IIT JEE 2008 Mathematics Question Paper with Answer and Solution

(B) The given equation is $(a x^2+b y^2+c)(x^2-5 x y+6 y^2)=0$.
This implies $a x^2+b y^2+c=0$ or $x^2-5 x y+6 y^2=0$.
The second part $x^2-5 x y+6 y^2=0$ can be factored as $(x-2 y)(x-3 y)=0$,which represents two straight lines passing through the origin.
For the first part $a x^2+b y^2+c=0$,if $a=b$ and $c$ has the opposite sign to $a$,then $x^2+y^2 = -c/a$,which represents a circle.
Thus,the equation represents two straight lines and a circle when $a=b$ and $c$ is of sign opposite to that of $a$.

(C) The function is $f(x) = |x-1|$. For $x < 1$,$f(x) = -(x-1) = -x+1$. The left-hand derivative at $x=1$ is $p = \frac{d}{dx}(-x+1) = -1$.
Given $\lim_{x \rightarrow 1^{+}} g(x) = p$,we have $\lim_{x \rightarrow 1^{+}} \frac{(x-1)^n}{\log \cos^m(x-1)} = -1$.
Let $h = x-1$. As $x \rightarrow 1^{+}$,$h \rightarrow 0^{+}$. The limit becomes $\lim_{h \rightarrow 0^{+}} \frac{h^n}{\log \cos^m h} = -1$.
Using the property $\log \cos^m h = m \log \cos h$,we have $\lim_{h \rightarrow 0^{+}} \frac{h^n}{m \log \cos h} = -1$.
Using $L$'$H$ôpital's rule,$\lim_{h \rightarrow 0^{+}} \frac{n h^{n-1}}{m (\frac{-\sin h}{\cos h})} = \lim_{h \rightarrow 0^{+}} \frac{n h^{n-1}}{-m \tan h} = -\frac{n}{m} \lim_{h \rightarrow 0^{+}} \frac{h^{n-1}}{\tan h} = -1$.
For this limit to be a non-zero constant,we must have $n-1 = 1$,so $n=2$. Then the limit is $-\frac{2}{m} \lim_{h \rightarrow 0^{+}} \frac{h}{\tan h} = -\frac{2}{m} (1) = -1$,which implies $m=2$.
Thus,$n=2$ and $m=2$.

(B) To find the points of intersection,substitute $y^2 = 4x$ into the equation of the circle $x^2 + y^2 - 6x + 1 = 0$.
This gives $x^2 + 4x - 6x + 1 = 0$,which simplifies to $x^2 - 2x + 1 = 0$.
This is $(x - 1)^2 = 0$,so $x = 1$.
Substituting $x = 1$ into $y^2 = 4x$,we get $y^2 = 4$,which means $y = 2$ or $y = -2$.
Thus,the curves intersect at the points $(1, 2)$ and $(1, -2)$.
To check if they touch,we compare the slopes of the tangents at these points.
For the parabola $y^2 = 4x$,differentiating gives $2y \frac{dy}{dx} = 4$,so $\frac{dy}{dx} = \frac{2}{y}$. At $(1, 2)$,slope $m_1 = 1$. At $(1, -2)$,slope $m_1 = -1$.
For the circle $x^2 + y^2 - 6x + 1 = 0$,differentiating gives $2x + 2y \frac{dy}{dx} - 6 = 0$,so $\frac{dy}{dx} = \frac{3 - x}{y}$. At $(1, 2)$,slope $m_2 = \frac{3 - 1}{2} = 1$. At $(1, -2)$,slope $m_2 = \frac{3 - 1}{-2} = -1$.
Since the slopes are equal at both points,the curves touch each other at exactly two points.

(A) By the power of a point theorem for the circle,we have $PS \times ST = QS \times SR$.
Using the Arithmetic Mean-Geometric Mean ($AM$-$GM$) inequality for the terms $\frac{1}{PS}$ and $\frac{1}{ST}$,we have:
$\frac{\frac{1}{PS}+\frac{1}{ST}}{2} > \sqrt{\frac{1}{PS} \times \frac{1}{ST}}$
$\Rightarrow \frac{1}{PS}+\frac{1}{ST} > \frac{2}{\sqrt{PS \times ST}} = \frac{2}{\sqrt{QS \times SR}}$.
This proves that option $(B)$ is correct.
Next,by the $AM$-$GM$ inequality for $QS$ and $SR$,we have:
$\frac{QS+SR}{2} > \sqrt{QS \times SR}$
$\Rightarrow \frac{QR}{2} > \sqrt{QS \times SR}$
$\Rightarrow \frac{1}{\sqrt{QS \times SR}} > \frac{2}{QR}$.
Substituting this into the previous inequality:
$\frac{1}{PS}+\frac{1}{ST} > \frac{2}{\sqrt{QS \times SR}} > \frac{2 \times 2}{QR} = \frac{4}{QR}$.
This proves that option $(D)$ is correct.
Thus,the correct options are $(B)$ and $(D)$.

(C) The given ellipse is $\frac{x^2}{4} + \frac{y^2}{1} = 1$. Here $a^2 = 4$ and $b^2 = 1$.
The eccentricity $e$ is given by $b^2 = a^2(1 - e^2)$ $\Rightarrow 1 = 4(1 - e^2)$ $\Rightarrow 1 - e^2 = \frac{1}{4}$ $\Rightarrow e = \frac{\sqrt{3}}{2}$.
The coordinates of the foci are $(\pm ae, 0) = (\pm \sqrt{3}, 0)$.
Since $P$ and $Q$ are endpoints of the latus rectum with $y < 0$,we have $x = \pm \sqrt{3}$. Substituting into the ellipse equation: $3 + 4y^2 = 4$ $\Rightarrow 4y^2 = 1$ $\Rightarrow y = -\frac{1}{2}$ (as $y < 0$).
Thus,$P = (\sqrt{3}, -\frac{1}{2})$ and $Q = (-\sqrt{3}, -\frac{1}{2})$.
The length of the latus rectum $PQ = 2\sqrt{3}$.
For a parabola with latus rectum $PQ$,the length of the latus rectum $4a' = 2\sqrt{3} \Rightarrow a' = \frac{\sqrt{3}}{2}$.
The midpoint of $PQ$ is $R = (0, -\frac{1}{2})$. The axis of the parabola is the $y$-axis.
The vertex of the parabola is $(0, y_v)$,where $y_v = -\frac{1}{2} \pm a' = -\frac{1}{2} \pm \frac{\sqrt{3}}{2}$.
Case $1$: Vertex $(0, -\frac{1}{2} - \frac{\sqrt{3}}{2})$. The equation is $x^2 = 4a'(y - y_v)$ $\Rightarrow x^2 = 2\sqrt{3}(y + \frac{1}{2} + \frac{\sqrt{3}}{2})$ $\Rightarrow x^2 = 2\sqrt{3}y + \sqrt{3} + 3$ $\Rightarrow x^2 - 2\sqrt{3}y = 3 + \sqrt{3}$.
Case $2$: Vertex $(0, -\frac{1}{2} + \frac{\sqrt{3}}{2})$. The equation is $x^2 = -4a'(y - y_v)$ $\Rightarrow x^2 = -2\sqrt{3}(y + \frac{1}{2} - \frac{\sqrt{3}}{2})$ $\Rightarrow x^2 = -2\sqrt{3}y - \sqrt{3} + 3$ $\Rightarrow x^2 + 2\sqrt{3}y = 3 - \sqrt{3}$.
Thus,the equations are $x^2 - 2\sqrt{3}y = 3 + \sqrt{3}$ and $x^2 + 2\sqrt{3}y = 3 - \sqrt{3}$,which correspond to options $B$ and $C$.

(B) $1.$ The normal to the line $PQ: \sqrt{3}x + y - 6 = 0$ at $D\left(\frac{3\sqrt{3}}{2}, \frac{3}{2}\right)$ has slope $\frac{1}{\sqrt{3}}$.
The equation of the normal line $CD$ is $y - \frac{3}{2} = \frac{1}{\sqrt{3}}\left(x - \frac{3\sqrt{3}}{2}\right) \Rightarrow x - \sqrt{3}y = 0$.
Since the radius is $1$ and the centre $C(h, k)$ is at distance $1$ from $PQ$ and lies on $x - \sqrt{3}y = 0$,we find $C = (\sqrt{3}, 1)$.
Thus,the equation of circle $C$ is $(x - \sqrt{3})^2 + (y - 1)^2 = 1$. Correct option is $(D)$.
$2.$ The angle between the sides of an equilateral triangle is $60^\circ$. The lines $CE$ and $CF$ make angles of $120^\circ$ and $240^\circ$ with $CD$ respectively.
Using rotation or geometry,$F = (\sqrt{3}, 0)$ and $E = \left(\frac{\sqrt{3}}{2}, \frac{3}{2}\right)$. Correct option is $(A)$.
$3.$ The side $QR$ passes through $E\left(\frac{\sqrt{3}}{2}, \frac{3}{2}\right)$ with slope $\sqrt{3}$,so $y - \frac{3}{2} = \sqrt{3}\left(x - \frac{\sqrt{3}}{2}\right) \Rightarrow y = \sqrt{3}x$. The side $RP$ passes through $F(\sqrt{3}, 0)$ with slope $0$,so $y = 0$. Correct option is $(D)$.

(B, C, D) $1.$ $A$ represents the region $y \geq 1$. $B$ is a circle with center $(2, 1)$ and radius $3$. $C$ is the line $\operatorname{Re}((1-i)(x+iy)) = x+y = \sqrt{2}$.
Substituting $y = \sqrt{2}-x$ into the circle equation $(x-2)^2 + (y-1)^2 = 9$ gives $(x-2)^2 + (\sqrt{2}-x-1)^2 = 9$.
Expanding this: $(x^2 - 4x + 4) + (x^2 + 2x(1-\sqrt{2}) + (1-\sqrt{2})^2) = 9$.
$2x^2 + x(2-2\sqrt{2}-4) + 4 + 1 - 2\sqrt{2} + 2 = 9 \implies 2x^2 - (2+2\sqrt{2})x - 2 - 2\sqrt{2} = 0$.
Solving this quadratic,we find one point with $y \geq 1$. Thus,the number of elements is $1$.
$2.$ Let $z = x+iy$. The expression is $|(x+1)+i(y-1)|^2 + |(x-5)+i(y-1)|^2 = (x+1)^2 + (y-1)^2 + (x-5)^2 + (y-1)^2$.
Since $z$ is on the circle $(x-2)^2 + (y-1)^2 = 9$,we have $(y-1)^2 = 9 - (x-2)^2$.
Substituting this,the expression becomes $(x+1)^2 + (x-5)^2 + 2(9 - (x-2)^2) = x^2+2x+1 + x^2-10x+25 + 18 - 2(x^2-4x+4) = 2x^2-8x+26 + 18 - 2x^2+8x-8 = 36$.
Since $36$ is between $35$ and $39$,the answer is $(C)$.
$3.$ Since $|z-2-i|=3$ and $|w-2-i| < 3$,by the triangle inequality $|z-w| < |z-(2+i)| + |w-(2+i)| < 3+3 = 6$.
Also $|z-w| > ||z-(2+i)| - |w-(2+i)|| = |3 - |w-(2+i)|| > 0$.
Thus $0 < |z-w| < 6$. Using $||z|-|w|| \leq |z-w|$,we have $-6 < |z|-|w| < 6$.
Adding $3$,we get $-3 < |z|-|w|+3 < 9$.

(C) Let $P = (x_1, y_1) = (-\sin(\beta - \alpha), -\cos \beta)$ and $Q = (x_2, y_2) = (\cos(\beta - \alpha), \sin \beta)$.
We observe that $R = (\cos(\beta - \alpha + \theta), \sin(\beta - \theta))$.
Using trigonometric identities,$\cos(\beta - \alpha + \theta) = \cos(\beta - \alpha)\cos \theta - \sin(\beta - \alpha)\sin \theta = x_2 \cos \theta + x_1 \sin \theta$.
Similarly,$\sin(\beta - \theta) = \sin \beta \cos \theta - \cos \beta \sin \theta = y_2 \cos \theta + y_1 \sin \theta$.
Thus,$R = (x_2 \cos \theta + x_1 \sin \theta, y_2 \cos \theta + y_1 \sin \theta)$.
This shows that $R$ divides the line segment $PQ$ internally in the ratio $\sin \theta : \cos \theta$.
Since $0 < \theta < \frac{\pi}{4}$,both $\sin \theta$ and $\cos \theta$ are positive,so $R$ lies on the line segment $PQ$ (or $QP$).

(B) The given equation is $x^2 - 2y^2 - 2\sqrt{2}x - 4\sqrt{2}y - 6 = 0$.
Completing the square,we get $(x^2 - 2\sqrt{2}x + 2) - 2(y^2 + 2\sqrt{2}y + 2) = 6 + 2 - 4$,which simplifies to $(x - \sqrt{2})^2 - 2(y + \sqrt{2})^2 = 4$.
Dividing by $4$,we obtain the standard form: $\frac{(x - \sqrt{2})^2}{4} - \frac{(y + \sqrt{2})^2}{2} = 1$.
Here,$a^2 = 4 \Rightarrow a = 2$ and $b^2 = 2 \Rightarrow b = \sqrt{2}$.
The eccentricity $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{2}{4}} = \sqrt{\frac{3}{2}}$.
The vertex $A$ is $(\sqrt{2} + a, -\sqrt{2}) = (2 + \sqrt{2}, -\sqrt{2})$.
The focus $C$ nearest to $A$ is $(\sqrt{2} + ae, -\sqrt{2}) = (\sqrt{2} + 2\sqrt{\frac{3}{2}}, -\sqrt{2}) = (\sqrt{2} + \sqrt{6}, -\sqrt{2})$.
The distance $AC = ae - a = a(e - 1) = 2(\sqrt{\frac{3}{2}} - 1) = \sqrt{6} - 2$.
The length of the semi-latus rectum is $\frac{b^2}{a} = \frac{2}{2} = 1$.
The point $B$ is $(ae + \sqrt{2}, \frac{b^2}{a} - \sqrt{2}) = (\sqrt{6} + \sqrt{2}, 1 - \sqrt{2})$.
The area of the triangle $ABC$ with base $AC$ and height $BC$ is $\frac{1}{2} \times AC \times \frac{b^2}{a} = \frac{1}{2} \times (\sqrt{6} - 2) \times 1 = \frac{\sqrt{6}}{2} - 1 = \sqrt{\frac{6}{4}} - 1 = \sqrt{\frac{3}{2}} - 1$.

(D) The initial position is $z_0 = 1 + 2i$.
Moving horizontally by $5$ units: $z = (1 + 5) + 2i = 6 + 2i$.
Moving vertically by $3$ units: $z_1 = 6 + (2 + 3)i = 6 + 5i$.
From $z_1$,moving $\sqrt{2}$ units in the direction of $\hat{i} + \hat{j}$ (which is the direction of $1+i$): The unit vector is $\frac{1+i}{\sqrt{2}}$. The displacement is $\sqrt{2} \times \frac{1+i}{\sqrt{2}} = 1 + i$.
So,the position becomes $z' = (6 + 5i) + (1 + i) = 7 + 6i$.
Finally,rotating $z'$ by $\frac{\pi}{2}$ anticlockwise about the origin is equivalent to multiplying by $e^{i\pi/2} = i$.
$z_2 = (7 + 6i) \times i = 7i + 6i^2 = 7i - 6 = -6 + 7i$.

(C) The equation of the circle $C$ is $x^2 + 6x + 9 + y^2 - 10y + 25 = -30 + 9 + 25$,which simplifies to $(x + 3)^2 + (y - 5)^2 = 4$.
The center of the circle is $(-3, 5)$ and the radius $r = 2$.
For $L_1: 2x + 3y + p - 3 = 0$ to be a chord,the distance from the center $(-3, 5)$ to $L_1$ must be less than the radius $r=2$.
Distance $d_1 = \frac{|2(-3) + 3(5) + p - 3|}{\sqrt{2^2 + 3^2}} = \frac{|p + 6|}{\sqrt{13}} < 2 \Rightarrow |p + 6| < 2\sqrt{13}$.
For $L_2: 2x + 3y + p + 3 = 0$ to be a diameter,it must pass through the center $(-3, 5)$,so $2(-3) + 3(5) + p + 3 = 0 \Rightarrow p = -12$.
If $p = -12$,$L_1$ becomes $2x + 3y - 15 = 0$. The distance $d_1 = \frac{|-12 + 6|}{\sqrt{13}} = \frac{6}{\sqrt{13}} \approx 1.66 < 2$,so $L_1$ is a chord. Thus,$L_2$ can be a diameter,but it is not always one. So $STATEMENT-1$ is True.
For $STATEMENT-2$: If $L_1$ is a diameter,$2(-3) + 3(5) + p - 3 = 0 \Rightarrow p = -6$. Then $L_2$ is $2x + 3y - 3 = 0$. The distance from the center to $L_2$ is $d_2 = \frac{|2(-3) + 3(5) - 3|}{\sqrt{13}} = \frac{6}{\sqrt{13}} \approx 1.66 < 2$. Since $d_2 < r$,$L_2$ is a chord. Thus,$STATEMENT-2$ is False.

(B) For the first equation $x^2+2px+q=0$,the roots are $\alpha, \beta$. Thus,$\alpha+\beta = -2p$ and $\alpha\beta = q$. The discriminant is $D_1 = 4p^2-4q = 4(p^2-q)$.
For the second equation $ax^2+2bx+c=0$,the roots are $\alpha, \frac{1}{\beta}$. Thus,$\alpha+\frac{1}{\beta} = -\frac{2b}{a}$ and $\alpha\cdot\frac{1}{\beta} = \frac{c}{a}$. The discriminant is $D_2 = 4b^2-4ac = 4(b^2-ac)$.
Since $\alpha$ is a common root,we have $\alpha^2+2p\alpha+q=0$ and $a\alpha^2+2b\alpha+c=0$. Eliminating $\alpha^2$,we get $(2ap-2b)\alpha + (aq-c) = 0$. If $b=ap$ and $c=aq$,then the equations are identical,which contradicts $\beta^2 \neq 1$. Thus $b \neq ap$ or $c \neq aq$.
Since $\alpha$ is real,$p^2-q \geq 0$ and $b^2-ac \geq 0$ (as $\alpha$ is a root of both),their product is $\geq 0$. Thus $STATEMENT-1$ is True.
$STATEMENT-2$ is also True as shown by the contradiction of identical equations,but it is not the direct explanation for the inequality product. Hence,option $B$ is correct.

(C) Let the $G.P.$ be $a, ar, ar^2, ar^3$ where $r \neq 1$ and $r > 0$.
Then $b_1 = a, b_2 = a(1+r), b_3 = a(1+r+r^2), b_4 = a(1+r+r^2+r^3)$.
For $b_1, b_2, b_3, b_4$ to be in $A.P.$,$b_2-b_1 = b_3-b_2 = b_4-b_3$ must hold.
$b_2-b_1 = ar$,$b_3-b_2 = ar^2$,$b_4-b_3 = ar^3$.
Since $r \neq 1$ and $r > 0$,$ar \neq ar^2 \neq ar^3$,so they are not in $A.P.$
For $G.P.$,$\frac{b_2}{b_1} = 1+r$ and $\frac{b_3}{b_2} = \frac{1+r+r^2}{1+r}$. These are not equal for $r \neq 0, 1$.
For $H.P.$,the reciprocals must be in $A.P.$,which is clearly not true.
Thus,$STATEMENT-1$ is True and $STATEMENT-2$ is False.

(B) The lines are $L_1: x+3y-5=0$,$L_2: 3x-ky-1=0$,and $L_3: 5x+2y-12=0$.
$(A)$ For concurrency,the lines must intersect at a single point. Solving $L_1$ and $L_3$: $x+3y=5$ and $5x+2y=12$. Multiplying $L_1$ by $5$: $5x+15y=25$. Subtracting $L_3$: $13y=13 \Rightarrow y=1$. Then $x=2$. The point of intersection is $(2, 1)$. Substituting into $L_2$: $3(2)-k(1)-1=0 \Rightarrow 6-k-1=0 \Rightarrow k=5$. Thus,$(A) \rightarrow (s)$.
$(B)$ For lines to be parallel:
$L_1 \parallel L_2: \frac{1}{3} = \frac{3}{-k} \Rightarrow k=-9$.
$L_2 \parallel L_3: \frac{3}{5} = \frac{-k}{2} \Rightarrow k=-\frac{6}{5}$.
$L_1 \parallel L_3$ is not possible as slopes are $-1/3$ and $-5/2$. Thus,$(B) \rightarrow (p, q)$.
$(C)$ Lines form a triangle if they are not concurrent and no two lines are parallel. This happens when $k \neq 5, -9, -\frac{6}{5}$. Among the options,only $k=\frac{5}{6}$ satisfies this. Thus,$(C) \rightarrow (r)$.
$(D)$ Lines do not form a triangle if they are concurrent or parallel. This happens when $k=5, -9, -\frac{6}{5}$. Thus,$(D) \rightarrow (p, q, s)$.

(A) The word $ENDEANOEL$ has $9$ letters: $E, E, E, N, N, D, A, O, L$.
$(A)$ Treating $ENDEA$ as a single block,we have the block $(ENDEA)$ and the remaining letters $N, O, E, L$. Total items $= 5$. The number of permutations is $5! = 120$. This matches $(p)$.
$(B)$ Total letters $= 9$. $E$ occurs $3$ times. If $E$ is at the first and last position,we have $7$ positions left to fill with $E, N, N, D, A, O, L$. The number of ways is $\frac{7!}{2!} = \frac{5040}{2} = 2520 = 21 \times 120 = 21 \times 5!$. This matches $(s)$.
$(C)$ None of $D, L, N$ in the last $5$ positions means $D, L, N$ must be in the first $4$ positions. The letters are $E, E, E, N, N, D, A, O, L$. The first $4$ positions must contain $D, L, N$ and one $E$. Ways $= \frac{4!}{2!} = 12$. The last $5$ positions must contain the remaining $E, E, A, O$. Ways $= \frac{5!}{3!} = 20$. Total $= 12 \times 20 = 240 = 2 \times 120 = 2 \times 5!$. This matches $(q)$.
$(D)$ $A, E, O$ occur only in odd positions $(1, 3, 5, 7, 9)$. There are $5$ odd positions. We have $3$ $E$'s,$1$ $A$,$1$ $O$. Total $5$ letters. Ways to arrange these in $5$ odd positions $= \frac{5!}{3!} = 20$. The remaining $4$ letters $(N, N, D, L)$ in $4$ even positions $= \frac{4!}{2!} = 12$. Total $= 20 \times 12 = 240 = 2 \times 5!$. This matches $(q)$.

(C) To find the local maxima and local minima,we analyze the function $f(x)$ in its given intervals:
$1$. For $-3 < x \leq -1$,$f(x) = (2+x)^3$. The derivative is $f'(x) = 3(2+x)^2$. Since $f'(x) \geq 0$,the function is increasing on $(-3, -1]$. At $x = -1$,$f(-1) = (2-1)^3 = 1$. Since the function is increasing up to $x = -1$,this point acts as a local maximum.
$2$. For $-1 < x < 2$,$f(x) = x^{2/3}$. The derivative is $f'(x) = \frac{2}{3}x^{-1/3} = \frac{2}{3\sqrt[3]{x}}$.
$3$. At $x = 0$,$f'(x)$ is undefined. For $x < 0$,$f'(x) < 0$ (decreasing),and for $x > 0$,$f'(x) > 0$ (increasing). Thus,$x = 0$ is a local minimum with $f(0) = 0$.
$4$. Comparing the values,we have a local maximum at $x = -1$ and a local minimum at $x = 0$.
Therefore,the total number of local maxima and local minima is $2$.

(A) The volume of a parallelepiped defined by vectors $\hat{a}, \hat{b}, \hat{c}$ is given by the scalar triple product $|\hat{a} \cdot (\hat{b} \times \hat{c})|$.
This is equal to the square root of the determinant of the Gram matrix:
Volume $= \sqrt{\det \begin{bmatrix} \hat{a} \cdot \hat{a} & \hat{a} \cdot \hat{b} & \hat{a} \cdot \hat{c} \\ \hat{b} \cdot \hat{a} & \hat{b} \cdot \hat{b} & \hat{b} \cdot \hat{c} \\ \hat{c} \cdot \hat{a} & \hat{c} \cdot \hat{b} & \hat{c} \cdot \hat{c} \end{bmatrix}}$.
Given $\hat{a} \cdot \hat{a} = \hat{b} \cdot \hat{b} = \hat{c} \cdot \hat{c} = 1$ and $\hat{a} \cdot \hat{b} = \hat{b} \cdot \hat{c} = \hat{c} \cdot \hat{a} = 1/2$,we have:
Volume $= \sqrt{\det \begin{bmatrix} 1 & 1/2 & 1/2 \\ 1/2 & 1 & 1/2 \\ 1/2 & 1/2 & 1 \end{bmatrix}}$.
Calculating the determinant: $1(1 - 1/4) - 1/2(1/2 - 1/4) + 1/2(1/4 - 1/2) = 1(3/4) - 1/2(1/4) + 1/2(-1/4) = 3/4 - 1/8 - 1/8 = 3/4 - 2/8 = 3/4 - 1/4 = 1/2$.
Thus,the volume is $\sqrt{1/2} = \frac{1}{\sqrt{2}}$.

(B) Given $f(x) = f(1-x)$. Differentiating with respect to $x$,we get $f^{\prime}(x) = -f^{\prime}(1-x)$.
At $x = 1/2$,$f^{\prime}(1/2) = -f^{\prime}(1-1/2) = -f^{\prime}(1/2)$,which implies $2f^{\prime}(1/2) = 0$,so $f^{\prime}(1/2) = 0$. Thus,$(B)$ is correct.
Since $f(x) = f(1-x)$,the graph of $f(x)$ is symmetric about $x = 1/2$. Let $x = 1/2 + t$,then $f(1/2 + t) = f(1/2 - t)$,so $g(t) = f(1/2 + t)$ is an even function.
For $(C)$,$\int_{-1/2}^{1/2} f(x+1/2) \sin x dx$. Since $f(x+1/2)$ is even and $\sin x$ is odd,their product is an odd function. The integral of an odd function over a symmetric interval $[-a, a]$ is $0$. Thus,$(C)$ is correct.
For $(A)$,we have $f^{\prime}(1/4) = 0$. Since $f^{\prime}(x) = -f^{\prime}(1-x)$,$f^{\prime}(3/4) = -f^{\prime}(1/4) = 0$. Also $f^{\prime}(1/2) = 0$. By Rolle's Theorem,$f^{\prime\prime}(x)$ vanishes at least once in $(1/4, 1/2)$ and at least once in $(1/2, 3/4)$. Thus,$f^{\prime\prime}(x)$ vanishes at least twice in $[0, 1]$. Thus,$(A)$ is correct.
For $(D)$,let $I = \int_{1/2}^1 f(1-t) e^{\sin \pi t} dt$. Let $1-t = u$,then $dt = -du$. When $t=1/2, u=1/2$; when $t=1, u=0$. $I = \int_{1/2}^0 f(u) e^{\sin \pi (1-u)} (-du) = \int_0^{1/2} f(u) e^{\sin \pi u} du$. Thus,$(D)$ is correct.

(A,D) We can rewrite the sums as Riemann sums:
$S_n = \sum_{k=1}^n \frac{1}{n} \cdot \frac{1}{1 + (k/n) + (k/n)^2}$
$T_n = \sum_{k=0}^{n-1} \frac{1}{n} \cdot \frac{1}{1 + (k/n) + (k/n)^2}$
Let $f(x) = \frac{1}{1+x+x^2}$. Since $f(x)$ is a strictly decreasing function on $[0, 1]$,the right Riemann sum $S_n$ is an underestimate of the integral,and the left Riemann sum $T_n$ is an overestimate.
Thus,$S_n < \int_0^1 \frac{dx}{1+x+x^2} < T_n$.
Evaluating the integral: $\int_0^1 \frac{dx}{(x+1/2)^2 + 3/4} = \left[ \frac{2}{\sqrt{3}} \tan^{-1} \left( \frac{2x+1}{\sqrt{3}} \right) \right]_0^1 = \frac{2}{\sqrt{3}} (\tan^{-1}(\sqrt{3}) - \tan^{-1}(1/\sqrt{3})) = \frac{2}{\sqrt{3}} (\pi/3 - \pi/6) = \frac{2}{\sqrt{3}} \cdot \frac{\pi}{6} = \frac{\pi}{3\sqrt{3}}$.
Therefore,$S_n < \frac{\pi}{3\sqrt{3}}$ and $T_n > \frac{\pi}{3\sqrt{3}}$.

(B) The system of equations is homogeneous: $ax+by=0$ and $cx+dy=0$.
$A$ homogeneous system always has the trivial solution $(x=0, y=0)$.
Therefore,the probability that the system has a solution is $1$,which makes $Statement-2$ True.
For a unique solution,the determinant of the coefficient matrix must be non-zero: $\left|\begin{array}{ll}a & b \\ c & d\end{array}\right| \neq 0$,which implies $ad - bc \neq 0$.
Since $a, b, c, d \in \{0, 1\}$,the total number of possible outcomes is $2^4 = 16$.
The condition $ad - bc \neq 0$ implies $ad \neq bc$.
Possible cases for $(ad, bc)$ are $(1, 0)$ or $(0, 1)$.
If $ad=1$,then $a=1$ and $d=1$. $bc$ must be $0$. $bc=0$ occurs if $(b,c) \in \{(0,0), (0,1), (1,0)\}$. This gives $3$ cases.
If $ad=0$,then $bc$ must be $1$,so $b=1$ and $c=1$. $ad=0$ occurs if $(a,d) \in \{(0,0), (0,1), (1,0)\}$. This gives $3$ cases.
Total favorable cases $= 3 + 3 = 6$.
Probability $= 6/16 = 3/8$. Thus,$Statement-1$ is True.
Since $Statement-2$ states that the system always has a solution (which is true for any homogeneous system),it does not explain why the probability of a unique solution is $3/8$. Therefore,$Statement-2$ is not the correct explanation for $Statement-1$.

(A) The system of equations is given by $AX = B$,where $A = \begin{bmatrix} 1 & -2 & 3 \\ -1 & 1 & -2 \\ 1 & -3 & 4 \end{bmatrix}$.
First,calculate the determinant of the coefficient matrix $D = |A| = 1(4-6) + 2(-4+2) + 3(3-1) = 1(-2) + 2(-2) + 3(2) = -2 - 4 + 6 = 0$.
Since $D = 0$,the system either has no solution or infinitely many solutions.
Now,calculate $D_1 = \left|\begin{array}{ccc}-1 & -2 & 3 \\ k & 1 & -2 \\ 1 & -3 & 4\end{array}\right| = -1(4-6) + 2(4k+2) + 3(3k-1) = 2 + 8k + 4 + 9k - 3 = 17k + 3$.
Wait,let's re-evaluate $D_1$: $D_1 = -1(4-6) + 2(4k+2) + 3(3k-1) = 2 + 8k + 4 + 9k - 3 = 17k + 3$. Actually,let's use row operations: $R_3 \to R_3 - R_1$ and $R_2 \to R_2 + R_1$: $\left|\begin{array}{ccc}1 & -2 & 3 \\ 0 & -1 & 1 \\ 0 & -1 & 1\end{array}\right| = 0$.
For the system to have no solution,at least one of $D_1, D_2, D_3$ must be non-zero.
$D_1 = \left|\begin{array}{ccc}-1 & -2 & 3 \\ k & 1 & -2 \\ 1 & -3 & 4\end{array}\right| = -1(4-6) + 2(4k+2) + 3(-3k-1) = 2 + 8k + 4 - 9k - 3 = 3 - k$.
$D_1 = 0$ only if $k = 3$. If $k \neq 3$,$D_1 \neq 0$,so the system has no solution.
Thus,Statement-$1$ is True. Statement-$2$ is True as it correctly identifies $D=0$ which is the condition for the system to have no solution or infinite solutions.

(B) Given $f(x) = g(x) \sin x$.
First,find $f^{\prime}(x)$ using the product rule:
$f^{\prime}(x) = g^{\prime}(x) \sin x + g(x) \cos x$.
At $x = 0$,$f^{\prime}(0) = g^{\prime}(0) \sin(0) + g(0) \cos(0) = 0 + g(0) \cdot 1 = g(0)$.
Thus,$Statement-2$ is True.
Next,find $f^{\prime \prime}(x)$:
$f^{\prime \prime}(x) = g^{\prime \prime}(x) \sin x + g^{\prime}(x) \cos x + g^{\prime}(x) \cos x - g(x) \sin x = g^{\prime \prime}(x) \sin x + 2g^{\prime}(x) \cos x - g(x) \sin x$.
At $x = 0$,$f^{\prime \prime}(0) = g^{\prime \prime}(0) \cdot 0 + 2g^{\prime}(0) \cdot 1 - g(0) \cdot 0 = 2g^{\prime}(0) = 0$ (since $g^{\prime}(0) = 0$).
Now,evaluate the limit in $Statement-1$:
$L = \lim_{x \rightarrow 0} [g(x) \cot x - g(0) \operatorname{cosec} x] = \lim_{x \rightarrow 0} \frac{g(x) \cos x - g(0)}{\sin x}$.
Using $L$'$H$ôpital's rule (since it is a $0/0$ form):
$L = \lim_{x \rightarrow 0} \frac{g^{\prime}(x) \cos x - g(x) \sin x}{\cos x} = \frac{g^{\prime}(0) \cdot 1 - g(0) \cdot 0}{1} = g^{\prime}(0) = 0$.
Since $f^{\prime \prime}(0) = 0$ and $L = 0$,$Statement-1$ is True.
However,$Statement-2$ $(f^{\prime}(0) = g(0))$ is not used to derive the limit in $Statement-1$. Thus,$Statement-2$ is not the correct explanation for $Statement-1$.

(D) The normal vectors to the planes are $\vec{n}_1 = (1, -1, 1)$,$\vec{n}_2 = (1, 1, -1)$,and $\vec{n}_3 = (1, -3, 3)$.
The direction vector of line $L_1$ (intersection of $P_2$ and $P_3$) is $\vec{v}_1 = \vec{n}_2 \times \vec{n}_3 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -1 \\ 1 & -3 & 3 \end{vmatrix} = (0, -4, -4)$,which is parallel to $(0, 1, 1)$.
The direction vector of line $L_2$ (intersection of $P_3$ and $P_1$) is $\vec{v}_2 = \vec{n}_3 \times \vec{n}_1 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 3 \\ 1 & -1 & 1 \end{vmatrix} = (0, 2, 2)$,which is parallel to $(0, 1, 1)$.
The direction vector of line $L_3$ (intersection of $P_1$ and $P_2$) is $\vec{v}_3 = \vec{n}_1 \times \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{vmatrix} = (0, 2, 2)$,which is parallel to $(0, 1, 1)$.
Since all lines $L_1, L_2, L_3$ have the same direction vector $(0, 1, 1)$,they are all parallel to each other. Thus,$STATEMENT-1$ is False.
For $STATEMENT-2$,we check if the system has a common point by solving the equations. Adding $P_1$ and $P_2$ gives $2x = 0$,so $x = 0$. Substituting $x=0$ into $P_1$ and $P_2$ gives $-y+z=1$ and $y-z=-1$,which are identical. Substituting $x=0$ into $P_3$ gives $-3y+3z=2$,or $-y+z=2/3$. Since $1 \neq 2/3$,the system has no common point. Thus,$STATEMENT-2$ is True.

(A, D, B) $1.$ Differentiating $y^3-3y+x=0$ with respect to $x$,we get $3y^2y^{\prime}-3y^{\prime}+1=0$,which implies $y^{\prime} = \frac{-1}{3(y^2-1)}$.
At $x = -10\sqrt{2}$,$y = 2\sqrt{2}$,so $y^{\prime} = \frac{-1}{3((2\sqrt{2})^2-1)} = \frac{-1}{3(8-1)} = -\frac{1}{21}$.
Differentiating $3y^2y^{\prime}-3y^{\prime}+1=0$ again,we get $6yy^{\prime 2} + 3y^2y^{\prime\prime} - 3y^{\prime\prime} = 0$.
$y^{\prime\prime}(3y^2-3) = -6yy^{\prime 2} \Rightarrow y^{\prime\prime} = \frac{-2yy^{\prime 2}}{y^2-1}$.
Substituting $y=2\sqrt{2}$ and $y^{\prime}=-\frac{1}{21}$,we get $f^{\prime\prime}(-10\sqrt{2}) = \frac{-2(2\sqrt{2})(-1/21)^2}{8-1} = \frac{-4\sqrt{2}}{7 \times 441} = -\frac{4\sqrt{2}}{7^3 \times 3^2}$.
$2.$ The area is $\int_a^b |f(x)| dx$. Since $f(x) < -2$ for $x < -2$,the area is $-\int_a^b f(x) dx$.
Using integration by parts: $\int f(x) dx = xf(x) - \int xf^{\prime}(x) dx$.
Since $f^{\prime}(x) = \frac{-1}{3(f(x)^2-1)}$,the integral is $bf(b)-af(a) - \int_a^b x \left(\frac{-1}{3(f(x)^2-1)}\right) dx = bf(b)-af(a) + \int_a^b \frac{x}{3(f(x)^2-1)} dx$.
The area is $-\int_a^b f(x) dx = -bf(b)+af(a) - \int_a^b \frac{x}{3(f(x)^2-1)} dx$.
$3.$ $\int_{-1}^1 g^{\prime}(x) dx = g(1) - g(-1)$.
Since $g(x)^3 - 3g(x) + x = 0$,$g(-x)^3 - 3g(-x) - x = 0$. Let $h(x) = -g(-x)$,then $(-h(x))^3 - 3(-h(x)) - x = 0 \Rightarrow -h(x)^3 + 3h(x) - x = 0 \Rightarrow h(x)^3 - 3h(x) + x = 0$.
Thus $g(x) = -g(-x)$,so $g$ is an odd function. $g(-1) = -g(1)$.
Therefore,$g(1) - g(-1) = g(1) - (-g(1)) = 2g(1)$.

(D) Let $n(S) = 10$ be the total number of outcomes. Given $n(A) = 4$,so $P(A) = \frac{4}{10} = \frac{2}{5}$.
Let $n(B) = p$,so $P(B) = \frac{p}{10}$.
For $A$ and $B$ to be independent,$P(A \cap B) = P(A) \times P(B) = \frac{2}{5} \times \frac{p}{10} = \frac{2p}{50} = \frac{p}{25}$.
Since $P(A \cap B) = \frac{n(A \cap B)}{10}$,we have $\frac{n(A \cap B)}{10} = \frac{p}{25}$,which implies $n(A \cap B) = \frac{10p}{25} = \frac{2p}{5}$.
Since $n(A \cap B)$ must be an integer,$2p$ must be divisible by $5$. Since $2$ and $5$ are coprime,$p$ must be a multiple of $5$.
Given $B$ is non-empty,$p \in \{5, 10\}$.

(B) The area $A$ is given by $\int_0^{\pi/4} \left( \sqrt{\frac{1+\sin x}{\cos x}} - \sqrt{\frac{1-\sin x}{\cos x}} \right) dx$.
Using the half-angle identities $\sin x = \frac{2\tan(x/2)}{1+\tan^2(x/2)}$ and $\cos x = \frac{1-\tan^2(x/2)}{1+\tan^2(x/2)}$,we simplify the integrand.
Let $t = \tan(x/2)$,then $dt = \frac{1}{2} \sec^2(x/2) dx = \frac{1}{2}(1+t^2) dx$,so $dx = \frac{2}{1+t^2} dt$.
At $x=0$,$t=0$. At $x=\pi/4$,$t=\tan(\pi/8) = \sqrt{2}-1$.
The expression becomes $\int_0^{\sqrt{2}-1} \left( \sqrt{\frac{1+\frac{2t}{1+t^2}}{\frac{1-t^2}{1+t^2}}} - \sqrt{\frac{1-\frac{2t}{1+t^2}}{\frac{1-t^2}{1+t^2}}} \right) \frac{2}{1+t^2} dt$.
This simplifies to $\int_0^{\sqrt{2}-1} \left( \sqrt{\frac{(1+t)^2}{1-t^2}} - \sqrt{\frac{(1-t)^2}{1-t^2}} \right) \frac{2}{1+t^2} dt = \int_0^{\sqrt{2}-1} \frac{2(1+t - (1-t))}{\sqrt{1-t^2}} \frac{1}{1+t^2} dt = \int_0^{\sqrt{2}-1} \frac{4t}{(1+t^2)\sqrt{1-t^2}} dt$.

(C) Given $J = \int \frac{e^{-x}}{e^{-4x} + e^{-2x} + 1} dx$. Multiplying numerator and denominator by $e^{4x}$,we get $J = \int \frac{e^{3x}}{1 + e^{2x} + e^{4x}} dx$.
Now,$J - I = \int \frac{e^{3x} - e^x}{e^{4x} + e^{2x} + 1} dx = \int \frac{e^x(e^{2x} - 1)}{e^{4x} + e^{2x} + 1} dx$.
Let $z = e^x$,then $dz = e^x dx$. The integral becomes $\int \frac{z^2 - 1}{z^4 + z^2 + 1} dz$.
Divide numerator and denominator by $z^2$: $\int \frac{1 - 1/z^2}{z^2 + 1 + 1/z^2} dz = \int \frac{1 - 1/z^2}{(z + 1/z)^2 - 1} dz$.
Let $u = z + 1/z$,then $du = (1 - 1/z^2) dz$. The integral becomes $\int \frac{du}{u^2 - 1} = \frac{1}{2} \ln \left| \frac{u - 1}{u + 1} \right| + C$.
Substituting $u = e^x + e^{-x}$ back: $\frac{1}{2} \ln \left| \frac{e^x + e^{-x} - 1}{e^x + e^{-x} + 1} \right| + C = \frac{1}{2} \ln \left( \frac{e^{2x} - e^x + 1}{e^{2x} + e^x + 1} \right) + C$.

(A) Given $g(x) = \log(f(x))$ and $f(x+1) = x f(x)$.
Taking logarithm on both sides,$\log(f(x+1)) = \log(x) + \log(f(x))$.
This implies $g(x+1) = g(x) + \log(x)$,or $g(x+1) - g(x) = \log(x)$.
Differentiating twice with respect to $x$,we get $g^{\prime \prime}(x+1) - g^{\prime \prime}(x) = -\frac{1}{x^2}$.
We want to find $g^{\prime \prime}\left(N+\frac{1}{2}\right) - g^{\prime \prime}\left(\frac{1}{2}\right)$.
We can write this as a telescoping sum:
$g^{\prime \prime}\left(N+\frac{1}{2}\right) - g^{\prime \prime}\left(\frac{1}{2}\right) = \sum_{k=1}^{N} \left[ g^{\prime \prime}\left(k+\frac{1}{2}\right) - g^{\prime \prime}\left(k-\frac{1}{2}\right) \right]$.
Using the relation $g^{\prime \prime}(x+1) - g^{\prime \prime}(x) = -\frac{1}{x^2}$,we substitute $x = k - \frac{1}{2}$:
$g^{\prime \prime}\left(k+\frac{1}{2}\right) - g^{\prime \prime}\left(k-\frac{1}{2}\right) = -\frac{1}{(k-\frac{1}{2})^2} = -\frac{1}{(\frac{2k-1}{2})^2} = -\frac{4}{(2k-1)^2}$.
Summing from $k=1$ to $N$:
$\sum_{k=1}^{N} -\frac{4}{(2k-1)^2} = -4 \left[ \frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \ldots + \frac{1}{(2N-1)^2} \right]$.
This simplifies to $-4 \left\{ 1 + \frac{1}{9} + \frac{1}{25} + \ldots + \frac{1}{(2N-1)^2} \right\}$.

(A) The position vector is $\overline{OP} = \hat{a} \cos t + \hat{b} \sin t$.
The square of the length is $|\overline{OP}|^2 = (\hat{a} \cos t + \hat{b} \sin t) \cdot (\hat{a} \cos t + \hat{b} \sin t) = \cos^2 t + \sin^2 t + 2 \sin t \cos t (\hat{a} \cdot \hat{b}) = 1 + \sin(2t) (\hat{a} \cdot \hat{b})$.
Since $\hat{a}$ and $\hat{b}$ form an acute angle,$\hat{a} \cdot \hat{b} > 0$. Thus,$|\overline{OP}|$ is maximum when $\sin(2t) = 1$,i.e.,$2t = \frac{\pi}{2} \Rightarrow t = \frac{\pi}{4}$.
At $t = \frac{\pi}{4}$,$M = |\overline{OP}| = (1 + \hat{a} \cdot \hat{b})^{1/2}$.
The vector $\overline{OP}$ at $t = \frac{\pi}{4}$ is $\frac{1}{\sqrt{2}}(\hat{a} + \hat{b})$.
The unit vector $\hat{u}$ is $\frac{\overline{OP}}{|\overline{OP}|} = \frac{\frac{1}{\sqrt{2}}(\hat{a} + \hat{b})}{\frac{1}{\sqrt{2}}|\hat{a} + \hat{b}|} = \frac{\hat{a} + \hat{b}}{|\hat{a} + \hat{b}|}$.

(C) Given the function $g(u) = 2 \tan^{-1}(e^u) - \frac{\pi}{2}$.
To check for odd/even,we evaluate $g(-u)$:
$g(-u) = 2 \tan^{-1}(e^{-u}) - \frac{\pi}{2}$.
Using the identity $\tan^{-1}(x) + \cot^{-1}(x) = \frac{\pi}{2}$,we know $\tan^{-1}(e^{-u}) = \frac{\pi}{2} - \tan^{-1}(e^u)$.
Substituting this into $g(-u)$:
$g(-u) = 2 \left( \frac{\pi}{2} - \tan^{-1}(e^u) \right) - \frac{\pi}{2} = \pi - 2 \tan^{-1}(e^u) - \frac{\pi}{2} = \frac{\pi}{2} - 2 \tan^{-1}(e^u) = -g(u)$.
Since $g(-u) = -g(u)$,the function is odd.
To check for monotonicity,we find the derivative $g'(u)$:
$g'(u) = \frac{d}{du} (2 \tan^{-1}(e^u) - \frac{\pi}{2}) = 2 \cdot \frac{1}{1 + (e^u)^2} \cdot e^u = \frac{2e^u}{1 + e^{2u}}$.
Since $e^u > 0$ for all $u \in (-\infty, \infty)$,$g'(u) > 0$ for all $u$.
Therefore,$g$ is strictly increasing in $(-\infty, \infty)$.

(C) Given differential equation: $x \sqrt{x^2-1} dy = y \sqrt{y^2-1} dx$.
Separating variables: $\int \frac{dy}{y \sqrt{y^2-1}} = \int \frac{dx}{x \sqrt{x^2-1}}$.
Integrating both sides: $\sec^{-1} y = \sec^{-1} x + C$.
Using the condition $y(2) = \frac{2}{\sqrt{3}}$: $\sec^{-1} \left(\frac{2}{\sqrt{3}}\right) = \sec^{-1} (2) + C$.
$\frac{\pi}{6} = \frac{\pi}{3} + C \implies C = -\frac{\pi}{6}$.
Thus,$\sec^{-1} y = \sec^{-1} x - \frac{\pi}{6}$,which gives $y(x) = \sec \left(\sec^{-1} x - \frac{\pi}{6}\right)$. So,$STATEMENT-1$ is True.
Now,$\cos^{-1} \left(\frac{1}{y}\right) = \cos^{-1} \left(\frac{1}{x}\right) - \frac{\pi}{6}$.
Taking $\cos$ on both sides: $\frac{1}{y} = \cos \left(\cos^{-1} \frac{1}{x} - \frac{\pi}{6}\right) = \cos \left(\cos^{-1} \frac{1}{x}\right) \cos \left(\frac{\pi}{6}\right) + \sin \left(\cos^{-1} \frac{1}{x}\right) \sin \left(\frac{\pi}{6}\right)$.
$\frac{1}{y} = \frac{1}{x} \cdot \frac{\sqrt{3}}{2} + \sqrt{1 - \frac{1}{x^2}} \cdot \frac{1}{2} = \frac{\sqrt{3}}{2x} + \frac{1}{2} \sqrt{1 - \frac{1}{x^2}}$.
Comparing this with $STATEMENT-2$,we see $STATEMENT-2$ is False.

(A) $1.$ Given $f(x) = \frac{x^2-ax+1}{x^2+ax+1}$. Using quotient rule,$f^{\prime}(x) = \frac{(2x-a)(x^2+ax+1) - (x^2-ax+1)(2x+a)}{(x^2+ax+1)^2} = \frac{2ax^2-2a}{(x^2+ax+1)^2} = \frac{2a(x^2-1)}{(x^2+ax+1)^2}$.
Calculating $f^{\prime \prime}(x)$,we find $f^{\prime \prime}(1) = \frac{4a}{(2+a)^2}$ and $f^{\prime \prime}(-1) = \frac{-4a}{(2-a)^2}$.
Thus,$(2+a)^2 f^{\prime \prime}(1) + (2-a)^2 f^{\prime \prime}(-1) = 4a - 4a = 0$. So,$(A)$ is true.
$2.$ $f^{\prime}(x) = \frac{2a(x^2-1)}{(x^2+ax+1)^2}$. For $x \in (-1, 1)$,$x^2-1 < 0$,so $f^{\prime}(x) < 0$. Thus $f(x)$ is decreasing on $(-1, 1)$.
At $x=1$,$f^{\prime}(x)$ changes from negative to positive,so $f(x)$ has a local minimum at $x=1$. So,$(A)$ is true.
$3.$ $g(x) = \int_0^{e^x} \frac{f^{\prime}(t)}{1+t^2} dt$. By Leibniz rule,$g^{\prime}(x) = \frac{f^{\prime}(e^x)}{1+(e^x)^2} \cdot e^x = \frac{2a(e^{2x}-1)}{(e^{2x}+ae^x+1)^2} \cdot \frac{e^x}{1+e^{2x}}$.
For $x < 0$,$e^x < 1$,so $e^{2x}-1 < 0$,hence $g^{\prime}(x) < 0$.
For $x > 0$,$e^x > 1$,so $e^{2x}-1 > 0$,hence $g^{\prime}(x) > 0$.
Thus,$g^{\prime}(x)$ is negative on $(-\infty, 0)$ and positive on $(0, \infty)$. So,$(B)$ is true.

(B, D, C) $1.$ The direction vectors of $L_1$ and $L_2$ are $\vec{v_1} = 3\hat{i} + \hat{j} + 2\hat{k}$ and $\vec{v_2} = \hat{i} + 2\hat{j} + 3\hat{k}$.
The vector perpendicular to both is $\vec{n} = \vec{v_1} \times \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 2 \\ 1 & 2 & 3 \end{vmatrix} = \hat{i}(3-4) - \hat{j}(9-2) + \hat{k}(6-1) = -\hat{i} - 7\hat{j} + 5\hat{k}$.
The magnitude is $|\vec{n}| = \sqrt{(-1)^2 + (-7)^2 + 5^2} = \sqrt{1 + 49 + 25} = \sqrt{75} = 5\sqrt{3}$.
The unit vector is $\frac{-\hat{i} - 7\hat{j} + 5\hat{k}}{5\sqrt{3}}$. Thus,option $(B)$ is correct.
$2.$ The shortest distance between lines passing through $A(-1, -2, -1)$ and $B(2, -2, 3)$ with directions $\vec{v_1}, \vec{v_2}$ is $d = \frac{|(\vec{AB}) \cdot (\vec{v_1} \times \vec{v_2})|}{|\vec{v_1} \times \vec{v_2}|}$.
$\vec{AB} = (2 - (-1))\hat{i} + (-2 - (-2))\hat{j} + (3 - (-1))\hat{k} = 3\hat{i} + 0\hat{j} + 4\hat{k}$.
$d = \frac{|(3\hat{i} + 0\hat{j} + 4\hat{k}) \cdot (-\hat{i} - 7\hat{j} + 5\hat{k})|}{5\sqrt{3}} = \frac{|-3 + 0 + 20|}{5\sqrt{3}} = \frac{17}{5\sqrt{3}}$. Thus,option $(D)$ is correct.
$3.$ The plane passes through $(-1, -2, -1)$ with normal $\vec{n} = -\hat{i} - 7\hat{j} + 5\hat{k}$.
Equation: $-1(x+1) - 7(y+2) + 5(z+1) = 0 \Rightarrow -x - 1 - 7y - 14 + 5z + 5 = 0 \Rightarrow x + 7y - 5z + 10 = 0$.
The distance from $(1, 1, 1)$ is $d = \frac{|1 + 7(1) - 5(1) + 10|}{\sqrt{1^2 + 7^2 + (-5)^2}} = \frac{|1 + 7 - 5 + 10|}{\sqrt{1 + 49 + 25}} = \frac{13}{\sqrt{75}}$. Thus,option $(C)$ is correct.

(C) Let $y = \frac{x^2+2x+4}{x+2} = \frac{x(x+2)+4}{x+2} = x + \frac{4}{x+2} = (x+2) + \frac{4}{x+2} - 2$. By $AM$-$GM$ inequality,for $x > -2$,$(x+2) + \frac{4}{x+2} \geq 2\sqrt{(x+2) \cdot \frac{4}{x+2}} = 4$. Thus,$y \geq 4 - 2 = 2$. The minimum value is $2$ (Option $r$).
$(B)$ Given $(A+B)(A-B) = (A-B)(A+B) \Rightarrow A^2 - AB + BA - B^2 = A^2 + AB - BA - B^2 \Rightarrow 2BA = 2AB \Rightarrow AB = BA$. Since $A^t = A$ and $B^t = -B$,$(AB)^t = B^t A^t = -BA = -AB$. Thus,$(-1)^k AB = -AB$,which implies $(-1)^k = -1$. This holds for odd integers $k$. In the set ${1, 2, 3}$,$k$ can be $1$ or $3$ (Options $q, s$).
$(C)$ $a = \log_3 \log_3 2$. Note $3^{-a} = 3^{-\log_3 \log_3 2} = (\log_3 2)^{-1} = \log_2 3$. The inequality is $1 < 2^{-k + \log_2 3} < 2 \Rightarrow 1 < 2^{-k} \cdot 3 < 2 \Rightarrow \frac{1}{3} < 2^{-k} < \frac{2}{3} \Rightarrow \log_2(1/3) < -k < \log_2(2/3) \Rightarrow -\log_2 3 < -k < 1 - \log_2 3 \Rightarrow \log_2 3 - 1 < k < \log_2 3$. Since $1 < \log_2 3 < 2$,we have $0.58 < k < 1.58$. The integer $k=1$. $1$ is less than $2$ and $3$ (Options $r, s$).
$(D)$ $\sin \theta = \cos \phi \Rightarrow \cos(\frac{\pi}{2} - \theta) = \cos \phi \Rightarrow \frac{\pi}{2} - \theta = 2n\pi \pm \phi \Rightarrow \theta \pm \phi - \frac{\pi}{2} = -2n\pi$. Thus,$\frac{1}{\pi}(\theta \pm \phi - \frac{\pi}{2}) = -2n$,which are even integers. Among options,$0$ is even (Option $p$).

(C) Let $\cot ^{-1} x = \theta$,then $x = \cot \theta$.
Since $0 < x < 1$,we have $\frac{\pi}{4} < \theta < \frac{\pi}{2}$.
The expression is $\sqrt{1+x^2} [\{x \cos \theta + \sin \theta\}^2 - 1]^{\frac{1}{2}}$.
Substituting $x = \cot \theta = \frac{\cos \theta}{\sin \theta}$,we get:
$\sqrt{1+\cot^2 \theta} [\{\frac{\cos^2 \theta}{\sin \theta} + \sin \theta\}^2 - 1]^{\frac{1}{2}}$
$= \sqrt{\operatorname{cosec}^2 \theta} [\{\frac{\cos^2 \theta + \sin^2 \theta}{\sin \theta}\}^2 - 1]^{\frac{1}{2}}$
$= \operatorname{cosec} \theta [\{\frac{1}{\sin \theta}\}^2 - 1]^{\frac{1}{2}}$
$= \operatorname{cosec} \theta \sqrt{\operatorname{cosec}^2 \theta - 1}$
$= \operatorname{cosec} \theta \sqrt{\cot^2 \theta}$
$= \operatorname{cosec} \theta \cdot \cot \theta$ (since $\cot \theta > 0$ for $0 < \theta < \frac{\pi}{2}$)
$= \frac{1}{\sin \theta} \cdot \frac{\cos \theta}{\sin \theta} = \frac{\cos \theta}{\sin^2 \theta} = \frac{\cos \theta}{1 - \cos^2 \theta}$.
Alternatively,using $\sqrt{1+x^2} = \sqrt{1+\cot^2 \theta} = \operatorname{cosec} \theta$:
Expression $= \operatorname{cosec} \theta \sqrt{(\cot \theta \cos \theta + \sin \theta)^2 - 1}$
$= \operatorname{cosec} \theta \sqrt{(\frac{\cos^2 \theta}{\sin \theta} + \sin \theta)^2 - 1}$
$= \operatorname{cosec} \theta \sqrt{(\frac{1}{\sin \theta})^2 - 1} = \operatorname{cosec} \theta \sqrt{\operatorname{cosec}^2 \theta - 1} = \operatorname{cosec} \theta \cot \theta$.
Since $\operatorname{cosec} \theta = \sqrt{1+x^2}$ and $\cot \theta = x$,the result is $x \sqrt{1+x^2}$.