IIT JEE 2012 Mathematics Question Paper with Answer and Solution

(D) Given $z^2 + z + 1 - a = 0$.
Using the quadratic formula,$z = \frac{-1 \pm \sqrt{1^2 - 4(1)(1-a)}}{2} = \frac{-1 \pm \sqrt{4a - 3}}{2}$.
Since the imaginary part of $z$ is non-zero,the discriminant must be negative.
Thus,$4a - 3 < 0$,which implies $a < \frac{3}{4}$.
Among the given options,$\frac{3}{4}$ is not less than $\frac{3}{4}$,so $a$ cannot take the value $\frac{3}{4}$.

(B) To distribute $5$ distinct balls among $3$ distinct persons such that each person receives at least one ball,we use the principle of inclusion-exclusion or partition the balls into groups of sizes $(3, 1, 1)$ and $(2, 2, 1)$.
Case $1$: Group sizes $(3, 1, 1)$.
The number of ways to partition $5$ distinct balls into groups of sizes $3, 1, 1$ is $\frac{5!}{3!1!1! \cdot 2!} = \frac{120}{6 \cdot 2} = 10$.
Since the persons are distinct,we multiply by $3!$ to distribute these groups: $10 \times 6 = 60$.
Case $2$: Group sizes $(2, 2, 1)$.
The number of ways to partition $5$ distinct balls into groups of sizes $2, 2, 1$ is $\frac{5!}{2!2!1! \cdot 2!} = \frac{120}{4 \cdot 2} = 15$.
Since the persons are distinct,we multiply by $3!$ to distribute these groups: $15 \times 6 = 90$.
Total ways $= 60 + 90 = 150$.

(A) Let $P(t, \frac{4t - 20}{5})$ be a point on the line $4x - 5y = 20$.
The equation of the chord of contact of tangents drawn from $P(t, \frac{4t - 20}{5})$ to the circle $x^2 + y^2 = 9$ is given by $T = 0$:
$tx + (\frac{4t - 20}{5})y = 9$ --- $(1)$
Let $M(h, k)$ be the mid-point of this chord. The equation of a chord of a circle with mid-point $(h, k)$ is given by $T = S_1$:
$hx + ky = h^2 + k^2$ --- $(2)$
Comparing equations $(1)$ and $(2)$,we have:
$\frac{t}{h} = \frac{4t - 20}{5k} = \frac{9}{h^2 + k^2}$
From $\frac{t}{h} = \frac{9}{h^2 + k^2}$,we get $t = \frac{9h}{h^2 + k^2}$.
From $\frac{4t - 20}{5k} = \frac{9}{h^2 + k^2}$,we get $4t - 20 = \frac{45k}{h^2 + k^2}$.
Substituting $t$ in the second equation:
$4(\frac{9h}{h^2 + k^2}) - 20 = \frac{45k}{h^2 + k^2}$
$36h - 20(h^2 + k^2) = 45k$
$20(h^2 + k^2) - 36h + 45k = 0$
Replacing $(h, k)$ with $(x, y)$,the locus is:
$20(x^2 + y^2) - 36x + 45y = 0$

(C) The ellipse $E_1: \frac{x^2}{9}+\frac{y^2}{4}=1$ has vertices at $(\pm 3, 0)$ and $(0, \pm 2)$.
Since $E_1$ is inscribed in a rectangle $R$ with sides parallel to the coordinate axes,the vertices of the rectangle $R$ are $(\pm 3, \pm 2)$.
Let the equation of the ellipse $E_2$ be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.
Since $E_2$ passes through $(0, 4)$,we have $\frac{0}{a^2}+\frac{16}{b^2}=1$,which gives $b^2=16$.
Since $E_2$ circumscribes the rectangle $R$,it passes through the point $(3, 2)$.
Substituting $(3, 2)$ into the equation of $E_2$: $\frac{9}{a^2}+\frac{4}{b^2}=1$.
Substituting $b^2=16$: $\frac{9}{a^2}+\frac{4}{16}=1$ $\Rightarrow \frac{9}{a^2}+\frac{1}{4}=1$ $\Rightarrow \frac{9}{a^2}=\frac{3}{4}$ $\Rightarrow a^2=12$.
For the ellipse $E_2$,$b^2 > a^2$ (since $16 > 12$),so the eccentricity $e$ is given by $a^2=b^2(1-e^2)$.
$12=16(1-e^2) \Rightarrow 1-e^2=\frac{12}{16}=\frac{3}{4}$.
$e^2=1-\frac{3}{4}=\frac{1}{4} \Rightarrow e=\frac{1}{2}$.

(C) Given $\tan (2\pi - \theta) > 0$ $\Rightarrow -\tan \theta > 0$ $\Rightarrow \tan \theta < 0$.
Also,$-1 < \sin \theta < -\frac{\sqrt{3}}{2}$ implies $\theta \in (\frac{4\pi}{3}, \frac{3\pi}{2}) \cup (\frac{3\pi}{2}, \frac{5\pi}{3})$.
Since $\tan \theta < 0$,we have $\theta \in (\frac{3\pi}{2}, \frac{5\pi}{3})$.
Using the identity $\tan \frac{\theta}{2} + \cot \frac{\theta}{2} = \frac{2}{\sin \theta}$,the equation becomes:
$2 \cos \theta (1 - \sin \phi) = \sin^2 \theta \left(\frac{2}{\sin \theta}\right) \cos \phi - 1$
$2 \cos \theta - 2 \cos \theta \sin \phi = 2 \sin \theta \cos \phi - 1$
$2 \cos \theta + 1 = 2(\sin \theta \cos \phi + \cos \theta \sin \phi) = 2 \sin(\theta + \phi)$.
For $\theta \in (\frac{3\pi}{2}, \frac{5\pi}{3})$,$2 \cos \theta + 1 \in (1, 2)$.
Thus,$1 < 2 \sin(\theta + \phi) < 2 \Rightarrow \frac{1}{2} < \sin(\theta + \phi) < 1$.
This implies $\theta + \phi \in (\frac{\pi}{6} + 2k\pi, \frac{5\pi}{6} + 2k\pi)$.
For $k=0$,$\phi \in (\frac{\pi}{6} - \theta, \frac{5\pi}{6} - \theta)$. Since $\theta \in (\frac{3\pi}{2}, \frac{5\pi}{3})$,$\phi \in (\frac{\pi}{6} - \frac{5\pi}{3}, \frac{5\pi}{6} - \frac{3\pi}{2}) = (-\frac{3\pi}{2}, -\frac{2\pi}{3})$. Adjusting to $[0, 2\pi]$,$\phi \in (\frac{\pi}{2}, \frac{4\pi}{3})$.
For $k=1$,$\phi \in (\frac{13\pi}{6} - \theta, \frac{17\pi}{6} - \theta) = (\frac{13\pi}{6} - \frac{5\pi}{3}, \frac{17\pi}{6} - \frac{3\pi}{2}) = (\frac{\pi}{2}, \frac{4\pi}{3})$.
Thus,$\phi \in (\frac{\pi}{2}, \frac{4\pi}{3})$.
Therefore,$\phi$ cannot satisfy $(A), (C), (D)$.

(D) The slope of the line $2x - y = 1$ is $m = 2$.
The equation of the tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ with slope $m$ is $y = mx \pm \sqrt{a^2m^2 - b^2}$.
Here $a^2 = 9$ and $b^2 = 4$,so the tangent equations are $y = 2x \pm \sqrt{9(2)^2 - 4} = 2x \pm \sqrt{36 - 4} = 2x \pm \sqrt{32} = 2x \pm 4\sqrt{2}$.
The point of contact $(x_1, y_1)$ for a tangent $y = mx + c$ to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is given by $\left(\frac{a^2m}{c}, \frac{b^2}{c}\right)$.
For $c = 4\sqrt{2}$,the point is $\left(\frac{9(2)}{4\sqrt{2}}, \frac{4}{4\sqrt{2}}\right) = \left(\frac{18}{4\sqrt{2}}, \frac{1}{\sqrt{2}}\right) = \left(\frac{9}{2\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$,which is point $(A)$.
For $c = -4\sqrt{2}$,the point is $\left(\frac{9(2)}{-4\sqrt{2}}, \frac{4}{-4\sqrt{2}}\right) = \left(-\frac{9}{2\sqrt{2}}, -\frac{1}{\sqrt{2}}\right)$,which is point $(B)$.
Thus,the points of contact are $(A)$ and $(B)$.

(D) The focus of the parabola $y^2=8x$ is $S \equiv (2, 0)$.
To find the common chord,subtract the equation of the parabola from the equation of the circle:
$(x^2+y^2-2x-4y) - (y^2-8x) = 0$
$x^2+6x-4y = 0$
Since $y^2=8x$,we substitute $x = \frac{y^2}{8}$ into the equation of the chord:
$(\frac{y^2}{8})^2 + 6(\frac{y^2}{8}) - 4y = 0$
$\frac{y^4}{64} + \frac{3y^2}{4} - 4y = 0$
$y^4 + 48y^2 - 256y = 0$
$y(y^3 + 48y - 256) = 0$
One solution is $y=0$,which gives $x=0$. Thus,$P \equiv (0, 0)$.
For $y^3 + 48y - 256 = 0$,by inspection,$y=4$ is a root $(64 + 192 - 256 = 0)$.
If $y=4$,then $x = \frac{16}{8} = 2$. Thus,$Q \equiv (2, 4)$.
The vertices of $\triangle PQS$ are $P(0, 0)$,$Q(2, 4)$,and $S(2, 0)$.
Area $= \frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$
Area $= \frac{1}{2} |0(4-0) + 2(0-0) + 2(0-4)|$
Area $= \frac{1}{2} |0 + 0 - 8| = \frac{1}{2} |-8| = 4$.

(A) Let $t = \sqrt{4-\frac{1}{3\sqrt{2}}\sqrt{4-\frac{1}{3\sqrt{2}}\sqrt{4-\dots}}}$.
Then $t = \sqrt{4-\frac{1}{3\sqrt{2}}t}$.
Squaring both sides,we get $4-\frac{1}{3\sqrt{2}}t = t^2$,which implies $t^2 + \frac{1}{3\sqrt{2}}t - 4 = 0$.
Multiplying by $3\sqrt{2}$,we get $3\sqrt{2}t^2 + t - 12\sqrt{2} = 0$.
Using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we have $t = \frac{-1 \pm \sqrt{1 - 4(3\sqrt{2})(-12\sqrt{2})}}{2(3\sqrt{2})} = \frac{-1 \pm \sqrt{1 + 288}}{6\sqrt{2}} = \frac{-1 \pm 17}{6\sqrt{2}}$.
Since $t > 0$,we take $t = \frac{16}{6\sqrt{2}} = \frac{8}{3\sqrt{2}}$.
The expression becomes $6 + \log_{\frac{3}{2}}\left(\frac{1}{3\sqrt{2}} \cdot \frac{8}{3\sqrt{2}}\right) = 6 + \log_{\frac{3}{2}}\left(\frac{8}{18}\right) = 6 + \log_{\frac{3}{2}}\left(\frac{4}{9}\right)$.
Since $\frac{4}{9} = (\frac{2}{3})^2 = (\frac{3}{2})^{-2}$,the expression is $6 + \log_{\frac{3}{2}}\left((\frac{3}{2})^{-2}\right) = 6 - 2 = 4$.

(A) Given $a=2, b=\frac{7}{2}, c=\frac{5}{2}$.
Semi-perimeter $s = \frac{a+b+c}{2} = \frac{2 + 3.5 + 2.5}{2} = \frac{8}{2} = 4$.
The expression is $\frac{2 \sin P - \sin 2P}{2 \sin P + \sin 2P} = \frac{2 \sin P - 2 \sin P \cos P}{2 \sin P + 2 \sin P \cos P} = \frac{2 \sin P(1 - \cos P)}{2 \sin P(1 + \cos P)} = \frac{1 - \cos P}{1 + \cos P}$.
Using half-angle formulas,$\frac{1 - \cos P}{1 + \cos P} = \frac{2 \sin^2(P/2)}{2 \cos^2(P/2)} = \tan^2(P/2)$.
We know $\tan^2(P/2) = \frac{(s-b)(s-c)}{s(s-a)}$.
Substituting the values: $s-a = 4-2 = 2$,$s-b = 4-3.5 = 0.5 = 1/2$,$s-c = 4-2.5 = 1.5 = 3/2$.
$\tan^2(P/2) = \frac{(1/2)(3/2)}{4(2)} = \frac{3/4}{8} = \frac{3}{32}$.
Also,$\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{4 \times 2 \times 0.5 \times 1.5} = \sqrt{6}$.
Thus,$\left(\frac{3}{4 \Delta}\right)^2 = \frac{9}{16 \Delta^2} = \frac{9}{16 \times 6} = \frac{9}{96} = \frac{3}{32}$.
Therefore,the expression equals $\left(\frac{3}{4 \Delta}\right)^2$.

(D) Since $a_1, a_2, \ldots$ are in harmonic progression,their reciprocals $\frac{1}{a_1}, \frac{1}{a_2}, \ldots$ are in arithmetic progression.
Let the arithmetic progression be $b_n = \frac{1}{a_n} = A + (n-1)D$.
Given $a_1 = 5 \Rightarrow b_1 = \frac{1}{5}$ and $a_{20} = 25 \Rightarrow b_{20} = \frac{1}{25}$.
$b_{20} = b_1 + 19D \Rightarrow \frac{1}{25} = \frac{1}{5} + 19D$.
$19D = \frac{1}{25} - \frac{1}{5} = \frac{1-5}{25} = -\frac{4}{25}$.
$D = -\frac{4}{19 \times 25} = -\frac{4}{475}$.
We want $a_n < 0$,which implies $\frac{1}{a_n} < 0$ is not necessarily true,but since $a_1 > 0$ and $a_{20} > 0$,we look for when the term becomes negative.
$b_n = \frac{1}{5} - (n-1)\frac{4}{475} < 0$.
$\frac{1}{5} < (n-1)\frac{4}{475}$.
$\frac{475}{20} < n-1$.
$23.75 < n-1$.
$n > 24.75$.
The least positive integer $n$ is $25$.

(B) The given equation is $((1+a)^{1/3}-1)x^2 + ((1+a)^{1/2}-1)x + ((1+a)^{1/6}-1) = 0$.
Let $1+a = t^6$. As $a \rightarrow 0^{+}$,$t \rightarrow 1^{+}$.
The equation becomes $(t^2-1)x^2 + (t^3-1)x + (t-1) = 0$.
Dividing by $(t-1)$ (since $t \neq 1$ as $a > 0$):
$(t+1)x^2 + (t^2+t+1)x + 1 = 0$.
Taking the limit as $t \rightarrow 1$:
$(1+1)x^2 + (1^2+1+1)x + 1 = 0$.
$2x^2 + 3x + 1 = 0$.
Factoring the quadratic equation:
$(2x+1)(x+1) = 0$.
Thus,the roots are $x = -\frac{1}{2}$ and $x = -1$.

(A) For an $n$-digit integer with no consecutive $0$s:
If it ends in $1$,the previous digit can be $0$ or $1$. Thus,$b_n = b_{n-1} + c_{n-1} = a_{n-1}$.
If it ends in $0$,the previous digit must be $1$. Thus,$c_n = b_{n-1}$.
Since $a_n = b_n + c_n$,we have $a_n = a_{n-1} + a_{n-2}$.
$1.$ For $(A)$,$a_{17} = a_{16} + a_{15}$ is true by the recurrence relation.
For $(B)$,$c_{17} = c_{16} + c_{15}$ is true,so $c_{17} \neq c_{16} + c_{15}$ is false.
For $(C)$,$b_{17} = b_{16} + c_{16}$ is true,so $b_{17} \neq b_{16} + c_{16}$ is false.
For $(D)$,$a_{17} = b_{17} + c_{17} = (b_{16} + c_{16}) + c_{17} = a_{16} + c_{17}$. Since $a_{16} = b_{16} + c_{16}$,this does not simplify to $a_{17} = c_{17} + b_{16}$.
Thus,only $(A)$ is correct.
$2.$ We have $b_n = a_{n-1}$.
$a_1 = 1$ $(1)$
$a_2 = 2$ $(10, 11)$
$a_3 = 3$ $(101, 110, 111)$
$a_4 = 5$ $(1010, 1011, 1101, 1110, 1111)$
$a_5 = 8$
Therefore,$b_6 = a_5 = 8$.
Correct option is $(B)$.

(A) $1.$ The equation of the tangent to $x^2+y^2=4$ at $P(\sqrt{3}, 1)$ is $\sqrt{3}x + y = 4$.
The centers of the circles are $C_1(0,0)$ with radius $r_1=2$ and $C_2(3,0)$ with radius $r_2=1$.
The external center of similitude $B$ divides $C_1C_2$ in the ratio $r_1:r_2 = 2:1$ externally.
$B = \left(\frac{2(3)-1(0)}{2-1}, \frac{2(0)-1(0)}{2-1}\right) = (6,0)$.
Let the common tangent be $y-0 = m(x-6)$,or $mx - y - 6m = 0$.
The perpendicular distance from $C_1(0,0)$ to this line equals $r_1=2$.
$\left|\frac{-6m}{\sqrt{m^2+1}}\right| = 2$ $\Rightarrow 3|m| = \sqrt{m^2+1}$ $\Rightarrow 9m^2 = m^2+1$ $\Rightarrow m^2 = \frac{1}{8}$ $\Rightarrow m = \pm \frac{1}{2\sqrt{2}}$.
The equations are $x \pm 2\sqrt{2}y = 6$. Thus,$(D)$ is a common tangent.
$2.$ The slope of $PT$ $(\sqrt{3}x+y=4)$ is $m_1 = -\sqrt{3}$.
Since $L$ is perpendicular to $PT$,the slope of $L$ is $m_2 = -\frac{1}{m_1} = \frac{1}{\sqrt{3}}$.
Let the equation of $L$ be $x - \sqrt{3}y + k = 0$.
$L$ is tangent to $(x-3)^2+y^2=1$,so the distance from $(3,0)$ to $L$ is $1$.
$\left|\frac{3 - \sqrt{3}(0) + k}{\sqrt{1^2 + (-\sqrt{3})^2}}\right| = 1 \Rightarrow \left|\frac{3+k}{2}\right| = 1$.
$3+k = 2$ or $3+k = -2$,so $k = -1$ or $k = -5$.
The possible equations are $x - \sqrt{3}y - 1 = 0$ or $x - \sqrt{3}y - 5 = 0$.
Comparing with options,$x - \sqrt{3}y = 1$ is not listed,but $x - \sqrt{3}y = -1$ is option $(C)$.

(B) Given $\lim _{x \rightarrow \infty}\left(\frac{x^2+x+1}{x+1}-a x-b\right)=4$.
Simplify the expression inside the limit:
$\lim _{x \rightarrow \infty}\left(\frac{x^2+x+1-ax(x+1)-b(x+1)}{x+1}\right)=4$
$\lim _{x \rightarrow \infty}\left(\frac{x^2+x+1-ax^2-ax-bx-b}{x+1}\right)=4$
$\lim _{x \rightarrow \infty}\left(\frac{(1-a)x^2+(1-a-b)x+(1-b)}{x+1}\right)=4$
For the limit to be finite,the coefficient of $x^2$ must be zero:
$1-a=0 \Rightarrow a=1$.
Now,substitute $a=1$ into the expression:
$\lim _{x \rightarrow \infty}\left(\frac{(1-1-b)x+(1-b)}{x+1}\right)=4$
$\lim _{x \rightarrow \infty}\left(\frac{-bx+(1-b)}{x+1}\right)=4$
Divide numerator and denominator by $x$:
$\lim _{x \rightarrow \infty}\left(\frac{-b+\frac{1-b}{x}}{1+\frac{1}{x}}\right)=4$
$-b=4 \Rightarrow b=-4$.
Thus,$a=1$ and $b=-4$.

(A) The equation of the line $QR$ passing through $Q(2, 3, 5)$ and $R(1, -1, 4)$ is given by $\frac{x-2}{1-2} = \frac{y-3}{-1-3} = \frac{z-5}{4-5}$,which simplifies to $\frac{x-2}{-1} = \frac{y-3}{-4} = \frac{z-5}{-1} = \lambda$.
Thus,any point on the line is $P(2-\lambda, 3-4\lambda, 5-\lambda)$.
Since $P$ lies on the plane $5x - 4y - z = 1$,we have $5(2-\lambda) - 4(3-4\lambda) - (5-\lambda) = 1$.
$10 - 5\lambda - 12 + 16\lambda - 5 + \lambda = 1 \Rightarrow 12\lambda - 7 = 1 \Rightarrow 12\lambda = 8 \Rightarrow \lambda = \frac{2}{3}$.
Substituting $\lambda = \frac{2}{3}$,we get $P = (2-\frac{2}{3}, 3-\frac{8}{3}, 5-\frac{2}{3}) = (\frac{4}{3}, \frac{1}{3}, \frac{13}{3})$.
Now,for the foot of the perpendicular $S$ from $T(2, 1, 4)$ to $QR$,let $S = (2-\mu, 3-4\mu, 5-\mu)$.
The vector $\vec{TS} = (2-\mu-2, 3-4\mu-1, 5-\mu-4) = (-\mu, 2-4\mu, 1-\mu)$.
Since $\vec{TS}$ is perpendicular to the line direction vector $\vec{v} = (-1, -4, -1)$,their dot product is zero: $(-\mu)(-1) + (2-4\mu)(-4) + (1-\mu)(-1) = 0$.
$\mu - 8 + 16\mu - 1 + \mu = 0 \Rightarrow 18\mu = 9 \Rightarrow \mu = \frac{1}{2}$.
Thus,$S = (2-\frac{1}{2}, 3-2, 5-\frac{1}{2}) = (\frac{3}{2}, 1, \frac{9}{2})$.
The distance $PS = \sqrt{(\frac{4}{3}-\frac{3}{2})^2 + (\frac{1}{3}-1)^2 + (\frac{13}{3}-\frac{9}{2})^2} = \sqrt{(-\frac{1}{6})^2 + (-\frac{2}{3})^2 + (-\frac{1}{6})^2} = \sqrt{\frac{1}{36} + \frac{4}{9} + \frac{1}{36}} = \sqrt{\frac{2}{36} + \frac{16}{36}} = \sqrt{\frac{18}{36}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.

(A) Let $t = \sec x + \tan x$. Then $dt = (\sec x \tan x + \sec^2 x) dx = \sec x(\tan x + \sec x) dx = \sec x \cdot t \cdot dx$.
Thus,$\sec x dx = \frac{dt}{t}$.
Also,$\sec x - \tan x = \frac{1}{t}$.
Adding the two equations: $2 \sec x = t + \frac{1}{t} \implies \sec x = \frac{1}{2} \left( t + \frac{1}{t} \right)$.
Substituting into the integral:
$I = \int \frac{\sec x \cdot (\sec x dx)}{t^{9/2}} = \int \frac{\frac{1}{2}(t + \frac{1}{t}) \cdot \frac{dt}{t}}{t^{9/2}} = \frac{1}{2} \int \frac{t + t^{-1}}{t^{11/2}} dt = \frac{1}{2} \int (t^{-9/2} + t^{-13/2}) dt$.
Integrating:
$I = \frac{1}{2} \left[ \frac{t^{-7/2}}{-7/2} + \frac{t^{-11/2}}{-11/2} \right] + K = -\left[ \frac{1}{7 t^{7/2}} + \frac{1}{11 t^{11/2}} \right] + K$.
Factoring out $-\frac{1}{t^{11/2}}$:
$I = -\frac{1}{t^{11/2}} \left[ \frac{t^2}{7} + \frac{1}{11} \right] + K$.
Substituting $t = \sec x + \tan x$ back,we get option $A$.

(B) $(i)$ Check for differentiability at $x=0$:
$LHD = f'(0^-) = \lim_{h \to 0^+} \frac{f(0-h) - f(0)}{-h} = \lim_{h \to 0^+} \frac{(-h)^2 |\cos(-\pi/h)| - 0}{-h} = \lim_{h \to 0^+} -h |\cos(\pi/h)| = 0$.
$RHD = f'(0^+) = \lim_{h \to 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^+} \frac{h^2 |\cos(\pi/h)| - 0}{h} = \lim_{h \to 0^+} h |\cos(\pi/h)| = 0$.
Since $LHD = RHD = 0$,$f(x)$ is differentiable at $x=0$.
$(ii)$ Check for differentiability at $x=2$:
$f(2) = 2^2 |\cos(\pi/2)| = 0$.
$RHD = f'(2^+) = \lim_{h \to 0^+} \frac{f(2+h) - f(2)}{h} = \lim_{h \to 0^+} \frac{(2+h)^2 |\cos(\pi/(2+h))|}{h}$.
Since $\cos(\pi/(2+h)) > 0$ for small $h > 0$,$|\cos(\pi/(2+h))| = \cos(\pi/(2+h)) = \sin(\pi/2 - \pi/(2+h)) = \sin(\pi h / (2(2+h)))$.
$RHD = \lim_{h \to 0^+} \frac{(2+h)^2 \sin(\pi h / (2(2+h)))}{h} = 4 \cdot \frac{\pi}{4} = \pi$.
$LHD = f'(2^-) = \lim_{h \to 0^+} \frac{f(2-h) - f(2)}{-h} = \lim_{h \to 0^+} \frac{(2-h)^2 |\cos(\pi/(2-h))|}{-h}$.
Since $\cos(\pi/(2-h)) < 0$ for small $h > 0$,$|\cos(\pi/(2-h))| = -\cos(\pi/(2-h)) = -\sin(\pi/2 - \pi/(2-h)) = -\sin(-\pi h / (2(2-h))) = \sin(\pi h / (2(2-h)))$.
$LHD = \lim_{h \to 0^+} \frac{(2-h)^2 \sin(\pi h / (2(2-h)))}{-h} = 4 \cdot (-\pi/4) = -\pi$.
Since $LHD \neq RHD$,$f(x)$ is not differentiable at $x=2$.

(B) Given function: $f(x) = 2x^3 - 15x^2 + 36x + 1$ for $x \in [0, 3]$.
Step $1$: Check for one-one property.
Find the derivative: $f'(x) = 6x^2 - 30x + 36 = 6(x^2 - 5x + 6) = 6(x-2)(x-3)$.
Since $f'(x)$ changes sign at $x=2$ within the interval $[0, 3]$,the function is not monotonic. Specifically,$f(x)$ increases on $[0, 2]$,decreases on $[2, 3]$,and increases on $[3, \infty)$. Since it is not strictly monotonic on $[0, 3]$,it is not one-one (many-one).
Step $2$: Check for onto property.
Find the range of $f(x)$ on $[0, 3]$.
$f(0) = 2(0)^3 - 15(0)^2 + 36(0) + 1 = 1$.
$f(2) = 2(8) - 15(4) + 36(2) + 1 = 16 - 60 + 72 + 1 = 29$.
$f(3) = 2(27) - 15(9) + 36(3) + 1 = 54 - 135 + 108 + 1 = 28$.
Since the function is continuous on $[0, 3]$,the range is $[min(f(0), f(2), f(3)), max(f(0), f(2), f(3))] = [1, 29]$.
Since the range $[1, 29]$ is equal to the codomain $[1, 29]$,the function is onto.
Conclusion: The function is onto but not one-one.

(D) Given $P = [a_{ij}]_{3 \times 3}$ and $b_{ij} = 2^{i+j} a_{ij}$.
$Q = [b_{ij}]_{3 \times 3} = \begin{bmatrix} 2^{1+1}a_{11} & 2^{1+2}a_{12} & 2^{1+3}a_{13} \\ 2^{2+1}a_{21} & 2^{2+2}a_{22} & 2^{2+3}a_{23} \\ 2^{3+1}a_{31} & 2^{3+2}a_{32} & 2^{3+3}a_{33} \end{bmatrix} = \begin{bmatrix} 4a_{11} & 8a_{12} & 16a_{13} \\ 8a_{21} & 16a_{22} & 32a_{23} \\ 16a_{31} & 32a_{32} & 64a_{33} \end{bmatrix}$.
Taking common factors from each row:
Row $1$ has a factor of $4 = 2^2$.
Row $2$ has a factor of $8 = 2^3$.
Row $3$ has a factor of $16 = 2^4$.
$|Q| = (2^2 \cdot 2^3 \cdot 2^4) \begin{vmatrix} a_{11} & 2a_{12} & 4a_{13} \\ a_{21} & 2a_{22} & 4a_{23} \\ a_{31} & 2a_{32} & 4a_{33} \end{vmatrix}$.
Taking common factors from each column:
Column $2$ has a factor of $2 = 2^1$.
Column $3$ has a factor of $4 = 2^2$.
$|Q| = (2^2 \cdot 2^3 \cdot 2^4) \cdot (2^1 \cdot 2^2) \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix}$.
$|Q| = 2^{2+3+4+1+2} |P| = 2^{12} \cdot 2 = 2^{13}$.

(A) The given differential equation is $\frac{dy}{dx} - y \tan x = 2x \sec x$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = -\tan x$ and $Q = 2x \sec x$.
The integrating factor ($I$.$F$.) is $e^{\int P dx} = e^{-\int \tan x dx} = e^{-\ln(\sec x)} = \cos x$.
Multiplying both sides by the $I$.$F$.,we get $\frac{d}{dx}(y \cos x) = 2x \sec x \cdot \cos x = 2x$.
Integrating both sides,$y \cos x = x^2 + C$.
Given $y(0) = 0$,we have $0 = 0^2 + C$,so $C = 0$.
Thus,$y = x^2 \sec x$.
For option $(A)$: $y(\frac{\pi}{4}) = (\frac{\pi}{4})^2 \sec(\frac{\pi}{4}) = \frac{\pi^2}{16} \cdot \sqrt{2} = \frac{\pi^2}{8\sqrt{2}}$. This is true.
For option $(D)$: $y'(x) = 2x \sec x + x^2 \sec x \tan x$.
$y'(\frac{\pi}{3}) = 2(\frac{\pi}{3}) \sec(\frac{\pi}{3}) + (\frac{\pi}{3})^2 \sec(\frac{\pi}{3}) \tan(\frac{\pi}{3}) = 2(\frac{\pi}{3})(2) + \frac{\pi^2}{9}(2)(\sqrt{3}) = \frac{4\pi}{3} + \frac{2\pi^2\sqrt{3}}{9}$. This is true.
Therefore,options $(A)$ and $(D)$ are correct.

(A) Given probabilities: $P(X_1) = \frac{1}{2}, P(X_2) = \frac{1}{4}, P(X_3) = \frac{1}{4}$.
Let $X_1^c, X_2^c, X_3^c$ be the events that engines do not function,so $P(X_1^c) = \frac{1}{2}, P(X_2^c) = \frac{3}{4}, P(X_3^c) = \frac{3}{4}$.
The ship is operational $(X)$ if at least two engines function:
$P(X) = P(X_1 X_2 X_3^c) + P(X_1 X_2^c X_3) + P(X_1^c X_2 X_3) + P(X_1 X_2 X_3)$
$P(X) = (\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{3}{4}) + (\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{1}{4}) + (\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{4}) + (\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{4}) = \frac{3}{32} + \frac{3}{32} + \frac{1}{32} + \frac{1}{32} = \frac{8}{32} = \frac{1}{4}$.
$(A) P(X_1^c \mid X) = \frac{P(X_1^c \cap X)}{P(X)} = \frac{P(X_1^c X_2 X_3)}{P(X)} = \frac{\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{4}}{\frac{1}{4}} = \frac{1}{8} \neq \frac{3}{16}$. (False)
$(B) P(\text{Exactly two} \mid X) = \frac{P(X_1 X_2 X_3^c) + P(X_1 X_2^c X_3) + P(X_1^c X_2 X_3)}{P(X)} = \frac{\frac{3}{32} + \frac{3}{32} + \frac{1}{32}}{\frac{1}{4}} = \frac{7/32}{1/4} = \frac{7}{8}$. (True)
$(C) P(X \mid X_2) = \frac{P(X \cap X_2)}{P(X_2)} = \frac{P(X_1 X_2 X_3^c) + P(X_1^c X_2 X_3) + P(X_1 X_2 X_3)}{P(X_2)} = \frac{\frac{3}{32} + \frac{1}{32} + \frac{1}{32}}{\frac{1}{4}} = \frac{5/32}{1/4} = \frac{5}{8} \neq \frac{5}{16}$. (False)
$(D) P(X \mid X_1) = \frac{P(X \cap X_1)}{P(X_1)} = \frac{P(X_1 X_2 X_3^c) + P(X_1 X_2^c X_3) + P(X_1 X_2 X_3)}{P(X_1)} = \frac{\frac{3}{32} + \frac{3}{32} + \frac{1}{32}}{\frac{1}{2}} = \frac{7/32}{1/2} = \frac{7}{16}$. (True)

(B) The area $S$ is given by the integral $S = \int_0^1 e^{-x^2} dx$.
$1$. For $x \in [0, 1]$,we have $0 \leq x^2 \leq x \leq 1$. Thus,$-x^2 \geq -x$,which implies $e^{-x^2} \geq e^{-x}$.
Integrating both sides from $0$ to $1$:
$S = \int_0^1 e^{-x^2} dx \geq \int_0^1 e^{-x} dx = [-e^{-x}]_0^1 = 1 - \frac{1}{e}$.
Since $1 - \frac{1}{e} \approx 1 - 0.367 = 0.633$ and $\frac{1}{e} \approx 0.367$,we have $S \geq 1 - \frac{1}{e} > \frac{1}{e}$. Thus,$(A)$ and $(B)$ are correct.
$2$. Using the upper Riemann sum with two rectangles of width $\frac{1}{\sqrt{2}}$ and $1-\frac{1}{\sqrt{2}}$:
The function $f(x) = e^{-x^2}$ is decreasing on $[0, 1]$.
$S = \int_0^{1/\sqrt{2}} e^{-x^2} dx + \int_{1/\sqrt{2}}^1 e^{-x^2} dx$.
Using the maximum value of $f(x)$ on each sub-interval:
$S \leq \left(\frac{1}{\sqrt{2}}\right) f(0) + \left(1 - \frac{1}{\sqrt{2}}\right) f\left(\frac{1}{\sqrt{2}}\right)$
$S \leq \frac{1}{\sqrt{2}}(1) + \left(1 - \frac{1}{\sqrt{2}}\right) e^{-(1/\sqrt{2})^2} = \frac{1}{\sqrt{2}} + \left(1 - \frac{1}{\sqrt{2}}\right) \frac{1}{\sqrt{e}}$.
Thus,$(D)$ is correct.

(D) Since $p(x)$ has a local maximum at $x=1$ and a local minimum at $x=3$,$p^{\prime}(x)$ must have roots at $x=1$ and $x=3$.
Thus,$p^{\prime}(x) = \lambda(x-1)(x-3) = \lambda(x^2-4x+3)$.
Integrating $p^{\prime}(x)$,we get $p(x) = \lambda(\frac{x^3}{3} - 2x^2 + 3x) + \mu$.
Given $p(1) = 6$,we have $6 = \lambda(\frac{1}{3} - 2 + 3) + \mu = \frac{4}{3}\lambda + \mu$,which implies $18 = 4\lambda + 3\mu \quad \dots (i)$.
Given $p(3) = 2$,we have $2 = \lambda(\frac{27}{3} - 2(9) + 3(3)) + \mu = \lambda(9 - 18 + 9) + \mu = \mu$.
So,$\mu = 2$.
Substituting $\mu = 2$ into equation $(i)$,$18 = 4\lambda + 3(2) \implies 18 = 4\lambda + 6 \implies 4\lambda = 12 \implies \lambda = 3$.
Thus,$p^{\prime}(x) = 3(x-1)(x-3)$.
Evaluating at $x=0$,$p^{\prime}(0) = 3(0-1)(0-3) = 3(-1)(-3) = 9$.

(C) The function is given by $f(x) = |x| + |x^2 - 1|$.
We analyze the function in different intervals:
$1$. For $x < -1$: $f(x) = -x + x^2 - 1 = x^2 - x - 1$. The vertex is at $x = 1/2$,which is not in this interval. As $x \to -1^-$,$f(x) \to 1$. There is no local extremum here.
$2$. For $-1 \leq x < 0$: $f(x) = -x - (x^2 - 1) = -x^2 - x + 1$. The derivative is $f'(x) = -2x - 1$. Setting $f'(x) = 0$ gives $x = -1/2$. Since $f''(-1/2) = -2 < 0$,there is a local maximum at $x = -1/2$ with $f(-1/2) = 5/4$.
$3$. At $x = -1$: $f(-1) = 1$. Since $f(x)$ decreases to $1$ from the left and increases to $5/4$ from the right,$x = -1$ is a local minimum.
$4$. At $x = 0$: $f(0) = 1$. Since $f(x)$ decreases to $1$ from the left and increases to $1$ from the right,$x = 0$ is a local minimum.
$5$. For $0 < x < 1$: $f(x) = x - (x^2 - 1) = -x^2 + x + 1$. The derivative is $f'(x) = -2x + 1$. Setting $f'(x) = 0$ gives $x = 1/2$. Since $f''(1/2) = -2 < 0$,there is a local maximum at $x = 1/2$ with $f(1/2) = 5/4$.
$6$. At $x = 1$: $f(1) = 1$. Since $f(x)$ decreases to $1$ from the left and increases to $1$ from the right,$x = 1$ is a local minimum.
Summarizing the points:
- Local minima at $x = -1, 0, 1$ ($3$ points).
- Local maxima at $x = -1/2, 1/2$ ($2$ points).
Total number of points = $3 + 2 = 5$.

(B) Given that $\vec{a}, \vec{b}, \vec{c}$ are unit vectors,so $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1$.
Expanding the given equation: $|\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2=9$.
This becomes $(|\vec{a}|^2+|\vec{b}|^2-2\vec{a} \cdot \vec{b}) + (|\vec{b}|^2+|\vec{c}|^2-2\vec{b} \cdot \vec{c}) + (|\vec{c}|^2+|\vec{a}|^2-2\vec{c} \cdot \vec{a}) = 9$.
Since $|\vec{a}|^2 = |\vec{b}|^2 = |\vec{c}|^2 = 1$,we have $6 - 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 9$.
Thus,$\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -\frac{3}{2}$.
Now,consider $|\vec{a}+\vec{b}+\vec{c}|^2 = |\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 3 + 2(-\frac{3}{2}) = 0$.
This implies $\vec{a}+\vec{b}+\vec{c} = \vec{0}$,so $\vec{b}+\vec{c} = -\vec{a}$.
We need to find $|2 \vec{a}+5 \vec{b}+5 \vec{c}| = |2 \vec{a}+5(\vec{b}+\vec{c})| = |2 \vec{a}+5(-\vec{a})| = |-3 \vec{a}| = 3|\vec{a}| = 3(1) = 3$.

(A) The equation of any plane passing through the line of intersection of the planes $x+2y+3z-2=0$ and $x-y+z-3=0$ is given by $(x+2y+3z-2) + \lambda(x-y+z-3) = 0$.
This simplifies to $(1+\lambda)x + (2-\lambda)y + (3+\lambda)z - (2+3\lambda) = 0$.
The distance of this plane from the point $(3,1,-1)$ is given by $\frac{|(1+\lambda)(3) + (2-\lambda)(1) + (3+\lambda)(-1) - (2+3\lambda)|}{\sqrt{(1+\lambda)^2 + (2-\lambda)^2 + (3+\lambda)^2}} = \frac{2}{\sqrt{3}}$.
Simplifying the numerator: $|3+3\lambda + 2-\lambda - 3-\lambda - 2-3\lambda| = |-2\lambda|$.
Simplifying the denominator: $\sqrt{(1+2\lambda+\lambda^2) + (4-4\lambda+\lambda^2) + (9+6\lambda+\lambda^2)} = \sqrt{3\lambda^2+4\lambda+14}$.
Thus,$\frac{|-2\lambda|}{\sqrt{3\lambda^2+4\lambda+14}} = \frac{2}{\sqrt{3}}$.
Squaring both sides: $\frac{4\lambda^2}{3\lambda^2+4\lambda+14} = \frac{4}{3}$.
$3\lambda^2 = 3\lambda^2+4\lambda+14$,which gives $4\lambda = -14$,so $\lambda = -\frac{7}{2}$.
Substituting $\lambda = -\frac{7}{2}$ into the equation: $(1-\frac{7}{2})x + (2+\frac{7}{2})y + (3-\frac{7}{2})z - (2-\frac{21}{2}) = 0$.
$-\frac{5}{2}x + \frac{11}{2}y - \frac{1}{2}z + \frac{17}{2} = 0$.
Multiplying by $-2$,we get $5x-11y+z-17=0$,or $5x-11y+z=17$.

(A) Let $X$ be the number shown by $D_4$. The total number of outcomes for the four dice is $6^4 = 1296$.
Alternatively,consider the probability for a fixed value of $D_4$. For any specific number $k \in \{1, 2, 3, 4, 5, 6\}$ shown by $D_4$,the probability that $k$ does $NOT$ appear on any of $D_1, D_2, D_3$ is $(\frac{5}{6})^3 = \frac{125}{216}$.
Therefore,the probability that $k$ appears on at least one of $D_1, D_2, D_3$ is $1 - \frac{125}{216} = \frac{216 - 125}{216} = \frac{91}{216}$.
Since this probability is independent of the value shown by $D_4$,the overall probability is $\frac{91}{216}$.

(B) Let $I = \int_{-\pi / 2}^{\pi / 2} \left(x^2 + \ln \frac{\pi+x}{\pi-x}\right) \cos x \, dx$.
We can split this into two integrals: $I = \int_{-\pi / 2}^{\pi / 2} x^2 \cos x \, dx + \int_{-\pi / 2}^{\pi / 2} \ln \left(\frac{\pi+x}{\pi-x}\right) \cos x \, dx$.
Let $f(x) = \ln \left(\frac{\pi+x}{\pi-x}\right) \cos x$. Since $\ln \left(\frac{\pi+x}{\pi-x}\right)$ is an odd function and $\cos x$ is an even function,their product is an odd function. Thus,$\int_{-\pi / 2}^{\pi / 2} f(x) \, dx = 0$.
Now,$I = \int_{-\pi / 2}^{\pi / 2} x^2 \cos x \, dx$. Since $x^2 \cos x$ is an even function,$I = 2 \int_0^{\pi / 2} x^2 \cos x \, dx$.
Using integration by parts: $\int x^2 \cos x \, dx = x^2 \sin x - \int 2x \sin x \, dx = x^2 \sin x - 2 \left( -x \cos x + \int \cos x \, dx \right) = x^2 \sin x + 2x \cos x - 2 \sin x$.
Evaluating from $0$ to $\pi / 2$: $2 \left[ (x^2 \sin x + 2x \cos x - 2 \sin x) \right]_0^{\pi / 2} = 2 \left[ ((\pi/2)^2 \cdot 1 + 0 - 2(1)) - (0 + 0 - 0) \right] = 2 \left( \frac{\pi^2}{4} - 2 \right) = \frac{\pi^2}{2} - 4$.

(D) Given the equation $P^{\top} = 2P + I$.
Taking the transpose on both sides,we get $(P^{\top})^{\top} = (2P + I)^{\top}$.
Since $(P^{\top})^{\top} = P$,we have $P = 2P^{\top} + I$.
Substitute $P^{\top} = 2P + I$ into this equation:
$P = 2(2P + I) + I$.
$P = 4P + 2I + I$.
$P = 4P + 3I$.
Rearranging the terms,we get $3P = -3I$,which implies $P = -I$.
Therefore,for any column matrix $X \neq 0$,we have $PX = (-I)X = -X$.

(B) $1.$ Given $f(x)=(1-x)^2 \sin^2 x+x^2$ and $g(x)=\int_1^x \left(\frac{2(t-1)}{t+1}-\ln t\right) f(t) dt$.
By the Fundamental Theorem of Calculus,$g'(x) = \left(\frac{2(x-1)}{x+1}-\ln x\right) f(x)$.
Let $\phi(x) = \frac{2(x-1)}{x+1}-\ln x$. Then $\phi'(x) = \frac{2(x+1)-2(x-1)}{(x+1)^2} - \frac{1}{x} = \frac{4}{(x+1)^2} - \frac{1}{x} = \frac{4x - (x^2+2x+1)}{x(x+1)^2} = \frac{-(x-1)^2}{x(x+1)^2}$.
Since $\phi'(x) \leq 0$ for $x > 1$,$\phi(x)$ is decreasing. Since $\phi(1) = 0$,$\phi(x) < 0$ for $x > 1$. Also $f(x) = (1-x)^2 \sin^2 x + x^2 > 0$ for $x > 1$. Thus $g'(x) < 0$,so $g$ is decreasing on $(1, \infty)$.
$2.$ For $P$: $f(x)+2x = (1-x)^2 \sin^2 x + x^2 + 2x = 2(1+x^2) = 2+2x^2$.
$(1-x)^2 \sin^2 x = x^2 - 2x + 2 = (x-1)^2 + 1$.
Since $\sin^2 x \leq 1$,$(1-x)^2 \sin^2 x \leq (1-x)^2$,but $(1-x)^2 + 1 > (1-x)^2$. Thus $P$ is false.
For $Q$: Let $H(x) = 2f(x) + 1 - 2x(1+x)$. $H(0) = 2(1)^2(0) + 0 + 1 - 0 = 1$. $H(1) = 2(0+1) + 1 - 2(1)(2) = 3 - 4 = -1$. Since $H(x)$ is continuous and changes sign,there exists $x$ such that $H(x)=0$. Thus $Q$ is true.

(B) Given $P(X \mid Y) = \frac{P(X \cap Y)}{P(Y)} = \frac{1}{2}$.
Since $P(X \cap Y) = \frac{1}{6}$,we have $\frac{1/6}{P(Y)} = \frac{1}{2} \Rightarrow P(Y) = \frac{1}{3}$.
Given $P(Y \mid X) = \frac{P(X \cap Y)}{P(X)} = \frac{1}{3}$.
Since $P(X \cap Y) = \frac{1}{6}$,we have $\frac{1/6}{P(X)} = \frac{1}{3} \Rightarrow P(X) = \frac{1}{2}$.
Now,$P(X \cup Y) = P(X) + P(Y) - P(X \cap Y) = \frac{1}{2} + \frac{1}{3} - \frac{1}{6} = \frac{3+2-1}{6} = \frac{4}{6} = \frac{2}{3}$. Thus,$(A)$ is correct.
Check for independence: $P(X) \cdot P(Y) = \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6}$.
Since $P(X \cap Y) = \frac{1}{6} = P(X) \cdot P(Y)$,$X$ and $Y$ are independent. Thus,$(B)$ is correct.
$P(X^C \cap Y) = P(Y) - P(X \cap Y) = \frac{1}{3} - \frac{1}{6} = \frac{1}{6}$. Thus,$(D)$ is incorrect.

(C) Given $f(x)=\int_0^x e^{t^2}(t-2)(t-3) dt$.
By the Fundamental Theorem of Calculus,$f^{\prime}(x)=e^{x^2}(x-2)(x-3)$.
To find the critical points,set $f^{\prime}(x)=0$,which gives $x=2$ and $x=3$.
Analyzing the sign of $f^{\prime}(x)$:
For $x < 2$,$f^{\prime}(x) > 0$ (increasing).
For $2 < x < 3$,$f^{\prime}(x) < 0$ (decreasing).
For $x > 3$,$f^{\prime}(x) > 0$ (increasing).
Thus,$f$ has a local maximum at $x=2$ and a local minimum at $x=3$. This confirms $(A)$,$(B)$,and $(D)$ are correct.
Now,find $f^{\prime \prime}(x)$:
$f^{\prime \prime}(x) = \frac{d}{dx} [e^{x^2}(x^2-5x+6)] = e^{x^2}(2x)(x^2-5x+6) + e^{x^2}(2x-5) = e^{x^2}(2x^3-10x^2+12x+2x-5) = e^{x^2}(2x^3-10x^2+14x-5)$.
Let $g(x) = 2x^3-10x^2+14x-5$. Since $g(0) = -5$ and $g(1) = 2-10+14-5 = 1$,by the Intermediate Value Theorem,there exists $c \in (0, 1)$ such that $g(c) = 0$,implying $f^{\prime \prime}(c) = 0$. Thus,$(C)$ is also correct.
Therefore,all options $(A), (B), (C), (D)$ are correct.

(A) For $f$ to be continuous at $x = 2n$,the left-hand limit and right-hand limit must be equal to $f(2n)$.
$f(2n) = a_n + \sin(2n\pi) = a_n$.
$f(2n^+) = a_n + \sin(2n\pi) = a_n$.
$f(2n^-) = b_n + \cos(2n\pi) = b_n + 1$.
Equating $f(2n^+) = f(2n^-)$,we get $a_n = b_n + 1$,which implies $a_n - b_n = 1$. Thus,$(B)$ is correct.
For $f$ to be continuous at $x = 2n+1$,the left-hand limit and right-hand limit must be equal to $f(2n+1)$.
$f(2n+1) = a_n + \sin((2n+1)\pi) = a_n$.
$f((2n+1)^-) = a_n + \sin((2n+1)\pi) = a_n$.
$f((2n+1)^+) = b_{n+1} + \cos((2n+1)\pi) = b_{n+1} - 1$.
Equating $f((2n+1)^-) = f((2n+1)^+)$,we get $a_n = b_{n+1} - 1$,which implies $a_n - b_{n+1} = -1$. Replacing $n$ with $n-1$,we get $a_{n-1} - b_n = -1$. Thus,$(D)$ is correct.

(B) For two lines to be coplanar,the scalar triple product of the vector connecting a point on each line and the direction vectors of the lines must be zero: $[\vec{a}-\vec{c}, \vec{b}, \vec{d}] = 0$.
Given points $\vec{a} = (1, -1, 0)$ and $\vec{c} = (-1, -1, 0)$,the vector $\vec{a}-\vec{c} = (2, 0, 0)$.
The direction vectors are $\vec{b} = 2\hat{i} + k\hat{j} + 2\hat{k}$ and $\vec{d} = 5\hat{i} + 2\hat{j} + k\hat{k}$.
Setting the determinant to zero:
$\left|\begin{array}{lll} 2 & 0 & 0 \\ 2 & k & 2 \\ 5 & 2 & k \end{array}\right| = 0 \Rightarrow 2(k^2 - 4) = 0 \Rightarrow k^2 = 4 \Rightarrow k = \pm 2$.
Case $1$: If $k = 2$,the direction vectors are $\vec{b} = (2, 2, 2)$ and $\vec{d} = (5, 2, 2)$.
The normal vector is $\vec{n}_1 = \vec{b} \times \vec{d} = \left|\begin{array}{lll} \hat{i} & \hat{j} & \hat{k} \\ 2 & 2 & 2 \\ 5 & 2 & 2 \end{array}\right| = 0\hat{i} + 6\hat{j} - 6\hat{k}$.
The plane equation is $(\vec{r} - \vec{a}) \cdot \vec{n}_1 = 0 \Rightarrow 0(x-1) + 6(y+1) - 6(z-0) = 0 \Rightarrow y - z = -1$.
Case $2$: If $k = -2$,the direction vectors are $\vec{b} = (2, -2, 2)$ and $\vec{d} = (5, 2, -2)$.
The normal vector is $\vec{n}_2 = \vec{b} \times \vec{d} = \left|\begin{array}{lll} \hat{i} & \hat{j} & \hat{k} \\ 2 & -2 & 2 \\ 5 & 2 & -2 \end{array}\right| = 0\hat{i} + 14\hat{j} + 14\hat{k}$.
The plane equation is $(\vec{r} - \vec{a}) \cdot \vec{n}_2 = 0 \Rightarrow 0(x-1) + 14(y+1) + 14(z-0) = 0 \Rightarrow y + z = -1$.
Thus,the planes are $y-z=-1$ and $y+z=-1$,which correspond to options $(C)$ and $(B)$.

(A, D) Let $P$ be a $3 \times 3$ matrix. We are given that $\operatorname{adj}(P) = \begin{bmatrix} 1 & 4 & 4 \\ 2 & 1 & 7 \\ 1 & 1 & 3 \end{bmatrix}$.
We know the property $|\operatorname{adj}(P)| = |P|^{n-1}$,where $n$ is the order of the matrix.
Here,$n = 3$,so $|\operatorname{adj}(P)| = |P|^{3-1} = |P|^2$.
First,calculate the determinant of $\operatorname{adj}(P)$:
$|\operatorname{adj}(P)| = 1(1 \times 3 - 7 \times 1) - 4(2 \times 3 - 7 \times 1) + 4(2 \times 1 - 1 \times 1)$
$|\operatorname{adj}(P)| = 1(3 - 7) - 4(6 - 7) + 4(2 - 1)$
$|\operatorname{adj}(P)| = 1(-4) - 4(-1) + 4(1)$
$|\operatorname{adj}(P)| = -4 + 4 + 4 = 4$.
Now,equate this to $|P|^2$:
$|P|^2 = 4$
$|P| = \pm 2$.
Thus,the possible values of the determinant of $P$ are $2$ and $-2$,which correspond to options $(A)$ and $(D)$.

(AB) Given $f(\cos 4 \theta) = \frac{2}{2-\sec^2 \theta}$.
We know that $\sec^2 \theta = \frac{1}{\cos^2 \theta} = \frac{2}{1+\cos 2 \theta}$.
Substituting this into the expression for $f(\cos 4 \theta)$:
$f(\cos 4 \theta) = \frac{2}{2 - \frac{2}{1+\cos 2 \theta}} = \frac{2(1+\cos 2 \theta)}{2(1+\cos 2 \theta) - 2} = \frac{1+\cos 2 \theta}{\cos 2 \theta} = 1 + \frac{1}{\cos 2 \theta}$.
Now,let $\cos 4 \theta = \frac{1}{3}$.
Using the identity $\cos 4 \theta = 2 \cos^2 2 \theta - 1$,we have $2 \cos^2 2 \theta - 1 = \frac{1}{3} \Rightarrow 2 \cos^2 2 \theta = \frac{4}{3} \Rightarrow \cos^2 2 \theta = \frac{2}{3}$.
Thus,$\cos 2 \theta = \pm \sqrt{\frac{2}{3}}$.
Substituting this back into the expression for $f(\cos 4 \theta)$:
$f\left(\frac{1}{3}\right) = 1 + \frac{1}{\pm \sqrt{2/3}} = 1 \pm \sqrt{\frac{3}{2}}$.
Therefore,the values are $1+\sqrt{\frac{3}{2}}$ and $1-\sqrt{\frac{3}{2}}$.

(A) Given $\vec{a} \times(2 \hat{i}+3 \hat{j}+4 \hat{k})=(2 \hat{i}+3 \hat{j}+4 \hat{k}) \times \vec{b}$.
We know that $\vec{u} \times \vec{v} = -\vec{v} \times \vec{u}$,so $(2 \hat{i}+3 \hat{j}+4 \hat{k}) \times \vec{b} = -\vec{b} \times (2 \hat{i}+3 \hat{j}+4 \hat{k})$.
Substituting this into the equation,we get $\vec{a} \times(2 \hat{i}+3 \hat{j}+4 \hat{k}) + \vec{b} \times(2 \hat{i}+3 \hat{j}+4 \hat{k}) = \vec{0}$.
This implies $(\vec{a}+\vec{b}) \times(2 \hat{i}+3 \hat{j}+4 \hat{k}) = \vec{0}$.
This means the vector $(\vec{a}+\vec{b})$ is parallel to $(2 \hat{i}+3 \hat{j}+4 \hat{k})$.
Let $\vec{a}+\vec{b} = \lambda(2 \hat{i}+3 \hat{j}+4 \hat{k})$ for some scalar $\lambda$.
Taking the magnitude on both sides,$|\vec{a}+\vec{b}| = |\lambda| \sqrt{2^2+3^2+4^2} = |\lambda| \sqrt{4+9+16} = |\lambda| \sqrt{29}$.
Given $|\vec{a}+\vec{b}| = \sqrt{29}$,we have $|\lambda| \sqrt{29} = \sqrt{29}$,which gives $\lambda = \pm 1$.
Thus,$\vec{a}+\vec{b} = \pm(2 \hat{i}+3 \hat{j}+4 \hat{k})$.
Now,calculate the dot product: $(\vec{a}+\vec{b}) \cdot(-7 \hat{i}+2 \hat{j}+3 \hat{k}) = \pm(2 \hat{i}+3 \hat{j}+4 \hat{k}) \cdot(-7 \hat{i}+2 \hat{j}+3 \hat{k})$.
$= \pm((2)(-7) + (3)(2) + (4)(3)) = \pm(-14 + 6 + 12) = \pm 4$.
Therefore,a possible value is $4$.