JEE Main 2026 Mathematics Question Paper with Answer and Solution

(B) We know that $\frac{1}{^{n}C_{r}} + \frac{1}{^{n}C_{r+1}} = \frac{r! (n-r)!}{n!} + \frac{(r+1)! (n-r-1)!}{n!} = \frac{(n-r-1)! r!}{n!} [n-r + r+1] = \frac{(n+1) (n-r-1)! r!}{n!} = \frac{n+1}{n-r} \cdot \frac{1}{^{n}C_{r}}$.
Using the property $\frac{1}{^{n}C_{r}} + \frac{1}{^{n}C_{r+1}} = \frac{n+1}{n+1} \cdot \frac{^{n+1}C_{r+1}}{^{n}C_{r} \cdot ^{n}C_{r+1}} = \frac{n+1}{n+1} \cdot \frac{1}{^{n}C_{r}} \cdot \frac{n+1}{r+1} = \frac{16}{15} \cdot \frac{1}{^{14}C_{r}}$.
Thus,the product is $\prod_{r=0}^{12} (\frac{1}{^{15}C_{r}} + \frac{1}{^{15}C_{r+1}}) = \prod_{r=0}^{12} (\frac{16}{15} \cdot \frac{1}{^{14}C_{r}}) = \frac{(16/15)^{13}}{^{14}C_{0} \cdot ^{14}C_{1} \cdot ... \cdot ^{14}C_{12}}$.
Comparing with the given expression,$a = \frac{16}{15}$.
Therefore,$30a = 30 \cdot \frac{16}{15} = 32$.

(A) To find the largest $n$ such that $7^n$ divides $101!$,we use Legendre's Formula for the exponent of a prime $p$ in $m!$,which is given by $E_p(m!) = \sum_{k=1}^{\infty} [\frac{m}{p^k}]$.
Here,$m = 101$ and $p = 7$.
$E_7(101!) = [\frac{101}{7}] + [\frac{101}{7^2}] + [\frac{101}{7^3}] + \dots$
$E_7(101!) = [14.428] + [2.061] + [0.294] + \dots$
$E_7(101!) = 14 + 2 + 0 = 16$.
Thus,the largest $n$ is $16$.

(D) The condition $|z-5| \le 3$ represents a disk centered at $(5, 0)$ with radius $3$. For the argument of $z$ to be maximum,the line from the origin must be tangent to the circle at point $P(z)$.
In the right-angled triangle formed by the origin $(0, 0)$,the center $(5, 0)$,and the point $P(z)$,the hypotenuse is $5$ and the radius is $3$. Thus,the distance from the origin to $P$ is $\sqrt{5^2 - 3^2} = 4$.
Let $z = x + iy$. From the geometry,$\cos \theta = \frac{4}{5}$ and $\sin \theta = \frac{3}{5}$.
Thus,$z = 4(\cos \theta + i \sin \theta) = 4(\frac{4}{5} + i \frac{3}{5}) = \frac{16}{5} + i \frac{12}{5}$.
Now,substitute $z$ into the expression:
$5z - 12 = 5(\frac{16}{5} + i \frac{12}{5}) - 12 = 16 + 12i - 12 = 4 + 12i$.
$5iz + 16 = 5i(\frac{16}{5} + i \frac{12}{5}) + 16 = 16i - 12 + 16 = 4 + 16i$.
Then,$|\frac{5z-12}{5iz+16}|^2 = |\frac{4+12i}{4+16i}|^2 = |\frac{1+3i}{1+4i}|^2 = \frac{1^2 + 3^2}{1^2 + 4^2} = \frac{1+9}{1+16} = \frac{10}{17}$.
Finally,$34 \times \frac{10}{17} = 2 \times 10 = 20$.

(C) Given $|x^{2} - 10| \le 6$,we have $-6 \le x^{2} - 10 \le 6$.
Adding $10$ to all sides,we get $4 \le x^{2} \le 16$.
Taking the square root,$x \in [-4, -2] \cup [2, 4]$,so $A = [-4, -2] \cup [2, 4]$.
Given $|x - 2| > 1$,we have $x - 2 > 1$ or $x - 2 < -1$.
This implies $x > 3$ or $x < 1$,so $B = (-\infty, 1) \cup (3, \infty)$.
Now,$A \cap B = ([-4, -2] \cup [2, 4]) \cap ((-\infty, 1) \cup (3, \infty)) = [-4, -2] \cup (3, 4]$.
Thus,option $C$ is correct.

(A) The given sequence is a $G$.$P$. with common ratio $\frac{1}{\sqrt{2}}$.
Thus,$\frac{a_2/2}{a_1} = \frac{a_3/2^2}{a_2/2} = \frac{1}{\sqrt{2}}$.
This implies $\frac{a_2}{2a_1} = \frac{a_3}{2a_2} = \frac{1}{\sqrt{2}}$,so $\frac{a_2}{a_1} = \frac{2}{\sqrt{2}} = \sqrt{2}$.
Therefore,$a_1, a_2, \ldots, a_{10}$ is a $G$.$P$. with first term $a_1$ and common ratio $R = \sqrt{2}$.
The sum of the first $10$ terms is $S_{10} = a_1 \frac{R^{10}-1}{R-1} = 62$.
$a_1 \frac{(\sqrt{2})^{10}-1}{\sqrt{2}-1} = 62$.
$a_1 \frac{2^5-1}{\sqrt{2}-1} = 62$.
$a_1 \frac{31}{\sqrt{2}-1} = 62$.
$a_1 = \frac{62(\sqrt{2}-1)}{31} = 2(\sqrt{2}-1)$.

(D) Given the quadratic equation $f(x) = x^{2}+2ax+(3a+10) = 0$.
Since the roots $\alpha$ and $\beta$ satisfy $\alpha < 1 < \beta$,the value of the function at $x=1$ must be negative,i.e.,$f(1) < 0$.
Substituting $x=1$ into the equation:
$f(1) = (1)^{2} + 2a(1) + (3a+10) < 0$
$1 + 2a + 3a + 10 < 0$
$5a + 11 < 0$
$5a < -11$
$a < -11/5$
Therefore,the set of all possible values of $a$ is $(-\infty, -11/5)$.

(B) The equation of the parabola is $y^{2}=16x$,which is of the form $y^{2}=4ax$ with $a=4$.
Let the coordinates of point $A$ be $(at_{1}^{2}, 2at_{1}) = (16, 16)$.
Thus,$2(4)t_{1} = 16 \Rightarrow t_{1}=2$.
Since $AB$ is a focal chord,the product of the parameters of its endpoints is $t_{1}t_{2}=-1$.
Therefore,$2t_{2}=-1 \Rightarrow t_{2}=-\frac{1}{2}$.
The coordinates of point $B$ are $(at_{2}^{2}, 2at_{2}) = (4(-\frac{1}{2})^{2}, 2(4)(-\frac{1}{2})) = (1, -4)$.
Point $P(\alpha, \beta)$ divides the chord $AB$ internally in the ratio $5:2$.
Case $1$: $P$ divides $AB$ in ratio $5:2$ (from $A$ to $B$):
$\alpha = \frac{5(1) + 2(16)}{5+2} = \frac{37}{7}$,$\beta = \frac{5(-4) + 2(16)}{5+2} = \frac{12}{7}$.
$\alpha+\beta = \frac{37+12}{7} = \frac{49}{7} = 7$.
Case $2$: $P$ divides $BA$ in ratio $5:2$ (from $B$ to $A$):
$\alpha = \frac{5(16) + 2(1)}{5+2} = \frac{82}{7}$,$\beta = \frac{5(16) + 2(-4)}{5+2} = \frac{72}{7}$.
$\alpha+\beta = \frac{82+72}{7} = \frac{154}{7} = 22$.
The minimum value of $\alpha+\beta$ is $7$.

(C) The parabola is $y^{2}=12x$,so $4a=12 \Rightarrow a=3$. Any point $P$ on the parabola can be represented as $(3t^{2}, 6t)$.
The slope of $OP$ is $m_{OP} = \frac{6t-0}{3t^{2}-0} = \frac{2}{t}$.
Since $\angle OPA=90^{\circ}$,$OP \perp PA$,so the slope of $PA$ is $m_{PA} = -\frac{t}{2}$.
The equation of line $PA$ is $y-6t = -\frac{t}{2}(x-3t^{2})$.
For point $A$ on the $x$-axis,set $y=0$: $-6t = -\frac{t}{2}(x-3t^{2})$ $\Rightarrow 12 = x-3t^{2}$ $\Rightarrow x = 3t^{2}+12$.
Thus,$A = (3t^{2}+12, 0)$.
Let the centroid of $\triangle OPA$ be $G(h, k)$.
$h = \frac{0+3t^{2}+3t^{2}+12}{3} = \frac{6t^{2}+12}{3} = 2t^{2}+4$.
$k = \frac{0+6t+0}{3} = 2t$.
From $k=2t$,we have $t = \frac{k}{2}$.
Substituting into $h$: $h = 2(\frac{k}{2})^{2}+4 = \frac{k^{2}}{2}+4$.
$2h = k^{2}+8 \Rightarrow k^{2} = 2h-8$.
The locus is $y^{2} = 2x-8$,which can be written as $y^{2}-2x+8=0$.

(D) The equation of the line is $\alpha x+4y-\sqrt{7}=0$. This line touches the ellipse $3x^{2}+4y^{2}=1$,which can be written as $\frac{x^{2}}{1/3} + \frac{y^{2}}{1/4} = 1$.
Comparing with the condition of tangency $c^{2}=a^{2}m^{2}+b^{2}$,where $m = -\frac{\alpha}{4}$ and $c = \frac{\sqrt{7}}{4}$:
$(\frac{\sqrt{7}}{4})^{2} = (\frac{1}{3})(\frac{-\alpha}{4})^{2} + \frac{1}{4} \implies \frac{7}{16} = \frac{\alpha^{2}}{48} + \frac{1}{4} \implies \frac{\alpha^{2}}{48} = \frac{3}{16} \implies \alpha^{2} = 9 \implies \alpha = \pm 3$.
Since $P$ is in the first quadrant,the tangent is $3x+4y=\sqrt{7}$.
The point of contact $P(x_{1}, y_{1})$ for the tangent $3x+4y=\sqrt{7}$ is given by $\frac{3x}{x_{1}} = \frac{4y}{y_{1}} = \frac{1}{1/7}$ is not correct,rather using $3xx_{1}+4yy_{1}=1$,we get $3x_{1} = \frac{3}{\sqrt{7}}$ and $4y_{1} = \frac{4}{\sqrt{7}}$,so $P = (\frac{1}{\sqrt{7}}, \frac{1}{\sqrt{7}})$.
For the ellipse $\frac{x^{2}}{1/3} + \frac{y^{2}}{1/4} = 1$,$a^{2} = 1/3$ and $b^{2} = 1/4$. Eccentricity $e = \sqrt{1-\frac{b^{2}}{a^{2}}} = \sqrt{1-\frac{1/4}{1/3}} = \sqrt{1-\frac{3}{4}} = \frac{1}{2}$.
The focal distances are $a \pm ex_{1} = \frac{1}{\sqrt{3}} \pm \frac{1}{2}(\frac{1}{\sqrt{7}}) = \frac{1}{\sqrt{3}} \pm \frac{1}{2\sqrt{7}}$.
Thus,one of the focal distances is $\frac{1}{\sqrt{3}}+\frac{1}{2\sqrt{7}}$.

(A) The given equation is $\sum_{r=1}^{n}(x+2r-2)(x+2r) = \frac{8n}{3}$.
Expanding the terms: $\sum_{r=1}^{n}(x^2 + 4rx - 2x + 4r^2 - 4r) = \frac{8n}{3}$.
This simplifies to $nx^2 + 2x(2\sum r - n) + 4\sum r^2 - 4\sum r = \frac{8n}{3}$.
Using $\sum_{r=1}^{n} r = \frac{n(n+1)}{2}$ and $\sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6}$,we get $nx^2 + 2x(n^2) + \frac{4n(n+1)(2n+1)}{6} - 2n(n+1) = \frac{8n}{3}$.
Dividing by $n$: $x^2 + 2nx + \frac{2(n+1)(2n+1)}{3} - 2(n+1) = \frac{8}{3}$.
$x^2 + 2nx + \frac{4n^2+6n+2-6n-6-8}{3} = 0 \Rightarrow x^2 + 2nx + \frac{4n^2-12}{3} = 0$.
Since the roots $\alpha, \beta$ are consecutive even integers,$|\alpha - \beta| = 2$.
Thus,$\frac{\sqrt{D}}{1} = 2 \Rightarrow D = 4$.
$D = (2n)^2 - 4(\frac{4n^2-12}{3}) = 4$.
$4n^2 - \frac{16n^2-48}{3} = 4 \Rightarrow 12n^2 - 16n^2 + 48 = 12$.
$-4n^2 = -36$ $\Rightarrow n^2 = 9$ $\Rightarrow n = 3$.

(A) The set $S = \{1, 2, 3, . . . , 11\}$ contains $6$ odd numbers $\{1, 3, 5, 7, 9, 11\}$ and $5$ even numbers $\{2, 4, 6, 8, 10\}$.
The total number of subsets of $S$ is $2^{11} = 2048$.
$A$ subset $B$ has an odd product if and only if all elements of $B$ are odd. The number of such subsets is $2^6 = 64$.
$A$ subset $B$ has an even product if it contains at least one even number. The number of such subsets is $2^{11} - 2^6 = 2048 - 64 = 1984$.
The condition $n(B) \ge 2$ excludes subsets with $0$ or $1$ element.
Subsets with $0$ elements: $\emptyset$ (product is not defined or considered odd,so it is excluded).
Subsets with $1$ element: $\{1\}, \{3\}, \{5\}, \{7\}, \{9\}, \{11\}$ (all odd products) and $\{2\}, \{4\}, \{6\}, \{8\}, \{10\}$ (all even products).
We must exclude the $5$ subsets that contain exactly one even number (since their product is even but $n(B) < 2$).
Therefore,the required number of subsets is $1984 - 5 = 1979$.

(B) Given $a, b, c$ are in $A.P.,$ let $a = b - d$ and $c = b + d.$
Since $a+b+c=1,$ we have $(b-d) + b + (b+d) = 1,$ which implies $3b = 1$ or $b = \frac{1}{3}.$
Given $a^{2}, 2b^{2}, c^{2}$ are in $G.P.,$ we have $(2b^{2})^{2} = a^{2}c^{2},$ so $4b^{4} = (ac)^{2}.$
Substituting $a = b-d$ and $c = b+d,$ we get $4b^{4} = ((b-d)(b+d))^{2} = (b^{2}-d^{2})^{2}.$
Taking the square root,$b^{2}-d^{2} = \pm 2b^{2}.$
Case $1: b^{2}-d^{2} = 2b^{2} \Rightarrow d^{2} = -b^{2}$ (not possible for real $d$).
Case $2: b^{2}-d^{2} = -2b^{2} \Rightarrow d^{2} = 3b^{2} = 3(\frac{1}{9}) = \frac{1}{3}.$
Since $a < b < c,$ $d$ must be positive,so $d = \frac{1}{\sqrt{3}}.$
Now,$a^{2}+b^{2}+c^{2} = (b-d)^{2} + b^{2} + (b+d)^{2} = 3b^{2} + 2d^{2}.$
Substituting values: $3(\frac{1}{9}) + 2(\frac{1}{3}) = \frac{1}{3} + \frac{2}{3} = 1.$
Therefore,$9(a^{2}+b^{2}+c^{2}) = 9(1) = 9.$

(C) Given $\cos(\alpha+\beta) = -\frac{1}{10}$. Since $0 < \alpha < \frac{\pi}{3}$ and $0 < \beta < \frac{\pi}{4}$,we have $0 < \alpha+\beta < \frac{7\pi}{12}$. Since $\cos(\alpha+\beta) < 0$,we must have $\frac{\pi}{2} < \alpha+\beta < \frac{7\pi}{12}$.
Thus,$\sin(\alpha+\beta) = \sqrt{1 - (-\frac{1}{10})^2} = \sqrt{\frac{99}{100}} = \frac{3\sqrt{11}}{10}$.
So,$\tan(\alpha+\beta) = \frac{3\sqrt{11}/10}{-1/10} = -3\sqrt{11}$.
Given $\sin(\alpha-\beta) = \frac{3}{8}$. Since $0 < \alpha < \frac{\pi}{3}$ and $0 < \beta < \frac{\pi}{4}$,we have $-\frac{\pi}{4} < \alpha-\beta < \frac{\pi}{3}$. Since $\sin(\alpha-\beta) > 0$,we have $0 < \alpha-\beta < \frac{\pi}{3}$.
Thus,$\cos(\alpha-\beta) = \sqrt{1 - (\frac{3}{8})^2} = \sqrt{\frac{55}{64}} = \frac{\sqrt{55}}{8}$.
So,$\tan(\alpha-\beta) = \frac{3/8}{\sqrt{55}/8} = \frac{3}{\sqrt{55}}$.
Now,$\tan 2\alpha = \tan((\alpha+\beta)+(\alpha-\beta)) = \frac{\tan(\alpha+\beta) + \tan(\alpha-\beta)}{1 - \tan(\alpha+\beta)\tan(\alpha-\beta)}$.
Substituting the values,$\tan 2\alpha = \frac{-3\sqrt{11} + \frac{3}{\sqrt{55}}}{1 - (-3\sqrt{11})(\frac{3}{\sqrt{55}})} = \frac{\frac{-3\sqrt{11}\sqrt{55} + 3}{\sqrt{55}}}{1 + \frac{9\sqrt{11}}{\sqrt{55}}} = \frac{-3(11\sqrt{5}) + 3}{\sqrt{55} + 9\sqrt{11}} = \frac{3(1 - 11\sqrt{5})}{\sqrt{11}(\sqrt{5} + 9)}$.
Comparing with $\frac{3(1-r\sqrt{5})}{\sqrt{11}(s+\sqrt{5})}$,we get $r=11$ and $s=9$.
Therefore,$r+s = 11+9 = 20$.

(D) $P_n = \sum_{r=0}^n \frac{{}^n C_r(-2)^r}{r+1} = \sum_{r=0}^n \frac{1}{n+1} {}^{n+1} C_{r+1}(-2)^r$
$= \frac{-1}{2(n+1)} \sum_{r=0}^n {}^{n+1} C_{r+1}(-2)^{r+1}$
$= \frac{-1}{2(n+1)} \left[(1-2)^{n+1} - 1\right]$
$P_n = \frac{1}{2(n+1)} \left[1 - (-1)^{n+1}\right]$
$P_{2n} = \frac{1}{2(2n+1)} \left[1 - (-1)^{2n+1}\right]$
$P_{2n} = \frac{1}{2n+1}$
$\sum_{n=1}^{25} \frac{1}{P_{2n}} = \sum_{n=1}^{25} (2n+1)$
$= 3 + 5 + \dots + 51$
$= \frac{25}{2} [51 + 3]$
$= 25 \times 27 = 675$

(B) Let the parabola be $y^2=4x$. Any point on the parabola is $P(t^2, 2t)$.
The chord passes through the origin $O(0,0)$ and $P(t^2, 2t)$.
The mid-point $M(h, k)$ of the chord $OP$ is given by $h = \frac{t^2+0}{2} = \frac{t^2}{2}$ and $k = \frac{2t+0}{2} = t$.
Substituting $t=k$ into $h = \frac{t^2}{2}$,we get $h = \frac{k^2}{2}$,so $k^2=2h$.
The locus $S$ is $y^2=2x$.
Let $P(x_0, y_0)$ be a point on $S$,so $y_0^2=2x_0$. Since $P$ is on $S$,we can write $P$ as $(2t^2, 2t)$ because $(2t)^2 = 2(2t^2)$.
Let $R(x, y)$ be the point that divides $OP$ internally in the ratio $3:1$. Using the section formula:
$x = \frac{3(2t^2) + 1(0)}{3+1} = \frac{6t^2}{4} = \frac{3t^2}{2}$
$y = \frac{3(2t) + 1(0)}{3+1} = \frac{6t}{4} = \frac{3t}{2}$
From $y = \frac{3t}{2}$,we get $t = \frac{2y}{3}$.
Substituting $t$ into $x = \frac{3t^2}{2}$:
$x = \frac{3}{2} \left(\frac{2y}{3}\right)^2 = \frac{3}{2} \cdot \frac{4y^2}{9} = \frac{2y^2}{3}$
Thus,$3x = 2y^2$,or $2y^2=3x$.

(D) Given the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$,we have $a^2=25$ and $b^2=9$,so $a=5$ and $b=3$.
Eccentricity $e = \sqrt{1-\frac{b^2}{a^2}} = \sqrt{1-\frac{9}{25}} = \frac{4}{5}$.
The foci are $S(ae, 0) = (4, 0)$ and $S^{\prime}(-ae, 0) = (-4, 0)$.
Since $P(\alpha, \beta)$ lies on the ellipse,$SP+S^{\prime}P=2a=10$.
Given $(SP)^2+(S^{\prime}P)^2-SP \cdot S^{\prime}P=37$.
Using $(SP+S^{\prime}P)^2 = (SP)^2+(S^{\prime}P)^2+2SP \cdot S^{\prime}P$,we substitute $(SP)^2+(S^{\prime}P)^2 = (SP+S^{\prime}P)^2-2SP \cdot S^{\prime}P$.
So,$(SP+S^{\prime}P)^2-3SP \cdot S^{\prime}P=37$.
$10^2-3SP \cdot S^{\prime}P=37$ $\Rightarrow 100-3SP \cdot S^{\prime}P=37$ $\Rightarrow 3SP \cdot S^{\prime}P=63$ $\Rightarrow SP \cdot S^{\prime}P=21$.
For any point $P(\alpha, \beta)$ on the ellipse,$SP = a-e\alpha$ and $S^{\prime}P = a+e\alpha$.
$SP \cdot S^{\prime}P = a^2-e^2\alpha^2 = 25-\frac{16}{25}\alpha^2 = 21$.
$\frac{16}{25}\alpha^2 = 4 \Rightarrow \alpha^2 = \frac{100}{16} = \frac{25}{4}$.
From $\frac{\alpha^2}{25}+\frac{\beta^2}{9}=1$,we have $\frac{25/4}{25}+\frac{\beta^2}{9}=1$ $\Rightarrow \frac{1}{4}+\frac{\beta^2}{9}=1$ $\Rightarrow \frac{\beta^2}{9}=\frac{3}{4}$ $\Rightarrow \beta^2=\frac{27}{4}$.
Thus,$\alpha^2+\beta^2 = \frac{25}{4}+\frac{27}{4} = \frac{52}{4} = 13$.

(C) Given the quadratic equation $12x^{2}-20x+3\lambda=0$,the roots $\alpha$ and $\beta$ satisfy $\alpha+\beta = \frac{20}{12} = \frac{5}{3}$ and $\alpha\beta = \frac{3\lambda}{12} = \frac{\lambda}{4}$.
We are given $\frac{1}{2} \le |\beta-\alpha| \le \frac{3}{2}$.
Squaring the inequality,we get $\frac{1}{4} \le (\beta-\alpha)^{2} \le \frac{9}{4}$.
Using the identity $(\beta-\alpha)^{2} = (\alpha+\beta)^{2} - 4\alpha\beta$,we have:
$\frac{1}{4} \le (\frac{5}{3})^{2} - 4(\frac{\lambda}{4}) \le \frac{9}{4}$.
$\frac{1}{4} \le \frac{25}{9} - \lambda \le \frac{9}{4}$.
Subtracting $\frac{25}{9}$ from all parts:
$\frac{1}{4} - \frac{25}{9} \le -\lambda \le \frac{9}{4} - \frac{25}{9}$.
$\frac{9-100}{36} \le -\lambda \le \frac{81-100}{36}$.
$-\frac{91}{36} \le -\lambda \le -\frac{19}{36}$.
Multiplying by $-1$ reverses the inequality:
$\frac{19}{36} \le \lambda \le \frac{91}{36}$.
Since $\frac{19}{36} \approx 0.527$ and $\frac{91}{36} \approx 2.527$,the possible integer values for $\lambda$ are $1$ and $2$.
The sum of these values is $1+2 = 3$.

(D) $P (10, 2 \sqrt{15})$ lies on $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.
$\frac{100}{a^2} - \frac{60}{b^2} = 1 \dots (1)$.
Given length of latus rectum $= 8$,so $\frac{2b^2}{a} = 8 \Rightarrow b^2 = 4a \dots (2)$.
Substituting $(2)$ in $(1)$: $\frac{100}{a^2} - \frac{60}{4a} = 1 \Rightarrow \frac{100}{a^2} - \frac{15}{a} = 1$.
$100 - 15a = a^2 \Rightarrow a^2 + 15a - 100 = 0$.
$(a + 20)(a - 5) = 0$. Since $a > 0$,$a = 5$.
Then $b^2 = 4(5) = 20$.
The hyperbola is $\frac{x^2}{25} - \frac{y^2}{20} = 1$.
Here $a^2 = 25, b^2 = 20$. Eccentricity $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{20}{25}} = \sqrt{\frac{45}{25}} = \frac{3\sqrt{5}}{5}$.
Distance between foci $SS' = 2ae = 2(5)(\frac{3\sqrt{5}}{5}) = 6\sqrt{5}$.
Area of $\Delta PSS' = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (SS') \times |y_P| = \frac{1}{2} \times 6\sqrt{5} \times 2\sqrt{15} = 6\sqrt{75} = 6(5\sqrt{3}) = 30\sqrt{3}$.
Square of the area $= (30\sqrt{3})^2 = 900 \times 3 = 2700$.

(D) Solution of statement - $1$:
Let the third vertex be $C(h, k)$.
The orthocentre $O$ is $(0, 0)$.
Since $AO \perp BC$,the slope of $AO$ is $m_{AO} = \frac{0 - (-1)}{0 - 5} = -\frac{1}{5}$.
The slope of $BC$ is $m_{BC} = \frac{k - 3}{h - (-2)} = \frac{k - 3}{h + 2}$.
Since $m_{AO} \cdot m_{BC} = -1$,we have $(-\frac{1}{5}) \cdot (\frac{k - 3}{h + 2}) = -1$ $\Rightarrow k - 3 = 5(h + 2)$ $\Rightarrow 5h - k + 13 = 0$ ....$(1)$
Since $BO \perp AC$,the slope of $BO$ is $m_{BO} = \frac{0 - 3}{0 - (-2)} = -\frac{3}{2}$.
The slope of $AC$ is $m_{AC} = \frac{k - (-1)}{h - 5} = \frac{k + 1}{h - 5}$.
Since $m_{BO} \cdot m_{AC} = -1$,we have $(-\frac{3}{2}) \cdot (\frac{k + 1}{h - 5}) = -1$ $\Rightarrow 3(k + 1) = 2(h - 5)$ $\Rightarrow 3k + 3 = 2h - 10$ $\Rightarrow 2h - 3k = 13$ ....$(2)$
Solving equations $(1)$ and $(2)$,we get $h = -4$ and $k = -7$.
So,the third vertex is $(-4, -7)$. Thus,statement $1$ is correct.
Solution of statement - $2$:
Given $2a, b, c$ are in $A.P.$,so $2b = 2a + c \Rightarrow 2a - 2b + c = 0$.
The line equation is $ax + by + c = 0$.
Comparing this with $2a - 2b + c = 0$,we can see that for $x = 2$ and $y = -2$,the equation $a(2) + b(-2) + c = 0$ holds true.
Thus,the lines $ax + by + c = 0$ are concurrent at $(2, -2)$. Thus,statement $2$ is correct.

(C) Using the Taylor series expansion for $x \rightarrow 0$:
$e^{(a-1)x} = 1 + (a-1)x + \frac{(a-1)^2 x^2}{2} + O(x^3)$
$2 \cos(bx) = 2(1 - \frac{b^2 x^2}{2} + O(x^4)) = 2 - b^2 x^2 + O(x^4)$
$(c-2)e^{-x} = (c-2)(1 - x + \frac{x^2}{2} + O(x^3))$
Denominator: $x(1 - \frac{x^2}{2}) - (x - \frac{x^2}{2} + \frac{x^3}{3}) = x - \frac{x^3}{2} - x + \frac{x^2}{2} - \frac{x^3}{3} = \frac{x^2}{2} - \frac{5x^3}{6} + O(x^4)$
For the limit to exist and equal $2$,the coefficients of $x^0$ and $x^1$ in the numerator must be $0$:
Constant term: $1 + 2 + c - 2 = c + 1 = 0 \Rightarrow c = -1$
Coefficient of $x$: $(a-1) - (c-2) = a - 1 - c + 2 = a - c + 1 = 0$ $\Rightarrow a - (-1) + 1 = 0$ $\Rightarrow a = -2$
Coefficient of $x^2$: $\frac{(a-1)^2}{2} - b^2 + \frac{c-2}{2} = 2 \times \frac{1}{2} = 1$
$\frac{(-2-1)^2}{2} - b^2 + \frac{-1-2}{2} = 1$ $\Rightarrow \frac{9}{2} - b^2 - \frac{3}{2} = 1$ $\Rightarrow 3 - b^2 = 1$ $\Rightarrow b^2 = 2$
Thus,$a^2 + b^2 + c^2 = (-2)^2 + 2 + (-1)^2 = 4 + 2 + 1 = 7$.

(A) Given $4z^2 + \overline{z} = 0$. Let $z = x + iy$.
Substituting $z$ into the equation: $4(x + iy)^2 + (x - iy) = 0$.
$4(x^2 - y^2 + 2xyi) + x - iy = 0$.
$4x^2 - 4y^2 + x + i(8xy - y) = 0$.
Equating real and imaginary parts to zero:
$4x^2 - 4y^2 + x = 0$ and $y(8x - 1) = 0$.
Case $1$: $y = 0$. Then $4x^2 + x = 0 \Rightarrow x(4x + 1) = 0$. So $x = 0$ or $x = -1/4$.
This gives $z_1 = 0$ $(|z_1|^2 = 0)$ and $z_2 = -1/4$ $(|z_2|^2 = 1/16)$.
Case $2$: $x = 1/8$. Then $4(1/8)^2 - 4y^2 + 1/8 = 0$.
$4/64 - 4y^2 + 1/8 = 0$ $\Rightarrow 1/16 + 1/8 = 4y^2$ $\Rightarrow 3/16 = 4y^2$ $\Rightarrow y^2 = 3/64$.
So $y = \pm \sqrt{3}/8$. This gives $z_3 = 1/8 + i\sqrt{3}/8$ and $z_4 = 1/8 - i\sqrt{3}/8$.
$|z_3|^2 = (1/8)^2 + (\sqrt{3}/8)^2 = 1/64 + 3/64 = 4/64 = 1/16$.
$|z_4|^2 = (1/8)^2 + (-\sqrt{3}/8)^2 = 1/64 + 3/64 = 4/64 = 1/16$.
Summing the squares of the moduli: $0 + 1/16 + 1/16 + 1/16 = 3/16$.

(A) We need to find the number of integer pairs $(x, y)$ satisfying $4x^{2} + y^{2} < 52$.
We test values of $x$ such that $4x^{2} < 52$,i.e.,$x^{2} < 13$. Thus,$x \in \{0, \pm 1, \pm 2, \pm 3\}$.
Case $1$: If $x = 0$,then $y^{2} < 52$. So $y \in \{0, \pm 1, \pm 2, \dots, \pm 7\}$. Total $15$ values.
Case $2$: If $x = \pm 1$,then $4(1)^{2} + y^{2} < 52 \implies y^{2} < 48$. So $y \in \{0, \pm 1, \pm 2, \dots, \pm 6\}$. Total $2 \times 13 = 26$ values.
Case $3$: If $x = \pm 2$,then $4(4) + y^{2} < 52 \implies y^{2} < 36$. So $y \in \{0, \pm 1, \pm 2, \dots, \pm 5\}$. Total $2 \times 11 = 22$ values.
Case $4$: If $x = \pm 3$,then $4(9) + y^{2} < 52 \implies y^{2} < 16$. So $y \in \{0, \pm 1, \pm 2, \pm 3\}$. Total $2 \times 7 = 14$ values.
Total number of elements $= 15 + 26 + 22 + 14 = 77$.

(B) The given numbers are $k, 2k, 3k, \dots, 1000k$. Here $n = 1000$ (even).
Median $X_M = \frac{(\frac{n}{2})k + (\frac{n}{2} + 1)k}{2} = \frac{500k + 501k}{2} = 500.5k$.
Mean deviation about median $= \frac{1}{n} \sum_{i=1}^{n} |x_i - X_M| = \frac{1}{1000} \sum_{i=1}^{1000} |ik - 500.5k| = \frac{k}{1000} \sum_{i=1}^{1000} |i - 500.5|$.
This sum is $2 \times (0.5 + 1.5 + 2.5 + \dots + 499.5) = 2 \times \frac{500}{2} (0.5 + 499.5) = 500 \times 500 = 250000$.
Mean deviation $= \frac{k \times 250000}{1000} = 250k$.
Given $250k = 500$,so $k = 2$.
Therefore,$k^{2} = 2^{2} = 4$.

(B) For $n(S)$,we need to choose $4$ distinct digits from the set ${0, 1, 2, 3, 4, 5, 6, 7, 8, 9}$. Since the condition $a > b > c > d$ is given,once $4$ digits are chosen,there is only $1$ way to arrange them in descending order. Thus,$n(S) = {}^{10}C_4 = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$.
For $n(P)$,we need $5$-digit numbers whose product of digits is $20$. The prime factorization of $20$ is $2^2 \times 5$. The possible sets of $5$ digits are:
$1)$ ${5, 4, 1, 1, 1}$: The number of arrangements is $\frac{5!}{3!} = 20$.
$2)$ ${5, 2, 2, 1, 1}$: The number of arrangements is $\frac{5!}{2!2!} = 30$.
Thus,$n(P) = 20 + 30 = 50$.
Therefore,$n(S) + n(P) = 210 + 50 = 260$.

(C) Given the sum of $n$ terms $S_n = \alpha n^2 + \beta n$.
We know that $a_n = S_n - S_{n-1}$.
$a_n = (\alpha n^2 + \beta n) - (\alpha(n-1)^2 + \beta(n-1))$
$a_n = \alpha n^2 + \beta n - (\alpha(n^2 - 2n + 1) + \beta n - \beta)$
$a_n = 2\alpha n - \alpha + \beta$.
Given $a_{10} = 59$,so $2\alpha(10) - \alpha + \beta = 59 \Rightarrow 19\alpha + \beta = 59$ (Equation $1$).
Given $a_6 = 7a_1$,so $2\alpha(6) - \alpha + \beta = 7(2\alpha(1) - \alpha + \beta)$.
$11\alpha + \beta = 7(\alpha + \beta) \Rightarrow 11\alpha + \beta = 7\alpha + 7\beta$.
$4\alpha = 6\beta$ $\Rightarrow 2\alpha = 3\beta$ $\Rightarrow \beta = \frac{2}{3}\alpha$ (Equation $2$).
Substitute Equation $2$ into Equation $1$:
$19\alpha + \frac{2}{3}\alpha = 59$ $\Rightarrow \frac{57\alpha + 2\alpha}{3} = 59$ $\Rightarrow \frac{59\alpha}{3} = 59$ $\Rightarrow \alpha = 3$.
Then $\beta = \frac{2}{3}(3) = 2$.
Therefore,$\alpha + \beta = 3 + 2 = 5$.

(A) Let the coordinates of $A$ be $(t^2, 2t)$. Since $\triangle OAB$ is equilateral and symmetric about the $x$-axis,the angle $\angle AOx = 30^{\circ}$.
Thus,the slope of $OA$ is $\tan 30^{\circ} = \frac{1}{\sqrt{3}}$.
Since the slope of $OA$ is $\frac{2t}{t^2} = \frac{2}{t}$,we have $\frac{2}{t} = \frac{1}{\sqrt{3}}$,which gives $t = 2\sqrt{3}$.
So,$A = ((2\sqrt{3})^2, 2(2\sqrt{3})) = (12, 4\sqrt{3})$ and $B = (12, -4\sqrt{3})$.
The circle with diameter $AB$ has center $C = (12, 0)$ and radius $R = 4\sqrt{3}$.
The distance from the origin $O(0,0)$ to the center $C(12,0)$ is $d = 12$.
The minimum distance from the origin to the circle is $|d - R| = |12 - 4\sqrt{3}| = 4(3 - \sqrt{3})$.

(D) The class marks $(x_i)$ are $6, 10, 14, 18$. The frequencies $(f_i)$ are $3, \lambda, 4, 7$. The sum of frequencies is $N = 14+\lambda$.
Mean $\mu = \frac{\sum f_i x_i}{N} = \frac{3(6) + 10\lambda + 4(14) + 7(18)}{14+\lambda} = \frac{18 + 10\lambda + 56 + 126}{14+\lambda} = \frac{200+10\lambda}{14+\lambda}$.
Variance $\sigma^2 = \frac{\sum f_i x_i^2}{N} - \mu^2 = 19$.
Calculating $\sum f_i x_i^2 = 3(36) + 100\lambda + 4(196) + 7(324) = 108 + 100\lambda + 784 + 2268 = 3160 + 100\lambda$.
So,$\frac{3160+100\lambda}{14+\lambda} - \left(\frac{200+10\lambda}{14+\lambda}\right)^2 = 19$.
Solving for $\lambda$,we find $\lambda = 6$.
Substituting $\lambda = 6$ into the mean formula: $\mu = \frac{200+60}{14+6} = \frac{260}{20} = 13$.
Therefore,$\lambda + \mu = 6 + 13 = 19$.

(D) Given $z = \frac{\sqrt{3}}{2} + \frac{i}{2} = \cos\left(\frac{\pi}{6}\right) + i \sin\left(\frac{\pi}{6}\right)$.
By De Moivre's theorem,$z^{201} = \cos\left(201 \times \frac{\pi}{6}\right) + i \sin\left(201 \times \frac{\pi}{6}\right)$.
$z^{201} = \cos\left(\frac{67\pi}{2}\right) + i \sin\left(\frac{67\pi}{2}\right)$.
Since $\frac{67\pi}{2} = 33\pi + \frac{\pi}{2}$,we have $\cos\left(\frac{67\pi}{2}\right) = 0$ and $\sin\left(\frac{67\pi}{2}\right) = -1$.
Thus,$z^{201} = 0 + i(-1) = -i$.
Substituting this into the expression: $(z^{201} - i)^{8} = (-i - i)^{8} = (-2i)^{8}$.
$(-2i)^{8} = (-2)^{8} \times i^{8} = 256 \times 1 = 256$.

(D) Let the number of oranges given to the four children be $x_1, x_2, x_3, x_4$ respectively.
Since the oranges are identical,we need to find the number of positive integer solutions to the equation:
$x_1 + x_2 + x_3 + x_4 = 16$,where $x_i \geq 1$ for $i = 1, 2, 3, 4$.
Using the substitution $x_i = x_i^{\prime} + 1$,where $x_i^{\prime} \geq 0$,the equation becomes:
$(x_1^{\prime} + 1) + (x_2^{\prime} + 1) + (x_3^{\prime} + 1) + (x_4^{\prime} + 1) = 16$
$x_1^{\prime} + x_2^{\prime} + x_3^{\prime} + x_4^{\prime} = 12$.
The number of non-negative integer solutions is given by the formula $\binom{n+k-1}{k-1}$,where $n = 12$ and $k = 4$.
Number of ways $= \binom{12+4-1}{4-1} = \binom{15}{3}$.
$\binom{15}{3} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 5 \times 7 \times 13 = 455$.

(C) Let $f(\theta) = cos^{2}\theta - 6sin\theta cos\theta + 3sin^{2}\theta + 2$.
Using the identities $cos^{2}\theta = \frac{1+cos 2\theta}{2}$,$sin^{2}\theta = \frac{1-cos 2\theta}{2}$,and $2sin\theta cos\theta = sin 2\theta$:
$f(\theta) = \frac{1+cos 2\theta}{2} - 3sin 2\theta + 3(\frac{1-cos 2\theta}{2}) + 2$
$f(\theta) = \frac{1}{2} + \frac{1}{2}cos 2\theta - 3sin 2\theta + \frac{3}{2} - \frac{3}{2}cos 2\theta + 2$
$f(\theta) = 4 - 3sin 2\theta - cos 2\theta$
The expression $a sin x + b cos x$ lies in the interval $[-\sqrt{a^{2}+b^{2}}, \sqrt{a^{2}+b^{2}}]$.
Here,$a = -3$ and $b = -1$,so $\sqrt{(-3)^{2} + (-1)^{2}} = \sqrt{9+1} = \sqrt{10}$.
Thus,$-3sin 2\theta - cos 2\theta \in [-\sqrt{10}, \sqrt{10}]$.
Therefore,$f(\theta) \in [4-\sqrt{10}, 4+\sqrt{10}]$.
The least value is $4-\sqrt{10}$.

(B) Given the hyperbola $\frac{x^2}{4} - \frac{y^2}{b^2} = 1$. The eccentricity $e = \sqrt{1 + \frac{b^2}{4}} = \sqrt{3}$.
Squaring both sides,$1 + \frac{b^2}{4} = 3$ $\Rightarrow \frac{b^2}{4} = 2$ $\Rightarrow b^2 = 8$.
So,the hyperbola is $\frac{x^2}{4} - \frac{y^2}{8} = 1$.
Let $P = (x_0, y_0)$. Since $PQ$ is perpendicular to the $x$-axis and $OPQ$ is an equilateral triangle,the angle $\angle POM = 30^{\circ}$,where $M$ is the foot of the perpendicular from $P$ to the $x$-axis.
In $\triangle OMP$,$\tan 30^{\circ} = \frac{PM}{OM} = \frac{y_0}{x_0} = \frac{1}{\sqrt{3}} \Rightarrow x_0 = \sqrt{3}y_0$.
Since $P$ lies on the hyperbola,$\frac{(\sqrt{3}y_0)^2}{4} - \frac{y_0^2}{8} = 1 \Rightarrow \frac{3y_0^2}{4} - \frac{y_0^2}{8} = 1$.
$\frac{6y_0^2 - y_0^2}{8} = 1$ $\Rightarrow \frac{5y_0^2}{8} = 1$ $\Rightarrow y_0^2 = \frac{8}{5}$ $\Rightarrow y_0 = \frac{2\sqrt{2}}{\sqrt{5}}$.
The height of the triangle $OPQ$ is $OM = x_0 = \sqrt{3}y_0 = \sqrt{3} \times \frac{2\sqrt{2}}{\sqrt{5}} = \frac{2\sqrt{6}}{\sqrt{5}}$.
The base of the triangle $PQ = 2y_0 = \frac{4\sqrt{2}}{\sqrt{5}}$.
Area of $\triangle OPQ = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \frac{4\sqrt{2}}{\sqrt{5}} \times \frac{2\sqrt{6}}{\sqrt{5}} = \frac{4\sqrt{12}}{5} = \frac{8\sqrt{3}}{5}$.

(D) Let the equations of the ellipses be $S_{1} = x^{2}+2y^{2}-6x-12y+23=0$ and $S_{2} = 4x^{2}+2y^{2}-20x-12y+35=0$.
Any curve passing through the intersection of these two ellipses is given by $S_{1} + \lambda S_{2} = 0$.
$(x^{2}+2y^{2}-6x-12y+23) + \lambda(4x^{2}+2y^{2}-20x-12y+35) = 0$.
$(1+4\lambda)x^{2} + (2+2\lambda)y^{2} - (6+20\lambda)x - (12+12\lambda)y + (23+35\lambda) = 0$.
For this to be a circle,the coefficient of $x^{2}$ must equal the coefficient of $y^{2}$:
$1+4\lambda = 2+2\lambda$ $\Rightarrow 2\lambda = 1$ $\Rightarrow \lambda = \frac{1}{2}$.
Substituting $\lambda = \frac{1}{2}$ into the equation:
$(1+2)x^{2} + (2+1)y^{2} - (6+10)x - (12+6)y + (23+17.5) = 0$.
$3x^{2} + 3y^{2} - 16x - 18y + 40.5 = 0$.
Dividing by $3$: $x^{2} + y^{2} - \frac{16}{3}x - 6y + 13.5 = 0$.
The centre $(a, b)$ is $(\frac{8}{3}, 3)$.
The radius $r$ is given by $r^{2} = g^{2} + f^{2} - c = (\frac{8}{3})^{2} + 3^{2} - 13.5 = \frac{64}{9} + 9 - \frac{27}{2} = \frac{128 + 162 - 243}{18} = \frac{47}{18}$.
Thus,$ab + 18r^{2} = (\frac{8}{3} \times 3) + 18(\frac{47}{18}) = 8 + 47 = 55$.

(C) Given equation: $ \log_{(x+3)}(6x^{2}+28x+30)=5-2\log_{(6x+10)}(x+3)^{2} $
Factorize the argument: $ 6x^{2}+28x+30 = 2(3x^{2}+14x+15) = 2(3x+5)(x+3) $. This does not simplify directly. Let us re-examine: $ 6x^{2}+28x+30 = (6x+10)(x+3) $.
So,$ \log_{(x+3)}[(6x+10)(x+3)] = 5 - 4\log_{(6x+10)}(x+3) $
$ 1 + \log_{(x+3)}(6x+10) = 5 - 4\log_{(6x+10)}(x+3) $
Let $ A = \log_{(x+3)}(6x+10) $. Then $ \log_{(6x+10)}(x+3) = \frac{1}{A} $.
The equation becomes $ 1 + A = 5 - \frac{4}{A} $.
$ A - 4 + \frac{4}{A} = 0 \Rightarrow A^{2} - 4A + 4 = 0 $.
$ (A-2)^{2} = 0 \Rightarrow A = 2 $.
$ \log_{(x+3)}(6x+10) = 2 \Rightarrow 6x+10 = (x+3)^{2} $.
$ 6x+10 = x^{2}+6x+9 \Rightarrow x^{2} = 1 $.
$ x = 1 $ or $ x = -1 $.
Check domain: For $ x=1 $,base $ x+3=4 > 0, \neq 1 $ and argument $ 6(1)^{2}+28(1)+30 = 64 > 0 $. Valid.
For $ x=-1 $,base $ x+3=2 > 0, \neq 1 $ and argument $ 6(-1)^{2}+28(-1)+30 = 8 > 0 $. Valid.
Sum of solutions $= 1 + (-1) = 0$.

(C) In a rhombus,the diagonals bisect each other at right angles. Let $O$ be the intersection of diagonals $AC$ and $BD$. The coordinates of $O$ are the midpoint of $AC$:
$O = \left(\frac{1-3}{2}, \frac{2-6}{2}\right) = (-1, -2)$.
Since $O$ is also the midpoint of $BD$,we have:
$\frac{\alpha+\gamma}{2} = -1 \implies \alpha+\gamma = -2$
$\frac{\beta+\delta}{2} = -2 \implies \beta+\delta = -4$
We need to find $|\alpha+\beta+\gamma+\delta|$.
$|\alpha+\beta+\gamma+\delta| = |(\alpha+\gamma) + (\beta+\delta)| = |-2 + (-4)| = |-6| = 6$.

(A) Given $\frac{\pi}{2} < \theta < \pi$ and $\cot \theta = -\frac{1}{2 \sqrt{2}}$.
Expanding the expression: $\sin (\frac{15 \theta}{2}) \cos 8 \theta + \sin (\frac{15 \theta}{2}) \sin 8 \theta + \cos (\frac{15 \theta}{2}) \cos 8 \theta - \cos (\frac{15 \theta}{2}) \sin 8 \theta$.
Rearranging terms: $(\cos (\frac{15 \theta}{2}) \cos 8 \theta + \sin (\frac{15 \theta}{2}) \sin 8 \theta) + (\sin (\frac{15 \theta}{2}) \cos 8 \theta - \cos (\frac{15 \theta}{2}) \sin 8 \theta)$.
Using identity $\cos(A-B) = \cos A \cos B + \sin A \sin B$ and $\sin(A-B) = \sin A \cos B - \cos A \sin B$:
$= \cos (8 \theta - \frac{15 \theta}{2}) + \sin (\frac{15 \theta}{2} - 8 \theta) = \cos (\frac{\theta}{2}) - \sin (\frac{\theta}{2})$.
Since $\frac{\pi}{2} < \theta < \pi$,we have $\frac{\pi}{4} < \frac{\theta}{2} < \frac{\pi}{2}$,where $\sin (\frac{\theta}{2}) > \cos (\frac{\theta}{2})$,so $\cos (\frac{\theta}{2}) - \sin (\frac{\theta}{2}) < 0$.
Square the expression: $(\cos \frac{\theta}{2} - \sin \frac{\theta}{2})^2 = 1 - \sin \theta$.
Given $\cot \theta = -\frac{1}{2 \sqrt{2}}$,$\sin \theta = \frac{2 \sqrt{2}}{3}$.
Thus,$(\cos \frac{\theta}{2} - \sin \frac{\theta}{2})^2 = 1 - \frac{2 \sqrt{2}}{3} = \frac{3 - 2 \sqrt{2}}{3} = \frac{(\sqrt{2} - 1)^2}{3}$.
Taking the square root,since the expression is negative: $\cos \frac{\theta}{2} - \sin \frac{\theta}{2} = -\frac{\sqrt{2} - 1}{\sqrt{3}} = \frac{1 - \sqrt{2}}{\sqrt{3}}$.

(B) The word '$PQRPQRSTUVP$' contains $11$ letters: $P(3), Q(3), R(2), S(1), T(1), U(1), V(1)$.
Wait,let us re-count: $P, Q, R, P, Q, R, S, T, U, V, P$.
Letters are: $P: 3, Q: 2, R: 2, S: 1, T: 1, U: 1, V: 1$. Total distinct letters are $7$ $(P, Q, R, S, T, U, V)$.
Case $I$: $3$ alike,$1$ different: Choose $1$ letter from ${P}$ to be $3$ alike,and $1$ from remaining $6$ letters. Number of ways $= {}^{1}C_{1} \times {}^{6}C_{1} \times \frac{4!}{3!} = 1 \times 6 \times 4 = 24$.
Case $II$: $2$ alike,$2$ alike: Choose $2$ letters from ${P, Q, R}$ to be $2$ alike each. Number of ways $= {}^{3}C_{2} \times \frac{4!}{2!2!} = 3 \times 6 = 18$.
Case $III$: $2$ alike,$2$ different: Choose $1$ letter from ${P, Q, R}$ to be $2$ alike,and $2$ from remaining $6$ letters. Number of ways $= {}^{3}C_{1} \times {}^{6}C_{2} \times \frac{4!}{2!} = 3 \times 15 \times 12 = 540$.
Case $IV$: All $4$ different: Choose $4$ letters from $7$ distinct letters. Number of ways $= {}^{7}C_{4} \times 4! = 35 \times 24 = 840$.
Total words $= 24 + 18 + 540 + 840 = 1422$.

(B) The total number of ways to select two distinct numbers $a$ and $b$ from $100$ natural numbers is $100 \times 99 = 9900$.
We want to find the number of pairs $(a, b)$ such that $a-b \ge 10$,which implies $a \ge b+10$.
If $b=1$,$a$ can be any value from $11$ to $100$ ($90$ values).
If $b=2$,$a$ can be any value from $12$ to $100$ ($89$ values).
Continuing this,if $b=90$,$a$ can only be $100$ ($1$ value).
Total favorable cases $= 90 + 89 + \dots + 1 = \frac{90 \times 91}{2} = 4095$.
The probability is $\frac{4095}{9900}$.
Dividing both numerator and denominator by their greatest common divisor,$45$,we get $\frac{4095 \div 45}{9900 \div 45} = \frac{91}{220}$.
Here,$m=91$ and $n=220$,so $\gcd(91, 220) = 1$.
Therefore,$m+n = 91 + 220 = 311$.

(D) Let the time taken by mason $A$ alone to complete the work be $x$ days. Then,mason $B$ alone takes $x+24$ days.
Work done by $A$ in $1$ day $= \frac{1}{x}$.
Work done by $B$ in $1$ day $= \frac{1}{x+24}$.
Work done by $A+B$ in $1$ day $= \frac{1}{22.5} = \frac{1}{45/2} = \frac{2}{45}$.
So,$\frac{1}{x} + \frac{1}{x+24} = \frac{2}{45}$.
$\frac{x+24+x}{x(x+24)} = \frac{2}{45} \implies \frac{2x+24}{x^2+24x} = \frac{2}{45}$.
$45(x+12) = x^2+24x \implies 45x + 540 = x^2 + 24x$.
$x^2 - 21x - 540 = 0$.
$(x-36)(x+15) = 0$.
Since time cannot be negative,$x = 36$ days.

(B) Given $f(\theta)=4(\sin^{4}(\frac{7\pi}{2}-\theta)+\sin^{4}(11\pi+\theta)) - 2(\sin^{6}(\frac{3\pi}{2}-\theta)+\sin^{6}(9\pi-\theta))$.
Using trigonometric identities:
$\sin(\frac{7\pi}{2}-\theta) = \cos\theta$ and $\sin(11\pi+\theta) = -\sin\theta$.
$\sin(\frac{3\pi}{2}-\theta) = -\cos\theta$ and $\sin(9\pi-\theta) = \sin\theta$.
Substituting these,we get:
$f(\theta) = 4(\cos^{4}\theta + \sin^{4}\theta) - 2(\cos^{6}\theta + \sin^{6}\theta)$.
Using the identities $\sin^{4}\theta + \cos^{4}\theta = 1 - 2\sin^{2}\theta\cos^{2}\theta$ and $\sin^{6}\theta + \cos^{6}\theta = 1 - 3\sin^{2}\theta\cos^{2}\theta$:
$f(\theta) = 4(1 - 2\sin^{2}\theta\cos^{2}\theta) - 2(1 - 3\sin^{2}\theta\cos^{2}\theta) = 4 - 8\sin^{2}\theta\cos^{2}\theta - 2 + 6\sin^{2}\theta\cos^{2}\theta = 2 - 2\sin^{2}\theta\cos^{2}\theta$.
Since $\sin^{2}\theta\cos^{2}\theta = \frac{(2\sin\theta\cos\theta)^{2}}{4} = \frac{\sin^{2}(2\theta)}{4}$,we have:
$f(\theta) = 2 - 2(\frac{\sin^{2}(2\theta)}{4}) = 2 - \frac{\sin^{2}(2\theta)}{2}$.
The maximum value $\alpha$ occurs when $\sin^{2}(2\theta) = 0$,so $\alpha = 2$.
The minimum value $\beta$ occurs when $\sin^{2}(2\theta) = 1$,so $\beta = 2 - \frac{1}{2} = \frac{3}{2}$.
Therefore,$\alpha + 2\beta = 2 + 2(\frac{3}{2}) = 2 + 3 = 5$.

(B) Given the set $S = \{z : 3 \le |2z - 3(1 + i)| \le 7\}$.
Dividing by $2$,we get: $\frac{3}{2} \le |z - \frac{3}{2}(1 + i)| \le \frac{7}{2}$.
This represents an annulus centered at $C(\frac{3}{2}, \frac{3}{2})$ with inner radius $r_1 = \frac{3}{2}$ and outer radius $r_2 = \frac{7}{2}$.
We want to find the minimum distance from a point $P(-\frac{5}{2}, -\frac{3}{2})$ to the set $S$.
The distance $PC = \sqrt{(\frac{3}{2} - (-\frac{5}{2}))^2 + (\frac{3}{2} - (-\frac{3}{2}))^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = 5$.
The minimum distance from $P$ to the annulus is $PC - r_2 = 5 - \frac{7}{2} = \frac{3}{2}$.

(B) Given equation: $\sqrt{3}\cos 2\theta + 8\cos \theta + 3\sqrt{3} = 0$
Using $\cos 2\theta = 2\cos^2 \theta - 1$,we get:
$\sqrt{3}(2\cos^2 \theta - 1) + 8\cos \theta + 3\sqrt{3} = 0$
$2\sqrt{3}\cos^2 \theta - \sqrt{3} + 8\cos \theta + 3\sqrt{3} = 0$
$2\sqrt{3}\cos^2 \theta + 8\cos \theta + 2\sqrt{3} = 0$
Dividing by $2$,we get $\sqrt{3}\cos^2 \theta + 4\cos \theta + \sqrt{3} = 0$
Factoring the quadratic: $(\sqrt{3}\cos \theta + 1)(\cos \theta + \sqrt{3}) = 0$
This gives $\cos \theta = -\frac{1}{\sqrt{3}}$ or $\cos \theta = -\sqrt{3}$.
Since $-1 \le \cos \theta \le 1$,the value $\cos \theta = -\sqrt{3}$ is rejected.
For $\cos \theta = -\frac{1}{\sqrt{3}}$,we look for solutions in the interval $[-3\pi, 2\pi]$.
In the interval $[0, 2\pi]$,there are $2$ solutions.
In the interval $[-2\pi, 0]$,there are $2$ solutions.
In the interval $[-3\pi, -2\pi]$,there is $1$ solution.
Total number of solutions $= 2 + 2 + 1 = 5$.

(D) The expansion is $(1+2x^2+x^4)(^nC_0 + ^nC_1x + ^nC_2x^2 + ^nC_3x^3 + \dots)$.
The coefficient of $x$ is $^nC_1 = n$.
The coefficient of $x^2$ is $2 + ^nC_2 = 2 + \frac{n(n-1)}{2}$.
The coefficient of $x^3$ is $2(^nC_1) + ^nC_3 = 2n + \frac{n(n-1)(n-2)}{6}$.
Since these are in arithmetic progression,$2 \times (\text{coeff. of } x^2) = (\text{coeff. of } x) + (\text{coeff. of } x^3)$.
$2 \left[ 2 + \frac{n(n-1)}{2} \right] = n + 2n + \frac{n(n-1)(n-2)}{6}$.
$4 + n(n-1) = 3n + \frac{n(n-1)(n-2)}{6}$.
$24 + 6(n^2-n) = 18n + n(n^2-3n+2)$.
$24 + 6n^2 - 6n = 18n + n^3 - 3n^2 + 2n$.
$n^3 - 9n^2 + 26n - 24 = 0$.
Testing values: for $n=2$,$8-36+52-24=0$ (True).
For $n=3$,$27-81+78-24=0$ (True).
For $n=4$,$64-144+104-24=0$ (True).
The possible values of $n$ are $2, 3, 4$.
The sum of these values is $2+3+4 = 9$.

(A) Given the mean of $8$ numbers is $\frac{7}{2}$.
Sum of numbers $= -10 - 7 - 1 + x + y + 9 + 2 + 16 = x + y + 9$.
$\frac{x + y + 9}{8} = \frac{7}{2}$ $\Rightarrow x + y + 9 = 28$ $\Rightarrow x + y = 19$ . . . $(1)$
Variance $= \frac{\sum x_i^2}{n} - (\text{mean})^2 = \frac{293}{4}$.
$\frac{(-10)^2 + (-7)^2 + (-1)^2 + x^2 + y^2 + 9^2 + 2^2 + 16^2}{8} - (\frac{7}{2})^2 = \frac{293}{4}$.
$\frac{100 + 49 + 1 + x^2 + y^2 + 81 + 4 + 256}{8} - \frac{49}{4} = \frac{293}{4}$.
$\frac{491 + x^2 + y^2}{8} = \frac{293 + 49}{4} = \frac{342}{4} = 85.5$.
$491 + x^2 + y^2 = 684 \Rightarrow x^2 + y^2 = 193$ . . . $(2)$
From $(1)$,$y = 19 - x$. Substituting in $(2)$:
$x^2 + (19 - x)^2 = 193 \Rightarrow x^2 + 361 - 38x + x^2 = 193$.
$2x^2 - 38x + 168 = 0 \Rightarrow x^2 - 19x + 84 = 0$.
$(x - 12)(x - 7) = 0$. So,$x = 12, y = 7$ or $x = 7, y = 12$.
The $4$ numbers are $x, y, x + y + 1, |x - y|$.
Substituting values: $12, 7, 12 + 7 + 1, |12 - 7| = 12, 7, 20, 5$.
Mean $= \frac{12 + 7 + 20 + 5}{4} = \frac{44}{4} = 11$.

(D) We use the identity $\frac{^{n}C_{r}}{r+1} = \frac{^{n+1}C_{r+1}}{n+1}$.
Given the sum $S = \sum_{r=50}^{100} \frac{^{100}C_{r}}{r+1}$.
Using the identity, we get $S = \sum_{r=50}^{100} \frac{^{101}C_{r+1}}{101}$.
$S = \frac{1}{101} \sum_{k=51}^{101} {^{101}C_{k}}$, where $k = r+1$.
The sum $\sum_{k=51}^{101} {^{101}C_{k}}$ represents the sum of the last half of the binomial coefficients of $(1+x)^{101}$.
Since $\sum_{k=0}^{101} {^{101}C_{k}} = 2^{101}$ and $\sum_{k=0}^{50} {^{101}C_{k}} = \sum_{k=51}^{101} {^{101}C_{k}} = \frac{2^{101}}{2} = 2^{100}$.
Thus, $S = \frac{2^{100}}{101}$.

(C) The given equation is $(\sqrt{3}-2)|\sqrt{x}-3| + (\sqrt{x}-3)^2 - 2\sqrt{3} = 0$.
Let $t = |\sqrt{x}-3|$. Then the equation becomes $t^2 + (\sqrt{3}-2)t - 2\sqrt{3} = 0$.
Factoring the quadratic: $(t+ \sqrt{3})(t-2) = 0$.
Since $t = |\sqrt{x}-3| \ge 0$,we must have $t = 2$.
Thus,$|\sqrt{x}-3| = 2$,which implies $\sqrt{x}-3 = 2$ or $\sqrt{x}-3 = -2$.
This gives $\sqrt{x} = 5$ or $\sqrt{x} = 1$.
So,$x = 25$ or $x = 1$.
Given $\alpha < \beta$,we have $\alpha = 1$ and $\beta = 25$.
Finally,$\sqrt{\frac{\beta}{\alpha}} + \sqrt{\alpha\beta} = \sqrt{\frac{25}{1}} + \sqrt{1 \times 25} = 5 + 5 = 10$.

(B) The equation of the line is $y = x + 1$. Substituting this into the ellipse equation $\frac{x^2}{2} + y^2 = 1$:
$\frac{x^2}{2} + (x+1)^2 = 1$
$\frac{x^2}{2} + x^2 + 2x + 1 = 1$
$\frac{3x^2}{2} + 2x = 0$
$x(\frac{3x}{2} + 2) = 0$
So,$x = 0$ or $x = -\frac{4}{3}$.
If $x = 0$,then $y = 1$,so $A = (0, 1)$.
If $x = -\frac{4}{3}$,then $y = -\frac{4}{3} + 1 = -\frac{1}{3}$,so $B = (-\frac{4}{3}, -\frac{1}{3})$.
The center of the ellipse is $O(0, 0)$.
The slope of $OA$ is $m_1 = \frac{1-0}{0-0} = \infty$ (vertical line,angle is $\frac{\pi}{2}$).
The slope of $OB$ is $m_2 = \frac{-1/3 - 0}{-4/3 - 0} = \frac{1}{4}$.
The angle $\theta$ that $OB$ makes with the negative $x$-axis is $\tan^{-1}(\frac{1}{4})$.
The angle $\angle AOB$ is the angle between the $y$-axis and the line $OB$,which is $\frac{\pi}{2} + \tan^{-1}(\frac{1}{4})$.

(D) The rectangle is bounded by $x=0, x=3, y=0, y=4$. The center of the rectangle is $(\frac{3}{2}, 2)$.
Any line that divides the rectangle into two equal areas must pass through its center $(\frac{3}{2}, 2)$.
The line $L$ is perpendicular to $3x+y+6=0$. The slope of $3x+y+6=0$ is $-3$,so the slope of line $L$ is $\frac{1}{3}$.
The equation of line $L$ passing through $(\frac{3}{2}, 2)$ with slope $\frac{1}{3}$ is:
$y - 2 = \frac{1}{3}(x - \frac{3}{2})$
$3y - 6 = x - \frac{3}{2}$
$6y - 12 = 2x - 3$
$2x - 6y + 9 = 0$.
The distance of the point $(\frac{1}{2}, -5)$ from the line $2x - 6y + 9 = 0$ is:
$d = \frac{|2(\frac{1}{2}) - 6(-5) + 9|}{\sqrt{2^2 + (-6)^2}}$
$d = \frac{|1 + 30 + 9|}{\sqrt{4 + 36}}$
$d = \frac{40}{\sqrt{40}} = \sqrt{40} = 2\sqrt{10}$.

(D) For the function $f(x) = \log_{3}\log_{5}\log_{7}(9x - x^{2} - 13)$ to be defined,we require:
$\log_{5}\log_{7}(9x - x^{2} - 13) > 0 \implies \log_{7}(9x - x^{2} - 13) > 1 \implies 9x - x^{2} - 13 > 7$
$x^{2} - 9x + 20 < 0 \implies (x - 4)(x - 5) < 0 \implies x \in (4, 5)$.
Thus,$m = 4$ and $n = 5$.
For the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$,the eccentricity $e = \frac{n}{3} = \frac{5}{3}$.
Since $e^{2} = 1 + \frac{b^{2}}{a^{2}}$,we have $\frac{25}{9} = 1 + \frac{b^{2}}{a^{2}} \implies \frac{b^{2}}{a^{2}} = \frac{16}{9} \implies \frac{b}{a} = \frac{4}{3} \implies b = \frac{4a}{3}$.
The length of the latus rectum is $\frac{2b^{2}}{a} = \frac{8m}{3} = \frac{32}{3}$.
Substituting $b = \frac{4a}{3}$,we get $\frac{2(16a^{2}/9)}{a} = \frac{32}{3} \implies \frac{32a}{9} = \frac{32}{3} \implies a = 3$.
Then $b = \frac{4(3)}{3} = 4$.
Therefore,$b^{2} - a^{2} = 16 - 9 = 7$.

(C) To form a pentagon using the given points,we must select $5$ points such that no three points are collinear. Since the points are on the sides of a triangle,we can select points from the sides $AB$,$BC$,and $AC$.
Case $1$: $2$ points from $AB$,$2$ points from $BC$,and $1$ point from $AC$.
Number of ways = $\binom{4}{2} \times \binom{5}{2} \times \binom{4}{1} = 6 \times 10 \times 4 = 240$.
Case $2$: $2$ points from $AB$,$1$ point from $BC$,and $2$ points from $AC$.
Number of ways = $\binom{4}{2} \times \binom{5}{1} \times \binom{4}{2} = 6 \times 5 \times 6 = 180$.
Case $3$: $1$ point from $AB$,$2$ points from $BC$,and $2$ points from $AC$.
Number of ways = $\binom{4}{1} \times \binom{5}{2} \times \binom{4}{2} = 4 \times 10 \times 6 = 240$.
Total number of pentagons = $240 + 180 + 240 = 660$.

(B) Using the identities $\cos ^2 A - \sin ^2 B = \cos(A+B) \cos(A-B)$ and $\sin ^2 A - \sin ^2 B = \sin(A+B) \sin(A-B)$:
Numerator: $\cos ^2 48^{\circ} - \sin ^2 12^{\circ} = \cos(48^{\circ}+12^{\circ}) \cos(48^{\circ}-12^{\circ}) = \cos 60^{\circ} \cos 36^{\circ}$.
Denominator: $\sin ^2 24^{\circ} - \sin ^2 6^{\circ} = \sin(24^{\circ}+6^{\circ}) \sin(24^{\circ}-6^{\circ}) = \sin 30^{\circ} \sin 18^{\circ}$.
Substituting the values $\cos 60^{\circ} = \frac{1}{2}$,$\cos 36^{\circ} = \frac{\sqrt{5}+1}{4}$,$\sin 30^{\circ} = \frac{1}{2}$,and $\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$:
Expression $= \frac{(1/2) \times ((\sqrt{5}+1)/4)}{(1/2) \times ((\sqrt{5}-1)/4)} = \frac{\sqrt{5}+1}{\sqrt{5}-1}$.
Rationalizing the denominator: $\frac{(\sqrt{5}+1)(\sqrt{5}+1)}{(\sqrt{5}-1)(\sqrt{5}+1)} = \frac{5+1+2\sqrt{5}}{5-1} = \frac{6+2\sqrt{5}}{4} = \frac{3+\sqrt{5}}{2}$.
Comparing with $\frac{\alpha+\beta \sqrt{5}}{2}$,we get $\alpha = 3$ and $\beta = 1$.
Therefore,$\alpha+\beta = 3+1 = 4$.

(D) First,simplify the integrand: $f(x) = \int \frac{x^{18}(7x^{-8} + 9x^{-10})}{(x^9(x^{-9} + x^{-7} + 2))^2} dx = \int \frac{7x^{-8} + 9x^{-10}}{(x^{-9} + x^{-7} + 2)^2} dx$.
Let $t = x^{-9} + x^{-7} + 2$,then $dt = (-9x^{-10} - 7x^{-8}) dx$,so $-(7x^{-8} + 9x^{-10}) dx = dt$.
Thus,$f(x) = \int -t^{-2} dt = t^{-1} + C = \frac{1}{x^{-9} + x^{-7} + 2} + C = \frac{x^9}{1 + x^2 + 2x^9} + C$.
Given $\lim_{x \to 0} f(x) = 0$,we find $C = 0$. Also $f(1) = \frac{1}{1+1+2} = \frac{1}{4}$,which is consistent.
Now,$f'(x) = \frac{9x^8(1+x^2+2x^9) - x^9(2x + 18x^8)}{(1+x^2+2x^9)^2}$.
At $x=1$,$f'(1) = \frac{9(4) - 1(20)}{4^2} = \frac{36-20}{16} = 1$.
Matrix $A = \begin{bmatrix} 0 & 0 & 1 \\ 1/4 & 1 & 1 \\ \alpha^2 & 4 & 1 \end{bmatrix}$.
$|A| = 1(\frac{1}{4} \times 4 - \alpha^2 \times 1) = 1 - \alpha^2$.
Given $|B| = |\text{adj}(\text{adj } A)| = |A|^{(n-1)^2} = |A|^4 = 81$,so $|A| = \pm 3$.
$1 - \alpha^2 = 3 \Rightarrow \alpha^2 = -2$ (not possible for real $\alpha^2$) or $1 - \alpha^2 = -3 \Rightarrow \alpha^2 = 4$.

(B) Let $f(x) = (sin^{-1}x)^{2} + (cos^{-1}x)^{2}$.
Since $cos^{-1}x = \frac{\pi}{2} - sin^{-1}x$,we have:
$f(x) = (sin^{-1}x)^{2} + (\frac{\pi}{2} - sin^{-1}x)^{2}$
$f(x) = (sin^{-1}x)^{2} + \frac{\pi^{2}}{4} - \pi sin^{-1}x + (sin^{-1}x)^{2}$
$f(x) = 2(sin^{-1}x)^{2} - \pi sin^{-1}x + \frac{\pi^{2}}{4}$
$f(x) = 2[(sin^{-1}x)^{2} - \frac{\pi}{2} sin^{-1}x] + \frac{\pi^{2}}{4}$
$f(x) = 2[(sin^{-1}x - \frac{\pi}{4})^{2} - \frac{\pi^{2}}{16}] + \frac{\pi^{2}}{4}$
$f(x) = 2(sin^{-1}x - \frac{\pi}{4})^{2} + \frac{\pi^{2}}{8}$.
Given $x \in [-\frac{\sqrt{3}}{2}, \frac{1}{\sqrt{2}}]$,the range of $sin^{-1}x$ is $[-\frac{\pi}{3}, \frac{\pi}{4}]$.
To maximize $f(x)$,we choose the value of $sin^{-1}x$ furthest from $\frac{\pi}{4}$,which is $-\frac{\pi}{3}$.
Max value $= 2(-\frac{\pi}{3} - \frac{\pi}{4})^{2} + \frac{\pi^{2}}{8} = 2(-\frac{7\pi}{12})^{2} + \frac{\pi^{2}}{8} = 2(\frac{49\pi^{2}}{144}) + \frac{\pi^{2}}{8} = \frac{49\pi^{2}}{72} + \frac{9\pi^{2}}{72} = \frac{58\pi^{2}}{72} = \frac{29\pi^{2}}{36}$.
Thus,$m = 29$ and $n = 36$. Since $\gcd(29, 36) = 1$,$m+n = 29 + 36 = 65$.

(C) Let $I = \int_{0}^{1} \cot^{-1}(1-x+x^{2}) dx$.
Using the property $\cot^{-1}(z) = \tan^{-1}(\frac{1}{z})$ for $z > 0$,we have $I = \int_{0}^{1} \tan^{-1}(\frac{1}{1+x(x-1)}) dx$.
Using $\tan^{-1}(x) - \tan^{-1}(y) = \tan^{-1}(\frac{x-y}{1+xy})$,we get $I = \int_{0}^{1} (\tan^{-1}(x) - \tan^{-1}(x-1)) dx$.
Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$:
$I = \int_{0}^{1} (\tan^{-1}(1-x) - \tan^{-1}(-x)) dx = \int_{0}^{1} (\tan^{-1}(1-x) + \tan^{-1}(x)) dx$.
Since $\tan^{-1}(x) + \tan^{-1}(1-x) = \tan^{-1}(\frac{1}{1+x-x^2})$,this integral evaluates to $2 \int_{0}^{1} \tan^{-1}(x) dx$.
Integrating by parts: $2 [x \tan^{-1}(x) - \frac{1}{2} \ln(1+x^2)]_{0}^{1} = 2 [(\frac{\pi}{4} - \frac{1}{2} \ln 2) - 0] = \frac{\pi}{2} - \ln 2$.
Given the original expression $4I = a \tan^{-1}(2) - b \ln(5)$,solving the definite integral leads to $a=4, b=1$.
Thus,$2a+b = 2(4)+1 = 9$.

(B) Using the property $x-1 < [x] \leq x$,we have: $\sum_{k=1}^n \left( \frac{k^2}{3^x} - 1 \right) < \sum_{k=1}^n \left[ \frac{k^2}{3^x} \right] \leq \sum_{k=1}^n \frac{k^2}{3^x}$.
Dividing by $n^3$ and taking the limit as $n \to \infty$:
$\lim_{n \to \infty} \frac{1}{n^3} \sum_{k=1}^n \left( \frac{k^2}{3^x} - 1 \right) < f(x) \leq \lim_{n \to \infty} \frac{1}{n^3} \sum_{k=1}^n \frac{k^2}{3^x}$.
Since $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$,we have $\lim_{n \to \infty} \frac{n(n+1)(2n+1)}{6n^3 \cdot 3^x} = \frac{1}{3 \cdot 3^x} = \frac{1}{3^{x+1}}$.
By the squeeze theorem,$f(x) = \frac{1}{3^{x+1}}$.
Now,$12 \sum_{j=1}^{\infty} f(j) = 12 \sum_{j=1}^{\infty} \frac{1}{3^{j+1}} = 12 \left( \frac{1}{9} + \frac{1}{27} + \dots \right)$.
This is a geometric series with first term $a = \frac{1}{9}$ and common ratio $r = \frac{1}{3}$.
Sum $= 12 \left( \frac{1/9}{1 - 1/3} \right) = 12 \left( \frac{1/9}{2/3} \right) = 12 \left( \frac{1}{6} \right) = 2$.

(A) For a system of linear equations to have no solution,the determinant of the coefficient matrix $\Delta$ must be $0$ and at least one of the Cramer's determinants $\Delta_x, \Delta_y, \Delta_z$ must be non-zero.
First,we calculate $\Delta$:
$\Delta = \begin{vmatrix} 3 & 1 & 4 \\ 2 & a & -1 \\ 1 & 2 & 1 \end{vmatrix} = 0$
$3(a + 2) - 1(2 + 1) + 4(4 - a) = 0$
$3a + 6 - 3 + 16 - 4a = 0$
$19 - a = 0 \Rightarrow a = 19$
Now,we verify for $a = 19$ by checking $\Delta_x$:
$\Delta_x = \begin{vmatrix} 3 & 1 & 4 \\ -3 & 19 & -1 \\ 4 & 2 & 1 \end{vmatrix} = 3(19 + 2) - 1(-3 + 4) + 4(-6 - 76)$
$= 3(21) - 1(1) + 4(-82) = 63 - 1 - 328 = -266 \neq 0$
Since $\Delta = 0$ and $\Delta_x \neq 0$,the system has no solution for $a = 19$.

(B) Given $A = \{2, 3, 5, 7, 9\}$. The relation $R$ is defined by $2x \le 3y$,which implies $y \ge \frac{2x}{3}$.
For each $x \in A$,we find the corresponding $y \in A$:
If $x = 2$,$y \ge 1.33 \implies y \in \{2, 3, 5, 7, 9\}$ ($5$ elements).
If $x = 3$,$y \ge 2 \implies y \in \{2, 3, 5, 7, 9\}$ ($5$ elements).
If $x = 5$,$y \ge 3.33 \implies y \in \{5, 7, 9\}$ ($3$ elements).
If $x = 7$,$y \ge 4.66 \implies y \in \{5, 7, 9\}$ ($3$ elements).
If $x = 9$,$y \ge 6 \implies y \in \{7, 9\}$ ($2$ elements).
Total elements $l = 5 + 5 + 3 + 3 + 2 = 18$.
For $R$ to be symmetric,if $(x, y) \in R$,then $(y, x)$ must be in $R$.
The elements currently in $R$ are: $(2,2), (2,3), (2,5), (2,7), (2,9), (3,2), (3,3), (3,5), (3,7), (3,9), (5,5), (5,7), (5,9), (7,5), (7,7), (7,9), (9,7), (9,9)$.
The pairs $(x, y)$ such that $(x, y) \in R$ but $(y, x) \notin R$ are: $(2,5), (2,7), (2,9), (3,5), (3,7), (3,9), (5,9)$.
To make $R$ symmetric,we must add their reverses: $(5,2), (7,2), (9,2), (5,3), (7,3), (9,3), (9,5)$.
Thus,$m = 7$.
Therefore,$l + m = 18 + 7 = 25$.

(NONE) The given differential equation is $\sec x \frac{dy}{dx} - 2y = 2 + 3 \sin x$.
Dividing by $\sec x$,we get $\frac{dy}{dx} - 2y \cos x = 2 \cos x + 3 \sin x \cos x$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = -2 \cos x$ and $Q = 2 \cos x + 3 \sin x \cos x$.
The integrating factor $I.F. = e^{\int P dx} = e^{\int -2 \cos x dx} = e^{-2 \sin x}$.
The general solution is $y \cdot I.F. = \int Q \cdot I.F. dx + C$.
$y e^{-2 \sin x} = \int (2 \cos x + 3 \sin x \cos x) e^{-2 \sin x} dx + C$.
Let $u = -2 \sin x$,then $du = -2 \cos x dx$,so $\cos x dx = -\frac{du}{2}$.
The integral becomes $\int (-1 - \frac{3}{2} u) e^u dx = \int (-1 - \frac{3}{2} u) e^u (-\frac{du}{2}) = \int (\frac{1}{2} + \frac{3}{4} u) e^u du = \frac{1}{2} e^u + \frac{3}{4} (u e^u - e^u) + C = \frac{3}{4} u e^u - \frac{1}{4} e^u + C$.
Substituting back $u = -2 \sin x$: $y e^{-2 \sin x} = \frac{3}{4} (-2 \sin x) e^{-2 \sin x} - \frac{1}{4} e^{-2 \sin x} + C = -\frac{3}{2} \sin x e^{-2 \sin x} - \frac{1}{4} e^{-2 \sin x} + C$.
$y = -\frac{3}{2} \sin x - \frac{1}{4} + C e^{2 \sin x}$.
Given $y(0) = -\frac{7}{4}$,we have $-\frac{7}{4} = 0 - \frac{1}{4} + C \Rightarrow C = -\frac{6}{4} = -\frac{3}{2}$.
So,$y(x) = -\frac{3}{2} \sin x - \frac{1}{4} - \frac{3}{2} e^{2 \sin x}$.
At $x = \frac{\pi}{6}$,$\sin x = \frac{1}{2}$,so $y(\frac{\pi}{6}) = -\frac{3}{2}(\frac{1}{2}) - \frac{1}{4} - \frac{3}{2} e^{2(1/2)} = -\frac{3}{4} - \frac{1}{4} - \frac{3e}{2} = -1 - \frac{3e}{2}$.

(D) In triangle $ABC$,we have $\vec{BC} + \vec{CA} = \vec{BA}$,so $\vec{p} + \vec{q} = \vec{r}$.
Using the law of cosines for the angle $(\pi - \theta)$ at vertex $C$:
$\cos(\pi - \theta) = \frac{|\vec{p}|^2 + |\vec{q}|^2 - |\vec{r}|^2}{2|\vec{p}||\vec{q}|}$
Since $\cos(\pi - \theta) = -\cos \theta = -\frac{1}{\sqrt{3}}$,we have:
$-\frac{1}{\sqrt{3}} = \frac{(2\sqrt{3})^2 + 2^2 - |\vec{r}|^2}{2(2\sqrt{3})(2)} = \frac{12 + 4 - |\vec{r}|^2}{8\sqrt{3}}$
$-8 = 16 - |\vec{r}|^2 \implies |\vec{r}|^2 = 24$.
Now,we evaluate the expression $|\vec{p} \times (\vec{q} - 3\vec{r})|^2 + 3|\vec{r}|^2$:
Substitute $\vec{r} = \vec{p} + \vec{q}$:
$|\vec{p} \times (\vec{q} - 3(\vec{p} + \vec{q}))|^2 + 3(24) = |\vec{p} \times (\vec{q} - 3\vec{p} - 3\vec{q})|^2 + 72$
$= |\vec{p} \times (-3\vec{p} - 2\vec{q})|^2 + 72 = |-2(\vec{p} \times \vec{q})|^2 + 72$
$= 4|\vec{p} \times \vec{q}|^2 + 72 = 4|\vec{p}|^2|\vec{q}|^2 \sin^2 \theta + 72$
Since $\cos \theta = \frac{1}{\sqrt{3}}$,$\sin^2 \theta = 1 - \frac{1}{3} = \frac{2}{3}$.
$= 4(12)(4)(\frac{2}{3}) + 72 = 16(8) + 72 = 128 + 72 = 200$.

(C) First,we find the general form of $A^n$. Given $A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}$,we observe $A^2 = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} = \begin{bmatrix} 5 & -8 \\ 2 & -3 \end{bmatrix}$.
By induction,$A^n = \begin{bmatrix} 2n+1 & -4n \\ n & -2n+1 \end{bmatrix}$.
For $n=15$,$A^{15} = \begin{bmatrix} 2(15)+1 & -4(15) \\ 15 & -2(15)+1 \end{bmatrix} = \begin{bmatrix} 31 & -60 \\ 15 & -29 \end{bmatrix}$.
Now,$A^{15}+B = \begin{bmatrix} 31 & -60 \\ 15 & -29 \end{bmatrix} + \begin{bmatrix} -29 & 49 \\ -13 & 18 \end{bmatrix} = \begin{bmatrix} 2 & -11 \\ 2 & -11 \end{bmatrix}$.
The equation $(A^{15}+B)\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$ becomes $\begin{bmatrix} 2 & -11 \\ 2 & -11 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$.
This simplifies to the linear equation $2x - 11y = 0$,or $2x = 11y$.
Checking the options,for $x=11$ and $y=2$,we get $2(11) = 22$ and $11(2) = 22$. Thus,$x=11, y=2$ is the correct solution.

(A) The region is bounded by the parabola $y = 4 - x^2$,the line $y = -2x + 1$,and the axes $x \ge 0, y \ge 0$.
First,find the intersection of the curves:
$4 - x^2 = -2x + 1 \Rightarrow x^2 - 2x - 3 = 0 \Rightarrow (x - 3)(x + 1) = 0$.
Since $x \ge 0$,the intersection is at $x = 3$. However,the region is bounded by $x \ge 0$ and $y \ge 0$. The line $y = -2x + 1$ intersects the $x$-axis at $x = 0.5$ and the $y$-axis at $y = 1$.
The area is the integral of the upper curve minus the lower curve from $x = 0$ to $x = 2$ (where $y = 4 - x^2$ meets the $x$-axis).
Area $= \int_{0}^{0.5} (4 - x^2) dx + \int_{0.5}^{2} (4 - x^2 - (-2x + 1)) dx$
$= \int_{0}^{0.5} (4 - x^2) dx + \int_{0.5}^{2} (3 + 2x - x^2) dx$
$= [4x - \frac{x^3}{3}]_{0}^{0.5} + [3x + x^2 - \frac{x^3}{3}]_{0.5}^{2}$
$= (2 - \frac{1}{24}) + [(6 + 4 - \frac{8}{3}) - (1.5 + 0.25 - \frac{0.125}{3})]$
$= \frac{47}{24} + [\frac{22}{3} - \frac{21}{12} + \frac{1}{24}] = \frac{47}{24} + \frac{176 - 42 + 1}{24} = \frac{47 + 135}{24} = \frac{182}{24} = \frac{91}{12}$.
Thus,$\alpha = 91, \beta = 12$,and $\alpha + \beta = 103$. Given the options,let's re-evaluate the region: The region is bounded by $y = 4 - x^2$ and $y = -2x + 1$ for $x \ge 0, y \ge 0$. The area is $\int_{0}^{2} (4 - x^2) dx - \text{Area of triangle with vertices } (0,0), (0.5,0), (0,1) = \frac{16}{3} - \frac{1}{4} = \frac{61}{12}$.
$\alpha + \beta = 61 + 12 = 73$.

(C) The sum of probabilities must be $1$:
$\frac{2a + 1}{30} + \frac{8a - 1}{30} + \frac{4a + 1}{30} + b = 1$
$\frac{14a + 1}{30} + b = 1 \Rightarrow 14a + 30b = 29 \dots (1)$
We are given $\sigma^{2} + \mu^{2} = 2$. Since $\sigma^{2} = E[X^{2}] - \mu^{2}$,we have $E[X^{2}] = 2$.
$E[X^{2}] = \sum x_{i}^{2} p(x_{i}) = 0^{2} \cdot \frac{2a+1}{30} + 1^{2} \cdot \frac{8a-1}{30} + 2^{2} \cdot \frac{4a+1}{30} + 3^{2} \cdot b = 2$
$\frac{8a - 1 + 16a + 4}{30} + 9b = 2$
$\frac{24a + 3}{30} + 9b = 2$ $\Rightarrow 24a + 3 + 270b = 60$ $\Rightarrow 24a + 270b = 57$
Dividing by $3$: $8a + 90b = 19 \dots (2)$
From $(1)$,$30b = 29 - 14a$. Substitute into $(2)$:
$8a + 3(29 - 14a) = 19$
$8a + 87 - 42a = 19$
$-34a = -68 \Rightarrow a = 2$
Substituting $a = 2$ into $(1)$: $14(2) + 30b = 29$ $\Rightarrow 28 + 30b = 29$ $\Rightarrow 30b = 1$ $\Rightarrow b = \frac{1}{30}$
Therefore,$\frac{a}{b} = \frac{2}{1/30} = 60$.

(B) The equation of line $L_{1}$ is $\frac{x-2}{-3} = \frac{y-6}{2} = \frac{z-7}{4} = \lambda_{1}$. Thus,any point $C$ on $L_{1}$ is $(-3\lambda_{1}+2, 2\lambda_{1}+6, 4\lambda_{1}+7)$.
The equation of line $L_{2}$ is $\frac{x-4}{2} = \frac{y-3}{1} = \frac{z-5}{3} = \lambda_{2}$. Thus,any point $D$ on $L_{2}$ is $(2\lambda_{2}+4, \lambda_{2}+3, 3\lambda_{2}+5)$.
The vector $\overrightarrow{CD} = (2\lambda_{2}+3\lambda_{1}+2)\hat{i} + (\lambda_{2}-2\lambda_{1}-3)\hat{j} + (3\lambda_{2}-4\lambda_{1}-2)\hat{k}$.
Since $L_{3}$ is parallel to $-3\hat{i}+5\hat{j}+16\hat{k}$,the components of $\overrightarrow{CD}$ must be proportional to $(-3, 5, 16)$:
$\frac{2\lambda_{2}+3\lambda_{1}+2}{-3} = \frac{\lambda_{2}-2\lambda_{1}-3}{5} = \frac{3\lambda_{2}-4\lambda_{1}-2}{16} = k$.
Solving these equations,we get $\lambda_{1} = -3$ and $\lambda_{2} = 2$.
Substituting these values,we get $C = (11, 0, -5)$ and $D = (8, 5, 11)$.
Then $\overrightarrow{CD} = (8-11)\hat{i} + (5-0)\hat{j} + (11-(-5))\hat{k} = -3\hat{i} + 5\hat{j} + 16\hat{k}$.
Therefore,$|\overrightarrow{CD}|^2 = (-3)^2 + 5^2 + 16^2 = 9 + 25 + 256 = 290$.

(D) The line $L$ passes through $A(-3, 5, 2)$ and makes equal angles with the coordinate axes,so its direction ratios are $(1, 1, 1)$.
The equation of the line is $\frac{x+3}{1} = \frac{y-5}{1} = \frac{z-2}{1} = \lambda$.
Any general point $R$ on the line is $(\lambda-3, \lambda+5, \lambda+2)$.
Let $P = (-2, r, 1)$. The vector $\overrightarrow{PR} = ((\lambda-3) - (-2), (\lambda+5) - r, (\lambda+2) - 1) = (\lambda-1, \lambda+5-r, \lambda+1)$.
Since $PR$ is the perpendicular distance,$\overrightarrow{PR} \cdot \vec{d} = 0$,where $\vec{d} = (1, 1, 1)$.
$(\lambda-1)(1) + (\lambda+5-r)(1) + (\lambda+1)(1) = 0 \Rightarrow 3\lambda - r + 5 = 0 \Rightarrow \lambda = \frac{r-5}{3}$.
Substituting $\lambda$ back,$R = (\frac{r-5}{3}-3, \frac{r-5}{3}+5, \frac{r-5}{3}+2) = (\frac{r-14}{3}, \frac{r+10}{3}, \frac{r+1}{3})$.
The distance $PR = \sqrt{\frac{14}{3}}$,so $PR^2 = \frac{14}{3}$.
$PR^2 = (\frac{r-14}{3} + 2)^2 + (\frac{r+10}{3} - r)^2 + (\frac{r+1}{3} - 1)^2 = \frac{14}{3}$.
$(\frac{r-8}{3})^2 + (\frac{10-2r}{3})^2 + (\frac{r-2}{3})^2 = \frac{14}{3}$.
$\frac{r^2-16r+64 + 100-40r+4r^2 + r^2-4r+4}{9} = \frac{14}{3}$.
$6r^2 - 60r + 168 = 42 \Rightarrow 6r^2 - 60r + 126 = 0$.
$r^2 - 10r + 21 = 0 \Rightarrow (r-3)(r-7) = 0$.
The possible values of $r$ are $3$ and $7$.
The sum of all possible values of $r$ is $3 + 7 = 10$.

(D) Given $f(x) = x^{3} + x^{2}f^{\prime}(1) + 2x f^{\prime\prime}(2) + f^{\prime\prime\prime}(3)$.
First derivative: $f^{\prime}(x) = 3x^{2} + 2x f^{\prime}(1) + 2f^{\prime\prime}(2)$.
Second derivative: $f^{\prime\prime}(x) = 6x + 2f^{\prime}(1)$.
Third derivative: $f^{\prime\prime\prime}(x) = 6$.
Now,evaluate the constants:
$f^{\prime\prime}(2) = 6(2) + 2f^{\prime}(1) = 12 + 2f^{\prime}(1)$.
$f^{\prime\prime\prime}(3) = 6$.
Substitute these into the expression for $f^{\prime}(x)$:
$f^{\prime}(x) = 3x^{2} + 2x f^{\prime}(1) + 2(12 + 2f^{\prime}(1)) = 3x^{2} + 2x f^{\prime}(1) + 24 + 4f^{\prime}(1)$.
Set $x = 1$ to find $f^{\prime}(1)$:
$f^{\prime}(1) = 3(1)^{2} + 2(1)f^{\prime}(1) + 24 + 4f^{\prime}(1)$.
$f^{\prime}(1) = 3 + 2f^{\prime}(1) + 24 + 4f^{\prime}(1) = 27 + 6f^{\prime}(1)$.
$-5f^{\prime}(1) = 27 \implies f^{\prime}(1) = -\frac{27}{5}$.
Now substitute $f^{\prime}(1)$ back into $f^{\prime}(x)$:
$f^{\prime}(x) = 3x^{2} + 2x(-\frac{27}{5}) + 24 + 4(-\frac{27}{5}) = 3x^{2} - \frac{54}{5}x + 24 - \frac{108}{5} = 3x^{2} - \frac{54}{5}x + \frac{12}{5}$.
Finally,calculate $f^{\prime}(5)$:
$f^{\prime}(5) = 3(5)^{2} - \frac{54}{5}(5) + \frac{12}{5} = 75 - 54 + \frac{12}{5} = 21 + \frac{12}{5} = \frac{105 + 12}{5} = \frac{117}{5}$.

(A) Given $g(x) = f((\tan x - 1)^{2} + a - 1)$.
Taking the derivative,$g^{\prime}(x) = f^{\prime}((\tan x - 1)^{2} + a - 1) \cdot 2(\tan x - 1) \cdot \sec^{2}x$.
Since $f^{\prime\prime}(x) > 0$,$f^{\prime}(x)$ is a strictly increasing function.
We are given $f^{\prime}(a-1) = 0$.
If $(\tan x - 1)^{2} > 0$,then $(\tan x - 1)^{2} + a - 1 > a - 1$,which implies $f^{\prime}((\tan x - 1)^{2} + a - 1) > f^{\prime}(a - 1) = 0$.
For $x \in (0, \frac{\pi}{4})$,$\tan x < 1$,so $(\tan x - 1) < 0$. Thus $g^{\prime}(x) = (\text{positive}) \cdot (\text{negative}) \cdot (\text{positive}) < 0$. So $g$ is decreasing in $(0, \frac{\pi}{4})$.
For $x \in (\frac{\pi}{4}, \frac{\pi}{2})$,$\tan x > 1$,so $(\tan x - 1) > 0$. Thus $g^{\prime}(x) = (\text{positive}) \cdot (\text{positive}) \cdot (\text{positive}) > 0$. So $g$ is increasing in $(\frac{\pi}{4}, \frac{\pi}{2})$.
Therefore,neither statement $(I)$ nor $(II)$ is true.

(B) First,we calculate $\alpha = \int_{0}^{64} (x^{1/3} - [x^{1/3}]) dx = \int_{0}^{64} x^{1/3} dx - \int_{0}^{64} [x^{1/3}] dx$.
$\int_{0}^{64} x^{1/3} dx = \left[ \frac{3}{4} x^{4/3} \right]_{0}^{64} = \frac{3}{4} \times 256 = 192$.
For $\int_{0}^{64} [x^{1/3}] dx$,we split the integral based on the values of $[x^{1/3}]$:
$= \int_{0}^{1} 0 dx + \int_{1}^{8} 1 dx + \int_{8}^{27} 2 dx + \int_{27}^{64} 3 dx = 0 + (8-1) + 2(27-8) + 3(64-27) = 7 + 38 + 111 = 156$.
Thus,$\alpha = 192 - 156 = 36$.
Now,we evaluate $E = \frac{1}{\pi} \int_{0}^{36\pi} \frac{\sin^2 \theta}{\sin^6 \theta + \cos^6 \theta} d\theta$.
Since the integrand has a period of $\pi$,$E = \frac{36}{\pi} \int_{0}^{\pi} \frac{\sin^2 \theta}{\sin^6 \theta + \cos^6 \theta} d\theta = \frac{36 \times 2}{\pi} \int_{0}^{\pi/2} \frac{\sin^2 \theta}{\sin^6 \theta + \cos^6 \theta} d\theta$.
Let $J = \int_{0}^{\pi/2} \frac{\sin^2 \theta}{\sin^6 \theta + \cos^6 \theta} d\theta$. By King's property,$J = \int_{0}^{\pi/2} \frac{\cos^2 \theta}{\sin^6 \theta + \cos^6 \theta} d\theta$.
Adding these,$2J = \int_{0}^{\pi/2} \frac{1}{\sin^6 \theta + \cos^6 \theta} d\theta = \int_{0}^{\pi/2} \frac{\sec^6 \theta}{\tan^6 \theta + 1} d\theta$.
Let $\tan \theta = t$,then $dt = \sec^2 \theta d\theta$. $2J = \int_{0}^{\infty} \frac{(1+t^2)^2}{t^6+1} dt = \int_{0}^{\infty} \frac{1+t^2}{t^4-t^2+1} dt = \pi$.
Thus $J = \pi/2$.
Finally,$E = \frac{72}{\pi} \times \frac{\pi}{2} = 36$.

(A) Given $\overrightarrow{a}=\sqrt{2}\hat{i}-\hat{j}+\lambda\hat{k}$ and $\overrightarrow{b}=-\lambda^{2}\hat{i}+4\sqrt{2}\hat{j}+4\sqrt{2}\hat{k}$.
Since $\overrightarrow{a}$ makes an angle $\theta$ with the positive $z$-axis,$\cos \theta = \frac{\overrightarrow{a} \cdot \hat{k}}{|\overrightarrow{a}|} = \frac{\lambda}{\sqrt{(\sqrt{2})^2+(-1)^2+\lambda^2}} = \frac{\lambda}{\sqrt{3+\lambda^2}}$.
Given $\frac{\pi}{6} < \theta < \frac{\pi}{2}$,so $\cos \frac{\pi}{2} < \cos \theta < \cos \frac{\pi}{6}$,which implies $0 < \frac{\lambda}{\sqrt{3+\lambda^2}} < \frac{\sqrt{3}}{2}$.
Squaring the inequality,$0 < \frac{\lambda^2}{3+\lambda^2} < \frac{3}{4}$.
Since $\lambda > 0$,the left part is always true. For the right part,$4\lambda^2 < 9 + 3\lambda^2 \Rightarrow \lambda^2 < 9 \Rightarrow \lambda < 3$. So $\lambda \in (0, 3)$....$(1)$
Since $\overrightarrow{a}$ makes an obtuse angle with $\overrightarrow{b}$,$\overrightarrow{a} \cdot \overrightarrow{b} < 0$.
$\overrightarrow{a} \cdot \overrightarrow{b} = (\sqrt{2})(-\lambda^2) + (-1)(4\sqrt{2}) + (\lambda)(4\sqrt{2}) = -\sqrt{2}(\lambda^2 - 4\lambda + 4) = -\sqrt{2}(\lambda-2)^2 < 0$.
Since $\sqrt{2} > 0$,we must have $(\lambda-2)^2 > 0$,which implies $\lambda \neq 2$....$(2)$
From $(1)$ and $(2)$,$\lambda \in (0, 3) - \{2\}$.
Thus,$\alpha=0, \beta=3, \gamma=2$.
Therefore,$\alpha+\beta+\gamma = 0+3+2 = 5$.

(B) Given $f(x+y)=f(x)f(y)$ with $f(1)=7$. This is a functional equation of the form $f(x)=a^x$. Since $f(1)=7$,we have $a^1=7$,so $f(x)=7^x$.
Given $g(x+y)=g(xy)$ with $g(1)=1$. Putting $y=1$,we get $g(x+1)=g(x)$. Since $g(1)=1$,it follows that $g(x)=1$ for all $x \in \mathbb{N}$.
The given summation is $\sum_{x=1}^{n} \frac{f(x)}{g(x)} = 19607$.
Substituting the functions,we get $\sum_{x=1}^{n} \frac{7^x}{1} = 19607$.
This is a geometric progression: $7 + 7^2 + \dots + 7^n = 19607$.
The sum of a $GP$ is $S_n = a\frac{r^n-1}{r-1}$. Here $a=7$ and $r=7$.
$7 \left(\frac{7^n-1}{7-1}\right) = 19607$.
$7 \left(\frac{7^n-1}{6}\right) = 19607$.
$7^n-1 = \frac{19607 \times 6}{7} = 2801 \times 6 = 16806$.
$7^n = 16807$.
Since $7^5 = 16807$,we have $n=5$.

(B) Given $f(x) = [x]^2 - ([x] + 3) - 3 = [x]^2 - [x] - 6$.
Factoring the expression,we get $f(x) = ([x] - 3)([x] + 2)$.
$(1)$ For $f(x) > 0$,we need $[x] > 3$ or $[x] < -2$.
This implies $x \in [4, \infty)$ or $x \in (-\infty, -2)$. Thus,option $A$ is incorrect.
$(2)$ For $f(x) < 0$,we need $-2 < [x] < 3$,which means $[x] \in \{-1, 0, 1, 2\}$.
This implies $x \in [-1, 3)$. Thus,option $B$ is correct.
$(3)$ Calculating the integral: $\int_0^2 f(x) dx = \int_0^1 f(x) dx + \int_1^2 f(x) dx$.
For $x \in [0, 1)$,$[x] = 0$,so $f(x) = 0^2 - 0 - 6 = -6$.
For $x \in [1, 2)$,$[x] = 1$,so $f(x) = 1^2 - 1 - 6 = -6$.
Thus,$\int_0^2 f(x) dx = \int_0^1 (-6) dx + \int_1^2 (-6) dx = -6 - 6 = -12$. Option $C$ is incorrect.
$(4)$ For $f(x) = 0$,we need $[x] = 3$ or $[x] = -2$.
This implies $x \in [3, 4)$ or $x \in [-2, -1)$,which contains infinitely many values. Option $D$ is incorrect.

(C) The function $g(x) = |x|[x^2]$ is discontinuous where $[x^2]$ is discontinuous,provided $|x| \neq 0$.
$[x^2]$ is discontinuous at $x^2 \in \mathbb{Z}$.
For $x \in (-2, 2)$,$x^2 \in [0, 4)$.
The integers in $[0, 4)$ are $0, 1, 2, 3$.
Thus,$x^2 = 1, 2, 3$ gives $x = \pm 1, \pm \sqrt{2}, \pm \sqrt{3}$.
At $x=0$,$g(x) = |x|[x^2] = 0 \cdot [0] = 0$,and $\lim_{x \to 0} |x|[x^2] = 0$,so $g(x)$ is continuous at $x=0$.
Thus,$S = \{-1, 1, -\sqrt{2}, \sqrt{2}, -\sqrt{3}, \sqrt{3}\}$.
Now,$f(x) = \min \{\sqrt{2}x, x^2\}$.
$f(-1) = \min \{-\sqrt{2}, 1\} = -\sqrt{2}$.
$f(1) = \min \{\sqrt{2}, 1\} = 1$.
$f(-\sqrt{2}) = \min \{-2, 2\} = -2$.
$f(\sqrt{2}) = \min \{2, 2\} = 2$.
$f(-\sqrt{3}) = \min \{-\sqrt{6}, 3\} = -\sqrt{6}$.
$f(\sqrt{3}) = \min \{\sqrt{6}, 3\} = \sqrt{6}$.
Summing these values: $\sum_{x \in S} f(x) = -\sqrt{2} + 1 - 2 + 2 - \sqrt{6} + \sqrt{6} = 1 - \sqrt{2}$.

(C) For the first term $\log_3 \log_5 (7 - \log_2 (x^2 - 10 x + 85))$,we require $\log_5 (7 - \log_2 (x^2 - 10 x + 85)) > 0$,which implies $7 - \log_2 (x^2 - 10 x + 85) > 1$,so $\log_2 (x^2 - 10 x + 85) < 6$. This gives $x^2 - 10 x + 85 < 2^6 = 64$,so $x^2 - 10 x + 21 < 0$. Factoring gives $(x - 3)(x - 7) < 0$,so $x \in (3, 7)$.
For the second term $\sin^{-1} ( | \frac{3 x - 7}{17 - x} | )$,we require $0 \leq | \frac{3 x - 7}{17 - x} | \leq 1$. The condition $| \frac{3 x - 7}{17 - x} | \leq 1$ implies $(3x - 7)^2 \leq (17 - x)^2$,so $9x^2 - 42x + 49 \leq 289 - 34x + x^2$. This simplifies to $8x^2 - 8x - 240 \leq 0$,or $x^2 - x - 30 \leq 0$. Factoring gives $(x - 6)(x + 5) \leq 0$,so $x \in [-5, 6]$.
Combining the domains $(3, 7)$ and $[-5, 6]$,we get $x \in (3, 6]$.
Thus,$\alpha = 3$ and $\beta = 6$.
Therefore,$\alpha + \beta = 3 + 6 = 9$.

(B) The given region is bounded by the ellipse $4x^2 + y^2 = 8$ and the parabola $y^2 = 4x$.
First,find the intersection points by substituting $y^2 = 4x$ into $4x^2 + y^2 = 8$:
$4x^2 + 4x - 8 = 0 \implies x^2 + x - 2 = 0 \implies (x+2)(x-1) = 0$.
Since $x \ge 0$ for the parabola,we have $x = 1$. At $x=1$,$y^2 = 4$,so $y = \pm 2$.
The area $A$ is given by $2 \int_0^1 \sqrt{4x} dx + 2 \int_1^{\sqrt{2}} \sqrt{8-4x^2} dx$.
$= 4 \int_0^1 \sqrt{x} dx + 4 \int_1^{\sqrt{2}} \sqrt{2-x^2} dx$.
$= 4 \left[ \frac{2}{3} x^{3/2} \right]_0^1 + 4 \left[ \frac{x}{2} \sqrt{2-x^2} + \frac{2}{2} \sin^{-1} \left( \frac{x}{\sqrt{2}} \right) \right]_1^{\sqrt{2}}$.
$= \frac{8}{3} + 4 \left[ (0 + \sin^{-1}(1)) - (\frac{1}{2} \sqrt{1} + \sin^{-1}(\frac{1}{\sqrt{2}})) \right]$.
$= \frac{8}{3} + 4 \left[ \frac{\pi}{2} - \frac{1}{2} - \frac{\pi}{4} \right] = \frac{8}{3} + 2\pi - 2 - \pi = \pi + \frac{2}{3}$ sq. units.

(A) Let any point on the line $L$ be $(a, b, c) = (2k-1, 3k-1, 6k-3)$.
Given that the distance of this point from the point $P(-1, -1, -9)$ is $7$.
Using the distance formula: $\sqrt{(2k-1 - (-1))^2 + (3k-1 - (-1))^2 + (6k-3 - (-9))^2} = 7$.
$\sqrt{(2k)^2 + (3k)^2 + (6k+6)^2} = 7$.
Squaring both sides: $4k^2 + 9k^2 + (6k+6)^2 = 49$.
$13k^2 + 36k^2 + 72k + 36 = 49$.
$49k^2 + 72k - 13 = 0$.
This is a quadratic equation in $k$. Let the roots be $k_1$ and $k_2$.
The sum of the coordinates for a point $(a, b, c)$ is $a+b+c = (2k-1) + (3k-1) + (6k-3) = 11k - 5$.
For the two points in $S$,the sum is $(11k_1 - 5) + (11k_2 - 5) = 11(k_1 + k_2) - 10$.
From the quadratic equation,$k_1 + k_2 = -\frac{72}{49}$.
Sum $= 11(-\frac{72}{49}) - 10 = -\frac{792}{49} - 10 = -\frac{1282}{49}$.

(D) We know that $X = A^{-1}B = \frac{\text{adj } A}{|A|} B$.
First,we find the determinant $|A|$. We know that $|\text{adj } A| = |A|^{n-1}$,where $n=3$.
$|\text{adj } A| = 4(0 - (-10)) - 2(-15 - 5) + 2(10 - 0) = 4(10) - 2(-20) + 2(10) = 40 + 40 + 20 = 100$.
So,$|A|^2 = 100$,which implies $|A| = \pm 10$.
Now,$X = \pm \frac{1}{10} \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix} \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix} = \pm \frac{1}{10} \begin{bmatrix} 16 + 0 + 4 \\ -20 + 0 + 10 \\ 4 + 0 + 6 \end{bmatrix} = \pm \frac{1}{10} \begin{bmatrix} 20 \\ -10 \\ 10 \end{bmatrix} = \pm \begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix}$.
Thus,$x = \pm 2, y = \mp 1, z = \pm 1$.
Then $|x + y + z| = |\pm(2 - 1 + 1)| = |\pm 2| = 2$.

(C) Given $\overrightarrow{a}=2 \hat{i}-\hat{j}+\hat{k}$ and $\overrightarrow{b}=\lambda \hat{j}+2 \hat{k}$.
$\overrightarrow{c}=\overrightarrow{a} \times \overrightarrow{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 0 & \lambda & 2 \end{vmatrix} = (-2-\lambda) \hat{i} - 4 \hat{j} + 2 \lambda \hat{k}$.
Given $|\overrightarrow{c}|=\sqrt{53}$,so $|\overrightarrow{c}|^2 = 53$.
$(-2-\lambda)^2 + (-4)^2 + (2\lambda)^2 = 53$
$4 + 4\lambda + \lambda^2 + 16 + 4\lambda^2 = 53$
$5\lambda^2 + 4\lambda - 33 = 0$.
Solving for $\lambda$: $\lambda = \frac{-4 \pm \sqrt{16 - 4(5)(-33)}}{10} = \frac{-4 \pm \sqrt{16 + 660}}{10} = \frac{-4 \pm \sqrt{676}}{10} = \frac{-4 \pm 26}{10}$.
Since $\lambda \in \mathbb{Z}$,we take $\lambda = -3$ (as $\frac{22}{10}$ is not an integer).
Thus,$\overrightarrow{c} = (-2 - (-3))\hat{i} - 4\hat{j} + 2(-3)\hat{k} = \hat{i} - 4\hat{j} - 6\hat{k}$.
Let $\overrightarrow{d} = y\hat{j} + z\hat{k}$ be a vector in the $yz$-plane with $|\overrightarrow{d}|=2$,so $y^2 + z^2 = 4$.
Then $\overrightarrow{c} \cdot \overrightarrow{d} = (\hat{i} - 4\hat{j} - 6\hat{k}) \cdot (y\hat{j} + z\hat{k}) = -4y - 6z$.
We want to maximize $(-4y - 6z)^2 = (4y + 6z)^2$.
By Cauchy-Schwarz inequality,$(4y + 6z)^2 \leq (4^2 + 6^2)(y^2 + z^2) = (16 + 36)(4) = 52 \times 4 = 208$.

(A) Given the differential equation: $16(\sqrt{x+9\sqrt{x}})(4+\sqrt{9+\sqrt{x}}) \cos y \, dy = (1+2 \sin y) \, dx$.
Separating the variables,we get: $\int \frac{\cos y}{1+2 \sin y} \, dy = \int \frac{dx}{16(\sqrt{x+9\sqrt{x}})(4+\sqrt{9+\sqrt{x}})}$.
Let $u = 1+2 \sin y$,then $du = 2 \cos y \, dy$,so $\int \frac{\cos y}{1+2 \sin y} \, dy = \frac{1}{2} \ln |1+2 \sin y|$.
For the $RHS$,let $t = 4+\sqrt{9+\sqrt{x}}$. Then $t-4 = \sqrt{9+\sqrt{x}}$. Squaring both sides: $(t-4)^2 = 9+\sqrt{x}$,so $\sqrt{x} = (t-4)^2 - 9$.
Differentiating $t = 4+\sqrt{9+\sqrt{x}}$,we get $dt = \frac{1}{2\sqrt{9+\sqrt{x}}} \cdot \frac{1}{2\sqrt{x}} \, dx = \frac{dx}{4\sqrt{x(9+\sqrt{x})}} = \frac{dx}{4\sqrt{x+9\sqrt{x}}}$.
Thus,$\frac{dx}{\sqrt{x+9\sqrt{x}}} = 4 \, dt$.
The integral becomes: $\frac{1}{2} \ln |1+2 \sin y| = \int \frac{4 \, dt}{16t} = \frac{1}{4} \ln |t| + C = \frac{1}{4} \ln |4+\sqrt{9+\sqrt{x}}| + C$.
Using $y(256) = \frac{\pi}{2}$: $\frac{1}{2} \ln(1+2 \sin \frac{\pi}{2}) = \frac{1}{4} \ln(4+\sqrt{9+\sqrt{256}}) + C \implies \frac{1}{2} \ln 3 = \frac{1}{4} \ln(4+\sqrt{9+16}) + C = \frac{1}{4} \ln 9 + C = \frac{1}{2} \ln 3 + C$. Thus $C = 0$.
Now,for $y(49) = \alpha$: $\frac{1}{2} \ln(1+2 \sin \alpha) = \frac{1}{4} \ln(4+\sqrt{9+\sqrt{49}}) = \frac{1}{4} \ln(4+\sqrt{16}) = \frac{1}{4} \ln 8 = \frac{1}{4} \ln(2^3) = \frac{3}{4} \ln 2$.
So $\ln(1+2 \sin \alpha) = \frac{3}{2} \ln 2 = \ln(2^{3/2}) = \ln(2\sqrt{2})$.
Therefore,$1+2 \sin \alpha = 2\sqrt{2}$,which gives $2 \sin \alpha = 2\sqrt{2}-1$.

(D) The system of equations has a unique solution if the determinant of the coefficient matrix $D \neq 0$.
$D = \begin{vmatrix} 1 & -n & 1 \\ 1 & n-2 & n+1 \\ 0 & n-1 & 1 \end{vmatrix}$
Expanding along the first column:
$D = 1((n-2)(1) - (n+1)(n-1)) - 1((-n)(1) - (1)(n-1)) + 0$
$D = (n-2 - (n^2-1)) - (-n - n + 1)$
$D = (n-2 - n^2 + 1) - (-2n + 1)$
$D = -n^2 + n - 1 + 2n - 1 = -n^2 + 3n - 2$
For a unique solution,$D \neq 0$,so $-n^2 + 3n - 2 \neq 0$,which implies $n^2 - 3n + 2 \neq 0$.
$(n-1)(n-2) \neq 0$,so $n \neq 1$ and $n \neq 2$.
Since $n$ is the outcome of a fair die,$n \in \{1, 2, 3, 4, 5, 6\}$.
The values of $n$ for which the system has a unique solution are $n \in \{3, 4, 5, 6\}$.
The number of such values is $4$,so the probability is $\frac{4}{6}$.
Thus,$k = 4$.
The sum of $k$ and all possible values of $n$ is $4 + (3 + 4 + 5 + 6) = 4 + 18 = 22$.

(C) Given $A = \begin{bmatrix} 0 & 2 & -3 \\ -2 & 0 & 1 \\ 3 & -1 & 0 \end{bmatrix}$. Note that $A^T = -A$,so $A$ is a skew-symmetric matrix.
Given $B(I - A) = I + A$,we have $B = (I + A)(I - A)^{-1}$.
Then $B^T = ((I + A)(I - A)^{-1})^T = ((I - A)^{-1})^T (I + A)^T = (I - A^T)^{-1} (I + A^T)$.
Since $A^T = -A$,we get $B^T = (I - (-A))^{-1} (I + (-A)) = (I + A)^{-1} (I - A)$.
Now,$B^T B = (I + A)^{-1} (I - A) (I + A) (I - A)^{-1}$.
Since $A$ is skew-symmetric,$(I - A)$ and $(I + A)$ commute,i.e.,$(I - A)(I + A) = I^2 - A^2 = (I + A)(I - A)$.
Thus,$B^T B = (I + A)^{-1} (I + A) (I - A) (I - A)^{-1} = I \cdot I = I$.
The matrix $B^T B$ is the identity matrix $I_3 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
The sum of the diagonal elements (trace) is $1 + 1 + 1 = 3$.

(B) Let $I = \int_{0}^{x} t^{2} \sin(x-t) dt$. Using the property $\int_{0}^{x} f(t) dt = \int_{0}^{x} f(x-t) dt$,we have $I = \int_{0}^{x} (x-t)^{2} \sin(t) dt$.
Expanding this,$I = \int_{0}^{x} (x^{2} - 2xt + t^{2}) \sin(t) dt = x^{2} \int_{0}^{x} \sin(t) dt - 2x \int_{0}^{x} t \sin(t) dt + \int_{0}^{x} t^{2} \sin(t) dt$.
Evaluating the integrals:
$1. \int_{0}^{x} \sin(t) dt = [-\cos(t)]_{0}^{x} = 1 - \cos(x)$.
$2. \int_{0}^{x} t \sin(t) dt = [-t \cos(t)]_{0}^{x} + \int_{0}^{x} \cos(t) dt = -x \cos(x) + \sin(x)$.
$3. \int_{0}^{x} t^{2} \sin(t) dt = [-t^{2} \cos(t)]_{0}^{x} + 2 \int_{0}^{x} t \cos(t) dt = -x^{2} \cos(x) + 2(t \sin(t) + \cos(t))_{0}^{x} = -x^{2} \cos(x) + 2x \sin(x) + 2 \cos(x) - 2$.
Substituting these back: $I = x^{2}(1 - \cos(x)) - 2x(-x \cos(x) + \sin(x)) + (-x^{2} \cos(x) + 2x \sin(x) + 2 \cos(x) - 2) = x^{2} - x^{2} \cos(x) + 2x^{2} \cos(x) - 2x \sin(x) - x^{2} \cos(x) + 2x \sin(x) + 2 \cos(x) - 2 = x^{2} + 2 \cos(x) - 2$.
Given $I = x^{2}$,we have $x^{2} + 2 \cos(x) - 2 = x^{2}$,which simplifies to $2 \cos(x) = 2$,or $\cos(x) = 1$.
For $x \in [0, 100]$,$\cos(x) = 1$ implies $x = 2n\pi$ for $n = 0, 1, 2, \dots$.
Since $2n\pi \le 100$,$n \le \frac{100}{2\pi} \approx \frac{100}{6.28} \approx 15.92$.
Thus,$n$ can be $0, 1, 2, \dots, 15$,which gives $16$ values.

(C) Let the line $L_1$ be $\frac{x}{2} = \frac{y+a}{1} = \frac{z}{1} = \lambda$. Any point on $L_1$ is $(2\lambda, \lambda-a, \lambda)$.
Since $Q$ is the image of $P(a, 2, a)$ in $L_1$,the midpoint of $PQ$ lies on $L_1$ and $PQ$ is perpendicular to $L_1$.
The midpoint $M$ of $PQ$ is $(2\lambda, \lambda-a, \lambda)$. Thus,$Q = (4\lambda-a, 2\lambda-2a-2, 2\lambda-a)$.
The vector $\vec{PQ} = (3\lambda-2a, 2\lambda-2a-4, 2\lambda-2a)$. Since $\vec{PQ} \perp (2, 1, 1)$,
$2(3\lambda-2a) + 1(2\lambda-2a-4) + 1(2\lambda-2a) = 0 \Rightarrow 10\lambda - 8a - 4 = 0 \Rightarrow 5\lambda - 4a = 2$.
Similarly,for the second line $L_2: \frac{x-2b}{2} = \frac{y-a}{1} = \frac{z+2b}{-5} = \mu$,the image of $Q$ is $P$.
Following the same logic for $L_2$,we find $a=1$ and $b=2$.
Therefore,$a+b = 1+2 = 3$.

(A) The given differential equation is $(x^2-4)y^{\prime}-2xy = -2x(4-x^2)^2$.
Dividing by $(x^2-4)^2$,we get $\frac{(x^2-4)y^{\prime}-2xy}{(x^2-4)^2} = -2x$.
This is the derivative of the quotient: $\frac{d}{dx} \left( \frac{y}{x^2-4} \right) = -2x$.
Integrating both sides with respect to $x$: $\frac{y}{x^2-4} = -x^2 + C$.
So,$y = (-x^2+C)(x^2-4)$.
Given that the curve passes through $(3, 15)$,we substitute $x=3$ and $y=15$: $15 = (-9+C)(9-4) \Rightarrow 15 = 5(-9+C) \Rightarrow 3 = -9+C \Rightarrow C=12$.
Thus,$f(x) = (12-x^2)(x^2-4)$.
To find the local maximum,we set $f^{\prime}(x) = 0$: $f^{\prime}(x) = (-2x)(x^2-4) + (12-x^2)(2x) = -2x^3 + 8x + 24x - 2x^3 = -4x^3 + 32x = 0$.
$-4x(x^2-8) = 0$. Since $x>2$,we have $x^2=8$,so $x=2\sqrt{2}$.
The local maximum value is $f(2\sqrt{2}) = (12-8)(8-4) = 4 \times 4 = 16$.

(B) The set $A = \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$ has $10$ elements.
We partition $A$ into equivalence classes based on remainders modulo $3$:
$C_0 = \{0, 3, 6, 9\}$ (size $4$)
$C_1 = \{1, 4, 7\}$ (size $3$)
$C_2 = \{2, 5, 8\}$ (size $3$)
For $(x, y) \in R$,$|x - y|$ must be a multiple of $3$,meaning $x$ and $y$ must belong to the same equivalence class.
The number of elements in $R$ is $n(R) = |C_0|^2 + |C_1|^2 + |C_2|^2 = 4^2 + 3^2 + 3^2 = 16 + 9 + 9 = 34$.
Thus,Statement $I$ is incorrect $(34 \neq 36)$.
For Statement $II$:
$1$. Reflexive: $|x - x| = 0$,which is a multiple of $3$.
$2$. Symmetric: If $|x - y| = 3k$,then $|y - x| = 3k$.
$3$. Transitive: If $|x - y| = 3k$ and $|y - z| = 3m$,then $|x - z| = |(x - y) + (y - z)| = 3|k \pm m|$,which is a multiple of $3$.
Since $R$ is reflexive,symmetric,and transitive,it is an equivalence relation. Statement $II$ is correct.

(C) Given $I(x) = \int \frac{3dx}{(4x+6)\sqrt{4x^2+8x+3}}$.
Rewrite the integral as $I(x) = \int \frac{3dx}{(4x+6)\sqrt{(2x+2)^2-1}}$.
Let $2x+2 = \sec \theta$,then $2dx = \sec \theta \tan \theta d\theta$.
Also,$4x+6 = 2(2x+2)+2 = 2\sec \theta + 2 = 2(\sec \theta + 1)$.
Substituting these into the integral:
$I(x) = \int \frac{3 \cdot \frac{1}{2} \sec \theta \tan \theta d\theta}{2(\sec \theta + 1)\tan \theta} = \frac{3}{4} \int \frac{\sec \theta}{\sec \theta + 1} d\theta = \frac{3}{4} \int \frac{1}{\cos \theta + 1} d\theta = \frac{3}{4} \int \frac{1}{2 \cos^2(\theta/2)} d\theta = \frac{3}{8} \int \sec^2(\theta/2) d\theta$.
$I(x) = \frac{3}{8} \cdot 2 \tan(\theta/2) + C = \frac{3}{4} \tan(\theta/2) + C$.
Since $\sec \theta = 2x+2$,$\tan^2(\theta/2) = \frac{1-\cos \theta}{1+\cos \theta} = \frac{1-1/(2x+2)}{1+1/(2x+2)} = \frac{2x+1}{2x+3}$.
So,$I(x) = \frac{3}{4} \sqrt{\frac{2x+1}{2x+3}} + C$.
Given $I(0) = \frac{3}{4} \sqrt{\frac{1}{3}} + C = \frac{\sqrt{3}}{4} + C = \frac{\sqrt{3}}{4} + 20$,so $C=20$.
Thus,$I(x) = \frac{3}{4} \sqrt{\frac{2x+1}{2x+3}} + 20$.
For $x = 1/2$,$I(1/2) = \frac{3}{4} \sqrt{\frac{1+1}{1+3}} + 20 = \frac{3}{4} \sqrt{\frac{2}{4}} + 20 = \frac{3}{4} \cdot \frac{\sqrt{2}}{2} + 20 = \frac{3\sqrt{2}}{8} + 20$.
Here $a=3, b=8, c=20$. $gcd(3,8)=1$.
Therefore,$a+b+c = 3+8+20 = 31$.

(D) The given circles are $x^{2}+y^{2}=4$ (center $(0,0)$,radius $2$) and $x^{2}+(y-2)^{2}=4$ (center $(0,2)$,radius $2$).
Solving the equations: $x^{2}+y^{2}=4$ and $x^{2}+y^{2}-4y+4=4$.
Substituting $x^{2}+y^{2}=4$ into the second equation: $4-4y+4=4$,which gives $4y=4$,so $y=1$.
Substituting $y=1$ into $x^{2}+y^{2}=4$,we get $x^{2}+1=4$,so $x^{2}=3$,$x=\pm\sqrt{3}$.
The area of the intersection is symmetric about the $y$-axis.
The area $A = 2 \int_{0}^{\sqrt{3}} [\sqrt{4-x^{2}} - (2 - \sqrt{4-x^{2}})] dx = 2 \int_{0}^{\sqrt{3}} (2\sqrt{4-x^{2}} - 2) dx = 4 \int_{0}^{\sqrt{3}} (\sqrt{4-x^{2}} - 1) dx$.
Using the formula $\int \sqrt{a^{2}-x^{2}} dx = \frac{x}{2}\sqrt{a^{2}-x^{2}} + \frac{a^{2}}{2}\sin^{-1}(\frac{x}{a})$:
$A = 4 [\frac{x}{2}\sqrt{4-x^{2}} + 2\sin^{-1}(\frac{x}{2}) - x]_{0}^{\sqrt{3}}$
$A = 4 [(\frac{\sqrt{3}}{2}\sqrt{4-3} + 2\sin^{-1}(\frac{\sqrt{3}}{2})) - \sqrt{3}]$
$A = 4 [\frac{\sqrt{3}}{2} + 2(\frac{\pi}{3}) - \sqrt{3}] = 4 [\frac{2\pi}{3} - \frac{\sqrt{3}}{2}] = \frac{8\pi}{3} - 2\sqrt{3} = \frac{2}{3}(4\pi - 3\sqrt{3})$.

(B) Let $W_B$ be the event of picking a white ball from bag $B$,and $B_B$ be the event of picking a black ball from bag $B$.
$P(W_B) = \frac{6}{6+4} = \frac{6}{10} = \frac{3}{5}$
$P(B_B) = \frac{4}{6+4} = \frac{4}{10} = \frac{2}{5}$
If a white ball is transferred to bag $A$,bag $A$ now contains $10$ white and $8$ black balls (total $18$). The probability of drawing a white ball from bag $A$ is $P(W_A | W_B) = \frac{10}{18}$.
If a black ball is transferred to bag $A$,bag $A$ now contains $9$ white and $9$ black balls (total $18$). The probability of drawing a white ball from bag $A$ is $P(W_A | B_B) = \frac{9}{18}$.
Using the law of total probability:
$P(W_A) = P(W_B) \times P(W_A | W_B) + P(B_B) \times P(W_A | B_B)$
$P(W_A) = \frac{3}{5} \times \frac{10}{18} + \frac{2}{5} \times \frac{9}{18}$
$P(W_A) = \frac{30}{90} + \frac{18}{90} = \frac{48}{90} = \frac{8}{15}$
Thus,$p=8$ and $q=15$. Since $gcd(8,15)=1$,$p+q = 8+15 = 23$.

(D) Given $\vec{a} \times \vec{b} = 2(\vec{a} \times \vec{c})$.
This implies $\vec{a} \times (\vec{b} - 2\vec{c}) = 0$.
Therefore,$\vec{b} - 2\vec{c} = \lambda \vec{a}$ for some scalar $\lambda$.
Taking the magnitude squared on both sides: $|\vec{b} - 2\vec{c}|^2 = \lambda^2 |\vec{a}|^2$.
$|\vec{b}|^2 + 4|\vec{c}|^2 - 4(\vec{b} \cdot \vec{c}) = \lambda^2 (1)^2$.
Given $|\vec{b}| = 4, |\vec{c}| = 2$,and the angle between $\vec{b}$ and $\vec{c}$ is $60^{\circ}$,$\vec{b} \cdot \vec{c} = |\vec{b}| |\vec{c}| \cos(60^{\circ}) = 4 \times 2 \times \frac{1}{2} = 4$.
So,$16 + 4(4) - 4(4) = \lambda^2 \Rightarrow \lambda^2 = 16 \Rightarrow \lambda = \pm 4$.
Now,$\vec{b} - 2\vec{c} = \pm 4\vec{a}$.
Taking the dot product with $\vec{c}$ on both sides:
$(\vec{b} - 2\vec{c}) \cdot \vec{c} = \pm 4(\vec{a} \cdot \vec{c})$.
$\vec{b} \cdot \vec{c} - 2|\vec{c}|^2 = \pm 4(\vec{a} \cdot \vec{c})$.
$4 - 2(4) = \pm 4(\vec{a} \cdot \vec{c}) \Rightarrow -4 = \pm 4(\vec{a} \cdot \vec{c})$.
$|\vec{a} \cdot \vec{c}| = |\frac{-4}{\pm 4}| = 1$.

(B) The system of equations is given by:
$x + y + z = 6$
$2x + 5y + az = 36$
$x + 2y + 3z = b$
First,calculate the determinant $D$ of the coefficient matrix:
$D = \begin{vmatrix} 1 & 1 & 1 \\ 2 & 5 & a \\ 1 & 2 & 3 \end{vmatrix} = 1(15 - 2a) - 1(6 - a) + 1(4 - 5) = 15 - 2a - 6 + a - 1 = 8 - a$.
For the system to have infinitely many solutions or no solution,we must have $D = 0$,which implies $a = 8$.
Now,calculate $D_3$ with $a = 8$:
$D_3 = \begin{vmatrix} 1 & 1 & 6 \\ 2 & 5 & 36 \\ 1 & 2 & b \end{vmatrix} = 1(5b - 72) - 1(2b - 36) + 6(4 - 5) = 5b - 72 - 2b + 36 - 6 = 3b - 42$.
For $D_3 = 0$,we get $3b = 42$,so $b = 14$.
When $a = 8$ and $b = 14$,we check $D_1$ and $D_2$:
$D_1 = \begin{vmatrix} 6 & 1 & 1 \\ 36 & 5 & 8 \\ 14 & 2 & 3 \end{vmatrix} = 6(15 - 16) - 1(108 - 112) + 1(72 - 70) = -6 + 4 + 2 = 0$.
$D_2 = \begin{vmatrix} 1 & 6 & 1 \\ 2 & 36 & 8 \\ 1 & 14 & 3 \end{vmatrix} = 1(108 - 112) - 6(6 - 8) + 1(28 - 36) = -4 + 12 - 8 = 0$.
Since $D = D_1 = D_2 = D_3 = 0$ for $a = 8$ and $b = 14$,the system has infinitely many solutions.

(A) First,we find the elements of set $A$:
$|(|x - 3| - 3)| \leq 1 \implies -1 \leq |x - 3| - 3 \leq 1$
$2 \leq |x - 3| \leq 4$
This implies $2 \leq x - 3 \leq 4$ or $-4 \leq x - 3 \leq -2$
$5 \leq x \leq 7$ or $-1 \leq x \leq 1$
Since $x \in \mathbb{Z}$,$A = \{-1, 0, 1, 5, 6, 7\}$. The number of elements in $A$ is $n(A) = 6$.
Next,we find the elements of set $B$:
$\frac{(x - 2)(x - 4)}{x - 1} \log_{e}(|x - 2|) = 0$
This implies $(x - 2)(x - 4) = 0$ or $\log_{e}(|x - 2|) = 0$.
If $(x - 2)(x - 4) = 0$,then $x = 2$ or $x = 4$. Since $x \in \mathbb{R} - \{1, 2\}$,we accept $x = 4$.
If $\log_{e}(|x - 2|) = 0$,then $|x - 2| = 1$,so $x - 2 = 1$ or $x - 2 = -1$.
This gives $x = 3$ or $x = 1$. Since $x \in \mathbb{R} - \{1, 2\}$,we accept $x = 3$.
Thus,$B = \{3, 4\}$,and the number of elements in $B$ is $n(B) = 2$.
The number of onto functions from a set with $n$ elements to a set with $m$ elements is given by $m^n - \binom{m}{1}(m-1)^n + \binom{m}{2}(m-2)^n - \dots$
For $n = 6$ and $m = 2$,the number of onto functions is $2^6 - 2 = 64 - 2 = 62$.

(D) Given $\vec{a}=\hat{i}-2\hat{j}+3\hat{k}$,$\vec{b}=2\hat{i}+\hat{j}-\hat{k}$,and $\vec{c}=\lambda\hat{i}+\hat{j}+\hat{k}$.
First,calculate $\vec{v} = \vec{a} \times \vec{b}$:
$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 2 & 1 & -1 \end{vmatrix} = \hat{i}(2-3) - \hat{j}(-1-6) + \hat{k}(1+4) = -\hat{i} + 7\hat{j} + 5\hat{k}$.
Given $\vec{v} \cdot \vec{c} = 11$,so $(-\hat{i} + 7\hat{j} + 5\hat{k}) \cdot (\lambda\hat{i} + \hat{j} + \hat{k}) = 11$.
$-\lambda + 7 + 5 = 11 \Rightarrow -\lambda + 12 = 11 \Rightarrow \lambda = 1$.
Now,$\vec{c} = \hat{i} + \hat{j} + \hat{k}$. The length of the projection of $\vec{b}$ on $\vec{c}$ is $p = \left| \vec{b} \cdot \frac{\vec{c}}{|\vec{c}|} \right|$.
$|\vec{c}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$.
$p = \left| (2\hat{i} + \hat{j} - \hat{k}) \cdot \frac{(\hat{i} + \hat{j} + \hat{k})}{\sqrt{3}} \right| = \left| \frac{2 + 1 - 1}{\sqrt{3}} \right| = \frac{2}{\sqrt{3}}$.
Therefore,$9p^2 = 9 \times \left( \frac{2}{\sqrt{3}} \right)^2 = 9 \times \frac{4}{3} = 12$.

(B) For $f(x)$ to be continuous at $x=0$,we must have $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = b$.
First,consider the right-hand limit $(x > 0)$:
$\lim_{x \to 0^+} \frac{ax + x^2 - 2\sin x \cos x}{x} = \lim_{x \to 0^+} \left( a + x - \frac{\sin(2x)}{x} \right) = a + 0 - 2 = a - 2$.
Next,consider the left-hand limit $(x < 0)$:
Let $x = -h$ where $h > 0$. As $x \to 0^-$,$h \to 0^+$.
$\lim_{h \to 0^+} \frac{a|-h| + (-h)^2 - 2\sin|-h|\cos|-h|}{-h} = \lim_{h \to 0^+} \frac{ah + h^2 - 2\sin h \cos h}{-h} = \lim_{h \to 0^+} \left( -a - h + \frac{\sin(2h)}{h} \right) = -a - 0 + 2 = -a + 2$.
For continuity,$a - 2 = -a + 2 = b$.
From $a - 2 = -a + 2$,we get $2a = 4$,so $a = 2$.
Substituting $a = 2$ into $b = a - 2$,we get $b = 2 - 2 = 0$.
Therefore,$a + b = 2 + 0 = 2$.

(B) Given the equation: $(f(x))^2 = 25 + \int_{0}^{x} ((f(t))^2 + (f'(t))^2) dt$.
Differentiating both sides with respect to $x$ using the Leibniz rule:
$2 f(x) f'(x) = (f(x))^2 + (f'(x))^2$.
Rearranging the terms: $(f(x))^2 - 2 f(x) f'(x) + (f'(x))^2 = 0$.
This simplifies to: $(f(x) - f'(x))^2 = 0$,which implies $f'(x) = f(x)$.
Solving this differential equation: $\frac{f'(x)}{f(x)} = 1 \Rightarrow \ln(f(x)) = x + C \Rightarrow f(x) = A e^x$.
At $x = 0$,$(f(0))^2 = 25 + 0 \Rightarrow f(0) = 5$ (since $f$ is non-negative).
Thus,$A e^0 = 5 \Rightarrow A = 5$,so $f(x) = 5 e^x$.
We need the mean of $f(\ln 1), f(\ln 2), \ldots, f(\ln 625)$.
Since $f(\ln n) = 5 e^{\ln n} = 5n$,the mean is:
$\text{Mean} = \frac{1}{625} \sum_{n=1}^{625} 5n = \frac{5}{625} \times \frac{625 \times 626}{2} = \frac{5 \times 626}{2} = 5 \times 313 = 1565$.

(B) The area $A$ is bounded by $y=\max\{\sin x, \cos x\}$ from $x=0$ to $x=\frac{3\pi}{2}$ and the x-axis.
We split the integral based on the intervals where $\sin x$ or $\cos x$ is greater:
$A = \int_{0}^{\pi/4} \cos x \, dx + \int_{\pi/4}^{\pi} \sin x \, dx + \int_{\pi}^{5\pi/4} |\sin x| \, dx + \int_{5\pi/4}^{3\pi/2} |\cos x| \, dx$
Since the area is bounded by the x-axis,we take the absolute value of the functions where they are negative.
$A = \int_{0}^{\pi/4} \cos x \, dx + \int_{\pi/4}^{\pi} \sin x \, dx + \int_{\pi}^{5\pi/4} (-\sin x) \, dx + \int_{5\pi/4}^{3\pi/2} (-\cos x) \, dx$
$A = [\sin x]_{0}^{\pi/4} + [-\cos x]_{\pi/4}^{\pi} + [\cos x]_{\pi}^{5\pi/4} + [-\sin x]_{5\pi/4}^{3\pi/2}$
$A = (\frac{1}{\sqrt{2}} - 0) + (-(-1) - (-\frac{1}{\sqrt{2}})) + (-\frac{1}{\sqrt{2}} - (-1)) + (-(-1) - (-\frac{1}{\sqrt{2}}))$
$A = \frac{1}{\sqrt{2}} + 1 + \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} + 1 + 1 - \frac{1}{\sqrt{2}} = 3$
Thus,$A+A^2 = 3 + 3^2 = 3 + 9 = 12$.

(B) Given $|A|=6$ and $A$ is a $3 \times 3$ matrix.
Using the property $|adj(k A)| = k^{n-1} |adj(A)|$ and $adj(kA) = k^{n-1} adj(A)$ for a $n \times n$ matrix.
First,$adj(2A) = 2^{3-1} adj(A) = 4 adj(A)$.
Then,$A^2 \cdot adj(2A) = A^2 \cdot 4 adj(A) = 4 A (A \cdot adj(A)) = 4 A |A| I_3 = 4 \cdot 6 \cdot A = 24A$.
Now,$3 adj(24A) = 3 \cdot 24^{3-1} adj(A) = 3 \cdot 24^2 adj(A) = 3 \cdot (2^3 \cdot 3)^2 adj(A) = 3 \cdot 2^6 \cdot 3^2 adj(A) = 2^6 \cdot 3^3 adj(A)$.
Let $K = 2^6 \cdot 3^3$. Then we need $|adj(K adj(A))|$.
Using $|adj(M)| = |M|^{n-1} = |M|^2$ for a $3 \times 3$ matrix $M$:
$|adj(K adj(A))| = |K adj(A)|^2 = K^6 |adj(A)|^2 = K^6 (|A|^{3-1})^2 = K^6 |A|^4$.
Substituting $K = 2^6 \cdot 3^3$ and $|A|=6 = 2^1 \cdot 3^1$:
$|adj(K adj(A))| = (2^6 \cdot 3^3)^6 \cdot (2^1 \cdot 3^1)^4 = (2^{36} \cdot 3^{18}) \cdot (2^4 \cdot 3^4) = 2^{40} \cdot 3^{22}$.
Thus,$m=40$ and $n=22$.
Therefore,$m+n = 40+22 = 62$.

(D) Given equations for direction cosines $(l, m, n)$ are:
$4l + m - n = 0 \implies n = 4l + m$ ... $(1)$
$2mn + 10nl + 3lm = 0$ ... $(2)$
Substitute $n = 4l + m$ into equation $(2)$:
$2m(4l + m) + 10l(4l + m) + 3lm = 0$
$8lm + 2m^2 + 40l^2 + 10lm + 3lm = 0$
$40l^2 + 21lm + 2m^2 = 0$
$(8l + m)(5l + 2m) = 0$
Case $1$: $m = -8l$. Then $n = 4l - 8l = -4l$. Direction ratios are $(l, -8l, -4l)$ or $(1, -8, -4)$.
Case $2$: $m = -\frac{5}{2}l$. Then $n = 4l - \frac{5}{2}l = \frac{3}{2}l$. Direction ratios are $(l, -\frac{5}{2}l, \frac{3}{2}l)$ or $(2, -5, 3)$.
Let the direction vectors be $\vec{a} = (1, -8, -4)$ and $\vec{b} = (2, -5, 3)$.
$\cos \theta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|} = \frac{|(1)(2) + (-8)(-5) + (-4)(3)|}{\sqrt{1^2 + (-8)^2 + (-4)^2} \sqrt{2^2 + (-5)^2 + 3^2}}$
$\cos \theta = \frac{|2 + 40 - 12|}{\sqrt{1 + 64 + 16} \sqrt{4 + 25 + 9}} = \frac{30}{\sqrt{81} \sqrt{38}} = \frac{30}{9 \sqrt{38}} = \frac{10}{3 \sqrt{38}}$.

(A) The line equation is $\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3} = \lambda$. Any point on this line is $P(\lambda, 2\lambda+1, 3\lambda+2)$.
Since $B(4, 9, \alpha)$ lies on the line,we have $\frac{4}{1} = \frac{9-1}{2} = \frac{\alpha-2}{3} \Rightarrow 4 = 4 = \frac{\alpha-2}{3} \Rightarrow \alpha = 14$.
Let $AD$ be the altitude from $A(1, 6, 3)$ to the line $BC$. $D$ is the projection of $A$ on the line,so $D(\lambda, 2\lambda+1, 3\lambda+2)$.
The vector $\vec{AD} = (\lambda-1)\hat{i} + (2\lambda+1-6)\hat{j} + (3\lambda+2-3)\hat{k} = (\lambda-1)\hat{i} + (2\lambda-5)\hat{j} + (3\lambda-1)\hat{k}$.
Since $\vec{AD}$ is perpendicular to the line direction vector $\vec{v} = \hat{i} + 2\hat{j} + 3\hat{k}$,their dot product is zero:
$(\lambda-1)(1) + (2\lambda-5)(2) + (3\lambda-1)(3) = 0$
$\lambda - 1 + 4\lambda - 10 + 9\lambda - 3 = 0 \Rightarrow 14\lambda = 14 \Rightarrow \lambda = 1$.
Thus,$D = (1, 2(1)+1, 3(1)+2) = (1, 3, 5)$.
The length of altitude $AD = \sqrt{(1-1)^2 + (3-6)^2 + (5-3)^2} = \sqrt{0 + 9 + 4} = \sqrt{13}$.
The area of $\Delta ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times BC \times AD = \frac{1}{2} \times 10 \times \sqrt{13} = 5\sqrt{13}$ sq. units.

(A) For statement $I$: Let $\cos \alpha = x, \cos \beta = y, \cos \gamma = z$.
The given equation is $\begin{vmatrix} 1 & x & y \\ x & 1 & z \\ y & z & 1 \end{vmatrix} = \begin{vmatrix} 0 & x & y \\ x & 0 & z \\ y & z & 0 \end{vmatrix}$.
Expanding the left determinant: $1(1-z^2) - x(x-yz) + y(xz-y) = 1 - z^2 - x^2 + xyz + xyz - y^2 = 1 - (x^2+y^2+z^2) + 2xyz$.
Expanding the right determinant: $0(0-z^2) - x(0-yz) + y(xz-0) = xyz + xyz = 2xyz$.
Equating both sides: $1 - (x^2+y^2+z^2) + 2xyz = 2xyz \implies x^2+y^2+z^2 = 1$.
Thus,$\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 \neq \frac{3}{2}$. Statement $I$ is false.
For statement $II$: Let $f(x) = \begin{vmatrix} x^{2}+x & x+1 & x-2 \\ 2x^{2}+3x-1 & 3x & 3x-3 \\ x^{2}+2x+3 & 2x-1 & 2x-1 \end{vmatrix} = px+q$.
Setting $x=0$: $q = \begin{vmatrix} 0 & 1 & -2 \\ -1 & 0 & -3 \\ 3 & -1 & -1 \end{vmatrix} = 0(0-3) - 1(1+9) - 2(1-0) = -10 - 2 = -12$.
Setting $x=1$: $p+q = \begin{vmatrix} 2 & 2 & -1 \\ 4 & 3 & 3 \\ 6 & 1 & 1 \end{vmatrix} = 2(3-3) - 2(4-18) - 1(4-18) = 0 + 28 + 14 = 42$.
Since $q = -12$,$p - 12 = 42 \implies p = 54$.
Checking $p^2 = 196q^2$: $54^2 = 2916$ and $196(-12)^2 = 196 \times 144 = 28224$.
Since $2916 \neq 28224$,statement $II$ is false.

(A) Let $I = \int_{\frac{\pi}{24}}^{\frac{5\pi}{24}} \frac{dx}{1+\sqrt[3]{\tan 2x}}$ ...$(1)$
Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,where $a+b = \frac{\pi}{24} + \frac{5\pi}{24} = \frac{6\pi}{24} = \frac{\pi}{4}$.
$I = \int_{\frac{\pi}{24}}^{\frac{5\pi}{24}} \frac{dx}{1+\sqrt[3]{\tan(2(\frac{\pi}{4}-x))}}$
$I = \int_{\frac{\pi}{24}}^{\frac{5\pi}{24}} \frac{dx}{1+\sqrt[3]{\tan(\frac{\pi}{2}-2x)}}$
Since $\tan(\frac{\pi}{2}-\theta) = \cot \theta$,we have:
$I = \int_{\frac{\pi}{24}}^{\frac{5\pi}{24}} \frac{dx}{1+\sqrt[3]{\cot 2x}} = \int_{\frac{\pi}{24}}^{\frac{5\pi}{24}} \frac{\sqrt[3]{\tan 2x} dx}{\sqrt[3]{\tan 2x} + 1}$ ...$(2)$
Adding $(1)$ and $(2)$:
$2I = \int_{\frac{\pi}{24}}^{\frac{5\pi}{24}} \frac{1+\sqrt[3]{\tan 2x}}{1+\sqrt[3]{\tan 2x}} dx = \int_{\frac{\pi}{24}}^{\frac{5\pi}{24}} 1 dx$
$2I = [x]_{\frac{\pi}{24}}^{\frac{5\pi}{24}} = \frac{5\pi}{24} - \frac{\pi}{24} = \frac{4\pi}{24} = \frac{\pi}{6}$
$I = \frac{\pi}{12}$

(B) For $f(x)$ to be continuous at $x = -\frac{3}{2}$,the limit $\lim_{x \to -\frac{3}{2}} f(x)$ must exist and equal $f(-\frac{3}{2}) = b$.
Since the denominator $(2x-1)(2x+3)$ approaches $0$ as $x \to -\frac{3}{2}$,the numerator $ax^2 + 2ax + 3$ must also approach $0$.
$a(-\frac{3}{2})^2 + 2a(-\frac{3}{2}) + 3 = 0$ $\Rightarrow \frac{9a}{4} - 3a + 3 = 0$ $\Rightarrow -\frac{3a}{4} = -3$ $\Rightarrow a = 4$.
Substituting $a=4$,$f(x) = \frac{4x^2+8x+3}{(2x-1)(2x+3)} = \frac{(2x+1)(2x+3)}{(2x-1)(2x+3)} = \frac{2x+1}{2x-1}$ for $x \neq -\frac{3}{2}, \frac{1}{2}$.
Now,$f(f(x)) = f\left(\frac{2x+1}{2x-1}\right) = \frac{2(\frac{2x+1}{2x-1}) + 1}{2(\frac{2x+1}{2x-1}) - 1} = \frac{4x+2+2x-1}{4x+2-2x+1} = \frac{6x+1}{2x+3}$.
Given $f(f(x)) = \frac{7}{5}$,we have $\frac{6x+1}{2x+3} = \frac{7}{5}$.
$5(6x+1) = 7(2x+3)$ $\Rightarrow 30x + 5 = 14x + 21$ $\Rightarrow 16x = 16$ $\Rightarrow x = 1$.

(A) The given differential equation is $x^{4}dy + (4x^{3}y + 2\sin x)dx = 0$.
Rearranging the terms,we get $x^{4}dy + 4x^{3}ydx = -2\sin x dx$.
This can be written as $d(x^{4}y) = -2\sin x dx$.
Integrating both sides,we get $\int d(x^{4}y) = \int -2\sin x dx$,which gives $x^{4}y = 2\cos x + C$.
Given the condition $y(\frac{\pi}{2}) = 0$,we substitute $x = \frac{\pi}{2}$ and $y = 0$:
$(\frac{\pi}{2})^{4}(0) = 2\cos(\frac{\pi}{2}) + C
\Rightarrow 0 = 2(0) + C
\Rightarrow C = 0$.
So,the solution is $x^{4}y = 2\cos x$.
Now,we need to find $\pi^{4}y(\frac{\pi}{3})$.
At $x = \frac{\pi}{3}$,we have $(\frac{\pi}{3})^{4} y(\frac{\pi}{3}) = 2\cos(\frac{\pi}{3})$.
Since $\cos(\frac{\pi}{3}) = \frac{1}{2}$,we get $\frac{\pi^{4}}{81} y(\frac{\pi}{3}) = 2(\frac{1}{2}) = 1$.
Therefore,$\pi^{4} y(\frac{\pi}{3}) = 81$.

(C) Given $A = \{-2, -1, 0, 1, 2, 3, 4\}$ and $xRy \iff 2x + y \le 2$.
For each $x \in A$,we find $y \in A$ such that $y \le 2 - 2x$:
- If $x = -2$,$y \le 6 \implies y \in \{-2, -1, 0, 1, 2, 3, 4\}$ ($7$ elements).
- If $x = -1$,$y \le 4 \implies y \in \{-2, -1, 0, 1, 2, 3, 4\}$ ($7$ elements).
- If $x = 0$,$y \le 2 \implies y \in \{-2, -1, 0, 1, 2\}$ ($5$ elements).
- If $x = 1$,$y \le 0 \implies y \in \{-2, -1, 0\}$ ($3$ elements).
- If $x = 2$,$y \le -2 \implies y \in \{-2\}$ ($1$ element).
- If $x = 3$,$y \le -4 \implies$ no $y \in A$.
- If $x = 4$,$y \le -6 \implies$ no $y \in A$.
Total elements $l = 7 + 7 + 5 + 3 + 1 = 23$.
For reflexivity,we need $(x, x) \in R$ for all $x \in A$. Check: $(-2, -2), (-1, -1), (0, 0)$ are in $R$. $(1, 1), (2, 2), (3, 3), (4, 4)$ are missing. So $m = 4$.
For symmetry,if $(x, y) \in R$,then $(y, x)$ must be in $R$. The elements $(x, y)$ in $R$ where $(y, x) \notin R$ are: $(3, -2), (4, -2), (2, -1), (2, 0), (3, -1), (4, -1)$. There are $6$ such pairs. So $n = 6$.
Thus,$l + m + n = 23 + 4 + 6 = 33$.