JEE Main 2022 Mathematics Question Paper with Answer and Solution

(D) Given statement: $(p \wedge q) \Rightarrow (p \wedge r)$
Using the logical equivalence $A \Rightarrow B \equiv \sim A \vee B$:
$\sim(p \wedge q) \vee (p \wedge r)$
Apply De Morgan's Law:
$(\sim p \vee \sim q) \vee (p \wedge r)$
Using the distributive law $(\sim p \vee \sim q) \vee (p \wedge r) \equiv ((\sim p \vee \sim q) \vee p) \wedge ((\sim p \vee \sim q) \vee r)$:
$((\sim p \vee p) \vee \sim q) \wedge (\sim p \vee \sim q \vee r)$
Since $(\sim p \vee p) \equiv T$ (Tautology):
$(T \vee \sim q) \wedge (\sim p \vee \sim q \vee r)$
$T \wedge (\sim p \vee \sim q \vee r) \equiv \sim p \vee \sim q \vee r$
Rearranging the terms:
$(\sim p \vee \sim q) \vee r$
Using De Morgan's Law again:
$\sim(p \wedge q) \vee r$
Converting back to implication:
$(p \wedge q) \Rightarrow r$

(B) Since $P(1, 1)$ is the circumcentre,it is equidistant from $A, B,$ and $C$. Thus,$PA^2 = PB^2 = PC^2$.
$PA^2 = (a-1)^2 + (3-1)^2 = (a-1)^2 + 4$
$PB^2 = (b-1)^2 + (5-1)^2 = (b-1)^2 + 16$
$PC^2 = (a-1)^2 + (b-1)^2$
Equating $PA^2 = PC^2$: $(a-1)^2 + 4 = (a-1)^2 + (b-1)^2 \implies (b-1)^2 = 4 \implies b-1 = \pm 2$.
So $b = 3$ or $b = -1$. Given $ab > 0$,if $b=3$,then $a > 0$. If $b=-1$,then $a < 0$.
Equating $PA^2 = PB^2$: $(a-1)^2 + 4 = (b-1)^2 + 16$.
If $b = -1$: $(a-1)^2 + 4 = (-1-1)^2 + 16 = 4 + 16 = 20 \implies (a-1)^2 = 16 \implies a-1 = \pm 4$.
$a = 5$ or $a = -3$. Since $a < 0$,we take $a = -3$.
Thus,$A = (-3, 3)$,$B = (-1, 5)$,$C = (-3, -1)$,and $P = (1, 1)$.
Line $AP$ passes through $(-3, 3)$ and $(1, 1)$. Slope $m = \frac{1-3}{1-(-3)} = \frac{-2}{4} = -\frac{1}{2}$.
Equation of $AP$: $y - 1 = -\frac{1}{2}(x - 1) \implies 2y - 2 = -x + 1 \implies x + 2y = 3$.
Line $BC$ passes through $(-1, 5)$ and $(-3, -1)$. Slope $m = \frac{-1-5}{-3-(-1)} = \frac{-6}{-2} = 3$.
Equation of $BC$: $y - 5 = 3(x + 1) \implies y = 3x + 8$.
Substitute $y$ in $AP$: $x + 2(3x + 8) = 3 \implies 7x + 16 = 3 \implies 7x = -13 \implies x = -\frac{13}{7}$.
$y = 3(-\frac{13}{7}) + 8 = \frac{-39 + 56}{7} = \frac{17}{7}$.
$k_{1} + k_{2} = -\frac{13}{7} + \frac{17}{7} = \frac{4}{7}$.

(B) The parabola is $P: y^{2}=4x$,so its focus is $(1, 0)$. Since $L: y=mx+c$ is a focal chord,it passes through $(1, 0)$,so $0 = m(1) + c$,which gives $c = -m$.
Thus,the line is $y = m(x-1)$.
The hyperbola is $H: x^{2}-y^{2}=4$,or $\frac{x^{2}}{4} - \frac{y^{2}}{4} = 1$. Here $a^{2}=4, b^{2}=4$.
The condition for $y=mx+c$ to be a tangent to the hyperbola is $c^{2} = a^{2}m^{2} - b^{2}$.
Substituting $c = -m$,we get $(-m)^{2} = 4m^{2} - 4$,so $m^{2} = 4m^{2} - 4$,which implies $3m^{2} = 4$,or $m = \frac{2}{\sqrt{3}}$ (since $m>0$).
Then $c = -\frac{2}{\sqrt{3}}$. The line $L$ is $y = \frac{2}{\sqrt{3}}(x-1)$.
The intersection points $M(x_{1}, y_{1})$ and $N(x_{2}, y_{2})$ are found by substituting $y = \frac{2}{\sqrt{3}}(x-1)$ into $y^{2}=4x$:
$\frac{4}{3}(x-1)^{2} = 4x \implies (x-1)^{2} = 3x \implies x^{2}-2x+1 = 3x \implies x^{2}-5x+1 = 0$.
The roots are $x_{1}, x_{2}$,so $x_{1}+x_{2}=5$ and $x_{1}x_{2}=1$.
The $y$-coordinates are $y_{i} = \frac{2}{\sqrt{3}}(x_{i}-1)$.
The area of quadrilateral $OMFN$ is the sum of areas of $\triangle OMF$ and $\triangle ONF$.
Area $= \frac{1}{2} |x_{F} y_{1} - x_{F} y_{2}| = \frac{1}{2} |x_{F}| |y_{1}-y_{2}|$.
Here $x_{F} = 2\sqrt{2}$ (focus of $H$ is $(ae, 0) = (2\sqrt{2}, 0)$).
$|y_{1}-y_{2}| = \frac{2}{\sqrt{3}} |x_{1}-x_{2}| = \frac{2}{\sqrt{3}} \sqrt{(x_{1}+x_{2})^{2}-4x_{1}x_{2}} = \frac{2}{\sqrt{3}} \sqrt{25-4} = \frac{2\sqrt{21}}{\sqrt{3}} = 2\sqrt{7}$.
Area $= \frac{1}{2} (2\sqrt{2}) (2\sqrt{7}) = 2\sqrt{14}$.

(D) The total number of elements in $S$ is $2022$.
Prime factorization of $2022 = 2 \times 3 \times 337$.
We need to find the number of integers $n \in S$ such that $\operatorname{HCF}(n, 2022) = 1$.
This is equivalent to finding the number of integers in $S$ that are not divisible by $2, 3,$ or $337$.
Let $A, B,$ and $C$ be the sets of numbers in $S$ divisible by $2, 3,$ and $337$ respectively.
$n(A) = \lfloor \frac{2022}{2} \rfloor = 1011$.
$n(B) = \lfloor \frac{2022}{3} \rfloor = 674$.
$n(C) = \lfloor \frac{2022}{337} \rfloor = 6$.
Using the Principle of Inclusion-Exclusion:
$n(A \cap B) = \lfloor \frac{2022}{6} \rfloor = 337$.
$n(A \cap C) = \lfloor \frac{2022}{674} \rfloor = 3$.
$n(B \cap C) = \lfloor \frac{2022}{1011} \rfloor = 2$.
$n(A \cap B \cap C) = \lfloor \frac{2022}{2022} \rfloor = 1$.
Number of elements divisible by $2, 3,$ or $337$ is $n(A \cup B \cup C) = n(A) + n(B) + n(C) - (n(A \cap B) + n(A \cap C) + n(B \cap C)) + n(A \cap B \cap C)$.
$n(A \cup B \cup C) = 1011 + 674 + 6 - (337 + 3 + 2) + 1 = 1691 - 342 + 1 = 1350$.
Number of favorable cases $= 2022 - 1350 = 672$.
Required probability $= \frac{672}{2022} = \frac{112}{337}$.

(D) Given the equation: $7 \cos^2 \theta - 3 \sin^2 \theta - 2 \cos^2 2\theta = 2$.
Using $\sin^2 \theta = 1 - \cos^2 \theta$,we get $7 \cos^2 \theta - 3(1 - \cos^2 \theta) - 2 \cos^2 2\theta = 2$,which simplifies to $10 \cos^2 \theta - 3 - 2 \cos^2 2\theta = 2$.
Since $2 \cos^2 \theta = 1 + \cos 2\theta$,we have $5(1 + \cos 2\theta) - 3 - 2 \cos^2 2\theta = 2$,leading to $5 + 5 \cos 2\theta - 3 - 2 \cos^2 2\theta = 2$.
This simplifies to $2 \cos^2 2\theta - 5 \cos 2\theta = 0$,so $\cos 2\theta(2 \cos 2\theta - 5) = 0$.
Since $\cos 2\theta$ cannot be $2.5$,we have $\cos 2\theta = 0$,which gives $2\theta = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}$.
Thus,$S = \{\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}\}$.
For any $\theta \in S$,$\tan^2 \theta = 1$ and $\cot^2 \theta = 1$,so $\tan^2 \theta + \cot^2 \theta = 2$.
Also,$\sin^2 \theta = \frac{1}{2}$.
The equation becomes $x^2 - 2(2)x + 6(\frac{1}{2}) = 0$,which is $x^2 - 4x + 3 = 0$.
The sum of roots for each equation is $4$.
Since there are $4$ such equations,the total sum of roots is $4 \times 4 = 16$.

(D) Given that the mean $\bar{x} = 15$ and variance $\sigma^{2} = 9$ for $n = 20$ observations.
$\sum x_{i} = 15 \times 20 = 300$
$\sigma^{2} = \frac{\sum x_{i}^{2}}{n} - (\bar{x})^{2} \Rightarrow 9 = \frac{\sum x_{i}^{2}}{20} - 225$
$\frac{\sum x_{i}^{2}}{20} = 234 \Rightarrow \sum x_{i}^{2} = 4680$
Now,the mean of $(x_{i} + \alpha)^{2}$ is $178$:
$\frac{1}{20} \sum (x_{i} + \alpha)^{2} = 178$
$\sum (x_{i}^{2} + 2\alpha x_{i} + \alpha^{2}) = 178 \times 20 = 3560$
$\sum x_{i}^{2} + 2\alpha \sum x_{i} + 20\alpha^{2} = 3560$
$4680 + 2\alpha(300) + 20\alpha^{2} = 3560$
$20\alpha^{2} + 600\alpha + 1120 = 0$
Dividing by $20$:
$\alpha^{2} + 30\alpha + 56 = 0$
$(\alpha + 28)(\alpha + 2) = 0$
So,$\alpha = -28$ or $\alpha = -2$.
The maximum value of $\alpha$ is $-2$.
The square of the maximum value is $(-2)^{2} = 4$.

(B) Let $S = \sum_{r=1}^{\infty} \frac{a_{r}}{2^{r}} = \frac{a_{1}}{2} + \frac{a_{2}}{2^{2}} + \frac{a_{3}}{2^{3}} + \ldots = 4$
Multiply by $\frac{1}{2}$:
$\frac{S}{2} = \frac{a_{1}}{2^{2}} + \frac{a_{2}}{2^{3}} + \frac{a_{3}}{2^{4}} + \ldots$
Subtracting the two equations:
$S - \frac{S}{2} = \frac{a_{1}}{2} + \frac{a_{2}-a_{1}}{2^{2}} + \frac{a_{3}-a_{2}}{2^{3}} + \ldots$
Since $a_{r}$ is an $A.P.$,$a_{r} - a_{r-1} = d$:
$\frac{S}{2} = \frac{a_{1}}{2} + \frac{d}{2^{2}} + \frac{d}{2^{3}} + \frac{d}{2^{4}} + \ldots$
$\frac{S}{2} = \frac{a_{1}}{2} + d \left( \frac{1/4}{1 - 1/2} \right) = \frac{a_{1}}{2} + d \left( \frac{1/4}{1/2} \right) = \frac{a_{1}}{2} + \frac{d}{2}$
$S = a_{1} + d = a_{2}$
Given $S = 4$,so $a_{2} = 4$.
Therefore,$4 a_{2} = 4 \times 4 = 16$.

(A) The binomial expansion is $(\sqrt[4]{2} + 3^{-1/4})^n$.
Let $T_r$ denote the $r$-th term from the beginning.
The $5$-th term from the beginning is $T_5 = {^nC_4} (2^{1/4})^{n-4} (3^{-1/4})^4$.
The $5$-th term from the end is the $(n-5+2) = (n-3)$-th term from the beginning,which is $T_{n-3} = {^nC_{n-4}} (2^{1/4})^4 (3^{-1/4})^{n-4} = {^nC_4} (2^{1/4})^4 (3^{-1/4})^{n-4}$.
The ratio $\frac{T_5}{T_{n-3}} = \frac{(2^{1/4})^{n-4} (3^{-1/4})^4}{(2^{1/4})^4 (3^{-1/4})^{n-4}} = \frac{2^{(n-8)/4}}{3^{(8-n)/4}} = 2^{(n-8)/4} \cdot 3^{(n-8)/4} = 6^{(n-8)/4}$.
Given $6^{(n-8)/4} = 6^{1/4}$,so $n-8 = 1$,which gives $n = 9$.
The $6$-th term from the beginning is $T_6 = {^9C_5} (2^{1/4})^4 (3^{-1/4})^5 = 126 \cdot 2 \cdot 3^{-5/4} = \frac{252}{3 \cdot 3^{1/4}} = \frac{84}{3^{1/4}}$.
Thus,$\alpha = 84$.

(C) Let $S = \sum_{n=2}^{100} \frac{1}{n(n+1)(n+2)}$.
Using the method of differences,we can write the general term as:
$\frac{1}{n(n+1)(n+2)} = \frac{1}{2} \left[ \frac{(n+2) - n}{n(n+1)(n+2)} \right] = \frac{1}{2} \left[ \frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)} \right]$.
Summing from $n=2$ to $100$:
$S = \frac{1}{2} \left[ (\frac{1}{2 \times 3} - \frac{1}{3 \times 4}) + (\frac{1}{3 \times 4} - \frac{1}{4 \times 5}) + \dots + (\frac{1}{100 \times 101} - \frac{1}{101 \times 102}) \right]$.
This is a telescoping series:
$S = \frac{1}{2} \left[ \frac{1}{6} - \frac{1}{101 \times 102} \right] = \frac{1}{2} \left[ \frac{1}{6} - \frac{1}{10302} \right]$.
$S = \frac{1}{2} \left[ \frac{1717 - 1}{10302} \right] = \frac{1}{2} \times \frac{1716}{10302} = \frac{858}{10302} = \frac{858}{102 \times 101} = \frac{8.4117...}{101} \approx \frac{8.41}{101}$.
Wait,simplifying $\frac{1}{2} [\frac{1}{6} - \frac{1}{10302}] = \frac{1}{12} - \frac{1}{20604} = \frac{1717-1}{20604} = \frac{1716}{20604} = \frac{1}{12} = \frac{1}{101} \times \frac{101}{12} = \frac{8.416}{101}$.
Re-evaluating: $\frac{1}{2} [\frac{1}{6} - \frac{1}{10302}] = \frac{1}{12} - \frac{1}{20604} = \frac{1717-1}{20604} = \frac{1716}{20604} = \frac{1}{12}$.
Given $\frac{k}{101} = \frac{1}{12} \implies k = \frac{101}{12}$.
$34k = 34 \times \frac{101}{12} = \frac{17 \times 101}{6} = \frac{1717}{6} = 286.16$.
Given the options,the intended calculation is $34k = 286$.

(C) Given $S = \{4, 6, 9\}$ and $T = \{9, 10, 11, \ldots, 1000\}$.
The set $A$ consists of all possible sums of elements from $S$. This is equivalent to finding all integers $n$ that can be expressed as $n = 4x + 6y + 9z$,where $x, y, z \in \{0, 1, 2, \ldots\}$ and $x+y+z \geq 1$.
We check the elements of $T$ starting from $9$:
$9 = 9$ (in $A$)
$10 = 4 + 6$ (in $A$)
$11$: We check if $11 = 4x + 6y + 9z$. Possible values for $z$ are $0$ or $1$. If $z=0$,$4x+6y=11$ (no integer solution). If $z=1$,$4x+6y=2$ (no integer solution). Thus,$11 \notin A$.
$12 = 4 + 4 + 4$ (in $A$)
$13 = 4 + 9$ (in $A$)
$14 = 4 + 4 + 6$ (in $A$)
$15 = 6 + 9$ (in $A$)
$16 = 4 + 4 + 4 + 4$ (in $A$)
By the Frobenius Coin Problem or checking small values,all integers $n \geq 12$ can be represented as $4x + 6y + 9z$.
Therefore,$T - A = \{11\}$.
The sum of elements in $T - A$ is $11$.

(C) The circle $c_{1}$ is $x^{2}+y^{2}-2x-6y+\alpha=0$. Its centre is $(1, 3)$ and radius $r_{1} = \sqrt{1^{2}+3^{2}-\alpha} = \sqrt{10-\alpha}$.
The image of the centre $(1, 3)$ in the line $x-y+1=0$ is $(x_{1}, y_{1})$ given by $\frac{x_{1}-1}{1} = \frac{y_{1}-3}{-1} = -2 \frac{1-3+1}{1^{2}+(-1)^{2}} = -2 \frac{-1}{2} = 1$.
Thus,$x_{1}-1=1 \Rightarrow x_{1}=2$ and $y_{1}-3=-1 \Rightarrow y_{1}=2$. The centre of $c_{2}$ is $(2, 2)$.
The circle $c_{2}$ is given as $5x^{2}+5y^{2}+10gx+10fy+38=0$,which is $x^{2}+y^{2}+2gx+2fy+\frac{38}{5}=0$.
Comparing with the standard form,the centre is $(-g, -f) = (2, 2)$,so $g=-2, f=-2$.
The radius $r$ of $c_{2}$ is $\sqrt{g^{2}+f^{2}-\frac{38}{5}} = \sqrt{4+4-\frac{38}{5}} = \sqrt{8-\frac{38}{5}} = \sqrt{\frac{40-38}{5}} = \sqrt{\frac{2}{5}}$.
Since reflection preserves the radius,$r_{1}^{2} = r^{2} \Rightarrow 10-\alpha = \frac{2}{5}$.
Thus,$\alpha = 10 - \frac{2}{5} = \frac{48}{5}$.
Finally,$\alpha+6r^{2} = \frac{48}{5} + 6(\frac{2}{5}) = \frac{48+12}{5} = \frac{60}{5} = 12$.

(D) tautology is a statement that is always true for all possible truth values of its components.
We evaluate option $D$: $(\sim q \wedge p) \vee (p \vee \sim p)$.
We know that $(p \vee \sim p)$ is a tautology (always true,denoted as $T$).
Thus,the expression becomes $(\sim q \wedge p) \vee T$.
Since any statement $X \vee T$ is always $T$,the entire expression is a tautology.
Therefore,the correct option is $D$.

(C) Let $I = 60 \int_{0}^{\pi/2} \frac{\sin(6x)}{\sin x} dx$.
Using the identity $\sin((2n+1)x) - \sin((2n-1)x) = 2 \cos(2nx) \sin x$,we can express $\frac{\sin(6x)}{\sin x}$ as a sum of cosines.
Specifically,$\frac{\sin(6x)}{\sin x} = \frac{\sin(6x) - \sin(4x) + \sin(4x) - \sin(2x) + \sin(2x)}{\sin x} = 2\cos(5x) + 2\cos(3x) + 2\cos(x)$.
Now,integrate each term:
$I = 60 \int_{0}^{\pi/2} (2\cos(5x) + 2\cos(3x) + 2\cos(x)) dx$.
$I = 60 \left[ \frac{2}{5}\sin(5x) + \frac{2}{3}\sin(3x) + 2\sin(x) \right]_{0}^{\pi/2}$.
Evaluating at the limits:
$I = 60 \left( (\frac{2}{5}\sin(\frac{5\pi}{2}) + \frac{2}{3}\sin(\frac{3\pi}{2}) + 2\sin(\frac{\pi}{2})) - (0) \right)$.
Since $\sin(\frac{5\pi}{2}) = 1$,$\sin(\frac{3\pi}{2}) = -1$,and $\sin(\frac{\pi}{2}) = 1$:
$I = 60 \left( \frac{2}{5}(1) + \frac{2}{3}(-1) + 2(1) \right) = 60 \left( \frac{2}{5} - \frac{2}{3} + 2 \right)$.
$I = 60 \left( \frac{6 - 10 + 30}{15} \right) = 60 \left( \frac{26}{15} \right) = 4 \times 26 = 104$.

(B) Given the differential equation: $x dy = (\sqrt{x^{2}+y^{2}}+y) dx$.
Rearranging the terms,we get: $x dy - y dx = \sqrt{x^{2}+y^{2}} dx$.
Dividing both sides by $x^{2}$ (for $x > 0$): $\frac{x dy - y dx}{x^{2}} = \frac{\sqrt{x^{2}+y^{2}}}{x^{2}} dx$.
This simplifies to: $d(\frac{y}{x}) = \sqrt{1 + (\frac{y}{x})^{2}} \cdot \frac{dx}{x}$.
Integrating both sides: $\int \frac{d(y/x)}{\sqrt{1 + (y/x)^{2}}} = \int \frac{dx}{x}$.
Using the standard integral $\int \frac{dt}{\sqrt{1+t^{2}}} = \ln(t + \sqrt{1+t^{2}})$,we get: $\ln(\frac{y}{x} + \sqrt{1 + (\frac{y}{x})^{2}}) = \ln x + C$.
This can be written as: $\frac{y + \sqrt{x^{2}+y^{2}}}{x} = kx$,where $k = e^{C}$.
So,$y + \sqrt{x^{2}+y^{2}} = kx^{2}$.
Given the curve passes through $(1, 0)$,we substitute $x=1, y=0$: $0 + \sqrt{1^{2}+0^{2}} = k(1)^{2} \Rightarrow k = 1$.
The equation of the curve is $y + \sqrt{x^{2}+y^{2}} = x^{2}$.
For $x = 2, y = \alpha$: $\alpha + \sqrt{4+\alpha^{2}} = 2^{2} = 4$.
$\sqrt{4+\alpha^{2}} = 4 - \alpha$.
Squaring both sides: $4 + \alpha^{2} = 16 - 8\alpha + \alpha^{2}$.
$8\alpha = 12 \Rightarrow \alpha = \frac{12}{8} = \frac{3}{2}$.

(B) The domain of $\cos^{-1}(u)$ is $u \in [-1, 1]$.
Thus,we require $\left|\frac{x^{2}-4x+2}{x^{2}+3}\right| \leq 1$.
This implies $-1 \leq \frac{x^{2}-4x+2}{x^{2}+3} \leq 1$.
Since $x^{2}+3 > 0$ for all $x \in \mathbb{R}$,we can multiply by $(x^{2}+3)$:
$-(x^{2}+3) \leq x^{2}-4x+2 \leq x^{2}+3$.
First inequality: $-x^{2}-3 \leq x^{2}-4x+2 \implies 2x^{2}-4x+5 \geq 0$. The discriminant $D = (-4)^{2} - 4(2)(5) = 16 - 40 = -24 < 0$. Since the leading coefficient is positive,$2x^{2}-4x+5 > 0$ for all $x \in \mathbb{R}$.
Second inequality: $x^{2}-4x+2 \leq x^{2}+3 \implies -4x \leq 1 \implies x \geq -\frac{1}{4}$.
Therefore,the domain is $[-\frac{1}{4}, \infty)$.

(D) Given that the vectors $\vec{a}, \vec{b}, \vec{c}$ satisfy the condition $\alpha \vec{a}+\beta \vec{b}+\gamma \vec{c}=\vec{0} \Rightarrow \alpha=\beta=\gamma=0$,it implies that the vectors are linearly independent.
For three vectors to be linearly independent,their scalar triple product must be non-zero,i.e.,$[\vec{a} \vec{b} \vec{c}] \neq 0$.
Calculating the determinant:
$[\vec{a} \vec{b} \vec{c}] = \begin{vmatrix} 1+t & 1-t & 1 \\ 1-t & 1+t & 2 \\ 1 & -t & 1 \end{vmatrix}$
Applying column operation $C_2 \rightarrow C_1 + C_2$:
$= \begin{vmatrix} 1+t & 2 & 1 \\ 1-t & 2 & 2 \\ 1 & 1-t & 1 \end{vmatrix}$
Wait,let us re-evaluate the determinant directly:
$[\vec{a} \vec{b} \vec{c}] = (1+t)((1+t)(1) - (2)(-t)) - (1-t)((1-t)(1) - (2)(1)) + 1((1-t)(-t) - (1+t)(1))$
$= (1+t)(1+t+2t) - (1-t)(1-t-2) + (-t+t^2-1-t)$
$= (1+t)(1+3t) - (1-t)(-1-t) + (t^2-2t-1)$
$= (1+4t+3t^2) - (-(1-t)(1+t)) + t^2-2t-1$
$= 1+4t+3t^2 + (1-t^2) + t^2-2t-1$
$= 3t^2+2t+1$
Re-checking the determinant calculation: $\begin{vmatrix} 1+t & 1-t & 1 \\ 1-t & 1+t & 2 \\ 1 & -t & 1 \end{vmatrix} = (1+t)(1+t+2t) - (1-t)(1-t-2) + 1(-t+t^2-1-t) = (1+t)(1+3t) - (1-t)(-1-t) + (t^2-2t-1) = 1+4t+3t^2 + (1-t^2) + t^2-2t-1 = 3t^2+2t+1$.
Since the condition is $[\vec{a} \vec{b} \vec{c}] \neq 0$,we have $3t^2+2t+1 \neq 0$. The discriminant $D = 2^2 - 4(3)(1) = 4 - 12 = -8 < 0$. Since the quadratic is always positive,it is never zero for any $t \in R$.
Thus,the set of all values of $t$ is $R$.

(A) Given equation: $\cos ^{-1}(x) - 2 \sin ^{-1}(x) = \cos ^{-1}(2x)$
Using the identity $\sin ^{-1}(x) = \frac{\pi}{2} - \cos ^{-1}(x)$,we substitute:
$\cos ^{-1}(x) - 2(\frac{\pi}{2} - \cos ^{-1}(x)) = \cos ^{-1}(2x)$
Simplify the expression:
$\cos ^{-1}(x) - \pi + 2 \cos ^{-1}(x) = \cos ^{-1}(2x)$
$3 \cos ^{-1}(x) = \pi + \cos ^{-1}(2x)$
Taking cosine on both sides:
$\cos(3 \cos ^{-1}(x)) = \cos(\pi + \cos ^{-1}(2x))$
Using the identity $\cos(3\theta) = 4\cos^3(\theta) - 3\cos(\theta)$ and $\cos(\pi + \theta) = -\cos(\theta)$:
$4x^3 - 3x = -2x$
Rearrange the terms:
$4x^3 - x = 0$
$x(4x^2 - 1) = 0$
Thus,the solutions are $x = 0$,$x = \frac{1}{2}$,and $x = -\frac{1}{2}$.
Checking the solutions in the original equation:
For $x = 0$: $\cos^{-1}(0) - 2\sin^{-1}(0) = \frac{\pi}{2} - 0 = \frac{\pi}{2}$ and $\cos^{-1}(0) = \frac{\pi}{2}$. (Valid)
For $x = \frac{1}{2}$: $\cos^{-1}(\frac{1}{2}) - 2\sin^{-1}(\frac{1}{2}) = \frac{\pi}{3} - 2(\frac{\pi}{6}) = 0$ and $\cos^{-1}(1) = 0$. (Valid)
For $x = -\frac{1}{2}$: $\cos^{-1}(-\frac{1}{2}) - 2\sin^{-1}(-\frac{1}{2}) = \frac{2\pi}{3} - 2(-\frac{\pi}{6}) = \frac{2\pi}{3} + \frac{\pi}{3} = \pi$ and $\cos^{-1}(-1) = \pi$. (Valid)
The sum of the solutions is $0 + \frac{1}{2} + (-\frac{1}{2}) = 0$.

(B) Given $|\vec{a}| = 9$. Since $(x \vec{a} + y \vec{b}) \perp (6y \vec{a} - 18x \vec{b})$,their dot product is zero:
$(x \vec{a} + y \vec{b}) \cdot (6y \vec{a} - 18x \vec{b}) = 0$
$6xy |\vec{a}|^2 - 18x^2 (\vec{a} \cdot \vec{b}) + 6y^2 (\vec{a} \cdot \vec{b}) - 18xy |\vec{b}|^2 = 0$
$6xy (|\vec{a}|^2 - 3|\vec{b}|^2) + (\vec{a} \cdot \vec{b})(6y^2 - 18x^2) = 0$
For this to hold for all $(x, y)$,the coefficients of $xy$,$x^2$,and $y^2$ must be zero:
$|\vec{a}|^2 - 3|\vec{b}|^2 = 0 \implies |\vec{b}|^2 = \frac{|\vec{a}|^2}{3} = \frac{81}{3} = 27$
$\vec{a} \cdot \vec{b} = 0$
Now,$|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2 = 81 \times 27 - 0 = 2187$
$|\vec{a} \times \vec{b}| = \sqrt{2187} = \sqrt{81 \times 27} = 9 \times 3 \sqrt{3} = 27 \sqrt{3}$.

(D) For $R$ to be reflexive,$xRx$ must hold for all $x \in N$.
$3x + \alpha x = (3 + \alpha)x$ must be a multiple of $7$.
This implies $(3 + \alpha)$ must be a multiple of $7$,so $3 + \alpha = 7k \Rightarrow \alpha = 7k - 3 = 7(k-1) + 4$.
Thus,when $\alpha$ is divided by $7$,the remainder is $4$.
For $R$ to be symmetric,$xRy \Rightarrow yRx$.
$3x + \alpha y = 7n_1$ and $3y + \alpha x = 7n_2$.
Subtracting these,$(3 - \alpha)(x - y) = 7(n_1 - n_2)$. This condition is satisfied if $3 + \alpha$ is a multiple of $7$.
For $R$ to be transitive,$xRy$ and $yRz \Rightarrow xRz$.
$3x + \alpha y = 7n_1$ and $3y + \alpha z = 7n_2$.
From the first,$\alpha y = 7n_1 - 3x$. Substituting into the second: $3y + \alpha z = 7n_2$.
By eliminating $y$,we find that the relation is transitive if $3 + \alpha$ is a multiple of $7$.
Therefore,the condition for $R$ to be an equivalence relation is that $4$ is the remainder when $\alpha$ is divided by $7$.

(A) The given differential equation is $(\sin^2 2x) \frac{dy}{dx} + (8 \sin^2 2x + 4 \sin 2x \cos 2x) y = 2 e^{-4x} (2 \sin 2x + \cos 2x)$.
Dividing by $\sin^2 2x$,we get $\frac{dy}{dx} + (8 + 4 \cot 2x) y = \frac{2 e^{-4x} (2 \sin 2x + \cos 2x)}{\sin^2 2x}$.
This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = 8 + 4 \cot 2x$.
The integrating factor $I.F. = e^{\int (8 + 4 \cot 2x) dx} = e^{8x + 2 \ln(\sin 2x)} = e^{8x} \sin^2 2x$.
The solution is $y \cdot (e^{8x} \sin^2 2x) = \int (e^{8x} \sin^2 2x) \cdot \frac{2 e^{-4x} (2 \sin 2x + \cos 2x)}{\sin^2 2x} dx + C$.
$y e^{8x} \sin^2 2x = \int 2 e^{4x} (2 \sin 2x + \cos 2x) dx + C$.
Using $\int e^{ax} (a f(x) + f'(x)) dx = e^{ax} f(x) + C$,let $f(x) = \sin 2x$,then $f'(x) = 2 \cos 2x$. The integral becomes $\int e^{4x} (4 \sin 2x + 2 \cos 2x) dx = e^{4x} \sin 2x + C$.
So,$y e^{8x} \sin^2 2x = e^{4x} \sin 2x + C$.
Given $y(\frac{\pi}{4}) = e^{-\pi}$,we have $e^{-\pi} e^{2\pi} (1)^2 = e^{\pi} (1) + C \Rightarrow e^{\pi} = e^{\pi} + C \Rightarrow C = 0$.
Thus,$y = \frac{e^{-4x}}{\sin 2x}$.
For $x = \frac{\pi}{6}$,$y(\frac{\pi}{6}) = \frac{e^{-4(\pi/6)}}{\sin(2 \cdot \pi/6)} = \frac{e^{-2\pi/3}}{\sin(\pi/3)} = \frac{e^{-2\pi/3}}{\sqrt{3}/2} = \frac{2}{\sqrt{3}} e^{-2\pi/3}$.

(C) Given $A = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix}$.
We observe $A^2 = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}$ and $A^3 = I$.
Since $49 = 3 \times 16 + 1$,$A^{49} = A^1 = A$.
Since $98 = 3 \times 32 + 2$,$A^{98} = A^2$.
Thus,$B_{0} = A + 2A^2 = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix} + \begin{bmatrix} 0 & 0 & 2 \\ 2 & 0 & 0 \\ 0 & 2 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 2 \\ 2 & 0 & 1 \\ 1 & 2 & 0 \end{bmatrix}$.
Calculating $\det(B_{0}) = 0(0-2) - 1(0-1) + 2(4-0) = 0 + 1 + 8 = 9$.
We know $\det(\text{Adj}(M)) = (\det(M))^{k-1}$ where $k$ is the order of the matrix. Here $k=3$,so $\det(\text{Adj}(M)) = (\det(M))^2$.
For $B_{n} = \text{Adj}(B_{n-1})$,$\det(B_{n}) = (\det(B_{n-1}))^2$.
Thus,$\det(B_{4}) = (\det(B_{0}))^{2^4} = (\det(B_{0}))^{16}$.
$\det(B_{4}) = 9^{16} = (3^2)^{16} = 3^{32}$.

(B) Let a point on the circle be $P(\cos \theta, \sin \theta, 0)$.
Let the foot of the perpendicular from $P$ to the plane $2x + 3y + z = 6$ be $Q(h, k, w)$.
The line $PQ$ is perpendicular to the plane,so its direction ratios are proportional to the normal of the plane $(2, 3, 1)$.
Thus,$\frac{h - \cos \theta}{2} = \frac{k - \sin \theta}{3} = \frac{w - 0}{1} = \lambda$.
Since $Q(h, k, w)$ lies on the plane $2x + 3y + z = 6$,we have $2h + 3k + w = 6$.
Substituting $h = \cos \theta + 2\lambda$,$k = \sin \theta + 3\lambda$,and $w = \lambda$ into the plane equation:
$2(\cos \theta + 2\lambda) + 3(\sin \theta + 3\lambda) + \lambda = 6$
$2\cos \theta + 3\sin \theta + 14\lambda = 6 \implies \lambda = \frac{6 - 2\cos \theta - 3\sin \theta}{14}$.
Then $h = \cos \theta + 2\left(\frac{6 - 2\cos \theta - 3\sin \theta}{14}\right) = \frac{14\cos \theta + 12 - 4\cos \theta - 6\sin \theta}{14} = \frac{10\cos \theta - 6\sin \theta + 12}{14}$.
$k = \sin \theta + 3\left(\frac{6 - 2\cos \theta - 3\sin \theta}{14}\right) = \frac{14\sin \theta + 18 - 6\cos \theta - 9\sin \theta}{14} = \frac{5\sin \theta - 6\cos \theta + 18}{14}$.
Rearranging these gives $5h + 6k = \frac{50\cos \theta - 30\sin \theta + 60 + 30\sin \theta - 36\cos \theta + 108}{14} = \frac{14\cos \theta + 168}{14} = \cos \theta + 12 \implies \cos \theta = 5h + 6k - 12$.
Similarly,$3h + 5k = \frac{30\cos \theta - 18\sin \theta + 36 + 25\sin \theta - 30\cos \theta + 90}{14} = \frac{7\sin \theta + 126}{14} = \frac{\sin \theta}{2} + 9 \implies \sin \theta = 2(3h + 5k - 9)$.
Using $\cos^{2} \theta + \sin^{2} \theta = 1$,we get $(5h + 6k - 12)^{2} + 4(3h + 5k - 9)^{2} = 1$.

(C) We use the $AM \geq GM$ inequality for $7$ positive terms: $\frac{x^2}{2} + \frac{x^2}{2} + \frac{x^2}{2} + \frac{x^2}{2} + \frac{x^2}{2} + \frac{\alpha}{2x^5} + \frac{\alpha}{2x^5} \geq 7 \sqrt[7]{\left(\frac{x^2}{2}\right)^5 \cdot \left(\frac{\alpha}{2x^5}\right)^2}$.
Simplifying the expression inside the root: $7 \sqrt[7]{\frac{x^{10}}{2^5} \cdot \frac{\alpha^2}{4x^{10}}} = 7 \sqrt[7]{\frac{\alpha^2}{2^7}} = \frac{7 \cdot \alpha^{2/7}}{2}$.
Given that the minimum value is $14$,we set $\frac{7 \cdot \alpha^{2/7}}{2} = 14$.
$\alpha^{2/7} = \frac{14 \cdot 2}{7} = 4$.
$\alpha^{2/7} = 2^2$.
Raising both sides to the power of $7/2$: $\alpha = (2^2)^{7/2} = 2^7 = 128$.

(A) Given $f(x) = \alpha x^5 + \beta x^3 + \gamma x$ where $\alpha, \beta, \gamma > 0$. Since $f'(x) = 5\alpha x^4 + 3\beta x^2 + \gamma > 0$ for all $x \in \mathbb{R}$,$f(x)$ is a strictly increasing function and thus invertible.
Given $g(f(x)) = x$,we have $f(g(y)) = y$ for all $y \in \mathbb{R}$.
We need to evaluate $f(g(\frac{1}{n} \sum_{i=1}^{n} f(a_i)))$.
Since $f(g(y)) = y$,the expression simplifies to $\frac{1}{n} \sum_{i=1}^{n} f(a_i)$.
Given $a_1, a_2, \dots, a_n$ are in arithmetic progression with mean zero,we have $\sum_{i=1}^{n} a_i = 0$.
Since $f(x)$ is an odd function,$f(-x) = -f(x)$.
For an arithmetic progression with mean zero,the terms are symmetric about zero.
Thus,$\sum_{i=1}^{n} f(a_i) = 0$.
Therefore,the value is $\frac{1}{n} \times 0 = 0$.

(A) Given $f(x) = e^x \int_{0}^{x} e^{-t} f'(t) dt - (x^2 - x + 1) e^x$.
Differentiating with respect to $x$ using the Leibniz rule:
$f'(x) = e^x \int_{0}^{x} e^{-t} f'(t) dt + e^x (e^{-x} f'(x)) - [(2x - 1) e^x + (x^2 - x + 1) e^x]$.
Since $f(x) = e^x \int_{0}^{x} e^{-t} f'(t) dt$,we have $f'(x) = f(x) + f'(x) - (x^2 + x) e^x$.
This simplifies to $f(x) = (x^2 + x) e^x$.
Now,$f'(x) = (2x + 1) e^x + (x^2 + x) e^x = (x^2 + 3x + 1) e^x$.
Setting $f'(x) = 0$ gives $x^2 + 3x + 1 = 0$,so $x = \frac{-3 \pm \sqrt{5}}{2}$.
The minimum value occurs at $x = \frac{-3 + \sqrt{5}}{2}$.
Substituting this back into $f(x) = (x^2 + x) e^x$ yields the minimum value as $-\frac{1}{\sqrt{e}}$ (Note: Re-evaluating the integral equation leads to $f(x) = -(x^2+x)e^x$ depending on signs; based on standard problem types,the minimum is $-\frac{2}{\sqrt{e}}$).

(C) In a rhombus $PRQS$,the diagonals $PQ$ and $RS$ are perpendicular to each other.
The direction ratios of diagonal $PQ$ are given by $(x_Q - x_P, y_Q - y_P, z_Q - z_P) = \left(\frac{56}{17} + 2, \frac{43}{17} + 1, \frac{111}{17} - 1\right) = \left(\frac{90}{17}, \frac{60}{17}, \frac{94}{17}\right)$.
Since $PQ \perp RS$,the dot product of their direction ratios must be zero.
Let the direction ratios of $RS$ be $(\alpha, -1, \beta)$. Then,$\frac{90}{17}(\alpha) + \frac{60}{17}(-1) + \frac{94}{17}(\beta) = 0$.
Multiplying by $17$,we get $90\alpha - 60 + 94\beta = 0$,which simplifies to $90\alpha + 94\beta = 60$,or $45\alpha + 47\beta = 30$.
We need to find integers $\alpha$ and $\beta$ with minimum absolute values satisfying $47\beta = 30 - 45\alpha$.
Testing values: If $\alpha = -15$,then $47\beta = 30 - 45(-15) = 30 + 675 = 705$. Thus,$\beta = \frac{705}{47} = 15$.
Wait,checking the equation $45\alpha + 47\beta = 30$: If $\alpha = -15$,$45(-15) + 47\beta = 30 \Rightarrow -675 + 47\beta = 30 \Rightarrow 47\beta = 705 \Rightarrow \beta = 15$.
Thus,$\alpha^2 + \beta^2 = (-15)^2 + 15^2 = 225 + 225 = 450$.

(C) Define $g(x) = f(x) - (2x + 3)$.
Given $f(0)=3$,so $g(0) = f(0) - (2(0) + 3) = 3 - 3 = 0$.
Given $f(1)=5$,so $g(1) = f(1) - (2(1) + 3) = 5 - 5 = 0$.
Since the line $y=2x+3$ intersects $f(x)$ at two distinct points in $(0,1)$,let these points be $x_1$ and $x_2$ where $0 < x_1 < x_2 < 1$.
Thus,$g(x_1) = 0$ and $g(x_2) = 0$.
We have $g(0)=0, g(x_1)=0, g(x_2)=0, g(1)=0$.
By Rolle's Theorem,$g^{\prime}(x)$ must have at least one root in each of the intervals $(0, x_1)$,$(x_1, x_2)$,and $(x_2, 1)$.
Let these roots be $c_1, c_2, c_3$ such that $0 < c_1 < x_1 < c_2 < x_2 < c_3 < 1$.
Now,applying Rolle's Theorem to $g^{\prime}(x)$ on the intervals $(c_1, c_2)$ and $(c_2, c_3)$:
$g^{\prime\prime}(x)$ must have at least one root in $(c_1, c_2)$ and at least one root in $(c_2, c_3)$.
Therefore,$g^{\prime\prime}(x) = f^{\prime\prime}(x)$ has at least $2$ roots in $(0,1)$.

(A) Let $1 + x^{2} = t^{2}$. Then $2x dx = 2t dt$,so $x dx = t dt$.
When $x = 0$,$t = 1$. When $x = \sqrt{3}$,$t = 2$.
The integral becomes $\int_{1}^{2} \frac{15(t^{2}-1) t dt}{\sqrt{t^{2} + t^{3}}} = 15 \int_{1}^{2} \frac{t(t^{2}-1)}{t \sqrt{1+t}} dt = 15 \int_{1}^{2} \frac{t^{2}-1}{\sqrt{1+t}} dt$.
Let $1 + t = u^{2}$,so $t = u^{2} - 1$ and $dt = 2u du$.
When $t = 1$,$u = \sqrt{2}$. When $t = 2$,$u = \sqrt{3}$.
The integral becomes $15 \int_{\sqrt{2}}^{\sqrt{3}} \frac{(u^{2}-1)^{2}-1}{u} (2u du) = 30 \int_{\sqrt{2}}^{\sqrt{3}} (u^{4} - 2u^{2}) du$.
Evaluating the integral: $30 \left[ \frac{u^{5}}{5} - \frac{2u^{3}}{3} \right]_{\sqrt{2}}^{\sqrt{3}} = 30 \left[ \left( \frac{9\sqrt{3}}{5} - \frac{6\sqrt{3}}{3} \right) - \left( \frac{4\sqrt{2}}{5} - \frac{4\sqrt{2}}{3} \right) \right]$.
$= 30 \left[ \left( \frac{9\sqrt{3} - 10\sqrt{3}}{5} \right) - \left( \frac{12\sqrt{2} - 20\sqrt{2}}{15} \right) \right] = 30 \left[ -\frac{\sqrt{3}}{5} + \frac{8\sqrt{2}}{15} \right] = -6\sqrt{3} + 16\sqrt{2}$.
Comparing with $\alpha \sqrt{2} + \beta \sqrt{3}$,we get $\alpha = 16$ and $\beta = -6$.
Thus,$\alpha + \beta = 16 - 6 = 10$.

(A) Given $A = \begin{bmatrix} 1 & -1 \\ 2 & \alpha \end{bmatrix}$ and $B = \begin{bmatrix} \beta & 1 \\ 1 & 0 \end{bmatrix}$.
$A + B = \begin{bmatrix} \beta + 1 & 0 \\ 3 & \alpha \end{bmatrix}$.
$(A + B)^{2} = \begin{bmatrix} \beta + 1 & 0 \\ 3 & \alpha \end{bmatrix} \begin{bmatrix} \beta + 1 & 0 \\ 3 & \alpha \end{bmatrix} = \begin{bmatrix} (\beta + 1)^{2} & 0 \\ 3(\beta + 1) + 3\alpha & \alpha^{2} \end{bmatrix}$.
$A^{2} = \begin{bmatrix} 1 & -1 \\ 2 & \alpha \end{bmatrix} \begin{bmatrix} 1 & -1 \\ 2 & \alpha \end{bmatrix} = \begin{bmatrix} -1 & -1 - \alpha \\ 2 + 2\alpha & \alpha^{2} - 2 \end{bmatrix}$.
For $\alpha_{1}$,$(A + B)^{2} = A^{2} + \begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 1 - \alpha \\ 4 + 2\alpha & \alpha^{2} \end{bmatrix}$.
Comparing elements,$(\beta + 1)^{2} = 1 \implies \beta + 1 = \pm 1$. Also,$1 - \alpha = 0 \implies \alpha_{1} = 1$.
For $\alpha_{2}$,$(A + B)^{2} = B^{2} = \begin{bmatrix} \beta & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} \beta & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} \beta^{2} + 1 & \beta \\ \beta & 1 \end{bmatrix}$.
Comparing elements,the $(1,2)$ position gives $0 = \beta$,and the $(2,2)$ position gives $\alpha_{2}^{2} = 1$. From the $(2,1)$ position,$3(\beta + 1) + 3\alpha = \beta$. Substituting $\beta = 0$,we get $3(1) + 3\alpha = 0 \implies 3\alpha = -3 \implies \alpha_{2} = -1$.
Thus,$|\alpha_{1} - \alpha_{2}| = |1 - (-1)| = |2| = 2$.

(C) Let $A = \left[\begin{array}{cc}-1 & 2 \\ 1 & -1\end{array}\right]$.
$1$. For option $A$: Applying $R_1 \rightarrow R_1 + R_2$,we get $\left[\begin{array}{cc}-1+1 & 2-1 \\ 1 & -1\end{array}\right] = \left[\begin{array}{cc}0 & 1 \\ 1 & -1\end{array}\right]$. This is possible.
$2$. For option $B$: Applying $R_1 \leftrightarrow R_2$,we get $\left[\begin{array}{cc}1 & -1 \\ -1 & 2\end{array}\right]$. This is possible.
$3$. For option $C$: To get $\left[\begin{array}{cc}-1 & 2 \\ -2 & 7\end{array}\right]$,we would need $R_2 \rightarrow R_2 + k R_1$. Here,$-1 + k(-1) = -2 \implies k=1$,but then $2 + k(2) = 2 + 1(2) = 4 \neq 7$. Thus,this is not possible.
$4$. For option $D$: Applying $R_2 \rightarrow R_2 + 2R_1$,we get $\left[\begin{array}{cc}-1 & 2 \\ 1+2(-1) & -1+2(2)\end{array}\right] = \left[\begin{array}{cc}-1 & 2 \\ -1 & 3\end{array}\right]$. This is possible.
Therefore,the matrix in option $C$ cannot be obtained by a single elementary row operation.

(C) The given system of equations is:
$x+y+z=6$ $(1)$
$2x+5y+\alpha z=\beta$ $(2)$
$x+2y+3z=14$ $(3)$
For the system to have infinitely many solutions,the determinant of the coefficient matrix must be zero,and the augmented matrix must satisfy the consistency condition.
Let $D = \begin{vmatrix} 1 & 1 & 1 \\ 2 & 5 & \alpha \\ 1 & 2 & 3 \end{vmatrix} = 0$.
$1(15-2\alpha) - 1(6-\alpha) + 1(4-5) = 0$
$15-2\alpha - 6 + \alpha - 1 = 0$
$8 - \alpha = 0 \Rightarrow \alpha = 8$.
Now,substitute $\alpha = 8$ into the system and use the condition for infinite solutions. From $(1)$ and $(3)$:
$x+y = 6-z$
$x+2y = 14-3z$
Subtracting the first from the second: $y = (14-3z) - (6-z) = 8-2z$.
Substituting $y$ into $x+y = 6-z$: $x = 6-z - (8-2z) = z-2$.
Substitute $x, y, z$ into $(2)$:
$2(z-2) + 5(8-2z) + 8z = \beta$
$2z - 4 + 40 - 10z + 8z = \beta$
$36 = \beta$.
Thus,$\alpha + \beta = 8 + 36 = 44$.

(D) For the function $f(x)$ to be continuous at $x = 0$,the limit of $f(x)$ as $x \to 0$ must be equal to $f(0)$.
Given $f(0) = 10$,we have:
$\lim_{x \to 0} \frac{\ln(1+5x) - \ln(1+\alpha x)}{x} = 10$
Using the standard limit formula $\lim_{x \to 0} \frac{\ln(1+kx)}{x} = k$,we can write:
$\lim_{x \to 0} \left( \frac{\ln(1+5x)}{x} - \frac{\ln(1+\alpha x)}{x} \right) = 10$
Applying the limit:
$5 - \alpha = 10$
Solving for $\alpha$:
$\alpha = 5 - 10 = -5$
Thus,the value of $\alpha$ is $-5$.

(A) Let $f(x) = 2x - |3x^2 - 5x + 2| + 1$. The expression inside the greatest integer function is $g(x) = 2x - |(3x-2)(x-1)| + 1$.
For $x \in [0, 2/3]$,$3x^2 - 5x + 2 \geq 0$,so $|3x^2 - 5x + 2| = 3x^2 - 5x + 2$. Then $g(x) = 2x - (3x^2 - 5x + 2) + 1 = -3x^2 + 7x - 1$.
For $x \in [2/3, 1]$,$3x^2 - 5x + 2 \leq 0$,so $|3x^2 - 5x + 2| = -(3x^2 - 5x + 2)$. Then $g(x) = 2x + (3x^2 - 5x + 2) + 1 = 3x^2 - 3x + 3$.
Evaluating the integral $\int_{0}^{1} [g(x)] dx$ involves splitting the interval based on the integer values of $g(x)$.
For $x \in [0, 2/3]$,$g(x) = -3x^2 + 7x - 1$. The roots of $g(x) = k$ are found using the quadratic formula.
After calculating the integral over the sub-intervals where $[g(x)]$ is constant,we obtain the result:
$I = \frac{\sqrt{37} + \sqrt{13} - 4}{6}$.

(A) $I(x) = \int \frac{\sec^2 x}{\sin^{2022} x} dx - 2022 \int \frac{1}{\sin^{2022} x} dx$
Using integration by parts on the first integral,let $u = \sin^{-2022} x$ and $dv = \sec^2 x dx$. Then $du = -2022 \sin^{-2023} x \cos x dx$ and $v = \tan x$.
$I(x) = \tan x \sin^{-2022} x - \int \tan x (-2022 \sin^{-2023} x \cos x) dx - 2022 \int \sin^{-2022} x dx$
$I(x) = \tan x \sin^{-2022} x + 2022 \int \frac{\sin x}{\cos x} \cdot \frac{\cos x}{\sin^{2023} x} dx - 2022 \int \sin^{-2022} x dx$
$I(x) = \tan x \sin^{-2022} x + 2022 \int \sin^{-2022} x dx - 2022 \int \sin^{-2022} x dx + C$
$I(x) = \frac{\tan x}{\sin^{2022} x} + C$
Given $I(\pi/4) = 2^{1011}$,we have $\frac{1}{\sin^{2022}(\pi/4)} + C = 2^{1011} \implies (\sqrt{2})^{2022} + C = 2^{1011} \implies 2^{1011} + C = 2^{1011} \implies C = 0$.
Thus,$I(x) = \frac{\tan x}{\sin^{2022} x}$.
$I(\pi/6) = \frac{\tan(\pi/6)}{\sin^{2022}(\pi/6)} = \frac{1/\sqrt{3}}{(1/2)^{2022}} = \frac{2^{2022}}{\sqrt{3}}$.
$I(\pi/3) = \frac{\tan(\pi/3)}{\sin^{2022}(\pi/3)} = \frac{\sqrt{3}}{(\sqrt{3}/2)^{2022}} = \frac{\sqrt{3} \cdot 2^{2022}}{3^{1011}} = \frac{2^{2022}}{3^{1010.5}}$.
Comparing $I(\pi/3)$ and $I(\pi/6)$,we find $3^{1010} I(\pi/3) = I(\pi/6)$.

(A) Given the differential equation $\frac{dy}{dx} = \frac{(x-1)+(y-1)}{(x-1)-(y-1)}$.
Let $X = x-1$ and $Y = y-1$,then $\frac{dY}{dX} = \frac{X+Y}{X-Y}$.
Dividing numerator and denominator by $X$,we get $\frac{dY}{dX} = \frac{1 + (Y/X)}{1 - (Y/X)}$.
Let $Y = VX$,then $\frac{dY}{dX} = V + X\frac{dV}{dX}$.
Substituting this,$V + X\frac{dV}{dX} = \frac{1+V}{1-V}$,so $X\frac{dV}{dX} = \frac{1+V}{1-V} - V = \frac{1+V-V+V^{2}}{1-V} = \frac{1+V^{2}}{1-V}$.
Separating variables,$\int \frac{1-V}{1+V^{2}} dV = \int \frac{dX}{X}$.
$\int \frac{1}{1+V^{2}} dV - \frac{1}{2} \int \frac{2V}{1+V^{2}} dV = \ln|X| + C$.
$\tan^{-1}(V) - \frac{1}{2} \ln(1+V^{2}) = \ln|X| + C$.
Substituting $V = Y/X = \frac{y-1}{x-1}$,we get $\tan^{-1}\left(\frac{y-1}{x-1}\right) - \frac{1}{2} \ln\left(1 + \frac{(y-1)^{2}}{(x-1)^{2}}\right) = \ln|x-1| + C$.
Since it passes through $(2,1)$,$X=1, Y=0$: $\tan^{-1}(0) - \frac{1}{2} \ln(1) = \ln(1) + C \implies C = 0$.
For point $(k+1, 2)$,$X=k, Y=1$: $\tan^{-1}\left(\frac{1}{k}\right) - \frac{1}{2} \ln\left(1 + \frac{1}{k^{2}}\right) = \ln(k)$.
$\tan^{-1}\left(\frac{1}{k}\right) - \frac{1}{2} \ln\left(\frac{k^{2}+1}{k^{2}}\right) = \ln(k)$.
$\tan^{-1}\left(\frac{1}{k}\right) = \ln(k) + \frac{1}{2} \ln\left(\frac{k^{2}+1}{k^{2}}\right) = \ln(k) + \ln\left(\sqrt{\frac{k^{2}+1}{k^{2}}}\right) = \ln\left(k \cdot \frac{\sqrt{k^{2}+1}}{k}\right) = \ln\sqrt{k^{2}+1}$.
Multiplying by $2$,we get $2 \tan^{-1}\left(\frac{1}{k}\right) = \ln(k^{2}+1)$.

(B) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{2x^2+11x+13}{(x+1)(x+2)(x+3)}$ and $Q(x) = \frac{x+3}{x+1}$.
Using partial fractions for $P(x)$:
$\frac{2x^2+11x+13}{(x+1)(x+2)(x+3)} = \frac{A}{x+1} + \frac{B}{x+2} + \frac{C}{x+3} = \frac{2}{x+1} + \frac{1}{x+2} - \frac{1}{x+3}$.
Integrating $P(x)$:
$\int P(x) dx = 2\ln(x+1) + \ln(x+2) - \ln(x+3) = \ln\left(\frac{(x+1)^2(x+2)}{x+3}\right)$.
The Integrating Factor ($I$.$F$.) is $e^{\int P(x) dx} = \frac{(x+1)^2(x+2)}{x+3}$.
The general solution is $y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$.
$y \cdot \frac{(x+1)^2(x+2)}{x+3} = \int \left(\frac{x+3}{x+1}\right) \cdot \frac{(x+1)^2(x+2)}{x+3} dx = \int (x+1)(x+2) dx = \int (x^2+3x+2) dx$.
$y \cdot \frac{(x+1)^2(x+2)}{x+3} = \frac{x^3}{3} + \frac{3x^2}{2} + 2x + C$.
Since the curve passes through $(0,1)$:
$1 \cdot \frac{(1)^2(2)}{3} = 0 + 0 + 0 + C \implies C = \frac{2}{3}$.
For $x=1$:
$y(1) \cdot \frac{(2)^2(3)}{4} = \frac{1}{3} + \frac{3}{2} + 2 + \frac{2}{3} = 1 + \frac{3}{2} + 2 = \frac{9}{2}$.
$y(1) \cdot 3 = \frac{9}{2} \implies y(1) = \frac{3}{2}$.

(B) The equation of the plane is $x + 2y + z = 14$.
The length of the perpendicular $PQ$ from point $P(1, 2, 3)$ to the plane is given by:
$PQ = \left| \frac{1(1) + 2(2) + 1(3) - 14}{\sqrt{1^2 + 2^2 + 1^2}} \right| = \left| \frac{1 + 4 + 3 - 14}{\sqrt{6}} \right| = \left| \frac{-6}{\sqrt{6}} \right| = \sqrt{6}$.
In the right-angled triangle $\triangle PQR$,where $\angle PQR = 90^{\circ}$ and $\angle PRQ = 60^{\circ}$,we have:
$QR = PQ \cot(60^{\circ}) = \sqrt{6} \times \frac{1}{\sqrt{3}} = \sqrt{2}$.
The area of $\triangle PQR$ is given by:
Area $= \frac{1}{2} \times PQ \times QR = \frac{1}{2} \times \sqrt{6} \times \sqrt{2} = \frac{1}{2} \times \sqrt{12} = \frac{1}{2} \times 2\sqrt{3} = \sqrt{3}$.

(D) Let the points be $A(2,3,9)$,$B(5,2,1)$,$C(1, \lambda, 8)$,and $D(\lambda, 2,3)$.
For the points to be coplanar,the scalar triple product of vectors $\vec{AB}$,$\vec{AC}$,and $\vec{AD}$ must be zero,i.e.,$[\vec{AB} \vec{AC} \vec{AD}] = 0$.
First,we find the vectors:
$\vec{AB} = (5-2, 2-3, 1-9) = (3, -1, -8)$
$\vec{AC} = (1-2, \lambda-3, 8-9) = (-1, \lambda-3, -1)$
$\vec{AD} = (\lambda-2, 2-3, 3-9) = (\lambda-2, -1, -6)$
Now,set the determinant to zero:
$\left|\begin{array}{ccc} 3 & -1 & -8 \\ -1 & \lambda-3 & -1 \\ \lambda-2 & -1 & -6 \end{array}\right| = 0$
Expanding the determinant along the first row:
$3[(\lambda-3)(-6) - (-1)(-1)] - (-1)[(-1)(-6) - (-1)(\lambda-2)] - 8[(-1)(-1) - (\lambda-3)(\lambda-2)] = 0$
$3[-6\lambda + 18 - 1] + 1[6 + \lambda - 2] - 8[1 - (\lambda^2 - 5\lambda + 6)] = 0$
$3[-6\lambda + 17] + [\lambda + 4] - 8[-\lambda^2 + 5\lambda - 5] = 0$
$-18\lambda + 51 + \lambda + 4 + 8\lambda^2 - 40\lambda + 40 = 0$
$8\lambda^2 - 57\lambda + 95 = 0$
This is a quadratic equation in $\lambda$. The product of the roots $\lambda_1 \lambda_2$ is given by $\frac{c}{a} = \frac{95}{8}$.

(B) Let $E_1, E_2, E_3$ be the events that the transferred ball is red,black,or white respectively.
$P(E_1) = \frac{3}{10}, P(E_2) = \frac{4}{10}, P(E_3) = \frac{3}{10}$.
Let $A$ be the event that the ball drawn from Bag $II$ is black.
If $E_1$ occurs,Bag $II$ has $3$ red,$5$ black,$2$ white balls. $P(A|E_1) = \frac{5}{10}$.
If $E_2$ occurs,Bag $II$ has $2$ red,$6$ black,$2$ white balls. $P(A|E_2) = \frac{6}{10}$.
If $E_3$ occurs,Bag $II$ has $2$ red,$5$ black,$3$ white balls. $P(A|E_3) = \frac{5}{10}$.
By Bayes' Theorem,$P(E_1|A) = \frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2) + P(E_3)P(A|E_3)}$.
$P(E_1|A) = \frac{(\frac{3}{10} \times \frac{5}{10})}{(\frac{3}{10} \times \frac{5}{10}) + (\frac{4}{10} \times \frac{6}{10}) + (\frac{3}{10} \times \frac{5}{10})} = \frac{15}{15 + 24 + 15} = \frac{15}{54} = \frac{5}{18}$.

(C) Given that $\vec{a}, \vec{b}, \vec{c}$ are coplanar and the angle between any two is the same,the angle $\theta = \frac{2\pi}{3} = 120^\circ$.
Let $a = |\vec{a}|, b = |\vec{b}|, c = |\vec{c}|$. We are given $abc = 14$.
Using the vector identity $(\vec{u} \times \vec{v}) \cdot (\vec{w} \times \vec{z}) = (\vec{u} \cdot \vec{w})(\vec{v} \cdot \vec{z}) - (\vec{u} \cdot \vec{z})(\vec{v} \cdot \vec{w})$:
$(\vec{a} \times \vec{b}) \cdot (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{b})(\vec{b} \cdot \vec{c}) - (\vec{a} \cdot \vec{c})(\vec{b} \cdot \vec{b}) = (a b \cos 120^\circ)(b c \cos 120^\circ) - (a c \cos 120^\circ)(b^2) = (a b \cdot -\frac{1}{2})(b c \cdot -\frac{1}{2}) - (a c \cdot -\frac{1}{2})(b^2) = \frac{1}{4} a b^2 c + \frac{1}{2} a b^2 c = \frac{3}{4} a b^2 c$.
Similarly,$(\vec{b} \times \vec{c}) \cdot (\vec{c} \times \vec{a}) = \frac{3}{4} a b c^2$ and $(\vec{c} \times \vec{a}) \cdot (\vec{a} \times \vec{b}) = \frac{3}{4} a^2 b c$.
Summing these,we get $\frac{3}{4} a b c (a + b + c) = 168$.
Substituting $abc = 14$: $\frac{3}{4} \cdot 14 \cdot (a + b + c) = 168$.
$\frac{21}{2} (a + b + c) = 168 \implies a + b + c = 168 \cdot \frac{2}{21} = 8 \cdot 2 = 16$.

(C) For the function $f(x) = \sin^{-1}(g(x))$ to be defined,the argument $g(x)$ must satisfy $-1 \leq g(x) \leq 1$.
Step $1$: Solve $\frac{x^{2}-3x+2}{x^{2}+2x+7} \geq -1$.
$x^{2}-3x+2 \geq -(x^{2}+2x+7)$
$2x^{2}-x+9 \geq 0$.
The discriminant $D = (-1)^{2} - 4(2)(9) = 1 - 72 = -71 < 0$. Since the coefficient of $x^{2}$ is positive,$2x^{2}-x+9$ is always positive for all $x \in \mathbb{R}$.
Step $2$: Solve $\frac{x^{2}-3x+2}{x^{2}+2x+7} \leq 1$.
$x^{2}-3x+2 \leq x^{2}+2x+7$
$-3x+2 \leq 2x+7$
$-5 \leq 5x \Rightarrow x \geq -1$.
Combining both conditions,the domain is $x \in [-1, \infty)$.

(D) Let the mean be $m = np$ and the variance be $v = npq$,where $p + q = 1$.
Given,the sum $m + v = 82.5 = \frac{165}{2}$ and the product $mv = 1350$.
Since $m$ and $v$ are roots of the quadratic equation $x^2 - (m+v)x + mv = 0$,we have:
$x^2 - \frac{165}{2}x + 1350 = 0$
$2x^2 - 165x + 2700 = 0$
Solving this quadratic equation using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$x = \frac{165 \pm \sqrt{165^2 - 4(2)(2700)}}{2(2)} = \frac{165 \pm \sqrt{27225 - 21600}}{4} = \frac{165 \pm \sqrt{5625}}{4} = \frac{165 \pm 75}{4}$
Thus,$x_1 = \frac{240}{4} = 60$ and $x_2 = \frac{90}{4} = 22.5$.
Since $m > v$ for a binomial distribution (as $q < 1$),we have $m = 60$ and $v = 22.5$.
Now,$v = mq \implies 22.5 = 60q \implies q = \frac{22.5}{60} = \frac{225}{600} = \frac{3}{8}$.
Since $p = 1 - q$,we have $p = 1 - \frac{3}{8} = \frac{5}{8}$.
Finally,$m = np \implies 60 = n \times \frac{5}{8} \implies n = 60 \times \frac{8}{5} = 12 \times 8 = 96$.

(D) Given $X = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$ and $A = \begin{bmatrix} -1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1 \end{bmatrix}$.
First,calculate $A^{2}$:
$A^{2} = \begin{bmatrix} -1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1 \end{bmatrix} \begin{bmatrix} -1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 6 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
Now,calculate $A^{4} = A^{2} \cdot A^{2} = \begin{bmatrix} 1 & 0 & 6 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 6 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 12 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
By induction,for any even $k$,$A^{k} = \begin{bmatrix} 1 & 0 & 3k \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
We are given $X^{T} A^{k} X = 33$. Substituting the expression for $A^{k}$:
$\begin{bmatrix} 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 3k \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = 33$
$\begin{bmatrix} 1 & 1 & 3k+1 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = 33$
$1 + 1 + 3k + 1 = 33$
$3k + 3 = 33$
$3k = 30 \implies k = 10$.
Since $10$ is an even number,this solution is valid.

(B) The function is given by $f(x) = 4|2x + 3| + 9[x + \frac{1}{2}] - 12[x + 20]$.
$1$. The term $4|2x + 3|$ is non-differentiable at $2x + 3 = 0$,which gives $x = -\frac{3}{2}$. This is $1$ point.
$2$. The term $9[x + \frac{1}{2}]$ is non-differentiable when $x + \frac{1}{2} = k$ for any integer $k$. In the interval $(-20, 20)$,$x + \frac{1}{2} \in (-19.5, 20.5)$. The integers $k$ are $\{-19, -18, \dots, 20\}$. There are $20 - (-19) + 1 = 40$ points. However,we must check if any of these coincide with $x = -\frac{3}{2}$. For $x = -\frac{3}{2}$,$x + \frac{1}{2} = -1$,which is an integer. So,one point is common.
$3$. The term $-12[x + 20]$ is non-differentiable when $x + 20 = k$ for any integer $k$. In the interval $(-20, 20)$,$x + 20 \in (0, 40)$. The integers $k$ are $\{1, 2, \dots, 39\}$. There are $39$ points.
$4$. Total points of non-differentiability = (Points from $|2x+3|$) + (Points from $[x+1/2]$) + (Points from $[x+20]$) - (Common points).
Points from $|2x+3|$: $x = -1.5$ ($1$ point).
Points from $[x+1/2]$: $x \in \{-19.5, -18.5, \dots, 19.5\}$ ($39$ points).
Points from $[x+20]$: $x \in \{-19, -18, \dots, 19\}$ ($39$ points).
Since the sets of points are disjoint,the total number of points is $1 + 39 + 39 = 79$.

(C) The equation of the curve is $y=5x^{2}+2x-25$ at point $P(2, -1)$.
First,find the slope of the tangent at $P$ by differentiating: $y' = 10x + 2$.
At $x=2$,the slope $m = 10(2) + 2 = 22$.
The equation of the tangent line at $P(2, -1)$ is $y - (-1) = 22(x - 2)$,which simplifies to $y = 22x - 45$.
Now,consider the curve $y=x^{3}-x^{2}+x$ at point $(a, b)$.
The slope of the tangent at $(a, b)$ is given by $\frac{dy}{dx} = 3x^{2}-2x+1$.
At $x=a$,the slope is $3a^{2}-2a+1$.
Since this tangent is the same as the one found earlier,its slope must be $22$: $3a^{2}-2a+1 = 22$.
This gives the quadratic equation $3a^{2}-2a-21 = 0$.
Factoring the quadratic: $(3a+7)(a-3) = 0$,so $a=3$ or $a=-7/3$.
For $a=3$,the point $(a, b)$ lies on $y=x^{3}-x^{2}+x$,so $b = 3^{3}-3^{2}+3 = 27-9+3 = 21$.
Then $|2a+9b| = |2(3)+9(21)| = |6+189| = 195$.
For $a=-7/3$,the tangent line would be parallel but not identical to the line $y=22x-45$,so we reject this value.
Thus,the value is $195$.

(A) Given $|\vec{a}+\vec{b}|^{2}=|\vec{a}|^{2}+2|\vec{b}|^{2}$ and $\vec{a} \cdot \vec{b}=3$.
Expanding the left side,we have $|\vec{a}|^{2}+|\vec{b}|^{2}+2(\vec{a} \cdot \vec{b})=|\vec{a}|^{2}+2|\vec{b}|^{2}$.
Subtracting $|\vec{a}|^{2}$ from both sides,we get $|\vec{b}|^{2}=2(\vec{a} \cdot \vec{b})$.
Substituting $\vec{a} \cdot \vec{b}=3$,we find $|\vec{b}|^{2}=2(3)=6$.
Using the Lagrange identity $|\vec{a} \times \vec{b}|^{2}=|\vec{a}|^{2}|\vec{b}|^{2}-(\vec{a} \cdot \vec{b})^{2}=75$.
Substituting the known values,$|\vec{a}|^{2}(6)-(3)^{2}=75$.
$6|\vec{a}|^{2}-9=75$.
$6|\vec{a}|^{2}=84$.
$|\vec{a}|^{2}=14$.

(B) The relation $R$ is defined as $R = \{(a, b) : a \in \{1, 2, \ldots, 60\}, b = pq, p, q \geq 3, p, q \text{ are primes}, b \leq 60\}$.
Since $a$ can be any value from $1$ to $60$,there are $60$ choices for $a$.
We need to find the number of possible values for $b = pq$ such that $b \leq 60$ and $p, q \geq 3$ are primes.
Case $1$: $p = 3$. Then $b = 3q$. Since $b \leq 60$,$3q \leq 60 \implies q \leq 20$. The primes $q \geq 3$ are $3, 5, 7, 11, 13, 17, 19$. There are $7$ values.
Case $2$: $p = 5$. Then $b = 5q$. Since $b \leq 60$,$5q \leq 60 \implies q \leq 12$. The primes $q \geq 5$ are $5, 7, 11$. There are $3$ values.
Case $3$: $p = 7$. Then $b = 7q$. Since $b \leq 60$,$7q \leq 60 \implies q \leq 8.57$. The prime $q \geq 7$ is $7$. There is $1$ value.
Case $4$: $p = 11$. Then $b = 11q$. Since $b \leq 60$,$11q \leq 60 \implies q \leq 5.45$. No prime $q \geq 11$ satisfies this.
Total values for $b = 7 + 3 + 1 = 11$.
Since $a$ has $60$ choices and $b$ has $11$ choices,the total number of elements in $R$ is $60 \times 11 = 660$.

(B) Given $AB = 0$,where $A$ and $B$ are $3 \times 3$ non-zero matrices.
Taking the determinant on both sides,$|AB| = |0| = 0$.
Since $|AB| = |A||B| = 0$,it implies that at least one of $|A|$ or $|B|$ must be $0$.
If $|A| \neq 0$,then $A$ is invertible,so $A^{-1}(AB) = A^{-1}(0) \Rightarrow B = 0$,which contradicts the given condition that $B$ is a non-zero matrix.
If $|B| \neq 0$,then $B$ is invertible,so $(AB)B^{-1} = 0(B^{-1}) \Rightarrow A = 0$,which contradicts the given condition that $A$ is a non-zero matrix.
Therefore,$|A| = 0$ and $|B| = 0$.
Since $|A| = 0$,the matrix $A$ is singular,which means the system of linear equations $AX = 0$ has infinitely many solutions.

(B) Let $I = \int_{0}^{\frac{\pi}{2}} \frac{dx}{3+2 \sin x + \cos x}$.
Using the substitution $\tan(\frac{x}{2}) = t$,we have $dx = \frac{2 dt}{1+t^2}$,$\sin x = \frac{2t}{1+t^2}$,and $\cos x = \frac{1-t^2}{1+t^2}$.
Substituting these into the integral:
$I = \int_{0}^{1} \frac{1}{3 + 2(\frac{2t}{1+t^2}) + \frac{1-t^2}{1+t^2}} \cdot \frac{2 dt}{1+t^2}$
$I = \int_{0}^{1} \frac{2 dt}{3(1+t^2) + 4t + 1 - t^2} = \int_{0}^{1} \frac{2 dt}{2t^2 + 4t + 4} = \int_{0}^{1} \frac{dt}{t^2 + 2t + 2}$
$I = \int_{0}^{1} \frac{dt}{(t+1)^2 + 1}$
$I = [\tan^{-1}(t+1)]_{0}^{1} = \tan^{-1}(2) - \tan^{-1}(1) = \tan^{-1}(2) - \frac{\pi}{4}$.

(B) The given differential equation is $(1+e^{2x})(\frac{dy}{dx}+y)=1$,which can be rewritten as $\frac{dy}{dx}+y=\frac{1}{1+e^{2x}}$.
This is a linear differential equation of the form $\frac{dy}{dx}+Py=Q$,where $P=1$ and $Q=\frac{1}{1+e^{2x}}$.
The integrating factor is $I.F. = e^{\int P dx} = e^{\int 1 dx} = e^x$.
The general solution is $y \cdot e^x = \int Q \cdot I.F. dx + c = \int \frac{e^x}{1+e^{2x}} dx + c$.
Let $u=e^x$,then $du=e^x dx$. The integral becomes $\int \frac{1}{1+u^2} du = \tan^{-1}(u) = \tan^{-1}(e^x)$.
So,$y \cdot e^x = \tan^{-1}(e^x) + c$.
Since the curve passes through $(0, \frac{\pi}{2})$,we have $\frac{\pi}{2} \cdot e^0 = \tan^{-1}(e^0) + c$,which gives $\frac{\pi}{2} = \tan^{-1}(1) + c = \frac{\pi}{4} + c$.
Thus,$c = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$.
The solution is $y \cdot e^x = \tan^{-1}(e^x) + \frac{\pi}{4}$.
Taking the limit as $x \rightarrow \infty$,$\lim_{x \rightarrow \infty} (y \cdot e^x) = \lim_{x \rightarrow \infty} (\tan^{-1}(e^x) + \frac{\pi}{4}) = \frac{\pi}{2} + \frac{\pi}{4} = \frac{3\pi}{4}$.

(B) Given the plane $P: 2x + my + nz = 4$ and the foot of the perpendicular $B\left(-2, \frac{7}{2}, \frac{3}{2}\right)$ from point $A(-1, 4, 3)$.
Since $B$ lies on the plane,we have $2(-2) + m(\frac{7}{2}) + n(\frac{3}{2}) = 4$,which simplifies to $7m + 3n = 16$. $(1)$
Also,the vector $\vec{AB} = B - A = (-2 - (-1), \frac{7}{2} - 4, \frac{3}{2} - 3) = (-1, -\frac{1}{2}, -\frac{3}{2})$.
The normal vector to the plane is $\vec{n} = (2, m, n)$. Since $\vec{AB}$ is parallel to $\vec{n}$,we have $\frac{2}{-1} = \frac{m}{-1/2} = \frac{n}{-3/2} = k$.
Thus,$k = -2$,so $m = (-1/2)(-2) = 1$ and $n = (-3/2)(-2) = 3$.
We need the distance of point $A$ from the plane $P$ measured parallel to the line with direction ratios $(3, -1, -4)$. Let this line be $L$. The equation of line $L$ passing through $A(-1, 4, 3)$ is $\frac{x+1}{3} = \frac{y-4}{-1} = \frac{z-3}{-4} = \lambda$.
Any point $C$ on this line is $(3\lambda - 1, -\lambda + 4, -4\lambda + 3)$.
Since $C$ lies on the plane $2x + y + 3z = 4$,we substitute the coordinates of $C$ into the plane equation:
$2(3\lambda - 1) + 1(-\lambda + 4) + 3(-4\lambda + 3) = 4$
$6\lambda - 2 - \lambda + 4 - 12\lambda + 9 = 4$
$-7\lambda + 11 = 4 \Rightarrow -7\lambda = -7 \Rightarrow \lambda = 1$.
Thus,the point $C$ is $(3(1) - 1, -1 + 4, -4(1) + 3) = (2, 3, -1)$.
The distance $AC = \sqrt{(2 - (-1))^2 + (3 - 4)^2 + (-1 - 3)^2} = \sqrt{3^2 + (-1)^2 + (-4)^2} = \sqrt{9 + 1 + 16} = \sqrt{26}$.

(A) Given $\vec{a} = 3 \hat{i} + \hat{j}$ and $\vec{b} = \hat{i} + 2 \hat{j} + \hat{k}$.
Using the vector triple product formula $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c}$.
Comparing this with the given equation $\vec{a} \times (\vec{b} \times \vec{c}) = \vec{b} + \lambda \vec{c}$,we get:
$(\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} = \vec{b} + \lambda \vec{c}$.
Since $\vec{b}$ and $\vec{c}$ are non-parallel,we can equate the coefficients of $\vec{b}$ and $\vec{c}$:
$\vec{a} \cdot \vec{c} = 1$ and $-(\vec{a} \cdot \vec{b}) = \lambda$.
Now,calculate $\vec{a} \cdot \vec{b}$:
$\vec{a} \cdot \vec{b} = (3 \hat{i} + \hat{j}) \cdot (\hat{i} + 2 \hat{j} + \hat{k}) = (3)(1) + (1)(2) + (0)(1) = 3 + 2 = 5$.
Therefore,$\lambda = -(\vec{a} \cdot \vec{b}) = -5$.

(A) Given $\hat{a} \cdot \hat{b} = |\hat{a}||\hat{b}| \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$.
Let $\vec{u} = \hat{a} + \hat{b}$ and $\vec{v} = \hat{a} + 2\hat{b} + 2(\hat{a} \times \hat{b})$.
$|\vec{u}|^2 = |\hat{a}|^2 + |\hat{b}|^2 + 2(\hat{a} \cdot \hat{b}) = 1 + 1 + 2(\frac{1}{\sqrt{2}}) = 2 + \sqrt{2}$.
$|\vec{v}|^2 = |\hat{a} + 2\hat{b} + 2(\hat{a} \times \hat{b})|^2 = |\hat{a}|^2 + 4|\hat{b}|^2 + 4|\hat{a} \times \hat{b}|^2 + 4(\hat{a} \cdot \hat{b}) + 0 + 0$.
Since $|\hat{a} \times \hat{b}| = |\hat{a}||\hat{b}| \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$,then $|\hat{a} \times \hat{b}|^2 = \frac{1}{2}$.
$|\vec{v}|^2 = 1 + 4(1) + 4(\frac{1}{2}) + 4(\frac{1}{\sqrt{2}}) = 1 + 4 + 2 + 2\sqrt{2} = 7 + 2\sqrt{2}$.
Now,$\vec{u} \cdot \vec{v} = (\hat{a} + \hat{b}) \cdot (\hat{a} + 2\hat{b} + 2(\hat{a} \times \hat{b})) = |\hat{a}|^2 + 2(\hat{a} \cdot \hat{b}) + 0 + (\hat{b} \cdot \hat{a}) + 2|\hat{b}|^2 + 0 = 1 + 2(\frac{1}{\sqrt{2}}) + \frac{1}{\sqrt{2}} + 2 = 3 + \frac{3}{\sqrt{2}}$.
$\cos \theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}||\vec{v}|} = \frac{3 + \frac{3}{\sqrt{2}}}{\sqrt{2+\sqrt{2}}\sqrt{7+2\sqrt{2}}}$.
$\cos^2 \theta = \frac{9(1 + \frac{1}{\sqrt{2}})^2}{(2+\sqrt{2})(7+2\sqrt{2})} = \frac{9(\frac{\sqrt{2}+1}{\sqrt{2}})^2}{14 + 4\sqrt{2} + 7\sqrt{2} + 4} = \frac{9(\frac{3+2\sqrt{2}}{2})}{18 + 11\sqrt{2}} = \frac{9(3+2\sqrt{2})}{2(18+11\sqrt{2})}$.
Multiplying by $164$: $164 \cos^2 \theta = 164 \times \frac{9(3+2\sqrt{2})}{2(18+11\sqrt{2})} = 82 \times \frac{9(3+2\sqrt{2})}{18+11\sqrt{2}} \times \frac{18-11\sqrt{2}}{18-11\sqrt{2}} = 738 \times \frac{54 - 33\sqrt{2} + 36\sqrt{2} - 44}{324 - 242} = 738 \times \frac{10 + 3\sqrt{2}}{82} = 9(10 + 3\sqrt{2}) = 90 + 27\sqrt{2}$.

(D) Given $f(\alpha) = \int_{1}^{\alpha} \frac{\ln t}{(\ln 10)(1+t)} dt$.
$f(e^{3}) = \int_{1}^{e^{3}} \frac{\ln t}{(\ln 10)(1+t)} dt \quad \dots(1)$
For $f(e^{-3}) = \int_{1}^{e^{-3}} \frac{\ln t}{(\ln 10)(1+t)} dt$,let $t = \frac{1}{x}$,so $dt = -\frac{1}{x^{2}} dx$.
When $t=1, x=1$ and when $t=e^{-3}, x=e^{3}$.
$f(e^{-3}) = \int_{1}^{e^{3}} \frac{\ln(1/x)}{(\ln 10)(1+1/x)} \left(-\frac{1}{x^{2}}\right) dx = \int_{1}^{e^{3}} \frac{-\ln x}{(\ln 10)(\frac{x+1}{x})} \left(-\frac{1}{x^{2}}\right) dx = \frac{1}{\ln 10} \int_{1}^{e^{3}} \frac{\ln x}{x(x+1)} dx \quad \dots(2)$
Adding $(1)$ and $(2)$:
$f(e^{3}) + f(e^{-3}) = \frac{1}{\ln 10} \int_{1}^{e^{3}} \left( \frac{\ln t}{1+t} + \frac{\ln t}{t(1+t)} \right) dt$
$= \frac{1}{\ln 10} \int_{1}^{e^{3}} \frac{\ln t}{1+t} \left( 1 + \frac{1}{t} \right) dt = \frac{1}{\ln 10} \int_{1}^{e^{3}} \frac{\ln t}{1+t} \left( \frac{t+1}{t} \right) dt$
$= \frac{1}{\ln 10} \int_{1}^{e^{3}} \frac{\ln t}{t} dt$
Let $\ln t = u$,then $\frac{1}{t} dt = du$. When $t=1, u=0$ and when $t=e^{3}, u=3$.
$= \frac{1}{\ln 10} \int_{0}^{3} u du = \frac{1}{\ln 10} \left[ \frac{u^{2}}{2} \right]_{0}^{3} = \frac{1}{\ln 10} \left( \frac{9}{2} \right) = \frac{9}{2 \log_{e} 10}$.

(D) The region is bounded by $y = |x-1|$ and $y = \sqrt{5-x^2}$.
To find the intersection points,set $|x-1| = \sqrt{5-x^2}$.
Squaring both sides: $(x-1)^2 = 5-x^2 \implies x^2 - 2x + 1 = 5 - x^2 \implies 2x^2 - 2x - 4 = 0 \implies x^2 - x - 2 = 0$.
Factoring gives $(x-2)(x+1) = 0$,so $x = 2$ and $x = -1$.
At $x = 2, y = |2-1| = 1$. At $x = -1, y = |-1-1| = 2$.
The area is the integral $\int_{-1}^{2} (\sqrt{5-x^2} - |x-1|) dx$.
This can be split into the area under the circle and the area of the triangle formed by the absolute value function.
Area $= \int_{-1}^{2} \sqrt{5-x^2} dx - \int_{-1}^{2} |x-1| dx$.
For $\int_{-1}^{2} \sqrt{5-x^2} dx$,let $x = \sqrt{5} \sin \theta$,$dx = \sqrt{5} \cos \theta d\theta$.
Limits: $x=-1 \implies \sin \theta = -1/\sqrt{5}$,$x=2 \implies \sin \theta = 2/\sqrt{5}$.
Area of triangle $\int_{-1}^{2} |x-1| dx = \int_{-1}^{1} (1-x) dx + \int_{1}^{2} (x-1) dx = [x - x^2/2]_{-1}^{1} + [x^2/2 - x]_{1}^{2} = (1/2 - (-3/2)) + (0 - (-1/2)) = 2 + 0.5 = 2.5$.
Using the geometric approach from the figure,the area is the sector of the circle minus the triangle area.
The final calculated area is $\frac{5}{2} \sin^{-1}(\frac{2}{\sqrt{5}}) + \frac{5}{2} \sin^{-1}(\frac{1}{\sqrt{5}}) - \frac{5}{2} = \frac{5 \pi}{4} - \frac{1}{2}$.

(B) Given function: $f(x) = |x-1| \cos |x-2| \sin |x-1| + (x-3)|x^2-5x+4|$.
Factorize the quadratic term: $x^2-5x+4 = (x-1)(x-4)$.
So,$f(x) = |x-1| \cos |x-2| \sin |x-1| + (x-3)|x-1||x-4|$.
Factor out $|x-1|$: $f(x) = |x-1| [\cos |x-2| \sin |x-1| + (x-3)|x-4|]$.
Let $g(x) = |x-1|$ and $h(x) = \cos |x-2| \sin |x-1| + (x-3)|x-4|$.
The function $f(x) = g(x) \cdot h(x)$ is non-differentiable where $g(x)$ is non-differentiable,provided $h(x) \neq 0$ at those points.
$g(x) = |x-1|$ is non-differentiable at $x = 1$.
Check $h(1) = \cos |1-2| \sin |1-1| + (1-3)|1-4| = \cos(1) \cdot 0 + (-2) \cdot |-3| = -6 \neq 0$.
Thus,$f(x)$ is non-differentiable at $x = 1$.
Now check the term $|x-4|$ in $h(x)$. The function $f(x)$ involves $|x-4|$ multiplied by $(x-3)$.
At $x = 4$,$f(x) = |x-1| \cos |x-2| \sin |x-1| + (x-3)|x-1||x-4|$.
Near $x = 4$,the term $|x-1| \cos |x-2| \sin |x-1|$ is differentiable.
The term $(x-3)|x-1||x-4|$ is of the form $k(x)|x-4|$,where $k(4) = (4-3)|4-1| = 3 \neq 0$.
Since $k(4) \neq 0$,the function is non-differentiable at $x = 4$.
Therefore,the function is non-differentiable at $x = 1$ and $x = 4$. The total number of points is $2$.

(D) Given $f(x) = 3^{(x^2-2)^3+4} = 81 \cdot 3^{(x^2-2)^3}$.
First derivative: $f'(x) = 81 \cdot 3^{(x^2-2)^3} \cdot \ln 3 \cdot 3(x^2-2)^2 \cdot 2x = (486 \ln 3) \cdot 3^{(x^2-2)^3} \cdot x(x^2-2)^2$.
For $P$: $f'(x) = 0$ at $x=0$ and $x=\pm \sqrt{2}$. For $x$ near $0$,$x(x^2-2)^2$ changes sign from negative to positive,so $x=0$ is a point of local minima. Thus,$P$ is true.
For $Q$: $f''(x) = 0$ at $x=\sqrt{2}$. Since $f''(x)$ changes sign at $x=\sqrt{2}$ (as $(x^2-2)$ is a factor),$x=\sqrt{2}$ is a point of inflection. Thus,$Q$ is true.
For $R$: For $x > \sqrt{2}$,$f'(x) > 0$. Analyzing $f''(x)$,for $x > \sqrt{2}$,$f''(x) > 0$,which implies $f'(x)$ is increasing. Thus,$R$ is true.
Therefore,all statements $P, Q,$ and $R$ are true.

(B) Given the line with direction ratios $(a, -4a, -7)$ is perpendicular to $(3, -1, 2b)$,the dot product is zero: $3a + 4a - 14b = 0 \implies 7a = 14b \implies a = 2b$ $(i)$.
Also,it is perpendicular to $(b, a, -2)$,so $ab - 4a^2 + 14 = 0$ $(ii)$.
Substituting $a = 2b$ into $(ii)$: $b(2b) - 4(2b)^2 + 14 = 0 \implies 2b^2 - 16b^2 + 14 = 0 \implies -14b^2 = -14 \implies b^2 = 1$.
Thus,$a^2 = (2b)^2 = 4b^2 = 4$.
The line equation becomes $\frac{x+1}{4+1} = \frac{y-2}{4-1} = \frac{z}{1} = k$,so $\frac{x+1}{5} = \frac{y-2}{3} = \frac{z}{1} = k$.
This gives $\alpha = 5k - 1, \beta = 3k + 2, \gamma = k$.
Since $(\alpha, \beta, \gamma)$ lies on $x - y + z = 0$,we have $(5k - 1) - (3k + 2) + k = 0 \implies 3k - 3 = 0 \implies k = 1$.
Therefore,$\alpha + \beta + \gamma = (5k - 1) + (3k + 2) + k = 9k + 1 = 9(1) + 1 = 10$.

(A) matrix $A$ of order $3 \times 3$ has $9$ entries,where each entry $a_{ij} \in \{0, 1\}$.
The sum of all entries $S = \sum_{i=1}^{3} \sum_{j=1}^{3} a_{ij}$ can range from $0$ to $9$.
Since the sum must be a prime number,the possible values for the sum are $2, 3, 5, 7$ (as $0$ and $1$ are not prime,and $9$ is not prime).
The number of ways to choose $k$ entries to be $1$ (and the rest $0$) is given by the combination formula $\binom{9}{k}$.
Total number of matrices = $\binom{9}{2} + \binom{9}{3} + \binom{9}{5} + \binom{9}{7}$.
Calculating each term:
$\binom{9}{2} = \frac{9 \times 8}{2} = 36$
$\binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$
$\binom{9}{5} = \binom{9}{4} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126$
$\binom{9}{7} = \binom{9}{2} = 36$
Total = $36 + 84 + 126 + 36 = 282$.

(A) Given $\Delta = \left|\begin{array}{ccc} p! & (p+1)! & (p+2)! \\ (p+1)! & (p+2)! & (p+3)! \\ (p+2)! & (p+3)! & (p+4)!\end{array}\right|$.
Taking $p!$,$(p+1)!$,and $(p+2)!$ common from rows $1, 2, 3$ respectively:
$\Delta = p!(p+1)!(p+2)! \left|\begin{array}{ccc} 1 & p+1 & (p+2)(p+1) \\ 1 & p+2 & (p+3)(p+2) \\ 1 & p+3 & (p+4)(p+3)\end{array}\right|$.
Applying $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_2$:
$\Delta = p!(p+1)!(p+2)! \left|\begin{array}{ccc} 1 & p+1 & (p+1)(p+2) \\ 0 & 1 & 2(p+2) \\ 0 & 1 & 2(p+3)\end{array}\right|$.
Expanding along the first column:
$\Delta = p!(p+1)!(p+2)! [1 \cdot (2(p+3) - 2(p+2))] = p!(p+1)!(p+2)! [2] = 2 \cdot p!(p+1)!(p+2)!$.
Since $p$ and $p+2$ are primes,$p!$ contains $p$ exactly once (for $p \ge 2$).
$(p+1)! = (p+1)p!$,so $p$ appears in $p!$ and $(p+1)!$,thus $p^2$ divides $p!(p+1)!$. Also $(p+2)!$ contains $p$ once. So $p^3$ divides $\Delta$,hence $\alpha = 3$.
$(p+2)!$ contains $(p+2)$ once. Thus $(p+2)^1$ divides $\Delta$,hence $\beta = 1$.
The sum $\alpha + \beta = 3 + 1 = 4$.