JEE Main 2018 Mathematics Question Paper with Answer and Solution

(B) Let $t = \sqrt{x}$,where $t \ge 0$. The equation becomes $2|t - 3| + t(t - 6) + 6 = 0$.
Case $I$: $0 \le t < 3$ (i.e.,$0 \le x < 9$)
$2(3 - t) + t^2 - 6t + 6 = 0$
$6 - 2t + t^2 - 6t + 6 = 0$
$t^2 - 8t + 12 = 0$
$(t - 6)(t - 2) = 0$
$t = 6$ or $t = 2$.
Since $0 \le t < 3$,we have $t = 2$,which implies $x = 4$.
Case $II$: $t \ge 3$ (i.e.,$x \ge 9$)
$2(t - 3) + t^2 - 6t + 6 = 0$
$2t - 6 + t^2 - 6t + 6 = 0$
$t^2 - 4t = 0$
$t(t - 4) = 0$
$t = 0$ or $t = 4$.
Since $t \ge 3$,we have $t = 4$,which implies $x = 16$.
Thus,$S = \{4, 16\}$,which contains exactly two elements.

(D) Let the height of the tower $MN = h$.
In $\Delta QMN$,we have $\tan 30^o = \frac{MN}{QM}$.
$\therefore QM = \frac{h}{\tan 30^o} = h\sqrt{3}$. Since $M$ is the midpoint of $QR$,$QM = MR = h\sqrt{3}$.
In $\Delta MNP$,the angle of elevation of the top of the tower at $P$ is $45^o$,so $\tan 45^o = \frac{MN}{PM}$.
$\therefore PM = \frac{h}{\tan 45^o} = h$.
In $\Delta PMQ$,since $PM \perp MQ$ (as the tower is vertical to the ground plane),we have $PQ^2 = PM^2 + QM^2$.
Substituting the values: $(200)^2 = h^2 + (h\sqrt{3})^2$.
$40000 = h^2 + 3h^2$.
$40000 = 4h^2$.
$h^2 = 10000$.
$h = 100 \ m$.

(A) Given equation: $8 \cos x \left\{ \cos \left( \frac{\pi}{6} + x \right) \cos \left( \frac{\pi}{6} - x \right) - \frac{1}{2} \right\} = 1$
Using the identity $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$:
$4 \cos x \left\{ \cos \left( \frac{\pi}{3} \right) + \cos(2x) - 1 \right\} = 1$
Since $\cos(\frac{\pi}{3}) = \frac{1}{2}$:
$4 \cos x \left\{ \frac{1}{2} + \cos 2x - 1 \right\} = 1$
$4 \cos x \left\{ \cos 2x - \frac{1}{2} \right\} = 1$
$4 \cos x \left\{ 2 \cos^2 x - 1 - \frac{1}{2} \right\} = 1$
$8 \cos^3 x - 6 \cos x = 1$
$2(4 \cos^3 x - 3 \cos x) = 1$
$2 \cos 3x = 1 \Rightarrow \cos 3x = \frac{1}{2}$
For $x \in [0, \pi]$,$3x \in [0, 3\pi]$.
Solutions for $3x$: $3x = \frac{\pi}{3}, 2\pi - \frac{\pi}{3}, 2\pi + \frac{\pi}{3} = \frac{\pi}{3}, \frac{5\pi}{3}, \frac{7\pi}{3}$.
Thus,$x = \frac{\pi}{9}, \frac{5\pi}{9}, \frac{7\pi}{9}$.
Sum of solutions $= \frac{\pi + 5\pi + 7\pi}{9} = \frac{13\pi}{9}$.
Given sum $= k\pi$,so $k = \frac{13}{9}$.

(B) The roots of the equation $x^2 - x + 1 = 0$ are given by $x = \frac{1 \pm \sqrt{1 - 4}}{2} = \frac{1 \pm i\sqrt{3}}{2}$.
These are $-\omega$ and $-\omega^2$,where $\omega$ is a complex cube root of unity.
Let $\alpha = -\omega$ and $\beta = -\omega^2$.
Then $\alpha^{101} + \beta^{107} = (-\omega)^{101} + (-\omega^2)^{107}$.
$= -\omega^{101} - \omega^{214}$.
Since $\omega^3 = 1$,we have $\omega^{101} = \omega^{3 \times 33 + 2} = \omega^2$ and $\omega^{214} = \omega^{3 \times 71 + 1} = \omega$.
Thus,$\alpha^{101} + \beta^{107} = -(\omega^2 + \omega)$.
Since $1 + \omega + \omega^2 = 0$,we have $\omega + \omega^2 = -1$.
Therefore,$\alpha^{101} + \beta^{107} = -(-1) = 1$.

(C) Step $1$: Select $4$ novels from $6$ different novels in $^6C_4$ ways.
$^6C_4 = \frac{6 \times 5}{2 \times 1} = 15$ ways.
Step $2$: Select $1$ dictionary from $3$ different dictionaries in $^3C_1$ ways.
$^3C_1 = 3$ ways.
Step $3$: Arrange the $4$ selected novels and $1$ dictionary in a row such that the dictionary is always in the middle.
Since the dictionary is fixed in the middle,we only need to arrange the $4$ novels in the remaining $4$ positions.
The number of ways to arrange $4$ novels is $4! = 24$.
Step $4$: Total number of arrangements = $^6C_4 \times ^3C_1 \times 4! = 15 \times 3 \times 24 = 1080$.

(C) Let $f(x) = (x + \sqrt{x^3 - 1})^5 + (x - \sqrt{x^3 - 1})^5$.
Using the binomial expansion formula $(a+b)^n + (a-b)^n = 2[\binom{n}{0}a^n + \binom{n}{2}a^{n-2}b^2 + \binom{n}{4}a^{n-4}b^4 + \dots]$,we get:
$f(x) = 2[\binom{5}{0}x^5 + \binom{5}{2}x^3(\sqrt{x^3-1})^2 + \binom{5}{4}x(\sqrt{x^3-1})^4]$
$f(x) = 2[x^5 + 10x^3(x^3-1) + 5x(x^3-1)^2]$
$f(x) = 2[x^5 + 10x^6 - 10x^3 + 5x(x^6 - 2x^3 + 1)]$
$f(x) = 2[x^5 + 10x^6 - 10x^3 + 5x^7 - 10x^4 + 5x]$
$f(x) = 10x^7 + 20x^6 + 2x^5 - 20x^4 - 20x^3 + 10x$.
The odd degree terms are $10x^7$,$2x^5$,$-20x^3$,and $10x$.
The sum of the coefficients of these terms is $10 + 2 - 20 + 10 = 2$.

(A) The series is $a_n = n^2$ if $n$ is odd,and $a_n = 2n^2$ if $n$ is even.
$B - 2A = \sum_{n=1}^{40} a_n - 2\sum_{n=1}^{20} a_n = \sum_{n=21}^{40} a_n - \sum_{n=1}^{20} a_n$.
For $n \in [21, 40]$,let $k = n-20$,so $n = k+20$. The terms are $(k+20)^2$ if $k$ is odd,and $2(k+20)^2$ if $k$ is even.
$B - 2A = \sum_{k=1}^{20} (a_{k+20} - a_k)$.
If $k$ is odd,$a_{k+20} - a_k = (k+20)^2 - k^2 = 40k + 400$.
If $k$ is even,$a_{k+20} - a_k = 2(k+20)^2 - 2k^2 = 2(40k + 400) = 80k + 800$.
Summing these for $k=1$ to $20$ ($10$ odd and $10$ even values of $k$):
$B - 2A = \sum_{k \in \text{odd}} (40k + 400) + \sum_{k \in \text{even}} (80k + 800)$.
$= 40 \sum_{k \in \text{odd}} k + 4000 + 80 \sum_{k \in \text{even}} k + 8000$.
$= 40(1+3+\dots+19) + 80(2+4+\dots+20) + 12000$.
$= 40(100) + 80(110) + 12000 = 4000 + 8800 + 12000 = 24800$.
Since $B - 2A = 100\lambda$,$100\lambda = 24800$,so $\lambda = 248$.

(B) Given $\sum_{k = 0}^{12} {a_{4k + 1}} = 416$. This is a sum of $13$ terms in $A.P.$ with first term $a_1$ and common difference $4d$.
$\frac{13}{2} [2a_1 + (13-1)4d] = 416$ $\Rightarrow \frac{13}{2} [2a_1 + 48d] = 416$ $\Rightarrow a_1 + 24d = 32 \dots (1)$
Given ${a_9} + {a_{43}} = 66$ $\Rightarrow (a_1 + 8d) + (a_1 + 42d) = 66$ $\Rightarrow 2a_1 + 50d = 66$ $\Rightarrow a_1 + 25d = 33 \dots (2)$
Subtracting $(1)$ from $(2)$,we get $d = 1$. Substituting $d=1$ in $(1)$,$a_1 + 24 = 32 \Rightarrow a_1 = 8$.
Now,$\sum_{r = 1}^{17} a_r^2 = \sum_{r = 1}^{17} [8 + (r-1)1]^2 = \sum_{r = 1}^{17} (r+7)^2 = \sum_{r = 1}^{17} (r^2 + 14r + 49) = 140m$.
Using summation formulas: $\sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6}$,$\sum_{r=1}^{n} r = \frac{n(n+1)}{2}$.
For $n=17$: $\frac{17 \times 18 \times 35}{6} + 14 \times \frac{17 \times 18}{2} + 49 \times 17 = 1785 + 2142 + 833 = 4760$.
$140m = 4760 \Rightarrow m = \frac{4760}{140} = 34$.

(B) Given that the orthocentre $A$ is $(-3, 5)$ and the centroid $B$ is $(3, 3)$.
The distance $AB = \sqrt{(3 - (-3))^2 + (3 - 5)^2} = \sqrt{6^2 + (-2)^2} = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10}$.
The centroid $B$ divides the line segment joining the orthocentre $A$ and the circumcentre $C$ in the ratio $2:1$,such that $AB:BC = 2:1$.
This implies $AB = \frac{2}{3}AC$,or $AC = \frac{3}{2}AB$.
Substituting the value of $AB$,we get $AC = \frac{3}{2}(2\sqrt{10}) = 3\sqrt{10}$.
The radius of the circle with $AC$ as diameter is $r = \frac{AC}{2} = \frac{3\sqrt{10}}{2}$.
We can rewrite this as $r = 3\sqrt{\frac{10}{4}} = 3\sqrt{\frac{5}{2}}$.

(B) Let the coordinates of $P$ be $(h, 0)$ and $Q$ be $(0, k)$.
Since the rectangle $OPRQ$ is completed,the coordinates of $R$ are $(h, k)$.
The equation of the line passing through $P(h, 0)$ and $Q(0, k)$ is given by the intercept form:
$\frac{x}{h} + \frac{y}{k} = 1$
Since this line passes through the fixed point $(2, 3)$,we have:
$\frac{2}{h} + \frac{3}{k} = 1$
To find the locus of $R(h, k)$,we replace $h$ with $x$ and $k$ with $y$:
$\frac{2}{x} + \frac{3}{y} = 1$
Multiplying both sides by $xy$,we get:
$2y + 3x = xy$
Thus,the locus of $R$ is $3x + 2y = xy$.

(C) The equation of the curve is $x^2 = y - 6$,which can be written as $y = x^2 + 6$.
To find the tangent at $(1, 7)$,we differentiate with respect to $x$: $\frac{dy}{dx} = 2x$.
At $(1, 7)$,the slope $m = 2(1) = 2$.
The equation of the tangent is $y - 7 = 2(x - 1)$,which simplifies to $2x - y + 5 = 0$.
For this line to be a tangent to the circle $x^2 + y^2 + 16x + 12y + c = 0$,the perpendicular distance from the center $(-8, -6)$ to the line must equal the radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{(-8)^2 + (-6)^2 - c} = \sqrt{64 + 36 - c} = \sqrt{100 - c}$.
The perpendicular distance is $\frac{|2(-8) - (-6) + 5|}{\sqrt{2^2 + (-1)^2}} = \frac{|-16 + 6 + 5|}{\sqrt{5}} = \frac{|-5|}{\sqrt{5}} = \sqrt{5}$.
Equating the distance to the radius: $\sqrt{5} = \sqrt{100 - c}$.
Squaring both sides: $5 = 100 - c$,which gives $c = 95$.

(D) The equation of the hyperbola is $4x^2 - y^2 = 36$,which can be written as $\frac{x^2}{9} - \frac{y^2}{36} = 1$.
The line $PQ$ is the chord of contact of the tangents drawn from $T(0, 3)$.
The equation of the chord of contact from $(x_1, y_1)$ is $\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1$.
Substituting $(x_1, y_1) = (0, 3)$,$a^2 = 9$,and $b^2 = 36$,we get:
$\frac{x(0)}{9} - \frac{y(3)}{36} = 1$
$\Rightarrow -\frac{y}{12} = 1$
$\Rightarrow y = -12$.
To find the coordinates of $P$ and $Q$,substitute $y = -12$ into the hyperbola equation:
$4x^2 - (-12)^2 = 36$
$4x^2 - 144 = 36$
$4x^2 = 180$
$x^2 = 45$
$x = \pm 3\sqrt{5}$.
Thus,the points are $P(3\sqrt{5}, -12)$ and $Q(-3\sqrt{5}, -12)$.
The length of the base $PQ = |3\sqrt{5} - (-3\sqrt{5})| = 6\sqrt{5}$.
The height of the triangle $\Delta PTQ$ is the vertical distance from $T(0, 3)$ to the line $y = -12$,which is $h = |3 - (-12)| = 15$.
Area of $\Delta PTQ = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6\sqrt{5} \times 15 = 45\sqrt{5}$ sq. units.

(A) The equation of the parabola is ${y^2} = 16x$,so $4a = 16$,which implies $a = 4$.
For the point $P(16, 16)$,we have $y_1^2 = 16x_1$,so $16^2 = 16(16)$,which is satisfied.
The tangent at $P(x_1, y_1)$ is $yy_1 = 2a(x + x_1)$,so $16y = 8(x + 16)$,which simplifies to $2y = x + 16$.
Setting $y = 0$ for the intersection with the $x$-axis,we get $x = -16$,so $A = (-16, 0)$.
The normal at $P(x_1, y_1)$ is $y - y_1 = -\frac{y_1}{2a}(x - x_1)$,so $y - 16 = -\frac{16}{8}(x - 16)$,which simplifies to $y - 16 = -2(x - 16)$,or $y = -2x + 48$.
Setting $y = 0$ for the intersection with the $x$-axis,we get $2x = 48$,so $x = 24$,thus $B = (24, 0)$.
The circle passes through $A(-16, 0)$,$B(24, 0)$,and $P(16, 16)$. Since $A$ and $B$ lie on the $x$-axis,the center $C$ must have an $x$-coordinate of $\frac{-16 + 24}{2} = 4$.
Let $C = (4, k)$. Since $CA = CP$,we have $(4 - (-16))^2 + (k - 0)^2 = (4 - 16)^2 + (k - 16)^2$.
$20^2 + k^2 = (-12)^2 + (k - 16)^2 \Rightarrow 400 + k^2 = 144 + k^2 - 32k + 256$.
$400 = 400 - 32k \Rightarrow k = 0$. Thus,$C = (4, 0)$.
Slope of $PC$ $(m_1)$ = $\frac{16 - 0}{16 - 4} = \frac{16}{12} = \frac{4}{3}$.
Slope of $PB$ $(m_2)$ = $\frac{16 - 0}{16 - 24} = \frac{16}{-8} = -2$.
$\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{\frac{4}{3} - (-2)}{1 + (\frac{4}{3})(-2)} \right| = \left| \frac{\frac{10}{3}}{1 - \frac{8}{3}} \right| = \left| \frac{\frac{10}{3}}{-\frac{5}{3}} \right| = |-2| = 2$.

(A) The set $A$ is defined by $|a - 5| < 1$ and $|b - 5| < 1$. Let $x = a - 5$ and $y = b - 5$. Then $A$ represents the interior of a square centered at $(5, 5)$ with side length $2$,specifically $|x| < 1$ and $|y| < 1$.
The set $B$ is defined by $4(a - 6)^2 + 9(b - 5)^2 \le 36$. Substituting $x = a - 5$ and $y = b - 5$,we get $4(x - 1)^2 + 9y^2 \le 36$,which simplifies to $\frac{(x - 1)^2}{9} + \frac{y^2}{4} \le 1$. This represents the interior and boundary of an ellipse centered at $(1, 0)$ in the $(x, y)$ plane.
The vertices of the square $A$ in the $(x, y)$ plane are $(1, 1), (-1, 1), (-1, -1), (1, -1)$.
Checking if these points satisfy the ellipse inequality $\frac{(x - 1)^2}{9} + \frac{y^2}{4} \le 1$:
For $(1, 1): \frac{0}{9} + \frac{1}{4} = 0.25 \le 1$ (True)
For $(-1, 1): \frac{4}{9} + \frac{1}{4} = \frac{16+9}{36} = \frac{25}{36} \le 1$ (True)
For $(-1, -1): \frac{4}{9} + \frac{1}{4} = \frac{25}{36} \le 1$ (True)
For $(1, -1): \frac{0}{9} + \frac{1}{4} = 0.25 \le 1$ (True)
Since all vertices of the square lie within the ellipse,$A \subset B$.

(B) We know that $[t] = t - \{t\}$,where $\{t\}$ is the fractional part of $t$.
Thus,the expression becomes $\lim_{x \to 0^+} x \sum_{r=1}^{15} [\frac{r}{x}] = \lim_{x \to 0^+} x \sum_{r=1}^{15} (\frac{r}{x} - \{\frac{r}{x}\})$.
$= \lim_{x \to 0^+} (\sum_{r=1}^{15} r - x \sum_{r=1}^{15} \{\frac{r}{x}\})$.
Since $0 \le \{\frac{r}{x}\} < 1$,we have $0 \le x \{\frac{r}{x}\} < x$.
As $x \to 0^+$,by the Squeeze Theorem,$x \{\frac{r}{x}\} \to 0$.
Therefore,the limit is $\sum_{r=1}^{15} r = \frac{15 \times 16}{2} = 120$.

(B) Let $y_i = x_i - 5$. Then $\sum_{i=1}^9 y_i = 9$ and $\sum_{i=1}^9 y_i^2 = 45$.
The variance of a set of observations is invariant under change of origin. Therefore,the standard deviation of $x_i$ is the same as the standard deviation of $y_i$.
Variance $(\sigma^2) = \frac{1}{n} \sum_{i=1}^n y_i^2 - \left( \frac{1}{n} \sum_{i=1}^n y_i \right)^2$
$\sigma^2 = \frac{45}{9} - \left( \frac{9}{9} \right)^2$
$\sigma^2 = 5 - 1 = 4$
Standard deviation $(\sigma) = \sqrt{4} = 2$.

(D) Using De Morgan's Law,$\sim(p \vee q) \equiv (\sim p \wedge \sim q)$.
Substituting this into the expression:
$(\sim p \wedge \sim q) \vee (\sim p \wedge q)$
Applying the Distributive Law:
$\sim p \wedge (\sim q \vee q)$
Since $(\sim q \vee q) \equiv T$ (Tautology):
$\sim p \wedge T \equiv \sim p$.

(B) The median through $C$ is $x = 4$. Since $C$ lies on this median,let $C = (4, y)$.
The midpoint $D$ of $AC$ is $D = (\frac{1+4}{2}, \frac{2+y}{2}) = (2.5, \frac{2+y}{2})$.
Since $D$ lies on the median through $B$ $(x + y = 5)$,we have $2.5 + \frac{2+y}{2} = 5$.
$2.5 + 1 + \frac{y}{2} = 5$ $\Rightarrow \frac{y}{2} = 1.5$ $\Rightarrow y = 3$. Thus,$C = (4, 3)$.
The centroid $G$ is the intersection of the medians $x = 4$ and $x + y = 5$. Substituting $x = 4$ into $x + y = 5$,we get $4 + y = 5$,so $y = 1$. Thus,$G = (4, 1)$.
To find $B$,we use the centroid formula $G = (\frac{x_A+x_B+x_C}{3}, \frac{y_A+y_B+y_C}{3})$.
$4 = \frac{1+x_B+4}{3}$ $\Rightarrow 12 = 5 + x_B$ $\Rightarrow x_B = 7$.
$1 = \frac{2+y_B+3}{3}$ $\Rightarrow 3 = 5 + y_B$ $\Rightarrow y_B = -2$. Thus,$B = (7, -2)$.
The area of $\Delta ABC$ with vertices $A(1, 2)$,$B(7, -2)$,and $C(4, 3)$ is given by:
Area $= \frac{1}{2} |x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B)|$
Area $= \frac{1}{2} |1(-2 - 3) + 7(3 - 2) + 4(2 - (-2))|$
Area $= \frac{1}{2} |-5 + 7 + 16| = \frac{1}{2} |18| = 9$ sq. units.

(C) Let the first term be $b$ and the common ratio be $r$.
For an infinite $G.P.$,the sum $S$ is given by $S = \frac{b}{1 - r}$,where $|r| < 1$.
Given $S = 5$,we have $\frac{b}{1 - r} = 5$.
This implies $b = 5(1 - r)$.
Since $-1 < r < 1$,we can find the range for $b$:
If $r \to 1$,then $b \to 5(1 - 1) = 0$.
If $r \to -1$,then $b \to 5(1 - (-1)) = 5(2) = 10$.
Thus,for $-1 < r < 1$,the value of $b$ lies in the interval $(0, 10)$.

(B) Let the roots of the equation $x^{2} + (2 - \lambda)x + (10 - \lambda) = 0$ be $\alpha$ and $\beta$.
From the properties of roots,$\alpha + \beta = \lambda - 2$ and $\alpha\beta = 10 - \lambda$.
The sum of the cubes of the roots is $S = \alpha^{3} + \beta^{3} = (\alpha + \beta)^{3} - 3\alpha\beta(\alpha + \beta)$.
Substituting the values,$S = (\lambda - 2)^{3} - 3(10 - \lambda)(\lambda - 2)$.
$S = (\lambda - 2)[(\lambda - 2)^{2} - 3(10 - \lambda)] = (\lambda - 2)(\lambda^{2} - 4\lambda + 4 - 30 + 3\lambda) = (\lambda - 2)(\lambda^{2} - \lambda - 26)$.
$S = \lambda^{3} - \lambda^{2} - 26\lambda - 2\lambda^{2} + 2\lambda + 52 = \lambda^{3} - 3\lambda^{2} - 24\lambda + 52$.
To find the minimum,set $\frac{dS}{d\lambda} = 3\lambda^{2} - 6\lambda - 24 = 0$.
$3(\lambda^{2} - 2\lambda - 8) = 0 \implies 3(\lambda - 4)(\lambda + 2) = 0$.
For minimum,$\frac{d^{2}S}{d\lambda^{2}} = 6\lambda - 6$. At $\lambda = 4$,$6(4) - 6 = 18 > 0$,so it is a minimum.
The difference of the roots is $|\alpha - \beta| = \sqrt{(\alpha + \beta)^{2} - 4\alpha\beta} = \sqrt{(\lambda - 2)^{2} - 4(10 - \lambda)}$.
For $\lambda = 4$,$|\alpha - \beta| = \sqrt{(4 - 2)^{2} - 4(10 - 4)} = \sqrt{4 - 24} = \sqrt{-20}$.
The magnitude of the difference is $|\sqrt{-20}| = \sqrt{20} = 2\sqrt{5}$.

(D) Let the center of the circle be $(h, k)$.
Since the center lies on the line $y - 4x + 3 = 0$,we have $k - 4h + 3 = 0$,or $k = 4h - 3$.
The distance from the center $(h, k)$ to the points $(2, 3)$ and $(4, 5)$ must be equal (radius $r$).
$(h - 2)^2 + (k - 3)^2 = (h - 4)^2 + (k - 5)^2$
$h^2 - 4h + 4 + k^2 - 6k + 9 = h^2 - 8h + 16 + k^2 - 10k + 25$
$-4h - 6k + 13 = -8h - 10k + 41$
$4h + 4k = 28 \Rightarrow h + k = 7$.
Substituting $k = 4h - 3$ into $h + k = 7$:
$h + (4h - 3) = 7$ $\Rightarrow 5h = 10$ $\Rightarrow h = 2$.
Then $k = 4(2) - 3 = 5$.
The center is $(2, 5)$.
The radius $r$ is the distance between $(2, 5)$ and $(2, 3)$:
$r = \sqrt{(2 - 2)^2 + (5 - 3)^2} = \sqrt{0^2 + 2^2} = 2$.

(D) The equation of the hyperbola is $4y^2 - x^2 = 1$.
Let the point of tangency be $P(x_1, y_1)$. The equation of the tangent at $P$ is $4yy_1 - xx_1 = 1$.
This tangent intersects the $x$-axis at $A$ where $y=0$,so $-xx_1 = 1 \Rightarrow x = -1/x_1$. Thus,$A = (-1/x_1, 0)$.
This tangent intersects the $y$-axis at $B$ where $x=0$,so $4yy_1 = 1 \Rightarrow y = 1/(4y_1)$. Thus,$B = (0, 1/(4y_1))$.
Let the midpoint of $AB$ be $(h, k)$.
Then $h = -1/(2x_1) \Rightarrow x_1 = -1/(2h)$ and $k = 1/(8y_1) \Rightarrow y_1 = 1/(8k)$.
Since $(x_1, y_1)$ lies on the hyperbola $4y_1^2 - x_1^2 = 1$,we substitute the values:
$4(1/(8k))^2 - (-1/(2h))^2 = 1$
$4/(64k^2) - 1/(4h^2) = 1$
$1/(16k^2) - 1/(4h^2) = 1$
Multiplying by $16h^2k^2$:
$h^2 - 4k^2 = 16h^2k^2$
Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 - 4y^2 - 16x^2y^2 = 0$.

(A) Let $d_1$ be the common difference of the $A.P.$ $x_1, x_2, \dots, x_n$.
Since $x_8 - x_3 = 5d_1 = 20 - 8 = 12$,we have $d_1 = \frac{12}{5} = 2.4$.
Then $x_5 = x_3 + 2d_1 = 8 + 2(2.4) = 8 + 4.8 = 12.8$.
Let $d_2$ be the common difference of the $A.P.$ $\frac{1}{h_1}, \frac{1}{h_2}, \dots, \frac{1}{h_n}$.
Since $\frac{1}{h_7} - \frac{1}{h_2} = 5d_2 = \frac{1}{20} - \frac{1}{8} = \frac{2-5}{40} = -\frac{3}{40}$,we have $d_2 = -\frac{3}{200}$.
Now,$\frac{1}{h_{10}} = \frac{1}{h_7} + 3d_2 = \frac{1}{20} + 3\left(-\frac{3}{200}\right) = \frac{10-9}{200} = \frac{1}{200}$.
Thus,$h_{10} = 200$.
Finally,$x_5 \cdot h_{10} = 12.8 \times 200 = 2560$.

(B) Let the observations be $x_1, x_2, \dots, x_{30}$. The mean is given by $\bar{x} = \frac{1}{30} \sum_{i=1}^{30} x_i = 75$.
When each observation is multiplied by $\lambda$ and decreased by $25$,the new observations are $y_i = \lambda x_i - 25$.
The new mean is $\bar{y} = \frac{1}{30} \sum_{i=1}^{30} (\lambda x_i - 25) = \lambda \left( \frac{1}{30} \sum_{i=1}^{30} x_i \right) - 25 = 75\lambda - 25$.
According to the problem,the mean remains the same,so $\bar{y} = 75$.
Thus,$75\lambda - 25 = 75$.
$75\lambda = 100$.
$\lambda = \frac{100}{75} = \frac{4}{3}$.

(D) Let $P(x) = \left[\frac{1}{\sqrt{5 x^{3}+1}-\sqrt{5 x^{3}-1}}\right]^{8}+\left[\frac{1}{\sqrt{5 x^{3}+1}+\sqrt{5 x^{3}-1}}\right]^{8}$.
Rationalizing the terms,we get:
$P(x) = \left[\frac{\sqrt{5 x^{3}+1}+\sqrt{5 x^{3}-1}}{2}\right]^{8} + \left[\frac{\sqrt{5 x^{3}+1}-\sqrt{5 x^{3}-1}}{2}\right]^{8}$.
$P(x) = \frac{1}{2^8} \left[ (\sqrt{5x^3+1} + \sqrt{5x^3-1})^8 + (\sqrt{5x^3+1} - \sqrt{5x^3-1})^8 \right]$.
Using the expansion $(a+b)^8 + (a-b)^8 = 2 \sum_{k=0, 2, 4, 6, 8} \binom{8}{k} a^{8-k} b^k$,we have:
$P(x) = \frac{2}{2^8} \left[ \binom{8}{0} (5x^3+1)^4 + \binom{8}{2} (5x^3+1)^3(5x^3-1) + \binom{8}{4} (5x^3+1)^2(5x^3-1)^2 + \binom{8}{6} (5x^3+1)(5x^3-1)^3 + \binom{8}{8} (5x^3-1)^4 \right]$.
The highest power of $x$ is $x^{3 \times 4} = x^{12}$,so $n = 12$.
The coefficient of $x^{12}$ is given by $\frac{2}{2^8} \times 5^4 \times \left[ \binom{8}{0} + \binom{8}{2} + \binom{8}{4} + \binom{8}{6} + \binom{8}{8} \right]$.
Since $\sum_{k \text{ even}} \binom{8}{k} = 2^{8-1} = 2^7$,we have:
$m = \frac{2}{2^8} \times 5^4 \times 2^7 = \frac{2^8}{2^8} \times 5^4 \times 2^3 = 8 \times 10^4$.
Thus,$(n, m) = (12, 8(10)^4)$.

(NONE) Given the quadratic equation $3x^2 - 10x - 25 = 0$ with roots $\tan A$ and $\tan B$.
From the properties of roots,$\tan A + \tan B = \frac{10}{3}$ and $\tan A \tan B = -\frac{25}{3}$.
Using the formula $\tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,we get:
$\tan (A + B) = \frac{10/3}{1 - (-25/3)} = \frac{10/3}{28/3} = \frac{10}{28} = \frac{5}{14}$.
Let $S = 3 \sin^2 (A + B) - 10 \sin (A + B) \cos (A + B) - 25 \cos^2 (A + B)$.
Divide the entire expression by $\cos^2 (A + B)$:
$S = \cos^2 (A + B) [3 \tan^2 (A + B) - 10 \tan (A + B) - 25]$.
Since $\tan (A + B) = 5/14$,the term in the bracket is $3(5/14)^2 - 10(5/14) - 25 = 3(25/196) - 50/14 - 25 = 75/196 - 700/196 - 4900/196 = -5525/196$.
Alternatively,note that the expression is of the form $f(\tan(A+B)) \cdot \cos^2(A+B)$ where $f(x) = 3x^2 - 10x - 25$. Since $\tan(A+B) = 5/14$ is a root of $3x^2 - 10x - 25 = 0$,the value of the expression is $0$.

(C) Since the origin is the common vertex,let the equations of the two parabolas be $y^2 = 4ax$ and $x^2 = 4by$.
Given the length of the latus rectum is $3$,we have $4a = 3$ and $4b = 3$,so $a = b = \frac{3}{4}$.
The equations are $y^2 = 3x$ and $x^2 = 3y$.
Let the common tangent be $y = mx + c$.
For $y^2 = 3x$,substituting $y = mx + c$ gives $(mx + c)^2 = 3x$,or $m^2x^2 + (2mc - 3)x + c^2 = 0$.
Since it is a tangent,the discriminant is zero: $(2mc - 3)^2 - 4m^2c^2 = 0$,which simplifies to $9 - 12mc = 0$,so $c = \frac{3}{4m}$.
For $x^2 = 3y$,substituting $y = mx + c$ gives $x^2 = 3(mx + c)$,or $x^2 - 3mx - 3c = 0$.
Setting the discriminant to zero: $(-3m)^2 - 4(1)(-3c) = 0$,which gives $9m^2 + 12c = 0$,so $c = -\frac{3m^2}{4}$.
Equating the two expressions for $c$: $\frac{3}{4m} = -\frac{3m^2}{4}$ $\Rightarrow m^3 = -1$ $\Rightarrow m = -1$.
Then $c = \frac{3}{4(-1)} = -\frac{3}{4}$.
The equation of the tangent is $y = -x - \frac{3}{4}$,which simplifies to $4(x + y) + 3 = 0$.

(B) The equation of the ellipse is $x^2 + 3y^2 = 9$.
Differentiating with respect to $x$,we get $2x + 6y \frac{dy}{dx} = 0$,which implies $\frac{dy}{dx} = -\frac{x}{3y}$.
The slope of the normal at any point $(x, y)$ is $m = -\frac{dx}{dy} = \frac{3y}{x}$.
For the point $P_1 = (3\cos \theta, \sqrt{3} \sin \theta)$,the slope of the normal $m_1 = \frac{3(\sqrt{3} \sin \theta)}{3 \cos \theta} = \sqrt{3} \tan \theta$.
For the point $P_2 = (-3\sin \theta, \sqrt{3} \cos \theta)$,the slope of the normal $m_2 = \frac{3(\sqrt{3} \cos \theta)}{-3 \sin \theta} = -\sqrt{3} \cot \theta$.
The angle $\beta$ between the normals is given by $\tan \beta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$.
Substituting the values,$\tan \beta = \left| \frac{\sqrt{3} \tan \theta - (-\sqrt{3} \cot \theta)}{1 + (\sqrt{3} \tan \theta)(-\sqrt{3} \cot \theta)} \right| = \left| \frac{\sqrt{3}(\tan \theta + \cot \theta)}{1 - 3} \right| = \left| \frac{\sqrt{3}(\tan \theta + \cot \theta)}{-2} \right| = \frac{\sqrt{3}}{2} (\tan \theta + \cot \theta)$.
Since $\tan \theta + \cot \theta = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\frac{1}{2} \sin 2\theta} = \frac{2}{\sin 2\theta}$.
Thus,$\tan \beta = \frac{\sqrt{3}}{2} \cdot \frac{2}{\sin 2\theta} = \frac{\sqrt{3}}{\sin 2\theta}$.
Therefore,$\frac{1}{\cot \beta} = \frac{\sqrt{3}}{\sin 2\theta}$,which implies $\frac{\cot \beta}{\sin 2\theta} = \frac{1}{\sqrt{3}}$.
Hence,$\frac{2 \cot \beta}{\sin 2\theta} = \frac{2}{\sqrt{3}}$.

(A) Given $w = \frac{1 + (1 - 8\alpha)z}{1 - z}$. Since $w$ is purely imaginary,$w + \bar{w} = 0$.
$w + \bar{w} = \frac{1 + (1 - 8\alpha)z}{1 - z} + \frac{1 + (1 - 8\alpha)\bar{z}}{1 - \bar{z}} = 0$
$\Rightarrow (1 + (1 - 8\alpha)z)(1 - \bar{z}) + (1 + (1 - 8\alpha)\bar{z})(1 - z) = 0$
$\Rightarrow 1 - \bar{z} + (1 - 8\alpha)z - (1 - 8\alpha)z\bar{z} + 1 - z + (1 - 8\alpha)\bar{z} - (1 - 8\alpha)z\bar{z} = 0$
Since $|z| = 1$,we have $z\bar{z} = 1$.
$\Rightarrow 2 - (z + \bar{z}) + (1 - 8\alpha)(\bar{z} + z) - 2(1 - 8\alpha) = 0$
$\Rightarrow 2 - (z + \bar{z}) + (1 - 8\alpha)(z + \bar{z}) - 2 + 16\alpha = 0$
$\Rightarrow (z + \bar{z})(1 - 8\alpha - 1) + 16\alpha = 0$
$\Rightarrow (z + \bar{z})(-8\alpha) + 16\alpha = 0$
$\Rightarrow -8\alpha(z + \bar{z} - 2) = 0$
Since $\text{Re}(z) \neq 1$,$z + \bar{z} \neq 2$,so $z + \bar{z} - 2 \neq 0$.
Therefore,$-8\alpha = 0$,which implies $\alpha = 0$.

(D) Let the height of the aeroplane be $h = \sqrt{3} \ km$. Let $O$ be the point of observation on the ground.
Let $A$ be the first position of the aeroplane and $B$ be the second position after $5 \ seconds$.
In $\Delta OA A_1$,where $A A_1 = \sqrt{3} \ km$ and $\angle A O A_1 = 60^\circ$:
$O A_1 = \frac{A A_1}{\tan 60^\circ} = \frac{\sqrt{3}}{\sqrt{3}} = 1 \ km$.
In $\Delta OB B_1$,where $B B_1 = \sqrt{3} \ km$ and $\angle B O B_1 = 30^\circ$:
$O B_1 = \frac{B B_1}{\tan 30^\circ} = \frac{\sqrt{3}}{1/\sqrt{3}} = 3 \ km$.
The distance covered by the aeroplane is $AB = A_1 B_1 = O B_1 - O A_1 = 3 - 1 = 2 \ km$.
Time taken is $5 \ seconds = \frac{5}{3600} \ hours$.
Speed = $\frac{\text{Distance}}{\text{Time}} = \frac{2}{5/3600} = \frac{2 \times 3600}{5} = 1440 \ km/hr$.

(D) Each of the $n$ places in the $n$-digit number can be filled in $3$ ways (using digits $2, 5,$ or $7$).
Therefore,the total number of distinct $n$-digit numbers that can be formed is $3^n$.
We need to find the smallest integer $n$ such that $3^n \geq 900$.
Calculating powers of $3$:
$3^1 = 3$
$3^2 = 9$
$3^3 = 27$
$3^4 = 81$
$3^5 = 243$
$3^6 = 729$
$3^7 = 2187$
Since $3^6 = 729 < 900$ and $3^7 = 2187 \geq 900$,the smallest value of $n$ is $7$.

(A) We have $(1+x)^2 = 1 + 2x + x^2$.
$(1+x^2)^3 = 1 + 3x^2 + 3x^4 + x^6$.
$(1+x^3)^4 = 1 + 4x^3 + 6x^6 + 4x^9 + x^{12}$.
We need to find the coefficient of $x^{10}$ in the product $(1 + 2x + x^2)(1 + 3x^2 + 3x^4 + x^6)(1 + 4x^3 + 6x^6 + 4x^9 + x^{12})$.
Let $P_1 = (1 + 2x + x^2)$,$P_2 = (1 + 3x^2 + 3x^4 + x^6)$,and $P_3 = (1 + 4x^3 + 6x^6 + 4x^9 + x^{12})$.
Possible combinations of terms from $P_1, P_2, P_3$ that result in $x^{10}$ are:
$1$) $(2x) \cdot (x^0) \cdot (4x^9) = 8x^{10} \implies \text{coeff} = 8$.
$2$) $(2x) \cdot (3x^4) \cdot (6x^6) = 36x^{11}$ (Not $x^{10}$).
$3$) $(x^2) \cdot (3x^2) \cdot (6x^6) = 18x^{10} \implies \text{coeff} = 18$.
$4$) $(1) \cdot (3x^4) \cdot (4x^6)$ is not possible as $P_3$ has $x^6$ but not $x^4$.
Let's re-evaluate systematically:
Terms from $P_1$: $1, 2x, x^2$.
Terms from $P_2$: $1, 3x^2, 3x^4, x^6$.
Terms from $P_3$: $1, 4x^3, 6x^6, 4x^9, x^{12}$.
Combinations $(a, b, c)$ where $a \in P_1, b \in P_2, c \in P_3$ such that $a \cdot b \cdot c = x^{10}$:
- $(2x) \cdot (1) \cdot (4x^9) = 8x^{10}$.
- $(2x) \cdot (3x^4) \cdot (6x^6) = 36x^{11}$ (Invalid).
- $(x^2) \cdot (3x^2) \cdot (6x^6) = 18x^{10}$.
- $(1) \cdot (3x^4) \cdot (4x^6)$ is not possible. Wait,$P_2$ has $3x^4$ and $P_3$ has $6x^6$,so $1 \cdot 3x^4 \cdot 6x^6 = 18x^{10}$.
- $(2x) \cdot (3x^2) \cdot (4x^3)$ is not $x^{10}$.
- $(x^2) \cdot (1) \cdot (4x^9)$ is not $x^{10}$.
- $(1) \cdot (x^6) \cdot (4x^3)$ is not $x^{10}$.
- $(2x) \cdot (x^6) \cdot (4x^3) = 8x^{10}$.
Summing the coefficients: $8 + 18 + 18 + 8 = 52$.

(D) Let the quadratic expression be $f(x) = k(x - r_1)(x - r_2)$.
Given that $-1$ is a root,let $r_1 = -1$. Let the other root be $a$.
So,$f(x) = k(x + 1)(x - a) = k(x^2 + (1 - a)x - a)$.
We are given $f(1) + f(2) = 0$.
$f(1) = k(1 + 1)(1 - a) = 2k(1 - a) = 2k - 2ka$.
$f(2) = k(2 + 1)(2 - a) = 3k(2 - a) = 6k - 3ka$.
Summing these: $f(1) + f(2) = (2k - 2ka) + (6k - 3ka) = 8k - 5ka$.
Setting the sum to zero: $8k - 5ka = 0$.
Since $f(x)$ is a quadratic expression,$k \neq 0$,so we can divide by $k$.
$8 - 5a = 0 \Rightarrow 5a = 8 \Rightarrow a = \frac{8}{5}$.
Thus,the other root is $\frac{8}{5}$.

(D) The word $BARRACK$ consists of $7$ letters: $A, A, R, R, B, C, K$. The distinct letters are ${A, R, B, C, K}$.
We need to form $4$-letter words. The cases are:
$(i)$ All $4$ letters are distinct: We choose $4$ letters from ${A, R, B, C, K}$ in $^5C_4 = 5$ ways. Each set can be arranged in $4! = 24$ ways. Total $= 5 \times 24 = 120$.
(ii) $2$ pairs of identical letters: The pairs are ${A, A}$ and ${R, R}$. We choose both pairs in $^2C_2 = 1$ way. The number of arrangements is $\frac{4!}{2!2!} = 6$.
(iii) $2$ identical letters and $2$ distinct letters: We choose $1$ pair from ${A, A}$ or ${R, R}$ in $^2C_1 = 2$ ways. We then choose $2$ distinct letters from the remaining $4$ letters in $^4C_2 = 6$ ways. The number of arrangements for each selection is $\frac{4!}{2!} = 12$. Total $= 2 \times 6 \times 12 = 144$.
Total number of $4$-letter words $= 120 + 6 + 144 = 270$.

(D) Given $\sin 3x = \cos 2x$.
We can write this as $\sin 3x = \sin \left( \frac{\pi}{2} - 2x \right)$.
The general solution for $\sin A = \sin B$ is $A = n\pi + (-1)^n B$,where $n \in \mathbb{Z}$.
Applying this,$3x = n\pi + (-1)^n \left( \frac{\pi}{2} - 2x \right)$.
Case $1$: If $n$ is even,let $n = 2k$. Then $3x = 2k\pi + \frac{\pi}{2} - 2x$ $\Rightarrow 5x = 2k\pi + \frac{\pi}{2}$ $\Rightarrow x = \frac{2k\pi}{5} + \frac{\pi}{10} = \frac{(4k+1)\pi}{10}$.
For $k=1$,$x = \frac{5\pi}{10} = \frac{\pi}{2}$ (not in interval).
For $k=2$,$x = \frac{9\pi}{10}$ (in interval).
Case $2$: If $n$ is odd,let $n = 2k+1$. Then $3x = (2k+1)\pi - \left( \frac{\pi}{2} - 2x \right) = 2k\pi + \pi - \frac{\pi}{2} + 2x = 2k\pi + \frac{\pi}{2} + 2x$ $\Rightarrow x = 2k\pi + \frac{\pi}{2}$.
For $k=0$,$x = \frac{\pi}{2}$ (not in interval).
For $k=1$,$x = 2\pi + \frac{\pi}{2} = \frac{5\pi}{2}$ (not in interval).
Thus,the only solution in the interval $\left( \frac{\pi}{2}, \pi \right)$ is $x = \frac{9\pi}{10}$.
Therefore,the number of solutions is $1$.

(A) For statement $p$: $\sin 120^\circ = \frac{\sqrt{3}}{2}$,so $2\sin 120^\circ = \sqrt{3}$.
Substituting $\theta = 240^\circ$ in the $RHS$: $\sqrt{1 + \sin 240^\circ} - \sqrt{1 - \sin 240^\circ} = \sqrt{1 - \frac{\sqrt{3}}{2}} - \sqrt{1 + \frac{\sqrt{3}}{2}} = \sqrt{\frac{2 - \sqrt{3}}{2}} - \sqrt{\frac{2 + \sqrt{3}}{2}} = \frac{\sqrt{3}-1}{2} - \frac{\sqrt{3}+1}{2} = -1 \neq \sqrt{3}$. Thus,statement $p$ is False.
For statement $q$: In any quadrilateral $ABCD$,$A + B + C + D = 360^\circ$,so $\frac{A+C}{2} + \frac{B+D}{2} = 180^\circ = \pi$.
Let $\alpha = \frac{A+C}{2}$,then $\frac{B+D}{2} = \pi - \alpha$.
Thus,$\cos(\alpha) + \cos(\pi - \alpha) = \cos(\alpha) - \cos(\alpha) = 0$. Thus,statement $q$ is True.
Therefore,the truth values are $F, T$.

(B) The given equation $|z - (3 - 2i)| \leq 4$ represents a disk with center $C(3, -2)$ and radius $R = 4$.
$|z|$ represents the distance of a point $z$ from the origin $O(0, 0)$.
The greatest and least distances from the origin to the circle occur along the line passing through the origin and the center $C$.
Let $OC$ be the distance from the origin to the center $C(3, -2)$.
$OC = \sqrt{3^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13}$.
The least distance of $|z|$ from the origin is $|OC - R| = |\sqrt{13} - 4|$. Since $\sqrt{13} < 4$,the least distance is $4 - \sqrt{13}$.
The greatest distance of $|z|$ from the origin is $OC + R = \sqrt{13} + 4$.
The difference between the greatest and least values is $(4 + \sqrt{13}) - (4 - \sqrt{13}) = 2\sqrt{13}$.

(A) The equation of the line $3x + y = \lambda$ can be written in intercept form as $\frac{x}{\lambda/3} + \frac{y}{\lambda} = 1$.
Thus,the coordinates of $A$ are $(\frac{\lambda}{3}, 0)$ and $B$ are $(0, \lambda)$.
The foot of the perpendicular $P$ from the origin $(0,0)$ to the line $ax + by + c = 0$ is given by $(\frac{-ac}{a^2+b^2}, \frac{-bc}{a^2+b^2})$.
Here,$3x + y - \lambda = 0$,so $a=3, b=1, c=-\lambda$.
$P = (\frac{-3(-\lambda)}{3^2+1^2}, \frac{-1(-\lambda)}{3^2+1^2}) = (\frac{3\lambda}{10}, \frac{\lambda}{10})$.
Now,we calculate the distances $BP$ and $PA$:
$BP^2 = (\frac{3\lambda}{10} - 0)^2 + (\frac{\lambda}{10} - \lambda)^2 = \frac{9\lambda^2}{100} + \frac{81\lambda^2}{100} = \frac{90\lambda^2}{100} = \frac{9\lambda^2}{10}$.
$PA^2 = (\frac{\lambda}{3} - \frac{3\lambda}{10})^2 + (0 - \frac{\lambda}{10})^2 = (\frac{10\lambda - 9\lambda}{30})^2 + \frac{\lambda^2}{100} = \frac{\lambda^2}{900} + \frac{9\lambda^2}{900} = \frac{10\lambda^2}{900} = \frac{\lambda^2}{90}$.
Therefore,$\frac{BP^2}{PA^2} = \frac{9\lambda^2/10}{\lambda^2/90} = \frac{9}{10} \times 90 = 81$.
Thus,$\frac{BP}{PA} = \sqrt{81} = 9$.
So,the ratio $BP : PA$ is $9 : 1$.

(B) $A_n$ is a geometric progression with first term $a = \frac{3}{4}$ and common ratio $r = -\frac{3}{4}$.
The sum of $n$ terms is $A_n = \frac{a(1-r^n)}{1-r} = \frac{\frac{3}{4}(1 - (-\frac{3}{4})^n)}{1 - (-\frac{3}{4})} = \frac{\frac{3}{4}(1 - (-\frac{3}{4})^n)}{\frac{7}{4}} = \frac{3}{7} \left[ 1 - \left( -\frac{3}{4} \right)^n \right]$.
Given $B_n = 1 - A_n$,the condition $B_n > A_n$ implies $1 - A_n > A_n$,which simplifies to $1 > 2A_n$,or $A_n < \frac{1}{2}$.
Substituting $A_n$: $\frac{3}{7} \left[ 1 - \left( -\frac{3}{4} \right)^n \right] < \frac{1}{2} \implies 1 - \left( -\frac{3}{4} \right)^n < \frac{7}{6}$.
This gives $-\left( -\frac{3}{4} \right)^n < \frac{1}{6}$,or $\left( -\frac{3}{4} \right)^n > -\frac{1}{6}$.
For odd $n$,$\left( -\frac{3}{4} \right)^n = -\left( \frac{3}{4} \right)^n$. So,$-\left( \frac{3}{4} \right)^n > -\frac{1}{6}$,which means $\left( \frac{3}{4} \right)^n < \frac{1}{6}$.
Taking logarithms: $n \log(\frac{3}{4}) < \log(\frac{1}{6}) \implies n(\log 3 - \log 4) < -\log 6$.
$n(0.4771 - 0.6020) < -0.7781 \implies n(-0.1249) < -0.7781 \implies n > \frac{0.7781}{0.1249} \approx 6.23$.
The least odd natural number $p$ satisfying $n > 6.23$ is $p = 7$.

(C) The equation of the hyperbola is $4x^2 - 9y^2 = 36$,which can be written as $\frac{x^2}{9} - \frac{y^2}{4} = 1$.
Let the point on the hyperbola be $(x_0, y_0)$. The equation of the normal at $(x_0, y_0)$ is $\frac{a^2x}{x_0} + \frac{b^2y}{y_0} = a^2 + b^2$.
Here $a^2 = 9$ and $b^2 = 4$,so the equation is $\frac{9x}{x_0} + \frac{4y}{y_0} = 9 + 4 = 13$.
This normal meets the $x$-axis at $A(\frac{13x_0}{9}, 0)$ and the $y$-axis at $B(0, \frac{13y_0}{4})$.
Since $OABP$ is a parallelogram with $O(0,0)$,the coordinates of $P(x, y)$ are given by $P = A + B = (\frac{13x_0}{9}, \frac{13y_0}{4})$.
Thus,$x = \frac{13x_0}{9} \Rightarrow x_0 = \frac{9x}{13}$ and $y = \frac{13y_0}{4} \Rightarrow y_0 = \frac{4y}{13}$.
Since $(x_0, y_0)$ lies on the hyperbola $4x_0^2 - 9y_0^2 = 36$,we substitute the values:
$4(\frac{9x}{13})^2 - 9(\frac{4y}{13})^2 = 36$
$4 \cdot \frac{81x^2}{169} - 9 \cdot \frac{16y^2}{169} = 36$
$\frac{324x^2 - 144y^2}{169} = 36$
$324x^2 - 144y^2 = 36 \cdot 169$
Dividing by $36$,we get $9x^2 - 4y^2 = 169$.

(A) The equation of the circle $C_1$ is $x^2 + y^2 - 2x - 1 = 0$.
The equation of the tangent at point $(x_1, y_1) = (2, 1)$ is given by $xx_1 + yy_1 - (x + x_1) - 1 = 0$.
Substituting the values,we get $2x + y - (x + 2) - 1 = 0$,which simplifies to $x + y - 3 = 0$.
This line acts as a chord for the circle $C_2$ with center $(3, -2)$.
The perpendicular distance $d$ from the center $(3, -2)$ to the line $x + y - 3 = 0$ is $d = \frac{|3 - 2 - 3|}{\sqrt{1^2 + 1^2}} = \frac{|-2|}{\sqrt{2}} = \sqrt{2}$.
The length of the chord is $l = 4$,so half the length of the chord is $\frac{l}{2} = 2$.
The radius $r$ of the circle $C_2$ is given by $r = \sqrt{(\frac{l}{2})^2 + d^2}$.
Substituting the values,$r = \sqrt{2^2 + (\sqrt{2})^2} = \sqrt{4 + 2} = \sqrt{6}$.

(D) Let the distance between $T_1$ and $T_2$ be $x$.
From the figure,$EA = 60 \, m$ $(T_1)$ and $DB = 80 \, m$ $(T_2)$.
Let $C$ be a point on $T_2$ such that $EC$ is horizontal. Then $EC = AB = x$.
$DC = DB - CB = 80 - 60 = 20 \, m$.
Given $\angle DEC = \theta$ (angle of elevation of top of $T_2$) and $\angle BEC = 2\theta$ (angle of depression of foot of $T_2$).
In $\Delta DEC$,$\tan \theta = \frac{DC}{EC} = \frac{20}{x}$.
In $\Delta BEC$,$\tan 2\theta = \frac{BC}{EC} = \frac{60}{x}$.
Using the identity $\tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta}$,we have:
$\frac{60}{x} = \frac{2(\frac{20}{x})}{1 - (\frac{20}{x})^2}$.
$\frac{60}{x} = \frac{40/x}{1 - 400/x^2} = \frac{40x}{x^2 - 400}$.
$60(x^2 - 400) = 40x^2$.
$60x^2 - 24000 = 40x^2$.
$20x^2 = 24000$.
$x^2 = 1200$.
$x = \sqrt{1200} = 20\sqrt{3} \, m$.

(D) Let the coordinate of vertex $A$ be $(0, c)$.
The equations of the lines parallel to the sides are $x - y + 2 = 0$ and $7x - y + 3 = 0$.
The diagonals of a rhombus are the angle bisectors of the lines containing the sides.
The equations of the angle bisectors are given by $\frac{x - y + 2}{\sqrt{1^2 + (-1)^2}} = \pm \frac{7x - y + 3}{\sqrt{7^2 + (-1)^2}}$.
$\frac{x - y + 2}{\sqrt{2}} = \pm \frac{7x - y + 3}{5\sqrt{2}}$.
$5x - 5y + 10 = \pm (7x - y + 3)$.
Case $1$: $5x - 5y + 10 = 7x - y + 3 \Rightarrow 2x + 4y - 7 = 0$. The slope is $m_1 = -\frac{1}{2}$.
Case $2$: $5x - 5y + 10 = -7x + y - 3 \Rightarrow 12x - 6y + 13 = 0$. The slope is $m_2 = 2$.
The diagonals pass through $P(1, 2)$ and $A(0, c)$. The slope of the line $AP$ is $\frac{2 - c}{1 - 0} = 2 - c$.
If $2 - c = 2$,then $c = 0$,which corresponds to the origin (not allowed).
If $2 - c = -\frac{1}{2}$,then $c = 2 + \frac{1}{2} = \frac{5}{2}$.

(D) Let $L = \mathop {\lim }\limits_{x \to 0} \frac{{x\tan 2x - 2x\tan x}}{{{{\left( {1 - \cos 2x} \right)}^2}}}$.
Using the identity $\tan 2x = \frac{{2\tan x}}{{1 - {{\tan }^2}x}}$ and $1 - \cos 2x = 2\sin^2 x$,we have:
$L = \mathop {\lim }\limits_{x \to 0} \frac{{x\left( \frac{{2\tan x}}{{1 - {{\tan }^2}x}} \right) - 2x\tan x}}{{{{\left( {2\sin^2 x} \right)}^2}}}$
$L = \mathop {\lim }\limits_{x \to 0} \frac{{2x\tan x - 2x\tan x(1 - {{\tan }^2}x)}}{{(1 - {{\tan }^2}x) \cdot 4\sin^4 x}}$
$L = \mathop {\lim }\limits_{x \to 0} \frac{{2x\tan^3 x}}{{4\sin^4 x(1 - {{\tan }^2}x)}}$
Since $\tan x = \frac{\sin x}{\cos x}$,we get:
$L = \mathop {\lim }\limits_{x \to 0} \frac{{2x \cdot \frac{\sin^3 x}{\cos^3 x}}}{{4\sin^4 x(1 - {{\tan }^2}x)}}$
$L = \mathop {\lim }\limits_{x \to 0} \frac{{2x}}{{4\sin x \cdot \cos^3 x(1 - {{\tan }^2}x)}}$
$L = \mathop {\lim }\limits_{x \to 0} \left( \frac{x}{2\sin x} \right) \cdot \mathop {\lim }\limits_{x \to 0} \frac{1}{{\cos^3 x(1 - {{\tan }^2}x)}}$
$L = \frac{1}{2} \cdot \frac{1}{1(1 - 0)} = \frac{1}{2}$.

(D) Given $a, b, c$ are in $A.P.$,so $a+c = 2b$.
Given $a+b+c = \frac{3}{4}$,substituting $a+c=2b$ gives $3b = \frac{3}{4}$,so $b = \frac{1}{4}$.
Given $a^2, b^2, c^2$ are in $G.P.$,so $(b^2)^2 = a^2 c^2$,which implies $ac = \pm b^2 = \pm \frac{1}{16}$.
Since $a < b < c$,$ac$ must be negative,so $ac = -\frac{1}{16}$.
We have $a+c = 2b = \frac{1}{2}$ and $ac = -\frac{1}{16}$.
The quadratic equation for $a$ and $c$ is $x^2 - (a+c)x + ac = 0$,which is $x^2 - \frac{1}{2}x - \frac{1}{16} = 0$.
Multiplying by $16$,we get $16x^2 - 8x - 1 = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $x = \frac{8 \pm \sqrt{64 - 4(16)(-1)}}{32} = \frac{8 \pm \sqrt{128}}{32} = \frac{8 \pm 8\sqrt{2}}{32} = \frac{1}{4} \pm \frac{\sqrt{2}}{4} = \frac{1}{4} \pm \frac{1}{2\sqrt{2}}$.
Since $a < b$,we choose the smaller value: $a = \frac{1}{4} - \frac{1}{2\sqrt{2}}$.

(A) The equation of the parabola is $y^2 = 8x$,so $4a = 8$,which implies $a = 2$. The focus $F$ is $(2, 0)$.
The equation of the chord of contact $PQ$ for a point $(x_1, y_1) = (-8, 0)$ is given by $T = 0$,which is $yy_1 = 2a(x + x_1)$.
Substituting the values,we get $y(0) = 2(2)(x - 8)$,which simplifies to $0 = 4(x - 8)$,so $x = 8$.
For $x = 8$,$y^2 = 8(8) = 64$,so $y = \pm 8$. Thus,the points are $P(8, 8)$ and $Q(8, -8)$.
The triangle $PFQ$ has vertices $P(8, 8)$,$F(2, 0)$,and $Q(8, -8)$.
The base $PQ$ is a vertical line segment with length $|8 - (-8)| = 16$.
The height of the triangle from $F(2, 0)$ to the line $x = 8$ is $|8 - 2| = 6$.
Area of $\triangle PFQ = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 16 \times 6 = 48$ sq. units.

(A) Let the height of the tower be $h$ and the foot of the tower be $D$. Let the positions of the car be $B$ and $A$ at the angles of depression $30^\circ$ and $45^\circ$ respectively.
In $\Delta ODA$,$\angle OAD = 45^\circ$. Thus,$\tan(45^\circ) = \frac{h}{DA} \Rightarrow DA = h$.
In $\Delta ODB$,$\angle OBD = 30^\circ$. Thus,$\tan(30^\circ) = \frac{h}{DB} \Rightarrow DB = h\sqrt{3}$.
The distance covered by the car in $18 \text{ min}$ is $BA = DB - DA = h(\sqrt{3} - 1)$.
Speed of the car $v = \frac{\text{distance}}{\text{time}} = \frac{h(\sqrt{3} - 1)}{18}$.
Time taken to cover distance $DA$ is $t = \frac{DA}{v} = \frac{h}{h(\sqrt{3} - 1) / 18} = \frac{18}{\sqrt{3} - 1}$.
Rationalizing the denominator: $t = \frac{18(\sqrt{3} + 1)}{3 - 1} = \frac{18(\sqrt{3} + 1)}{2} = 9(\sqrt{3} + 1) \text{ min}$.

(C) Given that $\frac{1}{x_1}, \frac{1}{x_2}, \dots, \frac{1}{x_n}$ are in $A.P.$
Let $a = \frac{1}{x_1} = \frac{1}{4}$ and $d$ be the common difference.
We have $\frac{1}{x_{21}} = \frac{1}{20}$.
Using the $A.P.$ formula $\frac{1}{x_{21}} = a + 20d$,we get $\frac{1}{20} = \frac{1}{4} + 20d$.
$20d = \frac{1}{20} - \frac{1}{4} = \frac{1-5}{20} = -\frac{4}{20} = -\frac{1}{5}$.
So,$d = -\frac{1}{100}$.
The $n^{th}$ term of the $A.P.$ is $\frac{1}{x_n} = a + (n-1)d = \frac{1}{4} - \frac{n-1}{100} = \frac{25 - n + 1}{100} = \frac{26 - n}{100}$.
Thus,$x_n = \frac{100}{26 - n}$.
Given $x_n > 50$,we have $\frac{100}{26 - n} > 50$.
Since $x_n > 50$,$26 - n$ must be positive and $26 - n < 2$,so $n > 24$.
The least positive integer $n$ is $25$.
We need to find $\sum_{i=1}^{25} \frac{1}{x_i} = \frac{25}{2} \left[ 2a + (25-1)d \right]$.
$= \frac{25}{2} \left[ 2 \left( \frac{1}{4} \right) + 24 \left( -\frac{1}{100} \right) \right] = \frac{25}{2} \left[ \frac{1}{2} - \frac{6}{25} \right] = \frac{25}{2} \left[ \frac{25 - 12}{50} \right] = \frac{25}{2} \times \frac{13}{50} = \frac{13}{4}$.

(A) Given lines are:
$L_1: \sqrt{2}x - y + 4\sqrt{2}k = 0 \Rightarrow y = \sqrt{2}x + 4\sqrt{2}k \quad (i)$
$L_2: \sqrt{2}kx + ky - 4\sqrt{2} = 0 \quad (ii)$
Substituting $y$ from $(i)$ into $(ii)$:
$\sqrt{2}kx + k(\sqrt{2}x + 4\sqrt{2}k) - 4\sqrt{2} = 0$
$\sqrt{2}kx + \sqrt{2}kx + 4\sqrt{2}k^2 - 4\sqrt{2} = 0$
$2\sqrt{2}kx = 4\sqrt{2}(1 - k^2)$
$x = \frac{2(1 - k^2)}{k}$
Substituting $x$ back into $(i)$:
$y = \sqrt{2}(\frac{2(1 - k^2)}{k}) + 4\sqrt{2}k = \frac{2\sqrt{2} - 2\sqrt{2}k^2 + 4\sqrt{2}k^2}{k} = \frac{2\sqrt{2}(1 + k^2)}{k}$
From these,we observe the relationship between $x$ and $y$:
$\frac{y}{4\sqrt{2}} = \frac{1+k^2}{2k}$ and $\frac{x}{4} = \frac{1-k^2}{2k}$
$(\frac{y}{4\sqrt{2}})^2 - (\frac{x}{4})^2 = (\frac{1+k^2}{2k})^2 - (\frac{1-k^2}{2k})^2 = \frac{(1+k^2)^2 - (1-k^2)^2}{4k^2} = \frac{4k^2}{4k^2} = 1$
This is the equation of a hyperbola $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$ where $a = 4\sqrt{2}$.
The length of the transverse axis is $2a = 2(4\sqrt{2}) = 8\sqrt{2}$.

(A) number is divisible by $3$ if the sum of its digits is divisible by $3$.
The thousands place can be filled with $2, 3,$ or $4$ because the number must be between $2,000$ and $5,000$.
Case $1$: Thousands place is $2$.
Possible sets of remaining $3$ digits from ${0, 1, 3, 4}$ such that the sum is divisible by $3$:
- ${0, 1, 3}$ (sum $= 2+0+1+3 = 6$)
- ${0, 3, 4}$ (sum $= 2+0+3+4 = 9$)
Each set can be arranged in $3! = 6$ ways. Total $= 2 \times 6 = 12$.
Case $2$: Thousands place is $3$.
Possible sets of remaining $3$ digits from ${0, 1, 2, 4}$ such that the sum is divisible by $3$:
- ${0, 1, 2}$ (sum $= 3+0+1+2 = 6$)
- ${0, 2, 4}$ (sum $= 3+0+2+4 = 9$)
Each set can be arranged in $3! = 6$ ways. Total $= 2 \times 6 = 12$.
Case $3$: Thousands place is $4$.
Possible sets of remaining $3$ digits from ${0, 1, 2, 3}$ such that the sum is divisible by $3$:
- ${0, 2, 1}$ (sum $= 4+0+2+1 = 7$ - No)
- ${0, 2, 3}$ (sum $= 4+0+2+3 = 9$)
- ${1, 2, 3}$ (sum $= 4+1+2+3 = 10$ - No)
Only ${0, 2, 3}$ works. Total $= 1 \times 6 = 6$.
Total numbers $= 12 + 12 + 6 = 30$.

(A) The equation of the plane passing through the line of intersection of the planes $2x - 2y + 3z - 2 = 0$ and $x - y + z + 1 = 0$ is given by:
$(2x - 2y + 3z - 2) + \lambda(x - y + z + 1) = 0$
$x(\lambda + 2) - y(\lambda + 2) + z(\lambda + 3) + (\lambda - 2) = 0 \quad \dots(1)$
Since this plane contains the line ${L_2}$ formed by the intersection of $x + 2y - z - 3 = 0$ and $3x - y + 2z - 1 = 0$,the normal vector of the plane must be perpendicular to the direction vectors of the lines forming ${L_2}$. Alternatively,the plane must satisfy the condition that the determinant of the coefficients of the three planes is zero:
$\begin{vmatrix} \lambda + 2 & -(\lambda + 2) & \lambda + 3 \\ 1 & 2 & -1 \\ 3 & -1 & 2 \end{vmatrix} = 0$
Expanding the determinant:
$(\lambda + 2)(4 - 1) + (\lambda + 2)(2 + 3) + (\lambda + 3)(-1 - 6) = 0$
$3(\lambda + 2) + 5(\lambda + 2) - 7(\lambda + 3) = 0$
$8\lambda + 16 - 7\lambda - 21 = 0 \Rightarrow \lambda = 5$
Substituting $\lambda = 5$ into equation $(1)$:
$x(5 + 2) - y(5 + 2) + z(5 + 3) + (5 - 2) = 0$
$7x - 7y + 8z + 3 = 0$
The perpendicular distance from the origin $(0, 0, 0)$ to the plane $Ax + By + Cz + D = 0$ is given by $d = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}}$.
$d = \frac{|3|}{\sqrt{7^2 + (-7)^2 + 8^2}} = \frac{3}{\sqrt{49 + 49 + 64}} = \frac{3}{\sqrt{162}} = \frac{3}{9\sqrt{2}} = \frac{1}{3\sqrt{2}}$.

(C) Let the points be $A(4, -1, 3)$ and $B(5, -1, 4)$. The vector $\overrightarrow{AB} = (5-4)\hat{i} + (-1 - (-1))\hat{j} + (4-3)\hat{k} = \hat{i} + \hat{k}$.
The normal to the plane $x + y + z = 7$ is $\vec{n} = \hat{i} + \hat{j} + \hat{k}$.
The unit normal vector is $\hat{n} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}$.
The projection of $\overrightarrow{AB}$ on the normal $\vec{n}$ is $d = |\overrightarrow{AB} \cdot \hat{n}| = |(\hat{i} + \hat{k}) \cdot \frac{(\hat{i} + \hat{j} + \hat{k})}{\sqrt{3}}| = \frac{1+0+1}{\sqrt{3}} = \frac{2}{\sqrt{3}}$.
The length of the projection of the line segment $AB$ on the plane is given by $\sqrt{|\overrightarrow{AB}|^2 - d^2}$.
Here,$|\overrightarrow{AB}|^2 = 1^2 + 0^2 + 1^2 = 2$.
So,the length of the projection $= \sqrt{2 - (\frac{2}{\sqrt{3}})^2} = \sqrt{2 - \frac{4}{3}} = \sqrt{\frac{6-4}{3}} = \sqrt{\frac{2}{3}}$.

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

(D) Given the quadratic equation $18x^2 - 9\pi x + \pi^2 = 0$.
Factoring the quadratic: $(3x - \pi)(6x - \pi) = 0$.
Thus,the roots are $\alpha = \frac{\pi}{6}$ and $\beta = \frac{\pi}{3}$ (since $\alpha < \beta$).
Next,we find the composite function $(g \circ f)(x) = g(f(x)) = \cos((\sqrt{x})^2) = \cos(x)$.
The area $A$ bounded by $y = \cos(x)$,$x = \frac{\pi}{6}$,$x = \frac{\pi}{3}$,and $y = 0$ is given by the integral:
$A = \int_{\pi/6}^{\pi/3} \cos(x) \, dx$.
Evaluating the integral: $A = [\sin(x)]_{\pi/6}^{\pi/3} = \sin(\frac{\pi}{3}) - \sin(\frac{\pi}{6})$.
Substituting the values: $A = \frac{\sqrt{3}}{2} - \frac{1}{2} = \frac{\sqrt{3} - 1}{2}$ sq. units.

(B) Let $\Delta = \left| \begin{matrix} x - 4 & 2x & 2x \\ 2x & x - 4 & 2x \\ 2x & 2x & x - 4 \end{matrix} \right|$.
Applying the operation $R_1 \to R_1 + R_2 + R_3$,we get:
$\Delta = \left| \begin{matrix} 5x - 4 & 5x - 4 & 5x - 4 \\ 2x & x - 4 & 2x \\ 2x & 2x & x - 4 \end{matrix} \right| = (5x - 4) \left| \begin{matrix} 1 & 1 & 1 \\ 2x & x - 4 & 2x \\ 2x & 2x & x - 4 \end{matrix} \right|$.
Applying $C_2 \to C_2 - C_1$ and $C_3 \to C_3 - C_1$:
$\Delta = (5x - 4) \left| \begin{matrix} 1 & 0 & 0 \\ 2x & -x - 4 & 0 \\ 2x & 0 & -x - 4 \end{matrix} \right| = (5x - 4)(-x - 4)^2 = (5x - 4)(x + 4)^2$.
Comparing this with $(A + Bx)(x - A)^2$,we have $(A + Bx)(x - A)^2 = (5x - 4)(x - (-4))^2$.
Thus,$A = -4$ and $B = 5$.
Therefore,the ordered pair $(A, B) = (-4, 5)$.

(A) For a system of homogeneous linear equations to have a non-zero solution,the determinant of the coefficient matrix must be zero:
$\begin{vmatrix} 1 & k & 3 \\ 3 & k & -2 \\ 2 & 4 & -3 \end{vmatrix} = 0$
Expanding along the first row:
$1(-3k + 8) - k(-9 + 4) + 3(12 - 2k) = 0$
$-3k + 8 + 5k + 36 - 6k = 0$
$-4k + 44 = 0 \Rightarrow k = 11$
Substituting $k = 11$ into the equations:
$x + 11y + 3z = 0$ $(1)$
$3x + 11y - 2z = 0$ $(2)$
$2x + 4y - 3z = 0$ $(3)$
Subtracting $(1)$ from $(2)$:
$(3x - x) + (11y - 11y) + (-2z - 3z) = 0 \Rightarrow 2x - 5z = 0 \Rightarrow x = \frac{5}{2}z$
Substituting $x = \frac{5}{2}z$ into $(1)$:
$\frac{5}{2}z + 11y + 3z = 0 \Rightarrow 11y = -\frac{11}{2}z \Rightarrow y = -\frac{1}{2}z$
Now,calculate $\frac{xz}{y^2}$:
$\frac{xz}{y^2} = \frac{(\frac{5}{2}z)(z)}{(-\frac{1}{2}z)^2} = \frac{\frac{5}{2}z^2}{\frac{1}{4}z^2} = \frac{5}{2} \times 4 = 10$

(C) Given $h(x) = \frac{x^2 + \frac{1}{x^2}}{x - \frac{1}{x}}$.
We can rewrite the numerator as $(x - \frac{1}{x})^2 + 2$.
So,$h(x) = \frac{(x - \frac{1}{x})^2 + 2}{x - \frac{1}{x}} = (x - \frac{1}{x}) + \frac{2}{x - \frac{1}{x}}$.
Let $t = x - \frac{1}{x}$. Since $x \in R - \{-1, 1, 0\}$,$t$ can take any real value except $0$.
Then $h(t) = t + \frac{2}{t}$.
For $t > 0$,by the $AM$-$GM$ inequality,$t + \frac{2}{t} \ge 2\sqrt{t \cdot \frac{2}{t}} = 2\sqrt{2}$.
The equality holds when $t = \frac{2}{t}$,i.e.,$t^2 = 2$,so $t = \sqrt{2}$ (since $t > 0$).
For $t < 0$,let $u = -t$,where $u > 0$. Then $h(t) = -u - \frac{2}{u} = -(u + \frac{2}{u}) \le -2\sqrt{2}$.
Thus,the local minimum value of $h(x)$ is $2\sqrt{2}$.

(D) The function is given by $f(x) = |x-\pi|(e^{|x|}-1)\sin|x|$.
We check the differentiability of $f(x)$ at the points where the absolute value functions might not be differentiable,namely $x=0$ and $x=\pi$.
At $x=\pi$:
$f(\pi) = 0$.
$RHD = \lim_{h \to 0^+} \frac{|\pi+h-\pi|(e^{|\pi+h|}-1)\sin|\pi+h| - 0}{h} = \lim_{h \to 0^+} \frac{h(e^{\pi+h}-1)\sin(\pi+h)}{h} = (e^{\pi}-1)\sin(\pi) = 0$.
$LHD = \lim_{h \to 0^+} \frac{|\pi-h-\pi|(e^{|\pi-h|}-1)\sin|\pi-h| - 0}{-h} = \lim_{h \to 0^+} \frac{h(e^{\pi-h}-1)\sin(\pi-h)}{-h} = -(e^{\pi}-1)\sin(\pi) = 0$.
Since $RHD = LHD = 0$,$f(x)$ is differentiable at $x=\pi$.
At $x=0$:
$f(0) = 0$.
$RHD = \lim_{h \to 0^+} \frac{|h-\pi|(e^{|h|}-1)\sin|h| - 0}{h} = \lim_{h \to 0^+} \frac{|h-\pi|(e^h-1)\sin(h)}{h} = |-\pi| \cdot (1) \cdot (0) = 0$.
$LHD = \lim_{h \to 0^+} \frac{|-h-\pi|(e^{|-h|}-1)\sin|-h| - 0}{-h} = \lim_{h \to 0^+} \frac{|-h-\pi|(e^h-1)\sin(h)}{-h} = |-\pi| \cdot (1) \cdot (0) = 0$.
Since $RHD = LHD = 0$,$f(x)$ is differentiable at $x=0$.
Since $f(x)$ is differentiable at all points,the set $S$ of points where $f(x)$ is not differentiable is empty,i.e.,$S = \emptyset$.

(A) Let $I = \int \frac{\sin^2 x \cos^2 x}{(\sin^5 x + \cos^3 x \sin^2 x + \sin^3 x \cos^2 x + \cos^5 x)^2} dx$.
Factor the denominator:
$\sin^5 x + \cos^3 x \sin^2 x + \sin^3 x \cos^2 x + \cos^5 x = \sin^2 x(\sin^3 x + \cos^3 x) + \cos^2 x(\sin^3 x + \cos^3 x) = (\sin^2 x + \cos^2 x)(\sin^3 x + \cos^3 x) = (\sin^3 x + \cos^3 x)$.
Thus,$I = \int \frac{\sin^2 x \cos^2 x}{(\sin^3 x + \cos^3 x)^2} dx$.
Divide numerator and denominator by $\cos^6 x$:
$I = \int \frac{\tan^2 x \sec^2 x}{(\tan^3 x + 1)^2} dx$.
Let $t = 1 + \tan^3 x$,then $dt = 3 \tan^2 x \sec^2 x dx$,so $\tan^2 x \sec^2 x dx = \frac{dt}{3}$.
$I = \frac{1}{3} \int \frac{dt}{t^2} = \frac{1}{3} (-t^{-1}) + C = \frac{-1}{3(1 + \tan^3 x)} + C$.

(A) Let $R_1$ be the event of drawing a red ball in the first draw and $B_1$ be the event of drawing a black ball in the first draw.
Let $R_2$ be the event of drawing a red ball in the second draw.
Initially,the bag contains $4$ red and $6$ black balls,total $10$ balls.
$P(R_1) = \frac{4}{10} = \frac{2}{5}$ and $P(B_1) = \frac{6}{10} = \frac{3}{5}$.
If a red ball is drawn first,it is returned along with $2$ additional red balls. The bag now contains $4 + 2 = 6$ red balls and $6$ black balls,total $12$ balls.
So,$P(R_2 | R_1) = \frac{6}{12} = \frac{1}{2}$.
If a black ball is drawn first,it is returned along with $2$ additional black balls. The bag now contains $4$ red balls and $6 + 2 = 8$ black balls,total $12$ balls.
So,$P(R_2 | B_1) = \frac{4}{12} = \frac{1}{3}$.
Using the law of total probability,the probability that the second ball is red is:
$P(R_2) = P(R_1) \times P(R_2 | R_1) + P(B_1) \times P(R_2 | B_1)$
$P(R_2) = (\frac{4}{10} \times \frac{6}{12}) + (\frac{6}{10} \times \frac{4}{12})$
$P(R_2) = \frac{24}{120} + \frac{24}{120} = \frac{48}{120} = \frac{2}{5}$.

(D) Since $\vec{u}$ is coplanar with $\vec{a}$ and $\vec{b}$,we can write $\vec{u} = x\vec{a} + y\vec{b}$.
Given $\vec{u} \perp \vec{a}$,so $\vec{u} \cdot \vec{a} = 0$.
$(x\vec{a} + y\vec{b}) \cdot \vec{a} = 0 \implies x|\vec{a}|^2 + y(\vec{a} \cdot \vec{b}) = 0$.
Calculate $|\vec{a}|^2 = 2^2 + 3^2 + (-1)^2 = 4 + 9 + 1 = 14$.
Calculate $\vec{a} \cdot \vec{b} = (2)(0) + (3)(1) + (-1)(1) = 3 - 1 = 2$.
So,$14x + 2y = 0 \implies y = -7x$.
Thus,$\vec{u} = x\vec{a} - 7x\vec{b} = x(\vec{a} - 7\vec{b})$.
$vec{a} - 7\vec{b} = (2\hat{i} + 3\hat{j} - \hat{k}) - 7(\hat{j} + \hat{k}) = 2\hat{i} - 4\hat{j} - 8\hat{k}$.
Given $\vec{u} \cdot \vec{b} = 24$,so $x(\vec{a} - 7\vec{b}) \cdot \vec{b} = 24$.
$(\vec{a} - 7\vec{b}) \cdot \vec{b} = \vec{a} \cdot \vec{b} - 7|\vec{b}|^2 = 2 - 7(1^2 + 1^2) = 2 - 14 = -12$.
So,$x(-12) = 24 \implies x = -2$.
Therefore,$\vec{u} = -2(2\hat{i} - 4\hat{j} - 8\hat{k}) = -4\hat{i} + 8\hat{j} + 16\hat{k}$.
$|\vec{u}|^2 = (-4)^2 + 8^2 + 16^2 = 16 + 64 + 256 = 336$.

(C) Let the intercepts of the variable plane on the $x, y,$ and $z$ axes be $a, b,$ and $c$ respectively.
The equation of the plane is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.
Since the plane passes through the fixed point $(3, 2, 1)$,we have $\frac{3}{a} + \frac{2}{b} + \frac{1}{c} = 1$.
The plane through $A(a, 0, 0)$ parallel to the $yz$-plane is $x = a$.
The plane through $B(0, b, 0)$ parallel to the $zx$-plane is $y = b$.
The plane through $C(0, 0, c)$ parallel to the $xy$-plane is $z = c$.
The point of intersection of these three planes is $(a, b, c)$.
Let the point of intersection be $(x, y, z)$. Then $x = a, y = b,$ and $z = c$.
Substituting these into the equation $\frac{3}{a} + \frac{2}{b} + \frac{1}{c} = 1$,we get the locus $\frac{3}{x} + \frac{2}{y} + \frac{1}{z} = 1$.

(B) Let $y = \frac{x - 4}{x + 2}$.
Then $x - 4 = y(x + 2) \Rightarrow x - 4 = xy + 2y$.
$x(1 - y) = 2y + 4 \Rightarrow x = \frac{2y + 4}{1 - y}$.
Substituting $x$ into $f(y) = 2x + 1$,we get $f(y) = 2\left( \frac{2y + 4}{1 - y} \right) + 1$.
$f(y) = \frac{4y + 8 + 1 - y}{1 - y} = \frac{3y + 9}{1 - y}$.
Thus,$f(x) = \frac{3x + 9}{1 - x} = \frac{3(x - 1 + 4)}{1 - x} = \frac{3(x - 1)}{1 - x} + \frac{12}{1 - x} = -3 + \frac{12}{1 - x}$.
Now,$\int f(x) \,dx = \int \left( -3 + \frac{12}{1 - x} \right) \,dx$.
$= -3x + 12 \int \frac{1}{1 - x} \,dx$.
$= -3x + 12 \left( \frac{\log_e |1 - x|}{-1} \right) + C$.
$= -12 \log_e |1 - x| - 3x + C$.

(A) First,check for symmetry:
For $R_1$: $(c, a) \in R_1$ but $(a, c) \in R_1$. However,$(b, c) \in R_1$ but $(c, b) \notin R_1$. Thus,$R_1$ is not symmetric.
For $R_2$: $(a, b) \in R_2$ and $(b, a) \in R_2$. $(a, c) \in R_2$ but $(c, a) \in R_2$. $(c, a) \in R_2$ but $(a, c) \in R_2$. Thus,$R_2$ is symmetric.
Next,check for transitivity:
For $R_1$: $(b, c) \in R_1$ and $(c, a) \in R_1$,but $(b, a) \notin R_1$. Thus,$R_1$ is not transitive.
For $R_2$: $(b, a) \in R_2$ and $(a, c) \in R_2$,but $(b, c) \notin R_2$. Thus,$R_2$ is not transitive.
Conclusion: $R_2$ is symmetric but not transitive,and $R_1$ is neither symmetric nor transitive.

(A) Given the equation: $x^2 + y^2 + \sin y = 4$.
Differentiating with respect to $x$:
$2x + 2y \frac{dy}{dx} + \cos y \frac{dy}{dx} = 0$
$2x + (2y + \cos y) \frac{dy}{dx} = 0$
$\frac{dy}{dx} = -\frac{2x}{2y + \cos y}$.
At the point $(-2, 0)$:
$\frac{dy}{dx} = -\frac{2(-2)}{2(0) + \cos 0} = \frac{4}{1} = 4$.
Differentiating $2x + (2y + \cos y) \frac{dy}{dx} = 0$ with respect to $x$:
$2 + (2 \frac{dy}{dx} + \cos y \frac{dy}{dx}) \frac{dy}{dx} + (2y + \cos y) \frac{d^2y}{dx^2} = 0$
$2 + (2 + \cos y) \left(\frac{dy}{dx}\right)^2 + (2y + \cos y) \frac{d^2y}{dx^2} = 0$.
Substituting $x = -2, y = 0$,and $\frac{dy}{dx} = 4$:
$2 + (2 + \cos 0)(4)^2 + (2(0) + \cos 0) \frac{d^2y}{dx^2} = 0$
$2 + (2 + 1)(16) + (1) \frac{d^2y}{dx^2} = 0$
$2 + 3(16) + \frac{d^2y}{dx^2} = 0$
$2 + 48 + \frac{d^2y}{dx^2} = 0$
$\frac{d^2y}{dx^2} = -50$.
Wait,re-evaluating the derivative of the first equation:
$2x + 2y \frac{dy}{dx} + \cos y \frac{dy}{dx} = 0$ is correct.
Second derivative: $2 + 2(\frac{dy}{dx})^2 + 2y \frac{d^2y}{dx^2} - \sin y (\frac{dy}{dx})^2 + \cos y \frac{d^2y}{dx^2} = 0$.
At $(-2, 0)$ with $\frac{dy}{dx} = 4$:
$2 + 2(4)^2 + 2(0) \frac{d^2y}{dx^2} - \sin(0)(4)^2 + \cos(0) \frac{d^2y}{dx^2} = 0$
$2 + 32 + 0 - 0 + 1 \cdot \frac{d^2y}{dx^2} = 0$
$34 + \frac{d^2y}{dx^2} = 0 \Rightarrow \frac{d^2y}{dx^2} = -34$.

(D) Given that $A \cdot \begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix}$ is a scalar matrix. Let this scalar matrix be $K = \begin{bmatrix} k & 0 \\ 0 & k \end{bmatrix}$.
Then $A = \begin{bmatrix} k & 0 \\ 0 & k \end{bmatrix} \cdot \begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix}^{-1}$.
The inverse of $\begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix}$ is $\frac{1}{3} \begin{bmatrix} 3 & -2 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -2/3 \\ 0 & 1/3 \end{bmatrix}$.
Thus,$A = \begin{bmatrix} k & 0 \\ 0 & k \end{bmatrix} \begin{bmatrix} 1 & -2/3 \\ 0 & 1/3 \end{bmatrix} = \begin{bmatrix} k & -2k/3 \\ 0 & k/3 \end{bmatrix}$.
Given $|3A| = 108$. Since $A$ is a $2 \times 2$ matrix,$|3A| = 3^2 |A| = 9|A|$.
So,$9|A| = 108 \Rightarrow |A| = 12$.
Calculating $|A|$ from our expression: $|A| = (k)(k/3) - 0 = k^2/3$.
Equating the two: $k^2/3 = 12 \Rightarrow k^2 = 36 \Rightarrow k = \pm 6$.
For $k = 6$,$A = \begin{bmatrix} 6 & -4 \\ 0 & 2 \end{bmatrix}$,so $A^2 = \begin{bmatrix} 6 & -4 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} 6 & -4 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 36 & -32 \\ 0 & 4 \end{bmatrix}$.
For $k = -6$,$A = \begin{bmatrix} -6 & 4 \\ 0 & -2 \end{bmatrix}$,so $A^2 = \begin{bmatrix} -6 & 4 \\ 0 & -2 \end{bmatrix} \begin{bmatrix} -6 & 4 \\ 0 & -2 \end{bmatrix} = \begin{bmatrix} 36 & -32 \\ 0 & 4 \end{bmatrix}$.
In both cases,$A^2 = \begin{bmatrix} 36 & -32 \\ 0 & 4 \end{bmatrix}$.

(A) Given $f(x) = \left| \begin{array}{ccc} \cos x & x & 1 \\ 2\sin x & x^2 & 2x \\ \tan x & x & 1 \end{array} \right|$.
Expanding the determinant along the first row:
$f(x) = \cos x(x^2 - 2x^2) - x(2\sin x - 2x\tan x) + 1(2x\sin x - x^2\tan x)$
$f(x) = -x^2\cos x - 2x\sin x + 2x^2\tan x + 2x\sin x - x^2\tan x$
$f(x) = x^2\tan x - x^2\cos x = x^2(\tan x - \cos x)$.
Now,differentiate $f(x)$ with respect to $x$:
$f'(x) = 2x(\tan x - \cos x) + x^2(\sec^2 x + \sin x)$.
We need to find $\lim_{x \to 0} \frac{f'(x)}{x}$:
$\lim_{x \to 0} \frac{2x(\tan x - \cos x) + x^2(\sec^2 x + \sin x)}{x}$
$= \lim_{x \to 0} [2(\tan x - \cos x) + x(\sec^2 x + \sin x)]$
$= 2(0 - 1) + 0(1 + 0) = -2$.
Thus,the limit exists and is equal to $-2$.

(B) The given system of linear equations is:
$x + y + z = 2$
$2x + y - z = 3$
$3x + 2y + kz = 4$
$A$ system of linear equations has a unique solution if and only if the determinant of the coefficient matrix is non-zero.
Let $D$ be the determinant of the coefficient matrix:
$D = \begin{vmatrix} 1 & 1 & 1 \\ 2 & 1 & -1 \\ 3 & 2 & k \end{vmatrix}$
Expanding along the first row:
$D = 1(1 \cdot k - (-1) \cdot 2) - 1(2 \cdot k - (-1) \cdot 3) + 1(2 \cdot 2 - 1 \cdot 3)$
$D = 1(k + 2) - 1(2k + 3) + 1(4 - 3)$
$D = k + 2 - 2k - 3 + 1$
$D = -k$
For a unique solution,$D \neq 0$,which implies $-k \neq 0$,or $k \neq 0$.
Therefore,$S = R - \{0\}$.

(C) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^4 x \left( 1 + \log \left( \frac{2 + \sin x}{2 - \sin x} \right) \right) dx$.
Using the property $\int_{-a}^{a} f(x) dx = \int_{0}^{a} (f(x) + f(-x)) dx$,we have:
$I = \int_{0}^{\frac{\pi}{2}} \left[ \sin^4 x \left( 1 + \log \left( \frac{2 + \sin x}{2 - \sin x} \right) \right) + \sin^4(-x) \left( 1 + \log \left( \frac{2 + \sin(-x)}{2 - \sin(-x)} \right) \right) \right] dx$.
Since $\sin(-x) = -\sin x$,we get:
$I = \int_{0}^{\frac{\pi}{2}} \sin^4 x \left[ 1 + \log \left( \frac{2 + \sin x}{2 - \sin x} \right) + 1 + \log \left( \frac{2 - \sin x}{2 + \sin x} \right) \right] dx$.
$I = \int_{0}^{\frac{\pi}{2}} \sin^4 x \left[ 2 + \log \left( \frac{2 + \sin x}{2 - \sin x} \cdot \frac{2 - \sin x}{2 + \sin x} \right) \right] dx$.
$I = \int_{0}^{\frac{\pi}{2}} \sin^4 x [ 2 + \log(1) ] dx = \int_{0}^{\frac{\pi}{2}} 2 \sin^4 x dx = 2 \int_{0}^{\frac{\pi}{2}} \sin^4 x dx$.
Using Wallis' formula,$\int_{0}^{\frac{\pi}{2}} \sin^n x dx = \frac{(n-1)!!}{n!!} \cdot \frac{\pi}{2}$ for even $n$:
$I = 2 \cdot \left( \frac{3 \cdot 1}{4 \cdot 2} \cdot \frac{\pi}{2} \right) = 2 \cdot \frac{3\pi}{16} = \frac{3\pi}{8}$.

(D) The direction vector $\vec{v}$ of the line of intersection of the planes $3x + 4y + z - 1 = 0$ and $5x + 8y + 2z + 14 = 0$ is given by the cross product of their normal vectors $\vec{n_1} = (3, 4, 1)$ and $\vec{n_2} = (5, 8, 2)$.
$\vec{v} = \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 1 \\ 5 & 8 & 2 \end{vmatrix} = \hat{i}(8-8) - \hat{j}(6-5) + \hat{k}(24-20) = 0\hat{i} - 1\hat{j} + 4\hat{k}$.
The normal vector to the plane $x + y + z = 5$ is $\vec{n} = (1, 1, 1)$.
The angle $\theta$ between a line with direction vector $\vec{v}$ and a plane with normal vector $\vec{n}$ is given by $\sin \theta = \frac{|\vec{v} \cdot \vec{n}|}{|\vec{v}| |\vec{n}|}$.
$\vec{v} \cdot \vec{n} = (0)(1) + (-1)(1) + (4)(1) = 3$.
$|\vec{v}| = \sqrt{0^2 + (-1)^2 + 4^2} = \sqrt{17}$.
$|\vec{n}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$.
$\sin \theta = \frac{|3|}{\sqrt{17} \cdot \sqrt{3}} = \frac{3}{\sqrt{51}} = \frac{\sqrt{3} \cdot \sqrt{3}}{\sqrt{17} \cdot \sqrt{3}} = \sqrt{\frac{3}{17}}$.
Therefore,$\theta = \sin^{-1}\left(\sqrt{\frac{3}{17}}\right)$.

(A) Let $R = 3 \, cm$ be the radius of the sphere.
Let $h$ be the height and $b$ be the base radius of the inscribed cone.
From the geometry of the sphere,the relationship between $h, b,$ and $R$ is given by $(h-R)^2 + b^2 = R^2$.
Thus,$b^2 = R^2 - (h-R)^2 = 2hR - h^2$.
The volume of the cone is $V = \frac{1}{3} \pi b^2 h = \frac{1}{3} \pi (2hR - h^2) h = \frac{\pi}{3} (2Rh^2 - h^3)$.
To maximize volume,we find $\frac{dV}{dh} = \frac{\pi}{3} (4Rh - 3h^2) = 0$.
This gives $h(4R - 3h) = 0$. Since $h \neq 0$,we have $h = \frac{4R}{3} = \frac{4(3)}{3} = 4 \, cm$.
Now,$b^2 = 2(4)(3) - (4)^2 = 24 - 16 = 8$,so $b = \sqrt{8} = 2\sqrt{2} \, cm$.
The slant height $l = \sqrt{h^2 + b^2} = \sqrt{4^2 + (2\sqrt{2})^2} = \sqrt{16 + 8} = \sqrt{24} = 2\sqrt{6} \, cm$.
The curved surface area is $A = \pi b l = \pi (2\sqrt{2}) (2\sqrt{6}) = 4\pi \sqrt{12} = 4\pi (2\sqrt{3}) = 8\sqrt{3} \pi \, cm^2$.

(A) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = 2$ and $Q = f(x)$.
Case $1$: For $x \in [0, 1]$,$f(x) = 1$.
$\frac{dy}{dx} + 2y = 1$. The integrating factor is $IF = e^{\int 2 dx} = e^{2x}$.
The solution is $y \cdot e^{2x} = \int 1 \cdot e^{2x} dx + C_1 = \frac{1}{2}e^{2x} + C_1$.
So,$y(x) = \frac{1}{2} + C_1 e^{-2x}$.
Given $y(0) = 0$,we have $0 = \frac{1}{2} + C_1 \Rightarrow C_1 = -\frac{1}{2}$.
Thus,$y(x) = \frac{1}{2} - \frac{1}{2}e^{-2x}$ for $x \in [0, 1]$.
At $x = 1$,$y(1) = \frac{1}{2} - \frac{1}{2}e^{-2} = \frac{e^2 - 1}{2e^2}$.
Case $2$: For $x > 1$,$f(x) = 0$.
$\frac{dy}{dx} + 2y = 0 \Rightarrow \frac{dy}{y} = -2 dx$.
Integrating both sides,$\ln|y| = -2x + C_2 \Rightarrow y = C_3 e^{-2x}$.
Using the continuity of $y(x)$ at $x = 1$,$y(1) = C_3 e^{-2} = \frac{e^2 - 1}{2e^2}$.
$C_3 = \frac{e^2 - 1}{2e^2} \cdot e^2 = \frac{e^2 - 1}{2}$.
So,$y(x) = \left(\frac{e^2 - 1}{2}\right) e^{-2x}$ for $x > 1$.
For $x = \frac{3}{2}$,$y\left(\frac{3}{2}\right) = \left(\frac{e^2 - 1}{2}\right) e^{-2(\frac{3}{2})} = \left(\frac{e^2 - 1}{2}\right) e^{-3} = \frac{e^2 - 1}{2e^3}$.

(B) The given region is bounded by $y = \sqrt{x}$,$y = x - 2$,$x = 0$,and $y = 0$.
To find the intersection of $y = \sqrt{x}$ and $y = x - 2$,we set $\sqrt{x} = x - 2$.
Squaring both sides,we get $x = (x - 2)^2 = x^2 - 4x + 4$,which simplifies to $x^2 - 5x + 4 = 0$.
Factoring gives $(x - 4)(x - 1) = 0$,so $x = 4$ or $x = 1$.
Since $y = \sqrt{x}$ must be non-negative,at $x = 1$,$y = 1$ but $x - 2 = -1$,which is not equal. Thus,the intersection is at $(4, 2)$.
The area is the integral of the upper curve minus the lower curve.
For $0 \le x \le 2$,the region is bounded by $y = \sqrt{x}$ and $y = 0$. Area $A_1 = \int_{0}^{2} \sqrt{x} \, dx = [\frac{2}{3}x^{3/2}]_{0}^{2} = \frac{2}{3}(2\sqrt{2}) = \frac{4\sqrt{2}}{3}$.
For $2 \le x \le 4$,the region is bounded by $y = \sqrt{x}$ and $y = x - 2$. Area $A_2 = \int_{2}^{4} (\sqrt{x} - (x - 2)) \, dx = [\frac{2}{3}x^{3/2} - \frac{x^2}{2} + 2x]_{2}^{4}$.
$A_2 = (\frac{2}{3}(8) - \frac{16}{2} + 8) - (\frac{2}{3}(2\sqrt{2}) - \frac{4}{2} + 4) = (\frac{16}{3}) - (\frac{4\sqrt{2}}{3} + 2) = \frac{10}{3} - \frac{4\sqrt{2}}{3}$.
Total Area = $A_1 + A_2 = \frac{4\sqrt{2}}{3} + \frac{10}{3} - \frac{4\sqrt{2}}{3} = \frac{10}{3}$ sq. units.

(B) Given: $\vec{a} + 2\vec{b} + 2\vec{c} = \vec{0}$ and $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1$.
Rearranging the equation: $\vec{a} + 2\vec{c} = -2\vec{b}$.
Squaring both sides: $(\vec{a} + 2\vec{c}) \cdot (\vec{a} + 2\vec{c}) = (-2\vec{b}) \cdot (-2\vec{b})$.
$|\vec{a}|^2 + 4|\vec{c}|^2 + 4(\vec{a} \cdot \vec{c}) = 4|\vec{b}|^2$.
Substituting the magnitudes: $1 + 4(1) + 4(\vec{a} \cdot \vec{c}) = 4(1)$.
$5 + 4(\vec{a} \cdot \vec{c}) = 4 \Rightarrow 4(\vec{a} \cdot \vec{c}) = -1 \Rightarrow \vec{a} \cdot \vec{c} = -\frac{1}{4}$.
We know that $|\vec{a} \times \vec{c}|^2 = |\vec{a}|^2 |\vec{c}|^2 - (\vec{a} \cdot \vec{c})^2$.
$|\vec{a} \times \vec{c}|^2 = (1)(1) - (-\frac{1}{4})^2 = 1 - \frac{1}{16} = \frac{15}{16}$.
Therefore,$|\vec{a} \times \vec{c}| = \sqrt{\frac{15}{16}} = \frac{\sqrt{15}}{4}$.

(A) Given $f(t) = (\|\lambda\|e^{\|t\|} - \mu) \sin(2\|t\|)$.
Since $\|t\|$ is involved,we analyze $f(t)$ for $t > 0$ and $t < 0$.
For $t > 0$,$f(t) = (\|\lambda\|e^t - \mu) \sin(2t)$.
For $t < 0$,$f(t) = (\|\lambda\|e^{-t} - \mu) \sin(-2t) = -(\|\lambda\|e^{-t} - \mu) \sin(2t)$.
For $f(t)$ to be differentiable at $t=0$,it must be continuous at $t=0$.
$f(0) = (\|\lambda\| - \mu) \sin(0) = 0$.
Now,we check the derivative $f'(t)$ at $t=0$ using $LHD = RHD$.
$RHD = \lim_{t \to 0^+} \frac{f(t) - f(0)}{t} = \lim_{t \to 0^+} (\|\lambda\|e^t - \mu) \frac{\sin(2t)}{t} = (\|\lambda\| - \mu) \times 2 = 2(\|\lambda\| - \mu)$.
$LHD = \lim_{t \to 0^-} \frac{f(t) - f(0)}{t} = \lim_{t \to 0^-} -(\|\lambda\|e^{-t} - \mu) \frac{\sin(2t)}{t} = -(\|\lambda\| - \mu) \times 2 = -2(\|\lambda\| - \mu)$.
For differentiability,$LHD = RHD \implies 2(\|\lambda\| - \mu) = -2(\|\lambda\| - \mu)$.
$4(\|\lambda\| - \mu) = 0 \implies \|\lambda\| = \mu$.
Since $\|\lambda\| \ge 0$,we must have $\mu \ge 0$.
Thus,$S = \{(\lambda, \mu) : \mu = \|\lambda\|, \mu \ge 0, \lambda \in R\}$.
This set $S$ is a subset of $R \times [0, \infty)$.

(A) Let $E$ be the event that one white and one red ball are drawn. Let $H_A$ and $H_B$ be the events of selecting box $A$ and box $B$ respectively. $P(H_A) = P(H_B) = \frac{1}{2}$.
The probability of drawing one white and one red ball from box $A$ is $P(E|H_A) = \frac{^2C_1 \times ^3C_1}{^7C_2} = \frac{2 \times 3}{21} = \frac{6}{21} = \frac{2}{7}$.
The probability of drawing one white and one red ball from box $B$ is $P(E|H_B) = \frac{^4C_1 \times ^2C_1}{^9C_2} = \frac{4 \times 2}{36} = \frac{8}{36} = \frac{2}{9}$.
Using Bayes' Theorem,the probability that the balls were drawn from box $B$ given event $E$ is:
$P(H_B|E) = \frac{P(H_B)P(E|H_B)}{P(H_A)P(E|H_A) + P(H_B)P(E|H_B)}$
$P(H_B|E) = \frac{\frac{1}{2} \times \frac{2}{9}}{\frac{1}{2} \times \frac{2}{7} + \frac{1}{2} \times \frac{2}{9}} = \frac{\frac{2}{9}}{\frac{2}{7} + \frac{2}{9}} = \frac{\frac{2}{9}}{\frac{18+14}{63}} = \frac{2}{9} \times \frac{63}{32} = \frac{1}{1} \times \frac{7}{16} = \frac{7}{16}$.

(D) To check if the function is invertible,we need to show it is both one-to-one (injective) and onto (surjective).
$1$. One-to-one: Let $f(x_1) = f(x_2)$.
$\frac{x_1 - 1}{x_1 - 2} = \frac{x_2 - 1}{x_2 - 2}$
$(x_1 - 1)(x_2 - 2) = (x_2 - 1)(x_1 - 2)$
$x_1x_2 - 2x_1 - x_2 + 2 = x_1x_2 - 2x_2 - x_1 + 2$
$-2x_1 - x_2 = -2x_2 - x_1$
$x_2 = x_1$. Thus,$f$ is one-to-one.
$2$. Onto: Let $y = \frac{x - 1}{x - 2}$.
$y(x - 2) = x - 1$
$yx - 2y = x - 1$
$yx - x = 2y - 1$
$x(y - 1) = 2y - 1$
$x = \frac{2y - 1}{y - 1}$.
Since for every $y \in R - \{1\}$,there exists an $x \in R - \{2\}$,the function is onto.
Since $f$ is both one-to-one and onto,it is invertible,and $f^{-1}(y) = \frac{2y - 1}{y - 1}$.

(D) For a system of linear equations to have no solution,the determinant of the coefficient matrix $\Delta$ must be zero,and at least one of the Cramer's rule determinants $(\Delta_1, \Delta_2, \Delta_3)$ must be non-zero.
First,calculate $\Delta$:
$\Delta = \begin{vmatrix} 1 & a & 1 \\ 1 & 2 & 2 \\ 1 & 5 & 3 \end{vmatrix} = 1(6 - 10) - a(3 - 2) + 1(5 - 2) = -4 - a + 3 = -a - 1$.
Setting $\Delta = 0$,we get $-a - 1 = 0$,which implies $a = -1$.
Next,calculate $\Delta_2$ (replacing the second column with the constants $3, 6, b$):
$\Delta_2 = \begin{vmatrix} 1 & 3 & 1 \\ 1 & 6 & 2 \\ 1 & b & 3 \end{vmatrix} = 1(18 - 2b) - 3(3 - 2) + 1(b - 6) = 18 - 2b - 3 + b - 6 = 9 - b$.
For the system to have no solution,we require $\Delta_2 \neq 0$,so $9 - b \neq 0$,which implies $b \neq 9$.
Thus,the condition is $a = -1$ and $b \neq 9$.

(B) Given differential equation is $(x^2 - y^2) \, dx + 2xy \, dy = 0$.
This can be written as $\frac{dy}{dx} = \frac{y^2 - x^2}{2xy}$.
Since this is a homogeneous differential equation,we put $y = ux$,which implies $\frac{dy}{dx} = u + x \frac{du}{dx}$.
Substituting these into the equation: $u + x \frac{du}{dx} = \frac{u^2 x^2 - x^2}{2x(ux)} = \frac{u^2 - 1}{2u}$.
$x \frac{du}{dx} = \frac{u^2 - 1}{2u} - u = \frac{u^2 - 1 - 2u^2}{2u} = \frac{-(1 + u^2)}{2u}$.
Separating variables: $\int \frac{2u}{1 + u^2} \, du = - \int \frac{1}{x} \, dx$.
Integrating both sides: $\ln(1 + u^2) = -\ln|x| + \ln|C|$.
$\ln(1 + u^2) = \ln\left(\frac{C}{x}\right) \Rightarrow 1 + u^2 = \frac{C}{x}$.
Substituting $u = \frac{y}{x}$: $1 + \frac{y^2}{x^2} = \frac{C}{x} \Rightarrow \frac{x^2 + y^2}{x^2} = \frac{C}{x} \Rightarrow x^2 + y^2 = Cx$.
Since the curve passes through $(1, 1)$,we have $1^2 + 1^2 = C(1) \Rightarrow C = 2$.
Thus,the equation of the curve is $x^2 + y^2 = 2x$,which can be rewritten as $(x - 1)^2 + y^2 = 1$.
This represents a circle with center $(1, 0)$ and radius $1$.

(A) Let $P(X)$ be the probability that player $X$ wins and $P(Y)$ be the probability that player $Y$ wins.
Player $X$ uses a biased coin with $P(H) = p$ and $P(T) = 1-p$. Player $Y$ uses a fair coin with $P(H) = 1/2$ and $P(T) = 1/2$.
$X$ wins if $X$ gets $H$ on the $1^{st}$ throw,or $X$ gets $T$,$Y$ gets $T$,and $X$ gets $H$ on the $3^{rd}$ throw,and so on.
$P(X) = p + (1-p)(1/2)p + (1-p)^2(1/2)^2p + \dots = p \sum_{n=0}^{\infty} (\frac{1-p}{2})^n = \frac{p}{1 - \frac{1-p}{2}} = \frac{2p}{1+p}$.
Since the total probability is $1$,$P(Y) = 1 - P(X) = 1 - \frac{2p}{1+p} = \frac{1-p}{1+p}$.
Given $P(X) = P(Y)$,we have $\frac{2p}{1+p} = \frac{1-p}{1+p}$.
$2p = 1 - p \Rightarrow 3p = 1 \Rightarrow p = \frac{1}{3}$.

(A) To solve the integral $I = \int \frac{2x + 5}{\sqrt{7 - 6x - x^2}} dx$,we express the numerator in terms of the derivative of the quadratic expression inside the square root.
Let $f(x) = 7 - 6x - x^2$. Then $f'(x) = -6 - 2x$.
We write $2x + 5 = -( -2x - 6 ) - 1$.
Thus,$I = \int \frac{-( -2x - 6 ) - 1}{\sqrt{7 - 6x - x^2}} dx = -\int \frac{-2x - 6}{\sqrt{7 - 6x - x^2}} dx - \int \frac{1}{\sqrt{7 - 6x - x^2}} dx$.
For the first integral,let $u = 7 - 6x - x^2$,so $du = (-6 - 2x) dx$. Then $\int \frac{du}{\sqrt{u}} = 2\sqrt{u} = 2\sqrt{7 - 6x - x^2}$.
So,$-\int \frac{-2x - 6}{\sqrt{7 - 6x - x^2}} dx = -2\sqrt{7 - 6x - x^2}$.
For the second integral,complete the square: $7 - 6x - x^2 = 16 - (x^2 + 6x + 9) = 4^2 - (x + 3)^2$.
So,$\int \frac{1}{\sqrt{4^2 - (x + 3)^2}} dx = \sin^{-1}\left(\frac{x + 3}{4}\right)$.
Combining these,$I = -2\sqrt{7 - 6x - x^2} - \sin^{-1}\left(\frac{x + 3}{4}\right) + C$.
Comparing with $A\sqrt{7 - 6x - x^2} + B\sin^{-1}\left(\frac{x + 3}{4}\right) + C$,we get $A = -2$ and $B = -1$.

(A) Let the points be $A(1, 2, 3)$ and $B(-3, 4, 5)$.
Since the plane bisects the line segment $AB$ at right angles,it passes through the midpoint $M$ of $AB$.
$M = \left(\frac{1-3}{2}, \frac{2+4}{2}, \frac{3+5}{2}\right) = (-1, 3, 4)$.
The normal vector $\vec{n}$ to the plane is the vector $\vec{AB} = (-3-1, 4-2, 5-3) = (-4, 2, 2)$.
We can simplify the normal vector to $\vec{n}' = (-2, 1, 1)$.
The equation of the plane is $-2(x - (-1)) + 1(y - 3) + 1(z - 4) = 0$.
$-2x - 2 + y - 3 + z - 4 = 0 \Rightarrow -2x + y + z = 9$.
Now,check the options:
For $(-3, 2, 1)$: $-2(-3) + 2 + 1 = 6 + 2 + 1 = 9$. This satisfies the equation.
Thus,the plane passes through the point $(-3, 2, 1)$.

(B) Let the position vectors of vertices $A, B, C$ be $\vec{a} = 4\hat{i} + 7\hat{j} + 8\hat{k}$,$\vec{b} = 2\hat{i} + 3\hat{j} + 4\hat{k}$,and $\vec{c} = 2\hat{i} + 5\hat{j} + 7\hat{k}$.
The angle bisector theorem states that the bisector of $\angle A$ divides the opposite side $BC$ in the ratio of the adjacent sides $AB:AC$.
First,calculate the lengths of sides $AB$ and $AC$:
$\vec{AB} = \vec{b} - \vec{a} = (2-4)\hat{i} + (3-7)\hat{j} + (4-8)\hat{k} = -2\hat{i} - 4\hat{j} - 4\hat{k}$.
$|\vec{AB}| = \sqrt{(-2)^2 + (-4)^2 + (-4)^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$.
$\vec{AC} = \vec{c} - \vec{a} = (2-4)\hat{i} + (5-7)\hat{j} + (7-8)\hat{k} = -2\hat{i} - 2\hat{j} - 1\hat{k}$.
$|\vec{AC}| = \sqrt{(-2)^2 + (-2)^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.
The ratio $AB:AC = 6:3 = 2:1$.
The position vector of point $D$ dividing $BC$ in ratio $m:n$ is $\frac{m\vec{c} + n\vec{b}}{m+n}$.
Here $m=2, n=1$:
$\vec{D} = \frac{2\vec{c} + 1\vec{b}}{2+1} = \frac{2(2\hat{i} + 5\hat{j} + 7\hat{k}) + 1(2\hat{i} + 3\hat{j} + 4\hat{k})}{3}$.
$\vec{D} = \frac{(4+2)\hat{i} + (10+3)\hat{j} + (14+4)\hat{k}}{3} = \frac{6\hat{i} + 13\hat{j} + 18\hat{k}}{3} = \frac{1}{3}(6\hat{i} + 13\hat{j} + 18\hat{k})$.

(A) Given $f(x) = \sin^{-1}\left(\frac{2 \cdot 3^x}{1 + (3^x)^2}\right)$.
Let $3^x = \tan \theta$,then $\theta = \tan^{-1}(3^x)$.
$f(x) = \sin^{-1}\left(\frac{2 \tan \theta}{1 + \tan^2 \theta}\right) = \sin^{-1}(\sin 2\theta) = 2\theta = 2 \tan^{-1}(3^x)$.
Differentiating with respect to $x$:
$f'(x) = 2 \cdot \frac{1}{1 + (3^x)^2} \cdot \frac{d}{dx}(3^x) = \frac{2}{1 + 9^x} \cdot 3^x \ln 3$.
Now,evaluate at $x = -\frac{1}{2}$:
$f'(-\frac{1}{2}) = \frac{2 \cdot 3^{-1/2}}{1 + 9^{-1/2}} \ln 3 = \frac{2 / \sqrt{3}}{1 + 1/3} \ln 3 = \frac{2 / \sqrt{3}}{4/3} \ln 3 = \frac{2}{\sqrt{3}} \cdot \frac{3}{4} \ln 3 = \frac{\sqrt{3}}{2} \ln 3$.
Since $\ln 3 = 2 \ln \sqrt{3}$,we have $f'(-\frac{1}{2}) = \frac{\sqrt{3}}{2} \cdot 2 \ln \sqrt{3} = \sqrt{3} \ln \sqrt{3}$.

(D) Since $f(x)$ is a polynomial of degree $4$,let $f(x) = Ax^4 + Bx^3 + Cx^2 + Dx + E$.
Given $\lim_{x \to 0} \left( \frac{f(x)}{x^2} + 1 \right) = 3$,we have $\lim_{x \to 0} \left( \frac{Ax^4 + Bx^3 + Cx^2 + Dx + E}{x^2} + 1 \right) = 3$.
This implies $\lim_{x \to 0} \left( Ax^2 + Bx + C + \frac{D}{x} + \frac{E}{x^2} + 1 \right) = 3$.
For the limit to exist and be finite,we must have $D = 0$ and $E = 0$. Thus,$C + 1 = 3$,which gives $C = 2$.
Now,$f(x) = Ax^4 + Bx^3 + 2x^2$. The derivative is $f'(x) = 4Ax^3 + 3Bx^2 + 4x$.
Since $f(x)$ has extreme values at $x = 1$ and $x = 2$,$f'(1) = 0$ and $f'(2) = 0$.
$f'(1) = 4A + 3B + 4 = 0 \implies 4A + 3B = -4$ (Equation $1$).
$f'(2) = 4A(8) + 3B(4) + 4(2) = 32A + 12B + 8 = 0 \implies 8A + 3B = -2$ (Equation $2$).
Subtracting Equation $1$ from Equation $2$: $(8A + 3B) - (4A + 3B) = -2 - (-4) \implies 4A = 2 \implies A = \frac{1}{2}$.
Substituting $A = \frac{1}{2}$ into Equation $1$: $4(\frac{1}{2}) + 3B = -4 \implies 2 + 3B = -4 \implies 3B = -6 \implies B = -2$.
Thus,$f(x) = \frac{1}{2}x^4 - 2x^3 + 2x^2$.
Finally,$f(-1) = \frac{1}{2}(-1)^4 - 2(-1)^3 + 2(-1)^2 = \frac{1}{2} + 2 + 2 = \frac{9}{2}$.

(B) Given equations are:
$l + 3m + 5n = 0$ ....$(1)$
$5lm - 2mn + 6nl = 0$ ....$(2)$
From equation $(1)$,we have $l = -3m - 5n$.
Substituting the value of $l$ in equation $(2)$:
$5(-3m - 5n)m - 2mn + 6n(-3m - 5n) = 0$
$-15m^2 - 25mn - 2mn - 18mn - 30n^2 = 0$
$-15m^2 - 45mn - 30n^2 = 0$
Dividing by $-15$,we get:
$m^2 + 3mn + 2n^2 = 0$
$(m + n)(m + 2n) = 0$
So,$m = -n$ or $m = -2n$.
Case $1$: If $m = -n$,then $l = -3(-n) - 5n = 3n - 5n = -2n$. The direction ratios are $(-2n, -n, n)$,which simplifies to $(-2, -1, 1)$.
Case $2$: If $m = -2n$,then $l = -3(-2n) - 5n = 6n - 5n = n$. The direction ratios are $(n, -2n, n)$,which simplifies to $(1, -2, 1)$.
Let the direction ratios be $\vec{a} = (-2, -1, 1)$ and $\vec{b} = (1, -2, 1)$.
The angle $\theta$ between the lines is given by:
$\cos \theta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|} = \frac{|(-2)(1) + (-1)(-2) + (1)(1)|}{\sqrt{(-2)^2 + (-1)^2 + 1^2} \sqrt{1^2 + (-2)^2 + 1^2}}$
$\cos \theta = \frac{|-2 + 2 + 1|}{\sqrt{6} \sqrt{6}} = \frac{1}{6}$
Therefore,$\theta = \cos^{-1}\left(\frac{1}{6}\right)$.

(A) Given the characteristic equation $(A - 3I)(A - 5I) = O$.
Expanding this,we get $A^2 - 8A + 15I = O$.
Since $A$ is non-singular,we can multiply throughout by $A^{-1}$:
$A^{-1}(A^2 - 8A + 15I) = A^{-1}O$
$A - 8I + 15A^{-1} = O$
$A + 15A^{-1} = 8I$.
To match the form $\alpha A + \beta A^{-1} = 4I$,we divide the entire equation by $2$:
$\frac{1}{2}A + \frac{15}{2}A^{-1} = 4I$.
Comparing this with $\alpha A + \beta A^{-1} = 4I$,we find $\alpha = \frac{1}{2}$ and $\beta = \frac{15}{2}$.
Therefore,$\alpha + \beta = \frac{1}{2} + \frac{15}{2} = \frac{16}{2} = 8$.

(B) Let $I = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{x}{1+\sin x} dx$.
Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,where $a+b = \frac{\pi}{4} + \frac{3\pi}{4} = \pi$:
$I = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{\pi-x}{1+\sin(\pi-x)} dx = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{\pi-x}{1+\sin x} dx$.
Adding the two expressions for $I$:
$2I = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{x + \pi - x}{1+\sin x} dx = \pi \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{1}{1+\sin x} dx$.
Multiply numerator and denominator by $(1-\sin x)$:
$2I = \pi \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{1-\sin x}{\cos^2 x} dx = \pi \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} (\sec^2 x - \sec x \tan x) dx$.
Integrating:
$2I = \pi [\tan x - \sec x]_{\frac{\pi}{4}}^{\frac{3\pi}{4}}$.
Evaluating at the limits:
$2I = \pi [(\tan \frac{3\pi}{4} - \sec \frac{3\pi}{4}) - (\tan \frac{\pi}{4} - \sec \frac{\pi}{4})]$.
$2I = \pi [(-1 - (-\sqrt{2})) - (1 - \sqrt{2})] = \pi [\sqrt{2} - 1 - 1 + \sqrt{2}] = \pi [2\sqrt{2} - 2]$.
$I = \pi(\sqrt{2}-1)$.

(D) For $x \in (0, 1)$,we have $x^3 < x^2 < x$.
Multiplying by $-1$,we get $-x^3 > -x^2 > -x$.
Since the exponential function $f(t) = e^t$ is strictly increasing,we have $e^{-x^3} > e^{-x^2} > e^{-x}$.
Now,consider the integrands:
For $I_3$,the integrand is $f_3(x) = e^{-x^3}$.
For $I_2$,the integrand is $f_2(x) = e^{-x^2} \cos^2 x$. Since $0 \le \cos^2 x \le 1$,we have $e^{-x^2} \cos^2 x \le e^{-x^2}$.
For $I_1$,the integrand is $f_1(x) = e^{-x} \cos^2 x$. Since $0 \le \cos^2 x \le 1$,we have $e^{-x} \cos^2 x \le e^{-x}$.
Comparing the functions on $(0, 1)$:
$e^{-x^3} > e^{-x^2} \ge e^{-x^2} \cos^2 x$ implies $I_3 > I_2$.
Also,$e^{-x^2} \cos^2 x > e^{-x} \cos^2 x$ because $e^{-x^2} > e^{-x}$ for $x \in (0, 1)$.
Thus,$I_3 > I_2 > I_1$.

(C) Since $f(x)$ is continuous at $x = 2$,we have $\lim_{x \to 2} f(x) = f(2)$.
$\lim_{x \to 2} (x - 1)^{\frac{1}{2 - x}} = k$. This is a $1^{\infty}$ form.
Using the formula $\lim_{x \to a} [g(x)]^{h(x)} = e^{\lim_{x \to a} (g(x) - 1)h(x)}$,we get:
$k = e^{\lim_{x \to 2} (x - 1 - 1) \cdot \frac{1}{2 - x}}$
$k = e^{\lim_{x \to 2} \frac{x - 2}{2 - x}}$
$k = e^{\lim_{x \to 2} \frac{-(2 - x)}{2 - x}}$
$k = e^{-1}$.

(C) Given $A = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \end{bmatrix}$.
Calculating powers of $A$:
$A^2 = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{bmatrix}$
$A^3 = \begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 6 & 3 & 1 \end{bmatrix}$
$A^4 = \begin{bmatrix} 1 & 0 & 0 \\ 4 & 1 & 0 \\ 10 & 4 & 1 \end{bmatrix}$
By observing the pattern,for $A^n$,the first column elements are $1$,$n$,and $\frac{n(n+1)}{2}$.
Thus,$A^n = \begin{bmatrix} 1 & 0 & 0 \\ n & 1 & 0 \\ \frac{n(n+1)}{2} & n & 1 \end{bmatrix}$.
For $n = 20$:
$A^{20} = \begin{bmatrix} 1 & 0 & 0 \\ 20 & 1 & 0 \\ \frac{20(21)}{2} & 20 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 20 & 1 & 0 \\ 210 & 20 & 1 \end{bmatrix}$.
The sum of the elements of the first column is $1 + 20 + 210 = 231$.

(B) Given $x = \sqrt{2^{\csc^{-1} t}}$ and $y = \sqrt{2^{\sec^{-1} t}}$.
Squaring both sides,we get $x^2 = 2^{\csc^{-1} t}$ and $y^2 = 2^{\sec^{-1} t}$.
Taking the product,$x^2 y^2 = 2^{\csc^{-1} t + \sec^{-1} t}$.
Since $\csc^{-1} t + \sec^{-1} t = \frac{\pi}{2}$ for $|t| \ge 1$,we have $x^2 y^2 = 2^{\pi/2}$,which is a constant.
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(x^2 y^2) = \frac{d}{dx}(2^{\pi/2})$
$2x y^2 + x^2 (2y \frac{dy}{dx}) = 0$
$2xy(y + x \frac{dy}{dx}) = 0$
Since $x, y \neq 0$,we have $y + x \frac{dy}{dx} = 0$.
Therefore,$\frac{dy}{dx} = -\frac{y}{x}$.

(C) The equation of the plane passing through three points $(x_1, y_1, z_1)$,$(x_2, y_2, z_2)$,and $(x_3, y_3, z_3)$ is given by the determinant equation:
$\begin{vmatrix} x - x_1 & y - y_1 & z - z_1 \\ x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ x_3 - x_1 & y_3 - y_1 & z_3 - z_1 \end{vmatrix} = 0$
Substituting the points $(-2, -2, 2)$,$(1, -1, 2)$,and $(1, 1, 1)$:
$\begin{vmatrix} x + 2 & y + 2 & z - 2 \\ 1 - (-2) & -1 - (-2) & 2 - 2 \\ 1 - (-2) & 1 - (-2) & 1 - 2 \end{vmatrix} = 0$
$\Rightarrow \begin{vmatrix} x + 2 & y + 2 & z - 2 \\ 3 & 1 & 0 \\ 3 & 3 & -1 \end{vmatrix} = 0$
Expanding the determinant:
$(x + 2)(-1 - 0) - (y + 2)(-3 - 0) + (z - 2)(9 - 3) = 0$
$-x - 2 + 3y + 6 + 6z - 12 = 0$
$-x + 3y + 6z - 8 = 0$
$x - 3y - 6z = -8$
Dividing by $-8$ to get the intercept form $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$:
$\frac{x}{-8} + \frac{y}{8/3} + \frac{z}{8/6} = 1$
The intercepts are $a = -8$,$b = 8/3$,and $c = 4/3$.
Sum of intercepts $= -8 + \frac{8}{3} + \frac{4}{3} = -8 + \frac{12}{3} = -8 + 4 = -4$.

(D) The equation of an ellipse centered at the origin is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
Since the ellipse passes through $(0,3)$,we have $\frac{0^2}{a^2} + \frac{3^2}{b^2} = 1$,which gives $b^2 = 9$.
Thus,the equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{9} = 1$.
Differentiating with respect to $x$,we get $\frac{2x}{a^2} + \frac{2y}{9} y' = 0$,which simplifies to $\frac{x}{a^2} + \frac{y y'}{9} = 0$,or $\frac{1}{a^2} = -\frac{y y'}{9x}$.
Substituting $\frac{1}{a^2}$ back into the equation $\frac{x^2}{a^2} + \frac{y^2}{9} = 1$,we get $x^2(-\frac{y y'}{9x}) + \frac{y^2}{9} = 1$.
Multiplying by $9$,we get $-x y y' + y^2 = 9$,or $x y y' - y^2 + 9 = 0$.

(C) We need to find $P[(\bar{A} \cap \bar{B})|C]$.
By the definition of conditional probability,$P[(\bar{A} \cap \bar{B})|C] = \frac{P(\bar{A} \cap \bar{B} \cap C)}{P(C)}$.
Using De Morgan's Law,$\bar{A} \cap \bar{B} = \overline{A \cup B}$.
Thus,$\bar{A} \cap \bar{B} \cap C = C \setminus ((A \cap C) \cup (B \cap C))$.
Therefore,$P(\bar{A} \cap \bar{B} \cap C) = P(C) - P((A \cap C) \cup (B \cap C))$.
Using the inclusion-exclusion principle,$P((A \cap C) \cup (B \cap C)) = P(A \cap C) + P(B \cap C) - P(A \cap B \cap C)$.
Since $A, B, C$ are pairwise independent,$P(A \cap C) = P(A)P(C)$ and $P(B \cap C) = P(B)P(C)$.
Given $P(A \cap B \cap C) = 0$,we have $P((A \cap C) \cup (B \cap C)) = P(A)P(C) + P(B)P(C) - 0 = P(C)(P(A) + P(B))$.
Substituting this back,$P(\bar{A} \cap \bar{B} \cap C) = P(C) - P(C)(P(A) + P(B)) = P(C)(1 - P(A) - P(B))$.
Finally,$P[(\bar{A} \cap \bar{B})|C] = \frac{P(C)(1 - P(A) - P(B))}{P(C)} = 1 - P(A) - P(B)$.
Since $1 - P(A) = P(\bar{A})$,the expression becomes $P(\bar{A}) - P(B)$.

(C) The given system of linear equations is:
$(k + 2)x + 10y = k$
$kx + (k + 3)y = k - 1$
For a system of linear equations to have no solution,the determinant of the coefficient matrix must be zero,and the system must be inconsistent.
Let $A = \begin{bmatrix} k + 2 & 10 \\ k & k + 3 \end{bmatrix}$.
Setting $|A| = 0$:
$(k + 2)(k + 3) - 10k = 0$
$k^2 + 5k + 6 - 10k = 0$
$k^2 - 5k + 6 = 0$
$(k - 2)(k - 3) = 0$
So,$k = 2$ or $k = 3$.
Case $1$: If $k = 2$,the equations are $4x + 10y = 2$ and $2x + 5y = 1$. Dividing the first by $2$ gives $2x + 5y = 1$,which is identical to the second equation. Thus,there are infinitely many solutions.
Case $2$: If $k = 3$,the equations are $5x + 10y = 3$ and $3x + 6y = 2$. Multiplying the first by $3$ and the second by $5$ gives $15x + 30y = 9$ and $15x + 30y = 10$. Since $9 \neq 10$,the system is inconsistent and has no solution.
Therefore,there is only $1$ value of $k$ for which the system has no solution.

(D) The direction ratios of the first line $\frac{x}{2} = \frac{y}{2} = \frac{z}{1}$ are $\vec{v_1} = (2, 2, 1)$.
The second line is $\frac{-(x - 5)}{-2} = \frac{7(y - 2)}{p} = \frac{z - 3}{4}$,which simplifies to $\frac{x - 5}{2} = \frac{y - 2}{p/7} = \frac{z - 3}{4}$.
The direction ratios of the second line are $\vec{v_2} = (2, p/7, 4)$.
The angle $\theta$ between the lines is given by $\cos \theta = \frac{|\vec{v_1} \cdot \vec{v_2}|}{|\vec{v_1}| |\vec{v_2}|}$.
Given $\cos \theta = \frac{2}{3}$,we have $\frac{2}{3} = \frac{|2(2) + 2(p/7) + 1(4)|}{\sqrt{2^2 + 2^2 + 1^2} \sqrt{2^2 + (p/7)^2 + 4^2}}$.
$\frac{2}{3} = \frac{|8 + 2p/7|}{3 \sqrt{4 + p^2/49 + 16}} = \frac{|8 + 2p/7|}{3 \sqrt{20 + p^2/49}}$.
Canceling $3$ from the denominator,we get $\frac{2}{1} = \frac{|8 + 2p/7|}{\sqrt{20 + p^2/49}}$.
Squaring both sides: $4 = \frac{(8 + 2p/7)^2}{20 + p^2/49} = \frac{64 + 32p/7 + 4p^2/49}{20 + p^2/49}$.
$80 + 4p^2/49 = 64 + 32p/7 + 4p^2/49$.
$80 = 64 + 32p/7 \Rightarrow 16 = 32p/7 \Rightarrow p = \frac{16 \times 7}{32} = \frac{7}{2}$.

(A) Given $\vec{a} = \hat{i} + \hat{j} + \hat{k}$,so $|\vec{a}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$.
Given $\vec{c} = \hat{j} - \hat{k}$,so $|\vec{c}| = \sqrt{0^2 + 1^2 + (-1)^2} = \sqrt{2}$.
We have $\vec{a} \times \vec{b} = \vec{c}$. Taking the magnitude on both sides,$|\vec{a} \times \vec{b}| = |\vec{c}|$.
$|\vec{a}| |\vec{b}| \sin \theta = |\vec{c}|$,where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$.
$\sqrt{3} |\vec{b}| \sin \theta = \sqrt{2} \quad \dots (1)$
Also,$\vec{a} \cdot \vec{b} = 3$.
$|\vec{a}| |\vec{b}| \cos \theta = 3 \Rightarrow \sqrt{3} |\vec{b}| \cos \theta = 3 \quad \dots (2)$
Squaring and adding $(1)$ and $(2)$:
$(\sqrt{3} |\vec{b}| \sin \theta)^2 + (\sqrt{3} |\vec{b}| \cos \theta)^2 = (\sqrt{2})^2 + 3^2$
$3 |\vec{b}|^2 (\sin^2 \theta + \cos^2 \theta) = 2 + 9$
$3 |\vec{b}|^2 = 11$
$|\vec{b}|^2 = \frac{11}{3} \Rightarrow |\vec{b}| = \sqrt{\frac{11}{3}}$.

(A) Given $f(x) = \int_0^x t(\sin x - \sin t) dt = \sin x \int_0^x t dt - \int_0^x t \sin t dt$.
Evaluating the integrals:
$\int_0^x t dt = \frac{x^2}{2}$.
Using integration by parts for $\int t \sin t dt = -t \cos t + \sin t$.
So,$f(x) = \sin x (\frac{x^2}{2}) - [-t \cos t + \sin t]_0^x = \frac{x^2}{2} \sin x + x \cos x - \sin x$.
Now,find the derivatives:
$f'(x) = x \sin x + \frac{x^2}{2} \cos x + \cos x - x \sin x - \cos x = \frac{x^2}{2} \cos x$.
$f''(x) = x \cos x - \frac{x^2}{2} \sin x$.
$f'''(x) = \cos x - x \sin x - x \sin x - \frac{x^2}{2} \cos x = \cos x - 2x \sin x - \frac{x^2}{2} \cos x$.
Check $f'''(x) + f'(x)$:
$f'''(x) + f'(x) = (\cos x - 2x \sin x - \frac{x^2}{2} \cos x) + (\frac{x^2}{2} \cos x) = \cos x - 2x \sin x$.
Thus,option $A$ is correct.

(A) Given function is $f(x) = 2x^3 - 9x^2 + 12x + 5$.
To find local maxima and minima,we find the derivative $f'(x)$:
$f'(x) = 6x^2 - 18x + 12$.
Setting $f'(x) = 0$ for critical points:
$6(x^2 - 3x + 2) = 0 \Rightarrow 6(x - 1)(x - 2) = 0$.
So,the critical points are $x = 1$ and $x = 2$.
Now,we check the second derivative $f''(x) = 12x - 18$.
At $x = 1$,$f''(1) = 12(1) - 18 = -6 < 0$,so $x = 1$ is a local maximum.
$M = f(1) = 2(1)^3 - 9(1)^2 + 12(1) + 5 = 2 - 9 + 12 + 5 = 10$.
At $x = 2$,$f''(2) = 12(2) - 18 = 6 > 0$,so $x = 2$ is a local minimum.
$m = f(2) = 2(2)^3 - 9(2)^2 + 12(2) + 5 = 16 - 36 + 24 + 5 = 9$.
Therefore,$M - m = 10 - 9 = 1$.