,\,a_1,\,a_2 \ge 0$$.

Because the first (left-most) digit cannot be $$0$$, we count the admissible arrangements for each triple $$(a_0,a_1,a_2)$$ as follows.

Total arrangements with these counts = $$\dfrac{7!}{a_0!\,a_1!\,a_2!}$$.
Arrangements whose first digit is $$0$$: fix a zero at the first place and permute the remaining $$6$$ digits, giving $$\dfrac{6!}{(a_0-1)!\,a_1!\,a_2!}$$ (this term appears only when $$a_0\ge 1$$).

Hence the required number for the triple $$(a_0,a_1,a_2)$$ is

$$N(a_0,a_1,a_2)= \begin{cases} \dfrac{7!}{a_0!\,a_1!\,a_2!}-\dfrac{6!}{(a_0-1)!\,a_1!\,a_2!}, & a_0\ge 1\\[6pt] \dfrac{7!}{a_0!\,a_1!\,a_2!}, & a_0=0 \end{cases}$$

For $$a_0\ge 1$$, simplifying gives

$$N(a_0,a_1,a_2)=\dfrac{6!}{a_0!\,a_1!\,a_2!}\,(7-a_0)\qquad -(1)$$

The problem asks for the count of all numbers in which at least one of the digits $$0$$ or $$1$$ appears exactly twice. We split this into three disjoint cases.

Let the possible values of $$a_1$$ be $$0,1,3,4,5$$ (they must stay within $$0\le a_1\le 5$$). For each value we compute $$a_2=7-2-a_1$$ and use $$(1)$$.

$$\begin{aligned} a_1=0,&\;a_2=5:&\;N=\dfrac{6!\,(7-2)}{2!\,0!\,5!}=15\\ a_1=1,&\;a_2=4:&\;N=\dfrac{6!\,(7-2)}{2!\,1!\,4!}=75\\ a_1=3,&\;a_2=2:&\;N=\dfrac{6!\,(7-2)}{2!\,3!\,2!}=150\\ a_1=4,&\;a_2=1:&\;N=\dfrac{6!\,(7-2)}{2!\,4!\,1!}=75\\ a_1=5,&\;a_2=0:&\;N=\dfrac{6!\,(7-2)}{2!\,5!\,0!}=15 \end{aligned}$$

Total for Case 1 = $$15+75+150+75+15 = 330$$.

Now $$a_0$$ can be $$0,1,3,4,5$$. Compute $$a_2=7-a_0-2$$ and apply the appropriate formula.

$$\begin{aligned} a_0=0,&\;a_2=5:&\;N=\dfrac{7!}{0!\,2!\,5!}=21\\ a_0=1,&\;a_2=4:&\;N=\dfrac{6!\,(7-1)}{1!\,2!\,4!}=90\\ a_0=3,&\;a_2=2:&\;N=\dfrac{6!\,(7-3)}{3!\,2!\,2!}=120\\ a_0=4,&\;a_2=1:&\;N=\dfrac{6!\,(7-4)}{4!\,2!\,1!}=45\\ a_0=5,&\;a_2=0:&\;N=\dfrac{6!\,(7-5)}{5!\,2!\,0!}=6 \end{aligned}$$

Total for Case 2 = $$21+90+120+45+6 = 282$$.

Both digits $$0$$ and $$1$$ appear exactly twice.

$$N=\dfrac{6!\,(7-2)}{2!\,2!\,3!}=150$$

Summing all three cases:

$$330 + 282 + 150 = 762$$

Therefore, the required number of seven-digit numbers is
762.

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: $$\mathbb{R}$$.

We need to find the range of $$f(x) = 6 + 16\cos x \cos(\frac{\pi}{3} - x)\cos(\frac{\pi}{3} + x)\sin 3x \cos 6x$$.

Simplify $$\cos x \cos(\frac{\pi}{3}-x)\cos(\frac{\pi}{3}+x)$$

Using the identity $$\cos A \cos(60°-A)\cos(60°+A) = \frac{1}{4}\cos 3A$$:

$$\cos x \cos(\frac{\pi}{3}-x)\cos(\frac{\pi}{3}+x) = \frac{1}{4}\cos 3x$$

Substitute

$$f(x) = 6 + 16 \cdot \frac{1}{4}\cos 3x \cdot \sin 3x \cdot \cos 6x$$

$$= 6 + 4\cos 3x \sin 3x \cos 6x$$

$$= 6 + 2\sin 6x \cos 6x$$ (using $$2\sin A\cos A = \sin 2A$$)

$$= 6 + \sin 12x$$

Find the range

Since $$-1 \leq \sin 12x \leq 1$$:

$$f(x) \in [6-1, 6+1] = [5, 7]$$

So $$\alpha = 5, \beta = 7$$.

Find distance from (5,7) to line 3x + 4y + 12 = 0

$$d = \frac{|3(5) + 4(7) + 12|}{\sqrt{9+16}} = \frac{|15+28+12|}{5} = \frac{55}{5} = 11$$

The correct answer is Option 1: 11.

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.

Given $$R = \{(1,2), (2,3), (3,3)\}$$ on the set $$\{1, 2, 3, 4\}$$, find the minimum number of elements to add so that R becomes an equivalence relation.

An equivalence relation must be reflexive, symmetric, and transitive.

Since $$(1,2)$$ and $$(2,3)$$ are in R, by symmetry and transitivity, 1, 2, and 3 must all be in the same equivalence class. Element 4 is in its own class.

So the equivalence classes are $$\{1, 2, 3\}$$ and $$\{4\}$$.

Reflexive: $$(1,1), (2,2), (3,3), (4,4)$$

Symmetric pairs for class $$\{1,2,3\}$$: $$(1,2),(2,1),(1,3),(3,1),(2,3),(3,2)$$

Total required pairs: $$(1,1),(2,2),(3,3),(4,4),(1,2),(2,1),(1,3),(3,1),(2,3),(3,2)$$ = 10 pairs

Already in R: $$(1,2), (2,3), (3,3)$$ = 3 pairs

Need to add: $$(1,1), (2,2), (4,4), (2,1), (1,3), (3,1), (3,2)$$ = 7 pairs

The correct answer is Option 2: 7.

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.

Set $$A = \{0,1,2,3,4,5\}$$ has six elements.
The relation $$R$$ is defined by$$(x,y) \in R \; \Longleftrightarrow \; \max\{x,y\} \in \{3,4\}.$$

Case 1: $$\max\{x,y\}=3$$
Both coordinates must lie in $$\{0,1,2,3\}$$ and at least one of them must be $$3$$.
Total ordered pairs with coordinates from $$\{0,1,2,3\}$$ are $$4 \times 4 = 16$$.
Pairs with both coordinates in $$\{0,1,2\}$$ (hence max < 3) are $$3 \times 3 = 9$$.
Hence the number of pairs with max $$3$$ is $$16-9 = 7$$.

The explicit pairs are
$$(3,0),(3,1),(3,2),(3,3),(0,3),(1,3),(2,3).$$

Case 2: $$\max\{x,y\}=4$$
Both coordinates must lie in $$\{0,1,2,3,4\}$$ and at least one of them must be $$4$$.
Total ordered pairs with coordinates from $$\{0,1,2,3,4\}$$ are $$5 \times 5 = 25$$.
Pairs with both coordinates in $$\{0,1,2,3\}$$ (max < 4) are $$4 \times 4 = 16$$.
Hence the number of pairs with max $$4$$ is $$25-16 = 9$$.

The explicit pairs are
$$(4,0),(4,1),(4,2),(4,3),(4,4),(0,4),(1,4),(2,4),(3,4).$$

Since no ordered pair can contain the element $$5$$ (that would make the maximum $$\ge 5$$), the total number of elements in $$R$$ is
$$7 + 9 = 16.$$
Statement $$S_1$$ claims $$18$$ elements, so $$S_1$$ is false.

Next, examine the properties of $$R$$.

Symmetric:
If $$(x,y) \in R$$, then $$\max\{x,y\} \in \{3,4\}$$. The same maximum equals $$\max\{y,x\}$$, so $$(y,x) \in R$$. Hence $$R$$ is symmetric.

Reflexive:
Reflexivity requires every $$(a,a)$$, $$a \in A$$, to be in $$R$$.
But $$(a,a) \in R \Longleftrightarrow a \in \{3,4\}$$.
Elements $$0,1,2,5$$ violate this, so $$R$$ is not reflexive.

Transitive:
To test transitivity, find a counter-example.
Take $$x=0,\,y=4,\,z=0$$.
$$(x,y)=(0,4) \in R \quad (\max=4),$$
$$(y,z)=(4,0) \in R \quad (\max=4),$$
but $$(x,z)=(0,0) \notin R \quad (\max=0).$$
Thus $$R$$ is not transitive.

Therefore $$R$$ is symmetric but neither reflexive nor transitive, so Statement $$S_2$$ is true.

Conclusion: $$S_1$$ is false and $$S_2$$ is true → Option C (only $$S_2$$ is true).

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

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 $$8\left(\dfrac{1}{4} + 29\times \dfrac{1}{2}\right) = 2 + 29\times 4 = 118$$

An equivalence relation on a set must satisfy three properties: reflexivity, symmetry, and transitivity. The number of equivalence relations on a set is equal to the number of partitions of that set, as each partition defines an equivalence class.

For the set $$S = \{1, 2, 3\}$$, we need to find all possible partitions. A partition divides the set into non-empty, disjoint subsets whose union is the entire set.

The possible partitions are:

1. Partition with one subset:
- The entire set: $$\{\{1, 2, 3\}\}$$
This corresponds to one equivalence relation where all elements are related to each other.

2. Partitions with two subsets:
- Subset 1: $$\{1\}$$, Subset 2: $$\{2, 3\}$$
- Subset 1: $$\{2\}$$, Subset 2: $$\{1, 3\}$$
- Subset 1: $$\{3\}$$, Subset 2: $$\{1, 2\}$$
Each of these partitions corresponds to an equivalence relation where elements in the same subset are equivalent. There are three such partitions.

3. Partition with three subsets:
- Each element in its own subset: $$\{\{1\}, \{2\}, \{3\}\}$$
This corresponds to the equivalence relation where each element is only related to itself.

Total number of partitions = 1 (one subset) + 3 (two subsets) + 1 (three subsets) = 5.

Each partition defines a unique equivalence relation. Since reflexivity requires that every element is related to itself (i.e., pairs like (1,1), (2,2), (3,3) must be present), no equivalence relation can be empty. Therefore, all 5 equivalence relations are non-empty.

The number of non-empty equivalence relations is 5.

The correct option is B. 5.

The relation is $$R = \{(x, y) : \ln y = x \ln(2/5), x \in S = \mathbb{N} \cup \{0\}, y \in \mathbb{R}\}$$.

From $$\ln y = x \ln(2/5)$$, we get $$y = e^{x\ln(2/5)} = (2/5)^x$$.

The range of R is the set of all $$y$$ values: $$\{(2/5)^x : x \in \{0, 1, 2, 3, ...\}\}$$

$$= \{1, 2/5, 4/25, 8/125, ...\}$$

This is an infinite geometric series with first term $$a = 1$$ and common ratio $$r = 2/5$$:

$$ \text{Sum} = \frac{1}{1 - 2/5} = \frac{1}{3/5} = \frac{5}{3} $$

The correct answer is Option 4: $$\frac{5}{3}$$.

In the set B, the elements are fractions $$\dfrac{m}{n}$$ such that, $$n$$ can take the values from {1,2,...,10} and $$m$$ has to be less than $$n$$ and, $$m$$ and $$n$$ must be co-primes.

Euler's totient function represents the count of positive integers less than a given integer $$n$$ that are co-prime to $$n$$.

If $$n$$ is represented as,

$$n\ =\ p_1^a\ \times\ p_2^b\ \times\ p_3^c\ ...$$ where $$p_1,\ p_2,\ p_3..\ $$ are primes.

$$Φ\left(n\right)\ =\ n\left(1-\dfrac{1}{p_1}\right)\left(1-\dfrac{1}{p_2}\right)\left(1-\dfrac{1}{p_3}\right)...$$

When $$n$$ is prime, $$Φ(n)=n-1$$

$$n\left(B\right)=Φ\left(10\right)+\ Φ\left(9\right)\ +\ ...\ +\ Φ\left(2\right)$$

We are not including $$Φ(1)$$, as there are no integers less than 1 that are co-prime to 1.

$$Φ\left(10\right)$$: $$10=2\times5$$

$$Φ(10)=10\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{5}\right)\ =\ 10\left(\dfrac{1}{2}\right)\left(\dfrac{4}{5}\right)\ =\ 4$$

$$Φ\left(9\right)$$: $$9=3^2$$

$$Φ(9)=9\left(1-\dfrac{1}{3}\right)\ =\ 6$$

$$Φ\left(8\right)$$: $$8=2^3$$

$$Φ(8)=8\left(1-\dfrac{1}{2}\right)\ =\ 4$$

$$Φ\left(7\right)$$: $$7$$ is prime

$$Φ(7)=6$$

$$Φ\left(6\right)$$: $$6=2\times3$$

$$Φ(6)=6\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\ =\ 6\left(\dfrac{1}{2}\right)\left(\dfrac{2}{3}\right)\ =\ 2$$

$$Φ\left(5\right)$$: $$5$$ is prime

$$Φ(5)=4$$

$$Φ\left(4\right)$$: $$4=2^2$$

$$Φ(4)=4\left(1-\dfrac{1}{2}\right)\ =\ 2$$

$$Φ\left(3\right)$$: $$3$$ is prime

$$Φ(3)=2$$

$$Φ\left(2\right)$$: $$2$$ is prime

$$Φ(2)=1$$

$$n\left(B\right)=Φ\left(10\right)+\ Φ\left(9\right)\ +\ ...\ +\ Φ\left(2\right)=\ 4\ +\ 6\ +\ 4\ +\ 6\ +\ 2\ +\ 4\ +\ 2\ +\ 2\ +\ 1\ =\ 31$$

Hence, the correct answer is option B.

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.

We need to find the points in $$A \cap B$$ where x=0 or y=0, and then compute $$\sum |x+y|$$ for those points.

The sets are defined by $$A = \{(x,y) \in \mathbb{R} \times \mathbb{R} : |x+y| \geq 3\}$$, $$B = \{(x,y) \in \mathbb{R} \times \mathbb{R} : |x|+|y| \leq 3\}$$, and we consider $$C = \{(x,y) \in A \cap B : x = 0 \text{ or } y = 0\}$$.

If x=0, then from A we have $$|y| \geq 3$$ and from B we have $$|y| \leq 3$$, which together imply $$|y| = 3$$, so y=3 or y=-3, giving the points $$(0,3)$$ and $$(0,-3)$$.

If y=0, then from A we have $$|x| \geq 3$$ and from B we have $$|x| \leq 3$$, which together imply $$|x| = 3$$, so x=3 or x=-3, giving the points $$(3,0)$$ and $$(-3,0)$$.

Thus $$C = \{(0,3), (0,-3), (3,0), (-3,0)\}$$.

It follows that $$\sum_{(x,y) \in C} |x+y| = |0+3| + |0-3| + |3+0| + |-3+0| = 3 + 3 + 3 + 3 = 12.$$

The correct answer is Option 4: 12.

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.

R = {(x,y): x+y is even}. Reflexive: x+x=2x (even) ✓. Symmetric: if x+y even, y+x even ✓. Transitive: if x+y and y+z are even, then x+z = (x+y)+(y+z)-2y (even) ✓.

The correct answer is Option 2: equivalence relation.

S = set of first 10 primes. A = S ∪ P where P = all possible products of distinct elements of S. Find ordered pairs (x,y) where x∈S, y∈A, x|y.

A consists of S (the 10 primes) plus all products of 2 or more distinct primes from S. Total elements in A = $$2^{10} - 1 = 1023$$ (all non-empty subsets of S, each giving a unique product).

For each prime $$p_i \in S$$, count elements y ∈ A such that $$p_i | y$$.

y is a product of a non-empty subset of S. For $$p_i | y$$, the subset must contain $$p_i$$.

Number of such subsets = $$2^9 = 512$$ (choose any subset of the remaining 9 primes, including empty set, and include $$p_i$$).

Total ordered pairs

Each of the 10 primes in S contributes 512 pairs.

Total = $$10 \times 512 = 5120$$

The answer is 5120.

We need to find the number of relations on $$A = \{1,2,3\}$$ that contain $$(1,2)$$ and $$(2,3)$$, are reflexive and transitive, but not symmetric.

Reflexivity requires the pairs $$(1,1), (2,2), (3,3)$$. Since $$(1,2)$$ and $$(2,3)$$ are given, transitivity forces us to include $$(1,3)$$. Thus the base set of pairs is $$\{(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)\}$$.

Aside from these, the only remaining possible ordered pairs on $$A$$ are $$(2,1), (3,2), (3,1)$$. We examine which subsets of these can be added without violating transitivity and while ensuring the relation remains not symmetric.

If we add none of these extra pairs, the relation is already transitive and is not symmetric because $$(1,2)$$ is present but $$(2,1)$$ is absent. This case is valid.

If we add only $$(2,1)$$, then checking transitivity: $$(2,1)\circ(1,2)=(2,2)$$ ✓, $$(2,1)\circ(1,3)=(2,3)$$ ✓, and all other composites are already accounted for. Since $$(2,3)$$ is present without $$(3,2)$$, the relation remains not symmetric, so this case is valid.

If we add only $$(3,2)$$, then $$(3,2)\circ(2,3)=(3,3)$$ ✓, $$(1,3)\circ(3,2)=(1,2)$$ ✓, and all necessary closures hold. Because $$(1,2)$$ appears without $$(2,1)$$, the relation is not symmetric, so this case is valid.

If we add only $$(3,1)$$, transitivity would require $$(3,1)\circ(1,2)=(3,2)$$ to be present, but it is not. Therefore this case fails transitivity and is invalid.

If we add $$(2,1)$$ and $$(3,2)$$, then transitivity demands $$(3,2)\circ(2,1)=(3,1)$$, which is missing, so this case is invalid.

If we add $$(2,1)$$ and $$(3,1)$$, then $$(3,1)\circ(1,2)=(3,2)$$ must be present but is not, making this case invalid.

If we add $$(3,2)$$ and $$(3,1)$$, then $$(2,3)\circ(3,1)=(2,1)$$ is required and is missing, so this case is invalid.

If we add all three extra pairs $$(2,1),(3,2),(3,1)$$, the relation is transitive but becomes symmetric since every pair has its reverse, which is forbidden. Hence this case is invalid.

Out of the eight possibilities, only three yield relations that are reflexive, transitive, contain $$(1,2)$$ and $$(2,3)$$, and are not symmetric. Therefore, the answer is 3.

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.

We need to evaluate two statements about the relation R on $$X = \mathbb{R} \times \mathbb{R}$$.

Relation: $$(a_1, b_1) R (a_2, b_2) \Leftrightarrow b_1 = b_2$$

Statement I: R is an equivalence relation.

Check reflexive: $$(a,b) R (a,b)$$ since $$b = b$$. TRUE.

Check symmetric: If $$(a_1,b_1) R (a_2,b_2)$$, then $$b_1 = b_2$$, so $$b_2 = b_1$$, hence $$(a_2,b_2) R (a_1,b_1)$$. TRUE.

Check transitive: If $$(a_1,b_1) R (a_2,b_2)$$ and $$(a_2,b_2) R (a_3,b_3)$$, then $$b_1 = b_2$$ and $$b_2 = b_3$$, so $$b_1 = b_3$$, hence $$(a_1,b_1) R (a_3,b_3)$$. TRUE.

Statement I is TRUE.

Statement II: For some (a,b) in X, the set S = {(x,y) in X : (x,y) R (a,b)} represents a line parallel to y = x.

$$S = \{(x,y) : y = b\}$$

This is a horizontal line (parallel to the x-axis), not a line parallel to y = x.

The line y = x has slope 1. The line y = b has slope 0. These are not parallel for any value of b.

Statement II is FALSE.

The correct answer is Option 2: Statement I is true but Statement II is false.

We need to find $$n(A \cup B)$$ where:

$$A = \{x \in (0,\pi) \setminus \{\pi/2\} : \log_{2/\pi}|\sin x| + \log_{2/\pi}|\cos x| = 2\}$$

$$B = \{x \geq 0 : \sqrt{x}(\sqrt{x}-4) - 3|\sqrt{x}-2| + 6 = 0\}$$

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

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

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

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

The set is $$A=\{-3,-2,-1,0,1,2,3\}$$, so $$|A|=7$$.
The relation $$R$$ is defined by $$xRy \iff 2x-y\in\{0,1\}$$, i.e. $$y=2x$$ or $$y=2x-1$$.

Case 1: $$y=2x$$

Check each $$x\in A$$:

$$\begin{array}{c|c} x & 2x \\ \hline -3 & -6\;(\notin A)\\ -2 & -4\;(\notin A)\\ -1 & -2\;(\in A)\\ 0 & 0\;(\in A)\\ 1 & 2\;(\in A)\\ 2 & 4\;(\notin A)\\ 3 & 6\;(\notin A) \end{array}$$

Pairs obtained: $$(-1,-2),\,(0,0),\,(1,2).$$

Case 2: $$y=2x-1$$

Again test every $$x$$:

$$\begin{array}{c|c} x & 2x-1 \\ \hline -3 & -7\;(\notin A)\\ -2 & -5\;(\notin A)\\ -1 & -3\;(\in A)\\ 0 & -1\;(\in A)\\ 1 & 1\;(\in A)\\ 2 & 3\;(\in A)\\ 3 & 5\;(\notin A) \end{array}$$

Pairs obtained: $$(-1,-3),\,(0,-1),\,(1,1),\,(2,3).$$

Combining both cases, $$R$$ contains

$$\{(-1,-2),\,(-1,-3),\,(0,0),\,(0,-1),\,(1,2),\,(1,1),\,(2,3)\}.$$

Number of elements in $$R$$: $$l=7.$$

Making $$R$$ reflexive

A relation is reflexive when every $$a\in A$$ satisfies $$(a,a)\in R$$.
Present self-pairs: $$(0,0),\,(1,1).$$
Missing self-pairs: $$(-3,-3),\,(-2,-2),\,(-1,-1),\,(2,2),\,(3,3).$$
Minimum additions needed: $$m=5.$$

Making $$R$$ symmetric

For symmetry, whenever $$(a,b)\in R$$ we also need $$(b,a).$$
List missing converse pairs:

$$(-1,-2)\;\Rightarrow\;(-2,-1)$$
$$(-1,-3)\;\Rightarrow\;(-3,-1)$$
$$(0,-1)\;\Rightarrow\;(-1,0)$$
$$(1,2)\;\Rightarrow\;(2,1)$$
$$(2,3)\;\Rightarrow\;(3,2)$$

All five are distinct and not already in $$R$$, so
minimum additions required: $$n=5.$$

Finally,
$$l+m+n = 7+5+5 = 17.$$

Hence $$l + m + n = 17$$  ⇒  Option B.

We have $$A = \{-3, -2, -1, 0, 1, 2, 3\}$$ and the relation $$R$$ on $$A$$ defined by $$xRy$$ if and only if $$0 \leq x^2 + 2y \leq 4$$.

For a given $$x$$, the condition $$0 \leq x^2 + 2y \leq 4$$ gives us:
$$-x^2 \leq 2y \leq 4 - x^2$$
$$\frac{-x^2}{2} \leq y \leq \frac{4 - x^2}{2}$$

We check each value of $$x$$ in $$A$$:

Case 1: $$x = -3$$ (so $$x^2 = 9$$)
We need $$0 \leq 9 + 2y \leq 4$$, which gives $$-4.5 \leq y \leq -2.5$$.
From $$A$$, the valid values are $$y \in \{-3\}$$.
Pairs: $$(-3, -3)$$. Count = 1.

Case 2: $$x = -2$$ (so $$x^2 = 4$$)
We need $$0 \leq 4 + 2y \leq 4$$, which gives $$-2 \leq y \leq 0$$.
From $$A$$, the valid values are $$y \in \{-2, -1, 0\}$$.
Pairs: $$(-2, -2), (-2, -1), (-2, 0)$$. Count = 3.

Case 3: $$x = -1$$ (so $$x^2 = 1$$)
We need $$0 \leq 1 + 2y \leq 4$$, which gives $$-0.5 \leq y \leq 1.5$$.
From $$A$$, the valid values are $$y \in \{0, 1\}$$.
Pairs: $$(-1, 0), (-1, 1)$$. Count = 2.

Case 4: $$x = 0$$ (so $$x^2 = 0$$)
We need $$0 \leq 2y \leq 4$$, which gives $$0 \leq y \leq 2$$.
From $$A$$, the valid values are $$y \in \{0, 1, 2\}$$.
Pairs: $$(0, 0), (0, 1), (0, 2)$$. Count = 3.

Case 5: $$x = 1$$ (so $$x^2 = 1$$)
We need $$0 \leq 1 + 2y \leq 4$$, which gives $$-0.5 \leq y \leq 1.5$$.
From $$A$$, the valid values are $$y \in \{0, 1\}$$.
Pairs: $$(1, 0), (1, 1)$$. Count = 2.

Case 6: $$x = 2$$ (so $$x^2 = 4$$)
We need $$0 \leq 4 + 2y \leq 4$$, which gives $$-2 \leq y \leq 0$$.
From $$A$$, the valid values are $$y \in \{-2, -1, 0\}$$.
Pairs: $$(2, -2), (2, -1), (2, 0)$$. Count = 3.

Case 7: $$x = 3$$ (so $$x^2 = 9$$)
We need $$0 \leq 9 + 2y \leq 4$$, which gives $$-4.5 \leq y \leq -2.5$$.
From $$A$$, the valid values are $$y \in \{-3\}$$.
Pairs: $$(3, -3)$$. Count = 1.

So the total number of elements in $$R$$ is:
$$l = 1 + 3 + 2 + 3 + 2 + 3 + 1 = 15$$

Now we find $$m$$, the minimum number of elements to be added to make $$R$$ reflexive. For reflexivity, we need $$(x, x) \in R$$ for every $$x \in A$$. Let us check which diagonal pairs are already in $$R$$:

$$(-3, -3)$$: $$9 + 2(-3) = 3$$. Since $$0 \leq 3 \leq 4$$, this is in $$R$$. ✓
$$(-2, -2)$$: $$4 + 2(-2) = 0$$. Since $$0 \leq 0 \leq 4$$, this is in $$R$$. ✓
$$(-1, -1)$$: $$1 + 2(-1) = -1$$. Since $$-1 \lt 0$$, this is NOT in $$R$$. ✗
$$(0, 0)$$: $$0 + 2(0) = 0$$. Since $$0 \leq 0 \leq 4$$, this is in $$R$$. ✓
$$(1, 1)$$: $$1 + 2(1) = 3$$. Since $$0 \leq 3 \leq 4$$, this is in $$R$$. ✓
$$(2, 2)$$: $$4 + 2(2) = 8$$. Since $$8 \gt 4$$, this is NOT in $$R$$. ✗
$$(3, 3)$$: $$9 + 2(3) = 15$$. Since $$15 \gt 4$$, this is NOT in $$R$$. ✗

We need to add 3 pairs: $$(-1, -1), (2, 2), (3, 3)$$. So $$m = 3$$.

Therefore, $$l + m = 15 + 3 = 18$$.

Hence, the correct answer is Option D.

The given set is $$A=\{-2,-1,0,1,2,3\}$$, so $$|A|=6$$.

Relation $$R$$ is defined by  $$xRy \Longleftrightarrow y=\max\{x,1\}$$.

Step 1 - List the ordered pairs in $$R$$
 • If $$x\lt 1$$ (i.e. $$x=-2,-1,0$$), then $$\max\{x,1\}=1$$, hence $$y=1$$.
 • If $$x\ge 1$$ (i.e. $$x=1,2,3$$), then $$\max\{x,1\}=x$$, hence $$y=x$$.

Therefore
$$R=\{(-2,1),(-1,1),(0,1),(1,1),(2,2),(3,3)\}.$$

The number of elements in $$R$$ is
$$\ell = 6.$$

Step 2 - Make $$R$$ reflexive
A relation is reflexive if $$(a,a)\in R$$ for every $$a\in A$$.

Pairs already present: $$(1,1),(2,2),(3,3).$$
Missing reflexive pairs: $$(-2,-2),(-1,-1),(0,0).$$

Hence the minimum number of pairs to be added is
$$m = 3.$$

Step 3 - Make $$R$$ symmetric
A relation is symmetric if $$(a,b)\in R \Longrightarrow (b,a)\in R.$$

Check each pair in $$R$$:
 • $$(1,1),(2,2),(3,3)$$ are self-symmetric.
 • $$(-2,1)$$ needs $$(1,-2).$$
 • $$(-1,1)$$ needs $$(1,-1).$$
 • $(0,1)$$ needs $$(1,0).$$

These three required pairs are not in $$R$$, so we must add exactly three of them. Thus
$$n = 3.$$

Step 4 - Compute $$\ell + m + n$$
$$\ell + m + n = 6 + 3 + 3 = 12.$$

Therefore, $$\ell + m + n = 12$$, which corresponds to Option A.

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.

The underlying set is $$A=\{a,b,c,d,e,f\}$$ with $$|A|=6$$.

A relation on $$A$$ is a subset of $$A\times A$$. The required relations must satisfy two conditions:

1. Reflexive: for every $$x\in A$$, the ordered pair $$(x,x)$$ is in the relation.   Therefore the six diagonal pairs   $$\{(a,a),(b,b),(c,c),(d,d),(e,e),(f,f)\}$$   are compulsory.   So each admissible relation already contains $$6$$ pairs.

2. Symmetric: if $$(x,y)$$ is in the relation, then so is $$(y,x)$$.

The problem states that the relation must contain exactly $$10$$ ordered pairs. Since $$6$$ pairs are fixed by reflexivity, we still have to add $$10-6=4$$ more ordered pairs.

Because of symmetry, the non-diagonal pairs must be chosen in symmetric blocks: adding $$(x,y)$$ automatically forces us to add $$(y,x)$$. Consequently the extra $$4$$ pairs must come in $$\frac{4}{2}=2$$ symmetric blocks.

How many off-diagonal symmetric blocks exist? Each block corresponds to an unordered pair $$\{x,y\}$$ with $$x\neq y$$. The number of such unordered pairs in a 6-element set is $$\binom{6}{2}=15$$.

To create the relation we simply pick any $$2$$ of these $$15$$ unordered pairs; each chosen pair contributes its two ordered pairs and thus supplies the needed $$4$$ additional elements.

Hence the number of admissible relations is $$\binom{15}{2}=105$$.

Therefore, the set $$\mathcal{S}$$ contains $$\mathbf{105}$$ relations.

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: $$\frac{9}{11x} + \frac{13}{11}x^{10}$$

The set $$S$$ consists of all numbers of the form $$a+b\sqrt{2}$$ with $$a,b\in\mathbb{Z}$$, that is $$S=\mathbb{Z}[\sqrt{2}]$$, the smallest ring containing $$\mathbb{Z}$$ and $$\sqrt{2}$$.

Option A: $$\mathbb{Z}\cup T_1\cup T_2\subset S$$

• Every integer $$m\in\mathbb{Z}$$ can be written as $$m+0\sqrt{2}\in S$$.
• For $$T_1=\{(-1+\sqrt{2})^n:n\in\mathbb{N}\}$$ use the binomial theorem: $$(-1+\sqrt{2})^n=\sum_{k=0}^{n}\binom{n}{k}(-1)^{\,n-k}(\sqrt{2})^{\,k}$$ Separate even and odd $$k$$: $$=(-1)^n\!\!\sum_{\substack{k\text{ even}}}\binom{n}{k}2^{k/2}+(-1)^{n-1}\!\!\sum_{\substack{k\text{ odd}}}\binom{n}{k}2^{(k-1)/2}\sqrt{2}$$ The two sums are integers, so the whole expression is of the form $$a+b\sqrt{2}$$ with $$a,b\in\mathbb{Z}$$. Hence every element of $$T_1$$ lies in $$S$$.
• The same reasoning with the sign changed shows that every element of $$T_2=\{(1+\sqrt{2})^n:n\in\mathbb{N}\}$$ also belongs to $$S$$.
Therefore $$\mathbb{Z}\cup T_1\cup T_2\subset S$$ is TRUE.

Option B: $$T_1\cap\left(0,\dfrac1{2024}\right)=\varnothing$$

Put $$\alpha=\sqrt{2}-1\approx0.4142$$. Then $$(-1+\sqrt{2})^n=\alpha^{\,n}$$ because $$\alpha=(-1+\sqrt{2})$$. Since $$0\lt \alpha\lt 1$$, the sequence $$\alpha^{\,n}$$ decreases to $$0$$. Choose $$n$$ so that $$\alpha^{\,n}\lt\dfrac1{2024}$$. Solve: $$n\gt\dfrac{\ln(2024)}{-\ln\alpha}\approx\dfrac{7.61}{0.8814}\approx8.64$$ Taking $$n=9$$ gives $$\alpha^{\,9}\approx0.0032\lt\dfrac1{2024}\approx0.000494$$ false, let’s test $$n=13$$ gives $$\alpha^{\,13}\approx0.00020\lt0.000494$$. Hence an $$n$$ exists with $$\alpha^{\,n}\in\left(0,\dfrac1{2024}\right)$$, so the intersection is NOT empty. Therefore Option B is FALSE.

Option C: $$T_2\cap(2024,\infty)\neq\varnothing$$

Put $$\beta=1+\sqrt{2}\approx2.4142$$. The sequence $$\beta^{\,n}$$ is strictly increasing and unbounded. Estimate $$\beta^{\,9}=\beta^{\,8}\beta\,. $$ Now $$\beta^{\,4}\approx34.0$$, so $$\beta^{\,8}\approx34^{\,2}\approx1156$$. Then $$\beta^{\,9}\approx1156\times2.414\approx2790\gt2024$$. Thus $$\beta^{\,9}\in T_2$$ and $$\beta^{\,9}\gt2024$$, proving the intersection is non-empty. Option C is TRUE.

Option D: For $$a,b\in\mathbb{Z}$$, $$\cos\bigl(\pi(a+b\sqrt{2})\bigr)+i\sin\bigl(\pi(a+b\sqrt{2})\bigr)\in\mathbb{Z}$$ iff $$b=0$$.

Write the complex number in exponential form: $$e^{\,i\pi(a+b\sqrt{2})}=e^{\,i\pi a}\,e^{\,i\pi b\sqrt{2}}=(-1)^{a}\,e^{\,i\pi b\sqrt{2}}.$$ Its magnitude is $$1$$, so the only possible integer values are $$1$$ and $$-1$$. This requires $$e^{\,i\pi b\sqrt{2}}=\pm1\;$$, i.e. $$\pi b\sqrt{2}=k\pi$$ for some $$k\in\mathbb{Z}$$. Hence $$b\sqrt{2}=k$$. Because $$\sqrt{2}$$ is irrational, the only integer multiple of $$\sqrt{2}$$ that equals an integer is $$0$$, forcing $$b=0$$. Conversely, if $$b=0$$, the expression reduces to $$\cos(\pi a)+i\sin(\pi a)=(-1)^{a}\in\mathbb{Z}$$. Therefore the bi-implication holds and Option D is TRUE.

Correct statements: Option A, Option C, Option D.

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).

We need to find the number of integers in $$[100, 700]$$ that are neither multiples of 3 nor multiples of 4. From 100 to 700 inclusive there are $$700 - 100 + 1 = 601$$ integers. To determine how many are multiples of 3, note that the number of multiples of 3 up to n is $$\lfloor n/3 \rfloor$$, so in this range one finds $$\lfloor 700/3 \rfloor - \lfloor 99/3 \rfloor = 233 - 33 = 200$$ multiples of 3. Similarly, the count of multiples of 4 is given by $$\lfloor 700/4 \rfloor - \lfloor 99/4 \rfloor = 175 - 24 = 151$$. Since LCM(3, 4) = 12, the integers divisible by both 3 and 4 correspond to multiples of 12, namely $$\lfloor 700/12 \rfloor - \lfloor 99/12 \rfloor = 58 - 8 = 50$$.

Applying the inclusion-exclusion principle, the total number of integers that are multiples of 3 or 4 is $$|A \cup B| = |A| + |B| - |A \cap B| = 200 + 151 - 50 = 301$$. Therefore, the complementary set of integers that are neither multiples of 3 nor multiples of 4 has size $$601 - 301 = 300$$.

The correct answer is Option 3: 300.

 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:

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

Relation: $$(a, b)R(c, d)$$ if $$3ad - 7bc$$ is even.

1. Reflexive: $$(a, b)R(a, b) \implies 3ab - 7ba = -4ab$$, which is always even. Yes.
2. Symmetric: If $$3ad - 7bc = 2k$$, then $$3cb - 7da = -(3ad - 7bc) - 4ad + 4bc = -2k - 4ad + 4bc$$, which is also even. Yes.
3. Transitive: Let $(2, 1)R(6, 5)$ because $$3(2)(5) - 7(1)(6) = 30 - 42 = -12$$ (even).

Let $$(6, 5)R(3, 10)$$ because $$3(6)(10) - 7(5)(3) = 180 - 105 = 75$$ (odd).

Check a valid pair: $$(a,b)R(c,d)$$ and $$(c,d)R(e,f)$$. Due to the mix of multipliers (3 and 7), parity isn't preserved across three pairs. No.

Result: Reflexive and symmetric but not transitive. (Option B)

The relation $$R$$ on $$\mathbb{Z} \times \mathbb{Z}$$ is defined by $$(a, b)R(c, d)$$ if and only if $$ad - bc$$ is divisible by 5.

Check Reflexivity.

$$(a, b)R(a, b) \iff ab - ba = 0$$ is divisible by 5. $$\checkmark$$

$$R$$ is reflexive.

Check Symmetry.

If $$(a, b)R(c, d)$$, then $$5 \mid (ad - bc)$$.

We need to check: is $$5 \mid (cb - da)$$?

$$cb - da = -(ad - bc)$$. Since $$5 \mid (ad - bc)$$, we have $$5 \mid (-(ad-bc)) = cb - da$$. $$\checkmark$$

$$R$$ is symmetric.

Check Transitivity.

We need: if $$(a,b)R(c,d)$$ and $$(c,d)R(e,f)$$, then $$(a,b)R(e,f)$$.

Given: $$5 \mid (ad - bc)$$ and $$5 \mid (cf - de)$$.

Need to show: $$5 \mid (af - be)$$.

Counterexample:

Take $$(a, b) = (1, 1)$$, $$(c, d) = (5, 5)$$, $$(e, f) = (1, 2)$$.

Check $$(1,1)R(5,5)$$: $$ad - bc = 1 \cdot 5 - 1 \cdot 5 = 0$$. Divisible by 5. $$\checkmark$$

Check $$(5,5)R(1,2)$$: $$cf - de = 5 \cdot 2 - 5 \cdot 1 = 5$$. Divisible by 5. $$\checkmark$$

Check $$(1,1)R(1,2)$$: $$af - be = 1 \cdot 2 - 1 \cdot 1 = 1$$. NOT divisible by 5. $$\times$$

So transitivity fails.

$$R$$ is reflexive and symmetric but not transitive.

The correct answer is Option (1): $$\boxed{\text{Reflexive and symmetric but not transitive}}$$.

We need to find the number of elements in the smallest equivalence relation $$R$$ on $$\{1, 2, 3, 4\}$$ such that $$\{(1,2), (1,3)\} \subset R$$.

An equivalence relation must be reflexive, symmetric, and transitive.

Given: $$(1,2)$$ and $$(1,3)$$ must be in $$R$$.

Reflexive property requires: $$(1,1), (2,2), (3,3), (4,4)$$ must all be in $$R$$.

Since $$(1,2) \in R$$, we need $$(2,1) \in R$$.

Since $$(1,3) \in R$$, we need $$(3,1) \in R$$.

Since $$(2,1) \in R$$ and $$(1,3) \in R$$, we need $$(2,3) \in R$$.

By symmetry of $$(2,3)$$: $$(3,2) \in R$$.

Check: $$(3,1) \in R$$ and $$(1,2) \in R$$ gives $$(3,2) \in R$$ — already included.

The elements 1, 2, 3 are all equivalent to each other, and 4 is only equivalent to itself.

$$R = \{(1,1), (2,2), (3,3), (4,4), (1,2), (2,1), (1,3), (3,1), (2,3), (3,2)\}$$

$$|R| = 10$$

The correct answer is Option 1 — $$10$$.

The relation $$R$$ on $$A = \{1,2,3,4,5\}$$ is defined by $$xRy$$ iff $$4x \leq 5y$$, i.e., $$y \geq 4x/5$$.

Count elements m: For $$x=1$$: $$y \geq 0.8$$, so $$y \in \{1,2,3,4,5\}$$ (5 pairs). $$x=2$$: $$y \geq 1.6$$ (4). $$x=3$$: $$y \geq 2.4$$ (3). $$x=4$$: $$y \geq 3.2$$ (2). $$x=5$$: $$y \geq 4$$ (2). Total $$m = 16$$.

All 5 diagonal elements $$(i,i)$$ are in $$R$$. Off-diagonal pairs in R: 11.

Check symmetry: For each off-diagonal $$(x,y) \in R$$, check if $$(y,x) \in R$$. The only symmetric off-diagonal pair is $$\{4,5\}$$ (both $$(4,5)$$ and $$(5,4)$$ are in $$R$$). The remaining 9 off-diagonal pairs lack their reverse.

So $$n = 9$$ elements must be added.

$$m + n = 16 + 9 = 25$$.

The correct answer is Option 1: $$25$$.

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.

Determine which of $$R_1$$ and $$R_2$$ is an equivalence relation.

An equivalence relation must be reflexive, symmetric, and transitive.

Checking $$R_1$$: $$aR_1b \Leftrightarrow a^2 + b^2 = 1$$ for $$a, b \in \mathbb{R}$$.

Reflexive? We need $$aR_1a$$, i.e., $$a^2 + a^2 = 1 \Rightarrow 2a^2 = 1 \Rightarrow a = \pm\frac{1}{\sqrt{2}}$$. This fails for other values of $$a$$ (e.g., $$0^2 + 0^2 = 0 \neq 1$$). Not reflexive.

Since $$R_1$$ is not reflexive, it is NOT an equivalence relation.

Checking $$R_2$$: $$(a,b)R_2(c,d) \Leftrightarrow a + d = b + c$$ for $$(a,b),(c,d) \in \mathbb{N} \times \mathbb{N}$$.

Reflexive? $$(a,b)R_2(a,b) \Leftrightarrow a + b = b + a$$. True always. Reflexive.

Symmetric? If $$(a,b)R_2(c,d)$$, then $$a+d = b+c$$, which means $$c+b = d+a$$, so $$(c,d)R_2(a,b)$$. Symmetric.

Transitive? If $$(a,b)R_2(c,d)$$ and $$(c,d)R_2(e,f)$$, then $$a+d = b+c$$ and $$c+f = d+e$$. Adding: $$a+d+c+f = b+c+d+e$$, which simplifies to $$a+f = b+e$$, so $$(a,b)R_2(e,f)$$. Transitive.

$$R_2$$ is reflexive, symmetric, and transitive, so it IS an equivalence relation.

The correct answer is Option B: Only $$R_2$$ is an equivalence relation.

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 have a relation $$R$$ on $$\mathbb{N} \times \mathbb{N}$$ defined as: $$(x_1, y_1)\,R\,(x_2, y_2)$$ if and only if $$x_1 \leq x_2$$ or $$y_1 \leq y_2$$.

For any $$(a, b) \in \mathbb{N} \times \mathbb{N}$$, we need $$(a, b)\,R\,(a, b)$$, that is $$a \leq a$$ or $$b \leq b$$. Since $$a \leq a$$ is always true, $$R$$ is reflexive. ✓

To check symmetry, consider $$(1, 1)$$ and $$(2, 2)$$. We have $$(1, 1)\,R\,(2, 2)$$ because $$1 \leq 2$$, but $$(2, 2)\,R\,(1, 1)$$ fails since neither $$2 \leq 1$$ nor $$2 \leq 1$$ holds. Thus $$R$$ is not symmetric, so Statement (I) — “$$R$$ is reflexive but not symmetric” — is TRUE.

To test transitivity, assume $$(x_1, y_1)\,R\,(x_2, y_2)$$ and $$(x_2, y_2)\,R\,(x_3, y_3)$$ and check whether $$(x_1, y_1)\,R\,(x_3, y_3)$$ always holds. As a counterexample, take $$(5, 2)$$, $$(1, 5)$$, and $$(4, 1)$$. First, $$(5, 2)\,R\,(1, 5)$$ holds because $$5 \leq 1$$ is false but $$2 \leq 5$$ is true. Next, $$(1, 5)\,R\,(4, 1)$$ holds since $$1 \leq 4$$. However, $$(5, 2)\,R\,(4, 1)$$ fails because both $$5 \leq 4$$ and $$2 \leq 1$$ are false. Consequently, $$R$$ is not transitive, and Statement (II) — “$$R$$ is transitive” — is FALSE.

Therefore, only Statement (I) is correct, and the answer is Option D: Only (I) is correct.

We need to find M + N where M and N are the minimum elements to add to make $$R_1$$ and $$R_2$$ symmetric.

To find the elements of $$R_1$$, note that $$R_1 = \{(x,y) : 2x - 3y = 2\}$$ where $$x, y \in \{1, 2, \ldots, 20\}$$. Solving: $$x = \frac{3y + 2}{2}$$. For $$x$$ to be a positive integer, $$3y + 2$$ must be even, so $$y$$ must be even.

Checking even values of $$y$$: $$y = 2$$: $$x = 4$$ ✓ | $$y = 4$$: $$x = 7$$ ✓ | $$y = 6$$: $$x = 10$$ ✓ | $$y = 8$$: $$x = 13$$ ✓ | $$y = 10$$: $$x = 16$$ ✓ | $$y = 12$$: $$x = 19$$ ✓ | $$y = 14$$: $$x = 22 > 20$$ ✗.

Thus $$R_1 = \{(4,2), (7,4), (10,6), (13,8), (16,10), (19,12)\}$$ — 6 pairs.

For symmetry in $$R_1$$, if $$(a,b) \in R_1$$ then $$(b,a)$$ must also be in $$R_1$$. None of the reverse pairs $$(2,4), (4,7), (6,10), (8,13), (10,16), (12,19)$$ are in $$R_1$$, so all 6 reverse pairs must be added.

M = 6

Next, to find the elements of $$R_2$$, observe that $$R_2 = \{(x,y) : -5x + 4y = 0\}$$, i.e., $$y = \frac{5x}{4}$$. For $$y$$ to be a positive integer, $$x$$ must be a multiple of 4.

Checking: $$x = 4$$: $$y = 5$$ ✓ | $$x = 8$$: $$y = 10$$ ✓ | $$x = 12$$: $$y = 15$$ ✓ | $$x = 16$$: $$y = 20$$ ✓ | $$x = 20$$: $$y = 25 > 20$$ ✗.

Hence $$R_2 = \{(4,5), (8,10), (12,15), (16,20)\}$$ — 4 pairs.

For symmetry in $$R_2$$, the reverse pairs $$(5,4), (10,8), (15,12), (20,16)$$ are not present, so all 4 must be added.

N = 4

Finally, $$M + N = 6 + 4 = 10$$. The correct answer is Option 4: 10.

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

$$M$$ is the power set of $$S = \{1, 2, ..., 10\}$$. The relation $$R = \{(A, B) : A \cap B \neq \phi\}$$.

Reflexive? Is $$(A, A) \in R$$ for all $$A \in M$$? We need $$A \cap A \neq \phi$$, i.e., $$A \neq \phi$$. But $$\phi \in M$$ and $$\phi \cap \phi = \phi$$. So $$(\phi, \phi) \notin R$$. Not reflexive.

Symmetric? If $$A \cap B \neq \phi$$, then $$B \cap A \neq \phi$$. Yes, symmetric.

Transitive? Consider $$A = \{1\}$$, $$B = \{1, 2\}$$, $$C = \{2\}$$. Then $$A \cap B = \{1\} \neq \phi$$ and $$B \cap C = \{2\} \neq \phi$$, but $$A \cap C = \phi$$. Not transitive.

The relation is symmetric only. The answer corresponds to Option (4).

We need to find $$(g \circ g \circ g)(4)$$ where $$g(x) = (f \circ f)(x)$$ and $$f(x) = \frac{4x+3}{6x-4}$$. First, compute $$f(f(x))$$:

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

Simplify the numerator: $$\frac{4(4x+3) + 3(6x-4)}{6x-4} = \frac{16x+12+18x-12}{6x-4} = \frac{34x}{6x-4}$$ and 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, $$(g \circ g \circ g)(4) = g(g(g(4))) = g(g(4)) = g(4) = 4.$$

The correct answer is Option 4: 4.

Find $$\alpha^2 + \beta^3$$ for the domain $$(-\infty, \alpha) \cup [\beta, \infty)$$ of $$f(x) = \frac{\sqrt{x^2-25}}{4-x^2} + \log_{10}(x^2+2x-15)$$.

Requires $$x^2 - 25 \geq 0$$, i.e., $$x^2 \geq 25$$.

$$x \leq -5$$ 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.

We must find the second derivative of the function $$f(x)=\begin{cases}x^{3}\sin\!\left(\dfrac{1}{x}\right), & x\neq 0\\[4pt]0,&x=0\end{cases}$$ at two points: $$x=\dfrac{2}{\pi}$$ and $$x=0$$, and then match with the given options.

Case 1: Evaluation of $$f''\!\left(\dfrac{2}{\pi}\right)$$ (point away from zero, so usual differentiation rules apply).

Write $$f(x)=x^{3}\sin\!\left(\dfrac{1}{x}\right)$$ for $$x\neq 0$$.
First derivative (product rule):
$$f'(x)=\frac{d}{dx}\!\bigl[x^{3}\bigr]\sin\!\left(\frac{1}{x}\right)+x^{3}\,\frac{d}{dx}\!\left[\sin\!\left(\frac{1}{x}\right)\right]$$

Using $$\dfrac{d}{dx}\!\bigl[x^{3}\bigr]=3x^{2}$$ and $$\dfrac{d}{dx}\!\left[\sin\!\left(\frac{1}{x}\right)\right]=\cos\!\left(\frac{1}{x}\right)\!\left(-\frac{1}{x^{2}}\right)$$, we get
$$f'(x)=3x^{2}\sin\!\left(\frac{1}{x}\right)-x\cos\!\left(\frac{1}{x}\right).$$

Second derivative:
Differentiate each term separately.

• For $$3x^{2}\sin\!\left(\frac{1}{x}\right)$$:
$$\frac{d}{dx}\Bigl[3x^{2}\sin\!\left(\frac{1}{x}\right)\Bigr]=6x\sin\!\left(\frac{1}{x}\right)+3x^{2}\!\left[\cos\!\left(\frac{1}{x}\right)\!\left(-\frac{1}{x^{2}}\right)\right]$$ $$=6x\sin\!\left(\frac{1}{x}\right)-3\cos\!\left(\frac{1}{x}\right).$$

• For $$-x\cos\!\left(\frac{1}{x}\right)$$:
$$\frac{d}{dx}\Bigl[-x\cos\!\left(\frac{1}{x}\right)\Bigr]=-\,\Bigl[\cos\!\left(\frac{1}{x}\right)+x\!\left(\sin\!\left(\frac{1}{x}\right)\frac{1}{x^{2}}\right)\Bigr]$$ $$= -\cos\!\left(\frac{1}{x}\right)-\frac{\sin\!\left(\frac{1}{x}\right)}{x}.$$

Combine the two results to get
$$f''(x)=\bigl[6x\sin\!\left(\frac{1}{x}\right)-3\cos\!\left(\frac{1}{x}\right)\bigr]-\Bigl[\cos\!\left(\frac{1}{x}\right)+\frac{\sin\!\left(\frac{1}{x}\right)}{x}\Bigr]$$ $$=6x\sin\!\left(\frac{1}{x}\right)-\frac{\sin\!\left(\frac{1}{x}\right)}{x}-4\cos\!\left(\frac{1}{x}\right).$$

Factor the sine term:
$$f''(x)=\left(\frac{6x^{2}-1}{x}\right)\sin\!\left(\frac{1}{x}\right)-4\cos\!\left(\frac{1}{x}\right).$$

Now put $$x=\dfrac{2}{\pi}$$:

• $$\dfrac{1}{x}=\dfrac{\pi}{2}\quad\Longrightarrow\quad \sin\!\left(\frac{1}{x}\right)=\sin\!\left(\frac{\pi}{2}\right)=1,$$ $$\cos\!\left(\frac{1}{x}\right)=\cos\!\left(\frac{\pi}{2}\right)=0.$$ • $$x^{2}=\dfrac{4}{\pi^{2}}\quad\Longrightarrow\quad 6x^{2}-1=\frac{24}{\pi^{2}}-1.$$ • Evaluate the prefactor:
$$\frac{6x^{2}-1}{x}=\frac{\dfrac{24}{\pi^{2}}-1}{\dfrac{2}{\pi}}=\left(\frac{24}{\pi^{2}}-1\right)\!\left(\frac{\pi}{2}\right)=\frac{24}{2\pi}-\frac{\pi}{2}=\frac{12}{\pi}-\frac{\pi}{2}=\frac{24-\pi^{2}}{2\pi}.$$

Hence
$$f''\!\left(\frac{2}{\pi}\right)=\left(\frac{24-\pi^{2}}{2\pi}\right)\cdot 1-4\cdot 0=\frac{24-\pi^{2}}{2\pi}.$$
This matches Option A.

Case 2: Does $$f''(0)$$ exist?

First, find $$f'(0)$$ by definition:
$$f'(0)=\lim_{h\to 0}\frac{f(h)-f(0)}{h}=\lim_{h\to 0}\frac{h^{3}\sin\!\left(\frac{1}{h}\right)}{h}= \lim_{h\to 0}h^{2}\sin\!\left(\frac{1}{h}\right)=0,$$ because $$|h^{2}\sin(1/h)|\le h^{2}\to 0.$$

For the second derivative, use
$$f''(0)=\lim_{h\to 0}\frac{f'(h)-f'(0)}{h}=\lim_{h\to 0}\frac{f'(h)}{h}.$$

Recall $$f'(h)=3h^{2}\sin\!\left(\frac{1}{h}\right)-h\cos\!\left(\frac{1}{h}\right).$$ Hence
$$\frac{f'(h)}{h}=3h\sin\!\left(\frac{1}{h}\right)-\cos\!\left(\frac{1}{h}\right).$$

As $$h\to 0$$, the term $$3h\sin\!\left(\frac{1}{h}\right)\to 0,$$ but $$\cos\!\left(\frac{1}{h}\right)$$ oscillates between $$-1$$ and $$1$$ without settling at a single value. Therefore the limit does not exist, so $$f''(0)$$ is undefined.

Thus both statements “$$f''(0)=1$$” and “$$f''(0)=0$$” are false.

Only Option A is correct.

We are given $$f(x) = e^{x^3 - 3x + 1}$$ defined on $$(-\infty, -1]$$ mapping onto $$[a, b]$$ as a one-one and onto function. We need to find the distance of point $$P(2b+4, a+2)$$ from the line $$x + e^{-3}y = 4$$.

Let $$g(x) = x^3 - 3x + 1$$, so $$f(x) = e^{g(x)}$$. Then $$g'(x) = 3x^2 - 3 = 3(x-1)(x+1).$$ On $$(-\infty, -1)$$ we have $$g'(x) > 0$$ (both factors have the same sign), 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 $$b = e^3$$.

Substituting these values into the coordinates of $$P$$ gives $$P(2b + 4, a + 2) = P(2e^3 + 4, 2).$$

The line $$x + e^{-3}y = 4$$ can be rewritten as $$x + e^{-3}y - 4 = 0$$. 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 $$\frac{|2e^3 + 4 + 2e^{-3} - 4|}{\sqrt{1 + e^{-6}}} = \frac{|2e^3 + 2e^{-3}|}{\sqrt{1 + e^{-6}}} = \frac{2(e^3 + e^{-3})}{\sqrt{1 + e^{-6}}} = \frac{2e^{-3}(e^6 + 1)}{e^{-3}\sqrt{e^6 + 1}} = \frac{2(e^6+1)}{\sqrt{e^6+1}} = 2\sqrt{e^6 + 1} = 2\sqrt{1 + e^6}.$$

Option 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.

Using inclusion-exclusion: |M∪P∪C| = |M|+|P|+|C|-|M∩P|-|P∩C|-|M∩C|+|M∩P∩C|

Students studying at least one = 220 - 10 = 210

210 = |M|+|P|+|C| - 40 - 30 - 50 + |M∩P∩C|

|M|+|P|+|C| + |M∩P∩C| = 330

Min sum: 125+85+75 = 285, max sum: 130+95+90 = 315

Min |M∩P∩C| = 330-315 = 15, Max |M∩P∩C| = 330-285 = 45

But we also need each pairwise intersection ≥ triple intersection.

Max triple ≤ min(40,30,50) = 30

So m = 15, n = 30, m+n = 45

The answer is 45.

$$R=\{(1,2),(2,3),(1,4)\}$$. Equivalence relation $$S\supset R$$: must be reflexive, symmetric, transitive.

From $$(1,2)$$: need $$(2,1)$$. From $$(2,3)$$: need $$(3,2)$$. From $$(1,4)$$: need $$(4,1)$$.

Transitivity: $$(1,2),(2,3)\Rightarrow(1,3)$$, then $$(3,1)$$. $$(2,1),(1,4)\Rightarrow(2,4)$$, then $$(4,2)$$. $$(3,2),(2,4)\Rightarrow(3,4)$$, then $$(4,3)$$.

All elements {1,2,3,4} are in same equivalence class. So $$S=\{1,2,3,4\}^2$$, n=16.

The answer is $$\boxed{16}$$.

We have the set $$A = \{1,2,3,\dots,100\}$$ and the relation $$R$$ defined by $$(x,y)\in R\iff 2x=3y\,. $$

First we rewrite the equation $$2x=3y$$ in parametric form. We note that $$\gcd(2,3)=1$$, so there exists an integer $$k$$ such that

$$x=3k,\quad y=2k\,. $$

Since $$x\in A$$, we have $$1\le 3k\le 100\,$$ which gives

$$1\le k\le \frac{100}{3}\quad\Longrightarrow\quad 1\le k\le 33\,. $$

Also, since $$y\in A$$, we have $$1\le 2k\le 100\,$$ which gives

$$1\le k\le 50\,. $$

Combining these two bounds on $$k$$ gives

$$1\le k\le 33\,, $$ so there are exactly $$33$$ integer values of $$k$$. Therefore, the number of ordered pairs in $$R$$ is $$33\,. $$

Next, we consider a relation $$R_1$$ on $$A$$ such that $$R\subset R_1$$ and $$R_1$$ is symmetric. By definition of symmetry:

$$\text{if }(x,y)\in R_1\text{ then }(y,x)\in R_1\,. $$

Since $$R\subset R_1$$, all $$33$$ pairs of the form $$(3k,2k)$$ must lie in $$R_1$$. For each such pair, symmetry forces the inclusion of the converse pair $$(2k,3k)$$. Because $$2k\neq 3k$$ for all positive $$k$$, these converse pairs are all distinct from the original ones.

Thus the minimal symmetric relation $$R_1$$ contains the original $$33$$ pairs plus the $$33$$ converse pairs, for a total of

$$n = 33 + 33 = 66\,. $$

Therefore, the minimum value of $$n$$ is 66.

Find the number of elements in $$S = \{(x,y,z): x,y,z \in \mathbb{Z}, x+2y+3z = 42, x,y,z \geq 0\}$$. We seek nonnegative integer solutions to $$x + 2y + 3z = 42$$. For each integer $$z$$ with $$0 \le z \le 14$$ (since $$3z \le 42$$), we set $$R = 42 - 3z$$ and count the nonnegative solutions to $$x + 2y = R$$. Because $$y$$ may range from $$0$$ to $$\lfloor R/2 \rfloor$$, and for each $$y$$ the corresponding $$x = R - 2y$$ is nonnegative, there are $$\lfloor R/2 \rfloor + 1$$ solutions for each fixed $$z$$.

Specifically, when $$z=0$$, we have $$R=42$$ and count $$21+1=22$$; when $$z=1$$, $$R=39$$ giving $$19+1=20$$; when $$z=2$$, $$R=36$$ giving $$18+1=19$$; when $$z=3$$, $$R=33$$ giving $$16+1=17$$; when $$z=4$$, $$R=30$$ giving $$15+1=16$$; when $$z=5$$, $$R=27$$ giving $$13+1=14$$; when $$z=6$$, $$R=24$$ giving $$12+1=13$$; when $$z=7$$, $$R=21$$ giving $$10+1=11$$; when $$z=8$$, $$R=18$$ giving $$9+1=10$$; when $$z=9$$, $$R=15$$ giving $$7+1=8$$; when $$z=10$$, $$R=12$$ giving $$6+1=7$$; when $$z=11$$, $$R=9$$ giving $$4+1=5$$; when $$z=12$$, $$R=6$$ giving $$3+1=4$$; when $$z=13$$, $$R=3$$ giving $$1+1=2$$; and when $$z=14$$, $$R=0$$ giving $$0+1=1$$ solutions.

Summing these counts for $$z=0,1,\dots,14$$ yields $$22+20+19+17+16+14+13+11+10+8+7+5+4+2+1 = 169$$. Therefore, the number of elements in $$S$$ is 169.

We have $$A = \{1, 2, 3, \ldots, 20\}$$, $$R_1 = \{(a,b) : b \text{ is divisible by } a\}$$, and $$R_2 = \{(a,b) : a \text{ is an integral multiple of } b\}$$.

In this context, $$(a,b)\in R_1$$ means $$a\mid b$$, while $$(a,b)\in R_2$$ means $$b\mid a$$, so that $$R_2 = \{(a,b) : a \text{ is divisible by } b\}$$.

The intersection $$R_1 \cap R_2$$ consists of all pairs $$(a,b)$$ for which $$a\mid b$$ and $$b\mid a$$, forcing $$a=b$$. Hence $$R_1 \cap R_2 = \{(a,a) : a \in A\}$$, which has 20 elements.

To find $$|R_1|$$, note that for each $$a$$ the number of multiples of $$a$$ in $$\{1,\dots,20\}$$ is $$\lfloor 20/a \rfloor$$, giving

$$|R_1| = \sum_{a=1}^{20} \lfloor 20/a \rfloor = 20 + 10 + 6 + 5 + 4 + 3 + 2 + 2 + 2 + 2 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 66$$

Since $$R_1 - R_2 = R_1 \setminus (R_1 \cap R_2)$$, we obtain

$$|R_1 - R_2| = |R_1| - |R_1 \cap R_2| = 66 - 20 = 46$$

The answer is 46.

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.

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

We need to find the number of symmetric relations on the set $$\{1, 2, 3, 4\}$$ that are NOT reflexive.

A relation $$R$$ on a set $$A$$ is symmetric if whenever $$(a, b)\in R$$ then $$(b, a)\in R$$. For a set with $$n$$ elements, a relation is a subset of $$A\times A$$, which has $$n^2$$ possible ordered pairs. In a symmetric relation we make independent choices for each diagonal pair $$(i, i)$$ (there are $$n$$ such pairs) and for each unordered off-diagonal pair $$\{(i,j),(j,i)\}$$ (there are $$\binom{n}{2}$$ such pairs), since each off-diagonal pair is either included or excluded together.

When $$n=4$$, there are $$4$$ diagonal pairs and $$\binom{4}{2}=6$$ off-diagonal pairs, so there are $$4+6=10$$ independent choices overall. Hence the total number of symmetric relations on a four-element set is $$2^{10}=1024$$.

A relation is reflexive if $$(i,i)\in R$$ for all $$i\in A$$. To count symmetric relations that are also reflexive, we must include all four diagonal pairs, leaving only the six off-diagonal pairs as free choices. Thus there are $$2^6=64$$ reflexive symmetric relations.

Subtracting gives the number of symmetric relations that are not reflexive: $$1024-64=960$$.

The answer is $$\boxed{960}$$.

$$a = [a] + \{a\}$$. 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.

Let $$P(S)$$ denote the power set of $$S = \{1, 2, 3, \ldots, 10\}$$.

Analysis of $$R_1$$:

$$AR_1B$$ if $$(A \cap B^c) \cup (B \cap A^c) = \phi$$

The expression $$(A \cap B^c) \cup (B \cap A^c)$$ is the symmetric difference $$A \triangle B$$.

$$A \triangle B = \phi$$ if and only if $$A = B$$.

So $$R_1$$ is the equality relation: $$AR_1B \iff A = B$$.

Check equivalence for $$R_1$$:

- Reflexive: $$A = A$$ for all $$A \in P(S)$$. $$\checkmark$$

- Symmetric: If $$A = B$$, then $$B = A$$. $$\checkmark$$

- Transitive: If $$A = B$$ and $$B = C$$, then $$A = C$$. $$\checkmark$$

Therefore, $$R_1$$ is an equivalence relation.

Analysis of $$R_2$$:

$$AR_2B$$ if $$A \cup B^c = B \cup A^c$$

We check when this equality holds by examining elements of $$S$$:

$$x \in A \cup B^c \iff x \in A \text{ or } x \notin B$$

$$x \in B \cup A^c \iff x \in B \text{ or } x \notin A$$

If $$x \in A$$ but $$x \notin B$$: Then $$x \in A \cup B^c$$ (true, since $$x \in A$$), but for $$x \in B \cup A^c$$ we need $$x \in B$$ (false) or $$x \notin A$$ (false). So $$x \notin B \cup A^c$$. The sets differ.

Similarly, if $$x \in B$$ but $$x \notin A$$: Then $$x \in B \cup A^c$$ (true), but $$x \notin A \cup B^c$$ (false). The sets differ.

Therefore $$A \cup B^c = B \cup A^c$$ requires that no element belongs to exactly one of $$A$$ or $$B$$, meaning $$A = B$$.

So $$R_2$$ is also the equality relation: $$AR_2B \iff A = B$$.

Check equivalence for $$R_2$$:

- Reflexive: $$A = A$$ for all $$A \in P(S)$$. $$\checkmark$$

- Symmetric: If $$A = B$$, then $$B = A$$. $$\checkmark$$

- Transitive: If $$A = B$$ and $$B = C$$, then $$A = C$$. $$\checkmark$$

Therefore, $$R_2$$ is an equivalence relation.

Since both $$R_1$$ and $$R_2$$ are equivalence relations, the correct answer is Option A: both $$R_1$$ and $$R_2$$ are equivalence relations.

Given the relation $$R$$ on $$\mathbb{N}$$ defined by $$a R b$$ if $$2a + 3b$$ is a multiple of $$5$$.

First, we show that $$R$$ is reflexive by checking whether $$a R a$$ holds. We have $$2a + 3a = 5a$$, which is clearly divisible by $$5$$, so $$R$$ is reflexive.

Next, to verify that $$R$$ is symmetric, suppose $$a R b$$; that is, assume $$2a + 3b = 5k$$ for some integer $$k$$. Consider the expression $$2b + 3a$$. We can write

$$2b + 3a = 5(a + b) - (2a + 3b) = 5(a + b) - 5k = 5(a + b - k).$$

Since this is a multiple of $$5$$, it follows that $$b R a$$, and hence $$R$$ is symmetric.

Finally, to prove transitivity, suppose $$a R b$$ and $$b R c$$, so that $$2a + 3b = 5k$$ and $$2b + 3c = 5m$$ for some integers $$k, m$$. Adding these two equations yields

$$2a + 5b + 3c = 5(k + m).$$

Rewriting, we have

$$2a + 3c = 5(k + m) - 5b = 5(k + m - b),$$

which is a multiple of $$5$$. Therefore $$a R c$$, and $$R$$ is transitive.

Since $$R$$ is reflexive, symmetric, and transitive, it is an equivalence relation. The correct answer is Option D.

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:

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

Now, the negation is:

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 need to determine the properties of $$R = \{(a, b) : \gcd(a, b) = 1, 2a \neq b, a, b \in \mathbb{Z}\}$$.

Reflexivity: For $$(a, a) \in R$$: need $$\gcd(a, a) = 1$$, which requires $$|a| = 1$$. Since $$(2, 2) \notin R$$, R is not reflexive.

Symmetry: Consider $$(2, 1)$$: $$\gcd(2, 1) = 1$$ and $$4 \neq 1$$, so $$(2, 1) \in R$$. But $$(1, 2)$$: $$\gcd(1, 2) = 1$$ and $$2(1) = 2 = b$$, so $$(1, 2) \notin R$$. R is not symmetric.

Transitivity: Consider $$a = 4, b = 3, c = 8$$. $$(4,3) \in R$$ since $$\gcd(4,3)=1$$ and $$8 \neq 3$$. $$(3,8) \in R$$ since $$\gcd(3,8)=1$$ and $$6 \neq 8$$. But $$(4,8) \notin R$$ since $$\gcd(4,8)=4 \neq 1$$. R is not transitive.

The correct answer is Option 4: neither symmetric nor transitive.

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.

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.

Using the inclusion-exclusion principle:

$$|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |B \cap C| - |A \cap C| + |A \cap B \cap C|$$

We are given that

$$|A| = 48$$, $$|B| = 25$$, $$|C| = 18$$

$$|A \cup B \cup C| = 60$$ (total men)

$$|A \cap B \cap C| = 5$$ (men with medals in all three events)

Substituting:

$$60 = 48 + 25 + 18 - (|A \cap B| + |B \cap C| + |A \cap C|) + 5$$

$$60 = 96 - (|A \cap B| + |B \cap C| + |A \cap C|)$$

$$|A \cap B| + |B \cap C| + |A \cap C| = 36$$

The number of men who received medals in exactly two events:

$$= (|A \cap B| + |B \cap C| + |A \cap C|) - 3|A \cap B \cap C|$$

$$= 36 - 3(5) = 36 - 15 = 21$$

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

We need to find the minimum number of elements to add to $$R = \{(a,b), (b,c)\}$$ on $$\{a, b, c\}$$ to make it symmetric and transitive.

To begin,

For symmetry, if $$(x,y) \in R$$, then $$(y,x) \in R$$.

Add: $$(b,a)$$ and $$(c,b)$$.

$$R = \{(a,b), (b,a), (b,c), (c,b)\}$$

Next,

For transitivity, if $$(x,y) \in R$$ and $$(y,z) \in R$$, then $$(x,z) \in R$$.

From $$(a,b)$$ and $$(b,a)$$: need $$(a,a)$$ ✓ Add

From $$(a,b)$$ and $$(b,c)$$: need $$(a,c)$$ ✓ Add

From $$(b,a)$$ and $$(a,b)$$: need $$(b,b)$$ ✓ Add

From $$(c,b)$$ and $$(b,a)$$: need $$(c,a)$$ ✓ Add

From $$(c,b)$$ and $$(b,c)$$: need $$(c,c)$$ ✓ Add

We also need to check new pairs: $$(a,c)$$ and $$(c,a)$$ → $$(a,a)$$ ✓ already added. $$(a,c)$$ and $$(c,b)$$ → $$(a,b)$$ ✓ already in R.

Total elements added: $$(b,a), (c,b), (a,a), (a,c), (b,b), (c,a), (c,c) = 7$$

The correct answer is Option 2: $$7$$.

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)

Let $$R = \{(a, b) : 3a - 3b + \sqrt{7}$$ is an irrational number$$\}$$. We need to determine the properties of $$R$$.

To begin,

For $$(a, a)$$: $$3a - 3a + \sqrt{7} = \sqrt{7}$$, which is irrational. So $$(a, a) \in R$$ for all $$a \in \mathbb{R}$$. $$R$$ is reflexive.

Next,

If $$(a, b) \in R$$, then $$3a - 3b + \sqrt{7}$$ is irrational. For $$(b, a)$$: $$3b - 3a + \sqrt{7} = -(3a - 3b) + \sqrt{7}$$.

Take $$a = 0, b = \dfrac{\sqrt{7}}{3}$$: $$3(0) - 3\cdot\dfrac{\sqrt{7}}{3} + \sqrt{7} = -\sqrt{7} + \sqrt{7} = 0$$, which is rational. So $$(0, \dfrac{\sqrt{7}}{3}) \notin R$$.

Take $$a = \dfrac{\sqrt{7}}{3}, b = 0$$: $$3\cdot\dfrac{\sqrt{7}}{3} - 0 + \sqrt{7} = \sqrt{7} + \sqrt{7} = 2\sqrt{7}$$, which is irrational. So $$(\dfrac{\sqrt{7}}{3}, 0) \in R$$.

But $$(0, \dfrac{\sqrt{7}}{3}) \notin R$$. So $$R$$ is not symmetric.

From this,

Let $$a = \dfrac{\sqrt{7}}{3}, b = 0, c = \dfrac{-\sqrt{7}}{3}$$.

$$(a, b)$$: $$3 \cdot \dfrac{\sqrt{7}}{3} - 0 + \sqrt{7} = 2\sqrt{7}$$ (irrational) $$\checkmark$$

$$(b, c)$$: $$0 - 3\cdot\dfrac{-\sqrt{7}}{3} + \sqrt{7} = \sqrt{7} + \sqrt{7} = 2\sqrt{7}$$ (irrational) $$\checkmark$$

$$(a, c)$$: $$3\cdot\dfrac{\sqrt{7}}{3} - 3\cdot\dfrac{-\sqrt{7}}{3} + \sqrt{7} = \sqrt{7} + \sqrt{7} + \sqrt{7} = 3\sqrt{7}$$ (irrational) $$\checkmark$$

This particular case works, but let's try another: $$a = \dfrac{2\sqrt{7}}{3}, b = \dfrac{\sqrt{7}}{3}, c = 0$$.

$$(a, b)$$: $$2\sqrt{7} - \sqrt{7} + \sqrt{7} = 2\sqrt{7}$$ (irrational) $$\checkmark$$

$$(b, c)$$: $$\sqrt{7} - 0 + \sqrt{7} = 2\sqrt{7}$$ (irrational) $$\checkmark$$

$$(a, c)$$: $$2\sqrt{7} - 0 + \sqrt{7} = 3\sqrt{7}$$ (irrational) $$\checkmark$$

Let me find a counterexample. Take $$b = \dfrac{\sqrt{7}}{6}$$, $$a = 0$$, $$c = \dfrac{\sqrt{7}}{3}$$:

$$(a,b)$$: $$0 - \dfrac{\sqrt{7}}{2} + \sqrt{7} = \dfrac{\sqrt{7}}{2}$$ (irrational) $$\checkmark$$

$$(b,c)$$: $$\dfrac{\sqrt{7}}{2} - \sqrt{7} + \sqrt{7} = \dfrac{\sqrt{7}}{2}$$ (irrational) $$\checkmark$$

$$(a,c)$$: $$0 - \sqrt{7} + \sqrt{7} = 0$$ (rational) $$\times$$

So $$R$$ is not transitive.

The correct answer is Option A: Reflexive but neither symmetric nor transitive.

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.

$$R = \{((a_1,b_1),(a_2,b_2)) \in A \times B \times A \times B: a_1 | b_2 \text{ and } a_2 | b_1\}$$

For each pair $$(a_1, b_2)$$ where $$a_1 | b_2$$: A={2,3,4}, B={8,9,12}.

2|8✓, 2|9✗, 2|12✓, 3|8✗, 3|9✓, 3|12✓, 4|8✓, 4|9✗, 4|12✓

Count of $$(a_1,b_2)$$ with $$a_1|b_2$$: 6 pairs.

Similarly for $$(a_2,b_1)$$ with $$a_2|b_1$$: same 6 pairs.

Total elements = 6 × 6 = 36.

The correct answer is Option 1: 36.

$$f : \mathbb{R} - \{0, 1\} \to \mathbb{R}$$ 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 $$\left(\frac{9}{4}\right)$$

To solve this, we use the property of Cartesian products and combinations.

1. Find the total elements in $$A \times B$$

The number of elements in the Cartesian product $$A \times B$$ is:

$$n(A \times B) = n(A) \times n(B) = 5 \times 2 = 10$$

2. Set up the Combination Sum

We need the number of subsets having "at least 3 and at most 6" elements. This is the sum of choosing 3, 4, 5, and 6 elements out of 10:

$$\text{Number of subsets} = ^{10}C_3 + ^{10}C_4 + ^{10}C_5 + ^{10}C_6$$

3. Calculate each term

• $$^{10}C_3 = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$$
• $$^{10}C_4 = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$$
• $$^{10}C_5 = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252$$
• $$^{10}C_6 = ^{10}C_4 = 210$$

4. Final Result

$$\text{Sum} = 120 + 210 + 252 + 210 = \mathbf{792}$$

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.