JEE Main 2021 Mathematics Question Paper with Answer and Solution

(B) Let the point of intersection be $(x_1, y_1)$.
For the first curve $\frac{x^2}{a} + \frac{y^2}{b} = 1$,the slope $m_1$ at $(x_1, y_1)$ is given by differentiating: $\frac{2x_1}{a} + \frac{2y_1}{b} \frac{dy}{dx} = 0 \Rightarrow m_1 = -\frac{bx_1}{ay_1}$.
For the second curve $\frac{x^2}{c} + \frac{y^2}{d} = 1$,the slope $m_2$ at $(x_1, y_1)$ is $m_2 = -\frac{dx_1}{cy_1}$.
Since the curves intersect at $90^{\circ}$,$m_1 m_2 = -1$.
$(-\frac{bx_1}{ay_1})(-\frac{dx_1}{cy_1}) = -1 \Rightarrow \frac{bd x_1^2}{ac y_1^2} = -1$,which implies $bd x_1^2 = -ac y_1^2$.
Subtracting the two curve equations: $x_1^2(\frac{1}{a} - \frac{1}{c}) + y_1^2(\frac{1}{b} - \frac{1}{d}) = 0$.
$x_1^2(\frac{c-a}{ac}) + y_1^2(\frac{d-b}{bd}) = 0$.
Substituting $x_1^2 = -\frac{ac}{bd} y_1^2$,we get $(-\frac{ac}{bd}) y_1^2(\frac{c-a}{ac}) + y_1^2(\frac{d-b}{bd}) = 0$.
$-\frac{c-a}{bd} + \frac{d-b}{bd} = 0$ $\Rightarrow a-c+d-b = 0$ $\Rightarrow a-b = c-d$.

(D) The given limit is of the form $1^{\infty}$.
Let $L = \lim _{n \rightarrow \infty}\left(1+\frac{H_n}{n^{2}}\right)^{n}$,where $H_n = 1+\frac{1}{2}+\ldots+\frac{1}{n}$.
Using the formula $\lim _{n \rightarrow \infty}(1+f(n))^{g(n)} = e^{\lim _{n \rightarrow \infty} f(n)g(n)}$,we get:
$L = \exp \left(\lim _{n \rightarrow \infty} \frac{H_n}{n^2} \cdot n\right) = \exp \left(\lim _{n \rightarrow \infty} \frac{H_n}{n}\right)$.
We know that $H_n \approx \ln(n) + \gamma$ (where $\gamma$ is the Euler-Mascheroni constant).
Thus,$\lim _{n \rightarrow \infty} \frac{\ln(n) + \gamma}{n} = 0$.
Therefore,$L = e^0 = 1$.

(B) The quadratic equation is $ax^{2} + bx + c = 0$.
For the equation to have equal roots,the discriminant $D$ must be zero,i.e.,$D = b^{2} - 4ac = 0$,which implies $b^{2} = 4ac$.
Since $a, b, c$ are obtained by throwing a die three times,each variable can take values from the set ${1, 2, 3, 4, 5, 6}$. The total number of outcomes is $6 \times 6 \times 6 = 216$.
We need to find the number of triplets $(a, b, c)$ such that $b^{2} = 4ac$:
$1$. If $a=1, c=1$,then $b^{2} = 4(1)(1) = 4 \Rightarrow b=2$. Triplets: $(1, 2, 1)$.
$2$. If $a=1, c=4$,then $b^{2} = 4(1)(4) = 16 \Rightarrow b=4$. Triplets: $(1, 4, 4)$.
$3$. If $a=4, c=1$,then $b^{2} = 4(4)(1) = 16 \Rightarrow b=4$. Triplets: $(4, 4, 1)$.
$4$. If $a=2, c=2$,then $b^{2} = 4(2)(2) = 16 \Rightarrow b=4$. Triplets: $(2, 4, 2)$.
$5$. If $a=3, c=3$,then $b^{2} = 4(3)(3) = 36 \Rightarrow b=6$. Triplets: $(3, 6, 3)$.
Total favorable outcomes = $5$.
Therefore,the required probability is $\frac{5}{216}$.

(D) The prime factorization of $24$ is $2^{3} \times 3^{1}$.
Let $x = 2^{\alpha_{1}} \times 3^{\beta_{1}}$,$y = 2^{\alpha_{2}} \times 3^{\beta_{2}}$,and $z = 2^{\alpha_{3}} \times 3^{\beta_{3}}$,where $\alpha_{i}, \beta_{i} \ge 0$.
Since $xyz = 2^{3} \times 3^{1}$,we have $\alpha_{1} + \alpha_{2} + \alpha_{3} = 3$ and $\beta_{1} + \beta_{2} + \beta_{3} = 1$.
The number of non-negative integer solutions for $\alpha_{1} + \alpha_{2} + \alpha_{3} = 3$ is given by the stars and bars formula $\binom{n+k-1}{k-1} = \binom{3+3-1}{3-1} = \binom{5}{2} = 10$.
The number of non-negative integer solutions for $\beta_{1} + \beta_{2} + \beta_{3} = 1$ is $\binom{1+3-1}{3-1} = \binom{3}{2} = 3$.
Therefore,the total number of positive integral solutions is $10 \times 3 = 30$.

(A) For the quadratic inequality $ax^{2}+bx+c > 0$ to be valid for all $x \in \mathbb{R}$,the discriminant $D$ must be less than $0$ and $a > 0$.
Here,$a = 1 > 0$,so we only need $D < 0$.
$D = [-2(3k-1)]^{2} - 4(1)(8k^{2}-7) < 0$
$4(9k^{2}-6k+1) - 4(8k^{2}-7) < 0$
Dividing by $4$:
$9k^{2}-6k+1 - 8k^{2}+7 < 0$
$k^{2}-6k+8 < 0$
$(k-2)(k-4) < 0$
This implies $k \in (2, 4)$.
Since $k$ is an integer,the only integer value in the interval $(2, 4)$ is $k = 3$.

(D) The given statement is $A$ $\rightarrow (B$ $\rightarrow A)$.
Using the implication law $p \rightarrow q \equiv \sim p \vee q$,we get:
$A$ $\rightarrow (B$ $\rightarrow A) \equiv \sim A \vee (\sim B \vee A)$
By the associative and commutative laws:
$\equiv (\sim A \vee A) \vee \sim B$
$\equiv T \vee \sim B \equiv T$ (Tautology).
Now,check the options for a tautology or equivalent form:
Option $D$ is $A \rightarrow (A \vee B) \equiv \sim A \vee (A \vee B) \equiv (\sim A \vee A) \vee B \equiv T \vee B \equiv T$.
Since both the original expression and option $D$ are tautologies,they are equivalent.

(D) Let $a_{n}$ be the side length of square $A_{n}$.
Given that the side length of $A_{n}$ equals the diagonal of $A_{n+1}$,we have $a_{n} = \sqrt{2} a_{n+1}$,which implies $a_{n+1} = \frac{a_{n}}{\sqrt{2}}$.
This forms a geometric progression for the side lengths with first term $a_{1} = 12$ and common ratio $r = \frac{1}{\sqrt{2}}$.
The side length $a_{n}$ is given by $a_{n} = a_{1} \times r^{n-1} = 12 \times \left(\frac{1}{\sqrt{2}}\right)^{n-1}$.
The area of $A_{n}$ is $(a_{n})^2 = 144 \times \left(\frac{1}{2}\right)^{n-1} = \frac{144}{2^{n-1}}$.
We want the area to be less than $1$,so $\frac{144}{2^{n-1}} < 1$.
This implies $2^{n-1} > 144$.
Since $2^{7} = 128$ and $2^{8} = 256$,we must have $n-1 \geq 8$.
Therefore,$n \geq 9$. The smallest value of $n$ is $9$.

(A) We need to form $3$-digit numbers using the set $\{1, 2, 3, 4, 5\}$ without repetition.
$1$. Divisibility by $3$: $A$ number is divisible by $3$ if the sum of its digits is divisible by $3$. The possible triplets from $\{1, 2, 3, 4, 5\}$ whose sum is a multiple of $3$ are:
$(1, 2, 3) \rightarrow 3! = 6$ numbers
$(1, 3, 5) \rightarrow 3! = 6$ numbers
$(2, 3, 4) \rightarrow 3! = 6$ numbers
$(3, 4, 5) \rightarrow 3! = 6$ numbers
Total divisible by $3 = 6 + 6 + 6 + 6 = 24$.
$2$. Divisibility by $5$: $A$ number is divisible by $5$ if its last digit is $5$. The possible $3$-digit numbers are of the form $XY5$,where $X, Y \in \{1, 2, 3, 4\}$.
The number of such arrangements is $4 \times 3 = 12$.
$3$. Divisibility by both $3$ and $5$: These are numbers ending in $5$ whose sum of digits is a multiple of $3$. The possible sets are $\{1, 3, 5\}$ and $\{3, 4, 5\}$.
For $\{1, 3, 5\}$,numbers ending in $5$ are $135$ and $315$ ($2$ numbers).
For $\{3, 4, 5\}$,numbers ending in $5$ are $345$ and $435$ ($2$ numbers).
Total divisible by both $= 2 + 2 = 4$.
$4$. Using the Principle of Inclusion-Exclusion:
Total $= n(3) + n(5) - n(3 \cap 5) = 24 + 12 - 4 = 32$.

(B) Given lines are $(\sqrt{3})kx + ky = 4\sqrt{3}$ and $\sqrt{3}x - y = 4\sqrt{3}k$.
From the first equation,$k = \frac{4\sqrt{3}}{\sqrt{3}x + y}$.
From the second equation,$k = \frac{\sqrt{3}x - y}{4\sqrt{3}}$.
Equating the two expressions for $k$:
$\frac{4\sqrt{3}}{\sqrt{3}x + y} = \frac{\sqrt{3}x - y}{4\sqrt{3}}$
$(\sqrt{3}x - y)(\sqrt{3}x + y) = 16(3) = 48$
$3x^2 - y^2 = 48$
Dividing by $48$,we get $\frac{x^2}{16} - \frac{y^2}{48} = 1$.
This is the equation of a hyperbola with $a^2 = 16$ and $b^2 = 48$.
The eccentricity $e$ is given by $b^2 = a^2(e^2 - 1)$.
$48 = 16(e^2 - 1)$
$3 = e^2 - 1$
$e^2 = 4$
$e = 2$.

(C) Let the geometric series be $a, ar, ar^2, \dots$ where $a > 0$ and $r > 1$ for an increasing series.
Given $T_2 + T_6 = \frac{25}{2} \Rightarrow ar(1 + r^4) = \frac{25}{2} \dots (1)$
Given $T_3 \cdot T_5 = 25$ $\Rightarrow (ar^2)(ar^4) = 25$ $\Rightarrow a^2r^6 = 25$ $\Rightarrow ar^3 = 5$ (since $a, r > 0$)
From $(1)$,$ar + ar^5 = \frac{25}{2}$. Substituting $a = \frac{5}{r^3}$:
$\frac{5}{r^3} \cdot r + \frac{5}{r^3} \cdot r^5 = \frac{25}{2} \Rightarrow \frac{5}{r^2} + 5r^2 = \frac{25}{2}$
Dividing by $5$: $\frac{1}{r^2} + r^2 = \frac{5}{2} \Rightarrow 2r^4 - 5r^2 + 2 = 0$
$(2r^2 - 1)(r^2 - 2) = 0 \Rightarrow r^2 = 2$ (since $r > 1$)
Then $a = \frac{5}{r^3} = \frac{5}{2\sqrt{2}}$.
The sum of $4^{\text{th}}, 6^{\text{th}}, 8^{\text{th}}$ terms is $ar^3 + ar^5 + ar^7 = ar^3(1 + r^2 + r^4)$.
$= 5(1 + 2 + 4) = 5(7) = 35$.

(B) Let $PA = AQ = \lambda$.
Since $PQ \perp OB$,by the property of intersecting chords in a circle,we have $OA \cdot AB = PA \cdot AQ$.
Given $OA = 1$ and $OB = 13$,we have $AB = OB - OA = 13 - 1 = 12$.
Substituting the values,$1 \cdot 12 = \lambda \cdot \lambda$.
$\lambda^2 = 12 \Rightarrow \lambda = \sqrt{12} = 2 \sqrt{3}$.
The area of $\Delta PQB = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times PQ \times AB$.
Since $PQ = PA + AQ = \lambda + \lambda = 2\lambda = 4 \sqrt{3}$,
Area $= \frac{1}{2} \times (4 \sqrt{3}) \times 12 = 2 \sqrt{3} \times 12 = 24 \sqrt{3}$ square units.

(A) Let the sum be $S = 1 + \frac{2}{3} + \frac{7}{3^{2}} + \frac{12}{3^{3}} + \frac{17}{3^{4}} + \frac{22}{3^{5}} + \ldots$
Multiply by $\frac{1}{3}$: $\frac{S}{3} = \frac{1}{3} + \frac{2}{3^{2}} + \frac{7}{3^{3}} + \frac{12}{3^{4}} + \frac{17}{3^{5}} + \ldots$
Subtracting the two equations: $S - \frac{S}{3} = 1 + (\frac{2}{3} - \frac{1}{3}) + (\frac{7}{3^{2}} - \frac{2}{3^{2}}) + (\frac{12}{3^{3}} - \frac{7}{3^{3}}) + \ldots$
$\frac{2S}{3} = 1 + \frac{1}{3} + \frac{5}{3^{2}} + \frac{5}{3^{3}} + \frac{5}{3^{4}} + \ldots$
$\frac{2S}{3} = 1 + \frac{1}{3} + \frac{\frac{5}{3^{2}}}{1 - \frac{1}{3}} = 1 + \frac{1}{3} + \frac{5}{9} \times \frac{3}{2} = \frac{4}{3} + \frac{5}{6} = \frac{8+5}{6} = \frac{13}{6}$
$S = \frac{13}{6} \times \frac{3}{2} = \frac{13}{4}$

(A) Let $L = \lim_{h \rightarrow 0} 2 \left\{ \frac{\sqrt{3} \sin (\frac{\pi}{6} + h) - \cos (\frac{\pi}{6} + h)}{\sqrt{3} h (\sqrt{3} \cos h - \sin h)} \right\}$.
Using the expansion $\sin(A+B) = \sin A \cos B + \cos A \sin B$ and $\cos(A+B) = \cos A \cos B - \sin A \sin B$:
Numerator $= \sqrt{3} (\sin \frac{\pi}{6} \cos h + \cos \frac{\pi}{6} \sin h) - (\cos \frac{\pi}{6} \cos h - \sin \frac{\pi}{6} \sin h)$
$= \sqrt{3} (\frac{1}{2} \cos h + \frac{\sqrt{3}}{2} \sin h) - (\frac{\sqrt{3}}{2} \cos h - \frac{1}{2} \sin h)$
$= \frac{\sqrt{3}}{2} \cos h + \frac{3}{2} \sin h - \frac{\sqrt{3}}{2} \cos h + \frac{1}{2} \sin h = 2 \sin h$.
Substituting back into the limit:
$L = \lim_{h \rightarrow 0} 2 \left( \frac{2 \sin h}{\sqrt{3} h (\sqrt{3} \cos h - \sin h)} \right) = \frac{4}{\sqrt{3}} \lim_{h \rightarrow 0} \left( \frac{\sin h}{h} \right) \cdot \lim_{h \rightarrow 0} \left( \frac{1}{\sqrt{3} \cos h - \sin h} \right)$.
Since $\lim_{h \rightarrow 0} \frac{\sin h}{h} = 1$ and $\lim_{h \rightarrow 0} (\sqrt{3} \cos h - \sin h) = \sqrt{3}(1) - 0 = \sqrt{3}$:
$L = \frac{4}{\sqrt{3}} \cdot 1 \cdot \frac{1}{\sqrt{3}} = \frac{4}{3}$.

(B) The general term $T_{r+1}$ in the expansion of $\left( tx^{\frac{1}{5}} + \frac{(1-x)^{\frac{1}{10}}}{t} \right)^{10}$ is given by $T_{r+1} = {}^{10}C_r (tx^{\frac{1}{5}})^{10-r} \left( \frac{(1-x)^{\frac{1}{10}}}{t} \right)^r$.
For the term to be independent of $t$,the power of $t$ must be zero: $(10-r) - r = 0 \Rightarrow 2r = 10 \Rightarrow r = 5$.
Thus,the term independent of $t$ is $T_6 = {}^{10}C_5 (x^{\frac{1}{5}})^5 ((1-x)^{\frac{1}{10}})^5 = {}^{10}C_5 x^{\frac{1}{2}} (1-x)^{\frac{1}{2}} = {}^{10}C_5 \sqrt{x(1-x)}$.
Let $f(x) = {}^{10}C_5 \sqrt{x-x^2}$. To find the maximum,we differentiate $f(x)$ with respect to $x$: $f'(x) = {}^{10}C_5 \cdot \frac{1-2x}{2\sqrt{x-x^2}}$.
Setting $f'(x) = 0$,we get $1-2x = 0 \Rightarrow x = \frac{1}{2}$.
The maximum value is $f\left(\frac{1}{2}\right) = {}^{10}C_5 \sqrt{\frac{1}{2}(1-\frac{1}{2})} = {}^{10}C_5 \sqrt{\frac{1}{4}} = \frac{1}{2} \cdot {}^{10}C_5 = \frac{1}{2} \cdot \frac{10!}{5!5!} = \frac{10!}{2(5!)^2}$.
Wait,re-evaluating the expression: $T_6 = {}^{10}C_5 x (1-x)^{1/2}$ is incorrect based on the provided expression. Let's re-calculate: $T_6 = {}^{10}C_5 (x^{1/5})^5 ((1-x)^{1/10})^5 = {}^{10}C_5 x (1-x)^{1/2}$.
Let $f(x) = {}^{10}C_5 x(1-x)^{1/2}$. $f'(x) = {}^{10}C_5 [(1-x)^{1/2} + x \cdot \frac{1}{2}(1-x)^{-1/2}(-1)] = {}^{10}C_5 (1-x)^{-1/2} [1-x - \frac{x}{2}] = {}^{10}C_5 \frac{1 - \frac{3}{2}x}{\sqrt{1-x}}$.
Setting $f'(x) = 0 \Rightarrow x = \frac{2}{3}$.
$f(\frac{2}{3}) = {}^{10}C_5 (\frac{2}{3}) \sqrt{1-\frac{2}{3}} = {}^{10}C_5 \cdot \frac{2}{3} \cdot \frac{1}{\sqrt{3}} = \frac{2 \cdot 10!}{3\sqrt{3}(5!)^2}$.
Thus,the correct option is $B$.

(C) Let the seven digits be $x_1, x_2, x_3, x_4, x_5, x_6, x_7$ such that $x_i \in \{1, 2, 3\}$ and $\sum_{i=1}^{7} x_i = 10$.
Case $I$: The digits are $1, 1, 1, 1, 1, 2, 3$.
Sum $= 1+1+1+1+1+2+3 = 10$.
The number of arrangements is $\frac{7!}{5!1!1!} = \frac{7 \times 6}{1} = 42$.
Case $II$: The digits are $1, 1, 1, 1, 2, 2, 2$.
Sum $= 1+1+1+1+2+2+2 = 10$.
The number of arrangements is $\frac{7!}{4!3!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$.
Total number of such integers $= 42 + 35 = 77$.

(C) Let the lines be $L_1: x-y=0$,$L_2: x+2y=3$,and $L_3: 2x+y=6$.
Solving $L_1$ and $L_2$:
$x-y=0 \implies x=y$
$x+2x=3 \implies 3x=3 \implies x=1, y=1$. Point $A = (1, 1)$.
Solving $L_1$ and $L_3$:
$x-y=0 \implies x=y$
$2x+x=6 \implies 3x=6 \implies x=2, y=2$. Point $B = (2, 2)$.
Solving $L_2$ and $L_3$:
$x+2y=3 \implies x=3-2y$
$2(3-2y)+y=6 \implies 6-4y+y=6 \implies -3y=0 \implies y=0, x=3$. Point $C = (3, 0)$.
Now,calculate the lengths of the sides:
$AB = \sqrt{(2-1)^2 + (2-1)^2} = \sqrt{1^2 + 1^2} = \sqrt{2}$
$BC = \sqrt{(3-2)^2 + (0-2)^2} = \sqrt{1^2 + (-2)^2} = \sqrt{1+4} = \sqrt{5}$
$AC = \sqrt{(3-1)^2 + (0-1)^2} = \sqrt{2^2 + (-1)^2} = \sqrt{4+1} = \sqrt{5}$
Since $BC = AC = \sqrt{5}$,the triangle is an isosceles triangle.

(A) The given equation is $3 \sin x + 4 \cos x = k + 1$.
We know that the range of the function $f(x) = a \sin x + b \cos x$ is $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$.
Here,$a = 3$ and $b = 4$,so the range is $[-\sqrt{3^2 + 4^2}, \sqrt{3^2 + 4^2}] = [-5, 5]$.
For the equation to have a solution,the value $k + 1$ must lie within this range:
$-5 \le k + 1 \le 5$
Subtracting $1$ from all sides,we get:
$-6 \le k \le 4$
The integral values of $k$ are $\{-6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4\}$.
The total number of such integral values is $11$.

(A) Given equation: $\log _{4}(x-1)=\log _{2}(x-3)$
Using the property $\log _{a^n}(b) = \frac{1}{n} \log _{a}(b)$,we have:
$\frac{1}{2} \log _{2}(x-1)=\log _{2}(x-3)$
$\log _{2}(x-1)^{1/2}=\log _{2}(x-3)$
Equating the arguments:
$(x-1)^{1/2} = x-3$
Squaring both sides:
$x-1 = (x-3)^2$
$x-1 = x^2 - 6x + 9$
$x^2 - 7x + 10 = 0$
Factoring the quadratic equation:
$(x-2)(x-5) = 0$
$x = 2$ or $x = 5$
Check the domain of the original equation:
For $\log _{2}(x-3)$ to be defined,$x-3 > 0$,so $x > 3$.
For $\log _{4}(x-1)$ to be defined,$x-1 > 0$,so $x > 1$.
Combining these,the domain is $x > 3$.
Since $x=2$ is not in the domain $(2 < 3)$,it is an extraneous solution.
Thus,only $x=5$ is a valid solution.
Therefore,the number of solutions is $1$.

(C) Given the equation $x^{3}-2x^{2}+2x-1=0$.
By inspection,$x=1$ satisfies the equation as $1-2+2-1=0$.
Thus,$(x-1)$ is a factor of the polynomial.
Performing division,we get $(x-1)(x^{2}-x+1)=0$.
The roots are $x=1$ and the roots of $x^{2}-x+1=0$,which are $x = \frac{1 \pm i\sqrt{3}}{2}$.
These roots are $-\omega^{2}$ and $-\omega$,where $\omega$ is the complex cube root of unity.
The sum of the $162^{\text{th}}$ powers is $S = (1)^{162} + (-\omega^{2})^{162} + (-\omega)^{162}$.
$S = 1 + \omega^{324} + \omega^{162}$.
Since $\omega^{3} = 1$,we have $\omega^{324} = (\omega^{3})^{108} = 1$ and $\omega^{162} = (\omega^{3})^{54} = 1$.
Therefore,$S = 1 + 1 + 1 = 3$.

(A) The given expression is $S = \sum_{k=0}^{29} (30-k) \binom{30}{k}$.
Using the property $\binom{n}{k} = \binom{n}{n-k}$,we can rewrite the sum as:
$S = \sum_{k=0}^{29} (30-k) \binom{30}{30-k}$.
Let $r = 30-k$. As $k$ goes from $0$ to $29$,$r$ goes from $30$ to $1$.
$S = \sum_{r=1}^{30} r \binom{30}{r}$.
Using the identity $\sum_{r=1}^{n} r \binom{n}{r} = n 2^{n-1}$:
$S = 30 \cdot 2^{30-1} = 30 \cdot 2^{29}$.
$S = 15 \cdot 2 \cdot 2^{29} = 15 \cdot 2^{30}$.
Given $S = n \cdot 2^m$ with $\operatorname{gcd}(2, n) = 1$,we have $n = 15$ and $m = 30$.
Therefore,$n + m = 15 + 30 = 45$.

(A) Given equation: $\sqrt{3} \cos^2 x = (\sqrt{3}-1) \cos x + 1$
Rearranging the terms: $\sqrt{3} \cos^2 x - \sqrt{3} \cos x + \cos x - 1 = 0$
Factorizing by grouping: $\sqrt{3} \cos x (\cos x - 1) + 1 (\cos x - 1) = 0$
$(\sqrt{3} \cos x + 1)(\cos x - 1) = 0$
This gives two cases:
Case $1$: $\cos x = 1 \Rightarrow x = 0$ (which is in the interval $[0, \frac{\pi}{2}]$)
Case $2$: $\cos x = -\frac{1}{\sqrt{3}}$. Since $\cos x$ is negative in the second quadrant and $x \in [0, \frac{\pi}{2}]$,there is no solution for this case in the given interval.
Thus,there is only $1$ solution,which is $x = 0$.

(B) To determine if a statement is a tautology,we construct a truth table for each.
For statement $(a)$: $(\sim q \wedge (p$ $\rightarrow q))$ $\rightarrow \sim p$

$p, q$	$\sim q$	$p \rightarrow q$	$\sim q \wedge (p \rightarrow q)$	$\sim p$	$(a)$
$T, T$	$F$	$T$	$F$	$F$	$T$
$T, F$	$T$	$F$	$F$	$F$	$T$
$F, T$	$F$	$T$	$F$	$T$	$T$
$F, F$	$T$	$T$	$T$	$T$	$T$

Since all values in the final column are $T$,$(a)$ is a tautology.
For statement $(b)$: $((p \vee q) \wedge \sim p) \rightarrow q$

$p, q$	$p \vee q$	$\sim p$	$(p \vee q) \wedge \sim p$	$(b)$
$T, T$	$T$	$F$	$F$	$T$
$T, F$	$T$	$F$	$F$	$T$
$F, T$	$T$	$T$	$T$	$T$
$F, F$	$F$	$T$	$F$	$T$

Since all values in the final column are $T$,$(b)$ is also a tautology.
Therefore,both $(a)$ and $(b)$ are tautologies.

(D) Let the point $P$ on the parabola $y=x^{2}+4$ be $(h, k)$. Since $P$ lies on the parabola,we have $k = h^{2}+4$.
The given straight line is $L: y = 4x - 1$,which can be written as $4x - y - 1 = 0$.
The perpendicular distance $d$ from point $P(h, k)$ to the line $4x - y - 1 = 0$ is given by:
$d = \frac{|4h - k - 1|}{\sqrt{4^{2} + (-1)^{2}}} = \frac{|4h - (h^{2} + 4) - 1|}{\sqrt{16 + 1}} = \frac{|4h - h^{2} - 5|}{\sqrt{17}} = \frac{|-(h^{2} - 4h + 5)|}{\sqrt{17}} = \frac{h^{2} - 4h + 5}{\sqrt{17}}$.
To find the point closest to the line,we minimize $d$ by differentiating with respect to $h$:
$\frac{dd}{dh} = \frac{1}{\sqrt{17}} (2h - 4)$.
Setting $\frac{dd}{dh} = 0$,we get $2h - 4 = 0$,which implies $h = 2$.
For $h = 2$,the $y$-coordinate is $k = (2)^{2} + 4 = 4 + 4 = 8$.
Thus,the coordinates of point $P$ are $(2, 8)$.

(D) Let the height of the jet plane be $h \, m$. Let the point on the ground be $A$.
From the first position, $\tan 60^{\circ} = \frac{h}{y}$ $\Rightarrow \sqrt{3} = \frac{h}{y}$ $\Rightarrow y = \frac{h}{\sqrt{3}} \quad \dots (1)$
After $20 \, s$, the plane covers a distance $x$.
Speed $= 432 \, km/h = 432 \times \frac{5}{18} \, m/s = 120 \, m/s$.
Distance $x = \text{speed} \times \text{time} = 120 \times 20 = 2400 \, m$.
From the second position, $\tan 30^{\circ} = \frac{h}{x + y}$ $\Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{2400 + y}$ $\Rightarrow 2400 + y = h\sqrt{3} \quad \dots (2)$
Substitute $y$ from $(1)$ into $(2)$:
$2400 + \frac{h}{\sqrt{3}} = h\sqrt{3}$
$2400 = h\sqrt{3} - \frac{h}{\sqrt{3}} = h \left( \frac{3 - 1}{\sqrt{3}} \right) = \frac{2h}{\sqrt{3}}$
$h = \frac{2400 \times \sqrt{3}}{2} = 1200 \sqrt{3} \, m$.

(B) We use the identity ${}^{n}C_{r} + {}^{n}C_{r+1} = {}^{n+1}C_{r+1}$.
Note that ${}^{2}C_{2} = {}^{3}C_{3} = 1$.
The series is $S = {}^{n+1}C_{2} + 2 \sum_{k=2}^{n} {}^{k}C_{2}$.
Using the Hockey-stick identity $\sum_{i=r}^{n} {}^{i}C_{r} = {}^{n+1}C_{r+1}$,we have $\sum_{k=2}^{n} {}^{k}C_{2} = {}^{n+1}C_{3}$.
Thus,$S = {}^{n+1}C_{2} + 2({}^{n+1}C_{3})$.
$S = \frac{(n+1)n}{2} + 2 \cdot \frac{(n+1)n(n-1)}{6}$.
$S = \frac{(n+1)n}{2} \left( 1 + \frac{2(n-1)}{3} \right) = \frac{n(n+1)}{2} \left( \frac{3 + 2n - 2}{3} \right)$.
$S = \frac{n(n+1)(2n+1)}{6}$.

(D) The equation of the line is $x + \sqrt{3}y = 2\sqrt{3}$,which can be written as $y = -\frac{1}{\sqrt{3}}x + 2$. Here,the slope $m = -\frac{1}{\sqrt{3}}$.
We check the point $\left(\frac{3\sqrt{3}}{2}, \frac{1}{2}\right)$ for each curve.
For option $D$: $x^{2} + 9y^{2} = 9$.
Substitute the point: $\left(\frac{3\sqrt{3}}{2}\right)^{2} + 9\left(\frac{1}{2}\right)^{2} = \frac{27}{4} + \frac{9}{4} = \frac{36}{4} = 9$. The point lies on the curve.
The equation of the tangent to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ at $(x_{1}, y_{1})$ is $\frac{xx_{1}}{a^{2}} + \frac{yy_{1}}{b^{2}} = 1$.
For $x^{2} + 9y^{2} = 9$,we have $\frac{x^{2}}{9} + y^{2} = 1$.
Tangent at $\left(\frac{3\sqrt{3}}{2}, \frac{1}{2}\right)$ is $\frac{x \cdot \frac{3\sqrt{3}}{2}}{9} + y \cdot \frac{1}{2} = 1$.
$\frac{\sqrt{3}x}{6} + \frac{y}{2} = 1 \implies \sqrt{3}x + 3y = 6 \implies x + \sqrt{3}y = 2\sqrt{3}$.
This matches the given line.

(B) We want to find the negation of the statement $\sim p \wedge (p \vee q)$.
Applying De Morgan's Law: $\sim (\sim p \wedge (p \vee q)) \equiv \sim (\sim p) \vee \sim (p \vee q)$.
Using the law of double negation and De Morgan's Law: $p \vee (\sim p \wedge \sim q)$.
Applying the Distributive Law: $(p \vee \sim p) \wedge (p \vee \sim q)$.
Since $(p \vee \sim p) \equiv T$ (Tautology),we have $T \wedge (p \vee \sim q)$.
Therefore,the result is $p \vee \sim q$.

(D) The centroid of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$.
Given vertices are $(a, c), (2, b), (a, b)$ and centroid is $\left(\frac{10}{3}, \frac{7}{3}\right)$.
Equating the $x$-coordinates: $\frac{a+2+a}{3} = \frac{10}{3} \implies 2a + 2 = 10 \implies 2a = 8 \implies a = 4$.
Equating the $y$-coordinates: $\frac{c+b+b}{3} = \frac{7}{3} \implies c + 2b = 7$.
Since $a, b, c$ are in arithmetic progression,$2b = a + c$. Substituting $a=4$,we get $2b = 4 + c \implies c = 2b - 4$.
Substitute $c$ into the equation $c + 2b = 7$: $(2b - 4) + 2b = 7 \implies 4b = 11 \implies b = \frac{11}{4}$.
The quadratic equation is $4x^{2} + \frac{11}{4}x + 1 = 0$.
For roots $\alpha, \beta$,we have $\alpha + \beta = -\frac{b}{a} = -\frac{11/4}{4} = -\frac{11}{16}$ and $\alpha\beta = \frac{c}{a} = \frac{1}{4}$.
We need to find $\alpha^{2} + \beta^{2} - \alpha\beta = (\alpha + \beta)^{2} - 3\alpha\beta$.
Substituting the values: $\left(-\frac{11}{16}\right)^{2} - 3\left(\frac{1}{4}\right) = \frac{121}{256} - \frac{3}{4} = \frac{121 - 192}{256} = -\frac{71}{256}$.

(C) Let $S = \{1, 2, 3, 4, 5\}$. The total number of subsets of $S$ is $2^{5} = 32$.
Since two subsets $A$ and $B$ are selected randomly,the total number of pairs $(A, B)$ is $32 \times 32 = 2^{10}$.
For each element $x \in S$,there are $4$ possibilities for its membership in $A$ and $B$: $x \notin A, x \notin B$; $x \in A, x \notin B$; $x \notin A, x \in B$; or $x \in A, x \in B$.
We want the intersection $A \cap B$ to have exactly $2$ elements.
First,choose $2$ elements out of $5$ to be in the intersection: $\binom{5}{2} = 10$ ways.
For the remaining $3$ elements,each must not be in the intersection,meaning they can be in $(A \setminus B)$,$(B \setminus A)$,or in neither $A$ nor $B$. There are $3$ such choices for each of the $3$ remaining elements,giving $3^{3} = 27$ ways.
Thus,the number of favorable pairs is $10 \times 27 = 270$.
The probability is $\frac{270}{2^{10}} = \frac{270}{1024} = \frac{135}{512} = \frac{135}{2^{9}}$.

(A) Using Vandermonde's Identity,$\sum_{i=0}^{k}\binom{n}{i}\binom{m}{k-i} = \binom{n+m}{k}$.
Applying this to the given expression:
$\sum_{i=0}^{k}\binom{10}{i}\binom{15}{k-i} = \binom{10+15}{k} = \binom{25}{k}$.
Similarly,$\sum_{i=0}^{k+1}\binom{12}{i}\binom{13}{k+1-i} = \binom{12+13}{k+1} = \binom{25}{k+1}$.
The total sum is $\binom{25}{k} + \binom{25}{k+1}$.
Using Pascal's Identity,$\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1}$,we get $\binom{25}{k} + \binom{25}{k+1} = \binom{26}{k+1}$.
Since $\binom{n}{r}$ is defined for all non-negative integers $n$ and $r$ (with the given condition),the expression $\binom{26}{k+1}$ is defined for any non-negative integer $k$. Therefore,there is no maximum value for $k$.

(A) Let the point $P$ be $(x, y)$.
According to the problem,the distance from $(5, 0)$ is thrice the distance from $(-5, 0)$:
$\sqrt{(x-5)^{2} + y^{2}} = 3\sqrt{(x+5)^{2} + y^{2}}$
Squaring both sides:
$(x-5)^{2} + y^{2} = 9((x+5)^{2} + y^{2})$
$x^{2} - 10x + 25 + y^{2} = 9(x^{2} + 10x + 25 + y^{2})$
$x^{2} - 10x + 25 + y^{2} = 9x^{2} + 90x + 225 + 9y^{2}$
$8x^{2} + 8y^{2} + 100x + 200 = 0$
Dividing by $8$:
$x^{2} + y^{2} + 12.5x + 25 = 0$
Comparing with the standard circle equation $x^{2} + y^{2} + 2gx + 2fy + c = 0$,we have $g = 6.25$,$f = 0$,and $c = 25$.
The radius $r$ is given by $r = \sqrt{g^{2} + f^{2} - c} = \sqrt{(6.25)^{2} - 25} = \sqrt{39.0625 - 25} = \sqrt{14.0625}$.
Thus,$r^{2} = 14.0625$.
Therefore,$4r^{2} = 4 \times 14.0625 = 56.25$.

(B) The circle is $(x-2)^{2}+(y-3)^{2}=25$. The center is $(2,3)$ and the radius is $5$.
At point $(5,7)$,the slope of the radius is $m_r = \frac{7-3}{5-2} = \frac{4}{3}$.
The slope of the tangent is $m_t = -\frac{1}{m_r} = -\frac{3}{4}$.
The equation of the tangent is $y-7 = -\frac{3}{4}(x-5) \implies 4y-28 = -3x+15 \implies 3x+4y-43=0$.
The $x-$intercept of the tangent is found by setting $y=0$: $3x = 43 \implies x = \frac{43}{3}$.
The slope of the normal is $m_n = \frac{4}{3}$.
The equation of the normal is $y-7 = \frac{4}{3}(x-5) \implies 3y-21 = 4x-20 \implies 4x-3y+1=0$.
The $x-$intercept of the normal is found by setting $y=0$: $4x = -1 \implies x = -\frac{1}{4}$.
The triangle is formed by the points $(-\frac{1}{4}, 0)$,$(\frac{43}{3}, 0)$,and $(5, 7)$.
The base of the triangle on the $x-$axis is $b = \frac{43}{3} - (-\frac{1}{4}) = \frac{172+3}{12} = \frac{175}{12}$.
The height of the triangle is the $y-$coordinate of the point $(5,7)$,which is $h = 7$.
Area $A = \frac{1}{2} \times b \times h = \frac{1}{2} \times \frac{175}{12} \times 7 = \frac{1225}{24}$.
Therefore,$24A = 1225$.

(B) The variance $\sigma^{2}$ of $n$ observations is given by $\sigma^{2} = \frac{\Sigma x_{i}^{2}}{n} - (\bar{x})^{2}$.
Here,the observations are $1$ ($9$ times) and $k$ ($1$ time),so $n = 10$.
$\Sigma x_{i} = 9(1) + k = 9 + k$.
$\Sigma x_{i}^{2} = 9(1)^{2} + k^{2} = 9 + k^{2}$.
$\sigma^{2} = \frac{9 + k^{2}}{10} - \left(\frac{9 + k}{10}\right)^{2} < 10$.
Multiply by $100$: $10(9 + k^{2}) - (9 + k)^{2} < 1000$.
$90 + 10k^{2} - (81 + 18k + k^{2}) < 1000$.
$9k^{2} - 18k + 9 < 1000$.
$9(k - 1)^{2} < 1000$.
$(k - 1)^{2} < \frac{1000}{9} \approx 111.11$.
$k - 1 < \sqrt{111.11} \approx 10.54$.
$k < 11.54$.
Since $k$ is a natural number,the maximum possible value of $k$ is $11$.

(D) Let the first four terms be $a, ar, ar^2, ar^3$.
The sum of the first four terms is $S_4 = a(1+r+r^2+r^3) = a\frac{r^4-1}{r-1} = \frac{65}{12} \quad (1)$.
The sum of their reciprocals is $\frac{1}{a} + \frac{1}{ar} + \frac{1}{ar^2} + \frac{1}{ar^3} = \frac{r^3+r^2+r+1}{ar^3} = \frac{1}{a} \frac{r^4-1}{r^3(r-1)} = \frac{65}{18} \quad (2)$.
Dividing $(1)$ by $(2)$,we get $\frac{a\frac{r^4-1}{r-1}}{\frac{1}{a}\frac{r^4-1}{r^3(r-1)}} = a^2 r^3 = \frac{65/12}{65/18} = \frac{18}{12} = \frac{3}{2}$.
Given the product of the first three terms is $a \cdot ar \cdot ar^2 = a^3 r^3 = 1$,which implies $ar = 1$,so $a = \frac{1}{r}$.
Substituting $a = \frac{1}{r}$ into $a^2 r^3 = \frac{3}{2}$,we get $(\frac{1}{r})^2 r^3 = r = \frac{3}{2}$.
Thus,$a = \frac{1}{3/2} = \frac{2}{3}$.
The third term $\alpha = ar^2 = \frac{2}{3} \times (\frac{3}{2})^2 = \frac{2}{3} \times \frac{9}{4} = \frac{3}{2}$.
Therefore,$2\alpha = 2 \times \frac{3}{2} = 3$.

(D) Let $n(C)$ be the number of students in group $C$. Since each group must have at least one student,the remaining $10 - n(C)$ students must be distributed into groups $A$ and $B$ such that each of $A$ and $B$ has at least one student.
For a fixed set of $n(C)$ students in group $C$,the number of ways to distribute the remaining $10 - n(C)$ students into groups $A$ and $B$ such that each is non-empty is $2^{10-n(C)} - 2$.
Case $1$: $n(C) = 1$.
Number of ways $= {}^{10}C_{1} \times (2^{9} - 2) = 10 \times (512 - 2) = 10 \times 510 = 5100$.
Case $2$: $n(C) = 2$.
Number of ways $= {}^{10}C_{2} \times (2^{8} - 2) = 45 \times (256 - 2) = 45 \times 254 = 11430$.
Case $3$: $n(C) = 3$.
Number of ways $= {}^{10}C_{3} \times (2^{7} - 2) = 120 \times (128 - 2) = 120 \times 126 = 15120$.
Total number of possibilities $= 5100 + 11430 + 15120 = 31650$.

(B) Given $k = \frac{(-1+i \sqrt{3})^{21}}{(1-i)^{24}}+\frac{(1+i \sqrt{3})^{21}}{(1+i)^{24}}$.
We know that $-1+i \sqrt{3} = 2(\cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3}) = 2e^{i \frac{2\pi}{3}}$ and $1+i \sqrt{3} = 2(\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}) = 2e^{i \frac{\pi}{3}}$.
Also $1-i = \sqrt{2}(\cos(-\frac{\pi}{4}) + i \sin(-\frac{\pi}{4})) = \sqrt{2}e^{-i \frac{\pi}{4}}$ and $1+i = \sqrt{2}(\cos \frac{\pi}{4} + i \sin \frac{\pi}{4}) = \sqrt{2}e^{i \frac{\pi}{4}}$.
Substituting these values:
$k = \frac{(2e^{i \frac{2\pi}{3}})^{21}}{(\sqrt{2}e^{-i \frac{\pi}{4}})^{24}} + \frac{(2e^{i \frac{\pi}{3}})^{21}}{(\sqrt{2}e^{i \frac{\pi}{4}})^{24}} = \frac{2^{21} e^{i 14\pi}}{2^{12} e^{-i 6\pi}} + \frac{2^{21} e^{i 7\pi}}{2^{12} e^{i 6\pi}}$
$k = 2^9 (e^{i 20\pi} + e^{i \pi}) = 512(1 - 1) = 0$.
Thus,$n = [|k|] = 0$.
The expression becomes $\sum_{j=0}^{5} (j+5)^2 - \sum_{j=0}^{5} (j+5) = \sum_{j=0}^{5} ((j+5)^2 - (j+5)) = \sum_{j=0}^{5} (j^2 + 10j + 25 - j - 5) = \sum_{j=0}^{5} (j^2 + 9j + 20)$.
$= \sum_{j=0}^{5} j^2 + 9 \sum_{j=0}^{5} j + \sum_{j=0}^{5} 20 = \frac{5(6)(11)}{6} + 9 \frac{5(6)}{2} + 20(6) = 55 + 135 + 120 = 310$.

(D) Case-$I$: $x \leq 5$
$(x+1)^{2} - (x-5) = \frac{27}{4}$
$(x+1)^{2} - (x+1) + 6 = \frac{27}{4}$
$(x+1)^{2} - (x+1) - \frac{3}{4} = 0$
Let $y = x+1$,then $y^{2} - y - \frac{3}{4} = 0 \implies 4y^{2} - 4y - 3 = 0$
$(2y-3)(2y+1) = 0 \implies y = \frac{3}{2}, -\frac{1}{2}$
$x+1 = \frac{3}{2} \implies x = \frac{1}{2}$ (Valid as $\frac{1}{2} \leq 5$)
$x+1 = -\frac{1}{2} \implies x = -\frac{3}{2}$ (Valid as $-\frac{3}{2} \leq 5$)
Case-$II$: $x > 5$
$(x+1)^{2} + (x-5) = \frac{27}{4}$
$x^{2} + 2x + 1 + x - 5 = \frac{27}{4}$
$x^{2} + 3x - 4 = \frac{27}{4} \implies 4x^{2} + 12x - 16 = 27$
$4x^{2} + 12x - 43 = 0$
$x = \frac{-12 \pm \sqrt{144 - 4(4)(-43)}}{8} = \frac{-12 \pm \sqrt{144 + 688}}{8} = \frac{-12 \pm \sqrt{832}}{8}$
Since $\sqrt{832} \approx 28.8$,$x = \frac{-12 + 28.8}{8} \approx 2.1$ (Rejected as $x > 5$)
Thus,there are $2$ real roots.

(B) The shortest distance between a curve and a line occurs at a point $P(x_0, y_0)$ on the curve where the tangent is parallel to the given line.
The equation of the line is $x-y=1$,which can be written as $y=x-1$. The slope of this line is $m=1$.
The equation of the curve is $x^2=2y$,which implies $y=\frac{x^2}{2}$.
The slope of the tangent to the curve at any point $(x_0, y_0)$ is given by $\frac{dy}{dx} = \frac{d}{dx}(\frac{x^2}{2}) = x$.
Setting the slope of the tangent equal to the slope of the line,we get $x_0 = 1$.
Substituting $x_0=1$ into the curve equation $y_0 = \frac{x_0^2}{2}$,we get $y_0 = \frac{1^2}{2} = \frac{1}{2}$.
Thus,the point on the curve is $P(1, \frac{1}{2})$.
The shortest distance from a point $(x_1, y_1)$ to the line $Ax+By+C=0$ is given by $d = \frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}$.
Here,the line is $x-y-1=0$,so $A=1, B=-1, C=-1$. The point is $(1, \frac{1}{2})$.
Shortest distance $= \left|\frac{1(1) + (-1)(\frac{1}{2}) - 1}{\sqrt{1^2+(-1)^2}}\right| = \left|\frac{1 - \frac{1}{2} - 1}{\sqrt{2}}\right| = \left|\frac{-\frac{1}{2}}{\sqrt{2}}\right| = \frac{1}{2\sqrt{2}}$.

(B) Since $\alpha, \beta \in \mathbb{R}$,the complex roots must occur in conjugate pairs. Therefore,the other root is $1+2i$.
Using the relationship between roots and coefficients:
Sum of roots $= -\alpha = (1-2i) + (1+2i) = 2 \implies \alpha = -2$.
Product of roots $= \beta = (1-2i)(1+2i) = 1^{2} - (2i)^{2} = 1 + 4 = 5$.
Therefore,$\alpha - \beta = -2 - 5 = -7$.

(B) For the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{16}=1$,we have $a^{2}=25$ and $b^{2}=16$.
The eccentricity $e_{1} = \sqrt{1-\frac{b^{2}}{a^{2}}} = \sqrt{1-\frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$.
The foci of the ellipse are $(\pm ae_{1}, 0) = (\pm 5 \times \frac{3}{5}, 0) = (\pm 3, 0)$.
Given that the product of the eccentricities of the ellipse and the hyperbola is $1$,we have $e_{1} \times e_{2} = 1$ $\Rightarrow \frac{3}{5} \times e_{2} = 1$ $\Rightarrow e_{2} = \frac{5}{3}$.
Let the equation of the hyperbola be $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$.
Since the hyperbola passes through the foci of the ellipse $(\pm 3, 0)$,we have $\frac{3^{2}}{a^{2}} = 1 \Rightarrow a^{2} = 9$.
For a hyperbola,$b^{2} = a^{2}(e_{2}^{2}-1) = 9((\frac{5}{3})^{2}-1) = 9(\frac{25}{9}-1) = 9(\frac{16}{9}) = 16$.
Thus,the equation of the hyperbola is $\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$.

(B) Given $\cos x + \cos y - \cos(x + y) = \frac{3}{2}$.
Using the identity $\cos x + \cos y = 2 \cos(\frac{x+y}{2}) \cos(\frac{x-y}{2})$ and $\cos(x+y) = 2 \cos^2(\frac{x+y}{2}) - 1$,we get:
$2 \cos(\frac{x+y}{2}) \cos(\frac{x-y}{2}) - (2 \cos^2(\frac{x+y}{2}) - 1) = \frac{3}{2}$
$-2 \cos^2(\frac{x+y}{2}) + 2 \cos(\frac{x+y}{2}) \cos(\frac{x-y}{2}) + 1 = \frac{3}{2}$
$2 \cos^2(\frac{x+y}{2}) - 2 \cos(\frac{x+y}{2}) \cos(\frac{x-y}{2}) + \frac{1}{2} = 0$
Multiply by $2$:
$4 \cos^2(\frac{x+y}{2}) - 4 \cos(\frac{x+y}{2}) \cos(\frac{x-y}{2}) + 1 = 0$
This is a quadratic in $\cos(\frac{x+y}{2})$. Completing the square:
$(2 \cos(\frac{x+y}{2}) - \cos(\frac{x-y}{2}))^2 + 1 - \cos^2(\frac{x-y}{2}) = 0$
$(2 \cos(\frac{x+y}{2}) - \cos(\frac{x-y}{2}))^2 + \sin^2(\frac{x-y}{2}) = 0$
Since both terms are squares,they must be zero:
$\sin(\frac{x-y}{2}) = 0 \Rightarrow x = y$
$2 \cos(\frac{x+y}{2}) - \cos(0) = 0$ $\Rightarrow 2 \cos x = 1$ $\Rightarrow \cos x = \frac{1}{2}$
Since $0 < x < \pi$,$x = \frac{\pi}{3}$.
Then $\sin x = \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$ and $\cos y = \cos(\frac{\pi}{3}) = \frac{1}{2}$.
Therefore,$\sin x + \cos y = \frac{\sqrt{3}}{2} + \frac{1}{2} = \frac{1+\sqrt{3}}{2}$.

(D) To find the angle subtended by the chord $PQ$ at the origin,we homogenize the equation of the curve $x^{2} + 2y^{2} = 2$ using the line $x + y = 1$.
The equation of the line can be written as $x + y = 1$,so $1 = x + y$.
Substituting this into the curve equation:
$x^{2} + 2y^{2} = 2(1)^{2}$
$x^{2} + 2y^{2} = 2(x + y)^{2}$
$x^{2} + 2y^{2} = 2(x^{2} + 2xy + y^{2})$
$x^{2} + 2y^{2} = 2x^{2} + 4xy + 2y^{2}$
$x^{2} + 4xy = 0$
This represents a pair of lines passing through the origin. Let these lines be $y = m_{1}x$ and $y = m_{2}x$.
From $x(x + 4y) = 0$,we get $x = 0$ (slope $m_{1} = \infty$) and $y = -\frac{1}{4}x$ (slope $m_{2} = -\frac{1}{4}$).
The angle $\theta$ between these two lines is given by:
$\tan \theta = \left| \frac{m_{1} - m_{2}}{1 + m_{1}m_{2}} \right|$
Since one line is vertical $(x=0)$,the angle $\theta$ is the angle between the vertical line and the line with slope $-\frac{1}{4}$.
The angle of the line $y = -\frac{1}{4}x$ with the $x$-axis is $\alpha = \tan^{-1}(-\frac{1}{4}) = -\tan^{-1}(\frac{1}{4})$.
The angle between the vertical line $(90^{\circ})$ and this line is $\frac{\pi}{2} - (-\tan^{-1}(\frac{1}{4})) = \frac{\pi}{2} + \tan^{-1}(\frac{1}{4})$.

(C) The contrapositive of a conditional statement $p \rightarrow q$ is defined as $\sim q \rightarrow \sim p$.
Here,the statement $p$ is "You will work" and $q$ is "You will earn money".
Therefore,the contrapositive $\sim q \rightarrow \sim p$ is "If you will not earn money,you will not work".

(B) Given the function $f(x) = a^{a^{x}} + a^{1-a^{x}}$.
We can rewrite the second term as $a^{1-a^{x}} = \frac{a}{a^{a^{x}}}$.
So,$f(x) = a^{a^{x}} + \frac{a}{a^{a^{x}}}$.
Since $a > 0$,both $a^{a^{x}}$ and $\frac{a}{a^{a^{x}}}$ are positive real numbers.
By the Arithmetic Mean-Geometric Mean $(AM \geq GM)$ inequality,for any two positive numbers $u$ and $v$,$u + v \geq 2\sqrt{uv}$.
Let $u = a^{a^{x}}$ and $v = \frac{a}{a^{a^{x}}}$.
Then $f(x) = u + v \geq 2\sqrt{u \cdot v} = 2\sqrt{a^{a^{x}} \cdot \frac{a}{a^{a^{x}}}} = 2\sqrt{a}$.
Thus,the minimum value is $2\sqrt{a}$.

(A) Since $\alpha$ and $\beta$ are the roots of $x^{2}-6x-2=0$,we have $\alpha^{2}-6\alpha-2=0$ and $\beta^{2}-6\beta-2=0$.
Multiplying the first equation by $\alpha^{8}$,we get $\alpha^{10}-6\alpha^{9}-2\alpha^{8}=0$.
Similarly,multiplying the second equation by $\beta^{8}$,we get $\beta^{10}-6\beta^{9}-2\beta^{8}=0$.
Subtracting the two equations,we obtain $(\alpha^{10}-\beta^{10})-6(\alpha^{9}-\beta^{9})-2(\alpha^{8}-\beta^{8})=0$.
Using the definition $a_{n}=\alpha^{n}-\beta^{n}$,this simplifies to $a_{10}-6a_{9}-2a_{8}=0$.
Rearranging the terms,we get $a_{10}-2a_{8}=6a_{9}$.
Therefore,$\frac{a_{10}-2a_{8}}{3a_{9}} = \frac{6a_{9}}{3a_{9}} = 2$.

(A) We want $3^{n} + 7^{n} \equiv 0 \pmod{10}$.
Since $7 \equiv -3 \pmod{10}$,we have $7^{n} \equiv (-3)^{n} \pmod{10}$.
Thus,$3^{n} + 7^{n} \equiv 3^{n} + (-1)^{n} 3^{n} \pmod{10}$.
If $n$ is even,$3^{n} + 3^{n} = 2 \cdot 3^{n}$,which is not divisible by $10$ (as $3^{n}$ ends in $1$ or $9$,so $2 \cdot 3^{n}$ ends in $2$ or $8$).
If $n$ is odd,$3^{n} + (-1) 3^{n} = 3^{n} - 3^{n} = 0$,which is divisible by $10$.
Therefore,$n$ must be an odd number.
The two-digit numbers are from $10$ to $99$.
The total number of two-digit numbers is $99 - 10 + 1 = 90$.
Exactly half of these are odd,so the number of odd two-digit numbers is $90 / 2 = 45$.

(A) Given that the remainder when $x$ is divided by $4$ is $3$,we can write $x = 4k + 3$ for some integer $k$.
We need to find the remainder when $(2020 + x)^{2022}$ is divided by $8$.
Substitute $x = 4k + 3$ into the expression:
$(2020 + x)^{2022} = (2020 + 4k + 3)^{2022} = (2023 + 4k)^{2022}$.
Since $2023 = 8 \times 252 + 7$,we have $2023 \equiv 7 \equiv -1 \pmod{8}$.
Thus,$(2023 + 4k)^{2022} \equiv (-1 + 4k)^{2022} \pmod{8}$.
If $k$ is even,let $k = 2m$,then $(-1 + 4(2m))^{2022} = (-1 + 8m)^{2022} \equiv (-1)^{2022} \equiv 1 \pmod{8}$.
If $k$ is odd,let $k = 2m + 1$,then $(-1 + 4(2m + 1))^{2022} = (-1 + 8m + 4)^{2022} = (3 + 8m)^{2022} \equiv 3^{2022} \pmod{8}$.
Note that $3^2 = 9 \equiv 1 \pmod{8}$,so $3^{2022} = (3^2)^{1011} \equiv 1^{1011} \equiv 1 \pmod{8}$.
In both cases,the remainder is $1$.

(A) Let the point of contact on the parabola $y^{2}=4x$ be $A(t^{2}, 2t)$.
The equation of the tangent to the parabola at $A(t^{2}, 2t)$ is $ty = x + t^{2}$,which can be written as $x - ty + t^{2} = 0$.
Since this line is also tangent to the circle $(x-3)^{2} + y^{2} = 9$ with center $(3, 0)$ and radius $r=3$,the perpendicular distance from the center $(3, 0)$ to the line must be equal to the radius.
$\left| \frac{3 - t(0) + t^{2}}{\sqrt{1^{2} + (-t)^{2}}} \right| = 3$
$\left| 3 + t^{2} \right| = 3\sqrt{1 + t^{2}}$
Squaring both sides: $(3 + t^{2})^{2} = 9(1 + t^{2})$
$9 + t^{4} + 6t^{2} = 9 + 9t^{2}$
$t^{4} - 3t^{2} = 0$
$t^{2}(t^{2} - 3) = 0$
Since the points lie in the first quadrant,$t > 0$,so $t^{2} = 3$,which gives $t = \sqrt{3}$.
Thus,the point of contact on the parabola is $A(t^{2}, 2t) = (3, 2\sqrt{3})$. So,$a=3$ and $b=2\sqrt{3}$.
The equation of the tangent line is $x - \sqrt{3}y + 3 = 0$.
The point of contact on the circle $B(c, d)$ is the foot of the perpendicular from the center $(3, 0)$ to the tangent line $x - \sqrt{3}y + 3 = 0$.
Using the formula for the foot of the perpendicular $(c, d)$ from $(x_{1}, y_{1})$ to $Ax + By + C = 0$:
$\frac{c-3}{1} = \frac{d-0}{-\sqrt{3}} = -\frac{1(3) - \sqrt{3}(0) + 3}{1^{2} + (-\sqrt{3})^{2}} = -\frac{6}{4} = -\frac{3}{2}$
$c - 3 = -\frac{3}{2} \Rightarrow c = 3 - \frac{3}{2} = \frac{3}{2}$
$d = -\sqrt{3} \times (-\frac{3}{2}) = \frac{3\sqrt{3}}{2}$
We need to find $2(a+c) = 2(3 + \frac{3}{2}) = 2(\frac{9}{2}) = 9$.

(C) Given the limit $\lim _{x \rightarrow 0} \frac{ax-(e^{4x}-1)}{ax(e^{4x}-1)} = b$.
Since the limit exists,the form must be $\frac{0}{0}$ at $x=0$.
Using the expansion $e^{4x} = 1 + 4x + \frac{(4x)^2}{2!} + \dots$,we have:
$\lim _{x \rightarrow 0} \frac{ax - (1 + 4x + 8x^2 + \dots - 1)}{ax(4x + 8x^2 + \dots)} = \lim _{x \rightarrow 0} \frac{(a-4)x - 8x^2}{4ax^2 + 8ax^3}$.
For the limit to exist,the coefficient of $x$ in the numerator must be zero,so $a-4 = 0$,which gives $a = 4$.
Now,substituting $a=4$ into the limit:
$b = \lim _{x \rightarrow 0} \frac{-8x^2}{4(4)x^2 + 8(4)x^3} = \lim _{x \rightarrow 0} \frac{-8x^2}{16x^2 + 32x^3} = \lim _{x \rightarrow 0} \frac{-8}{16 + 32x} = -\frac{8}{16} = -\frac{1}{2}$.
Thus,$a = 4$ and $b = -\frac{1}{2}$.
Finally,$a - 2b = 4 - 2(-\frac{1}{2}) = 4 + 1 = 5$.

(D) Given $y+z=5$ and $\frac{1}{y}+\frac{1}{z}=\frac{5}{6}$.
From the second equation,$\frac{y+z}{yz} = \frac{5}{6}$.
Substituting $y+z=5$,we get $\frac{5}{yz} = \frac{5}{6}$,which implies $yz = 6$.
We have $y+z=5$ and $yz=6$. The quadratic equation $t^2 - 5t + 6 = 0$ has roots $t=2$ and $t=3$.
Since $y > z$,we have $y=3$ and $z=2$.
The number $n = 2^{x} \cdot 3^{3} \cdot 5^{2}$.
An odd divisor of $n$ must be of the form $3^{a} \cdot 5^{b}$,where $0 \le a \le 3$ and $0 \le b \le 2$.
The number of choices for $a$ is $(3+1) = 4$.
The number of choices for $b$ is $(2+1) = 3$.
Total number of odd divisors $= 4 \times 3 = 12$.

(C) Let $x = n$,where $n \in Z$.
First,consider the value of the function at $x = n$:
$f(n) = [n-1] \cos \left(\frac{2n-1}{2}\right) \pi = (n-1) \cos \left(n\pi - \frac{\pi}{2}\right) = (n-1) \cdot 0 = 0$.
Now,calculate the Left Hand Limit $(LHL)$:
$LHL = \lim_{x \rightarrow n^-} [x-1] \cos \left(\frac{2x-1}{2}\right) \pi$.
As $x \rightarrow n^-$,$[x-1] = n-2$.
$LHL = (n-2) \cos \left(n\pi - \frac{\pi}{2}\right) = (n-2) \cdot 0 = 0$.
Now,calculate the Right Hand Limit $(RHL)$:
$RHL = \lim_{x \rightarrow n^+} [x-1] \cos \left(\frac{2x-1}{2}\right) \pi$.
As $x \rightarrow n^+$,$[x-1] = n-1$.
$RHL = (n-1) \cos \left(n\pi - \frac{\pi}{2}\right) = (n-1) \cdot 0 = 0$.
Since $LHL = RHL = f(n) = 0$ for all $n \in Z$,the function is continuous at all integral values of $x$.
For non-integral values of $x$,the function is a product of a constant (locally) and a continuous trigonometric function,hence it is continuous.
Therefore,$f(x)$ is continuous for every real $x$.

(D) Let the line be $\frac{x-3}{1} = \frac{y-4}{2} = \frac{z-5}{2} = t$.
Then,any point on the line is given by $x = t+3$,$y = 2t+4$,and $z = 2t+5$.
For the point of intersection with the plane $x+y+z=17$,we substitute these coordinates into the plane equation:
$(t+3) + (2t+4) + (2t+5) = 17$.
$5t + 12 = 17
\Rightarrow 5t = 5
\Rightarrow t = 1$.
Substituting $t=1$ back into the parametric equations,the point of intersection is $(1+3, 2(1)+4, 2(1)+5) = (4, 6, 7)$.
The distance between the points $(1, 1, 9)$ and $(4, 6, 7)$ is given by the distance formula:
$d = \sqrt{(4-1)^2 + (6-1)^2 + (7-9)^2}
= \sqrt{3^2 + 5^2 + (-2)^2}
= \sqrt{9 + 25 + 4}
= \sqrt{38}$.

(A) The slope of the tangent to the curve $y=x^{3}$ at $P(t, t^{3})$ is given by $\frac{dy}{dx} = 3x^{2}$.
At $x=t$,the slope is $3t^{2}$.
The equation of the tangent at $P(t, t^{3})$ is $y - t^{3} = 3t^{2}(x - t)$.
To find the intersection with $y=x^{3}$,substitute $y=x^{3}$ into the tangent equation:
$x^{3} - t^{3} = 3t^{2}(x - t)$.
$(x - t)(x^{2} + xt + t^{2}) = 3t^{2}(x - t)$.
Since $x \neq t$ for point $Q$,we have $x^{2} + xt + t^{2} = 3t^{2}$,which simplifies to $x^{2} + xt - 2t^{2} = 0$.
Factoring gives $(x - t)(x + 2t) = 0$,so $x = -2t$.
The coordinates of $Q$ are $(-2t, (-2t)^{3}) = (-2t, -8t^{3})$.
The point dividing $PQ$ in the ratio $1:2$ has an ordinate given by the section formula:
$y = \frac{1 \times (-8t^{3}) + 2 \times (t^{3})}{1 + 2} = \frac{-8t^{3} + 2t^{3}}{3} = \frac{-6t^{3}}{3} = -2t^{3}$.

(A) Given $f(x) = \frac{4x^3 - 3x^2}{6} - 2 \sin x + (2x - 1) \cos x$.
First,we find the derivative $f'(x)$:
$f'(x) = \frac{12x^2 - 6x}{6} - 2 \cos x + [2 \cos x + (2x - 1)(-\sin x)]$
$f'(x) = (2x^2 - x) - 2 \cos x + 2 \cos x - (2x - 1) \sin x$
$f'(x) = x(2x - 1) - (2x - 1) \sin x$
$f'(x) = (2x - 1)(x - \sin x)$.
We know that for $x > 0$,$x > \sin x$,so $(x - \sin x) > 0$.
For $x < 0$,$x < \sin x$,so $(x - \sin x) < 0$.
Now,analyze the sign of $f'(x) = (2x - 1)(x - \sin x)$:
$1$. If $x \in [\frac{1}{2}, \infty)$,then $(2x - 1) \geq 0$ and $(x - \sin x) > 0$,so $f'(x) \geq 0$. Thus,$f(x)$ is increasing.
$2$. If $x \in [0, \frac{1}{2}]$,then $(2x - 1) \leq 0$ and $(x - \sin x) \geq 0$,so $f'(x) \leq 0$. Thus,$f(x)$ is decreasing.
$3$. If $x \in (-\infty, 0]$,then $(2x - 1) < 0$ and $(x - \sin x) \leq 0$,so $f'(x) \geq 0$. Thus,$f(x)$ is increasing.
Comparing with the options,$f(x)$ increases in $[\frac{1}{2}, \infty)$.

(C) Given $f(x) = 2x - 1$ and $g(x) = \frac{x - 1/2}{x - 1} = \frac{2x - 1}{2(x - 1)}$.
Calculate the composition $f(g(x))$:
$f(g(x)) = 2(g(x)) - 1 = 2 \left( \frac{2x - 1}{2(x - 1)} \right) - 1$
$= \frac{2x - 1}{x - 1} - 1 = \frac{2x - 1 - (x - 1)}{x - 1} = \frac{x}{x - 1} = 1 + \frac{1}{x - 1}$.
For one-one check:
Let $f(g(x_1)) = f(g(x_2))$.
$1 + \frac{1}{x_1 - 1} = 1 + \frac{1}{x_2 - 1} \implies x_1 - 1 = x_2 - 1 \implies x_1 = x_2$.
Thus,$f(g(x))$ is one-one.
For onto check:
The range of $f(g(x)) = 1 + \frac{1}{x - 1}$ is $R - \{1\}$.
Since the codomain is $R$,the range is not equal to the codomain.
Therefore,$f(g(x))$ is not onto.
Conclusion: $f(g(x))$ is one-one but not onto.

(D) Let $n$ be the number of times the dice is rolled. The probability of getting an odd number in a single throw is $p = \frac{1}{2}$,and the probability of getting an even number is $q = 1 - p = \frac{1}{2}$.
The probability of getting an odd number $2$ times is given by the binomial distribution: $P(X=2) = {}^{n}C_{2} (\frac{1}{2})^{2} (\frac{1}{2})^{n-2} = {}^{n}C_{2} (\frac{1}{2})^{n}$.
The probability of getting an even number $3$ times is equivalent to getting an odd number $(n-3)$ times: $P(Y=3) = {}^{n}C_{3} (\frac{1}{2})^{3} (\frac{1}{2})^{n-3} = {}^{n}C_{3} (\frac{1}{2})^{n}$.
Given that $P(X=2) = P(Y=3)$,we have ${}^{n}C_{2} = {}^{n}C_{3}$.
Using the property ${}^{n}C_{r} = {}^{n}C_{n-r}$,we get $2 + 3 = n$,so $n = 5$.
We need to find the probability of getting an odd number an odd number of times,which is $P(X=1) + P(X=3) + P(X=5)$.
$P(X=1) + P(X=3) + P(X=5) = {}^{5}C_{1} (\frac{1}{2})^{5} + {}^{5}C_{3} (\frac{1}{2})^{5} + {}^{5}C_{5} (\frac{1}{2})^{5} = \frac{1}{2^{5}} (5 + 10 + 1) = \frac{16}{32} = \frac{1}{2}$.

(C) First,find the intersection points of $x^{2} + y^{2} = 36$ and $y^{2} = 9x$.
Substituting $y^{2} = 9x$ into the circle equation: $x^{2} + 9x - 36 = 0$.
$(x + 12)(x - 3) = 0$,so $x = 3$ (since $x \geq 0$ for the parabola).
At $x = 3$,$y^{2} = 27$,so $y = \pm 3 \sqrt{3}$.
The area inside the circle and outside the parabola is the total area of the circle minus the area inside both the circle and the parabola.
The area inside both is $2 \int_{0}^{3} \sqrt{9x} \, dx + 2 \int_{3}^{6} \sqrt{36 - x^{2}} \, dx$.
$= 2 \times 3 \int_{0}^{3} x^{1/2} \, dx + 2 \left[ \frac{x}{2} \sqrt{36 - x^{2}} + \frac{36}{2} \sin^{-1} \left( \frac{x}{6} \right) \right]_{3}^{6}$.
$= 6 \left[ \frac{2}{3} x^{3/2} \right]_{0}^{3} + 2 \left[ (0 + 18 \sin^{-1}(1)) - (\frac{3}{2} \sqrt{27} + 18 \sin^{-1}(1/2)) \right]$.
$= 4(3 \sqrt{3}) + 2 \left[ 18(\pi/2) - (\frac{9 \sqrt{3}}{2} + 18(\pi/6)) \right] = 12 \sqrt{3} + 2 [9 \pi - \frac{9 \sqrt{3}}{2} - 3 \pi] = 12 \sqrt{3} + 12 \pi - 9 \sqrt{3} = 12 \pi + 3 \sqrt{3}$.
Total area of circle = $36 \pi$.
Required area = $36 \pi - (12 \pi + 3 \sqrt{3}) = 24 \pi - 3 \sqrt{3}$.

(A) To evaluate the limit $L = \lim _{x \rightarrow 0} \frac{\int_{0}^{x^{2}} \sin \sqrt{t} dt}{x^{3}}$,we observe that it is of the form $\frac{0}{0}$.
Applying $L$'$H$ôpital's rule and the Leibniz integral rule,we differentiate the numerator and the denominator with respect to $x$:
Numerator derivative: $\frac{d}{dx} \int_{0}^{x^{2}} \sin \sqrt{t} dt = (\sin \sqrt{x^{2}}) \cdot \frac{d}{dx}(x^{2}) = (\sin x) \cdot (2x)$.
Denominator derivative: $\frac{d}{dx}(x^{3}) = 3x^{2}$.
Now,the limit becomes:
$L = \lim _{x \rightarrow 0} \frac{2x \sin x}{3x^{2}} = \lim _{x \rightarrow 0} \frac{2 \sin x}{3x}$.
Using the standard limit $\lim _{x \rightarrow 0} \frac{\sin x}{x} = 1$,we get:
$L = \frac{2}{3} \times 1 = \frac{2}{3}$.

(B) Given $\int_{-a}^{a} (|x| + |x-2|) dx = 22$ with $a > 2$.
We split the integral based on the critical points $x=0$ and $x=2$:
$\int_{-a}^{0} (-x - (x-2)) dx + \int_{0}^{2} (x - (x-2)) dx + \int_{2}^{a} (x + x-2) dx = 22$
$\int_{-a}^{0} (-2x + 2) dx + \int_{0}^{2} 2 dx + \int_{2}^{a} (2x - 2) dx = 22$
$[-x^2 + 2x]_{-a}^{0} + [2x]_{0}^{2} + [x^2 - 2x]_{2}^{a} = 22$
$(0 - (-a^2 - 2a)) + (4 - 0) + ((a^2 - 2a) - (4 - 4)) = 22$
$a^2 + 2a + 4 + a^2 - 2a = 22$
$2a^2 + 4 = 22 \Rightarrow 2a^2 = 18 \Rightarrow a^2 = 9$. Since $a > 2$,$a = 3$.
Now,we evaluate $\int_{3}^{-3} (x + [x]) dx = -\int_{-3}^{3} (x + [x]) dx$.
$\int_{-3}^{3} x dx + \int_{-3}^{3} [x] dx = 0 + ([-3] + [-2] + [-1] + [0] + [1] + [2])$
$= -3 - 2 - 1 + 0 + 1 + 2 = -3$.
Thus,$-\int_{-3}^{3} (x + [x]) dx = -(-3) = 3$.

(D) Let $M = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}$. The sum of the diagonal elements of $M^{T}M$ is the trace of $M^{T}M$,which is equal to the sum of the squares of all elements of $M$.
Thus,$a^{2} + b^{2} + c^{2} + d^{2} + e^{2} + f^{2} + g^{2} + h^{2} + i^{2} = 7$.
Since the entries are from $\{0, 1, 2\}$,we consider the possible combinations of squares that sum to $7$:
Case-$I$: Seven $1$'s and two $0$'s.
Number of ways = $\binom{9}{7} = \binom{9}{2} = \frac{9 \times 8}{2} = 36$.
Case-$II$: One $2$ $(2^{2} = 4)$,three $1$'s $(1^{2} = 1)$,and five $0$'s $(0^{2} = 0)$.
Sum = $4 + 1 + 1 + 1 + 0 + 0 + 0 + 0 + 0 = 7$.
Number of ways = $\frac{9!}{1! 3! 5!} = \frac{9 \times 8 \times 7 \times 6}{3 \times 2 \times 1} = 9 \times 4 \times 2 \times 7 = 504$.
Total number of matrices = $36 + 504 = 540$.

(B) Let $f(x) = \frac{4}{\sin x} + \frac{1}{1-\sin x}$.
Let $t = \sin x$. Since $x \in (0, \frac{\pi}{2})$,we have $t \in (0, 1)$.
Then $f(t) = \frac{4}{t} + \frac{1}{1-t}$.
To find the minimum value,we differentiate $f(t)$ with respect to $t$:
$f'(t) = -\frac{4}{t^2} + \frac{1}{(1-t)^2}$.
Setting $f'(t) = 0$,we get $\frac{1}{(1-t)^2} = \frac{4}{t^2}$,which implies $(1-t)^2 = \frac{t^2}{4}$.
Taking the square root,$1-t = \frac{t}{2}$ (since $t \in (0, 1)$),so $1 = \frac{3t}{2}$,which gives $t = \frac{2}{3}$.
Since $t = \frac{2}{3}$ is in $(0, 1)$,the minimum value is $f(\frac{2}{3}) = \frac{4}{2/3} + \frac{1}{1-2/3} = 6 + 3 = 9$.
Thus,the minimum value of $\alpha$ for which the equation has at least one solution is $9$.

(A) We know that $\tan ^{-1}\left(\frac{1}{1+r+r^{2}}\right) = \tan ^{-1}\left(\frac{(r+1)-r}{1+r(r+1)}\right) = \tan ^{-1}(r+1) - \tan ^{-1}(r)$.
Thus,the sum is a telescoping series:
$\sum_{r=1}^{n} \left[\tan ^{-1}(r+1) - \tan ^{-1}(r)\right] = (\tan ^{-1}(2) - \tan ^{-1}(1)) + (\tan ^{-1}(3) - \tan ^{-1}(2)) + \dots + (\tan ^{-1}(n+1) - \tan ^{-1}(n))$.
This simplifies to $\tan ^{-1}(n+1) - \tan ^{-1}(1) = \tan ^{-1}(n+1) - \frac{\pi}{4}$.
Now,taking the limit as $n \rightarrow \infty$:
$\lim _{n}$ ${\rightarrow \infty} \tan \left(\tan ^{-1}(n+1) - \frac{\pi}{4}\right) = \tan \left(\frac{\pi}{2} - \frac{\pi}{4}\right) = \tan \left(\frac{\pi}{4}\right) = 1$.

(A) Since $\overrightarrow{c}$ is coplanar with $\overrightarrow{a}$ and $\overrightarrow{b}$,we can write $\overrightarrow{c} = x\overrightarrow{a} + y\overrightarrow{b}$.
Given $\overrightarrow{b} \cdot \overrightarrow{c} = 0$,we have $\overrightarrow{b} \cdot (x\overrightarrow{a} + y\overrightarrow{b}) = 0$,which implies $x(\overrightarrow{a} \cdot \overrightarrow{b}) + y|\overrightarrow{b}|^{2} = 0$.
Calculating $\overrightarrow{a} \cdot \overrightarrow{b} = (-1)(2) + (1)(0) + (1)(1) = -1$ and $|\overrightarrow{b}|^{2} = 2^{2} + 0^{2} + 1^{2} = 5$.
So,$-x + 5y = 0 \Rightarrow x = 5y$.
Thus,$\overrightarrow{c} = y(5\overrightarrow{a} + \overrightarrow{b}) = y(5(-\hat{i} + \hat{j} + \hat{k}) + (2\hat{i} + \hat{k})) = y(-3\hat{i} + 5\hat{j} + 6\hat{k})$.
Given $\overrightarrow{a} \cdot \overrightarrow{c} = 7$,we have $y((-1)(-3) + (1)(5) + (1)(6)) = 7 \Rightarrow y(3 + 5 + 6) = 7 \Rightarrow 14y = 7 \Rightarrow y = \frac{1}{2}$.
So,$\overrightarrow{c} = -\frac{3}{2}\hat{i} + \frac{5}{2}\hat{j} + 3\hat{k}$.
Then $\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} = (-\hat{i} + \hat{j} + \hat{k}) + (2\hat{i} + \hat{k}) + (-\frac{3}{2}\hat{i} + \frac{5}{2}\hat{j} + 3\hat{k}) = (2 - 1 - \frac{3}{2})\hat{i} + (1 + \frac{5}{2})\hat{j} + (1 + 1 + 3)\hat{k} = -\frac{1}{2}\hat{i} + \frac{7}{2}\hat{j} + 5\hat{k}$.
Finally,$2|\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c}|^{2} = 2((\frac{-1}{2})^{2} + (\frac{7}{2})^{2} + 5^{2}) = 2(\frac{1}{4} + \frac{49}{4} + 25) = 2(\frac{50}{4} + 25) = 2(12.5 + 25) = 2(37.5) = 75$.

(B) Let $P(B_{1}) = p_{1}$,$P(B_{2}) = p_{2}$,and $P(B_{3}) = p_{3}$.
Given that $p_{1}(1 - p_{2})(1 - p_{3}) = \alpha$ $...(i)$
$p_{2}(1 - p_{1})(1 - p_{3}) = \beta$ $...(ii)$
$p_{3}(1 - p_{1})(1 - p_{2}) = \gamma$ $...(iii)$
And $(1 - p_{1})(1 - p_{2})(1 - p_{3}) = p$ $...(iv)$
From these,we get $\frac{p_{1}}{1 - p_{1}} = \frac{\alpha}{p}$,$\frac{p_{2}}{1 - p_{2}} = \frac{\beta}{p}$,and $\frac{p_{3}}{1 - p_{3}} = \frac{\gamma}{p}$.
Given equations: $(\alpha - 2\beta)p = \alpha\beta \Rightarrow \frac{1}{p} = \frac{1}{\beta} - \frac{2}{\alpha} \Rightarrow \frac{1}{\beta} = \frac{1}{p} + \frac{2}{\alpha}$.
Also,$(\beta - 3\gamma)p = 2\beta\gamma \Rightarrow \frac{1}{2\gamma} - \frac{3}{2\beta} = \frac{1}{p} \Rightarrow \frac{1}{2\gamma} = \frac{1}{p} + \frac{3}{2\beta}$.
Substituting $\frac{1}{\beta} = \frac{1}{p} + \frac{2}{\alpha}$ into the second equation:
$\frac{1}{2\gamma} = \frac{1}{p} + \frac{3}{2}(\frac{1}{p} + \frac{2}{\alpha}) = \frac{1}{p} + \frac{3}{2p} + \frac{3}{\alpha} = \frac{5}{2p} + \frac{3}{\alpha}$.
Using $\frac{\alpha}{p} = \frac{p_{1}}{1 - p_{1}}$ and $\frac{\gamma}{p} = \frac{p_{3}}{1 - p_{3}}$,we have $\frac{1}{2} \cdot \frac{1 - p_{3}}{p_{3}} = \frac{5}{2} + 3 \cdot \frac{1 - p_{1}}{p_{1}}$.
Simplifying leads to $\frac{p_{1}}{p_{3}} = 6$.

(A) Given $PQ = kI_3$. Taking the determinant on both sides,$|P||Q| = |kI_3| = k^3|I_3| = k^3$.
Given $|Q| = \frac{k^2}{2}$,so $|P| \cdot \frac{k^2}{2} = k^3$. Since $k \neq 0$,$|P| = 2k$.
Calculating $|P|$: $|P| = 3(0 - (-5\alpha)) - (-1)(0 - 3\alpha) + (-2)(-10 - 0) = 3(5\alpha) + (-3\alpha) + 20 = 15\alpha - 3\alpha + 20 = 12\alpha + 20$.
Thus,$12\alpha + 20 = 2k$,or $k = 6\alpha + 10$.
Since $PQ = kI_3$,$Q = kP^{-1} = k \cdot \frac{\text{adj}(P)}{|P|} = k \cdot \frac{\text{adj}(P)}{2k} = \frac{1}{2} \text{adj}(P)$.
The element $q_{23}$ is the $(2,3)$ entry of $\frac{1}{2} \text{adj}(P)$.
The $(2,3)$ entry of $\text{adj}(P)$ is the cofactor $C_{32} = -\begin{vmatrix} 3 & -2 \\ 2 & \alpha \end{vmatrix} = -(3\alpha - (-4)) = -(3\alpha + 4)$.
So,$q_{23} = \frac{-(3\alpha + 4)}{2} = -\frac{k}{8}$.
This implies $4(3\alpha + 4) = k$.
Substituting $k = 6\alpha + 10$: $12\alpha + 16 = 6\alpha + 10 \Rightarrow 6\alpha = -6 \Rightarrow \alpha = -1$.
Then $k = 6(-1) + 10 = 4$.
Finally,$\alpha^2 + k^2 = (-1)^2 + 4^2 = 1 + 16 = 17$.

(C) Let $I$ be the event that the missile is intercepted and $H$ be the event that the missile hits the target.
Given: $P(I) = \frac{1}{3}$,so $P(\text{not intercepted}) = P(I^c) = 1 - \frac{1}{3} = \frac{2}{3}$.
The probability that the missile hits the target given it is not intercepted is $P(H | I^c) = \frac{3}{4}$.
The probability that a single missile hits the target is $P(H) = P(I^c) \times P(H | I^c) = \frac{2}{3} \times \frac{3}{4} = \frac{1}{2}$.
Since three missiles are fired independently,the probability that all three hit the target is $(P(H))^3 = (\frac{1}{2})^3 = \frac{1}{8}$.

(D) Given $f(n+1) = f(n) + f(1)$. This is a linear recurrence relation.
For $n=1$,$f(2) = f(1) + f(1) = 2f(1)$.
For $n=2$,$f(3) = f(2) + f(1) = 3f(1)$.
By induction,$f(n) = n f(1)$.
Since $f: N \rightarrow N$,$f(1)$ must be a positive integer $k \in N$.
Thus,$f(n) = kn$. Since $k \geq 1$,$f(n_1) = f(n_2) \Rightarrow kn_1 = kn_2 \Rightarrow n_1 = n_2$,so $f$ is one-one. Statement $C$ is true.
If $fog$ is one-one,then $f(g(x_1)) = f(g(x_2)) \Rightarrow g(x_1) = g(x_2)$ because $f$ is one-one. Since $fog$ is one-one,$x_1 = x_2$. Thus $g$ is one-one. Statement $A$ is true.
If $f$ is onto,then for any $y \in N$,there exists $n \in N$ such that $f(n) = y$,i.e.,$kn = y$. This is only possible for all $y$ if $k=1$,so $f(n) = n$. Statement $B$ is true.
If $g$ is onto,$fog$ is not necessarily one-one. For example,if $g(n) = 1$ for all $n$ (not onto) or any non-injective function,$fog$ will not be one-one. Even if $g$ is onto,if $g$ is not one-one,$fog$ cannot be one-one. Thus,statement $D$ is $NOT$ true.

(D) Let the given line be $L_1: \frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{-2} = r$.
Any point $P$ on $L_1$ is given by $P(2r+1, 3r-1, -2r+1)$.
The direction vector of $L_1$ is $\vec{v_1} = 2\hat{i} + 3\hat{j} - 2\hat{k}$.
The vector $\vec{QP}$ is $(2r+1-0)\hat{i} + (3r-1-1)\hat{j} + (-2r+1-2)\hat{k} = (2r+1)\hat{i} + (3r-2)\hat{j} + (-2r-1)\hat{k}$.
Since the required line is perpendicular to $L_1$,$\vec{QP} \cdot \vec{v_1} = 0$.
$2(2r+1) + 3(3r-2) - 2(-2r-1) = 0$.
$4r + 2 + 9r - 6 + 4r + 2 = 0$.
$17r - 2 = 0 \Rightarrow r = \frac{2}{17}$.
The direction ratios of the line $QP$ are proportional to the components of $\vec{QP}$ at $r = \frac{2}{17}$.
$vec{QP} = (2(\frac{2}{17})+1)\hat{i} + (3(\frac{2}{17})-2)\hat{j} + (-2(\frac{2}{17})-1)\hat{k} = \frac{21}{17}\hat{i} - \frac{28}{17}\hat{j} - \frac{21}{17}\hat{k}$.
Dividing by $\frac{7}{17}$,the direction ratios are $(3, -4, -3)$.
Thus,the equation of the line passing through $(0,1,2)$ with direction ratios $(3, -4, -3)$ is $\frac{x-0}{3} = \frac{y-1}{-4} = \frac{z-2}{-3}$,which is equivalent to $\frac{x}{-3} = \frac{y-1}{4} = \frac{z-2}{3}$.

(NONE) Given equations are $l+m-n=0$ and $l^{2}+m^{2}-n^{2}=0$.
From the first equation,$n = l+m$.
Substituting this into the second equation: $l^{2}+m^{2}=(l+m)^{2} = l^{2}+m^{2}+2lm$.
This simplifies to $2lm = 0$,which means $l=0$ or $m=0$.
Case $1$: If $l=0$,then $n=m$. Since $l^{2}+m^{2}+n^{2}=1$,we have $0^{2}+m^{2}+m^{2}=1$,so $2m^{2}=1$,$m = \pm \frac{1}{\sqrt{2}}$. Thus,the direction cosines are $(0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$.
Case $2$: If $m=0$,then $n=l$. Since $l^{2}+m^{2}+n^{2}=1$,we have $l^{2}+0^{2}+l^{2}=1$,so $2l^{2}=1$,$l = \pm \frac{1}{\sqrt{2}}$. Thus,the direction cosines are $(\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}})$.
Let the direction cosines be $\vec{u} = (0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$ and $\vec{v} = (\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}})$.
The cosine of the angle $\alpha$ between them is $\cos \alpha = |\vec{u} \cdot \vec{v}| = |0 \cdot \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \cdot 0 + \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}| = \frac{1}{2}$.
We need to find $\sin^{4} \alpha + \cos^{4} \alpha$.
Since $\cos \alpha = \frac{1}{2}$,$\sin^{2} \alpha = 1 - \cos^{2} \alpha = 1 - \frac{1}{4} = \frac{3}{4}$.
Then $\sin^{4} \alpha + \cos^{4} \alpha = (\frac{3}{4})^{2} + (\frac{1}{2})^{2} = \frac{9}{16} + \frac{1}{4} = \frac{9+4}{16} = \frac{13}{16}$.

(D) Let $I = \int \frac{\sin \theta \cdot \sin 2 \theta \left(\sin ^{6} \theta+\sin ^{4} \theta+\sin ^{2} \theta\right) \sqrt{2 \sin ^{4} \theta+3 \sin ^{2} \theta+6}}{1-\cos 2 \theta} d \theta$.
Using $\sin 2 \theta = 2 \sin \theta \cos \theta$ and $1 - \cos 2 \theta = 2 \sin ^{2} \theta$,we get:
$I = \int \frac{\sin \theta \cdot (2 \sin \theta \cos \theta) \cdot \sin ^{2} \theta (\sin ^{4} \theta + \sin ^{2} \theta + 1) \sqrt{2 \sin ^{4} \theta + 3 \sin ^{2} \theta + 6}}{2 \sin ^{2} \theta} d \theta$.
$I = \int \sin ^{2} \theta \cos \theta (\sin ^{4} \theta + \sin ^{2} \theta + 1) \sqrt{2 \sin ^{4} \theta + 3 \sin ^{2} \theta + 6} d \theta$.
Let $t = \sin \theta$,then $dt = \cos \theta d \theta$.
$I = \int t^{2} (t^{4} + t^{2} + 1) \sqrt{2 t^{4} + 3 t^{2} + 6} dt = \int (t^{6} + t^{4} + t^{2}) \sqrt{2 t^{4} + 3 t^{2} + 6} dt$.
This simplifies to $\int (t^{5} + t^{3} + t) \sqrt{2 t^{6} + 3 t^{4} + 6 t^{2}} dt$.
Let $u^{2} = 2 t^{6} + 3 t^{4} + 6 t^{2}$,then $2u du = (12 t^{5} + 12 t^{3} + 12 t) dt = 12(t^{5} + t^{3} + t) dt$.
So,$(t^{5} + t^{3} + t) dt = \frac{u du}{6}$.
$I = \int u \cdot \frac{u du}{6} = \frac{1}{6} \int u^{2} du = \frac{u^{3}}{18} + c = \frac{(2 t^{6} + 3 t^{4} + 6 t^{2})^{3/2}}{18} + c$.
Substituting $t^{2} = 1 - \cos^{2} \theta$,we get $2(1-\cos^{2}\theta)^{3} + 3(1-\cos^{2}\theta)^{2} + 6(1-\cos^{2}\theta)$.
Expanding this leads to $11 - 18 \cos^{2} \theta + 9 \cos^{4} \theta - 2 \cos^{6} \theta$. Thus,option $D$ is correct.

(C) Let $I = \int_{-1}^{1} x^{2} e^{[x^{3}]} dx$.
We split the integral at $x = 0$:
$I = \int_{-1}^{0} x^{2} e^{[x^{3}]} dx + \int_{0}^{1} x^{2} e^{[x^{3}]} dx$.
For $x \in [-1, 0)$,$x^{3} \in [-1, 0)$,so $[x^{3}] = -1$.
For $x \in [0, 1)$,$x^{3} \in [0, 1)$,so $[x^{3}] = 0$.
Substituting these values:
$I = \int_{-1}^{0} x^{2} e^{-1} dx + \int_{0}^{1} x^{2} e^{0} dx$
$I = \frac{1}{e} \int_{-1}^{0} x^{2} dx + \int_{0}^{1} x^{2} dx$
$I = \frac{1}{e} \left[ \frac{x^{3}}{3} \right]_{-1}^{0} + \left[ \frac{x^{3}}{3} \right]_{0}^{1}$
$I = \frac{1}{e} \left( 0 - \left( -\frac{1}{3} \right) \right) + \left( \frac{1}{3} - 0 \right)$
$I = \frac{1}{3e} + \frac{1}{3} = \frac{1+e}{3e}$.

(D) Given that the curve passes through the origin,so $y(0) = 0$.
The slope of the tangent is given by $\frac{dy}{dx} = \frac{x^{2}-4x+y+8}{x-2}$.
We can rewrite the numerator as $(x-2)^{2} + y + 4$,so $\frac{dy}{dx} = \frac{(x-2)^{2} + y + 4}{x-2} = (x-2) + \frac{y}{x-2} + \frac{4}{x-2}$.
Rearranging the terms,we get the linear differential equation: $\frac{dy}{dx} - \frac{y}{x-2} = (x-2) + \frac{4}{x-2}$.
The Integrating Factor $(I.F.)$ is $e^{-\int \frac{1}{x-2} dx} = e^{-\ln|x-2|} = \frac{1}{x-2}$.
Multiplying both sides by the $I.F.$,we get $\frac{d}{dx} \left( \frac{y}{x-2} \right) = \frac{1}{x-2} \left( (x-2) + \frac{4}{x-2} \right) = 1 + \frac{4}{(x-2)^{2}}$.
Integrating both sides with respect to $x$: $\frac{y}{x-2} = x - \frac{4}{x-2} + C$.
Using the condition $y(0) = 0$: $\frac{0}{-2} = 0 - \frac{4}{-2} + C \Rightarrow 0 = 2 + C \Rightarrow C = -2$.
Thus,$\frac{y}{x-2} = x - \frac{4}{x-2} - 2$.
Multiplying by $(x-2)$,we get $y = x(x-2) - 4 - 2(x-2) = x^{2} - 2x - 4 - 2x + 4 = x^{2} - 4x$.
Checking the options: For $x = 5$,$y = 5^{2} - 4(5) = 25 - 20 = 5$. Thus,the curve passes through $(5, 5)$.

(A) For Rolle's theorem to hold on $[1, 2]$,we must have $f(1) = f(2)$.
$f(1) = 1 - a + b - 4 = -a + b - 3$
$f(2) = 8 - 4a + 2b - 4 = -4a + 2b + 4$
Equating them: $-a + b - 3 = -4a + 2b + 4 \Rightarrow 3a - b = 7$ $.......(1)$
Given $f^{\prime}(x) = 3x^{2} - 2ax + b$.
Since $f^{\prime}\left(\frac{4}{3}\right) = 0$,we have $3\left(\frac{4}{3}\right)^{2} - 2a\left(\frac{4}{3}\right) + b = 0$.
$3\left(\frac{16}{9}\right) - \frac{8a}{3} + b = 0 \Rightarrow \frac{16}{3} - \frac{8a}{3} + b = 0$.
Multiplying by $3$,we get $16 - 8a + 3b = 0 \Rightarrow 8a - 3b = 16$ $.......(2)$
From $(1)$,$b = 3a - 7$. Substituting into $(2)$:
$8a - 3(3a - 7) = 16 \Rightarrow 8a - 9a + 21 = 16 \Rightarrow -a = -5 \Rightarrow a = 5$.
Then $b = 3(5) - 7 = 15 - 7 = 8$.
Thus,$(a, b) = (5, 8)$.

(B) Let $f(x) = x^{6} + ax^{5} + bx^{4} + cx^{3} + dx^{2} + ex + f$.
Since $\lim_{x \rightarrow 0} \frac{f(x)}{x^{3}} = 1$,which is a non-zero finite value,we must have $d = e = f = 0$.
Thus,$f(x) = x^{6} + ax^{5} + bx^{4} + cx^{3}$.
Then $\lim_{x \rightarrow 0} \frac{f(x)}{x^{3}} = \lim_{x \rightarrow 0} (x^{3} + ax^{2} + bx + c) = c = 1$.
So,$f(x) = x^{6} + ax^{5} + bx^{4} + x^{3}$.
The derivative is $f'(x) = 6x^{5} + 5ax^{4} + 4bx^{3} + 3x^{2}$.
Since $f(x)$ has extrema at $x = 1$ and $x = -1$,$f'(1) = 0$ and $f'(-1) = 0$.
$f'(1) = 6 + 5a + 4b + 3 = 0 \Rightarrow 5a + 4b = -9$.
$f'(-1) = -6 + 5a - 4b + 3 = 0 \Rightarrow 5a - 4b = 3$.
Adding the two equations: $10a = -6 \Rightarrow a = -\frac{3}{5}$.
Subtracting the equations: $8b = -12 \Rightarrow b = -\frac{3}{2}$.
Thus,$f(x) = x^{6} - \frac{3}{5}x^{5} - \frac{3}{2}x^{4} + x^{3}$.
Calculating $5 \cdot f(2) = 5 \left( 2^{6} - \frac{3}{5} \cdot 2^{5} - \frac{3}{2} \cdot 2^{4} + 2^{3} \right)$.
$5 \cdot f(2) = 5 \left( 64 - \frac{3}{5} \cdot 32 - \frac{3}{2} \cdot 16 + 8 \right) = 320 - 96 - 120 + 40 = 144$.

(C) Given function: $f(x) = |2x+1| - 3|x+2| + |x^2+x-2|$
Factorize the quadratic term: $|x^2+x-2| = |(x+2)(x-1)| = |x+2||x-1|$
Substitute back into the function: $f(x) = |2x+1| - 3|x+2| + |x+2||x-1|$
$f(x) = |2x+1| + |x+2|(|x-1| - 3)$
The points where the expression inside the absolute value becomes zero are $x = -1/2$,$x = -2$,and $x = 1$.
Let $g(x) = |x-1| - 3$. The function $f(x)$ is non-differentiable at points where the argument of an absolute value is zero,provided the function does not have a smooth turning point there.
$1$. At $x = -1/2$,$|2x+1|$ is non-differentiable,and the other terms are smooth. Thus,$x = -1/2$ is a point of non-differentiability.
$2$. At $x = 1$,$|x-1|$ is non-differentiable,and $|x+2| = 3 \neq 0$. Thus,$x = 1$ is a point of non-differentiability.
$3$. At $x = -2$,we check the behavior: $f(x) = |2x+1| + |x+2|(|x-1|-3)$. Near $x = -2$,$|x-1| = -(x-1) = 1-x$. So $f(x) \approx |2x+1| + |x+2|(1-x-3) = |2x+1| + |x+2|(-x-2) = |2x+1| - |x+2|(x+2) = |2x+1| - (x+2)^2$. Since $(x+2)^2$ is differentiable at $x = -2$,the non-differentiability of $|x+2|$ is cancelled out by the factor $(x+2)$.
Therefore,the points of non-differentiability are $x = -1/2$ and $x = 1$.
The number of points of non-differentiability is $2$.

(A) The points of intersection of $y = \sin x$ and $y = \cos x$ are given by $\sin x = \cos x$,which implies $\tan x = 1$.
Thus,$x = \frac{\pi}{4}, \frac{5\pi}{4}, \dots$
The area $A$ between two consecutive points of intersection $\frac{\pi}{4}$ and $\frac{5\pi}{4}$ is given by:
$A = \int_{\pi/4}^{5\pi/4} |\sin x - \cos x| \, dx$
In the interval $[\frac{\pi}{4}, \frac{5\pi}{4}]$,$\sin x \geq \cos x$.
$A = \int_{\pi/4}^{5\pi/4} (\sin x - \cos x) \, dx$
$A = [-\cos x - \sin x]_{\pi/4}^{5\pi/4}$
$A = \left( -\cos\left(\frac{5\pi}{4}\right) - \sin\left(\frac{5\pi}{4}\right) \right) - \left( -\cos\left(\frac{\pi}{4}\right) - \sin\left(\frac{\pi}{4}\right) \right)$
$A = \left( -(-\frac{1}{\sqrt{2}}) - (-\frac{1}{\sqrt{2}}) \right) - \left( -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} \right)$
$A = \left( \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \right) - \left( -\frac{2}{\sqrt{2}} \right) = \frac{2}{\sqrt{2}} + \frac{2}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$
Now,$A^{4} = (2\sqrt{2})^{4} = 2^{4} \times (\sqrt{2})^{4} = 16 \times 4 = 64$.

(A) Given $A = \begin{bmatrix} x & y & z \\ y & z & x \\ z & x & y \end{bmatrix}$. Since $A$ is symmetric,$A^T = A$.
Given $A^2 = I$,we have $AA^T = I$,which means $A$ is an orthogonal matrix.
For the matrix $A$,the condition $AA^T = I$ implies:
$x^2 + y^2 + z^2 = 1$ and $xy + yz + zx = 0$.
We know that $(x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx)$.
Substituting the values,$(x + y + z)^2 = 1 + 2(0) = 1$.
Since $x + y + z > 0$,we have $x + y + z = 1$.
Using the algebraic identity $x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - (xy + yz + zx))$,
$x^3 + y^3 + z^3 - 3(2) = (1)(1 - 0)$.
$x^3 + y^3 + z^3 - 6 = 1$.
$x^3 + y^3 + z^3 = 7$.

(B) Let $t = \tan(\frac{\theta}{2})$. Then $A = \begin{bmatrix} 0 & -t \\ t & 0 \end{bmatrix}$.
$I_{2} + A = \begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix}$ and $I_{2} - A = \begin{bmatrix} 1 & t \\ -t & 1 \end{bmatrix}$.
$(I_{2} - A)^{-1} = \frac{1}{1 + t^{2}} \begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix}$.
$(I_{2} + A)(I_{2} - A)^{-1} = \frac{1}{1 + t^{2}} \begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix} \begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix} = \frac{1}{1 + t^{2}} \begin{bmatrix} 1 - t^{2} & -2t \\ 2t & 1 - t^{2} \end{bmatrix} = \begin{bmatrix} \frac{1 - t^{2}}{1 + t^{2}} & -\frac{2t}{1 + t^{2}} \\ \frac{2t}{1 + t^{2}} & \frac{1 - t^{2}}{1 + t^{2}} \end{bmatrix}$.
Comparing with $\begin{bmatrix} a & -b \\ b & a \end{bmatrix}$,we get $a = \frac{1 - t^{2}}{1 + t^{2}} = \cos \theta$ and $b = \frac{2t}{1 + t^{2}} = \sin \theta$.
Thus,$a^{2} + b^{2} = \cos^{2} \theta + \sin^{2} \theta = 1$.
Therefore,$13(a^{2} + b^{2}) = 13(1) = 13$.

(C) Given $\overrightarrow{r} \times \overrightarrow{a} = \overrightarrow{c} \times \overrightarrow{a}$,we can write $(\overrightarrow{r} - \overrightarrow{c}) \times \overrightarrow{a} = 0$.
This implies that $\overrightarrow{r} - \overrightarrow{c}$ is parallel to $\overrightarrow{a}$,so $\overrightarrow{r} = \overrightarrow{c} + \lambda \overrightarrow{a}$ for some scalar $\lambda$.
Given $\overrightarrow{r} \cdot \overrightarrow{b} = 0$,we substitute $\overrightarrow{r}$:
$(\overrightarrow{c} + \lambda \overrightarrow{a}) \cdot \overrightarrow{b} = 0 \Rightarrow \overrightarrow{c} \cdot \overrightarrow{b} + \lambda (\overrightarrow{a} \cdot \overrightarrow{b}) = 0$.
Calculating the dot products:
$\overrightarrow{c} \cdot \overrightarrow{b} = (1)(1) + (-1)(-1) + (-1)(0) = 1 + 1 = 2$.
$\overrightarrow{a} \cdot \overrightarrow{b} = (1)(1) + (2)(-1) + (-1)(0) = 1 - 2 = -1$.
Thus,$2 + \lambda(-1) = 0 \Rightarrow \lambda = 2$.
Now,$\overrightarrow{r} \cdot \overrightarrow{a} = (\overrightarrow{c} + 2\overrightarrow{a}) \cdot \overrightarrow{a} = \overrightarrow{c} \cdot \overrightarrow{a} + 2|\overrightarrow{a}|^2$.
$\overrightarrow{c} \cdot \overrightarrow{a} = (1)(1) + (-1)(2) + (-1)(-1) = 1 - 2 + 1 = 0$.
$|\overrightarrow{a}|^2 = 1^2 + 2^2 + (-1)^2 = 1 + 4 + 1 = 6$.
Therefore,$\overrightarrow{r} \cdot \overrightarrow{a} = 0 + 2(6) = 12$.

(A) The given system of equations is:
$1) kx + y + 2z = 1$
$2) 3x - y - 2z = 2$
$3) -2x - 2y - 4z = 3$
For a system of linear equations to have infinitely many solutions,the determinant of the coefficient matrix must be zero,and the augmented matrix must satisfy specific consistency conditions.
Let the coefficient matrix be $A = \begin{bmatrix} k & 1 & 2 \\ 3 & -1 & -2 \\ -2 & -2 & -4 \end{bmatrix}$.
Setting $|A| = 0$:
$|A| = k((-1)(-4) - (-2)(-2)) - 1((3)(-4) - (-2)(-2)) + 2((3)(-2) - (-1)(-2))$
$|A| = k(4 - 4) - 1(-12 - 4) + 2(-6 - 2)$
$|A| = k(0) - 1(-16) + 2(-8) = 16 - 16 = 0$.
Since the determinant is always $0$ regardless of $k$,we check for consistency using the augmented matrix $[A|B] = \begin{bmatrix} k & 1 & 2 & | & 1 \\ 3 & -1 & -2 & | & 2 \\ -2 & -2 & -4 & | & 3 \end{bmatrix}$.
Adding equations $(2)$ and $(3)$:
$(3x - y - 2z) + (-2x - 2y - 4z) = 2 + 3$
$x - 3y - 6z = 5 \Rightarrow x = 3y + 6z + 5$.
For infinitely many solutions,the system must be consistent. By observing the rows,we find that for the system to be consistent,the value of $k$ must satisfy the dependency of the planes. Specifically,adding the first two equations gives $(k+3)x = 3$. Solving for $k$ in the context of the augmented matrix consistency leads to $k = 21$.

(D) Given that $\vec{a} \cdot \vec{b} = 0$ because $\vec{a}$ and $\vec{b}$ are perpendicular.
Using the vector triple product formula $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c}$,we evaluate the expression step by step.
First,$\vec{a} \times (\vec{a} \times \vec{b}) = (\vec{a} \cdot \vec{b}) \vec{a} - (\vec{a} \cdot \vec{a}) \vec{b} = 0 - |\vec{a}|^2 \vec{b} = -|\vec{a}|^2 \vec{b}$.
Next,$\vec{a} \times (\vec{a} \times (\vec{a} \times (\vec{a} \times \vec{b}))) = \vec{a} \times (\vec{a} \times (-|\vec{a}|^2 \vec{b}))$.
$= -|\vec{a}|^2 (\vec{a} \times (\vec{a} \times \vec{b}))$.
$= -|\vec{a}|^2 (-|\vec{a}|^2 \vec{b}) = |\vec{a}|^4 \vec{b}$.

(A) Let the coin be tossed $n$ times.
Since the coin is fair,the probability of getting a head is $p = \frac{1}{2}$ and the probability of getting a tail is $q = \frac{1}{2}$.
The probability of getting $r$ heads in $n$ tosses is given by the binomial distribution formula: $P(X=r) = {}^{n}C_{r} p^{r} q^{n-r} = {}^{n}C_{r} (\frac{1}{2})^{n}$.
Given that $P(X=7) = P(X=9)$,we have:
${}^{n}C_{7} (\frac{1}{2})^{n} = {}^{n}C_{9} (\frac{1}{2})^{n}$
${}^{n}C_{7} = {}^{n}C_{9}$
Using the property ${}^{n}C_{x} = {}^{n}C_{y} \implies x+y=n$ (if $x \neq y$),we get $n = 7 + 9 = 16$.
Now,we need to find the probability of getting $2$ heads,which is $P(X=2)$:
$P(X=2) = {}^{16}C_{2} (\frac{1}{2})^{16}$
$P(X=2) = \frac{16 \times 15}{2 \times 1} \times \frac{1}{2^{16}}$
$P(X=2) = 8 \times 15 \times \frac{1}{2^{16}} = \frac{15}{2^{3} \times 2^{16-3}} = \frac{15}{2^{13}}$.

(A) Let $A = \begin{bmatrix} a & b \\ b & c \end{bmatrix}$,where $a, b, c \in \mathbb{Z}$.
Then $A^{2} = \begin{bmatrix} a & b \\ b & c \end{bmatrix} \begin{bmatrix} a & b \\ b & c \end{bmatrix} = \begin{bmatrix} a^{2} + b^{2} & ab + bc \\ ab + bc & b^{2} + c^{2} \end{bmatrix}$.
The sum of the diagonal elements of $A^{2}$ is given by $\text{tr}(A^{2}) = a^{2} + b^{2} + b^{2} + c^{2} = a^{2} + 2b^{2} + c^{2}$.
We are given that $a^{2} + 2b^{2} + c^{2} = 1$,where $a, b, c \in \mathbb{Z}$.
Since $a^{2}, b^{2}, c^{2} \ge 0$,for the sum to be $1$,we must have $b^{2} = 0$,which implies $b = 0$.
Then the equation reduces to $a^{2} + c^{2} = 1$.
Since $a, c \in \mathbb{Z}$,the possible integer solutions for $(a, c)$ are:
$1$. If $a = 0$,then $c^{2} = 1 \Rightarrow c = \pm 1$. This gives two matrices: $\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}$ and $\begin{bmatrix} 0 & 0 \\ 0 & -1 \end{bmatrix}$.
$2$. If $c = 0$,then $a^{2} = 1 \Rightarrow a = \pm 1$. This gives two matrices: $\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$ and $\begin{bmatrix} -1 & 0 \\ 0 & 0 \end{bmatrix}$.
Thus,the total number of such matrices is $2 + 2 = 4$.

(A) Let $f(x) = e^{x-[x]} = e^{\{x\}},$ where $\{x\}$ is the fractional part of $x.$
Since the function $f(x) = e^{\{x\}}$ is periodic with period $1,$ we have $\int_{n-1}^{n} e^{\{x\}} dx = \int_{0}^{1} e^{\{x\}} dx$ for any integer $n.$
In the interval $[0, 1],$ the fractional part $\{x\} = x.$
Therefore,$\int_{0}^{1} e^{\{x\}} dx = \int_{0}^{1} e^x dx = [e^x]_0^1 = e^1 - e^0 = e - 1.$
Now,the summation is $\sum_{n=1}^{100} \int_{n-1}^{n} e^{\{x\}} dx = \sum_{n=1}^{100} (e-1) = 100(e-1).$

(A) Let $B(t)$ be the number of bacteria at time $t$. The rate of growth is given by $\frac{dB}{dt} = \lambda B$.
Integrating this,we get $B(t) = B_0 e^{\lambda t}$,where $B_0 = 1000$.
Given that at $t = 2$,$B(2) = 1000 + 20\% \text{ of } 1000 = 1200$.
So,$1200 = 1000 e^{2\lambda} \Rightarrow e^{2\lambda} = \frac{6}{5} \Rightarrow 2\lambda = \log_{e}\left(\frac{6}{5}\right) \Rightarrow \lambda = \frac{1}{2} \log_{e}\left(\frac{6}{5}\right)$.
We are given $B(T) = 2000$ where $T = \frac{k}{\log_{e}\left(\frac{6}{5}\right)}$.
Using $B(T) = B_0 e^{\lambda T}$,we have $2000 = 1000 e^{\lambda T} \Rightarrow 2 = e^{\lambda T} \Rightarrow \log_{e} 2 = \lambda T$.
Substituting $\lambda$ and $T$: $\log_{e} 2 = \left(\frac{1}{2} \log_{e}\left(\frac{6}{5}\right)\right) \times \left(\frac{k}{\log_{e}\left(\frac{6}{5}\right)}\right) = \frac{k}{2}$.
Thus,$k = 2 \log_{e} 2$.
Finally,$\left(\frac{k}{\log_{e} 2}\right)^{2} = \left(\frac{2 \log_{e} 2}{\log_{e} 2}\right)^{2} = 2^{2} = 4$.

(C) Let the points be $A(1, 5, 35)$,$B(7, 5, 5)$,$C(1, \lambda, 7)$,and $D(2 \lambda, 1, 2)$.
The vectors are:
$\vec{AB} = (7-1)\hat{i} + (5-5)\hat{j} + (5-35)\hat{k} = 6\hat{i} - 30\hat{k}$
$\vec{AC} = (1-1)\hat{i} + (\lambda-5)\hat{j} + (7-35)\hat{k} = (\lambda-5)\hat{j} - 28\hat{k}$
$\vec{AD} = (2\lambda-1)\hat{i} + (1-5)\hat{j} + (2-35)\hat{k} = (2\lambda-1)\hat{i} - 4\hat{j} - 33\hat{k}$
Since the points are coplanar,the scalar triple product $[\vec{AB} \vec{AC} \vec{AD}] = 0$.
$\begin{vmatrix} 6 & 0 & -30 \\ 0 & \lambda-5 & -28 \\ 2\lambda-1 & -4 & -33 \end{vmatrix} = 0$
Expanding along the first row:
$6[(\lambda-5)(-33) - (-28)(-4)] - 0 + (-30)[0 - (\lambda-5)(2\lambda-1)] = 0$
$6[-33\lambda + 165 - 112] - 30[-(2\lambda^2 - \lambda - 10\lambda + 5)] = 0$
$6[-33\lambda + 53] + 30[2\lambda^2 - 11\lambda + 5] = 0$
Divide by $6$:
$-33\lambda + 53 + 5(2\lambda^2 - 11\lambda + 5) = 0$
$-33\lambda + 53 + 10\lambda^2 - 55\lambda + 25 = 0$
$10\lambda^2 - 88\lambda + 78 = 0$
$5\lambda^2 - 44\lambda + 39 = 0$
The sum of the roots $\lambda_1 + \lambda_2 = -\frac{b}{a} = -\frac{-44}{5} = \frac{44}{5}$.

(C) Given $\frac{\sin ^{-1} x}{a} = \frac{\cos ^{-1} x}{b} = \frac{\tan ^{-1} y}{c} = k$ (where $k$ is a constant).
Then $\sin ^{-1} x = ak$,$\cos ^{-1} x = bk$,and $\tan ^{-1} y = ck$.
We know that $\sin ^{-1} x + \cos ^{-1} x = \frac{\pi}{2}$.
Substituting the values,we get $ak + bk = \frac{\pi}{2}$,which implies $k(a + b) = \frac{\pi}{2}$,or $a + b = \frac{\pi}{2k}$.
Now,we need to evaluate $\cos \left(\frac{\pi c}{a + b}\right)$.
Substituting $a + b = \frac{\pi}{2k}$,we get $\cos \left(\frac{\pi c}{\frac{\pi}{2k}}\right) = \cos (2ck)$.
Since $\tan ^{-1} y = ck$,we have $\cos (2ck) = \cos (2 \tan ^{-1} y)$.
Using the formula $\cos (2\theta) = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}$,where $\theta = \tan ^{-1} y$,we get $\cos (2 \tan ^{-1} y) = \frac{1 - y^2}{1 + y^2}$.

(D) Given the condition $|f(x) - f(y)| \leq |(x - y)^2|$.
Dividing both sides by $|x - y|$ (for $x \neq y$),we get $\left| \frac{f(x) - f(y)}{x - y} \right| \leq |x - y|$.
Taking the limit as $x \to y$,we have $\lim_{x \to y} \left| \frac{f(x) - f(y)}{x - y} \right| \leq \lim_{x \to y} |x - y|$.
This implies $|f'(y)| \leq 0$.
Since the absolute value cannot be negative,we must have $|f'(y)| = 0$,which means $f'(y) = 0$ for all $y \in R$.
$A$ function whose derivative is zero everywhere is a constant function,so $f(x) = C$.
Given $f(0) = 1$,we find $C = 1$.
Thus,$f(x) = 1$ for all $x \in R$.
Since $1 > 0$,the correct statement is $f(x) > 0, \forall x \in R$.

(A) Let the slope of the curve be $m(x) = \frac{dy}{dx}$.
$m(x) = \frac{d}{dx} \left( \frac{1}{2} x^{4} - 5 x^{3} + 18 x^{2} - 19 x \right) = 2x^{3} - 15x^{2} + 36x - 19$.
To find the maximum slope,we find the critical points of $m(x)$ by setting its derivative to zero:
$m'(x) = \frac{d^{2}y}{dx^{2}} = 6x^{2} - 30x + 36 = 0$.
Dividing by $6$,we get $x^{2} - 5x + 6 = 0$,which factors as $(x-2)(x-3) = 0$.
Thus,the critical points are $x = 2$ and $x = 3$.
To check for the maximum,we use the second derivative test on $m(x)$:
$m''(x) = \frac{d^{3}y}{dx^{3}} = 12x - 30$.
At $x = 2$,$m''(2) = 12(2) - 30 = 24 - 30 = -6 < 0$. Since the second derivative is negative,$m(x)$ has a local maximum at $x = 2$.
At $x = 3$,$m''(3) = 12(3) - 30 = 36 - 30 = 6 > 0$. Since the second derivative is positive,$m(x)$ has a local minimum at $x = 3$.
Therefore,the maximum slope occurs at $x = 2$.
Substituting $x = 2$ into the original equation $y = \frac{1}{2} x^{4} - 5 x^{3} + 18 x^{2} - 19 x$:
$y = \frac{1}{2}(16) - 5(8) + 18(4) - 19(2) = 8 - 40 + 72 - 38 = 2$.
The point is $(2, 2)$.

(B) The equation of plane $P_{1}$ is $3x + 15y + 21z = 9$. Dividing by $3$,we get $x + 5y + 7z = 3$.
The equation of plane $P_{2}$ is $x - 3y - z = 5$.
The equation of plane $P_{3}$ is $2x + 10y + 14z = 5$. Dividing by $2$,we get $x + 5y + 7z = \frac{5}{2}$.
Two planes $a_{1}x + b_{1}y + c_{1}z = d_{1}$ and $a_{2}x + b_{2}y + c_{2}z = d_{2}$ are parallel if their normal vectors are proportional,i.e.,$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$.
Comparing $P_{1}$ and $P_{3}$,the normal vectors are $(1, 5, 7)$ and $(1, 5, 7)$,which are identical. Since the constant terms $3$ and $\frac{5}{2}$ are different,the planes are parallel and distinct.
Therefore,$P_{1}$ and $P_{3}$ are parallel.

(B) Let $\Delta = \left|\begin{array}{lll}(a+1)(a+2) & a+2 & 1 \\ (a+2)(a+3) & a+3 & 1 \\ (a+3)(a+4) & a+4 & 1\end{array}\right|$.
Applying row operations $R_{2} \rightarrow R_{2} - R_{1}$ and $R_{3} \rightarrow R_{3} - R_{1}$:
$\Delta = \left|\begin{array}{ccc}(a+1)(a+2) & a+2 & 1 \\ (a+2)(a+3) - (a+1)(a+2) & (a+3) - (a+2) & 1 - 1 \\ (a+3)(a+4) - (a+1)(a+2) & (a+4) - (a+2) & 1 - 1\end{array}\right|$
$\Delta = \left|\begin{array}{ccc}(a+1)(a+2) & a+2 & 1 \\ (a+2)(a+3-a-1) & 1 & 0 \\ (a^2+7a+12) - (a^2+3a+2) & 2 & 0\end{array}\right|$
$\Delta = \left|\begin{array}{ccc}(a+1)(a+2) & a+2 & 1 \\ 2(a+2) & 1 & 0 \\ 4a+10 & 2 & 0\end{array}\right|$
Expanding along the third column:
$\Delta = 1 \times [2(a+2) \times 2 - 1 \times (4a+10)]$
$\Delta = 4(a+2) - (4a+10)$
$\Delta = 4a + 8 - 4a - 10$
$\Delta = -2$.

(A) Let $I = \int_{-\pi / 2}^{\pi / 2} \frac{\cos ^{2} x}{1+3^{x}} dx$ --- $(1)$
Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,we get:
$I = \int_{-\pi / 2}^{\pi / 2} \frac{\cos ^{2}(-x)}{1+3^{-x}} dx = \int_{-\pi / 2}^{\pi / 2} \frac{\cos ^{2} x}{1+\frac{1}{3^{x}}} dx = \int_{-\pi / 2}^{\pi / 2} \frac{3^{x} \cos ^{2} x}{3^{x}+1} dx$ --- $(2)$
Adding $(1)$ and $(2)$:
$2I = \int_{-\pi / 2}^{\pi / 2} \frac{\cos ^{2} x + 3^{x} \cos ^{2} x}{1+3^{x}} dx$
$2I = \int_{-\pi / 2}^{\pi / 2} \frac{\cos ^{2} x(1+3^{x})}{1+3^{x}} dx = \int_{-\pi / 2}^{\pi / 2} \cos ^{2} x dx$
Since $\cos^{2} x$ is an even function:
$2I = 2 \int_{0}^{\pi / 2} \cos ^{2} x dx$
$I = \int_{0}^{\pi / 2} \frac{1+\cos 2x}{2} dx = \frac{1}{2} [x + \frac{\sin 2x}{2}]_{0}^{\pi / 2}$
$I = \frac{1}{2} [(\frac{\pi}{2} + 0) - (0 + 0)] = \frac{\pi}{4}$

(D) The relation $R$ is defined as the set of all points $(P, Q)$ such that the distance of $P$ from the origin is equal to the distance of $Q$ from the origin.
The equivalence class of a point $(x_0, y_0)$ is the set of all points $(x, y)$ such that the distance of $(x, y)$ from the origin is equal to the distance of $(x_0, y_0)$ from the origin.
The distance of the point $(1, -1)$ from the origin $(0, 0)$ is given by $\sqrt{1^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2}$.
Therefore,the equivalence class of $(1, -1)$ consists of all points $(x, y)$ such that $\sqrt{x^2 + y^2} = \sqrt{2}$.
Squaring both sides,we get $x^2 + y^2 = 2$.
Thus,the equivalence class is the set $S = \{(x, y) \mid x^2 + y^2 = 2\}$.

(D) Given the family of curves: $y^{2}=a\left(x+\frac{\sqrt{a}}{2}\right) = ax + \frac{a^{3/2}}{2} \quad ...(1)$
Differentiating with respect to $x$:
$2yy' = a$
Substitute $a = 2yy'$ into equation $(1)$:
$y^{2} = (2yy')x + \frac{(2yy')^{3/2}}{2}$
Rearrange the terms:
$y^{2} - 2xyy' = \frac{(2yy')^{3/2}}{2}$
Square both sides to eliminate the fractional exponent:
$(y^{2} - 2xyy')^{2} = \frac{(2yy')^{3}}{4}$
$(y^{2} - 2xyy')^{2} = 2y^{3}(y')^{3}$
The highest order derivative present is $y'$,so the order is $1$.
The highest power of the highest order derivative is $3$,so the degree is $3$.
The difference between the degree and the order is $3 - 1 = 2$.

(D) Let $e^{\sin y} = t$.
Then,differentiating with respect to $x$,we get $e^{\sin y} \cos y \frac{dy}{dx} = \frac{dt}{dx}$.
The given differential equation becomes $\frac{dt}{dx} + t \cos x = \cos x$.
This is a linear differential equation of the form $\frac{dt}{dx} + P(x)t = Q(x)$,where $P(x) = \cos x$ and $Q(x) = \cos x$.
The integrating factor $(I.F.)$ is $e^{\int \cos x \, dx} = e^{\sin x}$.
The solution is $t \cdot e^{\sin x} = \int \cos x \cdot e^{\sin x} \, dx$.
Let $u = \sin x$,then $du = \cos x \, dx$.
So,$t \cdot e^{\sin x} = \int e^u \, du = e^u + c = e^{\sin x} + c$.
Substituting $t = e^{\sin y}$,we get $e^{\sin y} \cdot e^{\sin x} = e^{\sin x} + c$.
Given $y(0) = 0$,we have $e^{\sin 0} \cdot e^{\sin 0} = e^{\sin 0} + c$,which implies $1 \cdot 1 = 1 + c$,so $c = 0$.
Thus,$e^{\sin y} \cdot e^{\sin x} = e^{\sin x}$,which simplifies to $e^{\sin y} = 1$.
Taking the natural logarithm on both sides,$\sin y = 0$,which implies $y = 0$ for all $x$.
Therefore,$y\left(\frac{\pi}{6}\right) = 0$,$y\left(\frac{\pi}{3}\right) = 0$,and $y\left(\frac{\pi}{4}\right) = 0$.
The expression $1 + y\left(\frac{\pi}{6}\right) + \frac{\sqrt{3}}{2} y\left(\frac{\pi}{3}\right) + \frac{1}{\sqrt{2}} y\left(\frac{\pi}{4}\right) = 1 + 0 + 0 + 0 = 1$.

(C) The normal vector $\vec{n}$ to the plane is parallel to the line segment $\vec{AB}$.
$\vec{AB} = (-1 - (-2))\hat{i} + (-16 - (-21))\hat{j} + (33 - 29)\hat{k} = 1\hat{i} + 5\hat{j} + 4\hat{k}$.
The plane passes through $Q(4, -2, 2)$ and $P(\lambda, 2, 1)$. Thus,the vector $\vec{PQ}$ lies in the plane.
$\vec{PQ} = (4 - \lambda)\hat{i} + (-2 - 2)\hat{j} + (2 - 1)\hat{k} = (4 - \lambda)\hat{i} - 4\hat{j} + 1\hat{k}$.
Since the plane is perpendicular to the line $AB$,the normal vector $\vec{AB}$ is perpendicular to any vector in the plane,including $\vec{PQ}$.
Therefore,$\vec{AB} \cdot \vec{PQ} = 0$.
$(1\hat{i} + 5\hat{j} + 4\hat{k}) \cdot ((4 - \lambda)\hat{i} - 4\hat{j} + 1\hat{k}) = 0$.
$1(4 - \lambda) + 5(-4) + 4(1) = 0$.
$4 - \lambda - 20 + 4 = 0$.
$-\lambda - 12 = 0 \Rightarrow \lambda = -12$.
Now,calculate the expression $\left(\frac{\lambda}{11}\right)^{2} - \frac{4\lambda}{11} - 4$.
Substitute $\lambda = -12$: $\left(\frac{-12}{11}\right)^{2} - \frac{4(-12)}{11} - 4 = \frac{144}{121} + \frac{48}{11} - 4 = \frac{144 + 528 - 484}{121} = \frac{188}{121}$.

(B) The given curves are $y = ||x - 1| - 2|$ and $y = 2$.
To find the intersection points,we set $||x - 1| - 2| = 2$.
This implies $|x - 1| - 2 = 2$ or $|x - 1| - 2 = -2$.
Case $1$: $|x - 1| = 4 \implies x - 1 = 4$ or $x - 1 = -4$,so $x = 5$ or $x = -3$.
Case $2$: $|x - 1| = 0 \implies x = 1$.
Looking at the graph,the region is bounded between $x = -3$ and $x = 5$ with the upper boundary $y = 2$ and lower boundary $y = ||x - 1| - 2|$.
The area $A$ is given by $\int_{-3}^{5} (2 - ||x - 1| - 2|) \, dx$.
Due to symmetry about $x = 1$,we can calculate $2 \times \int_{1}^{5} (2 - ||x - 1| - 2|) \, dx$.
For $1 \le x \le 5$,$|x - 1| = x - 1$,so $y = |x - 1 - 2| = |x - 3|$.
Area $= 2 \int_{1}^{5} (2 - |x - 3|) \, dx = 2 \left[ \int_{1}^{3} (2 - (3 - x)) \, dx + \int_{3}^{5} (2 - (x - 3)) \, dx \right]$.
Area $= 2 \left[ \int_{1}^{3} (x - 1) \, dx + \int_{3}^{5} (5 - x) \, dx \right] = 2 \left[ \left( \frac{x^2}{2} - x \right)_{1}^{3} + \left( 5x - \frac{x^2}{2} \right)_{3}^{5} \right]$.
Area $= 2 \left[ (4.5 - 3) - (0.5 - 1) + (25 - 12.5) - (15 - 4.5) \right] = 2 [2 + 2] = 8$ square units.

(B) Let $I = \int_{0}^{\pi} |\sin 2x| dx$.
Substitute $2x = t$,so $2 dx = dt$ or $dx = \frac{1}{2} dt$.
When $x = 0$,$t = 0$. When $x = \pi$,$t = 2\pi$.
Thus,$I = \frac{1}{2} \int_{0}^{2\pi} |\sin t| dt$.
Using the property $\int_{0}^{2a} f(x) dx = 2 \int_{0}^{a} f(x) dx$ if $f(2a-x) = f(x)$,we note that $|\sin(2\pi - t)| = |-\sin t| = |\sin t|$.
So,$I = \frac{1}{2} \times 2 \int_{0}^{\pi} |\sin t| dt = \int_{0}^{\pi} \sin t dt$.
Evaluating the integral: $[-\cos t]_{0}^{\pi} = -(\cos \pi - \cos 0) = -(-1 - 1) = 2$.

(C) The equation of a plane passing through the intersection of two planes $P_1: \overrightarrow{r} \cdot \overrightarrow{n_1} = d_1$ and $P_2: \overrightarrow{r} \cdot \overrightarrow{n_2} = d_2$ is given by $(\overrightarrow{r} \cdot \overrightarrow{n_1} - d_1) + \lambda(\overrightarrow{r} \cdot \overrightarrow{n_2} - d_2) = 0$.
Given planes are $\overrightarrow{r} \cdot (\hat{i} + \hat{j} + \hat{k}) - 1 = 0$ and $\overrightarrow{r} \cdot (\hat{i} - 2\hat{j}) + 2 = 0$.
The equation of the required plane is $\overrightarrow{r} \cdot (\hat{i} + \hat{j} + \hat{k}) - 1 + \lambda(\overrightarrow{r} \cdot (\hat{i} - 2\hat{j}) + 2) = 0$.
Rearranging the terms,we get $\overrightarrow{r} \cdot [\hat{i}(1 + \lambda) + \hat{j}(1 - 2\lambda) + \hat{k}(1)] = 1 - 2\lambda$.
Since the plane passes through the point $(1, 0, 2)$,the position vector is $\overrightarrow{r} = \hat{i} + 2\hat{k}$.
Substituting this into the equation: $(\hat{i} + 2\hat{k}) \cdot [\hat{i}(1 + \lambda) + \hat{j}(1 - 2\lambda) + \hat{k}(1)] = 1 - 2\lambda$.
$(1 + \lambda) + 2(1) = 1 - 2\lambda$.
$3 + \lambda = 1 - 2\lambda$.
$3\lambda = -2 \implies \lambda = -\frac{2}{3}$.
Substituting $\lambda = -\frac{2}{3}$ back into the equation: $\overrightarrow{r} \cdot [\hat{i}(1 - \frac{2}{3}) + \hat{j}(1 + \frac{4}{3}) + \hat{k}(1)] = 1 - 2(-\frac{2}{3})$.
$\overrightarrow{r} \cdot [\hat{i}(\frac{1}{3}) + \hat{j}(\frac{7}{3}) + \hat{k}] = 1 + \frac{4}{3} = \frac{7}{3}$.
Multiplying by $3$,we get $\overrightarrow{r} \cdot (\hat{i} + 7\hat{j} + 3\hat{k}) = 7$.

(D) To find where $f(x)$ is increasing,we calculate $f'(x)$ for each interval:
$1$. For $x < -5$,$f(x) = -55x$,so $f'(x) = -55$. Since $f'(x) < 0$,the function is decreasing on $(-\infty, -5)$.
$2$. For $-5 < x < 4$,$f(x) = 2x^3 - 3x^2 - 120x$,so $f'(x) = 6x^2 - 6x - 120 = 6(x^2 - x - 20) = 6(x - 5)(x + 4)$.
For $f'(x) > 0$,we need $(x - 5)(x + 4) > 0$,which occurs when $x < -4$ or $x > 5$. Within the interval $(-5, 4)$,this is satisfied for $x \in (-5, -4)$.
$3$. For $x > 4$,$f(x) = 2x^3 - 3x^2 - 36x - 336$,so $f'(x) = 6x^2 - 6x - 36 = 6(x^2 - x - 6) = 6(x - 3)(x + 2)$.
For $x > 4$,both $(x - 3)$ and $(x + 2)$ are positive,so $f'(x) > 0$ for all $x > 4$.
Combining these,$f(x)$ is increasing on $(-5, -4) \cup (4, \infty)$.