IIT JEE 2016 Mathematics Question Paper with Answer and Solution

(C) Given $-\frac{\pi}{6} < \theta < -\frac{\pi}{12}$.
For the equation $x^2 - 2x \sec \theta + 1 = 0$,the roots are $\alpha_1, \beta_1 = \frac{2 \sec \theta \pm \sqrt{4 \sec^2 \theta - 4}}{2} = \sec \theta \pm \tan \theta$.
Since $\theta \in (-\frac{\pi}{6}, -\frac{\pi}{12})$,$\sec \theta > 0$ and $\tan \theta < 0$. Thus,$\sec \theta - \tan \theta > \sec \theta + \tan \theta$. Given $\alpha_1 > \beta_1$,we have $\alpha_1 = \sec \theta - \tan \theta$.
For the equation $x^2 + 2x \tan \theta - 1 = 0$,the roots are $\alpha_2, \beta_2 = \frac{-2 \tan \theta \pm \sqrt{4 \tan^2 \theta + 4}}{2} = -\tan \theta \pm \sec \theta$.
Since $\sec \theta > 0$,we have $-\tan \theta + \sec \theta > -\tan \theta - \sec \theta$. Given $\alpha_2 > \beta_2$,we have $\beta_2 = -\tan \theta - \sec \theta$.
Therefore,$\alpha_1 + \beta_2 = (\sec \theta - \tan \theta) + (-\tan \theta - \sec \theta) = -2 \tan \theta$.

(A) Case $1$: $0$ boys are included.
Selecting $4$ girls from $6$ girls is given by ${}^6C_4 = 15$.
Selecting $1$ captain from the $4$ selected members is given by ${}^4C_1 = 4$.
Total ways for Case $1 = 15 \times 4 = 60$.
Case $2$: $1$ boy is included.
Selecting $3$ girls from $6$ and $1$ boy from $4$ is given by ${}^6C_3 \times {}^4C_1 = 20 \times 4 = 80$.
Selecting $1$ captain from the $4$ selected members is given by ${}^4C_1 = 4$.
Total ways for Case $2 = 80 \times 4 = 320$.
Total number of ways $= 60 + 320 = 380$.

(D) Given $\frac{s-x}{4} = \frac{s-y}{3} = \frac{s-z}{2} = k$.
Then $s-x = 4k, s-y = 3k, s-z = 2k$.
Summing these: $3s - (x+y+z) = 9k \Rightarrow 3s - 2s = 9k \Rightarrow s = 9k$.
Thus,$x = 5k, y = 6k, z = 7k$.
Area of incircle $\pi r^2 = \frac{8\pi}{3} \Rightarrow r^2 = \frac{8}{3} \Rightarrow r = \sqrt{\frac{8}{3}} = 2\sqrt{\frac{2}{3}}$.
Using $\Delta = rs = \sqrt{s(s-x)(s-y)(s-z)}$,we have $r^2 s^2 = s(4k)(3k)(2k) = s(24k^3)$.
Since $s=9k$,$r^2 s^2 = 9k(24k^3) = 216k^4$. Also $r^2 s^2 = \frac{8}{3} (81k^2) = 216k^2$.
So $216k^4 = 216k^2 \Rightarrow k^2 = 1 \Rightarrow k = 1$.
Thus $s = 9, x = 5, y = 6, z = 7$.
Area $\Delta = rs = \sqrt{\frac{8}{3}} \times 9 = 3\sqrt{8} \times 3 = 6\sqrt{6}$. (Option $A$ is correct).
Circumradius $R = \frac{xyz}{4\Delta} = \frac{5 \times 6 \times 7}{4 \times 6\sqrt{6}} = \frac{35}{4\sqrt{6}} = \frac{35\sqrt{6}}{24}$. (Option $B$ is correct).
$\sin \frac{X}{2} \sin \frac{Y}{2} \sin \frac{Z}{2} = \frac{r}{4R} = \frac{\sqrt{8/3}}{4 \times (35\sqrt{6}/24)} = \frac{2\sqrt{2}/\sqrt{3}}{35\sqrt{6}/6} = \frac{2\sqrt{2}}{\sqrt{3}} \times \frac{6}{35\sqrt{6}} = \frac{12\sqrt{2}}{35\sqrt{18}} = \frac{12\sqrt{2}}{35 \times 3\sqrt{2}} = \frac{4}{35}$. (Option $C$ is correct).
$\sin^2 \left(\frac{X+Y}{2}\right) = \cos^2 \frac{Z}{2} = \frac{1+\cos Z}{2}$. Using $\cos Z = \frac{x^2+y^2-z^2}{2xy} = \frac{25+36-49}{2 \times 5 \times 6} = \frac{12}{60} = \frac{1}{5}$.
So $\sin^2 \left(\frac{X+Y}{2}\right) = \frac{1+1/5}{2} = \frac{6/5}{2} = \frac{3}{5}$. (Option $D$ is correct).

(A, B, C) The circle is $x^2+y^2=3$ and the parabola is $x^2=2y$. Substituting $x^2=2y$ into the circle equation gives $2y+y^2=3$,so $y^2+2y-3=0$,which factors as $(y+3)(y-1)=0$. Since $P$ is in the first quadrant,$y=1$,and $x^2=2(1)=2$,so $x=\sqrt{2}$. Thus,$P \equiv (\sqrt{2}, 1)$.
The tangent to $x^2+y^2=3$ at $(\sqrt{2}, 1)$ is $\sqrt{2}x + y = 3$. Let this line be $L$. The slope of $L$ is $m = -\sqrt{2}$.
Let $\theta$ be the angle the line $L$ makes with the positive $x$-axis,so $\tan \theta = -\sqrt{2}$. The angle $\alpha$ between the line $L$ and the $y$-axis satisfies $\tan \alpha = |\cot \theta| = \frac{1}{|\tan \theta|} = \frac{1}{\sqrt{2}}$.
Let $T$ be the intersection of $L$ with the $y$-axis. Setting $x=0$ in $\sqrt{2}x+y=3$ gives $T \equiv (0, 3)$.
For circles $C_2, C_3$ with radius $r=2\sqrt{3}$ tangent to $L$ at $R_2, R_3$ with centers $Q_2, Q_3$ on the $y$-axis,the distance $Q_3T = \frac{r}{\sin \alpha}$. Since $\tan \alpha = \frac{1}{\sqrt{2}}$,$\sin \alpha = \frac{1}{\sqrt{3}}$. Thus $Q_3T = \frac{2\sqrt{3}}{1/\sqrt{3}} = 6$. Since $Q_2, Q_3$ are symmetric about $T$ on the $y$-axis,$Q_2Q_3 = 2 \times 6 = 12$. (Option $A$ is correct).
$R_3T = \frac{r}{\tan \alpha} = \frac{2\sqrt{3}}{1/\sqrt{2}} = 2\sqrt{6}$. Thus $R_2R_3 = 2 \times 2\sqrt{6} = 4\sqrt{6}$. (Option $B$ is correct).
The perpendicular distance from $O(0,0)$ to $L$ is $d = \frac{|3|}{\sqrt{(\sqrt{2})^2+1^2}} = \frac{3}{\sqrt{3}} = \sqrt{3}$. Area of $\triangle OR_2R_3 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (4\sqrt{6}) \times \sqrt{3} = 2\sqrt{18} = 6\sqrt{2}$. (Option $C$ is correct).
Area of $\triangle PQ_2Q_3 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (Q_2Q_3) \times |x_P| = \frac{1}{2} \times 12 \times \sqrt{2} = 6\sqrt{2}$. (Option $D$ is incorrect).

(C) Given,$RS$ is the diameter of $x^2+y^2=1$. Let $P = (\cos \theta, \sin \theta)$.
The tangent at $P$ is $x \cos \theta + y \sin \theta = 1$. The tangent at $S(1,0)$ is $x = 1$.
Solving these,$Q = \left(1, \frac{1-\cos \theta}{\sin \theta}\right) = \left(1, \tan \frac{\theta}{2}\right)$.
The line through $Q$ parallel to $RS$ (the $x$-axis) is $y = \tan \frac{\theta}{2}$.
The normal at $P$ is $y = x \tan \theta$.
Let $E = (h, k)$. Then $k = \tan \frac{\theta}{2}$ and $k = h \tan \theta$.
Using $\tan \theta = \frac{2 \tan(\theta/2)}{1 - \tan^2(\theta/2)}$,we get $k = h \frac{2k}{1-k^2}$.
Thus,$1-k^2 = 2h$,or the locus is $2x = 1-y^2$.
Checking the points:
For $x = 1/3$,$1-y^2 = 2/3$ $\Rightarrow y^2 = 1/3$ $\Rightarrow y = \pm 1/\sqrt{3}$.
Thus,points $(1/3, 1/\sqrt{3})$ and $(1/3, -1/\sqrt{3})$ lie on the locus.
For $x = 1/4$,$1-y^2 = 2/4 = 1/2$ $\Rightarrow y^2 = 1/2$ $\Rightarrow y = \pm 1/\sqrt{2} \neq \pm 1/2$.
Therefore,the locus passes through points $A$ and $C$.

(C) The coefficient of $x^2$ in the expansion is given by the sum of the coefficients of $x^2$ in each term:
$= {^2C_2} + {^3C_2} + {^4C_2} + \cdots + {^{49}C_2} + {^{50}C_2} \cdot m^2$
Using the identity ${^nC_r} + {^nC_{r-1}} = {^{n+1}C_r}$,we know that ${^2C_2} = {^3C_3}$.
Thus,the sum becomes ${^3C_3} + {^3C_2} + {^4C_2} + \cdots + {^{49}C_2} + {^{50}C_2} \cdot m^2 = {^4C_3} + {^4C_2} + \cdots + {^{49}C_2} + {^{50}C_2} \cdot m^2$
Continuing this process,the sum of the first $49$ terms is ${^{50}C_3}$.
So,the total coefficient is ${^{50}C_3} + {^{50}C_2} \cdot m^2$.
We are given that this equals $(3n+1) \cdot {^{51}C_3}$.
Note that ${^{51}C_3} = {^{50}C_3} + {^{50}C_2}$.
So,${^{50}C_3} + {^{50}C_2} \cdot m^2 = (3n+1)({^{50}C_3} + {^{50}C_2})$.
${^{50}C_3} + {^{50}C_2} \cdot m^2 = (3n+1){^{50}C_3} + (3n+1){^{50}C_2}$.
Since ${^{50}C_3} = \frac{50-2}{3} {^{50}C_2} = \frac{48}{3} {^{50}C_2} = 16 \cdot {^{50}C_2}$,we substitute this:
$16 \cdot {^{50}C_2} + {^{50}C_2} \cdot m^2 = (3n+1) \cdot 16 \cdot {^{50}C_2} + (3n+1) \cdot {^{50}C_2}$.
Dividing by ${^{50}C_2}$:
$16 + m^2 = 16(3n+1) + (3n+1) = 17(3n+1) = 51n + 17$.
$m^2 = 51n + 1$.
For $n=1$,$m^2 = 52$ (not a square).
For $n=2$,$m^2 = 103$ (not a square).
For $n=3$,$m^2 = 154$ (not a square).
For $n=4$,$m^2 = 205$ (not a square).
For $n=5$,$m^2 = 256 = 16^2$,so $m=16$.
The smallest positive integer $m$ is $16$,which gives $n=5$.

(B) Given $\lim _{x \rightarrow 0} \frac{x^2 \sin (\beta x)}{\alpha x-\sin x}=1$.
Using the Taylor series expansion for $\sin(x) = x - \frac{x^3}{3!} + \dots$ and $\sin(\beta x) = \beta x - \frac{(\beta x)^3}{3!} + \dots$,we get:
$\lim _{x \rightarrow 0} \frac{x^2(\beta x - \frac{\beta^3 x^3}{6} + \dots)}{\alpha x - (x - \frac{x^3}{6} + \dots)} = 1$
$\lim _{x \rightarrow 0} \frac{\beta x^3 - \frac{\beta^3 x^5}{6} + \dots}{(\alpha - 1)x + \frac{x^3}{6} - \dots} = 1$
For the limit to be a finite non-zero value,the coefficient of $x$ in the denominator must be zero,so $\alpha - 1 = 0 \Rightarrow \alpha = 1$.
Substituting $\alpha = 1$ into the limit:
$\lim _{x \rightarrow 0} \frac{\beta x^3 - \frac{\beta^3 x^5}{6} + \dots}{\frac{x^3}{6} - \dots} = 1$
Dividing numerator and denominator by $x^3$:
$\lim _{x \rightarrow 0} \frac{\beta - \frac{\beta^3 x^2}{6} + \dots}{\frac{1}{6} - \dots} = 1$
$\frac{\beta}{1/6} = 1$ $\Rightarrow 6\beta = 1$ $\Rightarrow \beta = \frac{1}{6}$.
Therefore,$6(\alpha + \beta) = 6(1 + \frac{1}{6}) = 6 + 1 = 7$.

(B) Given $\log _e b_1, \log _e b_2, \ldots, \log _e b_{101}$ are in $A.P.$,it follows that $b_1, b_2, \ldots, b_{101}$ are in $G.P.$ with common ratio $r = e^{\log _e 2} = 2$.
Let $a_1 = b_1 = a$. Since $a_{51} = b_{51}$,we have $a + 50d = a \cdot r^{50} = a \cdot 2^{50}$,so $d = \frac{a(2^{50} - 1)}{50}$.
The sum $t$ of the first $51$ terms of the $G.P.$ is $t = b_1 \frac{r^{51} - 1}{r - 1} = a(2^{51} - 1)$.
The sum $s$ of the first $51$ terms of the $A.P.$ is $s = \frac{51}{2}(2a + 50d) = \frac{51}{2}(2a + a(2^{50} - 1)) = \frac{51}{2}a(2^{50} + 1)$.
Comparing $s$ and $t$: $s = \frac{51}{2}a(2^{50} + 1) \approx 25.5 \cdot a \cdot 2^{50}$ and $t = a(2^{51} - 1) \approx 2 \cdot a \cdot 2^{50}$. Thus,$s > t$.
For the $101^{st}$ terms: $a_{101} = a + 100d = a + 2a(2^{50} - 1) = a(2^{51} - 1)$.
$b_{101} = b_1 \cdot r^{100} = a \cdot 2^{100}$.
Since $2^{100} > 2^{51} - 1$,we have $b_{101} > a_{101}$.

(C) Let $S = \sum_{k=1}^{13} \frac{1}{\sin \left(\frac{\pi}{4}+(k-1) \frac{\pi}{6}\right) \sin \left(\frac{\pi}{4}+\frac{k \pi}{6}\right)}$.
Multiply and divide by $\sin(\frac{\pi}{6})$:
$S = \frac{1}{\sin(\frac{\pi}{6})} \sum_{k=1}^{13} \frac{\sin(\frac{\pi}{6})}{\sin \left(\frac{\pi}{4}+(k-1) \frac{\pi}{6}\right) \sin \left(\frac{\pi}{4}+\frac{k \pi}{6}\right)}$.
Since $\frac{\pi}{6} = (\frac{\pi}{4}+\frac{k \pi}{6}) - (\frac{\pi}{4}+(k-1) \frac{\pi}{6})$,we use the identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$:
$S = 2 \sum_{k=1}^{13} \left[ \cot \left(\frac{\pi}{4}+(k-1) \frac{\pi}{6}\right) - \cot \left(\frac{\pi}{4}+\frac{k \pi}{6}\right) \right]$.
This is a telescoping sum:
$S = 2 \left[ \cot \frac{\pi}{4} - \cot \left(\frac{\pi}{4} + \frac{13\pi}{6}\right) \right]$.
Since $\cot \frac{\pi}{4} = 1$ and $\cot(\frac{\pi}{4} + \frac{13\pi}{6}) = \cot(\frac{\pi}{4} + \frac{\pi}{6}) = \cot(\frac{5\pi}{12}) = 2-\sqrt{3}$:
$S = 2 [1 - (2-\sqrt{3})] = 2(\sqrt{3}-1)$.

(D) Given $z = \frac{1}{a+ibt}$.
$x+iy = \frac{a-ibt}{a^2+b^2t^2}$.
Equating real and imaginary parts,$x = \frac{a}{a^2+b^2t^2}$ and $y = \frac{-bt}{a^2+b^2t^2}$.
If $a \neq 0$ and $b \neq 0$,then $a^2+b^2t^2 = \frac{a}{x}$,so $b^2t^2 = \frac{a}{x} - a^2 = \frac{a(1-ax)}{x}$.
Also $y^2 = \frac{b^2t^2}{(a^2+b^2t^2)^2} = \frac{a(1-ax)/x}{(a/x)^2} = \frac{x(1-ax)}{a} = \frac{x}{a} - x^2$.
Rearranging gives $x^2 - \frac{x}{a} + y^2 = 0$,which is $(x - \frac{1}{2a})^2 + y^2 = (\frac{1}{2a})^2$. This represents a circle with radius $|\frac{1}{2a}|$ and center $(\frac{1}{2a}, 0)$.
If $b=0$,then $z = \frac{1}{a}$,so $y=0$,which is the $x$-axis.
If $a=0$,then $z = \frac{1}{ibt} = -i(\frac{1}{bt})$,so $x=0$,which is the $y$-axis.
Thus,options $A, C, D$ are correct.

(B) The center of the circle $x^2+y^2-4x-16y+64=0$ is $S(2, 8)$ and its radius $r$ is $\sqrt{2^2+8^2-64} = 2$.
Let $P = (t^2, 2t)$ be a point on the parabola $y^2=4x$.
For $P$ to be the closest point to $S$,the line $SP$ must be the normal to the parabola at $P$.
The slope of the tangent at $P$ is $\frac{1}{t}$,so the slope of the normal is $-t$.
The slope of $SP$ is $\frac{2t-8}{t^2-2}$.
Setting $\frac{2t-8}{t^2-2} = -t$,we get $2t-8 = -t^3+2t$,which implies $t^3=8$,so $t=2$.
Thus,$P = (4, 4)$.
$(A)$ $SP = \sqrt{(4-2)^2+(4-8)^2} = \sqrt{2^2+(-4)^2} = \sqrt{20} = 2\sqrt{5}$. This is correct.
$(C)$ The slope of the normal at $P(4, 4)$ is $-t = -2$. The equation of the normal is $y-4 = -2(x-4)$,or $y = -2x+12$. The $x$-intercept is found by setting $y=0$,giving $2x=12$,so $x=6$. This is correct.
$(D)$ Since $SP$ is the normal to the parabola at $P$,and $S$ is the center of the circle,the line $SP$ passes through the center of the circle. Thus,$SP$ is a normal to the circle at $Q$. The slope of $SP$ is $-2$. The tangent at $Q$ is perpendicular to the normal $SQ$,so its slope is $-\frac{1}{-2} = \frac{1}{2}$. This is correct.
$(B)$ $SQ$ is the radius of the circle,so $SQ=2$. $SP=2\sqrt{5}$. $QP = SP - SQ = 2\sqrt{5}-2$. Thus $SQ:QP = 2 : (2\sqrt{5}-2) = 1 : (\sqrt{5}-1) = (\sqrt{5}+1) : 4$. This is incorrect.
Therefore,the correct options are $A, C, D$.

(C) $1.$ The correct option is $A \left(-\frac{9}{10}, 0\right)$.
Equation of the ellipse: $\frac{x^2}{9}+\frac{y^2}{8}=1$. Here $a^2=9, b^2=8$. Eccentricity $e = \sqrt{1-\frac{8}{9}} = \frac{1}{3}$. Foci are $(\pm ae, 0) = (\pm 1, 0)$.
Equation of the parabola with vertex $(0,0)$ and focus $F_2(1,0)$ is $y^2 = 4x$.
Substituting $y^2=4x$ into the ellipse equation: $\frac{x^2}{9} + \frac{4x}{8} = 1 \Rightarrow 2x^2 + 9x - 18 = 0 \Rightarrow (2x-3)(x+6)=0$. Since $x>0$,$x=\frac{3}{2}$. Then $y^2 = 4(\frac{3}{2}) = 6$,so $y = \pm \sqrt{6}$. Thus $M = (\frac{3}{2}, \sqrt{6})$ and $N = (\frac{3}{2}, -\sqrt{6})$.
In $\triangle F_1 M N$,the altitude from $M$ to $F_1 N$ has slope $m_1 = \frac{\sqrt{6} - 0}{3/2 - (-1)} = \frac{\sqrt{6}}{5/2} = \frac{2\sqrt{6}}{5}$. The slope of $F_1 N$ is $\frac{-\sqrt{6}-0}{3/2 - (-1)} = \frac{-\sqrt{6}}{5/2} = -\frac{2\sqrt{6}}{5}$. The slope of the altitude from $M$ is $m_{alt} = -\frac{1}{-2\sqrt{6}/5} = \frac{5}{2\sqrt{6}}$.
Equation of altitude from $M$: $y - \sqrt{6} = \frac{5}{2\sqrt{6}}(x - \frac{3}{2})$. Since the orthocentre lies on the $x$-axis (as $F_1 M N$ is isosceles with $F_1 M = F_1 N$),set $y=0$: $-\sqrt{6} = \frac{5}{2\sqrt{6}}(x - \frac{3}{2}) \Rightarrow -12 = 5x - \frac{15}{2} \Rightarrow 5x = -12 + 7.5 = -4.5 \Rightarrow x = -\frac{9}{10}$.
$2.$ The tangent to the ellipse at $M(\frac{3}{2}, \sqrt{6})$ is $\frac{x(3/2)}{9} + \frac{y\sqrt{6}}{8} = 1 \Rightarrow \frac{x}{6} + \frac{y\sqrt{6}}{8} = 1$. For $y=0$,$x=6$,so $R(6,0)$.
The normal to the parabola $y^2=4x$ at $M(\frac{3}{2}, \sqrt{6})$: slope of tangent is $\frac{dy}{dx} = \frac{2}{y} = \frac{2}{\sqrt{6}}$. Slope of normal is $-\frac{\sqrt{6}}{2}$. Equation: $y - \sqrt{6} = -\frac{\sqrt{6}}{2}(x - \frac{3}{2})$. For $y=0$,$\sqrt{6} = \frac{\sqrt{6}}{2}(x - \frac{3}{2}) \Rightarrow 2 = x - \frac{3}{2} \Rightarrow x = \frac{7}{2}$. So $Q(\frac{7}{2}, 0)$.
Area $\triangle MQR = \frac{1}{2} \times |x_R - x_Q| \times y_M = \frac{1}{2} \times (6 - 3.5) \times \sqrt{6} = \frac{1}{2} \times 2.5 \times \sqrt{6} = \frac{5\sqrt{6}}{4}$.
Area quadrilateral $MF_1NF_2 = \frac{1}{2} \times |F_1 F_2| \times |y_M - y_N| = \frac{1}{2} \times 2 \times 2\sqrt{6} = 2\sqrt{6}$.
Ratio $= \frac{5\sqrt{6}/4}{2\sqrt{6}} = \frac{5}{8}$.

(C) Given equation: $\sqrt{3} \sec x+\operatorname{cosec} x+2(\tan x-\cot x)=0$
Dividing by $2$: $\frac{\sqrt{3}}{2} \sec x+\frac{1}{2} \operatorname{cosec} x=\cot x-\tan x$
Converting to $\sin x$ and $\cos x$: $\frac{\sqrt{3}}{2 \cos x}+\frac{1}{2 \sin x}=\frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}$
Multiplying by $\sin x \cos x$: $\frac{\sqrt{3}}{2} \sin x+\frac{1}{2} \cos x=\cos^2 x-\sin^2 x$
Using $\cos(A-B)$ and $\cos 2x$ formulas: $\cos(x-\frac{\pi}{6})=\cos 2x$
General solution: $2x = 2n\pi \pm (x-\frac{\pi}{6})$
Case $1$: $2x = 2n\pi + x - \frac{\pi}{6} \implies x = 2n\pi - \frac{\pi}{6}$
For $n=0, x=-\frac{\pi}{6} \in S$. For $n=1, x=\frac{11\pi}{6} \notin S$.
Case $2$: $2x = 2n\pi - (x - \frac{\pi}{6}) \implies 3x = 2n\pi + \frac{\pi}{6} \implies x = \frac{2n\pi}{3} + \frac{\pi}{18}$
For $n=0, x=\frac{\pi}{18} \in S$. For $n=1, x=\frac{13\pi}{18} \in S$. For $n=-1, x=-\frac{11\pi}{18} \in S$.
Sum of solutions: $-\frac{\pi}{6} + \frac{\pi}{18} + \frac{13\pi}{18} - \frac{11\pi}{18} = \frac{-3\pi + \pi + 13\pi - 11\pi}{18} = 0$.

(C) Let $T_1$ and $T_2$ be the events that the computer is produced in plant $T_1$ and $T_2$ respectively. Let $D$ be the event that the computer is defective.
Given $P(T_1) = 0.2$,$P(T_2) = 0.8$,and $P(D) = 0.07$.
We are given $P(D | T_1) = 10 P(D | T_2)$. Let $P(D | T_2) = p$,then $P(D | T_1) = 10p$.
Using the law of total probability: $P(D) = P(D | T_1)P(T_1) + P(D | T_2)P(T_2)$.
$0.07 = (10p)(0.2) + (p)(0.8) = 2p + 0.8p = 2.8p$.
$p = \frac{0.07}{2.8} = \frac{7}{280} = \frac{1}{40}$.
So,$P(D | T_2) = \frac{1}{40}$ and $P(D | T_1) = 10 \times \frac{1}{40} = \frac{1}{4}$.
The probability that a computer is not defective is $P(\bar{D}) = 1 - P(D) = 1 - 0.07 = 0.93$.
We need to find $P(T_2 | \bar{D}) = \frac{P(\bar{D} | T_2)P(T_2)}{P(\bar{D})}$.
$P(\bar{D} | T_2) = 1 - P(D | T_2) = 1 - \frac{1}{40} = \frac{39}{40}$.
$P(T_2 | \bar{D}) = \frac{(\frac{39}{40}) \times 0.8}{0.93} = \frac{0.78}{0.93} = \frac{78}{93}$.

(C) Let $f(x) = 4 \alpha x^2 + \frac{1}{x}$ for $x > 0$.
To find the minimum value of $f(x)$,we differentiate with respect to $x$:
$f'(x) = 8 \alpha x - \frac{1}{x^2}$.
Setting $f'(x) = 0$ for critical points:
$8 \alpha x = \frac{1}{x^2}$ $\Rightarrow x^3 = \frac{1}{8 \alpha}$ $\Rightarrow x = \frac{1}{2 \alpha^{1/3}}$.
Substituting this value into $f(x)$ to find the minimum:
$f\left(\frac{1}{2 \alpha^{1/3}}\right) = 4 \alpha \left(\frac{1}{4 \alpha^{2/3}}\right) + \frac{1}{1/(2 \alpha^{1/3})} = \alpha^{1/3} + 2 \alpha^{1/3} = 3 \alpha^{1/3}$.
Given $f(x) \geq 1$ for all $x > 0$,the minimum value must be at least $1$:
$3 \alpha^{1/3} \geq 1$ $\Rightarrow \alpha^{1/3} \geq \frac{1}{3}$ $\Rightarrow \alpha \geq \frac{1}{27}$.
Thus,the least value of $\alpha$ is $\frac{1}{27}$.

(B) The base $OPQR$ is a square in the $xy$-plane with $O(0,0,0)$,$P(3,0,0)$,$Q(3,3,0)$,and $R(0,3,0)$.
The mid-point $T$ of $OQ$ is $(\frac{3+0}{2}, \frac{3+0}{2}, 0) = (1.5, 1.5, 0)$.
Since $S$ is directly above $T$ with $TS=3$,the coordinates of $S$ are $(1.5, 1.5, 3)$.
$(A)$ Vector $\vec{OQ} = 3\hat{i} + 3\hat{j}$ and $\vec{OS} = 1.5\hat{i} + 1.5\hat{j} + 3\hat{k}$.
$|\vec{OQ}| = \sqrt{3^2 + 3^2} = 3\sqrt{2}$. $|\vec{OS}| = \sqrt{1.5^2 + 1.5^2 + 3^2} = \sqrt{2.25 + 2.25 + 9} = \sqrt{13.5} = \sqrt{\frac{27}{2}} = \frac{3\sqrt{3}}{\sqrt{2}}$.
$\vec{OQ} \cdot \vec{OS} = (3)(1.5) + (3)(1.5) + (0)(3) = 4.5 + 4.5 = 9$.
$\cos \theta = \frac{9}{(3\sqrt{2})(\frac{3\sqrt{3}}{\sqrt{2}})} = \frac{9}{9\sqrt{3}} = \frac{1}{\sqrt{3}}$. Thus,$\theta \neq \frac{\pi}{3}$. ($A$ is incorrect).
$(B)$ The plane containing $\triangle OQS$ passes through $O(0,0,0)$,$Q(3,3,0)$,and $S(1.5, 1.5, 3)$.
Normal vector $\vec{n} = \vec{OQ} \times \vec{OS} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 3 & 0 \\ 1.5 & 1.5 & 3 \end{vmatrix} = \hat{i}(9) - \hat{j}(9) + \hat{k}(0) = 9(\hat{i} - \hat{j})$.
The equation is $1(x-0) - 1(y-0) + 0(z-0) = 0$,i.e.,$x-y=0$. ($B$ is correct).
$(C)$ The perpendicular distance from $P(3,0,0)$ to $x-y=0$ is $d = \frac{|3-0|}{\sqrt{1^2 + (-1)^2}} = \frac{3}{\sqrt{2}}$. ($C$ is correct).
$(D)$ Line $RS$ passes through $R(0,3,0)$ and $S(1.5, 1.5, 3)$. Direction vector $\vec{v} = S-R = (1.5, -1.5, 3) = 1.5(\hat{i} - \hat{j} + 2\hat{k})$.
Line equation: $\vec{r} = (0,3,0) + \lambda(1, -1, 2)$.
Distance from $O(0,0,0)$ to line is $\frac{|\vec{OR} \times \vec{v}|}{|\vec{v}|} = \frac{|(0,3,0) \times (1,-1,2)|}{|(1,-1,2)|} = \frac{|(6, 0, -3)|}{\sqrt{1+1+4}} = \frac{\sqrt{36+9}}{\sqrt{6}} = \sqrt{\frac{45}{6}} = \sqrt{\frac{15}{2}}$. ($D$ is correct).

(A) Given the linear differential equation $f^{\prime}(x) + \frac{f(x)}{x} = 2$.
The integrating factor is $e^{\int \frac{1}{x} dx} = e^{\ln x} = x$.
Multiplying by the integrating factor,we get $\frac{d}{dx}(x f(x)) = 2x$.
Integrating both sides,$x f(x) = x^2 + c$,which gives $f(x) = x + \frac{c}{x}$ for all $x \in (0, \infty)$.
Given $f(1) \neq 1$,we have $1 + c \neq 1$,so $c \neq 0$.
Now,$f^{\prime}(x) = 1 - \frac{c}{x^2}$.
Evaluating option $A$: $\lim _{x \rightarrow 0+} f^{\prime}\left(\frac{1}{x}\right) = \lim _{x \rightarrow 0+} (1 - c x^2) = 1$.
Evaluating option $B$: $\lim _{x \rightarrow 0+} x f\left(\frac{1}{x}\right) = \lim _{x \rightarrow 0+} x \left(\frac{1}{x} + cx\right) = \lim _{x \rightarrow 0+} (1 + cx^2) = 1 \neq 2$.
Evaluating option $C$: $\lim _{x \rightarrow 0+} x^2 f^{\prime}(x) = \lim _{x \rightarrow 0+} x^2 (1 - \frac{c}{x^2}) = \lim _{x \rightarrow 0+} (x^2 - c) = -c \neq 0$.
Thus,option $A$ is correct.

(A) Given $PQ = kI$,we have $Q = kP^{-1}$.
Since $P^{-1} = \frac{1}{\det(P)} \text{adj}(P)$,we have $Q = \frac{k}{\det(P)} \text{adj}(P)$.
First,calculate $\det(P) = 3(0 - (-5\alpha)) - (-1)(0 - 3\alpha) + (-2)(-10 - 0) = 3(5\alpha) + (-3\alpha) + 20 = 15\alpha - 3\alpha + 20 = 12\alpha + 20$.
The element $q_{23}$ is the $(2,3)$-th entry of $Q$.
$q_{23} = \frac{k}{\det(P)} \times (\text{adj}(P))_{23} = \frac{k}{12\alpha + 20} \times (-1)^{2+3} M_{32} = \frac{k}{12\alpha + 20} \times (-1) \times (3\alpha - (-4)) = \frac{-k(3\alpha + 4)}{12\alpha + 20} = \frac{-k(3\alpha + 4)}{4(3\alpha + 5)}$.
Given $q_{23} = -\frac{k}{8}$,we have $\frac{3\alpha + 4}{4(3\alpha + 5)} = \frac{1}{8} \implies 2(3\alpha + 4) = 3\alpha + 5 \implies 6\alpha + 8 = 3\alpha + 5 \implies 3\alpha = -3 \implies \alpha = -1$.
Then $\det(P) = 12(-1) + 20 = 8$.
Since $Q = kP^{-1}$,$\det(Q) = k^3 \det(P^{-1}) = \frac{k^3}{\det(P)} = \frac{k^3}{8}$.
Given $\det(Q) = \frac{k^2}{2}$,we have $\frac{k^3}{8} = \frac{k^2}{2} \implies k = 4$.
Check options:
$(B)$ $4\alpha - k + 8 = 4(-1) - 4 + 8 = -4 - 4 + 8 = 0$. (Correct)
$(C)$ $\det(P \text{adj}(Q)) = \det(P) \det(\text{adj}(Q)) = \det(P) (\det(Q))^2 = 8 \times (\frac{4^2}{2})^2 = 8 \times 8^2 = 8^3 = 512 = 2^9$. (Correct)
$(D)$ $\det(Q \text{adj}(P)) = \det(Q) \det(\text{adj}(P)) = \det(Q) (\det(P))^2 = \frac{16}{2} \times 8^2 = 8 \times 64 = 512 = 2^9 \neq 2^{13}$. (Incorrect)

(D) The given differential equation is $(x+2)(x+2+y) \frac{dy}{dx}-y^2=0$.
Substitute $y=(x+2)t$,then $\frac{dy}{dx}=(x+2) \frac{dt}{dx}+t$.
Substituting this into the equation: $(x+2)(x+2+(x+2)t) \frac{dy}{dx}-((x+2)t)^2=0$.
$(x+2)^2(1+t) \frac{dy}{dx}-(x+2)^2t^2=0$.
$(1+t)((x+2) \frac{dt}{dx}+t)-t^2=0$.
$(1+t)(x+2) \frac{dt}{dx}+t+t^2-t^2=0$.
$(1+t)(x+2) \frac{dt}{dx}+t=0$.
Separating variables: $\frac{1+t}{t} dt = -\frac{dx}{x+2}$.
Integrating both sides: $\int (\frac{1}{t}+1) dt = -\int \frac{dx}{x+2}$.
$\ln|t|+t = -\ln|x+2|+C$.
Substituting $t=\frac{y}{x+2}$: $\ln(\frac{y}{x+2})+\frac{y}{x+2} = -\ln(x+2)+C$.
$\ln y - \ln(x+2) + \frac{y}{x+2} = -\ln(x+2)+C$.
$\ln y + \frac{y}{x+2} = C$.
Since the curve passes through $(1,3)$,$\ln 3 + \frac{3}{1+2} = C \Rightarrow \ln 3 + 1 = C$.
The solution curve is $\ln y + \frac{y}{x+2} = \ln 3 + 1$.
For option $(A)$,intersect $y=x+2$: $\ln(x+2) + \frac{x+2}{x+2} = \ln 3 + 1 \Rightarrow \ln(x+2) + 1 = \ln 3 + 1 \Rightarrow x+2=3 \Rightarrow x=1$. Only one point.
For option $(D)$,$y=(x+3)^2$: $\ln(x+3)^2 + \frac{(x+3)^2}{x+2} = \ln 3 + 1$. Let $f(x) = 2\ln(x+3) + \frac{(x+3)^2}{x+2} - \ln 3 - 1$. Since $f(x)$ is strictly increasing for $x>0$ and $f(0) > 0$,it does not intersect.

(D) Given $f(x)=x^3+3x+2$. Since $g(f(x))=x$,$g$ is the inverse of $f$.
Differentiating $g(f(x))=x$ with respect to $x$,we get $g'(f(x)) \cdot f'(x)=1$.
For $g'(2)$,we set $f(x)=2 \implies x^3+3x+2=2 \implies x^3+3x=0 \implies x(x^2+3)=0$,so $x=0$.
Thus $g'(2) \cdot f'(0)=1$. Since $f'(x)=3x^2+3$,$f'(0)=3$.
So $g'(2) = \frac{1}{3}$. Option $A$ is incorrect.
Given $h(g(g(x)))=x$. Since $g(f(x))=x$,$g(g(f(f(x))))=x$,which implies $g(g(x))=f(f(x))$.
Thus $h(f(f(x)))=x$. This means $h$ is the inverse of $f(f(x))$.
However,the problem states $h(g(g(x)))=x$. Since $g(g(x))=f^{-1}(f^{-1}(x))$,we have $h(f^{-1}(f^{-1}(x)))=x$.
Let $f^{-1}(x)=y$,then $f(y)=x$. The equation becomes $h(f^{-1}(y))=f(y)$.
Let $f^{-1}(y)=z$,then $y=f(z)$. So $h(z)=f(f(z))$.
Thus $h(x)=f(f(x))$.
Now,$h'(x)=f'(f(x)) \cdot f'(x)$.
$h'(1)=f'(f(1)) \cdot f'(1)$. Since $f(1)=1^3+3(1)+2=6$ and $f'(x)=3x^2+3$,$f'(1)=6$ and $f'(6)=3(6^2)+3=111$.
$h'(1)=111 \times 6=666$. Option $B$ is correct.
$h(0)=f(f(0))=f(2)=2^3+3(2)+2=16$. Option $C$ is correct.
$h(g(3))=f(f(g(3)))=f(3)=3^3+3(3)+2=38$. Option $D$ is incorrect.
Therefore,the correct options are $B$ and $C$.

(A) Given the determinant equation:
$\left|\begin{array}{ccc} x & x^2 & 1+x^3 \\ 2x & 4x^2 & 1+8x^3 \\ 3x & 9x^2 & 1+27x^3 \end{array}\right| = 10$
We can split the determinant using the property of determinants:
$\left|\begin{array}{ccc} x & x^2 & 1 \\ 2x & 4x^2 & 1 \\ 3x & 9x^2 & 1 \end{array}\right| + \left|\begin{array}{ccc} x & x^2 & x^3 \\ 2x & 4x^2 & 8x^3 \\ 3x & 9x^2 & 27x^3 \end{array}\right| = 10$
Taking $x$ common from $C_1$,$x^2$ from $C_2$,and $x^3$ from $C_3$ in the second determinant:
$x^3 \left|\begin{array}{ccc} 1 & 1 & 1 \\ 2 & 4 & 8 \\ 3 & 9 & 27 \end{array}\right| + x^6 \left|\begin{array}{ccc} 1 & 1 & 1 \\ 2 & 4 & 8 \\ 3 & 9 & 27 \end{array}\right| = 10$
Calculating the determinant value $\Delta = \left|\begin{array}{ccc} 1 & 1 & 1 \\ 2 & 4 & 8 \\ 3 & 9 & 27 \end{array}\right| = 1(108-72) - 1(54-24) + 1(18-12) = 36 - 30 + 6 = 12$.
So,$12x^3 + 12x^6 = 10 \Rightarrow 6x^6 + 6x^3 - 5 = 0$.
Let $t = x^3$. Then $6t^2 + 6t - 5 = 0$.
Using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-6 \pm \sqrt{36 - 4(6)(-5)}}{12} = \frac{-6 \pm \sqrt{36 + 120}}{12} = \frac{-6 \pm \sqrt{156}}{12} = \frac{-6 \pm 2\sqrt{39}}{12} = \frac{-3 \pm \sqrt{39}}{6}$.
Since $x^3 = t$,for any real value of $t$,there exists exactly one real value of $x = \sqrt[3]{t}$.
Since there are two distinct real values for $t$,there are $2$ distinct real values for $x$.

(A) Let $f(x) = \int_0^x \frac{t^2}{1+t^4} dt - 2x + 1$ for $x \in [0, 1]$.
Taking the derivative,$f'(x) = \frac{x^2}{1+x^4} - 2$.
Since $x^4 + 1 \geq 2x^2$ by $AM$-$GM$ inequality,we have $\frac{x^2}{1+x^4} \leq \frac{1}{2}$.
Thus,$f'(x) = \frac{x^2}{1+x^4} - 2 \leq \frac{1}{2} - 2 = -\frac{3}{2} < 0$.
Since $f'(x) < 0$ for all $x \in [0, 1]$,$f(x)$ is a strictly decreasing function.
We evaluate the endpoints:
$f(0) = \int_0^0 \frac{t^2}{1+t^4} dt - 2(0) + 1 = 1$.
$f(1) = \int_0^1 \frac{t^2}{1+t^4} dt - 2(1) + 1 = \int_0^1 \frac{t^2}{1+t^4} dt - 1$.
Since $\frac{t^2}{1+t^4} < 1$ for $t \in [0, 1]$,the integral $\int_0^1 \frac{t^2}{1+t^4} dt < 1$,so $f(1) < 0$.
Since $f(0) > 0$ and $f(1) < 0$ and $f(x)$ is continuous and strictly decreasing,by the Intermediate Value Theorem,there exists exactly one root $x \in [0, 1]$ such that $f(x) = 0$.

(C) Given $z = \frac{-1}{2} + i \frac{\sqrt{3}}{2} = \omega$, where $\omega$ is the cube root of unity. Thus, $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.
$P = \begin{bmatrix} (-1)^r \omega^r & \omega^{2s} \\ \omega^{2s} & \omega^r \end{bmatrix}$.
$P^2 = \begin{bmatrix} (-1)^r \omega^r & \omega^{2s} \\ \omega^{2s} & \omega^r \end{bmatrix} \begin{bmatrix} (-1)^r \omega^r & \omega^{2s} \\ \omega^{2s} & \omega^r \end{bmatrix} = \begin{bmatrix} (-1)^{2r} \omega^{2r} + \omega^{4s} & (-1)^r \omega^{r+2s} + \omega^{r+2s} \\ (-1)^r \omega^{r+2s} + \omega^{r+2s} & \omega^{4s} + \omega^{2r} \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}$.
For the off-diagonal elements to be zero: $((-1)^r + 1) \omega^{r+2s} = 0$. Since $\omega \neq 0$, we must have $(-1)^r + 1 = 0$, which implies $r$ must be odd. Thus, $r \in \{1, 3\}$.
Case $1$: $r = 1$. The diagonal elements give $\omega^2 + \omega^{4s} = -1$. Since $1 + \omega + \omega^2 = 0$, we have $\omega^2 + \omega = -1$. Thus, $\omega^{4s} = \omega$, which means $4s \equiv 1 \pmod 3$, so $s \equiv 1 \pmod 3$. For $s \in \{1, 2, 3\}$, $s = 1$.
Case $2$: $r = 3$. The diagonal elements give $\omega^6 + \omega^{4s} = -1$. Since $\omega^3 = 1$, $1 + \omega^{4s} = -1$, so $\omega^{4s} = -2$. This is impossible as $|\omega^{4s}| = 1$.
Thus, the only valid pair is $(r, s) = (1, 1)$. The total number of pairs is $1$.

(B) Given $P = \begin{bmatrix} 1 & 0 & 0 \\ 4 & 1 & 0 \\ 16 & 4 & 1 \end{bmatrix}$. Let $P = I + A$,where $A = \begin{bmatrix} 0 & 0 & 0 \\ 4 & 0 & 0 \\ 16 & 4 & 0 \end{bmatrix}$.
Note that $A^2 = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 16 & 0 & 0 \end{bmatrix}$ and $A^3 = O$ (zero matrix).
Using the Binomial Theorem,$P^n = (I+A)^n = I + nA + \frac{n(n-1)}{2}A^2$.
For $n=50$,$P^{50} = I + 50A + \frac{50 \times 49}{2}A^2 = I + 50A + 1225A^2$.
$P^{50} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} + 50 \begin{bmatrix} 0 & 0 & 0 \\ 4 & 0 & 0 \\ 16 & 4 & 0 \end{bmatrix} + 1225 \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 16 & 0 & 0 \end{bmatrix}$.
$P^{50} = \begin{bmatrix} 1 & 0 & 0 \\ 200 & 1 & 0 \\ 800 + 19600 & 200 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 200 & 1 & 0 \\ 20400 & 200 & 1 \end{bmatrix}$.
Since $P^{50}-Q=I$,we have $Q = P^{50}-I = \begin{bmatrix} 0 & 0 & 0 \\ 200 & 0 & 0 \\ 20400 & 200 & 0 \end{bmatrix}$.
Thus,$q_{31} = 20400$,$q_{32} = 200$,and $q_{21} = 200$.
Therefore,$\frac{q_{31}+q_{32}}{q_{21}} = \frac{20400+200}{200} = \frac{20600}{200} = 103$.

(A) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x^2 \cos x}{1+e^x} dx$ ... $(i)$
Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,we have $a+b = -\frac{\pi}{2} + \frac{\pi}{2} = 0$,so $f(x) \to f(-x)$.
$I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{(-x)^2 \cos(-x)}{1+e^{-x}} dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x^2 \cos x}{1+e^{-x}} dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x^2 \cos x \cdot e^x}{e^x+1} dx$ ... (ii)
Adding $(i)$ and (ii):
$2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x^2 \cos x (1+e^x)}{1+e^x} dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^2 \cos x dx$
Since $x^2 \cos x$ is an even function,$2I = 2 \int_{0}^{\frac{\pi}{2}} x^2 \cos x dx$.
$I = \int_{0}^{\frac{\pi}{2}} x^2 \cos x dx$.
Using integration by parts $\int u dv = uv - \int v du$ with $u=x^2, dv=\cos x dx$:
$I = [x^2 \sin x]_{0}^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}} 2x \sin x dx$
$I = (\frac{\pi^2}{4} \cdot 1 - 0) - 2 \int_{0}^{\frac{\pi}{2}} x \sin x dx$
$I = \frac{\pi^2}{4} - 2 [x(-\cos x) - \int 1(-\cos x) dx]_{0}^{\frac{\pi}{2}}$
$I = \frac{\pi^2}{4} - 2 [-x \cos x + \sin x]_{0}^{\frac{\pi}{2}}$
$I = \frac{\pi^2}{4} - 2 [(-0 + 1) - (0 + 0)] = \frac{\pi^2}{4} - 2$.

(C) The given region is bounded by $y \geq \sqrt{|x+3|}$ and $5y \leq x+9 \leq 15$.
From $5y \leq x+9 \leq 15$,we get $y \leq \frac{x+9}{5}$ and $-6 \leq x \leq 6$.
The intersection points of $y = \sqrt{x+3}$ and $y = \frac{x+9}{5}$ are found by solving $25(x+3) = (x+9)^2$,which gives $x^2 - 7x - 6 = 0$ (not integer roots) or checking the graph provided.
Based on the graph,the region is bounded by the line $y = \frac{x+9}{5}$ above and the curves $y = \sqrt{-x-3}$ (for $x \in [-4, -3]$) and $y = \sqrt{x+3}$ (for $x \in [-3, 1]$) below.
The area is given by $\int_{-4}^{1} \frac{x+9}{5} dx - \int_{-4}^{-3} \sqrt{-x-3} dx - \int_{-3}^{1} \sqrt{x+3} dx$.
Calculating the integral of the line: $\int_{-4}^{1} \frac{x+9}{5} dx = \frac{1}{5} [\frac{x^2}{2} + 9x]_{-4}^{1} = \frac{1}{5} [(\frac{1}{2} + 9) - (8 - 36)] = \frac{1}{5} [9.5 + 28] = \frac{37.5}{5} = 7.5 = \frac{15}{2}$.
Calculating the area under the curves: $\int_{-4}^{-3} \sqrt{-(x+3)} dx = [-\frac{2}{3}(-(x+3))^{3/2}]_{-4}^{-3} = 0 - (-\frac{2}{3}(1)) = \frac{2}{3}$.
$\int_{-3}^{1} \sqrt{x+3} dx = [\frac{2}{3}(x+3)^{3/2}]_{-3}^{1} = \frac{2}{3}(4^{3/2} - 0) = \frac{2}{3}(8) = \frac{16}{3}$.
Required Area $= \frac{15}{2} - \frac{2}{3} - \frac{16}{3} = \frac{15}{2} - \frac{18}{3} = \frac{15}{2} - 6 = \frac{3}{2}$.

(C) Let the given point be $Q(3, 1, 7)$. The line passing through $Q$ and perpendicular to the plane $x-y+z=3$ has the direction ratios $(1, -1, 1)$.
Its equation is $\frac{x-3}{1} = \frac{y-1}{-1} = \frac{z-7}{1} = \lambda$.
Any point on this line is $(3+\lambda, 1-\lambda, 7+\lambda)$.
This point lies on the plane $x-y+z=3$,so $(3+\lambda) - (1-\lambda) + (7+\lambda) = 3$.
$3+\lambda-1+\lambda+7+\lambda=3 \Rightarrow 3\lambda+9=3 \Rightarrow 3\lambda=-6 \Rightarrow \lambda=-2$.
The foot of the perpendicular $R$ is $(3-2, 1-(-2), 7-2) = (1, 3, 5)$.
Let $P(x_1, y_1, z_1)$ be the image of $Q$. Since $R$ is the midpoint of $PQ$,we have $\frac{x_1+3}{2}=1, \frac{y_1+1}{2}=3, \frac{z_1+7}{2}=5$.
$x_1 = -1, y_1 = 5, z_1 = 3$. So $P$ is $(-1, 5, 3)$.
The plane passes through $P(-1, 5, 3)$ and contains the line $\frac{x}{1}=\frac{y}{2}=\frac{z}{1}$.
The plane equation is $a(x+1) + b(y-5) + c(z-3) = 0$.
Since it contains the line,it passes through $(0, 0, 0)$,so $a(1) + b(-5) + c(-3) = 0 \Rightarrow a-5b-3c=0$.
Also,the normal vector $(a, b, c)$ is perpendicular to the line direction $(1, 2, 1)$,so $a+2b+c=0$.
Solving $a-5b-3c=0$ and $a+2b+c=0$ by subtraction: $-7b-4c=0 \Rightarrow b = -4c/7$.
Then $a = 5(-4c/7) + 3c = -20c/7 + 21c/7 = c/7$.
Taking $c=7$,we get $a=1, b=-4$. The equation is $1(x+1) - 4(y-5) + 7(z-3) = 0$.
$x+1-4y+20+7z-21 = 0 \Rightarrow x-4y+7z=0$.

(B,C) Taking the natural logarithm on both sides:
$\ln f(x) = \lim_{n}$ ${\rightarrow \infty} \frac{x}{n} \sum_{r=1}^n \ln \left( \frac{x + n/r}{x^2 + n^2/r^2} \cdot \frac{n}{r} \right)$
$= \lim_{n \rightarrow \infty} \frac{x}{n} \sum_{r=1}^n \ln \left( \frac{1 + \frac{r}{n}x}{1 + (\frac{r}{n}x)^2} \right)$
$= x \int_0^1 \ln \left( \frac{1 + tx}{1 + (tx)^2} \right) dt$
Let $u = tx$,then $du = x dt$:
$\ln f(x) = \int_0^x \ln \left( \frac{1+u}{1+u^2} \right) du$
By the Fundamental Theorem of Calculus:
$\frac{f^{\prime}(x)}{f(x)} = \ln \left( \frac{1+x}{1+x^2} \right)$
For $0 < x < 1$,$\frac{1+x}{1+x^2} > 1$,so $\frac{f^{\prime}(x)}{f(x)} > 0$,meaning $f(x)$ is increasing.
For $x > 1$,$\frac{1+x}{1+x^2} < 1$,so $\frac{f^{\prime}(x)}{f(x)} < 0$,meaning $f(x)$ is decreasing.
Thus,$f(1/3) < f(2/3)$ (Option $B$ is correct).
Since $f(x)$ is decreasing for $x > 1$,$f^{\prime}(2) < 0$ (Option $C$ is correct).
Therefore,the correct options are $B$ and $C$.

(C) Given $f(x) = a \cos (|x^3-x|) + b|x| \sin (|x^3+x|)$.
Since $\cos(\theta) = \cos(-\theta)$,we have $\cos(|x^3-x|) = \cos(x^3-x)$.
Also,$|x| \sin(|x^3+x|) = x \sin(x^3+x)$ for all $x \in R$ because if $x \ge 0$,$|x|=x$ and $|x^3+x|=x^3+x$,and if $x < 0$,$|x|=-x$ and $|x^3+x|=-(x^3+x)$,so $-x \sin(-(x^3+x)) = -x(-\sin(x^3+x)) = x \sin(x^3+x)$.
Thus,$f(x) = a \cos(x^3-x) + b x \sin(x^3+x)$ for all $x \in R$.
Since $\cos(x^3-x)$ and $x \sin(x^3+x)$ are differentiable functions on $R$,$f(x)$ is differentiable for all $x \in R$ for any $a, b \in R$.
Therefore,$f$ is differentiable at $x=0$ and $x=1$ for any $a, b \in R$.
Checking the options:
$(A)$ Differentiable at $x=0$ if $a=0, b=1$ (True).
$(B)$ Differentiable at $x=1$ if $a=1, b=0$ (True).
$(C)$ Not differentiable at $x=0$ if $a=1, b=0$ (False).
$(D)$ Not differentiable at $x=1$ if $a=1, b=1$ (False).
Thus,options $A$ and $B$ are correct.

(A,D) Given $\lim_{x \rightarrow 2} \frac{f(x) g(x)}{f^{\prime}(x) g^{\prime}(x)} = 1$. Since $f^{\prime}(2) = 0$ and $g(2) = 0$,this is a $\frac{0}{0}$ form. Applying $L$'Hopital's rule:
$\lim_{x \rightarrow 2} \frac{f^{\prime}(x) g(x) + f(x) g^{\prime}(x)}{f^{\prime \prime}(x) g^{\prime}(x) + f^{\prime}(x) g^{\prime \prime}(x)} = 1$.
Substituting $x=2$ and using $f^{\prime}(2) = g(2) = 0$:
$\frac{0 + f(2) g^{\prime}(2)}{f^{\prime \prime}(2) g^{\prime}(2) + 0} = 1 \implies \frac{f(2)}{f^{\prime \prime}(2)} = 1 \implies f(2) = f^{\prime \prime}(2)$.
Since $f: R \rightarrow (0, \infty)$,$f(2) > 0$,so $f^{\prime \prime}(2) > 0$.
By the second derivative test,$f$ has a local minimum at $x=2$ (Option $A$).
Consider $h(x) = f(x) e^{-x^2/2}$. This is not directly helpful,but consider $h(x) = f(x) e^{-x/2}$ or similar. Actually,consider $h(x) = f(x) - f^{\prime \prime}(x)$. Since $f(2) = f^{\prime \prime}(2)$,$h(2) = 0$. Thus,$f(x) - f^{\prime \prime}(x) = 0$ for at least $x=2$ (Option $D$).
Thus,both $A$ and $D$ are correct.

(B, C) The function $f(x) = [x^2 - 3]$ is discontinuous where $x^2 - 3$ is an integer.
In the interval $[-\frac{1}{2}, 2]$,$x^2$ ranges from $0$ to $4$,so $x^2 - 3$ ranges from $-3$ to $1$.
The values of $x^2 - 3$ are integers at $x^2 - 3 \in \{-3, -2, -1, 0, 1\}$,which means $x^2 \in \{0, 1, 2, 3, 4\}$.
Since $x \in [-\frac{1}{2}, 2]$,the points of discontinuity for $f(x)$ are $x = 0, 1, \sqrt{2}, \sqrt{3}, 2$. However,at $x=2$,the function is defined on a closed interval,so we check the interior points. The points are $x = 0, 1, \sqrt{2}, \sqrt{3}$. Thus,$f$ is discontinuous at $4$ points in $[-\frac{1}{2}, 2]$. Option $B$ is correct.
Now consider $g(x) = (|x| + |4x - 7|)f(x)$.
$f(x)$ is discontinuous at $x \in \{0, 1, \sqrt{2}, \sqrt{3}, 2\}$.
$|x|$ is non-differentiable at $x=0$.
$|4x-7|$ is non-differentiable at $x=7/4 = 1.75$.
At $x=0$,$f(x) = [-3] = -3 \neq 0$,so $g(x)$ is non-differentiable.
At $x=1, \sqrt{2}, \sqrt{3}$,$f(x)$ is discontinuous,so $g(x)$ is non-differentiable.
At $x=7/4$,$f(x) = [(7/4)^2 - 3] = [49/16 - 3] = [3.0625 - 3] = [0.0625] = 0$. Since $f(7/4) = 0$,the product $g(x) = (|x| + |4x-7|)f(x)$ becomes differentiable at $x=7/4$ because the jump discontinuity of $f$ is multiplied by $0$.
Thus,$g(x)$ is non-differentiable at $x \in \{0, 1, \sqrt{2}, \sqrt{3}\}$,which is $4$ points. Option $C$ is correct.

(C) The given system of equations is:
$ax + 2y = \lambda$
$3x - 2y = \mu$
The determinant of the coefficient matrix is:
$\Delta = \begin{vmatrix} a & 2 \\ 3 & -2 \end{vmatrix} = -2a - 6 = -2(a + 3)$.
For a unique solution,$\Delta \neq 0$,which implies $a \neq -3$. Thus,statement $(B)$ is correct.
If $a = -3$,then $\Delta = 0$. The system will have either no solution or infinitely many solutions.
We calculate $\Delta_1 = \begin{vmatrix} \lambda & 2 \\ \mu & -2 \end{vmatrix} = -2\lambda - 2\mu = -2(\lambda + \mu)$.
If $\lambda + \mu = 0$,then $\Delta_1 = 0$. Since $\Delta = 0$ and $\Delta_1 = 0$,the system has infinitely many solutions. Thus,statement $(C)$ is correct.
If $\lambda + \mu \neq 0$,then $\Delta_1 \neq 0$. Since $\Delta = 0$ and $\Delta_1 \neq 0$,the system has no solution. Thus,statement $(D)$ is correct.
Therefore,statements $(B)$,$(C)$,and $(D)$ are correct.

(D) Given $\hat{u} \times \vec{w} = \hat{v}$. Since $\hat{u}$ and $\vec{w}$ are unit vectors,$|\hat{u} \times \vec{w}| = |\hat{u}| |\vec{w}| \sin \theta = \sin \theta$. Also,$|\hat{v}| = \frac{1}{\sqrt{6}} \sqrt{1^2 + 1^2 + 2^2} = 1$. Thus,$\sin \theta = 1$,which implies $\theta = 90^{\circ}$.
Since $\hat{u} \times \vec{w} = \hat{v}$,$\hat{v}$ is perpendicular to both $\hat{u}$ and $\vec{w}$.
For a fixed $\hat{u}$,there are infinitely many vectors $\vec{w}$ perpendicular to $\hat{u}$ such that their cross product with $\hat{u}$ is $\hat{v}$. Thus,option $B$ is correct.
Since $\hat{v} \cdot \hat{u} = 0$,we have $\frac{1}{\sqrt{6}}(u_1 + u_2 + 2u_3) = 0$,so $u_1 + u_2 + 2u_3 = 0$.
If $\hat{u}$ lies in the $xy$-plane,$u_3 = 0$,then $u_1 + u_2 = 0$,so $|u_1| = |u_2|$. Thus,option $C$ is correct.
If $\hat{u}$ lies in the $xz$-plane,$u_2 = 0$,then $u_1 + 2u_3 = 0$,so $|u_1| = 2|u_3|$. Thus,option $D$ is incorrect as it states $2|u_1| = |u_3|$.

(C) Let $W, D, L$ denote win,draw,and loss for $T_1$ respectively. $P(W) = \frac{1}{2}, P(D) = \frac{1}{6}, P(L) = \frac{1}{3}$.
$(1)$ $X > Y$ occurs if $T_1$ wins more points than $T_2$. Possible outcomes for $(T_1, T_2)$ in two games:
- $T_1$ wins both: $(W, W) \implies X=6, Y=0, P = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$
- $T_1$ wins one,draws one: $(W, D) \text{ or } (D, W) \implies X=4, Y=1, P = (\frac{1}{2} \times \frac{1}{6}) + (\frac{1}{6} \times \frac{1}{2}) = \frac{1}{12} + \frac{1}{12} = \frac{1}{6}$
- $T_1$ wins one,loses one: $(W, L) \text{ or } (L, W) \implies X=3, Y=3$ (Not $X>Y$)
- $T_1$ draws both: $(D, D) \implies X=2, Y=2$ (Not $X>Y$)
- $T_1$ draws one,loses one: $(D, L) \text{ or } (L, D) \implies X=1, Y=4$ (Not $X>Y$)
Summing probabilities for $X>Y$: $\frac{1}{4} + \frac{1}{6} = \frac{3+2}{12} = \frac{5}{12}$. Correct option is $(B)$.
$(2)$ $X=Y$ occurs if:
- Both draw: $(D, D) \implies X=2, Y=2, P = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$
- $T_1$ wins one,loses one: $(W, L) \text{ or } (L, W) \implies X=3, Y=3, P = (\frac{1}{2} \times \frac{1}{3}) + (\frac{1}{3} \times \frac{1}{2}) = \frac{1}{6} + \frac{1}{6} = \frac{1}{3} = \frac{12}{36}$
Summing probabilities for $X=Y$: $\frac{1}{36} + \frac{12}{36} = \frac{13}{36}$. Correct option is $(C)$.