JEE Main 2020 Mathematics Question Paper with Answer and Solution

(B) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $a > b$.
Given the length of the minor axis is $2b = \frac{4}{\sqrt{3}}$,so $b = \frac{2}{\sqrt{3}}$ and $b^2 = \frac{4}{3}$.
The line $x + 6y = 8$ can be written as $y = -\frac{1}{6}x + \frac{4}{3}$.
The condition for the line $y = mx + c$ to be a tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $c^2 = a^2m^2 + b^2$.
Here $m = -\frac{1}{6}$ and $c = \frac{4}{3}$.
Substituting the values: $(\frac{4}{3})^2 = a^2(-\frac{1}{6})^2 + \frac{4}{3}$.
$\frac{16}{9} = \frac{a^2}{36} + \frac{4}{3}$.
$\frac{a^2}{36} = \frac{16}{9} - \frac{12}{9} = \frac{4}{9}$.
$a^2 = \frac{4 \times 36}{9} = 16$,so $a = 4$.
The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{4/3}{16}} = \sqrt{1 - \frac{1}{12}} = \sqrt{\frac{11}{12}} = \sqrt{\frac{11}{4 \times 3}} = \frac{1}{2} \sqrt{\frac{11}{3}}$.

(B) For the equation $a x^{2}-2 b x+5=0$,the roots are $\alpha, \alpha$.
Thus,the sum of roots $2\alpha = \frac{2b}{a} \Rightarrow \alpha = \frac{b}{a}$ and the product of roots $\alpha^{2} = \frac{5}{a}$.
From these,$b = a\alpha$ and $a = \frac{5}{\alpha^{2}}$. Substituting $a$,we get $b = \frac{5}{\alpha}$.
Since $\alpha$ is also a root of $x^{2}-2 b x-10=0$,we have $\alpha^{2}-2 b \alpha-10=0$.
Substituting $b = \frac{5}{\alpha}$ into this equation: $\alpha^{2}-2(\frac{5}{\alpha})\alpha-10=0$ $\Rightarrow \alpha^{2}-10-10=0$ $\Rightarrow \alpha^{2}=20$.
Now,for the equation $x^{2}-2 b x-10=0$,the product of roots $\alpha \beta = -10$.
Since $\alpha^{2} = 20$,we have $\alpha = \pm \sqrt{20}$.
Then $\beta = \frac{-10}{\alpha}$.
Thus,$\beta^{2} = \frac{100}{\alpha^{2}} = \frac{100}{20} = 5$.
Therefore,$\alpha^{2}+\beta^{2} = 20 + 5 = 25$.

(C) The expression for $x$ is a geometric series with first term $a = 1$ and common ratio $r = -\tan^2 \theta$.
Since $0 < \theta < \frac{\pi}{4}$,we have $0 < \tan^2 \theta < 1$,so the series converges to $x = \frac{1}{1 - (-\tan^2 \theta)} = \frac{1}{1 + \tan^2 \theta} = \frac{1}{\sec^2 \theta} = \cos^2 \theta$.
The expression for $y$ is a geometric series with first term $a = 1$ and common ratio $r = \cos^2 \theta$.
Since $0 < \theta < \frac{\pi}{4}$,we have $\frac{1}{2} < \cos^2 \theta < 1$,so the series converges to $y = \frac{1}{1 - \cos^2 \theta} = \frac{1}{\sin^2 \theta}$.
From $x = \cos^2 \theta$,we have $\sin^2 \theta = 1 - x$.
Substituting this into the expression for $y$,we get $y = \frac{1}{1 - x}$.
Therefore,$y(1 - x) = 1$.

(B) The equation of the parabola is $y^{2}=8x$,which is of the form $y^{2}=4ax$,where $4a=8$,so $a=2$.
The coordinates of any point on the parabola are $(at^{2}, 2at) = (2t^{2}, 4t)$.
For point $A\left(\frac{1}{2}, -2\right)$,we have $4t_{1}=-2$,which gives $t_{1}=-\frac{1}{2}$.
For a focal chord,the product of the parameters of the endpoints is $t_{1}t_{2}=-1$.
Substituting $t_{1}=-\frac{1}{2}$,we get $t_{2} = -\frac{1}{t_{1}} = -\frac{1}{-1/2} = 2$.
The coordinates of point $B$ are $(2t_{2}^{2}, 4t_{2}) = (2(2)^{2}, 4(2)) = (8, 8)$.
The equation of the tangent to the parabola $y^{2}=4ax$ at point $(x_{1}, y_{1})$ is $yy_{1}=2a(x+x_{1})$.
For point $B(8, 8)$ and $a=2$,the equation of the tangent is $y(8) = 2(2)(x+8)$.
$8y = 4(x+8)$ $\Rightarrow 2y = x+8$ $\Rightarrow x-2y+8=0$.

(C) Given $A = \{x \in R : |x| < 2\} = (-2, 2)$.
Given $B = \{x \in R : |x - 2| \geq 3\}$.
This implies $x - 2 \geq 3$ or $x - 2 \leq -3$.
So,$x \geq 5$ or $x \leq -1$.
Thus,$B = (-\infty, -1] \cup [5, \infty)$.
Now,$B - A$ is the set of elements in $B$ that are not in $A$.
$B - A = ((-\infty, -1] \cup [5, \infty)) - (-2, 2)$.
Since $(-2, 2)$ overlaps with $(-\infty, -1]$ only on the interval $(-2, -1]$,we remove this part.
$B - A = (-\infty, -2] \cup [5, \infty)$.
This can be written as $R - (-2, 5)$.
Therefore,the correct option is $C$.

(D) Let $z = x + iy.$ The given equation is $|x| + |y| = 4.$
This represents a square in the complex plane with vertices at $(4, 0), (0, 4), (-4, 0),$ and $(0, -4).$
$|z| = \sqrt{x^2 + y^2}$ represents the distance of a point $(x, y)$ from the origin $(0, 0).$
The minimum distance from the origin to the line segment connecting $(4, 0)$ and $(0, 4)$ is the perpendicular distance from $(0, 0)$ to the line $x + y = 4.$
Using the formula for the distance from a point $(x_0, y_0)$ to the line $Ax + By + C = 0,$ we get $d = \frac{|0 + 0 - 4|}{\sqrt{1^2 + 1^2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2} = \sqrt{8}.$
The maximum distance from the origin to the square occurs at the vertices,which is $4 = \sqrt{16}.$
Thus,$|z|$ must lie in the interval $[\sqrt{8}, \sqrt{16}].$
Since $\sqrt{7} < \sqrt{8},$ $|z|$ cannot be $\sqrt{7}.$

(B) The implication $p \rightarrow (p \wedge \neg q)$ is false only when the antecedent $p$ is $T$ and the consequent $(p \wedge \neg q)$ is $F$.
Since $p$ is $T$,the expression $(p \wedge \neg q)$ becomes $(T \wedge \neg q)$.
For $(T \wedge \neg q)$ to be $F$,$\neg q$ must be $F$,which implies $q$ is $T$.
Therefore,the truth values are $p = T$ and $q = T$.

(D) The general term is given by $T_{r+1} = ^{16}C_{r} \left(\frac{x}{\cos \theta}\right)^{16-r} \left(\frac{1}{x \sin \theta}\right)^{r}$.
Simplifying,$T_{r+1} = ^{16}C_{r} x^{16-2r} \frac{1}{(\cos \theta)^{16-r} (\sin \theta)^{r}}$.
For the term to be independent of $x$,we set $16-2r = 0$,which gives $r = 8$.
Thus,the independent term is $T_{9} = ^{16}C_{8} \frac{1}{\cos^{8} \theta \sin^{8} \theta} = ^{16}C_{8} \frac{2^{8}}{(\sin 2\theta)^{8}}$.
Let $f(\theta) = \frac{^{16}C_{8} \cdot 2^{8}}{(\sin 2\theta)^{8}}$.
For $\theta \in [\frac{\pi}{8}, \frac{\pi}{4}]$,$\sin 2\theta$ is increasing,so $f(\theta)$ is least when $\sin 2\theta$ is maximum,i.e.,at $\theta = \frac{\pi}{4}$. Thus,$\ell_{1} = f(\frac{\pi}{4}) = ^{16}C_{8} \cdot 2^{8}$.
For $\theta \in [\frac{\pi}{16}, \frac{\pi}{8}]$,$f(\theta)$ is least when $\sin 2\theta$ is maximum,i.e.,at $\theta = \frac{\pi}{8}$. Thus,$\ell_{2} = f(\frac{\pi}{8}) = ^{16}C_{8} \frac{2^{8}}{(\sin \frac{\pi}{4})^{8}} = ^{16}C_{8} \frac{2^{8}}{(1/\sqrt{2})^{8}} = ^{16}C_{8} \cdot 2^{8} \cdot 2^{4} = ^{16}C_{8} \cdot 2^{12}$.
The ratio $\frac{\ell_{2}}{\ell_{1}} = \frac{^{16}C_{8} \cdot 2^{12}}{^{16}C_{8} \cdot 2^{8}} = 2^{4} = 16$.

(D) Given that $a_n$ is a $G$.$P$. with common ratio $r$.
$\sum_{n=1}^{100} a_{2n+1} = a_3 + a_5 + \dots + a_{201} = 200$
This is a $G$.$P$. with first term $a_3 = ar^2$ and common ratio $r^2$ with $100$ terms.
So,$ar^2 \frac{(r^2)^{100} - 1}{r^2 - 1} = ar^2 \frac{r^{200} - 1}{r^2 - 1} = 200$
$\sum_{n=1}^{100} a_{2n} = a_2 + a_4 + \dots + a_{200} = 100$
This is a $G$.$P$. with first term $a_2 = ar$ and common ratio $r^2$ with $100$ terms.
So,$ar \frac{(r^2)^{100} - 1}{r^2 - 1} = ar \frac{r^{200} - 1}{r^2 - 1} = 100$
Dividing the two equations: $\frac{ar^2 \frac{r^{200} - 1}{r^2 - 1}}{ar \frac{r^{200} - 1}{r^2 - 1}} = \frac{200}{100} \Rightarrow r = 2$
We need to find $S = \sum_{n=1}^{200} a_n = a_1 + a_2 + \dots + a_{200}$.
Note that $\sum_{n=1}^{100} a_{2n+1} + \sum_{n=1}^{100} a_{2n} = a_2 + a_3 + a_4 + \dots + a_{201} = 300$.
Since $a_{k+1} = r a_k$,we have $a_2 + a_3 + \dots + a_{201} = r(a_1 + a_2 + \dots + a_{200}) = 300$.
Substituting $r = 2$: $2 \sum_{n=1}^{200} a_n = 300 \Rightarrow \sum_{n=1}^{200} a_n = 150$.

(D) The first $A$.$P$. is $A_1: 3, 7, 11, \ldots, 407$. Here,$a_1 = 3$ and $d_1 = 4$. The general term is $T_n = 3 + (n-1)4 = 4n - 1$.
The second $A$.$P$. is $A_2: 2, 9, 16, \ldots, 709$. Here,$a_2 = 2$ and $d_2 = 7$. The general term is $T_m = 2 + (m-1)7 = 7m - 5$.
For a common term,$4n - 1 = 7m - 5$,which implies $4n = 7m - 4$. This means $7m$ must be a multiple of $4$. Since $7$ is not divisible by $4$,$m$ must be a multiple of $4$. Let $m = 4k$.
Substituting $m = 4k$,we get $4n = 7(4k) - 4$,so $n = 7k - 1$.
The first common term is for $k=1$,which is $m=4$,$T_4 = 7(4)-5 = 23$.
The common difference of the new $A$.$P$. is $\text{lcm}(4, 7) = 28$.
The common terms are $23, 51, 79, \ldots$. The last term must be $\leq 407$.
$23 + (N-1)28 \leq 407$
$(N-1)28 \leq 384$
$N-1 \leq 13.71$
$N \leq 14.71$.
Thus,the number of common terms is $N = 14$.

(C) Let $S = \sum_{r=0}^{25} (4r + 1) \cdot ^{25}C_{r}$.
Using the property $^{n}C_{r} = ^{n}C_{n-r}$,we can write:
$S = 1 \cdot ^{25}C_{0} + 5 \cdot ^{25}C_{1} + 9 \cdot ^{25}C_{2} + \ldots + 101 \cdot ^{25}C_{25}$.
Reversing the order:
$S = 101 \cdot ^{25}C_{25} + 97 \cdot ^{25}C_{24} + \ldots + 1 \cdot ^{25}C_{0}$.
Adding the two expressions:
$2S = \sum_{r=0}^{25} (4r + 1 + 101 - 4r) \cdot ^{25}C_{r} = \sum_{r=0}^{25} (102) \cdot ^{25}C_{r}$.
$2S = 102 \sum_{r=0}^{25} {^{25}C_{r}} = 102 \cdot 2^{25}$.
$S = 51 \cdot 2^{25}$.
Comparing with $2^{25} \cdot k$,we get $k = 51$.

(B) The given equations of the circles are $S_{1}: x^{2}+y^{2}-6x+8=0$ and $S_{2}: x^{2}+y^{2}-8y+16-k=0$.
For $S_{1}$,the center $C_{1} = (3, 0)$ and radius $r_{1} = \sqrt{3^{2}+0^{2}-8} = \sqrt{1} = 1$.
For $S_{2}$,the center $C_{2} = (0, 4)$ and radius $r_{2} = \sqrt{0^{2}+4^{2}-(16-k)} = \sqrt{k}$.
The distance between the centers is $d = \sqrt{(3-0)^{2}+(0-4)^{2}} = \sqrt{9+16} = 5$.
Since the circles touch each other,the distance between centers must be equal to the sum or difference of the radii: $d = |r_{1} \pm r_{2}|$.
Case $1$: $d = r_{1}+r_{2}$ $\Rightarrow 5 = 1+\sqrt{k}$ $\Rightarrow \sqrt{k} = 4$ $\Rightarrow k = 16$.
Case $2$: $d = |r_{1}-r_{2}| \Rightarrow 5 = |1-\sqrt{k}|$.
This implies $1-\sqrt{k} = 5$ or $1-\sqrt{k} = -5$.
$1-\sqrt{k} = 5 \Rightarrow \sqrt{k} = -4$ (not possible as $k>0$).
$1-\sqrt{k} = -5$ $\Rightarrow \sqrt{k} = 6$ $\Rightarrow k = 36$.
Comparing the values,the largest value of $k$ is $36$.

(A) five-digit number is represented as $\_ \;\_\;\_\;\underline{2}\;\_$.
The $10^{\text{th}}$ place is fixed as $2$.
For the $10,000^{\text{th}}$ place,we cannot use $0$ or $2$,so there are $8$ choices $(1, 3, 4, 5, 6, 7, 8, 9)$.
For the $1,000^{\text{th}}$ place,we can use $0$ and any of the remaining $7$ digits,so there are $8$ choices.
For the $100^{\text{th}}$ place,there are $7$ choices remaining.
For the units place,there are $6$ choices remaining.
Total number of such five-digit numbers $= 8 \times 8 \times 7 \times 6 = 2688$.
Given that the number is $336k$,we have $336k = 2688$.
Therefore,$k = \frac{2688}{336} = 8$.

(C) Given $\left|\frac{z-i}{z+2i}\right|=1$,which implies $|z-i|=|z+2i|$.
This means $z$ lies on the perpendicular bisector of the points $(0, 1)$ and $(0, -2)$.
The perpendicular bisector is the line $\text{Im}(z) = -\frac{1}{2}$.
Let $z = x - \frac{i}{2}$.
Given $|z| = \frac{5}{2}$,we have $x^2 + (-\frac{1}{2})^2 = (\frac{5}{2})^2$.
$x^2 + \frac{1}{4} = \frac{25}{4} \Rightarrow x^2 = 6$.
Now,$|z+3i| = |x - \frac{i}{2} + 3i| = |x + \frac{5i}{2}|$.
$|z+3i| = \sqrt{x^2 + (\frac{5}{2})^2} = \sqrt{6 + \frac{25}{4}} = \sqrt{\frac{24+25}{4}} = \sqrt{\frac{49}{4}} = \frac{7}{2}$.

(D) Given equation: $e^{4x} + e^{3x} - 4e^{2x} + e^x + 1 = 0$.
Divide the entire equation by $e^{2x}$ (since $e^{2x} \neq 0$):
$e^{2x} + e^x - 4 + \frac{1}{e^x} + \frac{1}{e^{2x}} = 0$.
Rearrange the terms:
$(e^{2x} + \frac{1}{e^{2x}}) + (e^x + \frac{1}{e^x}) - 4 = 0$.
Using the identity $a^2 + \frac{1}{a^2} = (a + \frac{1}{a})^2 - 2$,we get:
$(e^x + \frac{1}{e^x})^2 - 2 + (e^x + \frac{1}{e^x}) - 4 = 0$.
Let $t = e^x + \frac{1}{e^x}$. Since $e^x > 0$,by $AM$-$GM$ inequality,$t = e^x + \frac{1}{e^x} \geq 2$.
The equation becomes $t^2 + t - 6 = 0$.
Factoring the quadratic: $(t + 3)(t - 2) = 0$.
This gives $t = -3$ or $t = 2$.
Since $t \geq 2$,we must have $t = 2$.
$e^x + \frac{1}{e^x} = 2$ $\Rightarrow e^{2x} - 2e^x + 1 = 0$ $\Rightarrow (e^x - 1)^2 = 0$.
$e^x = 1 \Rightarrow x = 0$.
Thus,there is only $1$ real root.

(B) Let $p$ be the statement: $\sqrt{5}$ is an integer.
Let $q$ be the statement: $5$ is irrational.
The given statement is $p \vee q$.
The negation of the statement is $\sim(p \vee q)$.
By De Morgan's Law,$\sim(p \vee q) \equiv \sim p \wedge \sim q$.
Thus,the negation is: $\sqrt{5}$ is not an integer $AND$ $5$ is not irrational.

(D) Let $y_{i} = x_{i} - 5$. Then $\sum_{i=1}^{10} y_{i} = 10$ and $\sum_{i=1}^{10} y_{i}^{2} = 40$.
Mean of $y_{i}$ is $\bar{y} = \frac{1}{10} \sum y_{i} = \frac{10}{10} = 1$.
Variance of $y_{i}$ is $\sigma_{y}^{2} = \frac{1}{10} \sum y_{i}^{2} - (\bar{y})^{2} = \frac{40}{10} - (1)^{2} = 4 - 1 = 3$.
Now,let $z_{i} = x_{i} - 3$. We can write $z_{i} = (x_{i} - 5) + 2 = y_{i} + 2$.
The mean $\mu$ of $z_{i}$ is $\bar{z} = \bar{y} + 2 = 1 + 2 = 3$.
The variance $\lambda$ of $z_{i}$ is $\text{Var}(y_{i} + 2) = \text{Var}(y_{i}) = 3$.
Thus,the ordered pair $(\mu, \lambda) = (3, 3)$.

(A) The given expression is $P = 2^{\frac{1}{4}} \cdot 4^{\frac{1}{16}} \cdot 8^{\frac{1}{48}} \cdot 16^{\frac{1}{128}} \cdot \dots \infty$.
Expressing all terms with base $2$:
$P = 2^{\frac{1}{4}} \cdot (2^2)^{\frac{1}{16}} \cdot (2^3)^{\frac{1}{48}} \cdot (2^4)^{\frac{1}{128}} \cdot \dots$
$P = 2^{\frac{1}{4}} \cdot 2^{\frac{2}{16}} \cdot 2^{\frac{3}{48}} \cdot 2^{\frac{4}{128}} \cdot \dots$
$P = 2^{\frac{1}{4}} \cdot 2^{\frac{1}{8}} \cdot 2^{\frac{1}{16}} \cdot 2^{\frac{1}{32}} \cdot \dots$
Using the property $a^m \cdot a^n = a^{m+n}$,we get:
$P = 2^{(\frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \dots)}$
The exponent is an infinite geometric series with first term $a = \frac{1}{4}$ and common ratio $r = \frac{1}{2}$.
The sum of an infinite geometric series is $S = \frac{a}{1-r}$.
$S = \frac{1/4}{1 - 1/2} = \frac{1/4}{1/2} = \frac{1}{2}$.
Therefore,$P = 2^{\frac{1}{2}}$.

(C) The equation of a circle touching the $y$-axis at $(0,4)$ is $(x-0)^2 + (y-4)^2 + \lambda x = 0$.
Since it passes through $(2,0)$,we substitute $x=2$ and $y=0$:
$(2-0)^2 + (0-4)^2 + \lambda(2) = 0$ $\Rightarrow 4 + 16 + 2\lambda = 0$ $\Rightarrow 2\lambda = -20$ $\Rightarrow \lambda = -10$.
Thus,the circle equation is $x^2 + (y-4)^2 - 10x = 0$,which simplifies to $x^2 + y^2 - 10x - 8y + 16 = 0$.
The center is $(5,4)$ and the radius $r = \sqrt{5^2 + 4^2 - 16} = \sqrt{25 + 16 - 16} = 5$.
$A$ line $ax + by + c = 0$ is a tangent if the perpendicular distance from the center $(5,4)$ to the line equals the radius $5$.
For $A: |3(5) - 4(4) - 24| / \sqrt{3^2 + (-4)^2} = |15 - 16 - 24| / 5 = |-25| / 5 = 5$ (Tangent).
For $B: |3(5) + 4(4) - 6| / \sqrt{3^2 + 4^2} = |15 + 16 - 6| / 5 = |25| / 5 = 5$ (Tangent).
For $C: |4(5) + 3(4) - 8| / \sqrt{4^2 + 3^2} = |20 + 12 - 8| / 5 = |24| / 5 = 4.8 \neq 5$ (Not a tangent).
For $D: |4(5) - 3(4) + 17| / \sqrt{4^2 + (-3)^2} = |20 - 12 + 17| / 5 = |25| / 5 = 5$ (Tangent).

(D) For the ellipse $\frac{x^{2}}{18}+\frac{y^{2}}{4}=1$,we have $a^{2}=18$ and $b^{2}=4$. The eccentricity $e_{1} = \sqrt{1-\frac{b^{2}}{a^{2}}} = \sqrt{1-\frac{4}{18}} = \sqrt{\frac{14}{18}} = \sqrt{\frac{7}{9}} = \frac{\sqrt{7}}{3}$.
For the hyperbola $\frac{x^{2}}{9}-\frac{y^{2}}{4}=1$,we have $a^{2}=9$ and $b^{2}=4$. The eccentricity $e_{2} = \sqrt{1+\frac{b^{2}}{a^{2}}} = \sqrt{1+\frac{4}{9}} = \sqrt{\frac{13}{9}} = \frac{\sqrt{13}}{3}$.
Since the point $(e_{1}, e_{2})$ lies on the ellipse $15x^{2}+3y^{2}=k$,we substitute the values of $e_{1}$ and $e_{2}$ into the equation:
$15(\frac{\sqrt{7}}{3})^{2} + 3(\frac{\sqrt{13}}{3})^{2} = k$
$15(\frac{7}{9}) + 3(\frac{13}{9}) = k$
$\frac{105}{9} + \frac{39}{9} = k$
$\frac{144}{9} = k$
$k = 16$.

(B) The centroid $C$ of the triangle with vertices $(x_1, y_1), (x_2, y_2),$ and $(x_3, y_3)$ is given by $(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}).$
For the given vertices $(3, -1), (1, 3),$ and $(2, 4),$ the centroid $C$ is $(\frac{3+1+2}{3}, \frac{-1+3+4}{3}) = (2, 2).$
Next,we find the intersection point $P$ of the lines $x + 3y - 1 = 0$ and $3x - y + 1 = 0.$
Solving the system:
$x + 3y = 1$
$3x - y = -1 \Rightarrow y = 3x + 1$
Substituting $y$ into the first equation: $x + 3(3x + 1) = 1$ $\Rightarrow x + 9x + 3 = 1$ $\Rightarrow 10x = -2$ $\Rightarrow x = -\frac{1}{5}.$
Then $y = 3(-\frac{1}{5}) + 1 = -\frac{3}{5} + 1 = \frac{2}{5}.$
So,$P = (-\frac{1}{5}, \frac{2}{5}).$
The line passing through $C(2, 2)$ and $P(-\frac{1}{5}, \frac{2}{5})$ has slope $m = \frac{\frac{2}{5} - 2}{-\frac{1}{5} - 2} = \frac{-\frac{8}{5}}{-\frac{11}{5}} = \frac{8}{11}.$
The equation of the line is $y - 2 = \frac{8}{11}(x - 2)$ $\Rightarrow 11y - 22 = 8x - 16$ $\Rightarrow 8x - 11y + 6 = 0.$
Checking the options,for $(-9, -6): 8(-9) - 11(-6) + 6 = -72 + 66 + 6 = 0.$
Thus,the line passes through $(-9, -6).$

(C) Let $\theta = \frac{\pi}{8}$. Then $\frac{3\pi}{8} = 3\theta$.
Given expression is $\cos^{3}(\theta) \cos(3\theta) + \sin^{3}(\theta) \sin(3\theta)$.
Using the identities $\cos(3\theta) = 4\cos^{3}(\theta) - 3\cos(\theta)$ and $\sin(3\theta) = 3\sin(\theta) - 4\sin^{3}(\theta)$,we can rewrite the expression.
Alternatively,note that $\cos(\frac{3\pi}{8}) = \sin(\frac{\pi}{8})$ and $\sin(\frac{3\pi}{8}) = \cos(\frac{\pi}{8})$.
Substituting these: $\cos^{3}(\frac{\pi}{8}) \sin(\frac{\pi}{8}) + \sin^{3}(\frac{\pi}{8}) \cos(\frac{\pi}{8})$.
Factor out $\sin(\frac{\pi}{8}) \cos(\frac{\pi}{8})$:
$= \sin(\frac{\pi}{8}) \cos(\frac{\pi}{8}) [\cos^{2}(\frac{\pi}{8}) + \sin^{2}(\frac{\pi}{8})]$.
Since $\cos^{2}(\theta) + \sin^{2}(\theta) = 1$,this simplifies to $\sin(\frac{\pi}{8}) \cos(\frac{\pi}{8})$.
Using $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$,we get $\frac{1}{2} \sin(2 \cdot \frac{\pi}{8}) = \frac{1}{2} \sin(\frac{\pi}{4})$.
$= \frac{1}{2} \cdot \frac{1}{\sqrt{2}} = \frac{1}{2\sqrt{2}}$.

(A) The expression is $(1+x+x^{2})^{10} = (1 + x(1+x))^{10}$.
Using the binomial expansion $(1+y)^{n} = \sum_{k=0}^{n} {}^{n}C_{k} y^{k}$,we get:
$(1+x+x^{2})^{10} = \sum_{k=0}^{10} {}^{10}C_{k} (x(1+x))^{k} = \sum_{k=0}^{10} {}^{10}C_{k} x^{k} (1+x)^{k}$.
To find the coefficient of $x^{4}$,we consider terms where $x^{k} (1+x)^{k}$ contributes to $x^{4}$:
For $k=2$: ${}^{10}C_{2} x^{2} (1+x)^{2} = {}^{10}C_{2} x^{2} (1 + 2x + x^{2}) = {}^{10}C_{2} x^{2} + 2({}^{10}C_{2}) x^{3} + {}^{10}C_{2} x^{4}$. Coefficient is ${}^{10}C_{2} = 45$.
For $k=3$: ${}^{10}C_{3} x^{3} (1+x)^{3} = {}^{10}C_{3} x^{3} (1 + 3x + \dots) = {}^{10}C_{3} x^{3} + 3({}^{10}C_{3}) x^{4} + \dots$. Coefficient is $3 \times {}^{10}C_{3} = 3 \times 120 = 360$.
For $k=4$: ${}^{10}C_{4} x^{4} (1+x)^{4} = {}^{10}C_{4} x^{4} (1 + \dots) = {}^{10}C_{4} x^{4} + \dots$. Coefficient is ${}^{10}C_{4} = 210$.
Total coefficient of $x^{4} = 45 + 360 + 210 = 615$.

(A) Given equation: $\log _{1 / 2}|\sin x|=2-\log _{1 / 2}|\cos x|$ for $x \in [0, 2\pi]$.
Rearranging the terms,we get: $\log _{1 / 2}|\sin x| + \log _{1 / 2}|\cos x| = 2$.
Using the property $\log_a m + \log_a n = \log_a (mn)$,we have: $\log _{1 / 2}(|\sin x \cos x|) = 2$.
Converting to exponential form: $|\sin x \cos x| = (1/2)^2 = 1/4$.
Multiplying by $2$ on both sides: $|2 \sin x \cos x| = 2 \times (1/4) = 1/2$.
Thus,$|\sin 2x| = 1/2$.
In the interval $x \in [0, 2\pi]$,the angle $2x$ lies in the interval $[0, 4\pi]$.
For $|\sin \theta| = 1/2$,there are $4$ solutions in each interval of length $2\pi$ (i.e.,$[0, 2\pi]$).
Since the interval for $2x$ is $[0, 4\pi]$,which covers two full periods of the sine function,the total number of solutions is $4 \times 2 = 8$.

(A) Given the series $S = (x+y)+(x^{2}+xy+y^{2})+(x^{3}+x^{2}y+xy^{2}+y^{3})+\ldots$
Multiplying and dividing by $(x-y)$:
$S = \frac{(x-y)(x+y)+(x-y)(x^{2}+xy+y^{2})+(x-y)(x^{3}+x^{2}y+xy^{2}+y^{3})+\ldots}{x-y}$
Using the identity $(x-y)(x^n + x^{n-1}y + \ldots + y^n) = x^{n+1} - y^{n+1}$:
$S = \frac{(x^{2}-y^{2})+(x^{3}-y^{3})+(x^{4}-y^{4})+\ldots}{x-y}$
$S = \frac{(x^{2}+x^{3}+x^{4}+\ldots) - (y^{2}+y^{3}+y^{4}+\ldots)}{x-y}$
Using the sum of an infinite geometric series formula $S_{\infty} = \frac{a}{1-r}$:
$S = \frac{\frac{x^{2}}{1-x} - \frac{y^{2}}{1-y}}{x-y}$
$S = \frac{x^{2}(1-y) - y^{2}(1-x)}{(1-x)(1-y)(x-y)}$
$S = \frac{x^{2} - x^{2}y - y^{2} + xy^{2}}{(1-x)(1-y)(x-y)}$
$S = \frac{(x^{2}-y^{2}) - xy(x-y)}{(1-x)(1-y)(x-y)}$
$S = \frac{(x-y)(x+y) - xy(x-y)}{(1-x)(1-y)(x-y)}$
$S = \frac{(x-y)(x+y-xy)}{(1-x)(1-y)(x-y)}$
$S = \frac{x+y-xy}{(1-x)(1-y)}$

(B) The general term $T_{r+1}$ in the expansion of $(\alpha x^{\frac{1}{9}} + \beta x^{-\frac{1}{6}})^{10}$ is given by:
$T_{r+1} = {}^{10}C_{r} (\alpha x^{\frac{1}{9}})^{10-r} (\beta x^{-\frac{1}{6}})^{r} = {}^{10}C_{r} \alpha^{10-r} \beta^{r} x^{\frac{10-r}{9} - \frac{r}{6}}$.
For the term to be independent of $x$,the exponent of $x$ must be zero:
$\frac{10-r}{9} - \frac{r}{6} = 0$ $\Rightarrow 2(10-r) - 3r = 0$ $\Rightarrow 20 - 5r = 0$ $\Rightarrow r = 4$.
The independent term is $T_{5} = {}^{10}C_{4} \alpha^{6} \beta^{4} = 210 \alpha^{6} \beta^{4}$.
We are given $\alpha^{3} + \beta^{2} = 4$. By $AM \geq GM$ inequality for positive numbers:
$\frac{\frac{\alpha^{3}}{2} + \frac{\alpha^{3}}{2} + \frac{\beta^{2}}{2} + \frac{\beta^{2}}{2}}{4} \geq \sqrt[4]{\frac{\alpha^{3}}{2} \cdot \frac{\alpha^{3}}{2} \cdot \frac{\beta^{2}}{2} \cdot \frac{\beta^{2}}{2}}$
$\frac{4}{4} \geq \sqrt[4]{\frac{\alpha^{6} \beta^{4}}{16}}$ $\Rightarrow 1 \geq \frac{\alpha^{6} \beta^{4}}{16}$ $\Rightarrow \alpha^{6} \beta^{4} \leq 16$.
The maximum value of $T_{5}$ is $210 \times 16 = 3360$.
Given $10k = 3360$,we get $k = 336$.

(C) Let $p$ and $q$ be the statements:
$p: \text{I reach the station in time.}$
$q: \text{I will catch the train.}$
The given statement is of the form $p \rightarrow q$.
The contrapositive of the implication $p \rightarrow q$ is defined as $\sim q \rightarrow \sim p$.
Here,$\sim q$ is "$I$ will not catch the train" and $\sim p$ is "$I$ do not reach the station in time."
Therefore,the contrapositive is: "If $I$ will not catch the train,then $I$ do not reach the station in time."
This matches option $C$.

(A) Given $X = \{1, 2, \dots, 17\}$. The mean of $X$ is $\bar{x} = \frac{1+17}{2} = 9$. The variance of $X$ is $\sigma_X^2 = \frac{17^2 - 1}{12} = \frac{288}{12} = 24$.
For $Y = aX + b$,the mean is $\bar{Y} = a\bar{x} + b = 9a + b = 17$ (Equation $1$).
The variance of $Y$ is $\sigma_Y^2 = a^2 \sigma_X^2 = a^2(24) = 216$.
$a^2 = \frac{216}{24} = 9$. Since $a > 0$,we have $a = 3$.
Substituting $a = 3$ into Equation $1$: $9(3) + b = 17$ $\Rightarrow 27 + b = 17$ $\Rightarrow b = -10$.
Thus,$a + b = 3 + (-10) = -7$.

(A) Since $\alpha$ and $\beta$ are roots of $5x^{2} + 6x - 2 = 0$,they satisfy the equation.
$5\alpha^{2} + 6\alpha - 2 = 0 \implies 5\alpha^{n+2} + 6\alpha^{n+1} - 2\alpha^{n} = 0$ (multiplying by $\alpha^{n}$).
Similarly,$5\beta^{n+2} + 6\beta^{n+1} - 2\beta^{n} = 0$.
Adding these two equations,we get $5(\alpha^{n+2} + \beta^{n+2}) + 6(\alpha^{n+1} + \beta^{n+1}) - 2(\alpha^{n} + \beta^{n}) = 0$.
This simplifies to $5S_{n+2} + 6S_{n+1} - 2S_{n} = 0$.
For $n = 4$,we have $5S_{6} + 6S_{5} - 2S_{4} = 0$,which implies $5S_{6} + 6S_{5} = 2S_{4}$.

(B) The domain of $R^{-1}$ is the range of the relation $R$. The range of $R$ consists of all possible integer values of $y$ such that there exists at least one integer $x$ satisfying $x^{2} + 3y^{2} \leq 8$.
Given the inequality $3y^{2} \leq 8 - x^{2}$,since $x^{2} \geq 0$,we must have $3y^{2} \leq 8$,which implies $y^{2} \leq \frac{8}{3} \approx 2.66$.
Since $y$ must be an integer,the possible values for $y$ are $y \in \{-1, 0, 1\}$.
Let us verify these values:
If $y = 0$,$x^{2} \leq 8 \Rightarrow x \in \{-2, -1, 0, 1, 2\}$.
If $y = 1$,$x^{2} + 3(1)^{2} \leq 8 \Rightarrow x^{2} \leq 5 \Rightarrow x \in \{-2, -1, 0, 1, 2\}$.
If $y = -1$,$x^{2} + 3(-1)^{2} \leq 8 \Rightarrow x^{2} \leq 5 \Rightarrow x \in \{-2, -1, 0, 1, 2\}$.
Thus,the set of all possible values for $y$ is $\{-1, 0, 1\}$.
Therefore,the domain of $R^{-1}$ is $\{-1, 0, 1\}$.

(D) Let the three terms of the $G.P.$ be $\frac{a}{r}, a, ar$.
Given the product is $27$,so $\frac{a}{r} \times a \times ar = 27$ $\Rightarrow a^3 = 27$ $\Rightarrow a = 3$.
The sum $S = \frac{3}{r} + 3 + 3r = 3(\frac{1}{r} + r + 1)$.
Case $1$: If $r > 0$,by $AM \geq GM$,$\frac{1}{r} + r \geq 2$. Thus,$S = 3(\frac{1}{r} + r + 1) \geq 3(2 + 1) = 9$.
Case $2$: If $r < 0$,let $r = -k$ where $k > 0$. Then $\frac{1}{r} + r = -(\frac{1}{k} + k) \leq -2$. Thus,$S = 3(\frac{1}{r} + r + 1) \leq 3(-2 + 1) = -3$.
Therefore,$S \in (-\infty, -3] \cup [9, \infty)$.

(B) The slope of the line $2x - y = 0$ is $2$. Since the tangent is parallel to this line,its slope is $m = 2$.
The equation of the tangent to the hyperbola $\frac{x^{2}}{4} - \frac{y^{2}}{2} = 1$ at the point $(x_{1}, y_{1})$ is given by $\frac{xx_{1}}{4} - \frac{yy_{1}}{2} = 1$.
The slope of this tangent is $\frac{x_{1}/4}{y_{1}/2} = \frac{x_{1}}{2y_{1}}$.
Equating the slopes: $\frac{x_{1}}{2y_{1}} = 2 \Rightarrow x_{1} = 4y_{1} \quad (1)$.
Since $(x_{1}, y_{1})$ lies on the hyperbola,we have $\frac{x_{1}^{2}}{4} - \frac{y_{1}^{2}}{2} = 1 \quad (2)$.
Substituting $(1)$ into $(2)$: $\frac{(4y_{1})^{2}}{4} - \frac{y_{1}^{2}}{2} = 1 \Rightarrow 4y_{1}^{2} - \frac{y_{1}^{2}}{2} = 1$.
$\frac{7y_{1}^{2}}{2} = 1 \Rightarrow y_{1}^{2} = \frac{2}{7}$.
Now,calculate $x_{1}^{2} + 5y_{1}^{2} = (4y_{1})^{2} + 5y_{1}^{2} = 16y_{1}^{2} + 5y_{1}^{2} = 21y_{1}^{2}$.
Substituting $y_{1}^{2} = \frac{2}{7}$: $21 \times \frac{2}{7} = 3 \times 2 = 6$.

(C) Given the limit: $\lim_{x \rightarrow 1} \frac{x+x^{2}+\ldots+x^{n}-n}{x-1}=820$
We can rewrite the numerator by subtracting $1$ from each term: $\lim_{x \rightarrow 1} \left(\frac{x-1}{x-1} + \frac{x^{2}-1}{x-1} + \ldots + \frac{x^{n}-1}{x-1}\right) = 820$
Using the standard limit formula $\lim_{x \rightarrow a} \frac{x^{n}-a^{n}}{x-a} = na^{n-1}$,each term becomes $k$ when $x \rightarrow 1$: $\sum_{k=1}^{n} k = 820$
The sum of the first $n$ natural numbers is given by $\frac{n(n+1)}{2} = 820$
$\Rightarrow n(n+1) = 1640$
$\Rightarrow n^{2}+n-1640 = 0$
Solving the quadratic equation,we find $n(n+1) = 40 \times 41$
Since $n \in N$,we get $n = 40$.

(D) The letters of the word $MOTHER$ are $E, H, M, O, R, T$ in alphabetical order.
Words starting with $E$: $5! = 120$
Words starting with $H$: $5! = 120$
Words starting with $ME$: $4! = 24$
Words starting with $MH$: $4! = 24$
Words starting with $MOE$: $3! = 6$
Words starting with $MOH$: $3! = 6$
Words starting with $MOR$: $3! = 6$
Words starting with $MOT E$: $2! = 2$
Words starting with $MOT H E R$: $1$
Total rank = $120 + 120 + 24 + 24 + 6 + 6 + 6 + 2 + 1 = 309$.

(A) The given circle is $x^{2} + y^{2} - 2x - 4y + 4 = 0$.
Completing the square,we get $(x - 1)^{2} + (y - 2)^{2} = 1$.
Thus,the centre is $(1, 2)$ and the radius $r = 1$.
For the line $3x + 4y - k = 0$ to intersect the circle at two distinct points,the perpendicular distance from the centre to the line must be less than the radius.
Distance $d = \frac{|3(1) + 4(2) - k|}{\sqrt{3^{2} + 4^{2}}} < 1$.
$\frac{|11 - k|}{5} < 1$.
$|11 - k| < 5$.
$-5 < 11 - k < 5$.
$-16 < -k < -6$.
$6 < k < 16$.
The integral values of $k$ are $7, 8, 9, 10, 11, 12, 13, 14, 15$.
The total number of such values is $9$.

(C) Let the parabola be $y^2 = 8x$. The vertex is at $O(0,0)$.
Let the equilateral triangle be $OAB$, where $A$ and $B$ lie on the parabola.
Let the coordinates of $A$ be $(2t^2, 4t)$ for some $t > 0$.
Since the triangle is equilateral and symmetric about the $x$-axis, the coordinates of $B$ are $(2t^2, -4t)$.
The side length of the triangle is $AB = 4t - (-4t) = 8t$.
The angle $\angle AOx = 30^{\circ}$ because the triangle is equilateral and the $x$-axis bisects the angle at the vertex.
Thus, $\tan 30^{\circ} = \frac{4t}{2t^2} = \frac{2}{t}$.
Since $\tan 30^{\circ} = \frac{1}{\sqrt{3}}$, we have $\frac{1}{\sqrt{3}} = \frac{2}{t}$, which gives $t = 2\sqrt{3}$.
The side length $s = 8t = 8(2\sqrt{3}) = 16\sqrt{3}$.
The area of an equilateral triangle is given by $\frac{\sqrt{3}}{4} s^2$.
Area $= \frac{\sqrt{3}}{4} (16\sqrt{3})^2 = \frac{\sqrt{3}}{4} (256 \times 3) = \frac{\sqrt{3}}{4} (768) = 192\sqrt{3}$ sq. units.

(C) The number of blue lines corresponds to the number of sides of the polygon formed by the $n$ stations,which is $n$.
The total number of ways to connect any two stations is given by the combination formula ${}^{n}C_{2}$.
The number of red lines is the total number of connections minus the number of blue lines (sides),which is the number of diagonals: ${}^{n}C_{2} - n$.
According to the problem,the number of red lines is $99$ times the number of blue lines:
${}^{n}C_{2} - n = 99n$
Substituting the formula for ${}^{n}C_{2}$:
$\frac{n(n-1)}{2} - n = 99n$
Dividing both sides by $n$ (since $n > 2$):
$\frac{n-1}{2} - 1 = 99$
$\frac{n-1}{2} = 100$
$n - 1 = 200$
$n = 201$

(C) Let the quadratic polynomial be $f(x) = a(x - 3)(x - \alpha)$,where $\alpha$ is the other root.
Given $f(2) = a(2 - 3)(2 - \alpha) = a(-1)(2 - \alpha) = a(\alpha - 2)$.
Given $f(-1) = a(-1 - 3)(-1 - \alpha) = a(-4)(-1 - \alpha) = 4a(1 + \alpha)$.
Since $f(-1) + f(2) = 0$,we have $4a(1 + \alpha) + a(\alpha - 2) = 0$.
Since $a \neq 0$,we can divide by $a$: $4 + 4\alpha + \alpha - 2 = 0$.
$5\alpha + 2 = 0$ $\Rightarrow 5\alpha = -2$ $\Rightarrow \alpha = -\frac{2}{5} = -0.4$.
Thus,the other root $\alpha = -0.4$ lies in the interval $(-1, 0)$.

(B) Given the sum of the first $11$ terms of the $A.P.$ is $0$:
$S_{11} = \frac{11}{2}(2a_{1} + 10d) = 0$
$11(a_{1} + 5d) = 0 \Rightarrow a_{1} = -5d$ or $d = -\frac{a_{1}}{5}$.
We need to find the sum of the $A.P.$ $a_{1}, a_{3}, a_{5}, \ldots, a_{23}$.
This is an $A.P.$ with $12$ terms,first term $A = a_{1}$ and common difference $D = 2d$.
Sum $= \frac{12}{2}(2A + (12-1)D) = 6(2a_{1} + 11(2d)) = 6(2a_{1} + 22d)$.
Substitute $d = -\frac{a_{1}}{5}$:
Sum $= 6(2a_{1} + 22(-\frac{a_{1}}{5})) = 6(2a_{1} - \frac{22a_{1}}{5}) = 6(\frac{10a_{1} - 22a_{1}}{5}) = 6(-\frac{12a_{1}}{5}) = -\frac{72}{5}a_{1}$.
Thus,$k = -\frac{72}{5}$.

(A) Let $z = (3+2 \sqrt{-54})^{1/2} - (3-2 \sqrt{-54})^{1/2}$.
First,simplify the expression inside the square root: $\sqrt{-54} = \sqrt{54}i = 3\sqrt{6}i$.
So,$3+2\sqrt{-54} = 3 + 2(3\sqrt{6}i) = 3 + 6\sqrt{6}i$. This does not form a perfect square of the form $(a+bi)^2$ easily.
Let $(3+2\sqrt{-54})^{1/2} = x+iy$. Then $x^2-y^2 = 3$ and $2xy = 2\sqrt{54} = 6\sqrt{6}$.
$x^2+y^2 = \sqrt{3^2 + (6\sqrt{6})^2} = \sqrt{9 + 216} = \sqrt{225} = 15$.
Adding the equations: $2x^2 = 18 \implies x^2 = 9 \implies x = \pm 3$.
Subtracting: $2y^2 = 12 \implies y^2 = 6 \implies y = \pm \sqrt{6}$.
Thus,$(3+2\sqrt{-54})^{1/2} = \pm(3+\sqrt{6}i)$ and $(3-2\sqrt{-54})^{1/2} = \pm(3-\sqrt{6}i)$.
Possible values for the expression are:
$1) (3+\sqrt{6}i) - (3-\sqrt{6}i) = 2\sqrt{6}i$
$2) (3+\sqrt{6}i) - (-(3-\sqrt{6}i)) = 6$
$3) -(3+\sqrt{6}i) - (3-\sqrt{6}i) = -6$
$4) -(3+\sqrt{6}i) - (-(3-\sqrt{6}i)) = -2\sqrt{6}i$.
The imaginary parts are $2\sqrt{6}$ and $-2\sqrt{6}$.
Comparing with options,the correct choice is $-2\sqrt{6}$.

(D) The given limit is of the form $1^{\infty}$.
We use the formula $\lim \limits_{x \rightarrow a} f(x)^{g(x)} = e^{\lim \limits_{x \rightarrow a} g(x)(f(x)-1)}$.
Here,$f(x) = \tan(\frac{\pi}{4} + x)$ and $g(x) = \frac{1}{x}$.
$\lim \limits_{x \rightarrow 0} \left(\tan \left(\frac{\pi}{4}+x\right)\right)^{\frac{1}{x}} = e^{\lim \limits_{x \rightarrow 0} \frac{1}{x} \left(\tan \left(\frac{\pi}{4}+x\right)-1\right)}$.
Using the identity $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,we have $\tan(\frac{\pi}{4}+x) = \frac{1+\tan x}{1-\tan x}$.
So,$\tan(\frac{\pi}{4}+x)-1 = \frac{1+\tan x - (1-\tan x)}{1-\tan x} = \frac{2\tan x}{1-\tan x}$.
Thus,the exponent becomes $\lim \limits_{x \rightarrow 0} \frac{2\tan x}{x(1-\tan x)} = \lim \limits_{x \rightarrow 0} 2 \cdot \frac{\tan x}{x} \cdot \frac{1}{1-\tan x} = 2 \cdot 1 \cdot 1 = 2$.
Therefore,the limit is $e^{2}$.

(B) Given the hyperbola equation: $x^{2} - y^{2} \sec^{2} \theta = 10 \Rightarrow \frac{x^{2}}{10} - \frac{y^{2}}{10 \cos^{2} \theta} = 1.$
For a hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1,$ the eccentricity $e_{H} = \sqrt{1 + \frac{b^{2}}{a^{2}}}.$
Thus,$e_{H} = \sqrt{1 + \frac{10 \cos^{2} \theta}{10}} = \sqrt{1 + \cos^{2} \theta}.$
Given the ellipse equation: $x^{2} \sec^{2} \theta + y^{2} = 5 \Rightarrow \frac{x^{2}}{5 \cos^{2} \theta} + \frac{y^{2}}{5} = 1.$
For an ellipse $\frac{x^{2}}{b^{2}} + \frac{y^{2}}{a^{2}} = 1$ (where $a > b$),$e_{E} = \sqrt{1 - \frac{b^{2}}{a^{2}}}.$
Here $a^{2} = 5$ and $b^{2} = 5 \cos^{2} \theta,$ so $e_{E} = \sqrt{1 - \frac{5 \cos^{2} \theta}{5}} = \sqrt{1 - \cos^{2} \theta} = \sin \theta.$
Given $e_{H} = \sqrt{5} e_{E},$ we have $\sqrt{1 + \cos^{2} \theta} = \sqrt{5} \sin \theta.$
Squaring both sides: $1 + \cos^{2} \theta = 5 \sin^{2} \theta = 5(1 - \cos^{2} \theta).$
$1 + \cos^{2} \theta = 5 - 5 \cos^{2} \theta$ $\Rightarrow 6 \cos^{2} \theta = 4$ $\Rightarrow \cos^{2} \theta = \frac{2}{3}.$
Then $\sin^{2} \theta = 1 - \frac{2}{3} = \frac{1}{3}.$
The length of the latus rectum of the ellipse is $\frac{2b^{2}}{a} = \frac{2(5 \cos^{2} \theta)}{\sqrt{5}} = \frac{10 \times (2/3)}{\sqrt{5}} = \frac{20}{3 \sqrt{5}} = \frac{4 \sqrt{5}}{3}.$

(A) To determine which expression is a tautology,we evaluate the truth table or simplify the logical expressions.
For option $A$: $(\sim p \wedge (p \vee q)) \rightarrow q$
Using the distributive law: $(\sim p \wedge p) \vee (\sim p \wedge q) \rightarrow q$
Since $(\sim p \wedge p) \equiv F$ (Contradiction),we have: $F \vee (\sim p \wedge q) \rightarrow q$
This simplifies to: $(\sim p \wedge q) \rightarrow q$
Using the implication law $a \rightarrow b \equiv \sim a \vee b$: $\sim (\sim p \wedge q) \vee q$
Applying De Morgan's law: $(p \vee \sim q) \vee q$
By associativity: $p \vee (\sim q \vee q)$
Since $(\sim q \vee q) \equiv T$ (Tautology): $p \vee T \equiv T$
Thus,option $A$ is a tautology.

(C) The series is given by $S = \sum_{n=1}^{9} [x^n + (k + 2(n-1))a]$.
Expanding the sum,we get $S = (x + x^2 + \ldots + x^9) + \sum_{n=1}^{9} (k + 2n - 2)a$.
The sum of the geometric progression is $\sum_{n=1}^{9} x^n = x \frac{x^9 - 1}{x - 1} = \frac{x^{10} - x}{x - 1}$.
The sum of the arithmetic part is $\sum_{n=1}^{9} (k + 2n - 2)a = a [9k + 2 \frac{9 \times 8}{2} - 18] = a [9k + 72 - 18] = a(9k + 54)$.
Wait,recalculating the arithmetic part: $\sum_{n=0}^{8} (k + 2n)a = 9ka + 2a \frac{8 \times 9}{2} = 9ka + 72a = a(9k + 72)$.
Thus,$S = \frac{x^{10} - x}{x - 1} + \frac{a(9k + 72)(x - 1)}{x - 1} = \frac{x^{10} - x + (9k + 72)a(x - 1)}{x - 1}$.
Comparing this with the given $S = \frac{x^{10} - x + 45a(x - 1)}{x - 1}$,we have $9k + 72 = 45$.
$9k = 45 - 72 = -27$.
$k = -3$.

(D) Two points $(x_1, y_1)$ and $(x_2, y_2)$ lie on the same side of the line $ax+by+c=0$ if the expressions $(ax_1+by_1+c)$ and $(ax_2+by_2+c)$ have the same sign,i.e.,$(ax_1+by_1+c)(ax_2+by_2+c) > 0$.
Given the line $x+y-1=0$ and points $(1, 2)$ and $(\sin \theta, \cos \theta)$.
First,evaluate the expression at $(1, 2)$:
$1+2-1 = 2$,which is positive.
Therefore,for the points to lie on the same side,the expression at $(\sin \theta, \cos \theta)$ must also be positive:
$\sin \theta + \cos \theta - 1 > 0$
$\Rightarrow \sin \theta + \cos \theta > 1$
Divide by $\sqrt{2}$:
$\frac{1}{\sqrt{2}} \sin \theta + \frac{1}{\sqrt{2}} \cos \theta > \frac{1}{\sqrt{2}}$
$\Rightarrow \sin \left(\theta + \frac{\pi}{4}\right) > \sin \left(\frac{\pi}{4}\right)$
Since $\theta \in (0, \pi)$,then $\theta + \frac{\pi}{4} \in \left(\frac{\pi}{4}, \frac{5\pi}{4}\right)$.
In this interval,$\sin \left(\theta + \frac{\pi}{4}\right) > \frac{1}{\sqrt{2}}$ holds when:
$\frac{\pi}{4} < \theta + \frac{\pi}{4} < \frac{3\pi}{4}$
Subtracting $\frac{\pi}{4}$ from all parts:
$0 < \theta < \frac{\pi}{2}$

(A) Let $a$ be the first term and $d$ be the common difference of the given $A.P.$ Since the $A.P.$ is increasing,$d > 0$.
The terms are $a, a+d, a+2d, \ldots, a+10d$.
The mean $\bar{X} = \frac{1}{11} \sum_{i=0}^{10} (a + id) = a + \frac{d}{11} \times \frac{10 \times 11}{2} = a + 5d$.
The variance is given by $\sigma^2 = \frac{1}{n} \sum (x_i - \bar{X})^2$.
$\sigma^2 = \frac{1}{11} \sum_{i=0}^{10} (a + id - (a + 5d))^2 = \frac{1}{11} \sum_{i=0}^{10} ((i-5)d)^2 = \frac{d^2}{11} \sum_{i=0}^{10} (i-5)^2$.
Calculating the sum: $(-5)^2 + (-4)^2 + (-3)^2 + (-2)^2 + (-1)^2 + 0^2 + 1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 25 + 16 + 9 + 4 + 1 + 0 + 1 + 4 + 9 + 16 + 25 = 110$.
So,$90 = \frac{d^2}{11} \times 110 = 10d^2$.
$d^2 = 9 \Rightarrow d = \pm 3$.
Since the $A.P.$ is increasing,$d = 3$.

(D) The expansion of $(1+\frac{1}{x})^n$ is $\sum_{k=0}^{n} {}^{n}C_k (\frac{1}{x})^k = \sum_{k=0}^{n} {}^{n}C_k x^{-k}$.
Since the expansion is in increasing powers of $x$,we consider the coefficients of $x^{-r-1}, x^{-r}, x^{-r+1}$ which are ${}^{n}C_{r+1}, {}^{n}C_r, {}^{n}C_{r-1}$.
Given the ratio ${}^{n}C_{r-1} : {}^{n}C_r : {}^{n}C_{r+1} = 2 : 5 : 12$.
From $\frac{{}^{n}C_{r-1}}{{}^{n}C_r} = \frac{2}{5}$,we get $\frac{r}{n-r+1} = \frac{2}{5}$ $\Rightarrow 5r = 2n - 2r + 2$ $\Rightarrow 7r = 2n + 2$ (Equation $1$).
From $\frac{{}^{n}C_r}{{}^{n}C_{r+1}} = \frac{5}{12}$,we get $\frac{r+1}{n-r} = \frac{5}{12}$ $\Rightarrow 12r + 12 = 5n - 5r$ $\Rightarrow 17r = 5n - 12$ (Equation $2$).
Multiplying Equation $1$ by $17$ and Equation $2$ by $7$:
$119r = 34n + 34$
$119r = 35n - 84$
Equating the two: $35n - 84 = 34n + 34 \Rightarrow n = 118$.

(A) The parabola is $y^{2} = 4x$. Comparing with $y^{2} = 4ax$,we get $a = 1$.
The length of the latus rectum is $4a = 4(1) = 4$.
The common chord of the two circles $C_{1}$ and $C_{2}$ is the latus rectum of the parabola,so its length is $4$.
Let $D$ be one endpoint of the latus rectum and $B$ be the midpoint of the common chord. Thus,$DB = \frac{4}{2} = 2$.
Let $r$ be the radius of the circles,$r = 2\sqrt{5}$.
In the right-angled triangle formed by the radius,the distance from the center to the chord,and half the chord length,let $x$ be the distance from the center of a circle to the common chord.
$x^{2} + DB^{2} = r^{2}$
$x^{2} + 2^{2} = (2\sqrt{5})^{2}$
$x^{2} + 4 = 20$
$x^{2} = 16 \implies x = 4$.
The distance between the centers $C_{1}$ and $C_{2}$ is $x + x = 4 + 4 = 8$.

(C) Total number of $5-digit$ numbers is $9 \times 10^{4}$.
Case $1$: The two digits chosen are both non-zero.
Number of ways to choose $2$ digits from ${1, 2, \dots, 9}$ is $^{9}C_{2} = 36$.
For each selection,the number of $5-digit$ numbers using both digits is $2^{5} - 2 = 30$.
Total for Case $1 = 36 \times 30 = 1080$.
Case $2$: One digit is zero and the other is non-zero.
Number of ways to choose $1$ non-zero digit from ${1, 2, \dots, 9}$ is $^{9}C_{1} = 9$.
The first digit must be non-zero (fixed in $1$ way),and the remaining $4$ positions can be filled by either $0$ or the chosen non-zero digit,excluding the case where all $4$ are non-zero. This gives $2^{4} - 1 = 15$ ways.
Total for Case $2 = 9 \times 15 = 135$.
Total favorable outcomes $= 1080 + 135 = 1215$.
Probability $= \frac{1215}{9 \times 10^{4}} = \frac{135}{10^{4}}$.

(C) Let the orthocentre be $H(x_0, y_0)$.
The slope of $BC$ is $m_{BC} = \frac{-5-1}{5-(-7)} = \frac{-6}{12} = -\frac{1}{2}$.
Since $AH \perp BC$,the slope of $AH$ is $m_{AH} = -\frac{1}{m_{BC}} = 2$.
The equation of altitude $AH$ passing through $A(-1, 7)$ is $y - 7 = 2(x + 1)$,which simplifies to $2x - y + 9 = 0$ ... $(1)$.
The slope of $AC$ is $m_{AC} = \frac{-5-7}{5-(-1)} = \frac{-12}{6} = -2$.
Since $BH \perp AC$,the slope of $BH$ is $m_{BH} = -\frac{1}{m_{AC}} = \frac{1}{2}$.
The equation of altitude $BH$ passing through $B(-7, 1)$ is $y - 1 = \frac{1}{2}(x + 7)$,which simplifies to $x - 2y + 9 = 0$ ... $(2)$.
Solving equations $(1)$ and $(2)$:
From $(1)$,$y = 2x + 9$. Substituting into $(2)$:
$x - 2(2x + 9) + 9 = 0$
$x - 4x - 18 + 9 = 0$
$-3x - 9 = 0 \Rightarrow x = -3$.
Then $y = 2(-3) + 9 = 3$.
Thus,the orthocentre is $(-3, 3)$.

(A) The given system of equations is homogeneous:
$7x + 6y - 2z = 0 \dots (1)$
$3x + 4y + 2z = 0 \dots (2)$
$x - 2y - 6z = 0 \dots (3)$
First,we calculate the determinant of the coefficient matrix $\Delta$:
$\Delta = \begin{vmatrix} 7 & 6 & -2 \\ 3 & 4 & 2 \\ 1 & -2 & -6 \end{vmatrix}$
$= 7(4(-6) - 2(-2)) - 6(3(-6) - 2(1)) - 2(3(-2) - 4(1))$
$= 7(-24 + 4) - 6(-18 - 2) - 2(-6 - 4)$
$= 7(-20) - 6(-20) - 2(-10)$
$= -140 + 120 + 20 = 0$
Since $\Delta = 0$,the system has infinitely many solutions.
Adding equations $(1)$ and $(2)$:
$(7x + 6y - 2z) + (3x + 4y + 2z) = 0 + 0$
$10x + 10y = 0 \Rightarrow y = -x$
Substituting $y = -x$ into equation $(1)$:
$7x + 6(-x) - 2z = 0$
$7x - 6x - 2z = 0$
$x - 2z = 0 \Rightarrow x = 2z$
Thus,the solutions satisfy $x = 2z$.

(D) Given $x = 2 \sin \theta - \sin 2 \theta$ and $y = 2 \cos \theta - \cos 2 \theta$.
First,find the derivatives with respect to $\theta$:
$\frac{dx}{d\theta} = 2 \cos \theta - 2 \cos 2 \theta = 2(\cos \theta - (2 \cos^2 \theta - 1)) = 2(1 + \cos \theta - 2 \cos^2 \theta) = 2(1 - \cos \theta)(1 + 2 \cos \theta)$.
Alternatively,using sum-to-product formulas: $\frac{dx}{d\theta} = 2(\cos \theta - \cos 2 \theta) = 4 \sin \frac{3\theta}{2} \sin \frac{\theta}{2}$.
$\frac{dy}{d\theta} = -2 \sin \theta + 2 \sin 2 \theta = 2(\sin 2 \theta - \sin \theta) = 2(2 \sin \theta \cos \theta - \sin \theta) = 2 \sin \theta (2 \cos \theta - 1)$.
Using sum-to-product: $\frac{dy}{d\theta} = 2(\sin 2 \theta - \sin \theta) = 4 \cos \frac{3\theta}{2} \sin \frac{\theta}{2}$.
Now,$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{4 \cos(3\theta/2) \sin(\theta/2)}{4 \sin(3\theta/2) \sin(\theta/2)} = \cot \frac{3\theta}{2}$.
Now,differentiate $\frac{dy}{dx}$ with respect to $x$:
$\frac{d^2y}{dx^2} = \frac{d}{d\theta} \left( \cot \frac{3\theta}{2} \right) \cdot \frac{d\theta}{dx} = -\frac{3}{2} \csc^2 \frac{3\theta}{2} \cdot \frac{1}{dx/d\theta}$.
At $\theta = \pi$,$\frac{dx}{d\theta} = 2 \cos \pi - 2 \cos 2 \pi = 2(-1) - 2(1) = -4$.
$\frac{d^2y}{dx^2} = -\frac{3}{2} \csc^2 \frac{3\pi}{2} \cdot \frac{1}{-4} = -\frac{3}{2} (-1)^2 \cdot \left( -\frac{1}{4} \right) = \frac{3}{8}$.

(B) The region is bounded by $x = \frac{1}{2}$ and $x = \frac{\sqrt{3}}{2}$. In this interval,$f(x) = 1-x$ and $g(x) = (x-\frac{1}{2})^2$.
The area $A$ is given by the integral:
$A = \int_{1/2}^{\sqrt{3}/2} (f(x) - g(x)) dx$
$A = \int_{1/2}^{\sqrt{3}/2} ((1-x) - (x-\frac{1}{2})^2) dx$
Let $u = x - \frac{1}{2}$,then $du = dx$. When $x = 1/2, u = 0$. When $x = \sqrt{3}/2, u = \frac{\sqrt{3}-1}{2}$.
$A = \int_{0}^{(\sqrt{3}-1)/2} (1 - (u + 1/2) - u^2) du = \int_{0}^{(\sqrt{3}-1)/2} (1/2 - u - u^2) du$
$A = [\frac{1}{2}u - \frac{u^2}{2} - \frac{u^3}{3}]_{0}^{(\sqrt{3}-1)/2}$
Substituting the upper limit:
$A = \frac{1}{2}(\frac{\sqrt{3}-1}{2}) - \frac{1}{2}(\frac{3+1-2\sqrt{3}}{4}) - \frac{1}{3}(\frac{3\sqrt{3}-9+3\sqrt{3}-1}{8})$
$A = \frac{\sqrt{3}-1}{4} - \frac{4-2\sqrt{3}}{8} - \frac{6\sqrt{3}-10}{24} = \frac{6\sqrt{3}-6 - 12+6\sqrt{3} - 6\sqrt{3}+10}{24} = \frac{6\sqrt{3}-8}{24} = \frac{\sqrt{3}}{4} - \frac{1}{3}$.

(B) The sum of all probabilities in a probability distribution must be equal to $1$.
$\sum P(X) = 1 \Rightarrow K^2 + 2K + K + 2K + 5K^2 = 1$
$6K^2 + 5K - 1 = 0$
Factoring the quadratic equation: $(6K - 1)(K + 1) = 0$
This gives $K = \frac{1}{6}$ or $K = -1$.
Since the probability $P(X)$ cannot be negative,we reject $K = -1$. Thus,$K = \frac{1}{6}$.
We need to find $P(X > 2) = P(X = 3) + P(X = 4) + P(X = 5)$.
$P(X > 2) = K + 2K + 5K^2 = 3K + 5K^2$.
Substituting $K = \frac{1}{6}$:
$P(X > 2) = 3(\frac{1}{6}) + 5(\frac{1}{6})^2 = \frac{1}{2} + \frac{5}{36} = \frac{18}{36} + \frac{5}{36} = \frac{23}{36}$.

(A) Given $F(x) = \int_{1}^{x} t^{2} g(t) dt$. By the Fundamental Theorem of Calculus,$F'(x) = x^{2} g(x) = x^{2} \int_{1}^{x} f(u) du$.
At $x=1$,$F'(1) = 1^{2} \int_{1}^{1} f(u) du = 0$. Thus,$x=1$ is a critical point.
Now,find the second derivative: $F''(x) = \frac{d}{dx} [x^{2} g(x)] = x^{2} g'(x) + 2x g(x)$.
Since $g(t) = \int_{1}^{t} f(u) du$,$g'(t) = f(t)$.
So,$F''(x) = x^{2} f(x) + 2x \int_{1}^{x} f(u) du$.
Evaluating at $x=1$: $F''(1) = 1^{2} f(1) + 2(1) \int_{1}^{1} f(u) du = f(1) + 0 = 3$.
Since $F'(1) = 0$ and $F''(1) = 3 > 0$,by the Second Derivative Test,$x=1$ is a point of local minima.

(C) Total ways to place $10$ distinct balls in $4$ distinct boxes is $4^{10}$.
To find the number of ways such that two boxes contain exactly $2$ and $3$ balls:
$1$. Select $2$ boxes out of $4$ to contain $2$ and $3$ balls respectively: $P(4, 2) = 4 \times 3 = 12$ ways.
$2$. Select $2$ balls for the first box and $3$ balls for the second box from $10$ balls: $\binom{10}{2} \times \binom{8}{3} = 45 \times 56 = 2520$ ways.
$3$. The remaining $5$ balls can be placed in the remaining $2$ boxes in $2^5 = 32$ ways.
Total favorable ways $= 12 \times 2520 \times 32 = 967680$.
Probability $= \frac{967680}{4^{10}} = \frac{967680}{1048576} = \frac{945}{1024} = \frac{945}{2^{10}}$.

(D) Given the differential equation $\frac{dy}{dx} = \frac{xy}{x^2 + y^2}$.
Substitute $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting into the equation: $v + x \frac{dv}{dx} = \frac{x(vx)}{x^2 + v^2 x^2} = \frac{v}{1 + v^2}$.
$x \frac{dv}{dx} = \frac{v}{1 + v^2} - v = \frac{v - v - v^3}{1 + v^2} = -\frac{v^3}{1 + v^2}$.
Separating variables: $\int \frac{1 + v^2}{v^3} dv = -\int \frac{dx}{x}$.
$\int (v^{-3} + v^{-1}) dv = -\int \frac{dx}{x}$.
Integrating both sides: $-\frac{1}{2v^2} + \ln|v| = -\ln|x| + C$.
Substitute $v = \frac{y}{x}$: $-\frac{x^2}{2y^2} + \ln|\frac{y}{x}| = -\ln|x| + C$.
$-\frac{x^2}{2y^2} + \ln|y| - \ln|x| = -\ln|x| + C \implies -\frac{x^2}{2y^2} + \ln|y| = C$.
Using $y(1) = 1$: $-\frac{1^2}{2(1)^2} + \ln(1) = C \implies C = -\frac{1}{2}$.
The equation is $-\frac{x^2}{2y^2} + \ln|y| = -\frac{1}{2}$.
For $y = e$: $-\frac{x^2}{2e^2} + \ln(e) = -\frac{1}{2} \implies -\frac{x^2}{2e^2} + 1 = -\frac{1}{2}$.
$-\frac{x^2}{2e^2} = -\frac{3}{2} \implies x^2 = 3e^2 \implies x = \sqrt{3}e$.

(C) Given the determinant $f(x) = \begin{vmatrix} x+a & x+2 & x+1 \\ x+b & x+3 & x+2 \\ x+c & x+4 & x+3 \end{vmatrix}$.
Applying the row operation $R_1 \rightarrow R_1 + R_3 - 2R_2$:
The first row becomes:
$(x+a) + (x+c) - 2(x+b) = x+a+x+c-2x-2b = a-2b+c = 1$.
$(x+2) + (x+4) - 2(x+3) = 2x+6-2x-6 = 0$.
$(x+1) + (x+3) - 2(x+2) = 2x+4-2x-4 = 0$.
Thus,$f(x) = \begin{vmatrix} 1 & 0 & 0 \\ x+b & x+3 & x+2 \\ x+c & x+4 & x+3 \end{vmatrix}$.
Expanding along the first row:
$f(x) = 1 \cdot ((x+3)(x+3) - (x+2)(x+4)) = (x^2+6x+9) - (x^2+6x+8) = 1$.
Since $f(x) = 1$ for all $x$,$f(50) = 1$.

(C) Given that $f \circ g$ is the identity function,we have $f(g(x)) = x$ for all $x \in R$.
Differentiating both sides with respect to $x$ using the chain rule,we get:
$f^{\prime}(g(x)) \cdot g^{\prime}(x) = 1$.
We are given that $g(a) = b$ and $g^{\prime}(a) = 5$.
Substituting $x = a$ into the differentiated equation:
$f^{\prime}(g(a)) \cdot g^{\prime}(a) = 1$.
Substituting the known values $g(a) = b$ and $g^{\prime}(a) = 5$:
$f^{\prime}(b) \cdot 5 = 1$.
Therefore,$f^{\prime}(b) = 1/5$.

(C) Given $\vec{b} \cdot \vec{c} = |\vec{b}| |\vec{c}| \cos(\frac{\pi}{3}) = 10$.
Since $|\vec{b}| = 5$,we have $5 |\vec{c}| (\frac{1}{2}) = 10$,which implies $|\vec{c}| = 4$.
Now,the magnitude of the cross product is $|\vec{b} \times \vec{c}| = |\vec{b}| |\vec{c}| \sin(\frac{\pi}{3}) = 5 \times 4 \times \frac{\sqrt{3}}{2} = 10\sqrt{3}$.
Since $\vec{a}$ is perpendicular to $\vec{b} \times \vec{c}$,the angle between $\vec{a}$ and $\vec{b} \times \vec{c}$ is $\frac{\pi}{2}$.
Thus,$|\vec{a} \times (\vec{b} \times \vec{c})| = |\vec{a}| |\vec{b} \times \vec{c}| \sin(\frac{\pi}{2}) = \sqrt{3} \times 10\sqrt{3} \times 1 = 30$.

(B) The plane containing the two lines must have a normal vector $\vec{n}$ perpendicular to the direction vectors of the lines $\vec{v_1} = (2, 4, 3)$ and $\vec{v_2} = (2, 6, \lambda)$.
Thus,$\vec{n} = \vec{v_1} \times \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 4 & 3 \\ 2 & 6 & \lambda \end{vmatrix} = (4\lambda - 18)\hat{i} - (2\lambda - 6)\hat{j} + (12 - 8)\hat{k} = (4\lambda - 18, 6 - 2\lambda, 4)$.
For the lines to be coplanar,the vector connecting points $(-1, 3, -1)$ and $(-3, -2, 1)$,which is $\vec{a} = (-2, -5, 2)$,must be perpendicular to $\vec{n}$.
So,$\vec{a} \cdot \vec{n} = -2(4\lambda - 18) - 5(6 - 2\lambda) + 2(4) = -8\lambda + 36 - 30 + 10\lambda + 8 = 2\lambda + 14 = 0$,which gives $\lambda = -7$.
Substituting $\lambda = -7$,$\vec{n} = (4(-7) - 18, 6 - 2(-7), 4) = (-46, 20, 4)$.
Dividing by $-2$,we get the normal vector $(23, -10, -2)$,which is parallel to the given plane $23x - 10y - 2z + 48 = 0$.
The plane containing the lines passes through $(-1, 3, -1)$,so its equation is $23(x+1) - 10(y-3) - 2(z+1) = 0 \Rightarrow 23x - 10y - 2z + 51 = 0$.
The distance between $23x - 10y - 2z + 48 = 0$ and $23x - 10y - 2z + 51 = 0$ is $d = \frac{|51 - 48|}{\sqrt{23^2 + (-10)^2 + (-2)^2}} = \frac{3}{\sqrt{529 + 100 + 4}} = \frac{3}{\sqrt{633}}$.
Comparing with $\frac{k}{\sqrt{633}}$,we get $k = 3$.

(A) Let $P(A) = \frac{10}{20} = \frac{1}{2}$ and $P(B) = \frac{10}{20} = \frac{1}{2}$.
We want the second $A$ to appear before the third $B$.
This means in the sequence of draws,we must have at most two $B$'s before the second $A$.
The possible favorable sequences are:
$1$. $AA$: Probability $= (\frac{1}{2})^2 = \frac{1}{4}$
$2$. $ABA, BAA$: Probability $= 2 \times (\frac{1}{2})^3 = \frac{2}{8} = \frac{1}{4}$
$3$. $ABBA, BABA, BBAA$: Probability $= 3 \times (\frac{1}{2})^4 = \frac{3}{16}$
Summing these probabilities: $\frac{1}{4} + \frac{1}{4} + \frac{3}{16} = \frac{4}{16} + \frac{4}{16} + \frac{3}{16} = \frac{11}{16}$.

(D) Let $I = \int_{0}^{2 \pi} \frac{x \sin^{8} x}{\sin^{8} x + \cos^{8} x} dx$.
Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we have:
$I = \int_{0}^{2 \pi} \frac{(2 \pi - x) \sin^{8}(2 \pi - x)}{\sin^{8}(2 \pi - x) + \cos^{8}(2 \pi - x)} dx = \int_{0}^{2 \pi} \frac{(2 \pi - x) \sin^{8} x}{\sin^{8} x + \cos^{8} x} dx$.
Adding the two expressions for $I$:
$2I = \int_{0}^{2 \pi} \frac{2 \pi \sin^{8} x}{\sin^{8} x + \cos^{8} x} dx = 2 \pi \int_{0}^{2 \pi} \frac{\sin^{8} x}{\sin^{8} x + \cos^{8} x} dx$.
$I = \pi \int_{0}^{2 \pi} \frac{\sin^{8} x}{\sin^{8} x + \cos^{8} x} dx$.
Using the property $\int_{0}^{2a} f(x) dx = 2 \int_{0}^{a} f(x) dx$ if $f(2a-x) = f(x)$:
$I = 2 \pi \int_{0}^{\pi} \frac{\sin^{8} x}{\sin^{8} x + \cos^{8} x} dx = 4 \pi \int_{0}^{\pi/2} \frac{\sin^{8} x}{\sin^{8} x + \cos^{8} x} dx$.
Using $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$:
$I = 4 \pi \int_{0}^{\pi/2} \frac{\cos^{8} x}{\cos^{8} x + \sin^{8} x} dx$.
Adding these two: $2I = 4 \pi \int_{0}^{\pi/2} 1 dx = 4 \pi \cdot \frac{\pi}{2} = 2 \pi^{2}$.
Thus,$I = \pi^{2}$.

(C) Given $f^{\prime}(x)=\tan^{-1}(\sec x+\tan x)$.
We simplify the expression inside the inverse tangent:
$f^{\prime}(x)=\tan^{-1}\left(\frac{1+\sin x}{\cos x}\right) = \tan^{-1}\left(\frac{(\cos \frac{x}{2} + \sin \frac{x}{2})^2}{\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}}\right) = \tan^{-1}\left(\frac{\cos \frac{x}{2} + \sin \frac{x}{2}}{\cos \frac{x}{2} - \sin \frac{x}{2}}\right)$
Dividing numerator and denominator by $\cos \frac{x}{2}$:
$f^{\prime}(x)=\tan^{-1}\left(\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}\right) = \tan^{-1}\left(\tan(\frac{\pi}{4}+\frac{x}{2})\right)$
Since $-\frac{\pi}{2} < x < \frac{\pi}{2}$,we have $0 < \frac{\pi}{4}+\frac{x}{2} < \frac{\pi}{2}$,so $f^{\prime}(x) = \frac{\pi}{4} + \frac{x}{2}$.
Integrating with respect to $x$:
$f(x) = \int (\frac{\pi}{4} + \frac{x}{2}) dx = \frac{\pi}{4}x + \frac{x^2}{4} + C$.
Given $f(0)=0$,we find $C=0$.
Thus,$f(x) = \frac{\pi x + x^2}{4}$.
For $x=1$,$f(1) = \frac{\pi(1) + (1)^2}{4} = \frac{\pi+1}{4}$.

(C) Given $A = \begin{bmatrix} 1 & 1 & 2 \\ 1 & 3 & 4 \\ 1 & -1 & 3 \end{bmatrix}$.
First,calculate the determinant of $A$:
$|A| = 1(3 \times 3 - 4 \times -1) - 1(1 \times 3 - 4 \times 1) + 2(1 \times -1 - 3 \times 1)$
$|A| = 1(9 + 4) - 1(3 - 4) + 2(-1 - 3)$
$|A| = 13 + 1 - 8 = 6$.
We are given $B = \operatorname{adj} A$ and $C = 3A$. We need to find $\frac{|\operatorname{adj} B|}{|C|}$.
Since $B = \operatorname{adj} A$,then $\operatorname{adj} B = \operatorname{adj}(\operatorname{adj} A)$.
Using the property $|\operatorname{adj} M| = |M|^{n-1}$ where $n$ is the order of the matrix:
$|\operatorname{adj} B| = |\operatorname{adj}(\operatorname{adj} A)| = |A|^{(n-1)^2} = |A|^{(3-1)^2} = |A|^4$.
For the denominator,$|C| = |3A| = 3^n |A| = 3^3 |A| = 27 |A|$.
Thus,$\frac{|\operatorname{adj} B|}{|C|} = \frac{|A|^4}{27 |A|} = \frac{|A|^3}{27}$.
Substituting $|A| = 6$:
$\frac{6^3}{27} = \frac{216}{27} = 8$.

(C) Given that $f^{\prime}(x) > 0$ and $f^{\prime \prime}(x) < 0$ for all $x \in (a, b)$,the function $f$ is strictly increasing and concave downwards on the interval $[a, b]$.
Let $m_1$ be the slope of the secant line passing through $(a, f(a))$ and $(c, f(c))$,and $m_2$ be the slope of the secant line passing through $(c, f(c))$ and $(b, f(b))$.
$m_1 = \frac{f(c)-f(a)}{c-a}$ and $m_2 = \frac{f(b)-f(c)}{b-c}$.
Since the function is concave downwards,the slope of the secant line decreases as we move to the right. Therefore,$m_1 > m_2$.
Substituting the expressions for $m_1$ and $m_2$,we get:
$\frac{f(c)-f(a)}{c-a} > \frac{f(b)-f(c)}{b-c}$.
Rearranging the terms to isolate the required ratio:
$\frac{f(c)-f(a)}{f(b)-f(c)} > \frac{c-a}{b-c}$.

(A) For the three planes to intersect in a line,the system of linear equations must have infinitely many solutions. This occurs when the determinant of the coefficient matrix $\Delta = 0$ and the augmented determinants $\Delta_x, \Delta_y, \Delta_z$ are also $0$.
First,we calculate the determinant of the coefficient matrix:
$\Delta = \begin{vmatrix} 1 & 4 & -2 \\ 1 & 7 & -5 \\ 1 & 5 & \alpha \end{vmatrix} = 1(7\alpha + 25) - 4(\alpha + 5) - 2(5 - 7) = 7\alpha + 25 - 4\alpha - 20 + 4 = 3\alpha + 9$.
Setting $\Delta = 0$,we get $3\alpha + 9 = 0$,which implies $\alpha = -3$.
Next,for the system to have infinitely many solutions,we must have $\Delta_z = 0$:
$\Delta_z = \begin{vmatrix} 1 & 4 & 1 \\ 1 & 7 & \beta \\ 1 & 5 & 5 \end{vmatrix} = 1(35 - 5\beta) - 4(5 - \beta) + 1(5 - 7) = 35 - 5\beta - 20 + 4\beta - 2 = 13 - \beta$.
Setting $\Delta_z = 0$,we get $13 - \beta = 0$,which implies $\beta = 13$.
With $\alpha = -3$ and $\beta = 13$,the system is consistent and represents a line. Thus,$\alpha + \beta = -3 + 13 = 10$.

(A) Let $I = \int \frac{dx}{(x+4)^{\frac{8}{7}}(x-3)^{\frac{6}{7}}}$.
We can rewrite the integrand as $I = \int \frac{dx}{\left(\frac{x+4}{x-3}\right)^{\frac{8}{7}}(x-3)^{\frac{8}{7}}(x-3)^{\frac{6}{7}}} = \int \frac{dx}{\left(\frac{x+4}{x-3}\right)^{\frac{8}{7}}(x-3)^{2}}$.
Let $t = \frac{x+4}{x-3}$. Then $dt = \frac{(x-3)(1) - (x+4)(1)}{(x-3)^2} dx = \frac{-7}{(x-3)^2} dx$,which implies $\frac{dx}{(x-3)^2} = -\frac{1}{7} dt$.
Substituting these into the integral,we get $I = \int \frac{1}{t^{\frac{8}{7}}} \left(-\frac{1}{7} dt\right) = -\frac{1}{7} \int t^{-\frac{8}{7}} dt$.
Integrating,we get $I = -\frac{1}{7} \left( \frac{t^{-\frac{8}{7} + 1}}{-\frac{8}{7} + 1} \right) + C = -\frac{1}{7} \left( \frac{t^{-\frac{1}{7}}}{-\frac{1}{7}} \right) + C = t^{-\frac{1}{7}} + C$.
Substituting $t = \frac{x+4}{x-3}$ back,we get $I = \left(\frac{x+4}{x-3}\right)^{-\frac{1}{7}} + C = \left(\frac{x-3}{x+4}\right)^{\frac{1}{7}} + C$.

(D) For the function $f(x)$ to be continuous at $x = 0$, the left-hand limit, right-hand limit, and the value of the function at $x = 0$ must be equal.
$1$. Left-hand limit $(LHL)$:
$\lim_{x \to 0^-} f(x) = \lim_{x \to 0} \left( \frac{\sin(a+2)x}{x} + \frac{\sin x}{x} \right) = (a+2) + 1 = a+3$.
$2$. Right-hand limit $(RHL)$:
$\lim_{x \to 0^+} f(x) = \lim_{x \to 0} \frac{(x+3x^2)^{1/3} - x^{1/3}}{x^{4/3}} = \lim_{x \to 0} \frac{x^{1/3}((1+3x)^{1/3} - 1)}{x^{4/3}} = \lim_{x \to 0} \frac{(1+3x)^{1/3} - 1}{x}$.
Using the binomial expansion $(1+u)^n \approx 1 + nu$ for small $u$:
$\lim_{x \to 0} \frac{(1 + \frac{1}{3}(3x)) - 1}{x} = \lim_{x \to 0} \frac{1+x-1}{x} = 1$.
$3$. Value at $x=0$:
$f(0) = b$.
For continuity, $a+3 = b = 1$.
Thus, $a = -2$ and $b = 1$.
Therefore, $a+2b = -2 + 2(1) = 0$.

(C) Given $f(x) = a + bx + cx^2$.
Calculating the definite integral:
$\int_{0}^{1} (a + bx + cx^2) dx = [ax + \frac{bx^2}{2} + \frac{cx^3}{3}]_{0}^{1} = a + \frac{b}{2} + \frac{c}{3}$.
Simplifying the expression:
$a + \frac{b}{2} + \frac{c}{3} = \frac{6a + 3b + 2c}{6}$.
Now,evaluate the terms in option $C$:
$f(0) = a$.
$f(1) = a + b + c$.
$f(\frac{1}{2}) = a + \frac{b}{2} + \frac{c}{4}$.
Substituting these into $\frac{1}{6} \{f(0) + f(1) + 4f(\frac{1}{2})\}$:
$= \frac{1}{6} \{a + (a + b + c) + 4(a + \frac{b}{2} + \frac{c}{4})\}$
$= \frac{1}{6} \{a + a + b + c + 4a + 2b + c\}$
$= \frac{1}{6} \{6a + 3b + 2c\} = a + \frac{b}{2} + \frac{c}{3}$.
Thus,the correct option is $C$.

(D) The given differential equation is $(x+1) dy = ((x+1)^{2} + y - 3) dx$.
Rearranging the terms,we get $(x+1) dy - y dx = ((x+1)^{2} - 3) dx$.
Dividing both sides by $(x+1)^{2}$,we obtain $\frac{(x+1) dy - y dx}{(x+1)^{2}} = \left(1 - \frac{3}{(x+1)^{2}}\right) dx$.
This is equivalent to $d\left(\frac{y}{x+1}\right) = \left(1 - \frac{3}{(x+1)^{2}}\right) dx$.
Integrating both sides,we get $\frac{y}{x+1} = x + \frac{3}{x+1} + C$.
Given $y(2) = 0$,we substitute $x=2$ and $y=0$: $0 = 2 + \frac{3}{3} + C \Rightarrow 0 = 3 + C \Rightarrow C = -3$.
Thus,the solution is $\frac{y}{x+1} = x + \frac{3}{x+1} - 3$.
Multiplying by $(x+1)$,we get $y = x(x+1) + 3 - 3(x+1) = x^{2} + x + 3 - 3x - 3 = x^{2} - 2x$.
For $y(3)$,we substitute $x=3$: $y(3) = 3^{2} - 2(3) = 9 - 6 = 3$.

(B) Given vectors are $\overrightarrow{p}=(a+1) \hat{i}+a \hat{j}+a \hat{k}$,$\overrightarrow{q}=a \hat{i}+(a+1) \hat{j}+a \hat{k}$,and $\overrightarrow{r}=a \hat{i}+a \hat{j}+(a+1) \hat{k}$.
Since $\overrightarrow{p}, \overrightarrow{q}, \overrightarrow{r}$ are coplanar,their scalar triple product is zero:
$\left|\begin{array}{ccc} a+1 & a & a \\ a & a+1 & a \\ a & a & a+1 \end{array}\right|=0$
Applying $R_1 \to R_1+R_2+R_3$,we get $(3a+1) \left|\begin{array}{ccc} 1 & 1 & 1 \\ a & a+1 & a \\ a & a & a+1 \end{array}\right|=0$,which simplifies to $(3a+1)(1)=0$,so $a = -\frac{1}{3}$.
Now,$\overrightarrow{p} \cdot \overrightarrow{q} = a^2 + a(a+1) + a^2 = 3a^2+a = 3(-\frac{1}{3})^2 + (-\frac{1}{3}) = \frac{1}{3} - \frac{1}{3} = 0$. Wait,let's recompute: $\overrightarrow{p} \cdot \overrightarrow{q} = a(a+1) + a(a+1) + a^2 = 2a^2+2a+a^2 = 3a^2+2a$. For $a=-1/3$,$\overrightarrow{p} \cdot \overrightarrow{q} = 3(1/9) - 2/3 = 1/3 - 2/3 = -1/3$.
Also,$|\overrightarrow{r}|^2 = |\overrightarrow{q}|^2 = a^2 + a^2 + (a+1)^2 = 3a^2+2a+1 = 3(1/9) - 2/3 + 1 = 1/3 - 2/3 + 1 = 2/3$.
$\overrightarrow{r} \cdot \overrightarrow{q} = a^2 + a(a+1) + a(a+1) = a^2 + 2a^2 + 2a = 3a^2+2a = -1/3$.
Using Lagrange's identity,$|\overrightarrow{r} \times \overrightarrow{q}|^2 = |\overrightarrow{r}|^2 |\overrightarrow{q}|^2 - (\overrightarrow{r} \cdot \overrightarrow{q})^2 = (2/3)(2/3) - (-1/3)^2 = 4/9 - 1/9 = 3/9 = 1/3$.
Given $3(\overrightarrow{p} \cdot \overrightarrow{q})^2 - \lambda |\overrightarrow{r} \times \overrightarrow{q}|^2 = 0$,we have $3(-1/3)^2 - \lambda(1/3) = 0 \Rightarrow 3(1/9) - \lambda/3 = 0 \Rightarrow 1/3 = \lambda/3 \Rightarrow \lambda = 1$.

(A) Let the points be $P(1, -1, 3)$ and $Q(2, -4, 11)$. The vector $\overrightarrow{PQ} = (2-1)\hat{i} + (-4 - (-1))\hat{j} + (11-3)\hat{k} = \hat{i} - 3\hat{j} + 8\hat{k}$.
Let the points on the line be $A(-1, 2, 3)$ and $B(3, -2, 10)$. The vector $\overrightarrow{AB} = (3 - (-1))\hat{i} + (-2-2)\hat{j} + (10-3)\hat{k} = 4\hat{i} - 4\hat{j} + 7\hat{k}$.
The magnitude of $\overrightarrow{AB}$ is $|\overrightarrow{AB}| = \sqrt{4^2 + (-4)^2 + 7^2} = \sqrt{16 + 16 + 49} = \sqrt{81} = 9$.
The projection of $\overrightarrow{PQ}$ on $\overrightarrow{AB}$ is given by the formula $\left| \frac{\overrightarrow{PQ} \cdot \overrightarrow{AB}}{|\overrightarrow{AB}|} \right|$.
Calculating the dot product: $\overrightarrow{PQ} \cdot \overrightarrow{AB} = (1)(4) + (-3)(-4) + (8)(7) = 4 + 12 + 56 = 72$.
Therefore,the projection is $\left| \frac{72}{9} \right| = 8$.

(D) Given $f(x)$ is continuous at $x=1$ and $x=3$.
For continuity at $x=1$: $\lim_{x \rightarrow 1^{-}} f(x) = \lim_{x \rightarrow 1^{+}} f(x) \Rightarrow a e + b e^{-1} = c(1)^2 \Rightarrow a e + b/e = c \Rightarrow b = c e - a e^2 \quad (1)$.
For continuity at $x=3$: $\lim_{x \rightarrow 3^{-}} f(x) = \lim_{x \rightarrow 3^{+}} f(x) \Rightarrow c(3)^2 = a(3)^2 + 2c(3) \Rightarrow 9c = 9a + 6c \Rightarrow 3c = 9a \Rightarrow c = 3a \quad (2)$.
Substitute $(2)$ into $(1)$: $b = (3a)e - a e^2 = a(3e - e^2) \quad (3)$.
Now,$f^{\prime}(x) = \begin{cases} a e^x - b e^{-x}, & -1 < x < 1 \\ 2cx, & 1 < x < 3 \\ 2ax + 2c, & 3 < x < 4 \end{cases}$.
Given $f^{\prime}(0) + f^{\prime}(2) = e$.
$f^{\prime}(0) = a e^0 - b e^0 = a - b$.
$f^{\prime}(2) = 2c(2) = 4c$.
So,$a - b + 4c = e$.
Substitute $b = 3ae - ae^2$ and $c = 3a$ into the equation:
$a - (3ae - ae^2) + 4(3a) = e$.
$a - 3ae + ae^2 + 12a = e$.
$a(e^2 - 3e + 13) = e$.
Therefore,$a = \frac{e}{e^2 - 3e + 13}$.

(A) Let $B_{1}$ be the event that Box-$I$ is selected and $B_{2}$ be the event that Box-$II$ is selected.
$P(B_{1}) = P(B_{2}) = \frac{1}{2}$.
Let $E$ be the event that the selected card is a non-prime number.
In Box-$I$ (cards $1$ to $30$),the prime numbers are $\{2, 3, 5, 7, 11, 13, 17, 19, 23, 29\}$ ($10$ primes). Thus,there are $30 - 10 = 20$ non-prime numbers. So,$P(E|B_{1}) = \frac{20}{30} = \frac{2}{3}$.
In Box-$II$ (cards $31$ to $50$),the prime numbers are $\{31, 37, 41, 43, 47\}$ ($5$ primes). Thus,there are $20 - 5 = 15$ non-prime numbers. So,$P(E|B_{2}) = \frac{15}{20} = \frac{3}{4}$.
Using Bayes' Theorem,the probability that the card was drawn from Box-$I$ given it is non-prime is:
$P(B_{1}|E) = \frac{P(B_{1})P(E|B_{1})}{P(B_{1})P(E|B_{1}) + P(B_{2})P(E|B_{2})}$
$P(B_{1}|E) = \frac{\frac{1}{2} \times \frac{2}{3}}{\frac{1}{2} \times \frac{2}{3} + \frac{1}{2} \times \frac{3}{4}} = \frac{\frac{1}{3}}{\frac{1}{3} + \frac{3}{8}} = \frac{\frac{1}{3}}{\frac{8+9}{24}} = \frac{1}{3} \times \frac{24}{17} = \frac{8}{17}$.

(B) The given equations are $\frac{|x|}{2}+\frac{|y|}{3}=1$ (a rhombus) and $\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$ (an ellipse).
The area of the ellipse is given by $A_{e} = \pi ab = \pi \times 2 \times 3 = 6\pi$.
The region $\frac{|x|}{2}+\frac{|y|}{3}=1$ represents a rhombus with vertices at $(\pm 2, 0)$ and $(0, \pm 3)$.
The area of this rhombus is $A_{r} = \frac{1}{2} \times d_{1} \times d_{2} = \frac{1}{2} \times 4 \times 6 = 12$.
The required area is the area inside the ellipse but outside the rhombus,which is $A = A_{e} - A_{r} = 6\pi - 12 = 6(\pi - 2)$.

(C) For a system of linear equations to have no solution,the determinant of the coefficient matrix $D$ must be $0$,and at least one of the Cramer's determinants $(D_x, D_y, D_z)$ must be non-zero.
The determinant $D$ is given by:
$D = \begin{vmatrix} 2 & -1 & 2 \\ 1 & -2 & \lambda \\ 1 & \lambda & 1 \end{vmatrix} = 2(-2 - \lambda^2) + 1(1 - \lambda) + 2(\lambda + 2) = -4 - 2\lambda^2 + 1 - \lambda + 2\lambda + 4 = -2\lambda^2 + \lambda + 1$
Setting $D = 0$:
$-2\lambda^2 + \lambda + 1 = 0 \Rightarrow 2\lambda^2 - \lambda - 1 = 0 \Rightarrow (2\lambda + 1)(\lambda - 1) = 0$
Thus,$\lambda = 1$ or $\lambda = -\frac{1}{2}$.
Now,check $D_x$ for these values:
$D_x = \begin{vmatrix} 2 & -1 & 2 \\ -4 & -2 & \lambda \\ 4 & \lambda & 1 \end{vmatrix} = 2(-2 - \lambda^2) + 1(-4 - 4\lambda) + 2(-4\lambda + 8) = -4 - 2\lambda^2 - 4 - 4\lambda - 8\lambda + 16 = -2\lambda^2 - 12\lambda + 8$
For $\lambda = 1$,$D_x = -2 - 12 + 8 = -6 \neq 0$.
For $\lambda = -\frac{1}{2}$,$D_x = -2(\frac{1}{4}) - 12(-\frac{1}{2}) + 8 = -0.5 + 6 + 8 = 13.5 \neq 0$.
Since $D=0$ and $D_x \neq 0$ for both values,the system has no solution for both $\lambda = 1$ and $\lambda = -\frac{1}{2}$.
Therefore,$S = \{1, -\frac{1}{2}\}$,which contains exactly two elements.

(D) Given $A$ is a $2 \times 2$ matrix with entries from $\{0, 1\}$ and $|A| \neq 0$.
For statement $(P)$: If $A \neq I_{2}$,then $|A| = -1$.
Consider $A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$. Here $A \neq I_{2}$ and $|A| = (1)(1) - (1)(0) = 1$.
Since we found a case where $A \neq I_{2}$ but $|A| = 1$,statement $(P)$ is false.
For statement $(Q)$: If $|A| = 1$,then $\operatorname{tr}(A) = 2$.
The possible $2 \times 2$ matrices with entries from $\{0, 1\}$ such that $|A| = 1$ are:
$A_1 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \implies \operatorname{tr}(A_1) = 1+1 = 2$
$A_2 = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \implies \operatorname{tr}(A_2) = 1+1 = 2$
$A_3 = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} \implies \operatorname{tr}(A_3) = 1+1 = 2$
In all cases where $|A| = 1$,the trace is $2$. Thus,statement $(Q)$ is true.
Therefore,$(P)$ is false and $(Q)$ is true.

(D) Given the differential equation: $\frac{2+\sin x}{y+1} \frac{dy}{dx} = -\cos x$.
Separating the variables,we get: $\frac{dy}{y+1} = \frac{-\cos x}{2+\sin x} dx$.
Integrating both sides: $\int \frac{dy}{y+1} = -\int \frac{\cos x}{2+\sin x} dx$.
This gives: $\ln(y+1) = -\ln(2+\sin x) + C$.
Using the initial condition $y(0) = 1$: $\ln(1+1) = -\ln(2+\sin 0) + C \Rightarrow \ln 2 = -\ln 2 + C \Rightarrow C = 2\ln 2 = \ln 4$.
Thus,$\ln(y+1) = \ln\left(\frac{4}{2+\sin x}\right)$,which implies $y+1 = \frac{4}{2+\sin x}$,or $y(x) = \frac{4}{2+\sin x} - 1$.
For $x = \pi$,$a = y(\pi) = \frac{4}{2+\sin \pi} - 1 = \frac{4}{2} - 1 = 1$.
Now,find $b = \frac{dy}{dx}$ at $x = \pi$: $\frac{dy}{dx} = \frac{-\cos x}{2+\sin x} (y+1) = \frac{-\cos x}{2+\sin x} \left(\frac{4}{2+\sin x}\right) = \frac{-4\cos x}{(2+\sin x)^2}$.
At $x = \pi$,$b = \frac{-4\cos \pi}{(2+\sin \pi)^2} = \frac{-4(-1)}{(2+0)^2} = \frac{4}{4} = 1$.
Therefore,the ordered pair $(a, b) = (1, 1)$.

(C) The slope of the line joining $\left(0, \frac{3}{2}\right)$ and $\left(\frac{1}{2}, 2\right)$ is $m = \frac{2 - \frac{3}{2}}{\frac{1}{2} - 0} = \frac{\frac{1}{2}}{\frac{1}{2}} = 1$.
Since the tangent at $(a, b)$ is parallel to this line,the slope of the tangent $\left. \frac{dy}{dx} \right|_{(a, b)} = 1$.
Differentiating the curve $y = x + \sin y$ with respect to $x$,we get $\frac{dy}{dx} = 1 + \cos y \cdot \frac{dy}{dx}$.
Substituting the point $(a, b)$ and the slope $1$,we have $1 = 1 + \cos b \cdot (1)$.
This implies $\cos b = 0$,which means $\sin b = \pm 1$.
Since $(a, b)$ lies on the curve,$b = a + \sin b$.
Therefore,$b - a = \sin b$.
Taking the absolute value on both sides,$|b - a| = |\sin b| = 1$.

(D) The point $P$ on the curve $y=x^{2}+7x+2$ nearest to the line $y=3x-3$ is the point where the tangent to the curve is parallel to the given line.
$1$. Find the slope of the tangent at any point $(x, y)$ on the curve:
$\frac{dy}{dx} = \frac{d}{dx}(x^{2}+7x+2) = 2x+7$
$2$. Since the tangent is parallel to the line $y=3x-3$,its slope must be equal to the slope of the line,which is $3$:
$2x+7 = 3$
$2x = -4$
$x = -2$
$3$. Find the $y$-coordinate of $P$ by substituting $x=-2$ into the curve equation:
$y = (-2)^{2} + 7(-2) + 2 = 4 - 14 + 2 = -8$
So,the point $P$ is $(-2, -8)$.
$4$. The normal at $P$ is perpendicular to the tangent. The slope of the tangent is $3$,so the slope of the normal is $-\frac{1}{3}$.
$5$. The equation of the normal at $P(-2, -8)$ with slope $m = -\frac{1}{3}$ is:
$y - (-8) = -\frac{1}{3}(x - (-2))$
$y + 8 = -\frac{1}{3}(x + 2)$
$3(y + 8) = -(x + 2)$
$3y + 24 = -x - 2$
$x + 3y + 26 = 0$

(B) The given line is $2x = 3y, z = 1$,which can be written as $\frac{x}{3} = \frac{y}{2}, z = 1$. The direction vector of this line is $\vec{v} = 3\hat{i} + 2\hat{j} + 0\hat{k}$.
The plane passes through points $A(1, 2, 1)$ and $B(2, 1, 2)$. The vector $\vec{AB} = (2-1)\hat{i} + (1-2)\hat{j} + (2-1)\hat{k} = \hat{i} - \hat{j} + \hat{k}$.
The normal vector $\vec{n}$ to the plane is given by the cross product of $\vec{AB}$ and $\vec{v}$:
$\vec{n} = \vec{AB} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 3 & 2 & 0 \end{vmatrix} = \hat{i}(0-2) - \hat{j}(0-3) + \hat{k}(2+3) = -2\hat{i} + 3\hat{j} + 5\hat{k}$.
The equation of the plane passing through $(1, 2, 1)$ with normal $\vec{n} = -2\hat{i} + 3\hat{j} + 5\hat{k}$ is:
$-2(x-1) + 3(y-2) + 5(z-1) = 0$
$-2x + 2 + 3y - 6 + 5z - 5 = 0$
$-2x + 3y + 5z - 9 = 0$ or $2x - 3y - 5z + 9 = 0$.
Now,check the options:
For $(-2, 0, 1)$: $2(-2) - 3(0) - 5(1) + 9 = -4 - 0 - 5 + 9 = 0$.
Thus,the plane passes through $(-2, 0, 1)$.

(A) For the function $f(x) = \sin^{-1}\left(\frac{|x|+5}{x^2+1}\right)$ to be defined,the argument must satisfy $-1 \leq \frac{|x|+5}{x^2+1} \leq 1$.
Since $|x|+5 > 0$ and $x^2+1 > 0$ for all $x \in \mathbb{R}$,the condition $\frac{|x|+5}{x^2+1} \geq -1$ is always satisfied.
Thus,we only need to solve $\frac{|x|+5}{x^2+1} \leq 1$.
Multiplying by $x^2+1$ (which is positive),we get $|x|+5 \leq x^2+1$.
Rearranging the terms,we have $x^2 - |x| - 4 \geq 0$.
Let $t = |x|$,where $t \geq 0$. Then $t^2 - t - 4 \geq 0$.
The roots of $t^2 - t - 4 = 0$ are $t = \frac{1 \pm \sqrt{1 - 4(1)(-4)}}{2} = \frac{1 \pm \sqrt{17}}{2}$.
Since $t = |x| \geq 0$,we must have $t \geq \frac{1+\sqrt{17}}{2}$.
Therefore,$|x| \geq \frac{1+\sqrt{17}}{2}$,which implies $x \in \left(-\infty, -\frac{1+\sqrt{17}}{2}\right] \cup \left[\frac{1+\sqrt{17}}{2}, \infty\right)$.
Comparing this with the given domain $(-\infty, -a] \cup [a, \infty)$,we find $a = \frac{1+\sqrt{17}}{2}$.

(D) Since $p(x)$ has relative extrema at $x=1$ and $x=2$,we have $p'(x) = 0$ at $x=1$ and $x=2$.
Thus,$p'(x) = A(x-1)(x-2) = A(x^2 - 3x + 2)$.
Integrating $p'(x)$,we get $p(x) = A(\frac{x^3}{3} - \frac{3x^2}{2} + 2x) + C$.
Given $p(1) = 8$,we have $8 = A(\frac{1}{3} - \frac{3}{2} + 2) + C = A(\frac{2-9+12}{6}) + C = \frac{5A}{6} + C$,so $48 = 5A + 6C$ (Equation $1$).
Given $p(2) = 4$,we have $4 = A(\frac{8}{3} - 6 + 4) + C = A(\frac{8-6}{3}) + C = \frac{2A}{3} + C$,so $12 = 2A + 3C$ (Equation $2$).
Multiplying Equation $2$ by $2$,we get $24 = 4A + 6C$.
Subtracting this from Equation $1$,we get $48 - 24 = (5A - 4A) + (6C - 6C)$,which gives $A = 24$.
Substituting $A = 24$ into Equation $2$,we get $12 = 2(24) + 3C$,so $12 = 48 + 3C$,which implies $3C = -36$,so $C = -12$.
Since $p(0) = C$,we have $p(0) = -12$.

(A) Let $f(x) = ||x-1|-x|$.
We analyze the expression inside the absolute value:
If $x \geq 1$,then $|x-1| = x-1$,so $f(x) = |(x-1)-x| = |-1| = 1$.
If $x < 1$,then $|x-1| = 1-x$,so $f(x) = |(1-x)-x| = |1-2x|$.
Thus,$f(x) = \begin{cases} |1-2x|, & 0 \leq x < 1 \\ 1, & 1 \leq x \leq 2 \end{cases}$.
Now,we evaluate the integral:
$\int_{0}^{2} f(x) dx = \int_{0}^{1} |1-2x| dx + \int_{1}^{2} 1 dx$.
For the first part,$|1-2x| = 1-2x$ when $x \leq 1/2$ and $2x-1$ when $x > 1/2$:
$\int_{0}^{1} |1-2x| dx = \int_{0}^{1/2} (1-2x) dx + \int_{1/2}^{1} (2x-1) dx$
$= [x-x^2]_{0}^{1/2} + [x^2-x]_{1/2}^{1}$
$= (1/2 - 1/4) - 0 + (1-1) - (1/4 - 1/2) = 1/4 + 1/4 = 1/2$.
For the second part:
$\int_{1}^{2} 1 dx = [x]_{1}^{2} = 2-1 = 1$.
Total integral = $1/2 + 1 = 1.5$.

(B) Given that $\vec{a}, \vec{b}, \vec{c}$ are unit vectors,so $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1$.
Expanding the given equation $|\vec{a}-\vec{b}|^{2}+|\vec{a}-\vec{c}|^{2}=8$:
$(|\vec{a}|^{2} + |\vec{b}|^{2} - 2\vec{a}\cdot\vec{b}) + (|\vec{a}|^{2} + |\vec{c}|^{2} - 2\vec{a}\cdot\vec{c}) = 8$
Since $|\vec{a}|^{2} = |\vec{b}|^{2} = |\vec{c}|^{2} = 1$,we have:
$(1 + 1 - 2\vec{a}\cdot\vec{b}) + (1 + 1 - 2\vec{a}\cdot\vec{c}) = 8$
$4 - 2(\vec{a}\cdot\vec{b} + \vec{a}\cdot\vec{c}) = 8$
$-2(\vec{a}\cdot\vec{b} + \vec{a}\cdot\vec{c}) = 4$
$\vec{a}\cdot\vec{b} + \vec{a}\cdot\vec{c} = -2$
Now,evaluate $|\vec{a}+2\vec{b}|^{2}+|\vec{a}+2\vec{c}|^{2}$:
$= (|\vec{a}|^{2} + 4|\vec{b}|^{2} + 4\vec{a}\cdot\vec{b}) + (|\vec{a}|^{2} + 4|\vec{c}|^{2} + 4\vec{a}\cdot\vec{c})$
$= (1 + 4 + 4\vec{a}\cdot\vec{b}) + (1 + 4 + 4\vec{a}\cdot\vec{c})$
$= 10 + 4(\vec{a}\cdot\vec{b} + \vec{a}\cdot\vec{c})$
Substitute the value $\vec{a}\cdot\vec{b} + \vec{a}\cdot\vec{c} = -2$:
$= 10 + 4(-2) = 10 - 8 = 2$.

(A) Given the functional equation $f(x+y)=f(x)+f(y)$,this is Cauchy's functional equation,which implies $f(x)=cx$ for some constant $c$.
Since $f(1)=2$,we have $c(1)=2$,so $c=2$. Thus,$f(x)=2x$.
Now,we calculate $g(n) = \sum_{k=1}^{n-1} f(k) = \sum_{k=1}^{n-1} 2k$.
Using the sum formula $\sum_{k=1}^{m} k = \frac{m(m+1)}{2}$,we get $g(n) = 2 \times \frac{(n-1)n}{2} = n(n-1)$.
We are given $g(n)=20$,so $n(n-1)=20$.
$n^2 - n - 20 = 0$.
$(n-5)(n+4) = 0$.
Since $n \in N$,we must have $n=5$.

(D) Given $A^{T} A = I$,the matrix $A$ is an orthogonal matrix.
For the matrix $A = \begin{bmatrix} a & b & c \\ b & c & a \\ c & a & b \end{bmatrix}$,the condition $A^{T} A = I$ implies:
$a^2 + b^2 + c^2 = 1$
$ab + bc + ca = 0$
We know the identity: $(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) = 1 + 2(0) = 1$.
Thus,$a+b+c = \pm 1$.
Using the algebraic identity: $a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - (ab + bc + ca))$.
Substituting the known values: $2 - 3abc = (\pm 1)(1 - 0) = \pm 1$.
Case $1$: $2 - 3abc = 1 \Rightarrow 3abc = 1 \Rightarrow abc = \frac{1}{3}$.
Case $2$: $2 - 3abc = -1 \Rightarrow 3abc = 3 \Rightarrow abc = 1$.
Comparing with the given options,the possible value is $\frac{1}{3}$.

(A) For $x \neq 0$,$f'(x) = \frac{d}{dx} \left( \frac{\ln(1+x)}{x} \right) = \frac{x \cdot \frac{1}{1+x} - \ln(1+x)}{x^2} = \frac{x - (1+x)\ln(1+x)}{x^2(1+x)}$.
Let $h(x) = x - (1+x)\ln(1+x)$.
Then $h'(x) = 1 - [\ln(1+x) + (1+x) \cdot \frac{1}{1+x}] = 1 - \ln(1+x) - 1 = -\ln(1+x)$.
For $x \in (-1, 0)$,$1+x \in (0, 1)$,so $\ln(1+x) < 0$,which implies $h'(x) > 0$.
For $x \in (0, \infty)$,$1+x > 1$,so $\ln(1+x) > 0$,which implies $h'(x) < 0$.
Since $h(0) = 0 - (1)\ln(1) = 0$,$h(x)$ increases on $(-1, 0)$ and decreases on $(0, \infty)$.
Thus,$h(x) < h(0) = 0$ for all $x \in (-1, \infty) \setminus \{0\}$.
Since $x^2(1+x) > 0$ for all $x \in (-1, \infty) \setminus \{0\}$,$f'(x) = \frac{h(x)}{x^2(1+x)} < 0$ for all $x \in (-1, \infty) \setminus \{0\}$.
Therefore,the function $f$ is strictly decreasing on $(-1, \infty)$.

(B) Given the equation of the curve $y=(1+x)^{2y}+\cos^{2}(\sin^{-1} x)$.
At $x=0$,$y=(1+0)^{2y}+\cos^{2}(\sin^{-1} 0) = 1+1 = 2$.
So,we need to find the normal at the point $(0, 2)$.
Rewrite the equation as $y=e^{2y \ln(1+x)} + (1-x^2)$.
Differentiating with respect to $x$:
$y' = e^{2y \ln(1+x)} \left[ 2y \cdot \frac{1}{1+x} + \ln(1+x) \cdot 2y' \right] - 2x$.
Substituting $x=0$ and $y=2$:
$y' = e^{2(2) \ln(1)} \left[ 2(2) \cdot \frac{1}{1+0} + \ln(1) \cdot 2y' \right] - 2(0)$.
$y' = e^0 [4 + 0] - 0 = 4$.
Thus,the slope of the tangent $m_t = 4$.
The slope of the normal $m_n = -\frac{1}{m_t} = -\frac{1}{4}$.
The equation of the normal at $(0, 2)$ is $y - 2 = -\frac{1}{4}(x - 0)$.
$4y - 8 = -x$,which simplifies to $x + 4y = 8$.

(B) The normal vector $\overrightarrow{n}$ to the plane is perpendicular to both lines. Thus,$\overrightarrow{n}$ is the cross product of the direction vectors of the two lines:
$\overrightarrow{n} = (\hat{i} - 2\hat{j} + 2\hat{k}) \times (2\hat{i} + 3\hat{j} - \hat{k})$
$\overrightarrow{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 2 \\ 2 & 3 & -1 \end{vmatrix} = \hat{i}(2 - 6) - \hat{j}(-1 - 4) + \hat{k}(3 + 4) = -4\hat{i} + 5\hat{j} + 7\hat{k}$
The equation of the plane passing through $(3, 1, 1)$ with normal vector $\overrightarrow{n} = -4\hat{i} + 5\hat{j} + 7\hat{k}$ is:
$-4(x - 3) + 5(y - 1) + 7(z - 1) = 0$
$-4x + 12 + 5y - 5 + 7z - 7 = 0$
$-4x + 5y + 7z = 0$
Since the plane passes through $(\alpha, -3, 5)$,we substitute these coordinates into the plane equation:
$-4(\alpha) + 5(-3) + 7(5) = 0$
$-4\alpha - 15 + 35 = 0$
$-4\alpha + 20 = 0$
$4\alpha = 20$
$\alpha = 5$

(A) Given that $E_{1}, E_{2}, E_{3}$ are pairwise independent events,we have $P(E_{1} \cap E_{2}) = P(E_{1})P(E_{2})$,$P(E_{2} \cap E_{3}) = P(E_{2})P(E_{3})$,and $P(E_{3} \cap E_{1}) = P(E_{3})P(E_{1})$.
Also,$P(E_{1} \cap E_{2} \cap E_{3}) = 0$.
We need to find $P(E_{2}^{C} \cap E_{3}^{C} | E_{1})$.
Using the definition of conditional probability,$P(E_{2}^{C} \cap E_{3}^{C} | E_{1}) = \frac{P(E_{1} \cap E_{2}^{C} \cap E_{3}^{C})}{P(E_{1})}$.
By De Morgan's Law,$E_{2}^{C} \cap E_{3}^{C} = (E_{2} \cup E_{3})^{C}$.
Thus,$E_{1} \cap (E_{2} \cup E_{3})^{C} = E_{1} \setminus (E_{1} \cap (E_{2} \cup E_{3})) = E_{1} \setminus ((E_{1} \cap E_{2}) \cup (E_{1} \cap E_{3}))$.
Using the inclusion-exclusion principle for probability,$P(E_{1} \cap (E_{2} \cup E_{3})) = P(E_{1} \cap E_{2}) + P(E_{1} \cap E_{3}) - P(E_{1} \cap E_{2} \cap E_{3})$.
Substituting the given values,$P(E_{1} \cap (E_{2} \cup E_{3})) = P(E_{1})P(E_{2}) + P(E_{1})P(E_{3}) - 0 = P(E_{1})(P(E_{2}) + P(E_{3}))$.
Therefore,$P(E_{1} \cap E_{2}^{C} \cap E_{3}^{C}) = P(E_{1}) - P(E_{1} \cap (E_{2} \cup E_{3})) = P(E_{1}) - P(E_{1})(P(E_{2}) + P(E_{3})) = P(E_{1})(1 - P(E_{2}) - P(E_{3}))$.
Finally,$P(E_{2}^{C} \cap E_{3}^{C} | E_{1}) = \frac{P(E_{1})(1 - P(E_{2}) - P(E_{3}))}{P(E_{1})} = 1 - P(E_{2}) - P(E_{3}) = (1 - P(E_{3})) - P(E_{2}) = P(E_{3}^{C}) - P(E_{2})$.

(B) Given $P = \begin{bmatrix} 1 & 2 & 1 \\ -2 & 3 & -4 \\ 1 & 9 & -1 \end{bmatrix}$.
First,we find the determinant of $P$:
$|P| = 1(-3 + 36) - 2(2 + 4) + 1(-18 - 3) = 33 - 12 - 21 = 0$.
Since $|P| = 0$,the system $PX = 0$ has non-trivial solutions.
The equations are:
$x + 2y + z = 0$ $(i)$
$-2x + 3y - 4z = 0$ (ii)
$x + 9y - z = 0$ (iii)
From $(i)$ and (iii),adding them gives $2x + 11y = 0 \Rightarrow x = -\frac{11}{2}y$.
Substituting $x$ into $(i)$: $-\frac{11}{2}y + 2y + z = 0 \Rightarrow z = \frac{7}{2}y$.
Let $y = \lambda$,then $x = -\frac{11}{2}\lambda$ and $z = \frac{7}{2}\lambda$.
Given $x^{2} + y^{2} + z^{2} = 1$,we substitute the values:
$(-\frac{11}{2}\lambda)^{2} + \lambda^{2} + (\frac{7}{2}\lambda)^{2} = 1$
$\frac{121}{4}\lambda^{2} + \lambda^{2} + \frac{49}{4}\lambda^{2} = 1$
$\frac{121 + 4 + 49}{4}\lambda^{2} = 1 \Rightarrow \frac{174}{4}\lambda^{2} = 1 \Rightarrow \lambda^{2} = \frac{4}{174} = \frac{2}{87}$.
Since $\lambda^{2} = \frac{2}{87}$,there are two possible values for $\lambda$ $(\lambda = \pm \sqrt{\frac{2}{87}})$.
Thus,there are exactly two solutions for $(x, y, z)$.

(D) The region $R$ is bounded by the parabola $y=x^{2}$ and the line $y=2x$. The intersection points are found by $x^{2}=2x$,which gives $x=0$ and $x=2$. Thus,the points are $(0,0)$ and $(2,4)$.
The total area $A$ of region $R$ is given by:
$A = \int_{0}^{2} (2x - x^{2}) dx = [x^{2} - \frac{x^{3}}{3}]_{0}^{2} = 4 - \frac{8}{3} = \frac{4}{3}$.
Alternatively,integrating with respect to $y$,the region is bounded by $x = \sqrt{y}$ (right) and $x = y/2$ (left) for $y \in [0, 4]$:
$A = \int_{0}^{4} (\sqrt{y} - \frac{y}{2}) dy = [\frac{2}{3}y^{3/2} - \frac{y^{2}}{4}]_{0}^{4} = \frac{2}{3}(8) - \frac{16}{4} = \frac{16}{3} - 4 = \frac{4}{3}$.
The line $y=\alpha$ divides the area into two equal parts. The area of the lower part (from $y=0$ to $y=\alpha$) is half of the total area:
$\int_{0}^{\alpha} (\sqrt{y} - \frac{y}{2}) dy = \frac{1}{2} \times \frac{4}{3} = \frac{2}{3}$.
Evaluating the integral:
$[\frac{2}{3}y^{3/2} - \frac{y^{2}}{4}]_{0}^{\alpha} = \frac{2}{3} \Rightarrow \frac{2}{3}\alpha^{3/2} - \frac{\alpha^{2}}{4} = \frac{2}{3}$.
Multiplying by $12$ to clear the denominators:
$8\alpha^{3/2} - 3\alpha^{2} = 8 \Rightarrow 3\alpha^{2} - 8\alpha^{3/2} + 8 = 0$.

(B) Given differential equation: $2 x^{2} dy = (2 xy + y^{2}) dx$
$\frac{dy}{dx} = \frac{2xy + y^{2}}{2x^{2}}$
This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.
Substituting these into the equation:
$v + x\frac{dv}{dx} = \frac{2x(vx) + (vx)^{2}}{2x^{2}} = \frac{2x^{2}v + x^{2}v^{2}}{2x^{2}} = v + \frac{v^{2}}{2}$
$x\frac{dv}{dx} = \frac{v^{2}}{2}$
Separating variables:
$\frac{2}{v^{2}} dv = \frac{1}{x} dx$
Integrating both sides:
$\int 2v^{-2} dv = \int \frac{1}{x} dx$
$-2v^{-1} = \ln|x| + C$
$-\frac{2}{v} = \ln|x| + C$
Since $v = \frac{y}{x}$,we have $-\frac{2x}{y} = \ln|x| + C$.
The curve passes through $(1, 2)$,so substitute $x=1, y=2$:
$-\frac{2(1)}{2} = \ln(1) + C \Rightarrow -1 = 0 + C \Rightarrow C = -1$.
Thus,$-\frac{2x}{y} = \ln|x| - 1$,which implies $\frac{2x}{y} = 1 - \ln x$.
$y = \frac{2x}{1 - \ln x} \Rightarrow f(x) = \frac{2x}{1 - \ln x}$.
Now,calculate $f\left(\frac{1}{2}\right)$:
$f\left(\frac{1}{2}\right) = \frac{2(\frac{1}{2})}{1 - \ln(\frac{1}{2})} = \frac{1}{1 - (\ln 1 - \ln 2)} = \frac{1}{1 - (0 - \ln 2)} = \frac{1}{1 + \ln 2}$.

(B) Let $\cos \alpha = \frac{3}{5}$ and $\sin \alpha = \frac{4}{5}$,where $0 < \alpha < \frac{\pi}{2}$.
The expression inside the $\cos^{-1}$ function is $\frac{3}{5} \cos kx - \frac{4}{5} \sin kx$.
Using the trigonometric identity $\cos(A+B) = \cos A \cos B - \sin A \sin B$,we get $\cos \alpha \cos kx - \sin \alpha \sin kx = \cos(\alpha + kx)$.
Thus,$y = \sum_{k=1}^{6} k \cos^{-1}(\cos(\alpha + kx))$.
For values of $x$ near $0$,$\cos^{-1}(\cos(\alpha + kx)) = \alpha + kx$.
Therefore,$y = \sum_{k=1}^{6} k(\alpha + kx) = \sum_{k=1}^{6} (k\alpha + k^2 x)$.
Differentiating with respect to $x$,we get $\frac{dy}{dx} = \sum_{k=1}^{6} k^2$.
At $x = 0$,$\frac{dy}{dx} = \sum_{k=1}^{6} k^2 = 1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2$.
Using the sum of squares formula $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$,for $n=6$:
$\frac{6(7)(13)}{6} = 91$.

(C) Using the section formula,the position vector of point $P$ is given by:
$\overrightarrow{OP} = \frac{\lambda(2\hat{i}+\hat{j}+3\hat{k}) + 1(\hat{i}+\hat{j}+\hat{k})}{\lambda+1} = \frac{2\lambda+1}{\lambda+1}\hat{i} + \frac{\lambda+1}{\lambda+1}\hat{j} + \frac{3\lambda+1}{\lambda+1}\hat{k}$
Now,calculate $\overrightarrow{OB} \cdot \overrightarrow{OP}$:
$\overrightarrow{OB} \cdot \overrightarrow{OP} = (2\hat{i}+\hat{j}+3\hat{k}) \cdot \left( \frac{2\lambda+1}{\lambda+1}\hat{i} + \hat{j} + \frac{3\lambda+1}{\lambda+1}\hat{k} \right) = \frac{4\lambda+2 + \lambda+1 + 9\lambda+3}{\lambda+1} = \frac{14\lambda+6}{\lambda+1}$
Next,calculate $\overrightarrow{OA} \times \overrightarrow{OP}$:
$\overrightarrow{OA} \times \overrightarrow{OP} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ \frac{2\lambda+1}{\lambda+1} & 1 & \frac{3\lambda+1}{\lambda+1} \end{vmatrix} = \left( \frac{3\lambda+1}{\lambda+1} - 1 \right)\hat{i} - \left( \frac{3\lambda+1}{\lambda+1} - \frac{2\lambda+1}{\lambda+1} \right)\hat{j} + \left( 1 - \frac{2\lambda+1}{\lambda+1} \right)\hat{k}$
$= \frac{2\lambda}{\lambda+1}\hat{i} - \frac{\lambda}{\lambda+1}\hat{j} - \frac{\lambda}{\lambda+1}\hat{k}$
$|\overrightarrow{OA} \times \overrightarrow{OP}|^{2} = \frac{4\lambda^{2} + \lambda^{2} + \lambda^{2}}{(\lambda+1)^{2}} = \frac{6\lambda^{2}}{(\lambda+1)^{2}}$
Substituting these into the given equation:
$\frac{14\lambda+6}{\lambda+1} - 3 \left( \frac{6\lambda^{2}}{(\lambda+1)^{2}} \right) = 6$
Multiply by $(\lambda+1)^{2}$:
$(14\lambda+6)(\lambda+1) - 18\lambda^{2} = 6(\lambda+1)^{2}$
$14\lambda^{2} + 20\lambda + 6 - 18\lambda^{2} = 6(\lambda^{2} + 2\lambda + 1)$
$-4\lambda^{2} + 20\lambda + 6 = 6\lambda^{2} + 12\lambda + 6$
$10\lambda^{2} - 8\lambda = 0$
Since $\lambda > 0$,we have $10\lambda = 8 \Rightarrow \lambda = 0.8$.

(A) Let $I = \int_{1}^{2} |2x - [3x]| dx$. Since $1 \le x \le 2$,we have $3 \le 3x \le 6$.
We divide the interval $[1, 2]$ based on the values of $[3x]$:
Case $1$: $1 \le x < 4/3$,then $3 \le 3x < 4$,so $[3x] = 3$. The integrand is $|2x - 3| = 3 - 2x$.
Case $2$: $4/3 \le x < 5/3$,then $4 \le 3x < 5$,so $[3x] = 4$. The integrand is $|2x - 4| = 4 - 2x$.
Case $3$: $5/3 \le x \le 2$,then $5 \le 3x \le 6$,so $[3x] = 5$. The integrand is $|2x - 5| = 5 - 2x$.
Now,$I = \int_{1}^{4/3} (3 - 2x) dx + \int_{4/3}^{5/3} (4 - 2x) dx + \int_{5/3}^{2} (5 - 2x) dx$.
Evaluating each integral:
$\int_{1}^{4/3} (3 - 2x) dx = [3x - x^2]_{1}^{4/3} = (4 - 16/9) - (3 - 1) = 20/9 - 2 = 2/9$.
$\int_{4/3}^{5/3} (4 - 2x) dx = [4x - x^2]_{4/3}^{5/3} = (20/3 - 25/9) - (16/3 - 16/9) = 35/9 - 32/9 = 3/9 = 1/3$.
$\int_{5/3}^{2} (5 - 2x) dx = [5x - x^2]_{5/3}^{2} = (10 - 4) - (25/3 - 25/9) = 6 - 50/9 = 4/9$.
Summing these: $I = 2/9 + 3/9 + 4/9 = 9/9 = 1$.

(A) Let the side length of the cube be $a$. The surface area $S$ of the cube is given by $S = 6a^2$.
Given that $\frac{dS}{dt} = 3.6 \text{ cm}^2/\text{sec}$.
Differentiating $S$ with respect to $t$,we get $\frac{dS}{dt} = 12a \frac{da}{dt}$.
Substituting the given values: $3.6 = 12(10) \frac{da}{dt} \Rightarrow 3.6 = 120 \frac{da}{dt} \Rightarrow \frac{da}{dt} = \frac{3.6}{120} = 0.03 \text{ cm}/\text{sec}$.
The volume $V$ of the cube is $V = a^3$.
Differentiating $V$ with respect to $t$,we get $\frac{dV}{dt} = 3a^2 \frac{da}{dt}$.
Substituting $a = 10$ and $\frac{da}{dt} = 0.03$: $\frac{dV}{dt} = 3(10)^2(0.03) = 3(100)(0.03) = 300 \times 0.03 = 9 \text{ cm}^3/\text{sec}$.

(A) Let $I = \int_{0}^{1/2} \frac{x^{2}}{(1-x^{2})^{3/2}} dx$.
We can rewrite the numerator as $x^{2} = (x^{2}-1) + 1$.
Then,$I = \int_{0}^{1/2} \frac{x^{2}-1}{(1-x^{2})^{3/2}} dx + \int_{0}^{1/2} \frac{1}{(1-x^{2})^{3/2}} dx$.
$I = -\int_{0}^{1/2} \frac{1}{(1-x^{2})^{1/2}} dx + \int_{0}^{1/2} \frac{1}{(1-x^{2})^{3/2}} dx$.
For the first integral,$\int \frac{1}{\sqrt{1-x^{2}}} dx = \sin^{-1}(x)$.
Evaluating from $0$ to $1/2$,we get $-[\sin^{-1}(1/2) - \sin^{-1}(0)] = -\pi/6$.
For the second integral,let $x = \sin(\theta)$,then $dx = \cos(\theta) d\theta$.
When $x=0, \theta=0$; when $x=1/2, \theta=\pi/6$.
$\int_{0}^{\pi/6} \frac{\cos(\theta)}{\cos^{3}(\theta)} d\theta = \int_{0}^{\pi/6} \sec^{2}(\theta) d\theta = [\tan(\theta)]_{0}^{\pi/6} = \tan(\pi/6) - \tan(0) = 1/\sqrt{3}$.
Thus,$I = \frac{1}{\sqrt{3}} - \frac{\pi}{6} = \frac{\sqrt{3}}{3} - \frac{\pi}{6} = \frac{2\sqrt{3}-\pi}{6}$.
Comparing this with $\frac{k}{6}$,we get $k = 2\sqrt{3}-\pi$.