IIT JEE 2022 Mathematics Question Paper with Answer and Solution

(C) We need to evaluate $\lim_{x \rightarrow \alpha^{+}} g(x)$.
Let $L = \lim_{x \rightarrow \alpha^{+}} \frac{2 \ln(\sqrt{x}-\sqrt{\alpha})}{\ln(e^{\sqrt{x}}-e^{\sqrt{\alpha}})}$. This is a $\frac{0}{0}$ indeterminate form.
Using $L$'$H$ôpital's rule:
$L = 2 \lim_{x}$ ${\rightarrow \alpha^{+}} \frac{\frac{1}{\sqrt{x}-\sqrt{\alpha}} \cdot \frac{1}{2\sqrt{x}}}{\frac{1}{e^{\sqrt{x}}-e^{\sqrt{\alpha}}} \cdot e^{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}}}$
$L = 2 \lim_{x \rightarrow \alpha^{+}} \frac{e^{\sqrt{x}}-e^{\sqrt{\alpha}}}{e^{\sqrt{x}}(\sqrt{x}-\sqrt{\alpha})}$
$L = \frac{2}{e^{\sqrt{\alpha}}} \lim_{x}$ ${\rightarrow \alpha^{+}} \frac{e^{\sqrt{x}}-e^{\sqrt{\alpha}}}{\sqrt{x}-\sqrt{\alpha}}$
Applying $L$'$H$ôpital's rule again to the limit part:
$L = \frac{2}{e^{\sqrt{\alpha}}} \lim_{x}$ ${\rightarrow \alpha^{+}} \frac{e^{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}}}{\frac{1}{2\sqrt{x}}} = \frac{2}{e^{\sqrt{\alpha}}} \cdot e^{\sqrt{\alpha}} = 2$.
Now,$\lim_{x}$ ${\rightarrow \alpha^{+}} f(g(x)) = f(2) = \sin\left(\frac{2\pi}{12}\right) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} = 0.50$.

(A) $n(U) = 900$
Let $A \equiv \text{Fever}$,$B \equiv \text{Cough}$,$C \equiv \text{Breathing problem}$.
Given: $n(A) = 190, n(B) = 220, n(C) = 220$,
$n(A \cup B) = 330, n(B \cup C) = 350, n(A \cup C) = 340, n(A \cap B \cap C) = 30$.
Using $n(A \cup B) = n(A) + n(B) - n(A \cap B)$:
$330 = 190 + 220 - n(A \cap B) \Rightarrow n(A \cap B) = 80$.
Similarly,$350 = 220 + 220 - n(B \cap C) \Rightarrow n(B \cap C) = 90$.
And $340 = 190 + 220 - n(A \cap C) \Rightarrow n(A \cap C) = 70$.
Now,$n(A \cup B \cup C) = (n(A) + n(B) + n(C)) - (n(A \cap B) + n(B \cap C) + n(A \cap C)) + n(A \cap B \cap C)$
$= (190 + 220 + 220) - (80 + 90 + 70) + 30 = 630 - 240 + 30 = 420$.
Number of persons without any symptom $= n(U) - n(A \cup B \cup C) = 900 - 420 = 480$.
Number of persons with exactly one symptom $= (n(A) + n(B) + n(C)) - 2(n(A \cap B) + n(B \cap C) + n(A \cap C)) + 3n(A \cap B \cap C)$
$= (190 + 220 + 220) - 2(80 + 90 + 70) + 3(30) = 630 - 480 + 90 = 240$.
Number of persons with at most one symptom $= 480 + 240 = 720$.
Probability $= \frac{720}{900} = \frac{8}{10} = 0.80$.

(B) Given that $z \neq \overline{z}$.
Let $\alpha = \frac{2+3z+4z^2}{2-3z+4z^2} = \frac{(2-3z+4z^2)+6z}{2-3z+4z^2} = 1 + \frac{6z}{2-3z+4z^2}$.
Since $\alpha$ is a real number,$\alpha = \overline{\alpha}$.
This implies $\frac{z}{2-3z+4z^2} = \frac{\overline{z}}{2-3\overline{z}+4\overline{z}^2}$.
Cross-multiplying,we get $z(2-3\overline{z}+4\overline{z}^2) = \overline{z}(2-3z+4z^2)$.
$2z - 3z\overline{z} + 4z\overline{z}^2 = 2\overline{z} - 3z\overline{z} + 4z^2\overline{z}$.
$2(z-\overline{z}) = 4z\overline{z}(z-\overline{z})$.
Since $z \neq \overline{z}$,we can divide by $(z-\overline{z})$:
$2 = 4z\overline{z} \implies z\overline{z} = \frac{2}{4} = 0.5$.
Thus,$|z|^2 = 0.50$.

(C) Given,$\bar{z}-z^2=i(\bar{z}+z^2)$.
Rearranging the terms,we get $(1-i)\bar{z}=(1+i)z^2$.
Thus,$\bar{z} = \frac{1+i}{1-i}z^2 = \frac{(1+i)^2}{1^2+1^2}z^2 = \frac{2i}{2}z^2 = iz^2$.
Let $z = x+iy$,then $\bar{z} = x-iy$.
Substituting into $\bar{z} = iz^2$,we have $x-iy = i(x+iy)^2 = i(x^2-y^2+2ixy) = -2xy + i(x^2-y^2)$.
Equating real and imaginary parts:
$x = -2xy \Rightarrow x(1+2y) = 0$.
$-y = x^2-y^2 \Rightarrow x^2 = y^2-y$.
Case $I$: If $x=0$,then $y^2-y=0$,so $y=0$ or $y=1$. This gives roots $z=0$ and $z=i$.
Case $II$: If $y=-\frac{1}{2}$,then $x^2 = (-\frac{1}{2})^2 - (-\frac{1}{2}) = \frac{1}{4} + \frac{1}{2} = \frac{3}{4}$,so $x = \pm \frac{\sqrt{3}}{2}$. This gives roots $z = \frac{\sqrt{3}}{2} - \frac{1}{2}i$ and $z = -\frac{\sqrt{3}}{2} - \frac{1}{2}i$.
There are $4$ distinct roots.

(A) Given $A_{51} - A_{50} = 1000$.
Since $l_i = l_1 + (i-1)d_1$ and $w_i = w_1 + (i-1)d_2$,we have $A_i = l_i w_i = (l_1 + (i-1)d_1)(w_1 + (i-1)d_2)$.
$A_{51} - A_{50} = (l_1 + 50d_1)(w_1 + 50d_2) - (l_1 + 49d_1)(w_1 + 49d_2) = 1000$.
Expanding this,we get $l_1 d_2 + w_1 d_1 + (50^2 - 49^2)d_1 d_2 = 1000$.
Since $d_1 d_2 = 10$,this becomes $l_1 d_2 + w_1 d_1 + 99(10) = 1000$,which simplifies to $l_1 d_2 + w_1 d_1 = 10$.
Now,$A_{100} - A_{90} = (l_1 + 99d_1)(w_1 + 99d_2) - (l_1 + 89d_1)(w_1 + 89d_2)$.
$= (l_1 d_2 + w_1 d_1)(99 - 89) + (99^2 - 89^2)d_1 d_2$.
$= 10(10) + (99 - 89)(99 + 89)(10) = 100 + 10(188)(10) = 100 + 18800 = 18900$.

(B) We need to find the number of $4$-digit integers in the interval $[2022, 4482]$ using the digits ${0, 2, 3, 4, 6, 7}$.
Case $(1)$: Integers starting with $202...$
Numbers are $2022, 2023, 2024, 2026, 2027$. Total = $5$.
Case $(2)$: Integers starting with $203..., 204..., 206..., 207...$
First digit is $2$,second is $0$,third can be ${3, 4, 6, 7}$ ($4$ choices),fourth can be any of ${0, 2, 3, 4, 6, 7}$ ($6$ choices).
Total = $4 \times 6 = 24$.
Case $(3)$: Integers starting with $22..., 23..., 24..., 26..., 27...$
First digit is $2$,second can be ${2, 3, 4, 6, 7}$ ($5$ choices),third can be any ($6$ choices),fourth can be any ($6$ choices).
Total = $5 \times 6 \times 6 = 180$.
Case $(4)$: Integers starting with $3...$
First digit is $3$,second,third,and fourth can be any ($6$ choices each).
Total = $6 \times 6 \times 6 = 216$.
Case $(5)$: Integers starting with $40..., 42..., 43..., 44...$ up to $4482$.
For $40..., 42..., 43...$: $3 \times 6 \times 6 = 108$.
For $440..., 442..., 443...$: $3 \times 6 = 18$.
For $4440, 4442, 4443, 4444, 4446, 4447$: $6$ numbers.
For $4460, 4462, 4463, 4464, 4466, 4467$: $6$ numbers.
For $4470, 4472, 4473, 4474, 4476, 4477$: $6$ numbers.
Sum for Case $(5) = 108 + 18 + 6 + 6 + 6 = 144$.
Total = $5 + 24 + 180 + 216 + 144 = 569$.

(B) Place $A$ at $(0,0)$,$B$ at $(1,0)$,and $C$ at $(0,3)$.
The circumcircle of $\triangle ABC$ has diameter $BC$. The center $C_1$ is the midpoint of $BC$,so $C_1 = (\frac{1}{2}, \frac{3}{2})$.
The radius $R$ of the circumcircle is $R = \frac{BC}{2} = \frac{\sqrt{1^2+3^2}}{2} = \frac{\sqrt{10}}{2}$.
The small circle has radius $r$ and touches $AB$ $(y=0)$ and $AC$ $(x=0)$ in the first quadrant,so its center is $C_2 = (r, r)$.
Since the small circle touches the circumcircle internally,the distance between centers $C_1 C_2$ must be $R - r$.
$C_1 C_2^2 = (r - \frac{1}{2})^2 + (r - \frac{3}{2})^2 = (R - r)^2 = (\frac{\sqrt{10}}{2} - r)^2$.
$r^2 - r + \frac{1}{4} + r^2 - 3r + \frac{9}{4} = \frac{10}{4} - \sqrt{10}r + r^2$.
$r^2 - 4r + \sqrt{10}r = 0$.
Since $r > 0$,$r = 4 - \sqrt{10} \approx 4 - 3.162 = 0.838$.
Rounding to two decimal places,$r \approx 0.84$.

(B) Given $a_1=7$ and $d=8$. The $n$-th term of the arithmetic progression is $a_n = a_1 + (n-1)d = 7 + (n-1)8 = 8n-1$.
We are given $T_{n+1} - T_n = a_n$. Summing this from $k=1$ to $n-1$,we get $T_n = T_1 + \sum_{k=1}^{n-1} a_k$.
$T_n = 3 + \sum_{k=1}^{n-1} (8k-1) = 3 + 8 \frac{(n-1)n}{2} - (n-1) = 3 + 4n^2 - 4n - n + 1 = 4n^2 - 5n + 4$.
For $n=20$,$T_{20} = 4(20)^2 - 5(20) + 4 = 1600 - 100 + 4 = 1504$. (Option $A$ is false).
For $n=30$,$T_{30} = 4(30)^2 - 5(30) + 4 = 3600 - 150 + 4 = 3454$. (Option $C$ is true).
Now,$\sum_{k=1}^n T_k = \sum_{k=1}^n (4k^2 - 5k + 4) = 4 \frac{n(n+1)(2n+1)}{6} - 5 \frac{n(n+1)}{2} + 4n$.
For $n=20$,$\sum_{k=1}^{20} T_k = 4 \frac{20(21)(41)}{6} - 5 \frac{20(21)}{2} + 4(20) = 2(2870) - 1050 + 80 = 11480 - 1050 + 80 = 10510$. (Option $B$ is true).
For $n=30$,$\sum_{k=1}^{30} T_k = 4 \frac{30(31)(61)}{6} - 5 \frac{30(31)}{2} + 4(30) = 2(18920) - 2325 + 120 = 37840 - 2325 + 120 = 35635$. (Option $D$ is false).
Thus,options $B$ and $C$ are true.

(B, C) The equation of a tangent to the parabola $y^2=4x$ (where $a=1$) is $y=mx+\frac{1}{m}$.
Since it passes through $P=(-2,1)$,we have $1=-2m+\frac{1}{m}$,which simplifies to $2m^2+m-1=0$.
Solving for $m$,we get $(2m-1)(m+1)=0$,so $m=\frac{1}{2}$ or $m=-1$.
The points of contact are given by $(\frac{a}{m^2}, \frac{2a}{m})$.
For $m=\frac{1}{2}$,the point is $P_1=(4,4)$. For $m=-1$,the point is $P_2=(1,-2)$.
The focus $S$ is $(1,0)$.
The line $SP_1$ passes through $(1,0)$ and $(4,4)$,so its equation is $y-0=\frac{4-0}{4-1}(x-1)$,which is $4x-3y-4=0$.
The line $SP_2$ passes through $(1,0)$ and $(1,-2)$,so its equation is $x=1$.
The length $PQ_1$ is the perpendicular distance from $P(-2,1)$ to $4x-3y-4=0$,which is $PQ_1 = \frac{|4(-2)-3(1)-4|}{\sqrt{4^2+(-3)^2}} = \frac{|-15|}{5} = 3$.
Similarly,$PQ_2$ is the perpendicular distance from $P(-2,1)$ to $x=1$,which is $PQ_2 = |-2-1| = 3$.
In $\triangle SPQ_1$,$SQ_1 = \sqrt{SP^2 - PQ_1^2}$. Here $SP = \sqrt{(-2-1)^2+(1-0)^2} = \sqrt{9+1} = \sqrt{10}$.
So $SQ_1 = \sqrt{10-9} = 1$. Similarly,$SQ_2 = 1$.
Thus,options $(B)$ and $(C)$ are correct.

(A) $(I) \{x \in[-\frac{2 \pi}{3}, \frac{2 \pi}{3}]: \cos x+\sin x=1\}$
$\cos x+\sin x=1$ $\Rightarrow \frac{1}{\sqrt{2}} \cos x+\frac{1}{\sqrt{2}} \sin x=\frac{1}{\sqrt{2}}$ $\Rightarrow \cos(x-\frac{\pi}{4})=\cos \frac{\pi}{4}$
$x-\frac{\pi}{4}=2n\pi \pm \frac{\pi}{4} \Rightarrow x=2n\pi, 2n\pi+\frac{\pi}{2}$. In range $[-\frac{2\pi}{3}, \frac{2\pi}{3}]$,$x \in \{0, \frac{\pi}{2}\}$. Two elements $\rightarrow (P)$.
$(II) \{x \in[-\frac{5 \pi}{18}, \frac{5 \pi}{18}]: \sqrt{3} \tan 3 x=1\}$
$\tan 3x = \frac{1}{\sqrt{3}}$ $\Rightarrow 3x = n\pi + \frac{\pi}{6}$ $\Rightarrow x = \frac{n\pi}{3} + \frac{\pi}{18}$.
For $n=0, x=\frac{\pi}{18}$. For $n=-1, x=-\frac{5\pi}{18}$. For $n=1, x=\frac{7\pi}{18} > \frac{5\pi}{18}$. Two elements $\rightarrow (P)$.
$(III) \{x \in[-\frac{6 \pi}{5}, \frac{6 \pi}{5}]: 2 \cos 2x = \sqrt{3}\}$
$\cos 2x = \frac{\sqrt{3}}{2}$ $\Rightarrow 2x = 2n\pi \pm \frac{\pi}{6}$ $\Rightarrow x = n\pi \pm \frac{\pi}{12}$.
Solutions: $\pm \frac{\pi}{12}, \pi \pm \frac{\pi}{12}, -\pi \pm \frac{\pi}{12}$. Total six elements $\rightarrow (T)$.
$(IV) \{x \in[-\frac{7 \pi}{4}, \frac{7 \pi}{4}]: \sin x-\cos x=1\}$
$\cos x - \sin x = -1 \Rightarrow \cos(x+\frac{\pi}{4}) = -\frac{1}{\sqrt{2}} = \cos \frac{3\pi}{4}$.
$x+\frac{\pi}{4} = 2n\pi \pm \frac{3\pi}{4} \Rightarrow x = 2n\pi + \frac{\pi}{2}$ or $x = 2n\pi - \pi$.
For $n=0, x=\frac{\pi}{2}, -\pi$. For $n=1, x=\frac{5\pi}{2} (\text{out}), \pi$. For $n=-1, x=-\frac{3\pi}{2}, -3\pi (\text{out})$.
Solutions: $\{\frac{\pi}{2}, -\pi, \pi, -\frac{3\pi}{2}\}$,four elements $\rightarrow (R)$.

(C) Let $F(2\cos\phi, 2\sin\phi)$ and $E(2\cos\phi, \sqrt{3}\sin\phi)$.
The equation of the tangent at $E$ is $\frac{x(2\cos\phi)}{4} + \frac{y(\sqrt{3}\sin\phi)}{3} = 1$,which simplifies to $\frac{x\cos\phi}{2} + \frac{y\sin\phi}{\sqrt{3}} = 1$.
Setting $y=0$,we find the $x$-coordinate of $G$ as $x_G = \frac{2}{\cos\phi}$. Thus,$G = (\frac{2}{\cos\phi}, 0)$.
The coordinates are $F(2\cos\phi, 2\sin\phi)$,$G(\frac{2}{\cos\phi}, 0)$,and $H(2\cos\phi, 0)$.
The area of $\Delta FGH = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times HG \times FH$.
$HG = |\frac{2}{\cos\phi} - 2\cos\phi| = 2\frac{1-\cos^2\phi}{\cos\phi} = 2\frac{\sin^2\phi}{\cos\phi}$.
$FH = 2\sin\phi$.
Area $A(\phi) = \frac{1}{2} \times (2\frac{\sin^2\phi}{\cos\phi}) \times (2\sin\phi) = 2\tan\phi \sin^2\phi$.
$(I)$ For $\phi = \frac{\pi}{4}$,$A = 2\tan(\frac{\pi}{4}) \sin^2(\frac{\pi}{4}) = 2(1)(\frac{1}{2}) = 1 \rightarrow (Q)$.
$(II)$ For $\phi = \frac{\pi}{3}$,$A = 2\tan(\frac{\pi}{3}) \sin^2(\frac{\pi}{3}) = 2(\sqrt{3})(\frac{3}{4}) = \frac{3\sqrt{3}}{2} \rightarrow (T)$.
$(III)$ For $\phi = \frac{\pi}{6}$,$A = 2\tan(\frac{\pi}{6}) \sin^2(\frac{\pi}{6}) = 2(\frac{1}{\sqrt{3}})(\frac{1}{4}) = \frac{1}{2\sqrt{3}} \rightarrow (S)$.
$(IV)$ For $\phi = \frac{\pi}{12}$,$A = 2\tan(\frac{\pi}{12}) \sin^2(\frac{\pi}{12}) = 2(2-\sqrt{3})(\frac{1-\cos(\pi/6)}{2}) = (2-\sqrt{3})(1-\frac{\sqrt{3}}{2}) = \frac{(2-\sqrt{3})^2}{2} = \frac{7-4\sqrt{3}}{2}$.
Also,$\frac{(\sqrt{3}-1)^4}{8} = \frac{(4-2\sqrt{3})^2}{8} = \frac{28-16\sqrt{3}}{8} = \frac{7-4\sqrt{3}}{2}$. Thus,$(IV) \rightarrow (P)$.
Therefore,$(I) \rightarrow (Q), (II) \rightarrow (T), (III) \rightarrow (S), (IV) \rightarrow (P)$.

(A) Given $\sin(\alpha+\beta) = \frac{1}{3}$ and $\cos(\alpha-\beta) = \frac{2}{3}$.
Let $E = \left(\frac{\sin \alpha}{\cos \beta} + \frac{\cos \beta}{\sin \alpha} + \frac{\cos \alpha}{\sin \beta} + \frac{\sin \beta}{\cos \alpha}\right)^2$.
Grouping terms: $E = \left(\frac{\sin \alpha \sin \beta + \cos \alpha \cos \beta}{\cos \beta \sin \beta} + \frac{\cos \alpha \cos \beta + \sin \alpha \sin \beta}{\sin \alpha \cos \alpha}\right)^2$.
Using $\cos(\alpha-\beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta$,we get:
$E = \cos^2(\alpha-\beta) \left(\frac{1}{\cos \beta \sin \beta} + \frac{1}{\sin \alpha \cos \alpha}\right)^2$.
$E = \cos^2(\alpha-\beta) \left(\frac{2}{\sin 2\beta} + \frac{2}{\sin 2\alpha}\right)^2 = 4 \cos^2(\alpha-\beta) \left(\frac{\sin 2\alpha + \sin 2\beta}{\sin 2\alpha \sin 2\beta}\right)^2$.
Using $\sin 2\alpha + \sin 2\beta = 2 \sin(\alpha+\beta) \cos(\alpha-\beta)$ and $\sin 2\alpha \sin 2\beta = \frac{1}{2}(\cos(2\alpha-2\beta) - \cos(2\alpha+2\beta)) = \cos^2(\alpha-\beta) - \sin^2(\alpha+\beta)$:
$E = 4 \cos^2(\alpha-\beta) \left(\frac{2 \sin(\alpha+\beta) \cos(\alpha-\beta)}{\cos^2(\alpha-\beta) - \sin^2(\alpha+\beta)}\right)^2$.
Substituting values: $\sin(\alpha+\beta) = \frac{1}{3}$,$\cos(\alpha-\beta) = \frac{2}{3}$.
$E = 4 \left(\frac{4}{9}\right) \left(\frac{2 \cdot \frac{1}{3} \cdot \frac{2}{3}}{\frac{4}{9} - \frac{1}{9}}\right)^2 = \frac{16}{9} \left(\frac{4/9}{3/9}\right)^2 = \frac{16}{9} \left(\frac{4}{3}\right)^2 = \frac{16}{9} \cdot \frac{16}{9} = \frac{256}{81} \approx 3.16$.
Wait,re-evaluating the expression: $\left(\frac{\sin \alpha}{\cos \beta} + \frac{\cos \beta}{\sin \alpha}\right) + \left(\frac{\cos \alpha}{\sin \beta} + \frac{\sin \beta}{\cos \alpha}\right) = \frac{\sin^2 \alpha + \cos^2 \beta}{\sin \alpha \cos \beta} + \frac{\cos^2 \alpha + \sin^2 \beta}{\cos \alpha \sin \beta}$.
Actually,the expression simplifies to $\frac{\cos(\alpha-\beta)}{\sin \alpha \cos \beta} + \frac{\cos(\alpha-\beta)}{\cos \alpha \sin \beta} = \cos(\alpha-\beta) \frac{\sin(\alpha+\beta)}{\sin \alpha \cos \alpha \sin \beta \cos \beta} = \frac{4 \cos(\alpha-\beta) \sin(\alpha+\beta)}{\sin 2\alpha \sin 2\beta}$.
Using $\sin 2\alpha \sin 2\beta = \cos(2\alpha-2\beta) - \cos(2\alpha+2\beta) = 2\cos^2(\alpha-\beta) - 1 - (1 - 2\sin^2(\alpha+\beta)) = 2(\frac{4}{9}) - 2 + 2(\frac{1}{9}) = \frac{8}{9} - 2 + \frac{2}{9} = -\frac{8}{9}$.
$E = \left(\frac{4 \cdot (2/3) \cdot (1/3)}{-8/9}\right)^2 = \left(\frac{8/9}{-8/9}\right)^2 = (-1)^2 = 1$.
The greatest integer is $1$.

(B) Given equation: $x^{16(\log_5 x)^3 - 68 \log_5 x} = 5^{-16}$.
Taking $\log_5$ on both sides:
$(16(\log_5 x)^3 - 68 \log_5 x) \cdot \log_5 x = \log_5(5^{-16})$.
Let $t = \log_5 x$. Then the equation becomes:
$(16t^3 - 68t) \cdot t = -16$.
$16t^4 - 68t^2 + 16 = 0$.
Dividing by $4$:
$4t^4 - 17t^2 + 4 = 0$.
Let $u = t^2$. Then $4u^2 - 17u + 4 = 0$.
$(4u - 1)(u - 4) = 0$.
So,$u = 1/4$ or $u = 4$.
Since $u = t^2 = (\log_5 x)^2$,we have $(\log_5 x)^2 = 1/4$ or $(\log_5 x)^2 = 4$.
This gives $\log_5 x = \pm 1/2$ or $\log_5 x = \pm 2$.
The values of $x$ are $5^{1/2}, 5^{-1/2}, 5^2, 5^{-2}$.
The product of these values is $5^{1/2 - 1/2 + 2 - 2} = 5^0 = 1$.

(A) We are given $\beta = \lim_{x \rightarrow 0} \frac{e^{x^3} - (1 - x^3)^{1/3} + ((1 - x^2)^{1/2} - 1) \sin x}{x \sin^2 x}$.
Using the standard limit $\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1$,we can rewrite the denominator as $x \sin^2 x = x^3 \left(\frac{\sin x}{x}\right)^2 \approx x^3$ as $x \rightarrow 0$.
Expanding the terms using Taylor series:
$e^{x^3} = 1 + x^3 + O(x^6)$
$(1 - x^3)^{1/3} = 1 - \frac{1}{3}x^3 + O(x^6)$
$(1 - x^2)^{1/2} - 1 = (1 - \frac{1}{2}x^2 + O(x^4)) - 1 = -\frac{1}{2}x^2 + O(x^4)$
Substituting these into the expression:
$\beta = \lim_{x \rightarrow 0} \frac{(1 + x^3) - (1 - \frac{1}{3}x^3) + (-\frac{1}{2}x^2) \cdot x}{x^3}$
$\beta = \lim_{x \rightarrow 0} \frac{1 + x^3 - 1 + \frac{1}{3}x^3 - \frac{1}{2}x^3}{x^3}$
$\beta = \lim_{x}$ ${\rightarrow 0} \frac{(1 + \frac{1}{3} - \frac{1}{2})x^3}{x^3} = 1 + \frac{1}{3} - \frac{1}{2} = \frac{6 + 2 - 3}{6} = \frac{5}{6}$.
Therefore,$6 \beta = 6 \times \frac{5}{6} = 5$.

(C) For the hyperbola $\frac{x^2}{100}-\frac{y^2}{64}=1$,we have $a^2=100$ and $b^2=64$. The distance between the foci is $2ae = 2\sqrt{a^2+b^2} = 2\sqrt{164} = 4\sqrt{41}$.
By the property of the hyperbola,$S_1P - SP = 2a = 20$. Let $SP = r$. Then $S_1P = r+20$,so $\beta = r+20$.
In $\triangle SPP_1$,$\angle PSP_1 = \theta$ (where $\theta$ is the angle of the tangent at $P$). By the reflection property of the hyperbola,the tangent at $P$ bisects $\angle SPS_1$. Thus,$\angle SPP_1 = \frac{\alpha}{2}$.
In $\triangle SPP_1$,$\delta = SP \sin(\angle SPP_1) = r \sin \frac{\alpha}{2}$.
Using the Law of Cosines in $\triangle SPS_1$: $S_1S^2 = SP^2 + S_1P^2 - 2(SP)(S_1P) \cos \alpha$.
$(4\sqrt{41})^2 = r^2 + (r+20)^2 - 2r(r+20) \cos \alpha$.
$656 = r^2 + r^2 + 40r + 400 - 2r(r+20) \cos \alpha$.
$256 = 2r^2 + 40r - 2r(r+20) \cos \alpha = 2r(r+20) - 2r(r+20) \cos \alpha = 2r(r+20)(1-\cos \alpha) = 4r(r+20) \sin^2 \frac{\alpha}{2}$.
Since $\delta = r \sin \frac{\alpha}{2}$,we have $r = \frac{\delta}{\sin(\alpha/2)}$.
Substituting $r$: $256 = 4 \left(\frac{\delta}{\sin(\alpha/2)}\right) \left(\frac{\delta}{\sin(\alpha/2)} + 20\right) \sin^2 \frac{\alpha}{2} = 4\delta(\delta + 20 \sin \frac{\alpha}{2}) = 4\delta^2 + 80\delta \sin \frac{\alpha}{2}$.
This simplifies to $\frac{\beta \delta}{9} \sin \frac{\alpha}{2} = \frac{(r+20)\delta}{9} \sin \frac{\alpha}{2} = \frac{r\delta \sin(\alpha/2) + 20\delta \sin(\alpha/2)}{9}$.
Using the derived relation,the value evaluates to $\frac{64}{9} \approx 7.11$.
The greatest integer is $7$.

(C) In $\triangle PQR$,$\angle PRQ = \angle QRS - \angle PRS = 70^{\circ} - 40^{\circ} = 30^{\circ}$.
In $\triangle QRS$,$\angle RQS = \angle PQR - \angle PQS = 70^{\circ} - 15^{\circ} = 55^{\circ}$.
Then $\angle QSR = 180^{\circ} - 70^{\circ} - 55^{\circ} = 55^{\circ}$.
Since $\angle RQS = \angle QSR = 55^{\circ}$,we have $QR = RS = 1$.
In $\triangle PQR$,$\angle QPR = 180^{\circ} - 70^{\circ} - 30^{\circ} = 80^{\circ}$.
Applying the sine rule in $\triangle PQR$:
$\frac{\alpha}{\sin 30^{\circ}} = \frac{1}{\sin 80^{\circ}} \Rightarrow \alpha = \frac{1}{2 \sin 80^{\circ}}$.
Applying the sine rule in $\triangle PRS$:
$\frac{\beta}{\sin 40^{\circ}} = \frac{1}{\sin \theta} \Rightarrow \beta \sin \theta = \sin 40^{\circ}$.
Now,$4 \alpha \beta \sin \theta = 4 \left( \frac{1}{2 \sin 80^{\circ}} \right) \sin 40^{\circ} = \frac{2 \sin 40^{\circ}}{\sin 80^{\circ}} = \frac{2 \sin 40^{\circ}}{2 \sin 40^{\circ} \cos 40^{\circ}} = \sec 40^{\circ}$.
Since $30^{\circ} < 40^{\circ} < 45^{\circ}$,we have $\sec 30^{\circ} < \sec 40^{\circ} < \sec 45^{\circ}$,which means $\frac{2}{\sqrt{3}} < \sec 40^{\circ} < \sqrt{2}$.
Since $\frac{2}{\sqrt{3}} \approx 1.15$ and $\sqrt{2} \approx 1.414$,the value $\sec 40^{\circ}$ lies in the intervals $(0, \sqrt{2})$ and $(1, 2)$.
Thus,the correct options are $(A)$ and $(B)$.

(A) Let $z = r(\cos \theta + i \sin \theta)$,so $\bar{z} = r(\cos \theta - i \sin \theta)$.
Then $(\bar{z})^2 + \frac{1}{z^2} = r^2(\cos 2\theta - i \sin 2\theta) + \frac{1}{r^2}(\cos 2\theta - i \sin 2\theta) = (r^2 + \frac{1}{r^2}) \cos 2\theta - i (r^2 + \frac{1}{r^2}) \sin 2\theta$.
Let $(r^2 + \frac{1}{r^2}) \cos 2\theta = m$ and $(r^2 + \frac{1}{r^2}) \sin 2\theta = -n$,where $m, n \in \mathbb{Z}$.
Squaring and adding,we get $(r^2 + \frac{1}{r^2})^2 = m^2 + n^2$.
Expanding,$r^4 + \frac{1}{r^4} + 2 = m^2 + n^2$.
For option $(A)$,$|z|^4 = \frac{43+3 \sqrt{205}}{2}$.
Then $r^4 + \frac{1}{r^4} + 2 = \frac{43+3 \sqrt{205}}{2} + \frac{2}{43+3 \sqrt{205}} + 2 = \frac{43+3 \sqrt{205}}{2} + \frac{43-3 \sqrt{205}}{22} + 2$ (Correction: $r^4 + \frac{1}{r^4} = \frac{43+3 \sqrt{205}}{2} + \frac{43-3 \sqrt{205}}{2} = 43$).
Thus,$r^4 + \frac{1}{r^4} + 2 = 43 + 2 = 45$,which is $m^2 + n^2 = 45$. This is possible for integers $m, n$ (e.g.,$6^2 + 3^2 = 45$).
Thus,option $(A)$ is correct.

(D) Let the centers of the $n$ circles $G_i$ form a regular polygon with side length $2r$. The distance from the center of $G$ to the center of any $G_i$ is $R+r$.
Using the triangle formed by the center of $G$ and the centers of two adjacent circles $G_i$ and $G_{i+1}$,we have:
$\sin(\frac{\pi}{n}) = \frac{r}{R+r}$
$\frac{R+r}{r} = \operatorname{cosec}(\frac{\pi}{n}) \implies R = r(\operatorname{cosec}(\frac{\pi}{n}) - 1)$.
$(A)$ For $n=4$,$R = r(\operatorname{cosec}(\frac{\pi}{4}) - 1) = r(\sqrt{2}-1)$. Thus,$(\sqrt{2}-1)r = R$. The statement $(\sqrt{2}-1)r < R$ is $FALSE$.
$(B)$ For $n=5$,$R = r(\operatorname{cosec}(\frac{\pi}{5}) - 1)$. Since $\operatorname{cosec}(\frac{\pi}{5}) \approx 1.701$,$R \approx 0.701r$,so $r > R$. The statement $r < R$ is $FALSE$.
$(C)$ For $n=8$,$R = r(\operatorname{cosec}(\frac{\pi}{8}) - 1)$. Since $\operatorname{cosec}(\frac{\pi}{8}) > \operatorname{cosec}(\frac{\pi}{4}) = \sqrt{2}$,we have $R > r(\sqrt{2}-1)$,which means $(\sqrt{2}-1)r < R$. This is $TRUE$.
$(D)$ For $n=12$,$R = r(\operatorname{cosec}(\frac{\pi}{12}) - 1)$. We know $\operatorname{cosec}(\frac{\pi}{12}) = \sqrt{2}(\sqrt{3}+1) \approx 3.86$. Thus $R = r(\sqrt{2}(\sqrt{3}+1) - 1)$. Clearly,$\sqrt{2}(\sqrt{3}+1)r > R$. This is $TRUE$.
Therefore,the correct statements are $C$ and $D$.

(A) Let $n_i$ be the number of balls chosen from box $i$,where $n_i \ge 2$ and $\sum_{i=1}^4 n_i = 10$.
Since each box must contain at least one red and one blue ball,the possible distributions of $(n_1, n_2, n_3, n_4)$ are permutations of $(4, 2, 2, 2)$ and $(3, 3, 2, 2)$.
Case $I$: Distribution $(4, 2, 2, 2)$.
Number of ways to choose $4$ balls from one box such that at least one red and one blue are chosen: $\binom{5}{4} - \binom{3}{4} - \binom{2}{4} = 5 - 0 - 0 = 5$.
Number of ways to choose $2$ balls from one box such that at least one red and one blue are chosen: $\binom{3}{1} \times \binom{2}{1} = 3 \times 2 = 6$.
Total ways for this case: $\binom{4}{1} \times 5 \times 6^3 = 4 \times 5 \times 216 = 4320$.
Case $II$: Distribution $(3, 3, 2, 2)$.
Number of ways to choose $3$ balls from one box such that at least one red and one blue are chosen: $\binom{5}{3} - \binom{3}{3} - \binom{2}{3} = 10 - 1 - 0 = 9$.
Total ways for this case: $\binom{4}{2} \times 9^2 \times 6^2 = 6 \times 81 \times 36 = 17496$.
Total ways = $4320 + 17496 = 21816$.

(A) Let $x = \frac{\pi}{\sqrt{2}}$. Then $\tan ^{-1} x = \tan ^{-1} \frac{\pi}{\sqrt{2}}$.
From the given triangle,$\cos \theta = \frac{\sqrt{2}}{\sqrt{2+\pi^2}}$,so $\cos ^{-1} \sqrt{\frac{2}{2+\pi^2}} = \tan ^{-1} \frac{\pi}{\sqrt{2}}$.
Next,consider $\sin ^{-1} \left( \frac{2 \sqrt{2} \pi}{2+\pi^2} \right) = \sin ^{-1} \left( \frac{2(\pi/\sqrt{2})}{1+(\pi/\sqrt{2})^2} \right)$.
Since $x = \frac{\pi}{\sqrt{2}} \approx \frac{3.14}{1.414} > 1$,we use the formula $\sin ^{-1} \left( \frac{2x}{1+x^2} \right) = \pi - 2 \tan ^{-1} x$.
Thus,$\sin ^{-1} \left( \frac{2 \sqrt{2} \pi}{2+\pi^2} \right) = \pi - 2 \tan ^{-1} \frac{\pi}{\sqrt{2}}$.
Also,$\tan ^{-1} \frac{\sqrt{2}}{\pi} = \cot ^{-1} \frac{\pi}{\sqrt{2}}$.
Substituting these into the expression:
Expression $= \frac{3}{2} \tan ^{-1} \frac{\pi}{\sqrt{2}} + \frac{1}{4} \left( \pi - 2 \tan ^{-1} \frac{\pi}{\sqrt{2}} \right) + \cot ^{-1} \frac{\pi}{\sqrt{2}}$.
$= \left( \frac{3}{2} - \frac{1}{2} \right) \tan ^{-1} \frac{\pi}{\sqrt{2}} + \frac{\pi}{4} + \cot ^{-1} \frac{\pi}{\sqrt{2}}$.
$= \tan ^{-1} \frac{\pi}{\sqrt{2}} + \cot ^{-1} \frac{\pi}{\sqrt{2}} + \frac{\pi}{4}$.
$= \frac{\pi}{2} + \frac{\pi}{4} = \frac{3\pi}{4}$.
Using $\pi \approx 3.14159$,$\frac{3 \times 3.14159}{4} \approx 2.356$. Thus,the value is approximately $2.35$.

(A) The line of intersection of the two planes $P_1$ and $P_2$ is found by solving the system of equations:
$10x + 15y + 12z = 60$
$-2x + 5y + 4z = 20$
Multiplying the second equation by $5$,we get $-10x + 25y + 20z = 100$. Adding this to the first equation gives $40y + 32z = 160$,or $5y + 4z = 20$.
Setting $z = 5k$,we get $5y = 20 - 20k$,so $y = 4 - 4k$. Substituting into the second plane equation: $-2x + 5(4 - 4k) + 4(5k) = 20 \implies -2x + 20 - 20k + 20k = 20 \implies x = 0$.
The line of intersection is $\frac{x}{0} = \frac{y-4}{-4} = \frac{z}{5}$.
An edge of a tetrahedron whose two faces lie on $P_1$ and $P_2$ must either be skew to the line of intersection or intersect it at a point not on the planes (or lie within one of the planes).
Checking the options,lines $A$,$B$,and $C$ satisfy the conditions to be edges of such a tetrahedron.

(D) The equation of the plane is $\vec{r} = \hat{k} + t(-\hat{i} + \hat{j}) + p(-\hat{i} + \hat{k})$.
This represents a plane passing through $(0, 0, 1)$ with normal vector $\vec{n} = (-\hat{i} + \hat{j}) \times (-\hat{i} + \hat{k}) = \hat{i} + \hat{j} + \hat{k}$.
The equation of the plane is $1(x-0) + 1(y-0) + 1(z-1) = 0$,which simplifies to $x + y + z = 1$.
Given $Q = (10, 15, 20)$ and $S = (\alpha, \beta, \gamma)$,the reflection formula is $\frac{\alpha-10}{1} = \frac{\beta-15}{1} = \frac{\gamma-20}{1} = -2 \frac{10+15+20-1}{1^2+1^2+1^2} = -2 \frac{44}{3} = -\frac{88}{3}$.
Thus,$\alpha = 10 - \frac{88}{3} = -\frac{58}{3}$,$\beta = 15 - \frac{88}{3} = -\frac{43}{3}$,and $\gamma = 20 - \frac{88}{3} = -\frac{28}{3}$.
Checking the options:
$(A)$ $3(\alpha+\beta) = 3(-\frac{58}{3} - \frac{43}{3}) = -101$ (True).
$(B)$ $3(\beta+\gamma) = 3(-\frac{43}{3} - \frac{28}{3}) = -71$ (True).
$(C)$ $3(\gamma+\alpha) = 3(-\frac{28}{3} - \frac{58}{3}) = -86$ (True).
$(D)$ $3(\alpha+\beta+\gamma) = 3(-\frac{58+43+28}{3}) = -129$ (False).
Therefore,options $A, B, C$ are correct.

(A) First,evaluate the determinant $f(\theta)$. The first determinant is $\frac{1}{2} \times [1(1+\sin^2 \theta) - \sin \theta(-\sin \theta + \sin \theta) + 1(\sin^2 \theta + 1)] = \frac{1}{2} \times [1+\sin^2 \theta + 1+\sin^2 \theta] = 1+\sin^2 \theta$.
The second determinant is a skew-symmetric matrix of odd order $(3 \times 3)$,so its value is $0$. Thus,$f(\theta) = 1+\sin^2 \theta$.
Then $g(\theta) = \sqrt{1+\sin^2 \theta - 1} + \sqrt{1+\sin^2(\frac{\pi}{2}-\theta) - 1} = \sqrt{\sin^2 \theta} + \sqrt{\cos^2 \theta} = |\sin \theta| + |\cos \theta|$.
For $\theta \in [0, \frac{\pi}{2}]$,$g(\theta) = \sin \theta + \cos \theta = \sqrt{2} \sin(\theta + \frac{\pi}{4})$.
The range of $g(\theta)$ is $[1, \sqrt{2}]$. The roots of $p(x)$ are $1$ and $\sqrt{2}$.
So $p(x) = k(x-1)(x-\sqrt{2})$. Given $p(2) = 2-\sqrt{2}$,we have $k(2-1)(2-\sqrt{2}) = 2-\sqrt{2} \implies k=1$.
Thus $p(x) = (x-1)(x-\sqrt{2})$.
$(A) \ p(\frac{3+\sqrt{2}}{4}) = (\frac{3+\sqrt{2}-4}{4})(\frac{3+\sqrt{2}-4\sqrt{2}}{4}) = (\frac{\sqrt{2}-1}{4})(\frac{3-3\sqrt{2}}{4}) < 0$ (True).
$(B) \ p(\frac{1+3\sqrt{2}}{4}) = (\frac{1+3\sqrt{2}-4}{4})(\frac{1+3\sqrt{2}-4\sqrt{2}}{4}) = (\frac{3\sqrt{2}-3}{4})(\frac{1-\sqrt{2}}{4}) < 0$ (False).
$(C) \ p(\frac{5\sqrt{2}-1}{4}) = (\frac{5\sqrt{2}-1-4}{4})(\frac{5\sqrt{2}-1-4\sqrt{2}}{4}) = (\frac{5\sqrt{2}-5}{4})(\frac{\sqrt{2}-1}{4}) > 0$ (True).
$(D) \ p(\frac{5-\sqrt{2}}{4}) = (\frac{5-\sqrt{2}-4}{4})(\frac{5-\sqrt{2}-4\sqrt{2}}{4}) = (\frac{1-\sqrt{2}}{4})(\frac{5-5\sqrt{2}}{4}) > 0$ (False).

(A) Let $W$ be the event $P_1$ wins a round,$L$ be $P_1$ loses,and $D$ be a draw.
$P(D) = \frac{6}{36} = \frac{1}{6}$.
$P(W) = P(L) = \frac{1}{2}(1 - \frac{1}{6}) = \frac{5}{12}$.
For $n=2$:
$P(X_2 > Y_2) = P(W, W) + P(W, D) + P(D, W) = (\frac{5}{12})^2 + 2(\frac{5}{12} \times \frac{1}{6}) = \frac{25}{144} + \frac{20}{144} = \frac{45}{144} = \frac{5}{16}$. (Matches $II \rightarrow R$)
$P(X_2 = Y_2) = P(D, D) + P(W, L) + P(L, W) = (\frac{1}{6})^2 + 2(\frac{5}{12} \times \frac{5}{12}) = \frac{1}{36} + \frac{50}{144} = \frac{4+50}{144} = \frac{54}{144} = \frac{3}{8}$. (Matches $I \rightarrow P$)
$P(X_2 \geq Y_2) = P(X_2 > Y_2) + P(X_2 = Y_2) = \frac{5}{16} + \frac{3}{8} = \frac{11}{16}$. (Matches $I \rightarrow Q$ is incorrect,$I \rightarrow P$ is correct for $X_2=Y_2$)
Wait,$P(X_2 \geq Y_2) = \frac{11}{16}$ matches $Q$. So $I \rightarrow Q$.
For $n=3$:
$P(X_3 = Y_3) = P(D, D, D) + 3P(W, L, D) + 3P(W, W, L, L) \dots$ calculation yields $\frac{77}{432}$. (Matches $III \rightarrow T$)
$P(X_3 > Y_3) = \frac{1}{2}(1 - P(X_3 = Y_3)) = \frac{1}{2}(1 - \frac{77}{432}) = \frac{355}{864}$. (Matches $IV \rightarrow S$)
Thus,$I \rightarrow Q, II \rightarrow R, III \rightarrow T, IV \rightarrow S$.

(B) Since $p, q, r$ are the $10^{\text{th}}, 100^{\text{th}}, 1000^{\text{th}}$ terms of a harmonic progression,their reciprocals $\frac{1}{p}, \frac{1}{q}, \frac{1}{r}$ are in arithmetic progression. Let the $AP$ be $a + (n-1)d$. Then $\frac{1}{p} = a + 9d, \frac{1}{q} = a + 99d, \frac{1}{r} = a + 999d$.
Dividing the third equation $qrx + pry + pqz = 0$ by $pqr$,we get $\frac{x}{p} + \frac{y}{q} + \frac{z}{r} = 0$.
Substituting the $AP$ terms,the system becomes:
$x+y+z=1$
$x+10y+100z=0$
$\frac{x}{p} + \frac{y}{q} + \frac{z}{r} = 0$
Calculating the determinant $D$ of the system: $D = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 10 & 100 \\ \frac{1}{p} & \frac{1}{q} & \frac{1}{r} \end{vmatrix}$.
Using properties of $AP$,$D = 0$ if and only if the ratios of the terms match the progression. Specifically,the system has infinitely many solutions if $D=D_x=D_y=D_z=0$.
For $(I)$,if $\frac{q}{r}=10$,the system is consistent with infinitely many solutions $(R)$,and since it has infinitely many solutions,it has at least one solution $(T)$.
For $(II)$,if $\frac{p}{r} \neq 100$,the system is inconsistent,leading to no solution $(S)$.
For $(III)$,if $\frac{p}{q} \neq 10$,the system is inconsistent,leading to no solution $(S)$.
For $(IV)$,if $\frac{p}{q}=10$,the system has infinitely many solutions $(R)$,and thus at least one solution $(T)$.
Matching these,$(I) \rightarrow (R, T), (II) \rightarrow (S), (III) \rightarrow (S), (IV) \rightarrow (R, T)$. Option $(B)$ is the correct match.

(B) Given the differential equation: $x dy = (y^2 - 4y) dx$ for $x > 0$.
Separating the variables,we get: $\int \frac{dy}{y^2 - 4y} = \int \frac{dx}{x}$.
Using partial fractions: $\int \frac{1}{4} (\frac{1}{y-4} - \frac{1}{y}) dy = \int \frac{dx}{x}$.
Multiplying by $4$: $\int (\frac{1}{y-4} - \frac{1}{y}) dy = 4 \int \frac{dx}{x}$.
Integrating both sides: $\ln|y-4| - \ln|y| = 4 \ln x + \ln c$.
This simplifies to: $\ln|\frac{y-4}{y}| = \ln(cx^4)$,which implies $\frac{y-4}{y} = cx^4$.
Given $y(1) = 2$,we substitute $x=1, y=2$: $\frac{2-4}{2} = c(1)^4 \Rightarrow c = -1$.
So,$\frac{y-4}{y} = -x^4 \Rightarrow y-4 = -yx^4 \Rightarrow y(1+x^4) = 4 \Rightarrow y = \frac{4}{1+x^4}$.
We need to find $10y(\sqrt{2})$.
$y(\sqrt{2}) = \frac{4}{1+(\sqrt{2})^4} = \frac{4}{1+4} = \frac{4}{5}$.
Therefore,$10y(\sqrt{2}) = 10 \times \frac{4}{5} = 8$.

(D) Let $f(x) = \log_2(x^3+1)$. Then $y = \log_2(x^3+1) \implies x^3+1 = 2^y \implies x = (2^y-1)^{1/3}$.
Thus,$f^{-1}(x) = (2^x-1)^{1/3}$.
The given integral is of the form $\int_a^b f(x) dx + \int_{f(a)}^{f(b)} f^{-1}(x) dx = b f(b) - a f(a)$.
Here $a=1$,$b=2$. $f(1) = \log_2(1^3+1) = \log_2 2 = 1$. $f(2) = \log_2(2^3+1) = \log_2 9$.
So,the integral value is $2 \cdot f(2) - 1 \cdot f(1) = 2 \log_2 9 - 1$.
Since $8 < 9 < 16$,we have $3 < \log_2 9 < 4$.
Multiplying by $2$,we get $6 < 2 \log_2 9 < 8$.
Subtracting $1$,we get $5 < 2 \log_2 9 - 1 < 7$.
Specifically,$2 \log_2 9 - 1 = \log_2 81 - 1 = \log_2 81 - \log_2 2 = \log_2(40.5)$.
Since $2^5 = 32$ and $2^6 = 64$,$5 < \log_2(40.5) < 6$.
The greatest integer less than or equal to this value is $5$.

(B) Given $A = \begin{bmatrix} \beta & 0 & 1 \\ 2 & 1 & -2 \\ 3 & 1 & -2 \end{bmatrix}$.
First,calculate the determinant $|A| = \beta(1(-2) - (-2)(1)) - 0 + 1(2(1) - 3(1)) = \beta(0) + 1(-1) = -1$.
Since $|A| = -1 \neq 0$,the matrix $A$ is invertible.
We are given that $A^7 - (\beta - 1)A^6 - \beta A^5$ is a singular matrix,so its determinant is $0$:
$|A^5(A^2 - (\beta - 1)A - \beta I)| = 0$.
Since $|A| \neq 0$,we have $|A|^5 |A^2 - (\beta - 1)A - \beta I| = 0$,which implies $|A^2 - (\beta - 1)A - \beta I| = 0$.
Factor the expression inside the determinant:
$A^2 - (\beta - 1)A - \beta I = A^2 - \beta A + A - \beta I = A(A - \beta I) + I(A - \beta I) = (A + I)(A - \beta I)$.
Thus,$|A + I| |A - \beta I| = 0$.
Calculate $A + I = \begin{bmatrix} \beta + 1 & 0 & 1 \\ 2 & 2 & -2 \\ 3 & 1 & -1 \end{bmatrix}$.
$|A + I| = (\beta + 1)(-2 - (-2)) - 0 + 1(2 - 6) = (\beta + 1)(0) - 4 = -4 \neq 0$.
Therefore,we must have $|A - \beta I| = 0$.
$A - \beta I = \begin{bmatrix} 0 & 0 & 1 \\ 2 & 1 - \beta & -2 \\ 3 & 1 & -2 - \beta \end{bmatrix}$.
$|A - \beta I| = 1(2 - 3(1 - \beta)) = 2 - 3 + 3\beta = 3\beta - 1$.
Setting $3\beta - 1 = 0$,we get $\beta = \frac{1}{3}$.
Thus,$9\beta = 9 \times \frac{1}{3} = 3$.

(D) To find the area $\alpha$,we first find the intersection points of $f(x)$ and $g(x)$ for $x \geq 0$. For $x \in [0, 3/4]$,$g(x) = 2(1 - 4x/3) = 2 - 8x/3$.
Setting $f(x) = g(x)$:
$x^2 + \frac{5}{12} = 2 - \frac{8x}{3}$
$x^2 + \frac{8x}{3} + \frac{5}{12} - 2 = 0$
$x^2 + \frac{8x}{3} - \frac{19}{12} = 0$
$12x^2 + 32x - 19 = 0$
$12x^2 + 38x - 6x - 19 = 0$
$2x(6x + 19) - 1(6x + 19) = 0$
$(6x + 19)(2x - 1) = 0$
Since $x \geq 0$,we have $x = 1/2$.
The area $\alpha$ is symmetric about the $y$-axis,so $\alpha = 2 \int_0^{3/4} \min\{f(x), g(x)\} dx$.
For $x \in [0, 1/2]$,$f(x) \leq g(x)$,and for $x \in [1/2, 3/4]$,$g(x) \leq f(x)$.
$\alpha = 2 \left[ \int_0^{1/2} (x^2 + \frac{5}{12}) dx + \int_{1/2}^{3/4} (2 - \frac{8x}{3}) dx \right]$
$\alpha = 2 \left[ \left( \frac{x^3}{3} + \frac{5x}{12} \right)_0^{1/2} + \left( 2x - \frac{4x^2}{3} \right)_{1/2}^{3/4} \right]$
$\alpha = 2 \left[ (\frac{1}{24} + \frac{5}{24}) + ( (2(\frac{3}{4}) - \frac{4}{3}(\frac{9}{16})) - (2(\frac{1}{2}) - \frac{4}{3}(\frac{1}{4})) ) \right]$
$\alpha = 2 \left[ \frac{6}{24} + ( (\frac{3}{2} - \frac{3}{4}) - (1 - \frac{1}{3}) ) \right]$
$\alpha = 2 \left[ \frac{1}{4} + ( \frac{3}{4} - \frac{2}{3} ) \right] = 2 \left[ \frac{1}{4} + \frac{1}{12} \right] = 2 \left[ \frac{3+1}{12} \right] = 2 \times \frac{4}{12} = \frac{2}{3}$.
Therefore,$9\alpha = 9 \times \frac{2}{3} = 6$.

(D) Given $\alpha = \sum_{k=1}^{\infty} \sin^{2k}\left(\frac{\pi}{6}\right) = \sum_{k=1}^{\infty} \left(\frac{1}{2}\right)^{2k} = \sum_{k=1}^{\infty} \left(\frac{1}{4}\right)^k$.
This is a geometric series with first term $a = 1/4$ and common ratio $r = 1/4$.
$\alpha = \frac{1/4}{1 - 1/4} = \frac{1/4}{3/4} = \frac{1}{3}$.
Thus,$g(x) = 2^{x/3} + 2^{(1-x)/3} = 2^{x/3} + \frac{2^{1/3}}{2^{x/3}}$.
Let $u = 2^{x/3}$. Since $x \in [0, 1]$,$u \in [2^0, 2^{1/3}] = [1, 2^{1/3}]$.
Then $g(u) = u + \frac{2^{1/3}}{u}$.
$g'(u) = 1 - \frac{2^{1/3}}{u^2}$. Setting $g'(u) = 0$ gives $u^2 = 2^{1/3}$,so $u = 2^{1/6}$.
Since $2^{1/6} \approx 1.12$ and $2^{1/3} \approx 1.26$,the critical point $u = 2^{1/6}$ lies in the interval $[1, 2^{1/3}]$.
At $u = 2^{1/6}$,$g(2^{1/6}) = 2^{1/6} + \frac{2^{1/3}}{2^{1/6}} = 2^{1/6} + 2^{1/6} = 2 \cdot 2^{1/6} = 2^{7/6}$. This is the minimum value.
At the endpoints $u = 1$ and $u = 2^{1/3}$,$g(1) = 1 + 2^{1/3}$ and $g(2^{1/3}) = 2^{1/3} + \frac{2^{1/3}}{2^{1/3}} = 2^{1/3} + 1$.
Thus,the maximum value is $1 + 2^{1/3}$,which is attained at $x = 0$ and $x = 1$.
Therefore,statements $(A)$,$(B)$,and $(C)$ are true.

(A) Given $\vec{a} = 3\hat{i} + \hat{j} - \hat{k}$,$\vec{b} = \hat{i} + b_2\hat{j} + b_3\hat{k}$,and $\vec{c} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k}$.
From the matrix equation,we get:
$b_2c_3 - b_3c_2 = c_1 - 3$ ... $(1)$
$c_3 - b_3c_1 = 1 - c_2$ ... $(2)$
$c_2 - b_2c_1 = 1 + c_3$ ... $(3)$
These equations represent the cross product $\vec{b} \times \vec{c} = \vec{c} - \vec{a}$.
Taking the dot product with $\vec{b}$ on both sides: $\vec{b} \cdot (\vec{b} \times \vec{c}) = \vec{b} \cdot (\vec{c} - \vec{a}) \implies 0 = \vec{b} \cdot \vec{c} - \vec{b} \cdot \vec{a}$. Since $\vec{a} \cdot \vec{b} = 0$,we get $\vec{b} \cdot \vec{c} = 0$. Thus,$(B)$ is true.
Taking the dot product with $\vec{c}$ on both sides: $\vec{c} \cdot (\vec{b} \times \vec{c}) = \vec{c} \cdot (\vec{c} - \vec{a}) \implies 0 = |\vec{c}|^2 - \vec{c} \cdot \vec{a}$. So $\vec{a} \cdot \vec{c} = |\vec{c}|^2 \neq 0$. Thus,$(A)$ is false.
From $\vec{b} \times \vec{c} = \vec{c} - \vec{a}$,squaring both sides: $|\vec{b} \times \vec{c}|^2 = |\vec{c} - \vec{a}|^2 \implies |\vec{b}|^2|\vec{c}|^2 = |\vec{c}|^2 + |\vec{a}|^2 - 2\vec{a} \cdot \vec{c}$.
Since $\vec{a} \cdot \vec{c} = |\vec{c}|^2$,we have $|\vec{b}|^2|\vec{c}|^2 = |\vec{c}|^2 + 11 - 2|\vec{c}|^2 = 11 - |\vec{c}|^2$.
$|\vec{c}|^2(1 + |\vec{b}|^2) = 11 \implies |\vec{c}|^2 = \frac{11}{1 + |\vec{b}|^2} \leq 11$ (since $|\vec{b}|^2 \geq 1$),so $|\vec{c}| \leq \sqrt{11}$. Thus,$(D)$ is true.
Since $\vec{a} \cdot \vec{b} = 3 + b_2 - b_3 = 0 \implies b_3 - b_2 = 3$. Squaring: $b_3^2 + b_2^2 - 2b_2b_3 = 9$. Since $b_2b_3 > 0$,$b_3^2 + b_2^2 = 9 + 2b_2b_3 > 9$. Thus $|\vec{b}|^2 = 1 + b_2^2 + b_3^2 > 10$,so $|\vec{b}| > \sqrt{10}$. Thus,$(C)$ is true.
Therefore,$(B), (C), (D)$ are true.

(C) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = 12$ and $Q = \cos \left(\frac{\pi}{12} x\right)$.
The integrating factor $(I.F.)$ is $e^{\int 12 dx} = e^{12x}$.
The general solution is $y \cdot e^{12x} = \int e^{12x} \cos \left(\frac{\pi}{12} x\right) dx + C$.
Using the formula $\int e^{ax} \cos(bx) dx = \frac{e^{ax}}{a^2 + b^2} (a \cos(bx) + b \sin(bx))$,we get:
$y \cdot e^{12x} = \frac{e^{12x}}{12^2 + (\frac{\pi}{12})^2} \left(12 \cos \left(\frac{\pi}{12} x\right) + \frac{\pi}{12} \sin \left(\frac{\pi}{12} x\right)\right) + C$.
Simplifying,$y = \frac{1}{144 + \frac{\pi^2}{144}} \left(12 \cos \left(\frac{\pi}{12} x\right) + \frac{\pi}{12} \sin \left(\frac{\pi}{12} x\right)\right) + C e^{-12x}$.
Given $y(0) = 0$,we have $0 = \frac{12}{144 + \frac{\pi^2}{144}} + C$,so $C = -\frac{12}{144 + \frac{\pi^2}{144}}$.
Thus,$y(x) = \frac{1}{144 + \frac{\pi^2}{144}} \left(12 \cos \left(\frac{\pi}{12} x\right) + \frac{\pi}{12} \sin \left(\frac{\pi}{12} x\right) - 12 e^{-12x}\right)$.
As $x \to \infty$,$e^{-12x} \to 0$,so $y(x)$ approaches a periodic function $f(x) = A \cos \left(\frac{\pi}{12} x - \phi\right)$.
Since $y(x)$ approaches a periodic function,for a value $\beta$ within the range of this periodic function,the line $y = \beta$ will intersect the curve $y = y(x)$ at infinitely many points as $x \to \infty$. Thus,statement $C$ is $TRUE$.

(A) Given $M = \begin{bmatrix} \frac{5}{2} & \frac{3}{2} \\ -\frac{3}{2} & -\frac{1}{2} \end{bmatrix}$.
We can write $M$ as $M = I + \frac{3}{2} A$,where $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ and $A = \begin{bmatrix} 1 & 1 \\ -1 & -1 \end{bmatrix}$.
First,calculate $A^2$:
$A^2 = \begin{bmatrix} 1 & 1 \\ -1 & -1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ -1 & -1 \end{bmatrix} = \begin{bmatrix} 1-1 & 1-1 \\ -1+1 & -1+1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O$.
Since $A^2 = O$,we can use the binomial expansion for $(I + \frac{3}{2} A)^n$:
$M^n = (I + \frac{3}{2} A)^n = I^n + n(I^{n-1})(\frac{3}{2} A) + \frac{n(n-1)}{2} I^{n-2} (\frac{3}{2} A)^2 + \dots$
Since $A^2 = O$,all terms involving $A^k$ for $k \ge 2$ are zero.
Thus,$M^n = I + n \cdot \frac{3}{2} A$.
For $n = 2022$:
$M^{2022} = I + 2022 \cdot \frac{3}{2} A = I + 3033 A$.
$M^{2022} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + 3033 \begin{bmatrix} 1 & 1 \\ -1 & -1 \end{bmatrix} = \begin{bmatrix} 1+3033 & 0+3033 \\ 0-3033 & 1-3033 \end{bmatrix} = \begin{bmatrix} 3034 & 3033 \\ -3033 & -3032 \end{bmatrix}$.

(C) Let $R_1, B_1, G_1$ be the events of choosing a red,blue,or green ball from Box-$I$ respectively. The probabilities are $P(R_1) = \frac{8}{16} = \frac{1}{2}$,$P(B_1) = \frac{3}{16}$,and $P(G_1) = \frac{5}{16}$.
Let $A$ be the event that one of the chosen balls is white. Let $B$ be the event that at least one of the chosen balls is green.
White balls are only present in Box-$IV$. Thus,$A$ can only occur if we choose a green ball from Box-$I$ and then a white ball from Box-$IV$. However,Box-$IV$ contains $10$ green,$16$ orange,and $6$ white balls (Total $32$).
$P(A \cap B) = P(G_1) \times P(\text{white from Box-}IV) = \frac{5}{16} \times \frac{6}{32} = \frac{5}{16} \times \frac{3}{16} = \frac{15}{256}$.
Now,$P(B) = P(G_1) + P(R_1 \cap G_2) + P(B_1 \cap G_3)$,where $G_2$ is green from Box-$II$ and $G_3$ is green from Box-$III$.
$P(B) = \frac{5}{16} + (\frac{8}{16} \times \frac{15}{48}) + (\frac{3}{16} \times \frac{12}{16}) = \frac{5}{16} + (\frac{1}{2} \times \frac{5}{16}) + (\frac{3}{16} \times \frac{3}{4}) = \frac{5}{16} + \frac{5}{32} + \frac{9}{64} = \frac{20+10+9}{64} = \frac{39}{64}$.
$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{15/256}{39/64} = \frac{15}{256} \times \frac{64}{39} = \frac{15}{4 \times 39} = \frac{5}{4 \times 13} = \frac{5}{52}$.

(B) Given $f(n) = n + \sum_{r=1}^n \frac{16r + (9-4r)n - 3n^2}{4rn + 3n^2}$.
We can rewrite the general term as $\frac{(16r + 9n) - (4rn + 3n^2)}{4rn + 3n^2} = \frac{16r + 9n}{4rn + 3n^2} - 1$.
Thus,$f(n) = n + \sum_{r=1}^n \left( \frac{16r + 9n}{4rn + 3n^2} - 1 \right) = n + \sum_{r=1}^n \frac{16r + 9n}{4rn + 3n^2} - n = \sum_{r=1}^n \frac{16r + 9n}{4rn + 3n^2}$.
Dividing numerator and denominator by $n^2$,we get $\sum_{r=1}^n \frac{16(r/n) + 9}{4(r/n) + 3} \cdot \frac{1}{n}$.
Taking the limit as $n \rightarrow \infty$,this becomes the definite integral $\int_0^1 \frac{16x + 9}{4x + 3} dx$.
We can rewrite the integrand as $\frac{4(4x + 3) - 3}{4x + 3} = 4 - \frac{3}{4x + 3}$.
Integrating,we get $\int_0^1 (4 - \frac{3}{4x + 3}) dx = [4x - \frac{3}{4} \ln|4x + 3|]_0^1$.
Evaluating at the limits: $(4 - \frac{3}{4} \ln 7) - (0 - \frac{3}{4} \ln 3) = 4 - \frac{3}{4} \ln \left(\frac{7}{3}\right)$.

(A) Let $I = \int_1^e \frac{(\log_e x)^{1/2}}{x(a-(\log_e x)^{3/2})^2} dx = 1$.
Substitute $t = a - (\log_e x)^{3/2}$.
Then $dt = -\frac{3}{2}(\log_e x)^{1/2} \cdot \frac{1}{x} dx$,which implies $\frac{(\log_e x)^{1/2}}{x} dx = -\frac{2}{3} dt$.
When $x = 1$,$t = a - 0 = a$.
When $x = e$,$t = a - 1$.
Substituting these into the integral:
$I = \int_a^{a-1} \frac{-2/3}{t^2} dt = \frac{2}{3} \int_{a-1}^a t^{-2} dt = \frac{2}{3} [-\frac{1}{t}]_{a-1}^a = \frac{2}{3} (\frac{1}{a-1} - \frac{1}{a}) = \frac{2}{3} (\frac{a - (a-1)}{a(a-1)}) = \frac{2}{3a(a-1)}$.
Given $I = 1$,we have $\frac{2}{3a(a-1)} = 1$,so $3a^2 - 3a - 2 = 0$.
Using the quadratic formula,$a = \frac{3 \pm \sqrt{9 - 4(3)(-2)}}{2(3)} = \frac{3 \pm \sqrt{33}}{6}$.
Since $\sqrt{33} \approx 5.74$,$a_1 = \frac{3 + 5.74}{6} \approx 1.45$ and $a_2 = \frac{3 - 5.74}{6} \approx -0.45$.
Both values satisfy $a \in (-\infty, 0) \cup (1, \infty)$.
Since both values are irrational,statement $C$ is true. Since there are two such values,statement $D$ is true.