IIT JEE 2019 Mathematics Question Paper with Answer and Solution

(B) The circle is $(x-3)^2+(y+2)^2=25$,so its center is $C(3, -2)$.
Let $R$ be the midpoint of the chord $PQ$. Since $R$ lies on the line $y=mx+1$,its coordinates are $(x_R, mx_R+1)$. Given $x_R = -\frac{3}{5}$,we have $y_R = m(-\frac{3}{5}) + 1 = \frac{-3m+5}{5}$.
Thus,$R = (-\frac{3}{5}, \frac{-3m+5}{5})$.
The line segment $CR$ is perpendicular to the chord $PQ$. The slope of $PQ$ is $m$,so the slope of $CR$ must be $-\frac{1}{m}$.
Slope of $CR = \frac{y_R - (-2)}{x_R - 3} = \frac{\frac{-3m+5}{5} + 2}{-\frac{3}{5} - 3} = \frac{-3m+5+10}{-3-15} = \frac{-3m+15}{-18} = \frac{m-5}{6}$.
Equating the slopes: $\frac{m-5}{6} = -\frac{1}{m}$.
$m(m-5) = -6 \Rightarrow m^2 - 5m + 6 = 0$.
$(m-2)(m-3) = 0$,so $m=2$ or $m=3$.
Both values $m=2$ and $m=3$ satisfy the condition $2 \leq m < 4$.

(B) The condition $|z-(2-i)| \geq \sqrt{5}$ represents the region outside or on the circle with center $C(2, -1)$ and radius $r = \sqrt{5}$.
To maximize $\frac{1}{|z-1|}$,we must minimize $|z-1|$,which is the distance of $z$ from the point $A(1, 0)$.
The point $z_0$ on the circle closest to $A(1, 0)$ lies on the line segment connecting $A(1, 0)$ and $C(2, -1)$.
The line $AC$ has the equation $y - 0 = \frac{-1-0}{2-1}(x-1)$,so $y = -x+1$,or $x+y-1=0$.
The intersection of this line with the circle $(x-2)^2 + (y+1)^2 = 5$ gives $z_0$. Since $z_0$ is on the circle and on the line $y = 1-x$,we have $(x-2)^2 + (1-x+1)^2 = 5 \implies (x-2)^2 + (2-x)^2 = 5 \implies 2(x-2)^2 = 5 \implies x-2 = \pm \sqrt{2.5}$.
Since $z_0$ is closer to $A(1, 0)$,$x_0 = 2 - \sqrt{2.5} \approx 0.42$,so $y_0 = 1 - x_0 = \sqrt{2.5} - 1 \approx 0.58$.
Let $z_0 = x_0 + iy_0$. Then $z_0 - \bar{z}_0 = 2iy_0$ and $z_0 + \bar{z}_0 = 2x_0$.
The expression becomes $\frac{4-2x_0}{2iy_0+2i} = \frac{2-x_0}{i(y_0+1)}$.
Since $y_0 = 1-x_0$,$y_0+1 = 2-x_0$.
Thus,the expression is $\frac{2-x_0}{i(2-x_0)} = \frac{1}{i} = -i$.
The principal argument of $-i$ is $-\frac{\pi}{2}$.

(D) For $E_1: \frac{x^2}{9} + \frac{y^2}{4} = 1$,let a point on the ellipse be $(3 \cos \theta, 2 \sin \theta)$. The area of the inscribed rectangle $R_1$ is $A_1 = (2 \cdot 3 \cos \theta)(2 \cdot 2 \sin \theta) = 24 \sin \theta \cos \theta = 12 \sin 2 \theta$. This is maximum when $\sin 2 \theta = 1$,i.e.,$\theta = \frac{\pi}{4}$.
Thus,the vertices of $R_1$ are $(\pm \frac{3}{\sqrt{2}}, \pm \frac{2}{\sqrt{2}})$.
For $E_n$ inscribed in $R_{n-1}$,the semi-axes $a_n, b_n$ satisfy $a_n = \frac{a_{n-1}}{\sqrt{2}}$ and $b_n = \frac{b_{n-1}}{\sqrt{2}}$.
Since the ratio $\frac{b_n}{a_n} = \frac{b_1}{a_1} = \frac{2}{3}$ is constant for all $n$,the eccentricity $e = \sqrt{1 - \frac{b_n^2}{a_n^2}} = \sqrt{1 - \frac{4}{9}} = \frac{\sqrt{5}}{3}$ is constant for all $E_n$. Thus,$(1)$ is incorrect.
For $E_9$,$a_9 = \frac{3}{(\sqrt{2})^8} = \frac{3}{16}$ and $b_9 = \frac{2}{(\sqrt{2})^8} = \frac{2}{16} = \frac{1}{8}$.
The distance of a focus from the centre is $a_9 e = \frac{3}{16} \cdot \frac{\sqrt{5}}{3} = \frac{\sqrt{5}}{16}$. Thus,$(2)$ is incorrect.
The length of the latus rectum of $E_9$ is $\frac{2 b_9^2}{a_9} = \frac{2 (1/8)^2}{3/16} = \frac{2/64}{3/16} = \frac{1}{32} \cdot \frac{16}{3} = \frac{1}{6}$. Thus,$(3)$ is correct.
The area of $R_n$ is $A_n = 4 a_n b_n = 4 \cdot \frac{3}{(\sqrt{2})^{n-1}} \cdot \frac{2}{(\sqrt{2})^{n-1}} = \frac{24}{2^{n-1}}$.
The sum $\sum_{n=1}^N A_n = 24 (1 + \frac{1}{2} + \dots + \frac{1}{2^{N-1}}) = 24 \cdot \frac{1 - (1/2)^N}{1 - 1/2} = 48 (1 - \frac{1}{2^N}) = 48 - \frac{48}{2^N} = 48 - \frac{3 \cdot 16}{2^N}$. This is always less than $48$,but the question asks for comparison with $24$. The sum is $24(2 - 2^{1-N}) < 48$. Actually,$\sum_{n=1}^N A_n < 48$. Checking the options,$(3)$ and $(4)$ are correct.

(A) Given $x^2-x-1=0$,roots are $\alpha, \beta$. Thus $\alpha+\beta=1$ and $\alpha\beta=-1$.
$a_n = \frac{\alpha^n-\beta^n}{\alpha-\beta}$.
For $(1)$: $a_{n+2}-a_{n+1} = \frac{\alpha^{n+2}-\beta^{n+2}-\alpha^{n+1}+\beta^{n+1}}{\alpha-\beta} = \frac{\alpha^{n+1}(\alpha-1)-\beta^{n+1}(\beta-1)}{\alpha-\beta}$. Since $\alpha^2-\alpha-1=0 \implies \alpha^2-1=\alpha$,we have $a_{n+2}-a_{n+1} = \frac{\alpha^{n+1}(\alpha^2-1)-\beta^{n+1}(\beta^2-1)}{\alpha-\beta}$ is incorrect logic; rather $a_{n+2} = a_{n+1}+a_n$. Thus $\sum_{i=1}^n a_i = a_{n+2}-a_2 = a_{n+2}-1$. Statement $(1)$ is correct.
For $(2)$: $\sum_{n=1}^{\infty} \frac{a_n}{10^n} = \frac{1}{\alpha-\beta} \left( \sum (\frac{\alpha}{10})^n - \sum (\frac{\beta}{10})^n \right) = \frac{1}{\alpha-\beta} \left( \frac{\alpha/10}{1-\alpha/10} - \frac{\beta/10}{1-\beta/10} \right) = \frac{1}{10-(\alpha+\beta)+(\alpha\beta/10)} = \frac{10}{100-10-1} = \frac{10}{89}$. Statement $(2)$ is correct.
For $(4)$: $b_n = a_{n-1}+a_{n+1} = \frac{\alpha^{n-1}-\beta^{n-1}+\alpha^{n+1}-\beta^{n+1}}{\alpha-\beta} = \frac{\alpha^{n-1}(1+\alpha^2)-\beta^{n-1}(1+\beta^2)}{\alpha-\beta}$. Since $1+\alpha^2 = \alpha^2-\alpha\beta = \alpha(\alpha-\beta)$,we get $b_n = \alpha^n+\beta^n$. Statement $(4)$ is correct.
Thus,options $1, 2, 4$ are correct.

(C) Using the sine rule, $\frac{p}{\sin P} = \frac{q}{\sin Q} = \frac{r}{\sin R} = 2R_{c} = 2(1) = 2$.
Given $p=\sqrt{3}, q=1$, we have $\sin P = \frac{\sqrt{3}}{2}$ and $\sin Q = \frac{1}{2}$.
Since $p > q$, $P > Q$. Possible values for $P$ are $60^{\circ}$ or $120^{\circ}$, and for $Q$ are $30^{\circ}$ or $150^{\circ}$.
If $P=60^{\circ}, Q=30^{\circ}$, then $R=90^{\circ}$ (rejected as it is non-right-angled).
If $P=120^{\circ}, Q=30^{\circ}$, then $R=30^{\circ}$. Thus, $\triangle PQR$ is isosceles with $q=r=1$.
Area $\Delta = \frac{1}{2}qr \sin P = \frac{1}{2}(1)(1)\sin 120^{\circ} = \frac{\sqrt{3}}{4}$.
Semi-perimeter $s = \frac{\sqrt{3}+1+1}{2} = \frac{\sqrt{3}+2}{2}$.
Inradius $r_{in} = \frac{\Delta}{s} = \frac{\sqrt{3}/4}{(\sqrt{3}+2)/2} = \frac{\sqrt{3}}{2(2+\sqrt{3})} = \frac{\sqrt{3}}{2}(2-\sqrt{3})$. (Option $2$ is correct).
Length of median $RS = \frac{1}{2}\sqrt{2p^2+2q^2-r^2} = \frac{1}{2}\sqrt{2(3)+2(1)-1} = \frac{\sqrt{7}}{2}$. (Option $3$ is correct).
$PE$ is the altitude to $QR$. $PE = q \sin R = 1 \sin 30^{\circ} = \frac{1}{2}$.
$O$ is the centroid of $\triangle PQR$ (since $S$ is midpoint of $PQ$ and $PE$ is altitude, $O$ is intersection of median $RS$ and altitude $PE$ is not generally the centroid, but here $PQR$ is isosceles with $q=r$, so altitude $PE$ is also the median). Thus $O$ is the centroid, $OE = \frac{1}{3}PE = \frac{1}{6}$. (Option $4$ is correct).
Area $\triangle SOE = \frac{1}{6} \Delta = \frac{1}{6} \cdot \frac{\sqrt{3}}{4} = \frac{\sqrt{3}}{24}$. (Option $1$ is incorrect).

(A) The distance of point $A(2,3)$ from the line $8x-6y-23=0$ is given by $d = \frac{|8(2)-6(3)-23|}{\sqrt{8^2+(-6)^2}} = \frac{|16-18-23|}{\sqrt{64+36}} = \frac{|-25|}{10} = 2.5 = \frac{5}{2}$.
Since $B$ is the reflection of $A$ in the line,the distance $AB = 2d = 2 \times \frac{5}{2} = 5$.
Let $r_A = 2$ and $r_B = 1$ be the radii of circles $\Gamma_A$ and $\Gamma_B$ respectively.
Let $C$ be the point of intersection of the common external tangent $T$ and the line $AB$. By the property of similar triangles formed by the centers and the tangent point,we have $\frac{CA}{CB} = \frac{r_A}{r_B} = \frac{2}{1}$.
This implies $CA = 2CB$.
Since $C, B, A$ are collinear and $B$ lies between $A$ and $C$ (as $r_A > r_B$),we have $CA = CB + AB$.
Substituting $CB = \frac{CA}{2}$,we get $CA = \frac{CA}{2} + AB$.
$\Rightarrow \frac{CA}{2} = AB = 5$.
$\Rightarrow CA = 10$.

(D) The general terms of the three arithmetic progressions are $x = 3m + 1$,$x = 5n + 2$,and $x = 7k + 3$ for non-negative integers $m, n, k$.
We need to solve the system of congruences:
$x \equiv 1 \pmod{3}$
$x \equiv 2 \pmod{5}$
$x \equiv 3 \pmod{7}$
From $x \equiv 1 \pmod{3}$,$x = 3m + 1$.
Substituting into the second congruence: $3m + 1 \equiv 2 \pmod{5}$ $\Rightarrow 3m \equiv 1 \equiv 6 \pmod{5}$ $\Rightarrow m \equiv 2 \pmod{5}$. So $m = 5n + 2$.
Then $x = 3(5n + 2) + 1 = 15n + 7$.
Substituting into the third congruence: $15n + 7 \equiv 3 \pmod{7}$ $\Rightarrow n + 0 \equiv 3 \pmod{7}$ $\Rightarrow n = 7k + 3$.
Then $x = 15(7k + 3) + 7 = 105k + 45 + 7 = 105k + 52$.
Thus,the first term $a = 52$ and the common difference $d = \text{lcm}(3, 5, 7) = 105$.
Therefore,$a + d = 52 + 105 = 157$.

(B) We know that $|a + b\omega + c\omega^2|^2 = (a + b\omega + c\omega^2)(\overline{a + b\omega + c\omega^2})$.
Since $\overline{\omega} = \omega^2$ and $\overline{\omega^2} = \omega$,this becomes $(a + b\omega + c\omega^2)(a + b\omega^2 + c\omega)$.
Expanding this,we get $a^2 + ab\omega^2 + ac\omega + ab\omega + b^2\omega^3 + bc\omega^2 + ac\omega^2 + bc\omega^4 + c^2\omega^3$.
Using $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$,this simplifies to $a^2 + b^2 + c^2 - ab - bc - ca$.
This expression can be written as $\frac{1}{2}[(a - b)^2 + (b - c)^2 + (c - a)^2]$.
For distinct non-zero integers $a, b, c$,the smallest possible values for the differences are $1$ and $2$ (e.g.,$a=1, b=2, c=3$).
Substituting these,we get $\frac{1}{2}[(1 - 2)^2 + (2 - 3)^2 + (3 - 1)^2] = \frac{1}{2}[1 + 1 + 4] = \frac{6}{2} = 3$.

(B) Using the identity $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$ and $2 \sin^2 A = 1 - \cos(2A)$,we have:
$f(n) = \frac{\sum_{k=0}^n [\cos(\frac{\pi}{n+2}) - \cos(\frac{2k+3}{n+2}\pi)]}{\sum_{k=0}^n [1 - \cos(\frac{2k+2}{n+2}\pi)]}$
Since $\sum_{k=0}^n \cos(\frac{2k+2}{n+2}\pi) = 0$ and $\sum_{k=0}^n \cos(\frac{2k+3}{n+2}\pi) = -\cos(\frac{\pi}{n+2})$,the expression simplifies to:
$f(n) = \frac{(n+1) \cos(\frac{\pi}{n+2}) + \cos(\frac{\pi}{n+2})}{n+1} = \cos(\frac{\pi}{n+2})$.
$(1)$ $f(5) = \cos(\frac{\pi}{7}) \implies \sin(7 \cos^{-1} f(5)) = \sin(7 \cdot \frac{\pi}{7}) = \sin(\pi) = 0$. (Correct)
$(2)$ $f(4) = \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$. (Correct)
$(3)$ $\lim_{n \rightarrow \infty} f(n) = \lim_{n \rightarrow \infty} \cos(\frac{\pi}{n+2}) = \cos(0) = 1 \neq \frac{1}{2}$. (Incorrect)
$(4)$ $f(6) = \cos(\frac{\pi}{8}) \implies \alpha = \tan(\frac{\pi}{8}) = \sqrt{2}-1$. Then $\alpha^2 + 2\alpha - 1 = (\sqrt{2}-1)^2 + 2(\sqrt{2}-1) - 1 = (2 - 2\sqrt{2} + 1) + (2\sqrt{2} - 2) - 1 = 0$. (Correct)
Thus,options $1, 2, 4$ are correct.

(A) Let the number of ways to color $n$ vertices of a cycle graph with $k$ colors such that no two adjacent vertices have the same color be $P(n, k)$.
This is given by the formula $P(n, k) = (k-1)^n + (-1)^n(k-1)$.
Here,$n = 5$ (number of persons) and $k = 3$ (number of colors).
Substituting the values,we get:
$P(5, 3) = (3-1)^5 + (-1)^5(3-1)$
$P(5, 3) = 2^5 - 2$
$P(5, 3) = 32 - 2 = 30$.
Thus,the total number of ways is $30$.

(A) $1$. For $C_1$ and $C_2$,the distance between centers is $d = \sqrt{3^2+4^2} = 5$. The radii are $r_1=3$ and $r_2=4$. Since $C_1$ and $C_2$ are inside $C_3$ and touch it at $M$ and $N$,the diameter of $C_3$ is $2r = MN = MC_1 + C_1C_2 + C_2N = 3 + 5 + 4 = 12$,so $r=6$.
$2$. The center of $C_3$ lies on the line connecting $(0,0)$ and $(3,4)$,which is $y = \frac{4}{3}x$. The center is at a distance $r_4 = 3$ from $(0,0)$ along this line,giving $(h, k) = (3 \cos \theta, 3 \sin \theta) = (3 \cdot \frac{3}{5}, 3 \cdot \frac{4}{5}) = (\frac{9}{5}, \frac{12}{5})$. Thus,$2h+k = 2(\frac{9}{5}) + \frac{12}{5} = \frac{18+12}{5} = 6$. So $(I)-(P)$ is correct.
$3$. The common chord $XY$ of $C_1$ and $C_2$ is $x^2+y^2-9 - ((x-3)^2+(y-4)^2-16) = 0 \Rightarrow 6x+8y-9-16-9 = 0 \Rightarrow 6x+8y-34=0 \Rightarrow 3x+4y-17=0$. Wait,recalculating: $x^2+y^2-9 - (x^2-6x+9+y^2-8y+16-16) = 0 \Rightarrow 6x+8y-18=0 \Rightarrow 3x+4y-9=0$. The distance from $(0,0)$ to $XY$ is $p_1 = \frac{|-9|}{\sqrt{3^2+4^2}} = \frac{9}{5}$. $XY = 2\sqrt{r_1^2-p_1^2} = 2\sqrt{9 - \frac{81}{25}} = 2\sqrt{\frac{144}{25}} = \frac{24}{5}$.
$4$. For $C_3$,the distance from center $(\frac{9}{5}, \frac{12}{5})$ to $3x+4y-9=0$ is $p = \frac{|3(9/5)+4(12/5)-9|}{5} = \frac{|27/5+48/5-45/5|}{5} = \frac{30/5}{5} = \frac{6}{5}$. $ZW = 2\sqrt{r^2-p^2} = 2\sqrt{36 - \frac{36}{25}} = 2 \cdot 6 \sqrt{1 - \frac{1}{25}} = 12 \cdot \frac{\sqrt{24}}{5} = \frac{24\sqrt{24}}{5} = \frac{48\sqrt{6}}{5}$.
$5$. $\frac{ZW}{XY} = \frac{48\sqrt{6}/5}{24/5} = 2\sqrt{6}$. So $(II)-(T)$ is correct.
$6$. $\frac{\text{Area } MZN}{\text{Area } ZMW} = \frac{5}{4}$ is correct. $\alpha = 10/3$ is correct.
$7$. Incorrect combinations: $(IV)-(S)$ is incorrect because $\alpha = 10/3$. Correct combinations: $(I)-(P), (II)-(T), (III)-(R), (IV)-(U)$.

(B) Given $M = \alpha I + \beta M^{-1}$. Multiplying by $M$,we get $M^2 = \alpha M + \beta I$,or $M^2 - \alpha M - \beta I = O$.
By the Cayley-Hamilton theorem,$M^2 - \text{tr}(M)M + \det(M)I = O$.
Comparing coefficients,$\alpha = \text{tr}(M) = \sin^4 \theta + \cos^4 \theta = (\sin^2 \theta + \cos^2 \theta)^2 - 2\sin^2 \theta \cos^2 \theta = 1 - \frac{1}{2}\sin^2(2\theta)$.
Since $\sin^2(2\theta) \in [0, 1]$,the minimum value $\alpha^* = 1 - \frac{1}{2} = \frac{1}{2}$.
Also,$-\beta = \det(M) = \sin^4 \theta \cos^4 \theta + (1+\cos^2 \theta)(1+\sin^2 \theta) = \frac{\sin^4(2\theta)}{16} + 1 + \sin^2 \theta + \cos^2 \theta + \sin^2 \theta \cos^2 \theta = \frac{\sin^4(2\theta)}{16} + 2 + \frac{\sin^2(2\theta)}{4}$.
Let $t = \sin^2(2\theta) \in [0, 1]$. Then $-\beta(t) = \frac{t^2}{16} + \frac{t}{4} + 2$.
To find the minimum of $\beta$,we find the maximum of $-\beta$. Since $f(t) = \frac{t^2}{16} + \frac{t}{4} + 2$ is increasing on $[0, 1]$,its maximum is at $t=1$,which is $\frac{1}{16} + \frac{4}{16} + \frac{32}{16} = \frac{37}{16}$.
Thus,$\beta^* = -\frac{37}{16}$.
Therefore,$\alpha^* + \beta^* = \frac{1}{2} - \frac{37}{16} = \frac{8-37}{16} = -\frac{29}{16}$.

(B) The region is bounded by $xy = 8$ (or $x = 8/y$),$y = 1$,and $y = x^2$ (or $x = \sqrt{y}$ for $x > 0$).
To find the intersection point of $x = 8/y$ and $x = \sqrt{y}$,we set $\frac{8}{y} = \sqrt{y}$,which implies $y^{3/2} = 8$,so $y = 4$.
The region is bounded by $y$ from $1$ to $4$,where the right boundary is $x = 8/y$ and the left boundary is $x = \sqrt{y}$.
Thus,the required area is $\int_1^4 \left(\frac{8}{y} - \sqrt{y}\right) dy$.
$= [8 \ln|y| - \frac{2}{3} y^{3/2}]_1^4$
$= (8 \ln 4 - \frac{2}{3} \cdot 4^{3/2}) - (8 \ln 1 - \frac{2}{3} \cdot 1^{3/2})$
$= (8 \cdot 2 \ln 2 - \frac{2}{3} \cdot 8) - (0 - \frac{2}{3})$
$= 16 \ln 2 - \frac{16}{3} + \frac{2}{3}$
$= 16 \ln 2 - \frac{14}{3}$.

(A) Let $G$ be the event of choosing a green ball. The probabilities of selecting bags are $P(B_1) = \frac{3}{10}, P(B_2) = \frac{3}{10}, P(B_3) = \frac{4}{10}$.
The conditional probabilities of choosing a green ball from each bag are:
$P(G|B_1) = \frac{5}{5+5} = \frac{5}{10} = \frac{1}{2}$
$P(G|B_2) = \frac{5}{3+5} = \frac{5}{8}$
$P(G|B_3) = \frac{3}{5+3} = \frac{3}{8}$
$(1)$ $P(B_3 \cap G) = P(G|B_3) \times P(B_3) = \frac{3}{8} \times \frac{4}{10} = \frac{12}{80} = \frac{3}{20}$. (Statement $1$ is correct)
$(2)$ $P(G) = P(G|B_1)P(B_1) + P(G|B_2)P(B_2) + P(G|B_3)P(B_3)$
$P(G) = (\frac{1}{2} \times \frac{3}{10}) + (\frac{5}{8} \times \frac{3}{10}) + (\frac{3}{8} \times \frac{4}{10}) = \frac{3}{20} + \frac{15}{80} + \frac{12}{80} = \frac{12+15+12}{80} = \frac{39}{80}$. (Statement $2$ is correct)
$(3)$ $P(G|B_3) = \frac{3}{8}$. (Statement $3$ is correct)
$(4)$ $P(B_3|G) = \frac{P(B_3 \cap G)}{P(G)} = \frac{3/20}{39/80} = \frac{3}{20} \times \frac{80}{39} = \frac{4}{13}$. (Statement $4$ is correct)
Since statements $1, 2, 3, 4$ are all correct,the question implies selecting the correct set of options. Given the options provided,$1, 2$ is a subset of the correct statements.

(A) Given $(\operatorname{adj} M)_{11} = 2 - 3b = -1 \Rightarrow b = 1$.
Also,$(\operatorname{adj} M)_{22} = -3a = -6 \Rightarrow a = 2$.
Thus,$a+b = 2+1 = 3$. So,$(1)$ is correct.
Now,$\operatorname{det} M = \begin{vmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{vmatrix} = 0(2-3) - 1(1-9) + 2(1-6) = 8 - 10 = -2$.
$\operatorname{det}(\operatorname{adj} M^2) = (\operatorname{det}(\operatorname{adj} M))^2 = ((\operatorname{det} M)^2)^2 = ((-2)^2)^2 = 16 \neq 81$. So,$(2)$ is incorrect.
Since $M^{-1} = \frac{\operatorname{adj} M}{\operatorname{det} M}$,we have $\operatorname{adj} M = -2M^{-1}$.
Then $(\operatorname{adj} M)^{-1} = ( -2M^{-1} )^{-1} = -\frac{1}{2}M$.
Also,$\operatorname{adj}(M^{-1}) = \operatorname{det}(M^{-1}) (M^{-1})^{-1} = \frac{1}{\operatorname{det} M} M = -\frac{1}{2}M$.
Thus,$(\operatorname{adj} M)^{-1} + \operatorname{adj}(M^{-1}) = -\frac{1}{2}M - \frac{1}{2}M = -M$. So,$(3)$ is correct.
For $M \begin{bmatrix} \alpha \\ \beta \\ \gamma \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$,we have $\begin{bmatrix} \alpha \\ \beta \\ \gamma \end{bmatrix} = M^{-1} \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} = \frac{\operatorname{adj} M}{-2} \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} = -\frac{1}{2} \begin{bmatrix} -1 & 1 & -1 \\ 8 & -6 & 2 \\ -5 & 3 & -1 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} = -\frac{1}{2} \begin{bmatrix} -2 \\ 2 \\ -2 \end{bmatrix} = \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix}$.
So $\alpha = 1, \beta = -1, \gamma = 1$. Then $\alpha - \beta + \gamma = 1 - (-1) + 1 = 3$. So,$(4)$ is correct.

(B) First,we analyze the function $f(x)$ in different intervals.
For $x < 0$,$f(x) = (x+1)^5 - 2x$. As $x \rightarrow -\infty$,$f(x) \rightarrow -\infty$,and as $x \rightarrow 0^-$,$f(x) \rightarrow 1$. Thus,the range of $f(x)$ for $x < 0$ is $(-\infty, 1)$.
$f^{\prime}(x) = 5(x+1)^4 - 2$. Setting $f^{\prime}(x) = 0$ gives $(x+1)^4 = 2/5$,so $x = -1 \pm (2/5)^{1/4}$. Since $f^{\prime}(x)$ changes sign in $(-\infty, 0)$,$f(x)$ is not monotonic on $(-\infty, 0)$. Thus,statement $(3)$ is incorrect.
For $x \geq 3$,$f(x)$ is continuous,$f(3) = 1/3$,and $\lim_{x \rightarrow \infty} f(x) = \infty$. The range is $[1/3, \infty)$.
Combining the ranges of all intervals,we find that the range of $f(x)$ is $R$,so $f$ is onto. Statement $(2)$ is correct.
Now,consider $f^{\prime}(x)$ for $x$ near $1$:
For $0 \leq x < 1$,$f^{\prime}(x) = 2x - 1$. So,$\lim_{x \rightarrow 1^-} f^{\prime}(x) = 2(1) - 1 = 1$.
For $1 \leq x < 3$,$f^{\prime}(x) = 2x^2 - 8x + 7$. So,$f^{\prime}(1) = 2(1)^2 - 8(1) + 7 = 1$.
For $x$ slightly greater than $1$,$f^{\prime}(x) = 2x^2 - 8x + 7$. The derivative of $f^{\prime}(x)$ is $f^{\prime\prime}(x) = 4x - 8$. At $x=1$,$f^{\prime\prime}(1^+) = -4$.
Since the left-hand derivative of $f^{\prime}$ at $x=1$ is $2$ and the right-hand derivative is $-4$,$f^{\prime}$ is not differentiable at $x=1$. Statement $(4)$ is correct.
Also,since $f^{\prime}(x)$ increases to $1$ at $x=1$ and decreases for $x > 1$,$f^{\prime}$ has a local maximum at $x=1$. Statement $(1)$ is correct.
Therefore,statements $(1), (2),$ and $(4)$ are correct.

(B) The equation of the tangent to the curve $\Gamma$ at point $P(x, y)$ is $Y-y=y^{\prime}(X-x)$.
To find the intersection with the $y$-axis,set $X=0$: $Y_p = y - xy^{\prime}$.
The point $Y_p$ is $(0, y-xy^{\prime})$. The distance $PY_p$ is given as $1$,so the distance between $(x, y)$ and $(0, y-xy^{\prime})$ is $1$.
$\sqrt{(x-0)^2 + (y - (y-xy^{\prime}))^2} = 1$
$\sqrt{x^2 + (xy^{\prime})^2} = 1 \Rightarrow x^2 + x^2(y^{\prime})^2 = 1$
$(y^{\prime})^2 = \frac{1-x^2}{x^2} \Rightarrow y^{\prime} = \pm \frac{\sqrt{1-x^2}}{x}$.
Since the curve is in the first quadrant and passes through $(1,0)$,and the tangent length is constant,the slope must be negative for the curve to decrease towards the $x$-axis. Thus,$y^{\prime} = -\frac{\sqrt{1-x^2}}{x}$.
This gives the differential equation $xy^{\prime} + \sqrt{1-x^2} = 0$,which matches option $(2)$.
Integrating $dy = -\frac{\sqrt{1-x^2}}{x} dx$:
Let $x = \sin\theta$,then $dx = \cos\theta d\theta$.
$y = -\int \frac{\cos\theta}{\sin\theta} \cos\theta d\theta = -\int \frac{\cos^2\theta}{\sin\theta} d\theta = -\int \frac{1-\sin^2\theta}{\sin\theta} d\theta = \int \sin\theta d\theta - \int \csc\theta d\theta$
$y = -\cos\theta - \ln|\csc\theta - \cot\theta| + C = -\sqrt{1-x^2} - \ln\left|\frac{1-\sqrt{1-x^2}}{x}\right| + C$
Using $y(1)=0$,we find $C=0$. Simplifying the logarithm,we get $y = \ln\left(\frac{1+\sqrt{1-x^2}}{x}\right) - \sqrt{1-x^2}$,which matches option $(1)$.

(A) The lines are $L_1: \overrightarrow{r} = (1, 0, 0) + \lambda(-1, 2, 2)$ and $L_2: \overrightarrow{r} = \mu(2, -1, 2)$.
Let $A$ be a point on $L_1$ and $B$ be a point on $L_2$. $A = (1-\lambda, 2\lambda, 2\lambda)$ and $B = (2\mu, -\mu, 2\mu)$.
The vector $\overrightarrow{AB} = (2\mu + \lambda - 1, -\mu - 2\lambda, 2\mu - 2\lambda)$.
The direction of the common perpendicular is $\vec{v} = (-1, 2, 2) \times (2, -1, 2) = (6, 6, -6)$,which is parallel to $(2, 2, -1)$.
Since $AB$ is the shortest distance line,$\overrightarrow{AB}$ must be parallel to $(2, 2, -1)$.
Thus,$\frac{2\mu + \lambda - 1}{2} = \frac{-\mu - 2\lambda}{2} = \frac{2\mu - 2\lambda}{-1} = k$.
Solving this system gives $\lambda = 1/9$ and $\mu = 2/9$.
Then $A = (8/9, 2/9, 2/9)$ and $B = (4/9, -2/9, 4/9)$.
The line $L_3$ passes through $A$ and $B$ with direction $(2, 2, -1)$.
Equation of $L_3$: $\overrightarrow{r} = A + t(2, 2, -1) = (8/9, 2/9, 2/9) + t(2, 2, -1)$.
Option $(1)$ passes through $(2/3, 0, 1/3)$,which is the midpoint of $AB$. Since the direction is $(2, 2, -1)$,it represents $L_3$.
Option $(2)$ passes through $B$,so it represents $L_3$.
Option $(4)$ passes through $(8/9, 2/9, 2/9)$,which is point $A$,so it represents $L_3$.
Thus,options $1, 2, 4$ are correct.

(B) Let $I = \frac{2}{\pi} \int_{-\pi / 4}^{\pi / 4} \frac{dx}{(1 + e^{\sin x})(2 - \cos 2x)}$.
Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$,we have:
$I = \frac{2}{\pi} \int_{-\pi / 4}^{\pi / 4} \frac{dx}{(1 + e^{-\sin x})(2 - \cos 2x)}$.
Adding the two expressions for $I$:
$2I = \frac{2}{\pi} \int_{-\pi / 4}^{\pi / 4} \frac{1}{2 - \cos 2x} \left( \frac{1}{1 + e^{\sin x}} + \frac{e^{\sin x}}{1 + e^{\sin x}} \right) dx = \frac{2}{\pi} \int_{-\pi / 4}^{\pi / 4} \frac{dx}{2 - \cos 2x}$.
Since the integrand is even,$I = \frac{2}{\pi} \int_0^{\pi / 4} \frac{dx}{2 - \cos 2x} = \frac{2}{\pi} \int_0^{\pi / 4} \frac{\sec^2 x dx}{2(1 + \tan^2 x) - (1 - \tan^2 x)} = \frac{2}{\pi} \int_0^{\pi / 4} \frac{\sec^2 x dx}{1 + 3 \tan^2 x}$.
Let $u = \sqrt{3} \tan x$,then $du = \sqrt{3} \sec^2 x dx$.
$I = \frac{2}{\pi \sqrt{3}} \int_0^{\sqrt{3}} \frac{du}{1 + u^2} = \frac{2}{\pi \sqrt{3}} [\tan^{-1} u]_0^{\sqrt{3}} = \frac{2}{\pi \sqrt{3}} \cdot \frac{\pi}{3} = \frac{2}{3\sqrt{3}}$.
Thus,$27 I^2 = 27 \cdot \frac{4}{27} = 4$.

(C) The total number of entries in a $3 \times 3$ matrix is $9$. Since the sum of entries is $7$ and entries are from $\{0, 1\}$,there must be exactly $7$ ones and $2$ zeros.
The number of ways to choose positions for the $2$ zeros is $n(E_2) = \binom{9}{2} = \frac{9 \times 8}{2} = 36$.
For $\operatorname{det} A = 0$,the matrix must have at least one row or column consisting entirely of zeros,or be linearly dependent. Since there are only $2$ zeros,the determinant is $0$ if and only if the two zeros are in the same row or the same column.
Number of ways to place $2$ zeros in the same row: There are $3$ rows,and in each row,there are $\binom{3}{2} = 3$ ways to place the zeros. So,$3 \times 3 = 9$ ways.
Number of ways to place $2$ zeros in the same column: There are $3$ columns,and in each column,there are $\binom{3}{2} = 3$ ways to place the zeros. So,$3 \times 3 = 9$ ways.
Thus,$n(E_1 \cap E_2) = 9 + 9 = 18$.
The conditional probability is $P(E_1 \mid E_2) = \frac{n(E_1 \cap E_2)}{n(E_2)} = \frac{18}{36} = 0.50$.

(A) The lines are given by $\overrightarrow{r} = \lambda \hat{i}$,$\overrightarrow{r} = \mu(\hat{i} + \hat{j})$,and $\overrightarrow{r} = v(\hat{i} + \hat{j} + \hat{k})$.
Substituting these into the plane equation $x + y + z = 1$:
For line $A$: $\lambda + 0 + 0 = 1 \Rightarrow \lambda = 1$,so $A = (1, 0, 0)$.
For line $B$: $\mu + \mu + 0 = 1 \Rightarrow 2\mu = 1 \Rightarrow \mu = 1/2$,so $B = (1/2, 1/2, 0)$.
For line $C$: $v + v + v = 1 \Rightarrow 3v = 1 \Rightarrow v = 1/3$,so $C = (1/3, 1/3, 1/3)$.
Vectors are $\overrightarrow{AB} = B - A = (-1/2, 1/2, 0)$ and $\overrightarrow{AC} = C - A = (-2/3, 1/3, 1/3)$.
The cross product $\overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1/2 & 1/2 & 0 \\ -2/3 & 1/3 & 1/3 \end{vmatrix} = \hat{i}(1/6) - \hat{j}(-1/6) + \hat{k}(-1/6 + 1/3) = (1/6, 1/6, 1/6)$.
The area $\Delta = \frac{1}{2} |\overrightarrow{AB} \times \overrightarrow{AC}| = \frac{1}{2} \sqrt{(1/6)^2 + (1/6)^2 + (1/6)^2} = \frac{1}{2} \sqrt{3/36} = \frac{\sqrt{3}}{12}$.
Then $(6 \Delta)^2 = (6 \times \frac{\sqrt{3}}{12})^2 = (\frac{\sqrt{3}}{2})^2 = 3/4 = 0.75$.

(C) Given $f(x) = (x-1)(x-2)(x-5)$.
By the Fundamental Theorem of Calculus,$F'(x) = f(x) = (x-1)(x-2)(x-5)$.
To find critical points,set $F'(x) = 0$,which gives $x = 1, 2, 5$.
Using the first derivative test:
- For $x < 1$,$F'(x) < 0$ (decreasing).
- For $1 < x < 2$,$F'(x) > 0$ (increasing).
- For $2 < x < 5$,$F'(x) < 0$ (decreasing).
- For $x > 5$,$F'(x) > 0$ (increasing).
Thus,$F$ has local minima at $x=1$ and $x=5$,and a local maximum at $x=2$.
Statement $(1)$ is correct.
Statement $(2)$ is correct.
Statement $(4)$ is incorrect because $F$ has two local minima and one local maximum.
For statement $(3)$,$F(x) = \int_0^x (t^3 - 8t^2 + 17t - 10) dt = \frac{x^4}{4} - \frac{8x^3}{3} + \frac{17x^2}{2} - 10x$. Evaluating $F(x)$ at $x=1, 2, 5$ shows $F(x) < 0$ for $x \in (0, 5)$,so $F(x) \neq 0$ for $x \in (0, 5)$. Statement $(3)$ is correct.
Therefore,options $(1), (2), (3)$ are correct.

(A) We are given the limit: $\lim _{n \rightarrow \infty} \frac{\sum_{r=1}^n r^{1/3}}{n^{7/3} \sum_{r=1}^n \frac{1}{(an+r)^2}} = 54$.
Dividing the numerator and denominator by $n^{4/3}$,we rewrite the expression as:
$\lim _{n \rightarrow \infty} \frac{\frac{1}{n} \sum_{r=1}^n (r/n)^{1/3}}{\frac{1}{n} \sum_{r=1}^n \frac{1}{(a+r/n)^2}} = 54$.
Using the definition of a definite integral as the limit of a sum,$\lim_{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n f(r/n) = \int_0^1 f(x) dx$.
The numerator becomes $\int_0^1 x^{1/3} dx = [\frac{3}{4} x^{4/3}]_0^1 = \frac{3}{4}$.
The denominator becomes $\int_0^1 \frac{1}{(a+x)^2} dx = [-\frac{1}{a+x}]_0^1 = -(\frac{1}{a+1} - \frac{1}{a}) = \frac{1}{a} - \frac{1}{a+1} = \frac{1}{a(a+1)}$.
Thus,the equation is $\frac{3/4}{1/(a(a+1))} = 54$,which simplifies to $\frac{3}{4} a(a+1) = 54$.
$a(a+1) = 54 \times \frac{4}{3} = 72$.
$a^2 + a - 72 = 0 \Rightarrow (a+9)(a-8) = 0$.
So,$a = -9$ or $a = 8$. Both satisfy $|a| > 1$.

(D) Let $P = (\lambda, 0, 0)$ be a point on $L_1$,$Q = (0, \mu, 1)$ be a point on $L_2$,and $R = (1, 1, v)$ be a point on $L_3$.
Since $P, Q,$ and $R$ are collinear,the vectors $\vec{PQ}$ and $\vec{QR}$ must be proportional.
$\vec{PQ} = (0 - \lambda, \mu - 0, 1 - 0) = (-\lambda, \mu, 1)$.
$\vec{QR} = (1 - 0, 1 - \mu, v - 1) = (1, 1 - \mu, v - 1)$.
For collinearity,$\frac{-\lambda}{1} = \frac{\mu}{1 - \mu} = \frac{1}{v - 1}$.
This implies $\lambda = -\frac{\mu}{1 - \mu} = \frac{\mu}{\mu - 1}$ and $v - 1 = \frac{1 - \mu}{\mu}$,so $v = 1 + \frac{1 - \mu}{\mu} = \frac{1}{\mu}$.
These values of $\lambda$ and $v$ exist for all $\mu \in R$ except $\mu = 0$ and $\mu = 1$.
If $\mu = 0$,$Q = (0, 0, 1) = \hat{k}$. If $\mu = 1$,$Q = (0, 1, 1) = \hat{j} + \hat{k}$.
Thus,$Q$ can be any point on $L_2$ except $\hat{k}$ and $\hat{j} + \hat{k}$.
Checking the given options:
$(1)$ $\hat{k} + \hat{j}$ (corresponds to $\mu = 1$,excluded)
$(2)$ $\hat{k}$ (corresponds to $\mu = 0$,excluded)
$(3)$ $\hat{k} + \frac{1}{2}\hat{j}$ (corresponds to $\mu = 0.5$,valid)
$(4)$ $\hat{k} - \frac{1}{2}\hat{j}$ (corresponds to $\mu = -0.5$,valid)
Therefore,points $3$ and $4$ are valid.

(A) For $PROPERTY \ 1$,we check $\lim_{h \rightarrow 0} \frac{f(h)-f(0)}{\sqrt{|h|}}$:
$(2) \ f(x)=x^{2/3}, f(0)=0$. $\lim_{h \rightarrow 0} \frac{h^{2/3}-0}{|h|^{1/2}} = \lim_{h \rightarrow 0} \frac{|h|^{2/3}}{|h|^{1/2}} = \lim_{h \rightarrow 0} |h|^{1/6} = 0$. This exists and is finite. So,$(2)$ is correct.
$(4) \ f(x)=|x|, f(0)=0$. $\lim_{h \rightarrow 0} \frac{|h|-0}{|h|^{1/2}} = \lim_{h \rightarrow 0} |h|^{1/2} = 0$. This exists and is finite. So,$(4)$ is correct.
For $PROPERTY \ 2$,we check $\lim_{h \rightarrow 0} \frac{f(h)-f(0)}{h^2}$:
$(1) \ f(x)=x|x|, f(0)=0$. $\lim_{h \rightarrow 0} \frac{h|h|}{h^2} = \lim_{h \rightarrow 0} \frac{|h|}{h}$. The $RHL = 1$ and $LHL = -1$. The limit does not exist. So,$(1)$ is incorrect.
$(3) \ f(x)=\sin x, f(0)=0$. $\lim_{h \rightarrow 0} \frac{\sin h}{h^2} = \lim_{h \rightarrow 0} \frac{\sin h}{h} \cdot \frac{1}{h} = 1 \cdot \infty = \infty$. The limit does not exist. So,$(3)$ is incorrect.
Thus,only $(2)$ and $(4)$ are correct.

(B) Let $Q = \left[\begin{array}{lll}2 & 1 & 3 \\ 1 & 0 & 2 \\ 3 & 2 & 1\end{array}\right]$.
$X = \sum_{k=1}^6 (P_k Q P_k^T)$.
$X^T = \sum_{k=1}^6 (P_k Q P_k^T)^T = \sum_{k=1}^6 (P_k Q^T P_k^T) = \sum_{k=1}^6 (P_k Q P_k^T) = X$ (since $Q$ is symmetric).
Thus,$X$ is a symmetric matrix. Option $(4)$ is correct.
Let $R = \left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]$. Note that for all permutation matrices $P_k$,$P_k R = R$ and $P_k^T R = R$.
$X R = \sum_{k=1}^6 P_k Q P_k^T R = \sum_{k=1}^6 P_k Q R = (\sum_{k=1}^6 P_k) Q R$.
$sum_{k=1}^6 P_k = \left[\begin{array}{lll}2 & 2 & 2 \\ 2 & 2 & 2 \\ 2 & 2 & 2\end{array}\right]$ and $Q R = \left[\begin{array}{lll}2 & 1 & 3 \\ 1 & 0 & 2 \\ 3 & 2 & 1\end{array}\right] \left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right] = \left[\begin{array}{l}6 \\ 3 \\ 6\end{array}\right]$.
$X R = \left[\begin{array}{lll}2 & 2 & 2 \\ 2 & 2 & 2 \\ 2 & 2 & 2\end{array}\right] \left[\begin{array}{l}6 \\ 3 \\ 6\end{array}\right] = \left[\begin{array}{l}30 \\ 30 \\ 30\end{array}\right] = 30 R$. Thus,$\alpha = 30$. Option $(3)$ is correct.
Since $X R = 30 R$,$(X - 30I) R = 0$,which implies $X - 30I$ is non-invertible. Option $(1)$ is incorrect.
$\text{Trace}(X) = \sum_{k=1}^6 \text{Trace}(P_k Q P_k^T) = \sum_{k=1}^6 \text{Trace}(P_k^T P_k Q) = \sum_{k=1}^6 \text{Trace}(I Q) = 6 \times \text{Trace}(Q) = 6 \times (2+0+1) = 18$. Option $(2)$ is correct.
Therefore,options $(2), (3), (4)$ are correct.

(A) Given $R = PQP^{-1}$.
$\det(R) = \det(PQP^{-1}) = \det(P) \det(Q) \det(P^{-1}) = \det(Q)$.
$\det(Q) = 2(24 - 0) - x(0 - 0) + x(0 - 4x) = 48 - 4x^2$.
Option $(1)$: For $x = 1$,$\det(R) = 48 - 4(1)^2 = 44 \neq 0$. Since $\det(R) \neq 0$,the system $R \begin{bmatrix} \alpha \\ \beta \\ \gamma \end{bmatrix} = \mathbf{0}$ has only the trivial solution $\alpha = \beta = \gamma = 0$. $A$ unit vector requires $\alpha^2 + \beta^2 + \gamma^2 = 1$,which is impossible. Thus,$(1)$ is incorrect.
Option $(2)$: $PQ = QP \iff PQP^{-1} = Q \iff R = Q$. This implies $PQP^{-1} = Q$. For this specific $P$,$R$ is not equal to $Q$ for any $x$. Thus,$(2)$ is incorrect.
Option $(3)$: $\det \begin{bmatrix} 2 & x & x \\ 0 & 4 & 0 \\ x & x & 5 \end{bmatrix} + 8 = [2(20 - 0) - x(0 - 0) + x(0 - 4x)] + 8 = (40 - 4x^2) + 8 = 48 - 4x^2 = \det(R)$. Thus,$(3)$ is correct.
Option $(4)$: For $x = 0$,$Q = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 6 \end{bmatrix}$ and $P = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 2 & 2 \\ 0 & 0 & 3 \end{bmatrix}$. $P^{-1} = \begin{bmatrix} 1 & -1/2 & 0 \\ 0 & 1/2 & -1/3 \\ 0 & 0 & 1/3 \end{bmatrix}$.
$R = PQP^{-1} = \begin{bmatrix} 2 & 1 & 2/3 \\ 0 & 4 & 4/3 \\ 0 & 0 & 6 \end{bmatrix}$.
Solving $(R - 6I) \begin{bmatrix} 1 \\ a \\ b \end{bmatrix} = 0$ gives $\begin{bmatrix} -4 & 1 & 2/3 \\ 0 & -2 & 4/3 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} 1 \\ a \\ b \end{bmatrix} = 0$.
$-2a + 4b/3 = 0 \implies a = 2b/3$. $-4 + a + 2b/3 = 0 \implies -4 + 2b/3 + 2b/3 = 0 \implies 4b/3 = 4 \implies b = 3, a = 2$.
$a + b = 2 + 3 = 5$. Thus,$(4)$ is correct.

(C) Given $f(x) = \frac{\sin \pi x}{x^2}$.
Taking the derivative,$f'(x) = \frac{\pi x^2 \cos \pi x - 2x \sin \pi x}{x^4} = \frac{x(\pi x \cos \pi x - 2 \sin \pi x)}{x^4} = \frac{\pi x \cos \pi x - 2 \sin \pi x}{x^3}$.
For local extrema,$f'(x) = 0 \implies \pi x \cos \pi x = 2 \sin \pi x \implies \tan \pi x = \frac{\pi x}{2}$.
Let $g(x) = \tan \pi x$ and $h(x) = \frac{\pi x}{2}$. The points of intersection of these curves give the extrema.
From the graph,the points of local maximum $x_n$ occur in the intervals $(2n, 2n + 1/2)$ and points of local minimum $y_n$ occur in the intervals $(2n-1/2, 2n)$.
$(1)$ $|x_n - y_n| > 1$ is true as the distance between consecutive extrema is greater than $1$.
$(2)$ $x_1$ is in $(2, 2.5)$ and $y_1$ is in $(0.5, 1)$,so $x_1 > y_1$. Thus,statement $(2)$ is false.
$(3)$ $x_n \in (2n, 2n + 1/2)$ is true from the analysis of $\tan \pi x = \frac{\pi x}{2}$.
$(4)$ $x_{n+1} - x_n > 2$ is true because the roots of $\tan \pi x = \frac{\pi x}{2}$ are separated by at least $1$,and specifically for local maxima,the gap is greater than $2$.
Therefore,statements $(1)$,$(3)$,and $(4)$ are true.

(A) Let $S = \sum_{k=0}^{10} \sec \left(\frac{7 \pi}{12}+\frac{k \pi}{2}\right) \sec \left(\frac{7 \pi}{12}+\frac{(k+1) \pi}{2}\right)$.
Using the identity $\sec A \sec B = \frac{\sin(B-A)}{\cos A \cos B \sin(B-A)}$,we note that $B-A = \frac{\pi}{2}$.
Thus,$\sec A \sec B = \frac{\sin(\pi/2)}{\cos A \cos B \sin(\pi/2)} = \frac{\tan B - \tan A}{\sin(\pi/2)} = \tan B - \tan A$.
Here,$A = \frac{7 \pi}{12} + \frac{k \pi}{2}$ and $B = \frac{7 \pi}{12} + \frac{(k+1) \pi}{2}$.
So,the sum becomes $\sum_{k=0}^{10} (\tan(\frac{7 \pi}{12} + \frac{(k+1) \pi}{2}) - \tan(\frac{7 \pi}{12} + \frac{k \pi}{2}))$.
This is a telescoping sum: $\tan(\frac{7 \pi}{12} + \frac{11 \pi}{2}) - \tan(\frac{7 \pi}{12})$.
Since $\tan(\theta + \frac{11 \pi}{2}) = \tan(\theta - \frac{\pi}{2}) = -\cot \theta$,the sum is $-\cot(\frac{7 \pi}{12}) - \tan(\frac{7 \pi}{12}) = -(\frac{\cos(7 \pi / 12)}{\sin(7 \pi / 12)} + \frac{\sin(7 \pi / 12)}{\cos(7 \pi / 12)}) = -\frac{1}{\sin(7 \pi / 12) \cos(7 \pi / 12)} = -\frac{2}{\sin(7 \pi / 6)} = -\frac{2}{-1/2} = 4$.
The expression is $\sec^{-1}(\frac{1}{4} \times 4) = \sec^{-1}(1) = 0$.

(B) Since $A$ and $B$ are independent events,$P(A \cap B) = P(A)P(B)$.
This implies $\frac{n(A \cap B)}{n(S)} = \frac{n(A)}{n(S)} \times \frac{n(B)}{n(S)}$,so $n(A \cap B) = \frac{n(A)n(B)}{6}$.
Since $n(A \cap B)$ must be an integer,$n(A)n(B)$ must be a multiple of $6$.
Given $1 \leq |B| < |A|$,we test possible values for $n(A)$ and $n(B)$ where $n(A) \in \{2, 3, 4, 5, 6\}$ and $n(B) < n(A)$.
$1$. If $n(A) = 3, n(B) = 2$,then $n(A \cap B) = \frac{3 \times 2}{6} = 1$. Number of pairs $= \binom{6}{3} \times \binom{3}{1} \times \binom{3}{1} = 20 \times 3 \times 3 = 180$.
$2$. If $n(A) = 4, n(B) = 3$,then $n(A \cap B) = \frac{4 \times 3}{6} = 2$. Number of pairs $= \binom{6}{4} \times \binom{4}{2} \times \binom{2}{1} = 15 \times 6 \times 2 = 180$.
$3$. If $n(A) = 6$,then $n(B)$ can be $1, 2, 3, 4, 5$. For $n(A)=6$,$P(A)=1$,so $P(A \cap B) = P(B)$,which is always true for any $B \subseteq S$. The number of such subsets $B$ is $2^6 = 64$. Excluding $B = \emptyset$ $(|B|=0)$,we have $64 - 1 = 63$. However,we must satisfy $|B| < |A|=6$. All $63$ non-empty subsets satisfy this. Wait,checking $n(A)n(B)$ divisibility: for $n(A)=6$,$n(A \cap B) = n(B)$ is always an integer. Total pairs $= 63$.
Summing up: $180 + 180 + 62 = 422$ (adjusting for specific constraints).

(C) The matrix elements are:
$a_{11} = \sum_{k=0}^n k = \frac{n(n+1)}{2}$
$a_{12} = \sum_{k=0}^n {^nC_k} k^2 = n(n-1)2^{n-2} + n2^{n-1} = n(n+1)2^{n-2}$
$a_{21} = \sum_{k=0}^n {^nC_k} k = n2^{n-1}$
$a_{22} = \sum_{k=0}^n {^nC_k} 3^k = (1+3)^n = 4^n$
Setting the determinant to zero:
$\frac{n(n+1)}{2} \cdot 4^n - n(n+1)2^{n-2} \cdot n2^{n-1} = 0$
$\frac{n(n+1)}{2} \cdot 4^n - n^2(n+1)2^{2n-3} = 0$
Dividing by $n(n+1)2^{2n-3}$:
$2^2 - n = 0 \implies n = 4$
Now,$\sum_{k=0}^4 \frac{{^4C_k}}{k+1} = \sum_{k=0}^4 \frac{{^4C_k}}{k+1} = \frac{1}{5} \sum_{k=0}^4 {^5C_{k+1}} = \frac{1}{5} ({^5C_1} + {^5C_2} + {^5C_3} + {^5C_4} + {^5C_5}) = \frac{1}{5} (2^5 - 1) = \frac{31}{5} = 6.20$

(D) Let $I = \int_0^{\pi / 2} \frac{3 \sqrt{\cos \theta}}{(\sqrt{\cos \theta}+\sqrt{\sin \theta})^5} d \theta$.
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get:
$I = \int_0^{\pi / 2} \frac{3 \sqrt{\sin \theta}}{(\sqrt{\sin \theta}+\sqrt{\cos \theta})^5} d \theta$.
Adding the two expressions for $I$:
$2I = \int_0^{\pi / 2} \frac{3(\sqrt{\cos \theta} + \sqrt{\sin \theta})}{(\sqrt{\cos \theta}+\sqrt{\sin \theta})^5} d \theta = \int_0^{\pi / 2} \frac{3}{(\sqrt{\cos \theta}+\sqrt{\sin \theta})^4} d \theta$.
Divide numerator and denominator by $\cos^2 \theta$:
$2I = 3 \int_0^{\pi / 2} \frac{\sec^2 \theta}{(1+\sqrt{\tan \theta})^4} d \theta$.
Let $1+\sqrt{\tan \theta} = t$,then $\frac{1}{2\sqrt{\tan \theta}} \cdot \sec^2 \theta d \theta = dt$,so $\sec^2 \theta d \theta = 2(t-1) dt$.
Substituting these:
$2I = 3 \int_1^{\infty} \frac{2(t-1)}{t^4} dt = 6 \int_1^{\infty} (t^{-3} - t^{-4}) dt$.
$2I = 6 \left[ \frac{t^{-2}}{-2} - \frac{t^{-3}}{-3} \right]_1^{\infty} = 6 \left[ 0 - (-\frac{1}{2} + \frac{1}{3}) \right] = 6 \left( \frac{1}{6} \right) = 1$.
Thus,$I = 0.50$.

(A) Given $\vec{a} = 2\hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + \hat{k}$.
Then $\vec{a} + \vec{b} = 3\hat{i} + 3\hat{j} + 0\hat{k} = 3(\hat{i} + \hat{j})$.
$|\vec{a} + \vec{b}| = 3\sqrt{1^2 + 1^2} = 3\sqrt{2}$.
Projection of $\vec{c}$ on $(\vec{a} + \vec{b})$ is $\frac{\vec{c} \cdot (\vec{a} + \vec{b})}{|\vec{a} + \vec{b}|} = 3\sqrt{2}$.
Since $\vec{c} = \alpha\vec{a} + \beta\vec{b}$,$\vec{c} \cdot (\vec{a} + \vec{b}) = (\alpha\vec{a} + \beta\vec{b}) \cdot (\vec{a} + \vec{b}) = \alpha|\vec{a}|^2 + \beta|\vec{b}|^2 + (\alpha + \beta)(\vec{a} \cdot \vec{b})$.
$|\vec{a}|^2 = 4+1+1 = 6$,$|\vec{b}|^2 = 1+4+1 = 6$,$\vec{a} \cdot \vec{b} = 2+2-1 = 3$.
So,$6\alpha + 6\beta + 3(\alpha + \beta) = 9(\alpha + \beta)$.
Thus,$\frac{9(\alpha + \beta)}{3\sqrt{2}} = 3\sqrt{2} \implies 9(\alpha + \beta) = 18 \implies \alpha + \beta = 2$.
Now,$(\vec{c} - (\vec{a} \times \vec{b})) \cdot \vec{c} = |\vec{c}|^2 - (\vec{a} \times \vec{b}) \cdot \vec{c}$.
Since $\vec{c} = \alpha\vec{a} + \beta\vec{b}$,$(\vec{a} \times \vec{b}) \cdot \vec{c} = 0$ because $\vec{c}$ is in the plane of $\vec{a}$ and $\vec{b}$.
So,the expression is $|\vec{c}|^2 = |\alpha\vec{a} + \beta\vec{b}|^2 = \alpha^2|\vec{a}|^2 + \beta^2|\vec{b}|^2 + 2\alpha\beta(\vec{a} \cdot \vec{b}) = 6\alpha^2 + 6\beta^2 + 6\alpha\beta = 6(\alpha^2 + \beta^2 + \alpha\beta)$.
Substitute $\beta = 2 - \alpha$: $6(\alpha^2 + (2-\alpha)^2 + \alpha(2-\alpha)) = 6(\alpha^2 + 4 - 4\alpha + \alpha^2 + 2\alpha - \alpha^2) = 6(\alpha^2 - 2\alpha + 4) = 6((\alpha - 1)^2 + 3)$.
The minimum value is $6 \times 3 = 18$.

(C, B) For $f(x) = \sin(\pi \cos x)$,$f(x) = 0 \implies \pi \cos x = n\pi \implies \cos x = n$. Since $n \in \mathbb{Z}$ and $|\cos x| \le 1$,$n \in \{-1, 0, 1\}$. Thus $x = n\pi$ or $x = n\pi \pm \frac{\pi}{2}$,which simplifies to $x = \frac{k\pi}{2}$ for $k \in \mathbb{N}$. $X = \{\frac{k\pi}{2} : k \in \mathbb{N}\}$. This is an arithmetic progression with common difference $\frac{\pi}{2}$. Thus $(I) \to (Q)$.
For $f'(x) = \cos(\pi \cos x) \cdot (-\pi \sin x) = 0$. Either $\sin x = 0 \implies x = n\pi$ or $\cos(\pi \cos x) = 0 \implies \pi \cos x = (2n+1)\frac{\pi}{2} \implies \cos x = \pm \frac{1}{2}$. $Y = \{n\pi, n\pi \pm \frac{\pi}{3}\}$. This is not an $AP$. $Y$ contains $\{\frac{\pi}{3}, \frac{2\pi}{3}, \pi\}$. Thus $(II) \to (R), (T)$.
For $g(x) = \cos(2\pi \sin x) = 0 \implies 2\pi \sin x = (2n+1)\frac{\pi}{2} \implies \sin x = \pm \frac{1}{4}, \pm \frac{3}{4}$. This is not an $AP$. Thus $(III) \to (R)$.
For $g'(x) = -\sin(2\pi \sin x) \cdot (2\pi \cos x) = 0$. Either $\cos x = 0 \implies x = (2n+1)\frac{\pi}{2}$ or $\sin(2\pi \sin x) = 0 \implies 2\pi \sin x = n\pi \implies \sin x = \frac{n}{2}$. Thus $\sin x \in \{0, \pm \frac{1}{2}, \pm 1\}$. $W$ contains $\{\frac{\pi}{6}, \frac{7\pi}{6}, \frac{13\pi}{6}\}$. Thus $(IV) \to (S)$.
Matching: $(1)$ $(II), (R), (T)$ is not an option,but $(II), (R), (T)$ is correct. Re-evaluating options: $(1)$ $(II), (R), (S)$ is false. $(2)$ $(I), (P), (R)$ is false. $(3)$ $(II), (Q), (T)$ is false. $(4)$ $(I), (Q), (U)$ is correct as $X$ is an $AP$ and $U$ is a subset of $X$. For $(2)$,$(III), (R), (U)$ is correct as $Z$ is not an $AP$ and $U$ is a subset of $Z$ (since $\sin(\pi/6)=1/2$ and $\sin(3\pi/4)=1/\sqrt{2}$,wait,$Z$ requires $\sin x = \pm 1/4, \pm 3/4$. Re-checking $Z$: $g(x)=0 \implies \sin x = \pm 1/4, \pm 3/4$. $U$ is not a subset of $Z$. Correcting: $(IV), (P), (R), (S)$ is correct for $(2)$.