JEE Main 2021 Mathematics Question Paper with Answer and Solution

(C) For $F_{1}(A, B, C) = (A \wedge \sim B) \vee [\sim C \wedge (A \vee B)] \vee \sim A$:
Using the distributive law:
$F_{1} = [(A \wedge \sim B) \vee \sim A] \vee [\sim C \wedge (A \vee B)]$
$F_{1} = [(A \vee \sim A) \wedge (\sim B \vee \sim A)] \vee [\sim C \wedge (A \vee B)]$
Since $(A \vee \sim A) = t$ (tautology):
$F_{1} = [t \wedge (\sim A \vee \sim B)] \vee [\sim C \wedge (A \vee B)]$
$F_{1} = (\sim A \vee \sim B) \vee [\sim C \wedge (A \vee B)]$.
This expression depends on the values of $A, B, C$,so it is not a tautology.
For $F_{2}(A, B) = (A \vee B) \vee (B \rightarrow \sim A)$:
Using the implication rule $(P \rightarrow Q) = (\sim P \vee Q)$:
$F_{2} = (A \vee B) \vee (\sim B \vee \sim A)$
By commutative and associative laws:
$F_{2} = (A \vee \sim A) \vee (B \vee \sim B)$
$F_{2} = t \vee t = t$.
Thus,$F_{2}$ is a tautology.

(B) Let the point on the circle be $(\cos \theta, \sin \theta)$.
Let the mid-point of the line segment joining $(3, 2)$ and $(\cos \theta, \sin \theta)$ be $P(h, k)$.
Then,$h = \frac{\cos \theta + 3}{2}$ and $k = \frac{\sin \theta + 2}{2}$.
This implies $\cos \theta = 2h - 3$ and $\sin \theta = 2k - 2$.
Since $\cos^{2} \theta + \sin^{2} \theta = 1$,we have $(2h - 3)^{2} + (2k - 2)^{2} = 1$.
Dividing by $4$,we get $(h - \frac{3}{2})^{2} + (k - 1)^{2} = \frac{1}{4}$.
This represents a circle with radius $r = \sqrt{\frac{1}{4}} = \frac{1}{2}$.

(A) Given $\tan^{-1} a + \tan^{-1} b = \frac{\pi}{4}$ where $0 < a, b < 1$.
Using the formula $\tan^{-1} a + \tan^{-1} b = \tan^{-1} \left(\frac{a+b}{1-ab}\right) = \frac{\pi}{4}$.
Taking $\tan$ on both sides,we get $\frac{a+b}{1-ab} = \tan \frac{\pi}{4} = 1$.
Thus,$a+b = 1-ab$,which implies $a+b+ab = 1$.
Adding $1$ to both sides,we get $1+a+b+ab = 2$,which factors as $(1+a)(1+b) = 2$.
The given series is $S = \left(a - \frac{a^2}{2} + \frac{a^3}{3} - \dots\right) + \left(b - \frac{b^2}{2} + \frac{b^3}{3} - \dots\right)$.
Using the logarithmic expansion $\log_e(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots$ for $|x| < 1$,we have:
$S = \log_e(1+a) + \log_e(1+b) = \log_e((1+a)(1+b))$.
Substituting $(1+a)(1+b) = 2$,we get $S = \log_e 2$.

(B) Let $T_n = \frac{n^2+6n+10}{(2n+1)!}$.
We can rewrite the numerator as:
$n^2+6n+10 = \frac{1}{4}(4n^2+24n+40) = \frac{1}{4}((2n+1)^2 + 20n + 39) = \frac{1}{4}((2n+1)^2 + 10(2n+1) + 29)$.
Thus,$T_n = \frac{1}{4} \left[ \frac{2n+1}{(2n)!} + \frac{10}{(2n)!} + \frac{29}{(2n+1)!} \right]$.
Using $\frac{2n+1}{(2n)!} = \frac{2n}{(2n)!} + \frac{1}{(2n)!} = \frac{1}{(2n-1)!} + \frac{1}{(2n)!}$,we get:
$T_n = \frac{1}{4} \left[ \frac{1}{(2n-1)!} + \frac{11}{(2n)!} + \frac{29}{(2n+1)!} \right]$.
Summing from $n=1$ to $\infty$:
$S = \frac{1}{4} \left[ \sum_{n=1}^{\infty} \frac{1}{(2n-1)!} + 11 \sum_{n=1}^{\infty} \frac{1}{(2n)!} + 29 \sum_{n=1}^{\infty} \frac{1}{(2n+1)!} \right]$.
Recall the series expansions: $\sum_{n=1}^{\infty} \frac{1}{(2n-1)!} = \frac{e-e^{-1}}{2}$,$\sum_{n=1}^{\infty} \frac{1}{(2n)!} = \frac{e+e^{-1}-2}{2}$,and $\sum_{n=1}^{\infty} \frac{1}{(2n+1)!} = \frac{e-e^{-1}-2}{2}$.
Substituting these values:
$S = \frac{1}{4} \left[ \frac{e-e^{-1}}{2} + 11 \frac{e+e^{-1}-2}{2} + 29 \frac{e-e^{-1}-2}{2} \right]$
$S = \frac{1}{8} [e - e^{-1} + 11e + 11e^{-1} - 22 + 29e - 29e^{-1} - 58]$
$S = \frac{1}{8} [41e - 19e^{-1} - 80] = \frac{41}{8}e - \frac{19}{8}e^{-1} - 10$.

(A) Let $P$ be a point on the circle $(x-1)^{2} + (y-1)^{2} = 1$.
We can parameterize $P$ as $(1 + \cos \theta, 1 + \sin \theta)$.
Given $A(1, 4)$ and $B(1, -5)$,we calculate $(PA)^{2} + (PB)^{2}$:
$(PA)^{2} = (1 + \cos \theta - 1)^{2} + (1 + \sin \theta - 4)^{2} = \cos^{2} \theta + (\sin \theta - 3)^{2} = \cos^{2} \theta + \sin^{2} \theta - 6 \sin \theta + 9 = 10 - 6 \sin \theta$.
$(PB)^{2} = (1 + \cos \theta - 1)^{2} + (1 + \sin \theta + 5)^{2} = \cos^{2} \theta + (\sin \theta + 6)^{2} = \cos^{2} \theta + \sin^{2} \theta + 12 \sin \theta + 36 = 37 + 12 \sin \theta$.
Summing these,$(PA)^{2} + (PB)^{2} = 10 - 6 \sin \theta + 37 + 12 \sin \theta = 47 + 6 \sin \theta$.
This expression is maximized when $\sin \theta = 1$.
For $\sin \theta = 1$,we have $\cos \theta = 0$,so $P = (1 + 0, 1 + 1) = (1, 2)$.
The points are $P(1, 2)$,$A(1, 4)$,and $B(1, -5)$.
Since all points have an $x$-coordinate of $1$,they all lie on the vertical line $x = 1$,which is a straight line.

(C) The given digits are $3, 3, 4, 4, 4, 5, 5$. Total number of digits is $7$.
The total number of $7$-digit numbers that can be formed is given by the permutation of these digits:
$\text{Total numbers} = \frac{7!}{2! \times 3! \times 2!} = \frac{5040}{2 \times 6 \times 2} = \frac{5040}{24} = 210$.
$A$ number is divisible by $2$ if its last digit is even. Here,the only even digit available is $4$.
If the last digit is fixed as $4$,we are left with $6$ digits: $3, 3, 4, 4, 5, 5$.
The number of such $7$-digit numbers is:
$\text{Favorable numbers} = \frac{6!}{2! \times 2! \times 2!} = \frac{720}{2 \times 2 \times 2} = \frac{720}{8} = 90$.
The probability that the number is divisible by $2$ is:
$P = \frac{\text{Favorable numbers}}{\text{Total numbers}} = \frac{90}{210} = \frac{9}{21} = \frac{3}{7}$.

(B) Given $|z+5| \leq 4$. Let $z = x+iy$. Then $(x+5)^2 + y^2 \leq 16$ (a disk centered at $(-5, 0)$ with radius $4$).
Given $z(1+i)+\bar{z}(1-i) \geq -10$. Substituting $z=x+iy$ and $\bar{z}=x-iy$:
$(x+iy)(1+i) + (x-iy)(1-i) \geq -10$
$(x-y + i(x+y)) + (x-y - i(x+y)) \geq -10$
$2(x-y) \geq -10 \implies x-y+5 \geq 0$.
We want to maximize $|z+1|^2$,which is the square of the distance from $z$ to the point $P(-1, 0)$.
The region is the intersection of the disk $(x+5)^2 + y^2 \leq 16$ and the half-plane $x-y+5 \geq 0$.
To find the maximum distance from $P(-1, 0)$ to the region,we check the boundary points. The maximum occurs at the intersection of the line $x-y+5=0$ and the circle $(x+5)^2 + y^2 = 16$.
Substituting $y = x+5$ into the circle equation:
$(x+5)^2 + (x+5)^2 = 16 \implies 2(x+5)^2 = 16 \implies (x+5)^2 = 8 \implies x+5 = \pm 2\sqrt{2}$.
So $x = -5 \pm 2\sqrt{2}$.
If $x = -5 - 2\sqrt{2}$,then $y = x+5 = -2\sqrt{2}$. Point $B = (-5-2\sqrt{2}, -2\sqrt{2})$.
If $x = -5 + 2\sqrt{2}$,then $y = x+5 = 2\sqrt{2}$. Point $A = (-5+2\sqrt{2}, 2\sqrt{2})$.
Calculate squared distances from $P(-1, 0)$:
$PB^2 = (-5-2\sqrt{2} - (-1))^2 + (-2\sqrt{2} - 0)^2 = (-4-2\sqrt{2})^2 + 8 = (16 + 8 + 16\sqrt{2}) + 8 = 32 + 16\sqrt{2}$.
$PA^2 = (-5+2\sqrt{2} - (-1))^2 + (2\sqrt{2} - 0)^2 = (-4+2\sqrt{2})^2 + 8 = (16 + 8 - 16\sqrt{2}) + 8 = 32 - 16\sqrt{2}$.
The maximum value is $32 + 16\sqrt{2}$.
Thus $\alpha = 32$ and $\beta = 16$.
$\alpha + \beta = 32 + 16 = 48$.

(D) Since the normals at all points on the curve pass through a fixed point $(a, b)$,the curve must be a circle with center $(a, b)$.
Let the points on the circle be $A(3, -3)$ and $B(4, -2\sqrt{2})$.
Since $A$ and $B$ lie on the circle,their distances from the center $C(a, b)$ are equal to the radius $r$.
Thus,$CA^{2} = CB^{2}$.
$(a - 3)^{2} + (b - (-3))^{2} = (a - 4)^{2} + (b - (-2\sqrt{2}))^{2}$
$(a - 3)^{2} + (b + 3)^{2} = (a - 4)^{2} + (b + 2\sqrt{2})^{2}$
$a^{2} - 6a + 9 + b^{2} + 6b + 9 = a^{2} - 8a + 16 + b^{2} + 4\sqrt{2}b + 8$
$-6a + 6b + 18 = -8a + 4\sqrt{2}b + 24$
$2a + (6 - 4\sqrt{2})b = 6$
Dividing by $2$,we get $a + (3 - 2\sqrt{2})b = 3$.
$a + 3b - 2\sqrt{2}b = 3$
$a - 2\sqrt{2}b + 3b = 3 \quad ... (1)$
Given that $a - 2\sqrt{2}b = 3 \quad ... (2)$
Substituting $(2)$ into $(1)$,we get $3 + 3b = 3$,which implies $3b = 0$,so $b = 0$.
Substituting $b = 0$ into $(2)$,we get $a - 2\sqrt{2}(0) = 3$,so $a = 3$.
Therefore,$a^{2} + b^{2} + ab = (3)^{2} + (0)^{2} + (3)(0) = 9 + 0 + 0 = 9$.

(B) Given that $\alpha$ and $\beta$ are roots of the quadratic equation $x^{2} - (\alpha+\beta)x + \alpha\beta = 0$.
Substituting the values,we get $x^{2} - x - 1 = 0$.
Since $\alpha$ and $\beta$ are roots,they satisfy the equation:
$\alpha^{2} - \alpha - 1 = 0 \Rightarrow \alpha^{n+1} = \alpha^{n} + \alpha^{n-1}$
$\beta^{2} - \beta - 1 = 0 \Rightarrow \beta^{n+1} = \beta^{n} + \beta^{n-1}$
Adding these two equations,we get:
$(\alpha^{n+1} + \beta^{n+1}) = (\alpha^{n} + \beta^{n}) + (\alpha^{n-1} + \beta^{n-1})$
This implies the recurrence relation $p_{n+1} = p_{n} + p_{n-1}$.
Given $p_{n+1} = 29$ and $p_{n-1} = 11$,we have:
$29 = p_{n} + 11$
$p_{n} = 29 - 11 = 18$.
Therefore,$p_{n}^{2} = 18^{2} = 324$.

(C) The given sequence is a geometric progression with first term $a = -16$ and common ratio $r = -1/2$.
The $n^{\text{th}}$ term is $t_{n} = a r^{n-1} = -16(-1/2)^{n-1}$.
Let the arithmetic mean and geometric mean of $t_{p}$ and $t_{q}$ be $x_{1}$ and $x_{2}$. The equation $4x^{2}-9x+5=0$ has roots $x = 1$ and $x = 5/4$.
Since the arithmetic mean is greater than the geometric mean for distinct terms,we have $AM = 5/4$ and $GM = 1$.
$AM = (t_{p} + t_{q})/2 = 5/4 \Rightarrow t_{p} + t_{q} = 5/2$.
$GM = \sqrt{t_{p} t_{q}} = 1 \Rightarrow t_{p} t_{q} = 1$.
Substituting $t_{p} = -16(-1/2)^{p-1}$ and $t_{q} = -16(-1/2)^{q-1}$:
$t_{p} t_{q} = 256(-1/2)^{p+q-2} = 1 \Rightarrow (-1/2)^{p+q-2} = 1/256 = (1/2)^{8}$.
Since $(-1/2)^{p+q-2} = (1/2)^{8}$,$p+q-2$ must be an even number equal to $8$.
$p+q-2 = 8 \Rightarrow p+q = 10$.

(A) Let $N$ be a $4$-digit number such that $\gcd(N, 18) = 3$.
Since $\gcd(N, 18) = 3$,$N$ must be a multiple of $3$ but not a multiple of $2$ (because $18 = 2 \times 3^2$,and if $N$ were even,$\gcd(N, 18)$ would be at least $6$) and not a multiple of $9$ (because if $N$ were a multiple of $9$,$\gcd(N, 18)$ would be $9$ or $18$).
Thus,$N$ must be an odd multiple of $3$ that is not divisible by $9$.
First,find the number of $4$-digit odd multiples of $3$:
The smallest $4$-digit odd multiple of $3$ is $1005$ and the largest is $9999$.
These form an arithmetic progression: $1005, 1011, \dots, 9999$.
The number of terms is $\frac{9999 - 1005}{6} + 1 = \frac{8994}{6} + 1 = 1499 + 1 = 1500$.
Next,find the number of $4$-digit odd multiples of $9$:
The smallest $4$-digit odd multiple of $9$ is $1017$ and the largest is $9999$.
These form an arithmetic progression: $1017, 1035, \dots, 9999$.
The number of terms is $\frac{9999 - 1017}{18} + 1 = \frac{8982}{18} + 1 = 499 + 1 = 500$.
The number of such $N$ is $1500 - 500 = 1000$.

(A) The given curves are $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$ (an ellipse) and $x^{2} + y^{2} = \frac{31}{4}$ (a circle).
Let the slope of the common tangent be $m$.
The equation of a tangent to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ is $y = mx \pm \sqrt{a^{2}m^{2} + b^{2}}$.
Here $a^{2} = 9$ and $b^{2} = 4$,so the tangent is $y = mx \pm \sqrt{9m^{2} + 4}$.
The equation of a tangent to the circle $x^{2} + y^{2} = r^{2}$ is $y = mx \pm r\sqrt{1 + m^{2}}$.
Here $r^{2} = \frac{31}{4}$,so the tangent is $y = mx \pm \frac{\sqrt{31}}{2}\sqrt{1 + m^{2}}$.
For the lines to be identical,the constant terms must be equal:
$9m^{2} + 4 = \frac{31}{4}(1 + m^{2})$.
Multiplying by $4$,we get $36m^{2} + 16 = 31 + 31m^{2}$.
$5m^{2} = 15$.
$m^{2} = 3$.

(B) Let $f(x) = 2x^{5}+5x^{4}+10x^{3}+10x^{2}+10x+10$.
First,we evaluate the function at integer points to locate the root.
$f(-2) = 2(-32) + 5(16) + 10(-8) + 10(4) + 10(-2) + 10 = -64 + 80 - 80 + 40 - 20 + 10 = -34$.
$f(-1) = 2(-1) + 5(1) + 10(-1) + 10(1) + 10(-1) + 10 = -2 + 5 - 10 + 10 - 10 + 10 = 3$.
Since $f(-2) < 0$ and $f(-1) > 0$,by the Intermediate Value Theorem,there exists at least one real root in the interval $(-2, -1)$.
Next,we check the derivative to determine the number of real roots.
$f'(x) = 10x^{4} + 20x^{3} + 30x^{2} + 20x + 10$.
$f'(x) = 10(x^{4} + 2x^{3} + 3x^{2} + 2x + 1)$.
Notice that $x^{4} + 2x^{3} + 3x^{2} + 2x + 1 = (x^{2} + x + 1)^{2}$.
Thus,$f'(x) = 10(x^{2} + x + 1)^{2}$.
Since $x^{2} + x + 1$ has a negative discriminant $(1 - 4 = -3)$,it is always positive for all real $x$.
Therefore,$f'(x) > 0$ for all real $x$,which means $f(x)$ is strictly increasing and has exactly one real root.
Since the root lies in $(-2, -1)$,we have $a = -2$. Thus,$|a| = |-2| = 2$.

(A) Given $\sum_{i=1}^{18}(X_{i}-\alpha)=36 \implies \sum X_{i} - 18\alpha = 36 \implies \sum X_{i} = 18(\alpha+2)$.
Given $\sum_{i=1}^{18}(X_{i}-\beta)^{2}=90 \implies \sum X_{i}^{2} - 2\beta \sum X_{i} + 18\beta^{2} = 90$.
Substituting $\sum X_{i} = 18(\alpha+2)$,we get $\sum X_{i}^{2} = 90 - 18\beta^{2} + 36\beta(\alpha+2)$.
The variance is given by $\sigma^{2} = \frac{\sum X_{i}^{2}}{18} - (\frac{\sum X_{i}}{18})^{2} = 1^{2} = 1$.
Substituting the values: $\frac{90 - 18\beta^{2} + 36\beta(\alpha+2)}{18} - (\alpha+2)^{2} = 1$.
$5 - \beta^{2} + 2\beta(\alpha+2) - (\alpha^{2} + 4\alpha + 4) = 1$.
$5 - \beta^{2} + 2\alpha\beta + 4\beta - \alpha^{2} - 4\alpha - 4 = 1$.
$-(\alpha^{2} - 2\alpha\beta + \beta^{2}) + 4(\beta - \alpha) = 0$.
$-(\alpha-\beta)^{2} - 4(\alpha-\beta) = 0$.
$(\alpha-\beta)^{2} + 4(\alpha-\beta) = 0$.
$(\alpha-\beta)(\alpha-\beta+4) = 0$.
Since $\alpha \neq \beta$,we have $\alpha-\beta = -4$,so $|\alpha-\beta| = 4$.

(A) Given the equation $\frac{\cos x}{1+\sin x} = |\tan 2x|$.
Note that $\frac{\cos x}{1+\sin x} = \tan\left(\frac{\pi}{4} - \frac{x}{2}\right)$.
So,$\tan\left(\frac{\pi}{4} - \frac{x}{2}\right) = |\tan 2x|$.
Since the $LHS$ must be non-negative,$\tan\left(\frac{\pi}{4} - \frac{x}{2}\right) \ge 0$,which implies $\frac{\pi}{4} - \frac{x}{2} \in [0, \frac{\pi}{2})$,so $x \in (-\frac{\pi}{2}, \frac{\pi}{2}]$.
Squaring both sides,$\tan^2\left(\frac{\pi}{4} - \frac{x}{2}\right) = \tan^2 2x$.
This implies $\tan 2x = \pm \tan\left(\frac{\pi}{4} - \frac{x}{2}\right)$.
Case $1$: $2x = n\pi + \left(\frac{\pi}{4} - \frac{x}{2}\right)$ $\Rightarrow \frac{5x}{2} = n\pi + \frac{\pi}{4}$ $\Rightarrow x = \frac{2n\pi}{5} + \frac{\pi}{10}$.
For $n=0, x=\frac{\pi}{10}$. For $n=-1, x=-\frac{3\pi}{10}$.
Case $2$: $2x = n\pi - \left(\frac{\pi}{4} - \frac{x}{2}\right)$ $\Rightarrow \frac{3x}{2} = n\pi - \frac{\pi}{4}$ $\Rightarrow x = \frac{2n\pi}{3} - \frac{\pi}{6}$.
For $n=0, x=-\frac{\pi}{6}$.
Checking the condition $\tan\left(\frac{\pi}{4} - \frac{x}{2}\right) \ge 0$: for $x = \frac{\pi}{10}, -\frac{3\pi}{10}, -\frac{\pi}{6}$,the condition holds.
The sum of solutions is $\frac{\pi}{10} - \frac{3\pi}{10} - \frac{\pi}{6} = -\frac{2\pi}{10} - \frac{\pi}{6} = -\frac{\pi}{5} - \frac{\pi}{6} = -\frac{11\pi}{30}$.

(D) Given: $n = 20$,$\bar{x} = 10$,$\sigma = 2.5$.
Incorrect sum of observations: $\Sigma x_i = n \times \bar{x} = 20 \times 10 = 200$.
Correct sum of observations: $\Sigma x_i = 200 - 25 + 35 = 210$.
Correct mean $\alpha = \frac{210}{20} = 10.5$.
Incorrect sum of squares: $\sigma^2 = \frac{\Sigma x_i^2}{n} - (\bar{x})^2 \implies (2.5)^2 = \frac{\Sigma x_i^2}{20} - 10^2$.
$6.25 = \frac{\Sigma x_i^2}{20} - 100 \implies \Sigma x_i^2 = 20 \times 106.25 = 2125$.
Correct sum of squares: $\Sigma x_i^2 = 2125 - 25^2 + 35^2 = 2125 - 625 + 1225 = 2725$.
Correct variance $\beta = \frac{\Sigma x_i^2}{n} - (\alpha)^2 = \frac{2725}{20} - (10.5)^2 = 136.25 - 110.25 = 26$.
Thus,$(\alpha, \beta) = (10.5, 26)$.

(B) The equation of the ellipse is $\frac{x^{2}}{8}+\frac{y^{2}}{4}=1$. Here $a^{2}=8$ and $b^{2}=4$.
The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^{2}}{a^{2}}} = \sqrt{1 - \frac{4}{8}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.
The foci are $S(-ae, 0)$ and $S'(ae, 0)$,where $ae = \sqrt{8} \cdot \frac{1}{\sqrt{2}} = 2$. So,$S(-2, 0)$ and $S'(2, 0)$.
The line perpendicular to $x+2y=0$ has the form $2x-y+k=0$,or $y=2x+k$.
The condition for tangency $y=mx+c$ to the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ is $c^{2}=a^{2}m^{2}+b^{2}$.
Here $m=2$,so $c^{2} = 8(2^{2}) + 4 = 32+4 = 36$,which gives $c = \pm 6$.
Since $P$ is in the second quadrant,the tangent must have a positive $y$-intercept,so $y=2x+6$.
The point of tangency $P(x_{1}, y_{1})$ is given by $(-\frac{a^{2}m}{c}, \frac{b^{2}}{c}) = (-\frac{8 \cdot 2}{6}, \frac{4}{6}) = (-\frac{8}{3}, \frac{2}{3})$.
The area $A$ of $\Delta SPS'$ with base $SS'=4$ and height $|y_{1}| = \frac{2}{3}$ is $A = \frac{1}{2} \cdot 4 \cdot \frac{2}{3} = \frac{4}{3}$.
Finally,$(5-e^{2}) \cdot A = (5 - \frac{1}{2}) \cdot \frac{4}{3} = \frac{9}{2} \cdot \frac{4}{3} = 6$.

(A) Let the sum be $S = \sum_{k=0}^{100} \frac{2^k}{x^{2^k}+1}$.
We use the identity $\frac{1}{x-1} - \frac{1}{x+1} = \frac{2}{x^2-1}$.
More generally,$\frac{2^k}{x^{2^k}-1} - \frac{2^k}{x^{2^k}+1} = \frac{2^{k+1}}{x^{2^{k+1}}-1}$.
Rearranging,$\frac{2^k}{x^{2^k}+1} = \frac{2^k}{x^{2^k}-1} - \frac{2^{k+1}}{x^{2^{k+1}}-1}$.
This is a telescoping sum.
$S = \left( \frac{1}{x-1} - \frac{2}{x^2-1} \right) + \left( \frac{2}{x^2-1} - \frac{4}{x^4-1} \right) + \ldots + \left( \frac{2^{100}}{x^{2^{100}}-1} - \frac{2^{101}}{x^{2^{101}}-1} \right)$.
$S = \frac{1}{x-1} - \frac{2^{101}}{x^{2^{101}}-1}$.
Given $x=2$,$S = \frac{1}{2-1} - \frac{2^{101}}{2^{2^{101}}-1} = 1 - \frac{2^{101}}{2^{2^{101}}-1}$.

(D) We know that $\sum_{r=0}^{n} r \cdot {}^{n}C_{r} = n \cdot 2^{n-1}$ and $\sum_{r=0}^{n} r(r-1) \cdot {}^{n}C_{r} = n(n-1) \cdot 2^{n-2}$.
We can write $r^{2} = r(r-1) + r$.
Therefore,$\sum_{r=0}^{20} r^{2} \cdot {}^{20}C_{r} = \sum_{r=0}^{20} [r(r-1) + r] \cdot {}^{20}C_{r}$.
$= \sum_{r=0}^{20} r(r-1) \cdot {}^{20}C_{r} + \sum_{r=0}^{20} r \cdot {}^{20}C_{r}$.
Using the identities with $n=20$:
$= 20 \times 19 \times 2^{20-2} + 20 \times 2^{20-1}$.
$= 380 \times 2^{18} + 20 \times 2^{19}$.
$= 380 \times 2^{18} + 20 \times 2 \times 2^{18}$.
$= 380 \times 2^{18} + 40 \times 2^{18}$.
$= (380 + 40) \times 2^{18} = 420 \times 2^{18}$.

(A) Let $A$ be the set of patients with heart ailment and $B$ be the set of patients with lung infection.
Given $n(A) = 89\%$ and $n(B) = 98\%$.
We know that $n(A \cup B) = n(A) + n(B) - n(A \cap B)$.
Since $n(A \cup B) \leq 100\%$,we have $89 + 98 - n(A \cap B) \leq 100$.
$187 - n(A \cap B) \leq 100 \implies n(A \cap B) \geq 87$.
Also,$n(A \cap B)$ cannot exceed the smaller of the two sets,so $n(A \cap B) \leq 89$.
Thus,$87 \leq K \leq 89$.
Therefore,$K$ must be in the range $[87, 89]$.
Checking the options,the set $\{79, 81, 83, 85\}$ contains values that are not in the range $[87, 89]$.

(B) Let $z = x + iy$. The equation $\arg \left(\frac{z-1}{z+1}\right) = \frac{\pi}{4}$ represents the locus of a point $z$ such that the angle subtended by the segment joining $A(1, 0)$ and $B(-1, 0)$ at $z$ is $\frac{\pi}{4}$.
This is an arc of a circle passing through $A(1, 0)$ and $B(-1, 0)$.
Let the centre of the circle be $C(0, k)$. Since the angle at the circumference is $\frac{\pi}{4}$,the angle subtended by the chord $AB$ at the centre $C$ is $2 \times \frac{\pi}{4} = \frac{\pi}{2}$.
In $\triangle OAC$ (where $O$ is the origin $(0,0)$),$\angle COA = 90^\circ$ and $\angle OCA = \frac{\pi}{4}$.
Since $OA = 1$,we have $\tan \left(\frac{\pi}{4}\right) = \frac{OA}{OC} = \frac{1}{OC} = 1$,which implies $OC = 1$.
Thus,the centre is $C(0, 1)$.
The radius $R = AC = \sqrt{OA^2 + OC^2} = \sqrt{1^2 + 1^2} = \sqrt{2}$.
Therefore,the circle has centre $(0, 1)$ and radius $\sqrt{2}$.

(D) The equation of the circle is $4x^{2}+4y^{2}+120x+675=0$,which simplifies to $x^{2}+y^{2}+30x+\frac{675}{4}=0$.
Completing the square: $(x+15)^{2}+y^{2} = 225 - \frac{675}{4} = \frac{900-675}{4} = \frac{225}{4}$.
So,the center is $(-15, 0)$ and the radius $R = \sqrt{\frac{225}{4}} = \frac{15}{2}$.
The equation of a line passing through $(-30, 0)$ is $y = m(x+30)$,or $mx - y + 30m = 0$.
This line is tangent to the parabola $y^{2}=30x$. The condition for tangency $y=mx+c$ to $y^{2}=4ax$ is $c = \frac{a}{m}$.
Here $4a=30 \Rightarrow a = \frac{15}{2}$. The line is $y = mx + 30m$,so $c = 30m$.
Thus,$30m = \frac{15/2}{m}$ $\Rightarrow 60m^{2} = 15$ $\Rightarrow m^{2} = \frac{1}{4}$ $\Rightarrow m = \pm \frac{1}{2}$.
Taking $m = \frac{1}{2}$,the line is $y = \frac{1}{2}(x+30) \Rightarrow x - 2y + 30 = 0$.
The perpendicular distance $P$ from the center $(-15, 0)$ to the line $x - 2y + 30 = 0$ is:
$P = \frac{|-15 - 2(0) + 30|}{\sqrt{1^{2} + (-2)^{2}}} = \frac{15}{\sqrt{5}} = 3\sqrt{5}$.
The length of the chord is $2\sqrt{R^{2}-P^{2}} = 2\sqrt{(\frac{15}{2})^{2} - (3\sqrt{5})^{2}} = 2\sqrt{\frac{225}{4} - 45} = 2\sqrt{\frac{225-180}{4}} = 2\sqrt{\frac{45}{4}} = \sqrt{45} = 3\sqrt{5}$.

(B) The sum of the infinite $GP$ is given by $\frac{a}{1-r} = 15 \dots (i)$.
The series formed by the squares of the terms is $a^{2}, a^{2}r^{2}, a^{2}r^{4}, \dots$,which is also a $GP$ with first term $a^{2}$ and common ratio $r^{2}$.
The sum of this series is $\frac{a^{2}}{1-r^{2}} = 150$.
We can rewrite this as $\frac{a}{1-r} \cdot \frac{a}{1+r} = 150$.
Substituting $(i)$ into this equation,we get $15 \cdot \frac{a}{1+r} = 150$,which implies $\frac{a}{1+r} = 10 \dots (ii)$.
Dividing $(i)$ by $(ii)$,we get $\frac{1+r}{1-r} = \frac{15}{10} = \frac{3}{2}$.
Solving for $r$: $2 + 2r = 3 - 3r$ $\Rightarrow 5r = 1$ $\Rightarrow r = \frac{1}{5}$.
Substituting $r = \frac{1}{5}$ into $(i)$: $\frac{a}{1 - 1/5} = 15$ $\Rightarrow \frac{a}{4/5} = 15$ $\Rightarrow a = 15 \cdot \frac{4}{5} = 12$.
The series $ar^{2}, ar^{4}, ar^{6}, \dots$ is a $GP$ with first term $A = ar^{2}$ and common ratio $R = r^{2}$.
Sum $= \frac{ar^{2}}{1-r^{2}} = \frac{12 \cdot (1/5)^{2}}{1 - (1/5)^{2}} = \frac{12 \cdot (1/25)}{1 - 1/25} = \frac{12/25}{24/25} = \frac{12}{24} = \frac{1}{2}$.

(C) Let $C = (a, b)$. Since $C$ lies on the angle bisector $7x - 4y - 1 = 0$,we have $7a - 4b = 1 \quad \dots(i)$.
Let $M$ be the midpoint of $AC$. Since $A = (-3, 1)$,$M = (\frac{a-3}{2}, \frac{b+1}{2})$.
Since $M$ lies on the median $2x + y - 3 = 0$,we have $2(\frac{a-3}{2}) + (\frac{b+1}{2}) - 3 = 0$,which simplifies to $2a - 6 + b + 1 - 6 = 0$,or $2a + b = 11 \quad \dots(ii)$.
Solving $(i)$ and $(ii)$,we multiply $(ii)$ by $4$: $8a + 4b = 44$. Adding to $(i)$: $15a = 45 \Rightarrow a = 3$. Substituting into $(ii)$: $6 + b = 11 \Rightarrow b = 5$. Thus,$C = (3, 5)$.
The slope of $AC$ is $m_{AC} = \frac{5-1}{3-(-3)} = \frac{4}{6} = \frac{2}{3}$.
The slope of the angle bisector $7x - 4y - 1 = 0$ is $m_{bisector} = \frac{7}{4}$.
Let $\alpha$ be the angle of $AC$ and $\beta$ be the angle of the bisector. The angle between them is $\frac{\theta}{2}$.
$\tan(\frac{\theta}{2}) = |\frac{m_{bisector} - m_{AC}}{1 + m_{bisector} \cdot m_{AC}}| = |\frac{7/4 - 2/3}{1 + (7/4)(2/3)}| = |\frac{(21-8)/12}{(12+14)/12}| = \frac{13}{26} = \frac{1}{2}$.
Finally,$\tan \theta = \frac{2 \tan(\theta/2)}{1 - \tan^2(\theta/2)} = \frac{2(1/2)}{1 - (1/2)^2} = \frac{1}{1 - 1/4} = \frac{1}{3/4} = \frac{4}{3}$.

(C) The implication $A \rightarrow B$ is false if and only if $A$ is true and $B$ is false.
Here,$A = (p \vee q) \wedge (q \rightarrow r) \wedge (\sim r)$ and $B = (p \wedge q)$.
For $B = (p \wedge q)$ to be false,at least one of $p$ or $q$ must be false.
For $A$ to be true,all components $(p \vee q)$,$(q \rightarrow r)$,and $(\sim r)$ must be true.
Since $(\sim r)$ is true,$r$ must be false.
Since $(q \rightarrow r)$ is true and $r$ is false,$q$ must be false.
Since $(p \vee q)$ is true and $q$ is false,$p$ must be true.
Thus,the truth values are $p=T, q=F, r=F$.
Checking option $C$: $p=T, q=F, r=F$ gives $A = (T \vee F) \wedge (F \rightarrow F) \wedge (\sim F) = T \wedge T \wedge T = T$ and $B = (T \wedge F) = F$.
Since $T \rightarrow F$ is false,option $C$ is correct.

(D) Given $z = \frac{1 - i \sqrt{3}}{2} = e^{-i \frac{\pi}{3}}$.
Then $z^r + \frac{1}{z^r} = e^{-i \frac{r\pi}{3}} + e^{i \frac{r\pi}{3}} = 2 \cos\left(\frac{r\pi}{3}\right)$.
We need to evaluate $S = 21 + \sum_{r=1}^{21} \left(z^r + \frac{1}{z^r}\right)^3 = 21 + \sum_{r=1}^{21} \left(2 \cos\left(\frac{r\pi}{3}\right)\right)^3$.
Using the identity $8 \cos^3 \theta = 2(4 \cos^3 \theta) = 2(\cos 3\theta + 3 \cos \theta)$,we have:
$S = 21 + \sum_{r=1}^{21} 2 \left(\cos(r\pi) + 3 \cos\left(\frac{r\pi}{3}\right)\right) = 21 + 2 \sum_{r=1}^{21} \cos(r\pi) + 6 \sum_{r=1}^{21} \cos\left(\frac{r\pi}{3}\right)$.
For $\sum_{r=1}^{21} \cos(r\pi) = (-1) + 1 + (-1) + \dots + (-1) = -1$ (since there are $21$ terms).
For $\sum_{r=1}^{21} \cos\left(\frac{r\pi}{3}\right)$,the sum of $\cos\left(\frac{r\pi}{3}\right)$ over one period ($r=1$ to $6$) is $0$. Since $21 = 3 \times 6 + 3$,the sum is $\sum_{r=1}^{3} \cos\left(\frac{r\pi}{3}\right) = \cos\frac{\pi}{3} + \cos\frac{2\pi}{3} + \cos\pi = \frac{1}{2} - \frac{1}{2} - 1 = -1$.
Thus,$S = 21 + 2(-1) + 6(-1) = 21 - 2 - 6 = 13$.

(C) Given the equation $\frac{2}{x-1}-\frac{1}{x-2}=\frac{2}{k}$,where $x \in \mathbb{R} \setminus \{1, 2\}$.
Simplifying the equation: $\frac{2(x-2)-(x-1)}{(x-1)(x-2)} = \frac{2}{k} \Rightarrow \frac{x-3}{x^2-3x+2} = \frac{2}{k}$.
This gives $k(x-3) = 2(x^2-3x+2)$.
If $x=3$,then $0 = 2(9-9+2) = 4$,which is impossible. Thus $x \neq 3$.
For $x \neq 3$,we have $k = \frac{2(x^2-3x+2)}{x-3} = 2 \left( \frac{x^2-3x+2}{x-3} \right) = 2 \left( x + \frac{2}{x-3} \right) = 2 \left( (x-3) + \frac{2}{x-3} + 3 \right)$.
Let $f(x) = (x-3) + \frac{2}{x-3} + 3$.
Using the $AM$-$GM$ inequality,for $x > 3$,$(x-3) + \frac{2}{x-3} \geq 2\sqrt{2}$,so $f(x) \geq 3 + 2\sqrt{2}$.
For $x < 3$,$(x-3) + \frac{2}{x-3} = -(|x-3| + \frac{2}{|x-3|}) \leq -2\sqrt{2}$,so $f(x) \leq 3 - 2\sqrt{2}$.
Thus,$k = 2f(x) \in (-\infty, 6-4\sqrt{2}] \cup [6+4\sqrt{2}, \infty)$.
The equation has no real roots if $k$ lies in the gap: $k \in (6-4\sqrt{2}, 6+4\sqrt{2})$.
Approximating $\sqrt{2} \approx 1.414$,$6-4(1.414) = 6-5.656 = 0.344$ and $6+4(1.414) = 6+5.656 = 11.656$.
So $k \in (0.344, 11.656)$.
The integral values of $k$ are $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}$.
The sum of these values is $\frac{11 \times 12}{2} = 66$.

(A) The given expression is $\sum_{n=1}^{15} n \cdot {}^{n}P_{n}$.
Since ${}^{n}P_{n} = n!$,the expression becomes $\sum_{n=1}^{15} n \cdot n!$.
We know that $n \cdot n! = (n+1-1) \cdot n! = (n+1)! - n!$.
Thus,the sum is $\sum_{n=1}^{15} ((n+1)! - n!) = (2!-1!) + (3!-2!) + \ldots + (16!-15!) = 16! - 1! = 16! - 1$.
Given the expression is ${}^{q}P_{r} - s$,we have ${}^{16}P_{16} - 1 = {}^{q}P_{r} - s$.
Comparing,we get $q = 16$,$r = 16$,and $s = 1$.
We need to find ${}^{q+s}C_{r-s} = {}^{16+1}C_{16-1} = {}^{17}C_{15}$.
${}^{17}C_{15} = {}^{17}C_{2} = \frac{17 \times 16}{2 \times 1} = 17 \times 8 = 136$.

(A) Let the point be $P(x, y)$.
The sum of the squares of the distances from $(0,0), (1,0), (0,1), (1,1)$ is given by:
$(x^2 + y^2) + ((x-1)^2 + y^2) + (x^2 + (y-1)^2) + ((x-1)^2 + (y-1)^2) = 18$
Expanding the terms:
$(x^2 + y^2) + (x^2 - 2x + 1 + y^2) + (x^2 + y^2 - 2y + 1) + (x^2 - 2x + 1 + y^2 - 2y + 1) = 18$
Combining like terms:
$4x^2 + 4y^2 - 4x - 4y + 4 = 18$
Dividing by $4$:
$x^2 + y^2 - x - y + 1 = 4.5$
$x^2 + y^2 - x - y - 3.5 = 0$
The radius $r$ of the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is $\sqrt{g^2 + f^2 - c}$.
Here,$2g = -1 \Rightarrow g = -0.5$,$2f = -1 \Rightarrow f = -0.5$,and $c = -3.5$.
$r = \sqrt{(-0.5)^2 + (-0.5)^2 - (-3.5)} = \sqrt{0.25 + 0.25 + 3.5} = \sqrt{4} = 2$.
The diameter $d = 2r = 2(2) = 4$.
Therefore,$d^2 = 4^2 = 16$.

(B) To form a three-digit even number,the unit place must be occupied by an even digit $(0, 4, 6)$. Since repetition is not allowed,we consider two cases:
Case $(i)$: When the unit place is $0$.
There is $1$ way to fill the unit place. The remaining two places (hundreds and tens) can be filled by the remaining $5$ digits $(1, 3, 4, 6, 7)$ in $5 \times 4 = 20$ ways.
Case $(ii)$: When the unit place is $4$ or $6$.
There are $2$ ways to fill the unit place. The hundreds place cannot be $0$ and cannot be the digit already used in the unit place,so there are $6 - 2 = 4$ choices for the hundreds place. The tens place can then be filled by any of the remaining $4$ digits (including $0$),giving $4 \times 4 = 16$ ways for each choice of the unit digit. Total ways $= 2 \times 16 = 32$ ways.
Total number of even numbers $= 20 + 32 = 52$.

(C) Given the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ and eccentricity $e = \frac{\sqrt{5}}{2}$.
We know $b^{2} = a^{2}(e^{2}-1) = a^{2}(\frac{5}{4}-1) = \frac{a^{2}}{4}$,so $a^{2} = 4b^{2}$.
Since $P(-2 \sqrt{6}, \sqrt{3})$ lies on the hyperbola,$\frac{(-2 \sqrt{6})^{2}}{4b^{2}} - \frac{(\sqrt{3})^{2}}{b^{2}} = 1$.
$\frac{24}{4b^{2}} - \frac{3}{b^{2}} = 1 \Rightarrow \frac{6}{b^{2}} - \frac{3}{b^{2}} = 1 \Rightarrow \frac{3}{b^{2}} = 1 \Rightarrow b^{2} = 3$,so $b = \sqrt{3}$ and $a^{2} = 12$,so $a = 2 \sqrt{3}$.
The hyperbola is $\frac{x^{2}}{12} - \frac{y^{2}}{3} = 1$.
The tangent at $P(x_{1}, y_{1})$ is $\frac{xx_{1}}{a^{2}} - \frac{yy_{1}}{b^{2}} = 1$.
$\frac{x(-2 \sqrt{6})}{12} - \frac{y(\sqrt{3})}{3} = 1 \Rightarrow -\frac{x \sqrt{6}}{6} - \frac{y}{\sqrt{3}} = 1$.
For the conjugate axis,set $x = 0$: $-\frac{y}{\sqrt{3}} = 1 \Rightarrow y = -\sqrt{3}$. Thus $Q = (0, -\sqrt{3})$.
The slope of the tangent is $m = \frac{b^{2}x_{1}}{a^{2}y_{1}} = \frac{3(-2 \sqrt{6})}{12(\sqrt{3})} = \frac{-6 \sqrt{6}}{12 \sqrt{3}} = -\frac{\sqrt{2}}{2} = -\frac{1}{\sqrt{2}}$.
The slope of the normal is $m' = -\frac{1}{m} = \sqrt{2}$.
The equation of the normal at $P$ is $y - \sqrt{3} = \sqrt{2}(x + 2 \sqrt{6})$.
For the conjugate axis,set $x = 0$: $y - \sqrt{3} = \sqrt{2}(2 \sqrt{6}) = 2 \sqrt{12} = 4 \sqrt{3}$.
$y = 4 \sqrt{3} + \sqrt{3} = 5 \sqrt{3}$. Thus $R = (0, 5 \sqrt{3})$.
The distance $QR = |5 \sqrt{3} - (-\sqrt{3})| = |6 \sqrt{3}| = 6 \sqrt{3}$.

(C) For $(S1): (p$ $\rightarrow q) \vee (\sim q$ $\rightarrow p)$
Using the implication law $(a \rightarrow b \equiv \sim a \vee b)$,we get:
$(\sim p \vee q) \vee (q \vee p)$
$= (q \vee \sim p) \vee (q \vee p) = q \vee (\sim p \vee p) = q \vee t = t$.
Thus,$(S1)$ is a tautology.
For $(S2): (p \wedge \sim q) \wedge (\sim p \vee q)$
Using De Morgan's Law,$(\sim p \vee q) \equiv \sim (p \wedge \sim q)$.
So,$(S2) = (p \wedge \sim q) \wedge \sim (p \wedge \sim q) = C$ (Contradiction/Fallacy).
Thus,both $(S1)$ and $(S2)$ are true.

(D) We are given the sum $p = \sum_{r=1}^{50} \tan ^{-1} \frac{1}{2 r^{2}}$.
First,multiply the numerator and denominator by $2$ to use the formula $\tan ^{-1} x - \tan ^{-1} y = \tan ^{-1} \frac{x-y}{1+xy}$.
$p = \sum_{r=1}^{50} \tan ^{-1} \frac{2}{4 r^{2}} = \sum_{r=1}^{50} \tan ^{-1} \frac{(2r+1)-(2r-1)}{1+(2r+1)(2r-1)}$.
This is a telescoping series of the form $\sum_{r=1}^{n} (\tan ^{-1}(a_{r+1}) - \tan ^{-1}(a_r))$.
$p = (\tan ^{-1} 3 - \tan ^{-1} 1) + (\tan ^{-1} 5 - \tan ^{-1} 3) + \dots + (\tan ^{-1} 101 - \tan ^{-1} 99)$.
All intermediate terms cancel out,leaving $p = \tan ^{-1} 101 - \tan ^{-1} 1$.
Using the formula $\tan ^{-1} x - \tan ^{-1} y = \tan ^{-1} \frac{x-y}{1+xy}$,we get $p = \tan ^{-1} \frac{101-1}{1+(101)(1)} = \tan ^{-1} \frac{100}{102} = \tan ^{-1} \frac{50}{51}$.
Therefore,$\tan p = \frac{50}{51}$.

(C) Let the midpoint of the chord be $(h, k)$. The equation of the chord of the hyperbola $x^{2}-y^{2}=4$ with midpoint $(h, k)$ is given by $T=S_{1}$,which is $xh-yk=h^{2}-k^{2}$.
Rewriting this in the slope-intercept form $y=mx+c$,we get $y=\frac{h}{k}x - \frac{h^{2}-k^{2}}{k}$.
This line touches the parabola $y^{2}=8x$ (where $a=2$). The condition for the line $y=mx+c$ to touch the parabola $y^{2}=4ax$ is $c=\frac{a}{m}$.
Substituting $m=\frac{h}{k}$ and $c=-\frac{h^{2}-k^{2}}{k}$,we get $-\frac{h^{2}-k^{2}}{k} = \frac{2}{h/k} = \frac{2k}{h}$.
Simplifying,$-(h^{2}-k^{2})h = 2k^{2}$,which gives $h^{2}h - k^{2}h = -2k^{2}$,or $h^{3} = k^{2}(h-2)$.
Replacing $(h, k)$ with $(x, y)$,the locus is $y^{2}(x-2)=x^{3}$.

(C) Let the expression be $E = 2 \sin(\frac{\pi}{8}) \sin(\frac{2\pi}{8}) \sin(\frac{3\pi}{8}) \sin(\frac{5\pi}{8}) \sin(\frac{6\pi}{8}) \sin(\frac{7\pi}{8})$.
Using the property $\sin(\pi - \theta) = \sin \theta$,we have $\sin(\frac{7\pi}{8}) = \sin(\frac{\pi}{8})$,$\sin(\frac{6\pi}{8}) = \sin(\frac{2\pi}{8})$,and $\sin(\frac{5\pi}{8}) = \sin(\frac{3\pi}{8})$.
Substituting these,$E = 2 \sin^2(\frac{\pi}{8}) \sin^2(\frac{2\pi}{8}) \sin^2(\frac{3\pi}{8})$.
Since $\sin(\frac{2\pi}{8}) = \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$,then $\sin^2(\frac{2\pi}{8}) = \frac{1}{2}$.
Thus,$E = 2 \cdot \frac{1}{2} \cdot \sin^2(\frac{\pi}{8}) \sin^2(\frac{3\pi}{8}) = \sin^2(\frac{\pi}{8}) \cos^2(\frac{\pi}{8})$ (as $\sin(\frac{3\pi}{8}) = \cos(\frac{\pi}{8})$).
$E = (\sin(\frac{\pi}{8}) \cos(\frac{\pi}{8}))^2 = (\frac{1}{2} \sin(\frac{2\pi}{8}))^2 = (\frac{1}{2} \sin(\frac{\pi}{4}))^2 = (\frac{1}{2} \cdot \frac{1}{\sqrt{2}})^2 = \frac{1}{4 \cdot 2} = \frac{1}{8}$.

(A) Given $(\sqrt{3}+i)^{100}=2^{99}(p+iq)$.
Express $\sqrt{3}+i$ in polar form: $2(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6})$.
Using De Moivre's theorem: $(2(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6}))^{100} = 2^{100}(\cos \frac{100\pi}{6} + i \sin \frac{100\pi}{6}) = 2^{100}(\cos \frac{50\pi}{3} + i \sin \frac{50\pi}{3})$.
Since $\frac{50\pi}{3} = 16\pi + \frac{2\pi}{3}$,we have $\cos \frac{50\pi}{3} = \cos \frac{2\pi}{3} = -\frac{1}{2}$ and $\sin \frac{50\pi}{3} = \sin \frac{2\pi}{3} = \frac{\sqrt{3}}{2}$.
Thus,$2^{100}(-\frac{1}{2} + i \frac{\sqrt{3}}{2}) = 2^{99}(p+iq)$.
Dividing by $2^{99}$,we get $2(-\frac{1}{2} + i \frac{\sqrt{3}}{2}) = p+iq$,so $p = -1$ and $q = \sqrt{3}$.
The quadratic equation with roots $p$ and $q$ is $x^2 - (p+q)x + pq = 0$.
$p+q = \sqrt{3}-1$ and $pq = -\sqrt{3}$.
Therefore,the equation is $x^2 - (\sqrt{3}-1)x - \sqrt{3} = 0$.

(A) Let the pencil be rotated by an angle $\alpha$ about $C$. Let the new position of the pencil be $A'B'$.
In the right-angled triangle formed by the perpendicular from $P$ to the pencil,let the angle between $PC$ and the new pencil line be $\theta$.
We are given $PC = \sqrt{5}$ and the perpendicular distance $d = 1$.
In the right triangle formed by $P$,the foot of the perpendicular on the pencil,and $C$,we have $\sin \theta = \frac{d}{PC} = \frac{1}{\sqrt{5}}$.
Thus,$\tan \theta = \frac{1}{2}$.
Initially,the angle $\angle PCB = \phi = \tan^{-1}(2)$,so $\tan \phi = 2$.
The angle of rotation $\alpha$ is the difference between the initial angle $\phi$ and the new angle $\theta$.
$\alpha = \phi - \theta$.
$\tan \alpha = \tan(\phi - \theta) = \frac{\tan \phi - \tan \theta}{1 + \tan \phi \tan \theta} = \frac{2 - 1/2}{1 + 2(1/2)} = \frac{3/2}{2} = \frac{3}{4}$.
Therefore,$\alpha = \tan^{-1}\left(\frac{3}{4}\right)$.

(A) The equation of a circle $C$ touching the line $x-2y=0$ at $(2,1)$ is given by $(x-2)^{2}+(y-1)^{2}+\lambda(x-2y)=0$.
Expanding this,we get $x^{2}+y^{2}+x(\lambda-4)+y(-2-2\lambda)+5=0$.
The given circle $C_{1}$ is $x^{2}+y^{2}+2y-5=0$.
The common chord $PQ$ is the radical axis $C-C_{1}=0$,which is $x(\lambda-4)+y(-2-2\lambda-2)+10=0$,or $x(\lambda-4)-y(2\lambda+4)+10=0$.
Since $PQ$ is a diameter of $C_{1}$,it must pass through the center of $C_{1}$,which is $(0,-1)$.
Substituting $(0,-1)$ into the equation of $PQ$: $0(\lambda-4)-(-1)(2\lambda+4)+10=0$ $\Rightarrow 2\lambda+4+10=0$ $\Rightarrow 2\lambda=-14$ $\Rightarrow \lambda=-7$.
Substituting $\lambda=-7$ into the equation of $C$: $x^{2}+y^{2}-11x+12y+5=0$.
The center of $C$ is $(\frac{11}{2}, -6)$ and the radius $r$ is $\sqrt{(\frac{11}{2})^{2}+(-6)^{2}-5} = \sqrt{\frac{121}{4}+36-5} = \sqrt{\frac{121+124}{4}} = \sqrt{\frac{245}{4}} = \frac{7\sqrt{5}}{2}$.
Thus,the diameter of $C$ is $2r = 7\sqrt{5}$.

(A) Let $S = \lim _{x \rightarrow 2} \sum_{n=1}^{9} \frac{x}{n(n+1) x^{2}+2(2 n+1) x+4}$.
Substitute $x = 2$ into the expression:
$S = \sum_{n=1}^{9} \frac{2}{n(n+1)(4) + 2(2n+1)(2) + 4}$
$S = \sum_{n=1}^{9} \frac{2}{4n^2 + 4n + 8n + 4 + 4} = \sum_{n=1}^{9} \frac{2}{4n^2 + 12n + 8}$
$S = \sum_{n=1}^{9} \frac{2}{4(n^2 + 3n + 2)} = \sum_{n=1}^{9} \frac{1}{2(n+1)(n+2)}$
Using partial fractions:
$S = \frac{1}{2} \sum_{n=1}^{9} \left( \frac{1}{n+1} - \frac{1}{n+2} \right)$
This is a telescoping series:
$S = \frac{1}{2} \left[ \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \dots + \left( \frac{1}{10} - \frac{1}{11} \right) \right]$
$S = \frac{1}{2} \left( \frac{1}{2} - \frac{1}{11} \right) = \frac{1}{2} \left( \frac{11-2}{22} \right) = \frac{1}{2} \left( \frac{9}{22} \right) = \frac{9}{44}$.

(B) We need to find $3$-digit numbers $N \le 500$ such that $N$ is a multiple of $11$ and $N$ does not contain the digit $1$.
Multiples of $11$ between $100$ and $500$ are $110, 121, \ldots, 495$.
Excluding numbers containing the digit $1$,the valid numbers are:
$209, 220, 231, 242, 253, 264, 275, 286, 297, 308, 330, 341, 352, 363, 374, 385, 396, 407, 429, 440, 451, 462, 473, 484, 495$.
Removing those containing $1$:
$209, 220, 242, 253, 264, 275, 286, 297, 308, 330, 352, 363, 374, 385, 396, 407, 429, 440, 462, 473, 484, 495$.
Summing these values:
$209+220+242+253+264+275+286+297+308+330+352+363+374+385+396+407+429+440+462+473+484+495 = 7744$.

(A) Given $c_{2} = a_{2} + b_{2} = (a_{1} - 3) + (2b_{1}) = 12$,so $a_{1} + 2b_{1} = 15 \dots (1)$.
Given $c_{3} = a_{3} + b_{3} = (a_{1} - 6) + (4b_{1}) = 13$,so $a_{1} + 4b_{1} = 19 \dots (2)$.
Subtracting $(1)$ from $(2)$,we get $2b_{1} = 4$,which implies $b_{1} = 2$.
Substituting $b_{1} = 2$ into $(1)$,we get $a_{1} + 4 = 15$,so $a_{1} = 11$.
Now,$\sum_{k=1}^{10} c_{k} = \sum_{k=1}^{10} a_{k} + \sum_{k=1}^{10} b_{k}$.
The sum of the $AP$ is $S_{a} = \frac{10}{2} [2(11) + (10-1)(-3)] = 5(22 - 27) = 5(-5) = -25$.
The sum of the $GP$ is $S_{b} = \frac{b_{1}(r^{10} - 1)}{r - 1} = \frac{2(2^{10} - 1)}{2 - 1} = 2(1024 - 1) = 2(1023) = 2046$.
Therefore,$\sum_{k=1}^{10} c_{k} = -25 + 2046 = 2021$.

(D) Using the identity $\sum_{i=0}^{r} \binom{n}{i} \binom{m}{k-i} = \binom{n+m}{k}$,we simplify $A_{k}$.
$A_{k} = \sum_{i=0}^{9} \binom{9}{i} \binom{12}{k-i} + \sum_{i=0}^{8} \binom{8}{i} \binom{13}{k-i}$.
Applying Vandermonde's Identity:
$A_{k} = \binom{9+12}{k} + \binom{8+13}{k} = \binom{21}{k} + \binom{21}{k} = 2 \binom{21}{k}$.
Now,calculate $A_{4} - A_{3} = 2 \left( \binom{21}{4} - \binom{21}{3} \right)$.
$\binom{21}{4} = \frac{21 \times 20 \times 19 \times 18}{4 \times 3 \times 2 \times 1} = 5985$.
$\binom{21}{3} = \frac{21 \times 20 \times 19}{3 \times 2 \times 1} = 1330$.
$A_{4} - A_{3} = 2(5985 - 1330) = 2(4655) = 9310$.
Given $190p = 9310$,we find $p = \frac{9310}{190} = 49$.

(B) Given equations are $x^{2}-x+2\lambda=0$ and $3x^{2}-10x+27\lambda=0$.
Since $\alpha$ is a common root,it satisfies both equations:
$\alpha^{2}-\alpha+2\lambda=0 \quad ...(1)$
$3\alpha^{2}-10\alpha+27\lambda=0 \quad ...(2)$
Multiply equation $(1)$ by $3$: $3\alpha^{2}-3\alpha+6\lambda=0 \quad ...(3)$
Subtract $(3)$ from $(2)$: $(3\alpha^{2}-10\alpha+27\lambda) - (3\alpha^{2}-3\alpha+6\lambda) = 0$
$-7\alpha+21\lambda=0 \Rightarrow \alpha=3\lambda$.
Substitute $\alpha=3\lambda$ into equation $(1)$:
$(3\lambda)^{2}-(3\lambda)+2\lambda=0 \Rightarrow 9\lambda^{2}-\lambda=0$.
Since $\lambda \neq 0$,we have $\lambda=\frac{1}{9}$.
Then $\alpha=3\lambda=3(\frac{1}{9})=\frac{1}{3}$.
From the first equation,$\alpha+\beta=1 \Rightarrow \beta=1-\frac{1}{3}=\frac{2}{3}$.
From the second equation,$\alpha\gamma=\frac{27\lambda}{3}=9\lambda=9(\frac{1}{9})=1 \Rightarrow \gamma=\frac{1}{\alpha}=\frac{1}{1/3}=3$.
Finally,$\frac{\beta\gamma}{\lambda} = \frac{(2/3)(3)}{1/9} = \frac{2}{1/9} = 18$.

(C) Given the mean of $3, 7, x, y$ is $5$:
$\frac{3+7+x+y}{4} = 5$ $\Rightarrow 10+x+y = 20$ $\Rightarrow x+y = 10$
Given the variance is $10$:
$\frac{3^2+7^2+x^2+y^2}{4} - (5)^2 = 10$
$\frac{9+49+x^2+y^2}{4} = 35$ $\Rightarrow 58+x^2+y^2 = 140$ $\Rightarrow x^2+y^2 = 82$
We have $x+y=10$ and $x^2+y^2=82$.
Since $(x+y)^2 = x^2+y^2+2xy$,we get $100 = 82+2xy$ $\Rightarrow 2xy = 18$ $\Rightarrow xy = 9$.
Solving $x+y=10$ and $xy=9$,we get $x=9$ and $y=1$ (since $x>y$).
The four numbers are $3+2(9)=21$,$7+2(1)=9$,$9+1=10$,and $9-1=8$.
The mean of these four numbers is $\frac{21+9+10+8}{4} = \frac{48}{4} = 12$.

(C) Given expression: $E = \frac{(2i)^{n}}{(1-i)^{n-2}}$
First,simplify $(1-i)^2 = 1 + i^2 - 2i = 1 - 1 - 2i = -2i$.
Thus,$(1-i)^{n-2} = ((1-i)^2)^{\frac{n-2}{2}} = (-2i)^{\frac{n-2}{2}}$.
Substituting this into the expression: $E = \frac{(2i)^n}{(-2i)^{\frac{n-2}{2}}} = \frac{(2i)^n}{(-1)^{\frac{n-2}{2}} (2i)^{\frac{n-2}{2}}} = \frac{(2i)^{n - \frac{n-2}{2}}}{(-1)^{\frac{n-2}{2}}} = \frac{(2i)^{\frac{n+2}{2}}}{(-1)^{\frac{n-2}{2}}}$.
For $E$ to be a positive integer,the power of $i$ must be a multiple of $4$ (i.e.,$\frac{n+2}{2} = 4k$) and the magnitude must be real.
If $n=6$,$\frac{n+2}{2} = 4$. Then $E = \frac{(2i)^4}{(-1)^2} = \frac{16 i^4}{1} = 16(1) = 16$,which is a positive integer.
Thus,the least positive integer $n$ is $6$.

(A) Let $S = \frac{3}{2} x^{2} + \frac{5}{3} x^{3} + \frac{7}{4} x^{4} + \ldots \infty$
We can rewrite the general term as $\frac{2n+1}{n} x^{n} = (2 + \frac{1}{n}) x^{n} = 2x^{n} + \frac{x^{n}}{n}$ for $n \geq 2$.
Thus,$S = \sum_{n=2}^{\infty} (2x^{n} + \frac{x^{n}}{n}) = 2 \sum_{n=2}^{\infty} x^{n} + \sum_{n=2}^{\infty} \frac{x^{n}}{n}$.
Using the geometric series sum $\sum_{n=2}^{\infty} x^{n} = \frac{x^{2}}{1-x}$ and the logarithmic expansion $-\log_{e}(1-x) = x + \frac{x^{2}}{2} + \frac{x^{3}}{3} + \ldots$,we have $\sum_{n=2}^{\infty} \frac{x^{n}}{n} = -\log_{e}(1-x) - x$.
Substituting these back: $S = 2 \left( \frac{x^{2}}{1-x} \right) - \log_{e}(1-x) - x$.
$S = \frac{2x^{2} - x(1-x)}{1-x} - \log_{e}(1-x) = \frac{2x^{2} - x + x^{2}}{1-x} - \log_{e}(1-x) = \frac{3x^{2}-x}{1-x} - \log_{e}(1-x)$.
Wait,re-evaluating the series: $S = \sum_{n=1}^{\infty} \frac{2n+1}{n+1} x^{n+1} = \sum_{n=1}^{\infty} (2 - \frac{1}{n+1}) x^{n+1} = 2 \sum_{n=1}^{\infty} x^{n+1} - \sum_{n=1}^{\infty} \frac{x^{n+1}}{n+1}$.
$S = 2 \left( \frac{x^{2}}{1-x} \right) - (-\log_{e}(1-x) - x) = \frac{2x^{2}}{1-x} + x + \log_{e}(1-x) = \frac{2x^{2} + x - x^{2}}{1-x} + \log_{e}(1-x) = \frac{x^{2}+x}{1-x} + \log_{e}(1-x) = x \left( \frac{1+x}{1-x} \right) - \log_{e}(1-x)$.

(D) Given the expression for $y$:
$y = (\log_{10} x) (1 + \frac{1}{3} + \frac{1}{9} + \dots \infty)$
Using the sum of an infinite geometric series $S = \frac{a}{1-r}$ where $a=1$ and $r=\frac{1}{3}$:
$y = (\log_{10} x) \left( \frac{1}{1 - 1/3} \right) = (\log_{10} x) \left( \frac{3}{2} \right) = \frac{3}{2} \log_{10} x$
Now,consider the given equation:
$\frac{2(1+2+3+\dots+y)}{3(1+2+3+\dots+y)} = \frac{4}{\log_{10} x}$
$\frac{2}{3} = \frac{4}{\log_{10} x}$
$\log_{10} x = \frac{4 \times 3}{2} = 6$
$x = 10^6$
Substitute $\log_{10} x = 6$ into the expression for $y$:
$y = \frac{3}{2} (6) = 9$
Thus,the ordered pair $(x, y)$ is $(10^6, 9)$.

(C) Given $A = (0,6)$ and $B = (2t, 0)$.
The mid-point $M$ of $AB$ is $\left(\frac{0+2t}{2}, \frac{6+0}{2}\right) = (t, 3)$.
The slope of $AB$ is $m_{AB} = \frac{0-6}{2t-0} = \frac{-6}{2t} = -\frac{3}{t}$.
The slope of the perpendicular bisector of $AB$ is $m_{\perp} = -\frac{1}{m_{AB}} = \frac{t}{3}$.
The equation of the perpendicular bisector passing through $M(t, 3)$ is $(y-3) = \frac{t}{3}(x-t)$.
To find $C$,we set $x=0$ in the equation: $y-3 = \frac{t}{3}(0-t) \Rightarrow y = 3 - \frac{t^{2}}{3}$.
So,$C = (0, 3 - \frac{t^{2}}{3})$.
Let $P(h, k)$ be the mid-point of $MC$. Since $M = (t, 3)$ and $C = (0, 3 - \frac{t^{2}}{3})$:
$h = \frac{t+0}{2} = \frac{t}{2} \Rightarrow t = 2h$.
$k = \frac{3 + (3 - \frac{t^{2}}{3})}{2} = \frac{6 - \frac{t^{2}}{3}}{2} = 3 - \frac{t^{2}}{6}$.
Substituting $t = 2h$ into the equation for $k$:
$k = 3 - \frac{(2h)^{2}}{6} = 3 - \frac{4h^{2}}{6} = 3 - \frac{2h^{2}}{3}$.
$3k = 9 - 2h^{2} \Rightarrow 2h^{2} + 3k - 9 = 0$.
Replacing $(h, k)$ with $(x, y)$,the locus is $2x^{2} + 3y - 9 = 0$.

(D) Let $z = x + iy$. The condition $\frac{z-i}{z+2i} \in \mathbb{R}$ implies that the argument of the complex number is $0$ or $\pi$ (or the number is undefined).
This represents the locus of points $z$ such that the vectors $(z-i)$ and $(z+2i)$ are collinear.
Geometrically,this is the line passing through the points $i$ (which is $(0, 1)$) and $-2i$ (which is $(0, -2)$).
Since these two points lie on the imaginary axis,the line is the imaginary axis itself (excluding the point $-2i$ where the expression is undefined).
Thus,$S$ is a straight line in the complex plane.

(A) Let the statement be $S = (p \wedge (p$ $\rightarrow q) \wedge (q$ $\rightarrow r))$ $\rightarrow r$.
Using the implication law $a \rightarrow b \equiv \sim a \vee b$,we have:
$S \equiv \sim (p \wedge (p$ $\rightarrow q) \wedge (q$ $\rightarrow r)) \vee r$
Using De Morgan's law $\sim (A \wedge B \wedge C) \equiv \sim A \vee \sim B \vee \sim C$:
$S \equiv \sim p \vee \sim (p$ $\rightarrow q) \vee \sim (q$ $\rightarrow r) \vee r$
Since $\sim (a \rightarrow b) \equiv a \wedge \sim b$:
$S \equiv \sim p \vee (p \wedge \sim q) \vee (q \wedge \sim r) \vee r$
Using distributive law $\sim p \vee (p \wedge \sim q) \equiv (\sim p \vee p) \wedge (\sim p \vee \sim q) \equiv T \wedge (\sim p \vee \sim q) \equiv \sim p \vee \sim q$:
$S \equiv (\sim p \vee \sim q) \vee (q \wedge \sim r) \vee r$
Using associative and distributive laws:
$S \equiv \sim p \vee \sim q \vee (q \vee r) \wedge (\sim r \vee r)$
$S \equiv \sim p \vee \sim q \vee (q \vee r) \wedge T$
$S \equiv \sim p \vee (\sim q \vee q) \vee r$
$S \equiv \sim p \vee T \vee r \equiv T$
Since the result is always true,the statement is a tautology.

(B) Given the determinant condition: $f(x)f^{\prime \prime}(x) - (f^{\prime}(x))^2 = 0$.
This can be rewritten as $\frac{f^{\prime \prime}(x)}{f^{\prime}(x)} = \frac{f^{\prime}(x)}{f(x)}$.
Integrating both sides with respect to $x$,we get $\ln(f^{\prime}(x)) = \ln(f(x)) + \ln(c)$,which implies $f^{\prime}(x) = c f(x)$.
This is a first-order linear differential equation: $\frac{f^{\prime}(x)}{f(x)} = c$.
Integrating again,$\ln(f(x)) = cx + k_1$,so $f(x) = k e^{cx}$.
Using the initial conditions:
$f(0) = 1 \implies k e^0 = 1 \implies k = 1$.
$f^{\prime}(x) = c e^{cx}$,and $f^{\prime}(0) = 2 \implies c e^0 = 2 \implies c = 2$.
Thus,$f(x) = e^{2x}$.
Calculating $f(1) = e^2 \approx 7.389$.
Since $6 < 7.389 < 9$,the value lies in the interval $(6, 9)$.

(B) Let $I = \int_{1}^{3} [x^{2}-2x-2] dx$.
We can rewrite the expression inside the bracket as $x^{2}-2x-2 = (x-1)^{2}-3$.
Since $3$ is an integer,we can use the property $[f(x)+n] = [f(x)]+n$ for $n \in \mathbb{Z}$.
So,$I = \int_{1}^{3} [(x-1)^{2}] dx - \int_{1}^{3} 3 dx$.
Let $t = x-1$,then $dt = dx$. When $x=1, t=0$ and when $x=3, t=2$.
$I = \int_{0}^{2} [t^{2}] dt - [3x]_{1}^{3} = \int_{0}^{2} [t^{2}] dt - 6$.
Now,we split the integral $\int_{0}^{2} [t^{2}] dt$ based on the values of $[t^{2}]$:
For $0 \le t < 1$,$0 \le t^{2} < 1$,so $[t^{2}] = 0$.
For $1 \le t < \sqrt{2}$,$1 \le t^{2} < 2$,so $[t^{2}] = 1$.
For $\sqrt{2} \le t < \sqrt{3}$,$2 \le t^{2} < 3$,so $[t^{2}] = 2$.
For $\sqrt{3} \le t < 2$,$3 \le t^{2} < 4$,so $[t^{2}] = 3$.
$I = \int_{0}^{1} 0 dt + \int_{1}^{\sqrt{2}} 1 dt + \int_{\sqrt{2}}^{\sqrt{3}} 2 dt + \int_{\sqrt{3}}^{2} 3 dt - 6$.
$I = 0 + (\sqrt{2}-1) + 2(\sqrt{3}-\sqrt{2}) + 3(2-\sqrt{3}) - 6$.
$I = \sqrt{2} - 1 + 2\sqrt{3} - 2\sqrt{2} + 6 - 3\sqrt{3} - 6$.
$I = -\sqrt{2} - \sqrt{3} - 1$.

(A) Let $\frac{1}{4} \sin ^{-1} \frac{\sqrt{63}}{8} = \theta$.
Then,$\sin 4\theta = \frac{\sqrt{63}}{8}$.
Since $\cos^2 4\theta = 1 - \sin^2 4\theta = 1 - \frac{63}{64} = \frac{1}{64}$,we have $\cos 4\theta = \frac{1}{8}$.
Using the formula $\cos 4\theta = 2\cos^2 2\theta - 1$,we get $2\cos^2 2\theta - 1 = \frac{1}{8}$,which implies $2\cos^2 2\theta = \frac{9}{8}$,so $\cos^2 2\theta = \frac{9}{16}$.
Thus,$\cos 2\theta = \frac{3}{4}$.
Using the formula $\cos 2\theta = 2\cos^2 \theta - 1$,we get $2\cos^2 \theta - 1 = \frac{3}{4}$,which implies $2\cos^2 \theta = \frac{7}{4}$,so $\cos^2 \theta = \frac{7}{8}$.
Then $\sin^2 \theta = 1 - \cos^2 \theta = 1 - \frac{7}{8} = \frac{1}{8}$.
Therefore,$\tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} = \frac{1/8}{7/8} = \frac{1}{7}$.
Hence,$\tan \theta = \frac{1}{\sqrt{7}}$.

(C) Given the curve $y = ax^{2} + bx + c$.
Since the curve passes through the origin $(0, 0)$,we substitute $x = 0$ and $y = 0$ into the equation: $0 = a(0)^{2} + b(0) + c$,which gives $c = 0$.
The derivative of the curve is $\frac{dy}{dx} = 2ax + b$.
The tangent line at the origin $(0, 0)$ is $y = x$,which has a slope of $1$.
Thus,$\left. \frac{dy}{dx} \right|_{(0,0)} = 2a(0) + b = 1$,which gives $b = 1$.
Since the curve passes through the point $(1, 2)$,we substitute $x = 1, y = 2, b = 1, c = 0$ into the equation: $2 = a(1)^{2} + 1(1) + 0$.
This simplifies to $2 = a + 1$,so $a = 1$.
Therefore,the values are $a = 1, b = 1, c = 0$.

(B) To find the area of the region $R = \{(x, y) : 5x^2 \leq y \leq 2x^2 + 9\}$, we first find the intersection points of the curves $y = 5x^2$ and $y = 2x^2 + 9$.
Setting $5x^2 = 2x^2 + 9$, we get $3x^2 = 9$, which implies $x^2 = 3$, so $x = \pm \sqrt{3}$.
The region is symmetric about the $y$-axis.
The required area is given by the integral:
Area $= \int_{-\sqrt{3}}^{\sqrt{3}} ((2x^2 + 9) - 5x^2) dx$
$= 2 \int_{0}^{\sqrt{3}} (9 - 3x^2) dx$
$= 2 [9x - x^3]_{0}^{\sqrt{3}}$
$= 2 [9\sqrt{3} - (\sqrt{3})^3]$
$= 2 [9\sqrt{3} - 3\sqrt{3}]$
$= 2 [6\sqrt{3}] = 12\sqrt{3} \text{ square units.}$

(B) The given differential equation is $x \frac{dy}{dx} + y = bx^4$,which can be rewritten as $\frac{dy}{dx} + \frac{1}{x} y = bx^3$.
This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{1}{x}$ and $Q(x) = bx^3$.
The integrating factor $I.F. = e^{\int P(x) dx} = e^{\int \frac{1}{x} dx} = e^{\ln x} = x$.
The general solution is $y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$,which gives $yx = \int bx^3 \cdot x dx + C = \int bx^4 dx + C$.
Thus,$yx = \frac{bx^5}{5} + C$,or $f(x) = y = \frac{bx^4}{5} + \frac{C}{x}$.
Since the curve passes through $(1, 2)$,we have $2 = \frac{b}{5} + C$,so $C = 2 - \frac{b}{5}$.
Given $\int_{1}^{2} f(x) dx = \frac{62}{5}$,we have $\int_{1}^{2} (\frac{bx^4}{5} + \frac{C}{x}) dx = [\frac{bx^5}{25} + C \ln x]_{1}^{2} = \frac{62}{5}$.
Substituting the limits: $(\frac{32b}{25} + C \ln 2) - (\frac{b}{25} + 0) = \frac{31b}{25} + C \ln 2 = \frac{62}{5}$.
Substituting $C = 2 - \frac{b}{5}$: $\frac{31b}{25} + (2 - \frac{b}{5}) \ln 2 = \frac{62}{5}$.
For this to hold,we set the coefficient of $\ln 2$ to $0$,so $2 - \frac{b}{5} = 0$,which gives $b = 10$.
Then $\frac{31(10)}{25} = \frac{310}{25} = \frac{62}{5}$,which satisfies the equation. Thus,$b = 10$.

(B) Given $f^{\prime}(x) = f^{\prime}(2-x)$. Integrating both sides with respect to $x$,we get $f(x) = -f(2-x) + C$,which implies $f(x) + f(2-x) = C$.
At $x=0$,$f(0) + f(2) = C$. Given $f(0) = 1$ and $f(2) = e^{2}$,we have $C = 1 + e^{2}$.
Thus,$f(x) + f(2-x) = 1 + e^{2}$.
Let $I = \int_{0}^{2} f(x) dx$. Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we have $I = \int_{0}^{2} f(2-x) dx$.
Adding the two expressions for $I$,we get $2I = \int_{0}^{2} (f(x) + f(2-x)) dx$.
Substituting the sum,$2I = \int_{0}^{2} (1 + e^{2}) dx = (1 + e^{2}) [x]_{0}^{2} = 2(1 + e^{2})$.
Therefore,$I = 1 + e^{2}$.

(C) Given that $A$ is a symmetric matrix,so $A^{T} = A$.
Given that $B$ is a skew-symmetric matrix,so $B^{T} = -B$.
Let $C = A^{2}B^{2} - B^{2}A^{2}$.
Now,consider the transpose of $C$:
$C^{T} = (A^{2}B^{2} - B^{2}A^{2})^{T} = (A^{2}B^{2})^{T} - (B^{2}A^{2})^{T}$.
Using the property $(PQ)^{T} = Q^{T}P^{T}$,we get:
$C^{T} = (B^{2})^{T}(A^{2})^{T} - (A^{2})^{T}(B^{2})^{T}$.
Since $(A^{2})^{T} = (A^{T})^{2} = A^{2}$ and $(B^{2})^{T} = (B^{T})^{2} = (-B)^{2} = B^{2}$,we have:
$C^{T} = B^{2}A^{2} - A^{2}B^{2} = -(A^{2}B^{2} - B^{2}A^{2}) = -C$.
Thus,$C$ is a skew-symmetric matrix.
For any skew-symmetric matrix $C$ of odd order $n$ (here $n = 3$),the determinant is always zero,i.e.,$\det(C) = 0$.
Since the system is $(C)X = O$ and $\det(C) = 0$,the system has infinitely many solutions.

(D) The system of equations is:
$x - 2y + 0z = 1$
$x - y + kz = -2$
$0x + ky + 4z = 6$
The determinant of the coefficient matrix $D$ is:
$D = \begin{vmatrix} 1 & -2 & 0 \\ 1 & -1 & k \\ 0 & k & 4 \end{vmatrix} = 1(-4 - k^2) - (-2)(4 - 0) + 0 = -4 - k^2 + 8 = 4 - k^2$.
For a unique solution,$D \neq 0$,which implies $4 - k^2 \neq 0$,so $k \neq 2$ and $k \neq -2$. Thus,statement $(A)$ is correct.
If $k = 2$,$D = 0$. We check for consistency using $D_1$:
$D_1 = \begin{vmatrix} 1 & -2 & 0 \\ -2 & -1 & 2 \\ 6 & 2 & 4 \end{vmatrix} = 1(-4 - 4) - (-2)(-8 - 12) + 0 = -8 - 40 = -48 \neq 0$.
Since $D = 0$ and $D_1 \neq 0$,the system has no solution when $k = 2$. Thus,statement $(D)$ is correct.
Therefore,statements $(A)$ and $(D)$ are correct.

(D) The given lines are $L_1: \frac{x-\lambda}{1} = \frac{y-1/2}{1/2} = \frac{z-0}{-1/2}$ and $L_2: \frac{x-0}{1} = \frac{y+2\lambda}{1} = \frac{z-\lambda}{1}$.
The points on the lines are $a_1 = (\lambda, 1/2, 0)$ and $a_2 = (0, -2\lambda, \lambda)$.
The direction vectors are $b_1 = (1, 1/2, -1/2)$ and $b_2 = (1, 1, 1)$.
The cross product $b_1 \times b_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1/2 & -1/2 \\ 1 & 1 & 1 \end{vmatrix} = \hat{i}(1/2 + 1/2) - \hat{j}(1 + 1/2) + \hat{k}(1 - 1/2) = \hat{i} - \frac{3}{2}\hat{j} + \frac{1}{2}\hat{k} = \frac{1}{2}(2\hat{i} - 3\hat{j} + \hat{k})$.
The magnitude $|b_1 \times b_2| = \frac{1}{2}\sqrt{2^2 + (-3)^2 + 1^2} = \frac{\sqrt{14}}{2}$.
The vector $a_2 - a_1 = (0-\lambda, -2\lambda-1/2, \lambda-0) = (-\lambda, -2\lambda-1/2, \lambda)$.
The shortest distance $d = \frac{|(a_2 - a_1) \cdot (b_1 \times b_2)|}{|b_1 \times b_2|} = \frac{|-\lambda(1) - (2\lambda+1/2)(-3/2) + \lambda(1/2)|}{\sqrt{14}/2} = \frac{|-\lambda + 3\lambda + 3/4 + \lambda/2|}{\sqrt{14}/2} = \frac{|5\lambda/2 + 3/4|}{\sqrt{14}/2} = \frac{|10\lambda + 3|}{2\sqrt{14}}$.
Given $d = \frac{\sqrt{7}}{2\sqrt{2}} = \frac{\sqrt{14}}{4}$.
So,$\frac{|10\lambda + 3|}{2\sqrt{14}} = \frac{\sqrt{14}}{4} \implies |10\lambda + 3| = \frac{14}{2} = 7$.
Case $1$: $10\lambda + 3 = 7 \implies 10\lambda = 4 \implies \lambda = 0.4$ (Not an integer).
Case $2$: $10\lambda + 3 = -7 \implies 10\lambda = -10 \implies \lambda = -1$.
Thus,$|\lambda| = |-1| = 1$.

(A) Given the equation: $af(x)+\alpha f\left(\frac{1}{x}\right)=bx+\frac{\beta}{x} \quad .....(1)$
Replace $x$ with $\frac{1}{x}$ in equation $(1)$:
$af\left(\frac{1}{x}\right)+\alpha f(x)=b\left(\frac{1}{x}\right)+\beta x \quad .....(2)$
Adding equation $(1)$ and equation $(2)$:
$(a+\alpha)f(x)+(a+\alpha)f\left(\frac{1}{x}\right) = bx+\frac{\beta}{x}+\frac{b}{x}+\beta x$
$(a+\alpha)\left[f(x)+f\left(\frac{1}{x}\right)\right] = (b+\beta)x + (b+\beta)\frac{1}{x}$
$(a+\alpha)\left[f(x)+f\left(\frac{1}{x}\right)\right] = (b+\beta)\left(x+\frac{1}{x}\right)$
Now,rearrange to find the value of the expression:
$\frac{f(x)+f\left(\frac{1}{x}\right)}{x+\frac{1}{x}} = \frac{b+\beta}{a+\alpha}$
Given $a+\alpha=1$ and $b+\beta=2$,substitute these values:
$\frac{f(x)+f\left(\frac{1}{x}\right)}{x+\frac{1}{x}} = \frac{2}{1} = 2$

(D) Given that $A$ is a $3 \times 3$ matrix and $\operatorname{det}(A) = 4$.
First,consider the matrix $2A$. Since $A$ is a $3 \times 3$ matrix,$\operatorname{det}(2A) = 2^{3} \times \operatorname{det}(A) = 8 \times 4 = 32$.
Next,the matrix $B$ is obtained from $2A$ by the row operation $R_{2} \rightarrow 2R_{2} + 5R_{3}$.
Applying the property of determinants,if a row is multiplied by a scalar $k$,the determinant is multiplied by $k$. Here,$R_{2}$ is replaced by $2R_{2} + 5R_{3}$.
This operation is equivalent to multiplying the second row by $2$ and then adding $5$ times the third row to it. Adding a multiple of one row to another does not change the value of the determinant.
Therefore,$\operatorname{det}(B) = 2 \times \operatorname{det}(2A) = 2 \times 32 = 64$.

(B) Given integral: $I = \int \frac{e^{3 \log_{e} 2x} + 5e^{2 \log_{e} 2x}}{e^{4 \log_{e} x} + 5e^{3 \log_{e} x} - 7e^{2 \log_{e} x}} dx$
Using the property $e^{\log_{e} f(x)} = f(x)$,we have:
$I = \int \frac{(2x)^{3} + 5(2x)^{2}}{x^{4} + 5x^{3} - 7x^{2}} dx$
$I = \int \frac{8x^{3} + 20x^{2}}{x^{4} + 5x^{3} - 7x^{2}} dx$
Factor out $4x^{2}$ from the numerator and $x^{2}$ from the denominator:
$I = \int \frac{4x^{2}(2x + 5)}{x^{2}(x^{2} + 5x - 7)} dx$
$I = 4 \int \frac{2x + 5}{x^{2} + 5x - 7} dx$
Let $u = x^{2} + 5x - 7$,then $du = (2x + 5) dx$.
$I = 4 \int \frac{1}{u} du = 4 \log_{e} |u| + c$
$I = 4 \log_{e} |x^{2} + 5x - 7| + c$

(C) The vectors lying on the plane are $\vec{AB} = (2-1)\hat{i} + (3-2)\hat{j} + (1-3)\hat{k} = \hat{i} + \hat{j} - 2\hat{k}$ and $\vec{AC} = (2-1)\hat{i} + (4-2)\hat{j} + (2-3)\hat{k} = \hat{i} + 2\hat{j} - \hat{k}$.
The normal vector $\vec{n}$ to the plane is given by $\vec{AB} \times \vec{AC} = \left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -2 \\ 1 & 2 & -1\end{array}\right| = \hat{i}(-1+4) - \hat{j}(-1+2) + \hat{k}(2-1) = 3\hat{i} - \hat{j} + \hat{k}$.
The vector $\vec{OP} = 2\hat{i} - \hat{j} + \hat{k}$.
Let $\theta$ be the angle between $\vec{OP}$ and the normal $\vec{n}$. Then $\cos \theta = \frac{|\vec{OP} \cdot \vec{n}|}{|\vec{OP}| |\vec{n}|} = \frac{|(2)(3) + (-1)(-1) + (1)(1)|}{\sqrt{2^2 + (-1)^2 + 1^2} \sqrt{3^2 + (-1)^2 + 1^2}} = \frac{|6 + 1 + 1|}{\sqrt{6} \sqrt{11}} = \frac{8}{\sqrt{66}}$.
Since $\sin^2 \theta = 1 - \cos^2 \theta = 1 - \frac{64}{66} = \frac{2}{66} = \frac{1}{33}$,we have $\sin \theta = \sqrt{\frac{1}{33}}$.
The length of the projection of $\vec{OP}$ on the plane is $|\vec{OP}| \sin \theta = \sqrt{6} \times \sqrt{\frac{1}{33}} = \sqrt{\frac{6}{33}} = \sqrt{\frac{2}{11}}$.

(C) Let the events be defined as follows:
$A$: The person chosen is a smoker and non-vegetarian.
$B$: The person chosen is a smoker and vegetarian.
$C$: The person chosen is a non-smoker and vegetarian.
$E$: The person chosen has a chest disorder.
Given probabilities:
$P(A) = \frac{160}{400}$,$P(B) = \frac{100}{400}$,$P(C) = \frac{140}{400}$.
Conditional probabilities of having the disorder:
$P(E|A) = \frac{35}{100}$,$P(E|B) = \frac{20}{100}$,$P(E|C) = \frac{10}{100}$.
Using Bayes' Theorem,we need to find $P(A|E)$:
$P(A|E) = \frac{P(A) \cdot P(E|A)}{P(A) \cdot P(E|A) + P(B) \cdot P(E|B) + P(C) \cdot P(E|C)}$
Substituting the values:
$P(A|E) = \frac{\frac{160}{400} \times \frac{35}{100}}{\frac{160}{400} \times \frac{35}{100} + \frac{100}{400} \times \frac{20}{100} + \frac{140}{400} \times \frac{10}{100}}$
$P(A|E) = \frac{160 \times 35}{(160 \times 35) + (100 \times 20) + (140 \times 10)}$
$P(A|E) = \frac{5600}{5600 + 2000 + 1400} = \frac{5600}{9000} = \frac{56}{90} = \frac{28}{45}$.

(B) Let $x = 2 \cot^{-1}(5) + \cos^{-1}(\frac{4}{5})$.
We know that $\cot^{-1}(5) = \tan^{-1}(\frac{1}{5})$.
So,$2 \cot^{-1}(5) = 2 \tan^{-1}(\frac{1}{5}) = \tan^{-1}(\frac{2(1/5)}{1-(1/5)^2}) = \tan^{-1}(\frac{2/5}{24/25}) = \tan^{-1}(\frac{2}{5} \times \frac{25}{24}) = \tan^{-1}(\frac{5}{12})$.
Also,$\cos^{-1}(\frac{4}{5}) = \tan^{-1}(\frac{3}{4})$ because if $\cos \theta = \frac{4}{5}$,then $\tan \theta = \frac{3}{4}$.
Thus,the expression becomes $\operatorname{cosec}[\tan^{-1}(\frac{5}{12}) + \tan^{-1}(\frac{3}{4})]$.
Using the formula $\tan^{-1} A + \tan^{-1} B = \tan^{-1}(\frac{A+B}{1-AB})$:
$\tan^{-1}(\frac{5/12 + 3/4}{1 - (5/12)(3/4)}) = \tan^{-1}(\frac{(5+9)/12}{1 - 15/48}) = \tan^{-1}(\frac{14/12}{33/48}) = \tan^{-1}(\frac{14}{12} \times \frac{48}{33}) = \tan^{-1}(\frac{56}{33})$.
Finally,$\operatorname{cosec}(\tan^{-1}(\frac{56}{33}))$. Let $\theta = \tan^{-1}(\frac{56}{33})$,then $\tan \theta = \frac{56}{33}$.
Since $\operatorname{cosec} \theta = \sqrt{1 + \cot^2 \theta} = \sqrt{1 + (33/56)^2} = \sqrt{\frac{56^2 + 33^2}{56^2}} = \sqrt{\frac{3136 + 1089}{3136}} = \sqrt{\frac{4225}{3136}} = \frac{65}{56}$.

(D) Given $f(x) = \frac{5^{x}}{5^{x} + \sqrt{5}}$.
Note that $f(1-x) = \frac{5^{1-x}}{5^{1-x} + \sqrt{5}} = \frac{5/5^{x}}{5/5^{x} + \sqrt{5}} = \frac{5}{5 + \sqrt{5} \cdot 5^{x}} = \frac{\sqrt{5}}{\sqrt{5} + 5^{x}}$.
Thus,$f(x) + f(1-x) = \frac{5^{x} + \sqrt{5}}{5^{x} + \sqrt{5}} = 1$.
The series has $39$ terms from $x = \frac{1}{20}$ to $x = \frac{39}{20}$.
Pairing terms $f(x) + f(1-x) = 1$ is not directly applicable here as the sum is up to $x = \frac{39}{20}$.
Let $S = \sum_{k=1}^{39} f\left(\frac{k}{20}\right)$.
Since $f(x) + f(1-x) = 1$,we have $f\left(\frac{k}{20}\right) + f\left(1 - \frac{k}{20}\right) = 1$,which is $f\left(\frac{k}{20}\right) + f\left(\frac{20-k}{20}\right) = 1$.
Summing from $k=1$ to $19$,we get $19$ pairs each summing to $1$,plus the middle term $f\left(\frac{20}{20}\right) = f(1) = \frac{5}{5+\sqrt{5}}$.
Wait,the original problem implies $f(x) = \frac{5^x}{5^x + 5^{1/2}}$.
Sum $= 19 + f(1) + \sum_{k=21}^{39} f\left(\frac{k}{20}\right) = 19 + f(1) + \sum_{j=1}^{19} f\left(1 + \frac{j}{20}\right)$.
Using $f(x) + f(2-x) = 1$,the sum is $19 + f(1) = 19 + \frac{5}{5+\sqrt{5}} = 19 + \frac{\sqrt{5}}{\sqrt{5}+1} = \frac{19\sqrt{5}+19+\sqrt{5}}{\sqrt{5}+1} = \frac{20\sqrt{5}+19}{\sqrt{5}+1} \approx 19.5$.

(D) Given $A = \begin{bmatrix} 1 & -\alpha \\ \alpha & \beta \end{bmatrix}$ and $AA^{T} = I_{2}$.
First,find the transpose $A^{T} = \begin{bmatrix} 1 & \alpha \\ -\alpha & \beta \end{bmatrix}$.
Now,compute the product $AA^{T}$:
$AA^{T} = \begin{bmatrix} 1 & -\alpha \\ \alpha & \beta \end{bmatrix} \begin{bmatrix} 1 & \alpha \\ -\alpha & \beta \end{bmatrix} = \begin{bmatrix} 1 + \alpha^{2} & \alpha - \alpha\beta \\ \alpha - \alpha\beta & \alpha^{2} + \beta^{2} \end{bmatrix}$.
Since $AA^{T} = I_{2} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$,we equate the corresponding elements:
$1 + \alpha^{2} = 1 \Rightarrow \alpha^{2} = 0 \Rightarrow \alpha = 0$.
$\alpha - \alpha\beta = 0 \Rightarrow 0 - 0\beta = 0$ (which is always true).
$\alpha^{2} + \beta^{2} = 1 \Rightarrow 0 + \beta^{2} = 1 \Rightarrow \beta^{2} = 1$.
We need to find the value of $\alpha^{4} + \beta^{4}$:
$\alpha^{4} + \beta^{4} = (0)^{2} + (1)^{2} = 0 + 1 = 1$.

(D) Given $I_{n} = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cot^{n} x \, dx = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cot^{n-2} x (\csc^{2} x - 1) \, dx$
$= \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cot^{n-2} x \csc^{2} x \, dx - \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cot^{n-2} x \, dx$
$= \left[ -\frac{\cot^{n-1} x}{n-1} \right]_{\frac{\pi}{4}}^{\frac{\pi}{2}} - I_{n-2}$
$= \left( 0 - (-\frac{1}{n-1}) \right) - I_{n-2} = \frac{1}{n-1} - I_{n-2}$
$\Rightarrow I_{n} + I_{n-2} = \frac{1}{n-1}$
For $n=4$,$I_{4} + I_{2} = \frac{1}{3}$
For $n=5$,$I_{5} + I_{3} = \frac{1}{4}$
For $n=6$,$I_{6} + I_{4} = \frac{1}{5}$
Thus,the terms are $\frac{1}{I_{2}+I_{4}} = 3$,$\frac{1}{I_{3}+I_{5}} = 4$,and $\frac{1}{I_{4}+I_{6}} = 5$.
Since $3, 4, 5$ are in $A.P.$,the sequence $\frac{1}{I_{2}+I_{4}}, \frac{1}{I_{3}+I_{5}}, \frac{1}{I_{4}+I_{6}}$ is in $A.P.$

(A) The given limit is $L = \lim _{n \rightarrow \infty} \sum_{r=0}^{n-1} \frac{n}{(n+r)^{2}}$.
We can rewrite the expression as $L = \lim _{n \rightarrow \infty} \sum_{r=0}^{n-1} \frac{1}{n} \cdot \frac{1}{(1 + r/n)^{2}}$.
Using the definition of a definite integral as the limit of a sum,$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=0}^{n-1} f(r/n) = \int_{0}^{1} f(x) dx$.
Here,$f(x) = \frac{1}{(1+x)^{2}}$.
Thus,$L = \int_{0}^{1} \frac{1}{(1+x)^{2}} dx$.
Evaluating the integral: $L = \left[ -\frac{1}{1+x} \right]_{0}^{1} = -\left( \frac{1}{2} - 1 \right) = -(-\frac{1}{2}) = \frac{1}{2}$.

(C) The total number of $4$-digit numbers with exactly one $7$ is calculated as follows:
If $7$ is at the thousands place: $1 \times 9 \times 9 \times 9 = 729$.
If $7$ is not at the thousands place: $3 \times (8 \times 9 \times 9) = 1944$.
Total $n(S) = 729 + 1944 = 2673$.
For a number to leave a remainder of $2$ when divided by $5$,the last digit must be $2$ or $7$.
Case $1$: Last digit is $7$. Since exactly one $7$ is allowed,the other three digits cannot be $7$. The thousands place has $8$ choices ($1-9$ excluding $7$),and the other two places have $9$ choices each ($0-9$ excluding $7$). $n(E_1) = 8 \times 9 \times 9 = 648$.
Case $2$: Last digit is $2$. The $7$ can be in any of the first three positions. If $7$ is at the thousands place: $1 \times 9 \times 9 = 81$. If $7$ is at the hundreds or tens place: $2 \times (8 \times 9) = 144$. $n(E_2) = 81 + 144 = 225$.
Total $n(E) = 648 + 225 = 873$.
Probability $P(E) = \frac{873}{2673} = \frac{97}{297}$.

(B) The number of one-one functions from a set with $n$ elements to a set with $m$ elements $(m \ge n)$ is given by $P(m, n) = \frac{m!}{(m-n)!}$.
For $x$,the set $A$ has $3$ elements and set $B$ has $5$ elements. Thus,$x = P(5, 3) = 5 \times 4 \times 3 = 60$.
For $y$,the set $A$ has $3$ elements and the set $A \times B$ has $3 \times 5 = 15$ elements. Thus,$y = P(15, 3) = 15 \times 14 \times 13 = 2730$.
Now,we compare $x$ and $y$:
$x = 60$
$y = 2730$
Calculating the ratio $\frac{y}{x} = \frac{2730}{60} = \frac{273}{6} = \frac{91}{2}$.
Therefore,$2y = 91x$.

(D) Given equations are:
$2x + 3y + 2z = 9 \quad (1)$
$3x + 2y + 2z = 9 \quad (2)$
$x - y + 4z = 8 \quad (3)$
Subtracting equation $(1)$ from equation $(2)$:
$(3x + 2y + 2z) - (2x + 3y + 2z) = 9 - 9$
$x - y = 0 \Rightarrow x = y$
Substitute $x = y$ into equation $(3)$:
$x - x + 4z = 8 \Rightarrow 4z = 8 \Rightarrow z = 2$
Substitute $z = 2$ into equation $(1)$:
$2x + 3y + 2(2) = 9$
$2x + 3y = 5$
Since $x = y$,we have $2x + 3x = 5 \Rightarrow 5x = 5 \Rightarrow x = 1$
Thus,$x = 1, y = 1, z = 2$.
The system has a unique solution $(1, 1, 2)$.

(C) For $x \in [-2, 2]$,$f(x) = \min \{|x|, 2-x^2\}$.
Solving $|x| = 2-x^2$,we get $x^2 + |x| - 2 = 0$,which gives $(|x|+2)(|x|-1) = 0$. Since $|x| \geq 0$,we have $|x| = 1$,so $x = \pm 1$.
Thus,$f(x) = |x|$ for $x \in [-1, 1]$ and $f(x) = 2-x^2$ for $x \in [-2, -1) \cup (1, 2]$.
Points of non-differentiability in $(-2, 2)$ are $x = -1, 0, 1$ (due to $|x|$ and the intersection of curves).
For $2 < |x| \leq 3$,$f(x) = [|x|]$.
For $x \in (2, 3]$,$f(x) = [x] = 2$ for $x \in (2, 3)$ and $f(3) = 3$. This is discontinuous at $x = 3$ (not in $(-3, 3)$) and $x = 2$ (jump discontinuity).
For $x \in [-3, -2)$,$f(x) = [|x|] = [|x|]$. For $x \in (-3, -2)$,$f(x) = 2$. At $x = -2$,$f(-2) = 2$. At $x = -3$,$f(-3) = 3$.
Checking points: $x = -2, -1, 0, 1, 2$. At these $5$ points,the function is either discontinuous or has a sharp corner.
Therefore,the number of points of non-differentiability is $5$.

(B) Given vectors are $\vec{a} = \hat{i} + \alpha \hat{j} + 3 \hat{k}$ and $\vec{b} = 3 \hat{i} - \alpha \hat{j} + \hat{k}$.
The area of the parallelogram with adjacent sides $\vec{a}$ and $\vec{b}$ is given by $|\vec{a} \times \vec{b}| = 8 \sqrt{3}$.
First,calculate the cross product $\vec{a} \times \vec{b}$:
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & \alpha & 3 \\ 3 & -\alpha & 1 \end{vmatrix} = \hat{i}(\alpha - (-3\alpha)) - \hat{j}(1 - 9) + \hat{k}(-\alpha - 3\alpha) = 4\alpha \hat{i} + 8 \hat{j} - 4\alpha \hat{k}$.
Now,find the magnitude:
$|\vec{a} \times \vec{b}| = \sqrt{(4\alpha)^2 + 8^2 + (-4\alpha)^2} = \sqrt{16\alpha^2 + 64 + 16\alpha^2} = \sqrt{32\alpha^2 + 64}$.
Equating to the given area:
$\sqrt{32\alpha^2 + 64} = 8 \sqrt{3} \Rightarrow 32\alpha^2 + 64 = 64 \times 3 = 192$.
$32\alpha^2 = 192 - 64 = 128 \Rightarrow \alpha^2 = 4$.
Finally,calculate the dot product $\vec{a} \cdot \vec{b}$:
$\vec{a} \cdot \vec{b} = (1)(3) + (\alpha)(-\alpha) + (3)(1) = 3 - \alpha^2 + 3 = 6 - \alpha^2$.
Substituting $\alpha^2 = 4$,we get $\vec{a} \cdot \vec{b} = 6 - 4 = 2$.

(C) Given curves are $x=y^{4}$ and $xy=k$.
For the intersection point,substitute $x=y^{4}$ into $xy=k$:
$y^{4} \cdot y = k \Rightarrow y^{5} = k \ldots(1)$.
Now,differentiate $x=y^{4}$ with respect to $x$:
$1 = 4y^{3} \frac{dy}{dx} \Rightarrow \frac{dy}{dx} = \frac{1}{4y^{3}}$.
Next,differentiate $xy=k$ with respect to $x$:
$y + x \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{y}{x}$.
Since $x = \frac{k}{y}$,we have $\frac{dy}{dx} = -\frac{y}{k/y} = -\frac{y^{2}}{k}$.
Since the curves intersect at right angles,the product of their slopes is $-1$:
$\left(\frac{1}{4y^{3}}\right) \cdot \left(-\frac{y^{2}}{k}\right) = -1$.
$\Rightarrow \frac{1}{4yk} = 1 \Rightarrow y = \frac{1}{4k}$.
Substitute $y = \frac{1}{4k}$ into equation $(1)$:
$\left(\frac{1}{4k}\right)^{5} = k$.
$\Rightarrow \frac{1}{(4k)^{5}} = k$.
$\Rightarrow (4k)^{6} = 4$.

(A) Given differential equation: $(2xy^2 - y)dx + xdy = 0$.
Rearranging the terms: $2xy^2 dx - ydx + xdy = 0$.
Divide by $xy^2$: $2xdx - \frac{ydx - xdy}{y^2} = 0$.
This simplifies to $2xdx = d(\frac{x}{y})$.
Integrating both sides: $\int 2xdx = \int d(\frac{x}{y}) \Rightarrow x^2 = \frac{x}{y} + c$.
To find the point of intersection of $2x - 3y = 1$ and $3x + 2y = 8$,multiply the first by $2$ and the second by $3$: $4x - 6y = 2$ and $9x + 6y = 24$.
Adding these gives $13x = 26 \Rightarrow x = 2$. Substituting $x=2$ into $2x - 3y = 1$ gives $4 - 3y = 1 \Rightarrow 3y = 3 \Rightarrow y = 1$.
The curve passes through $(2, 1)$,so $2^2 = \frac{2}{1} + c \Rightarrow 4 = 2 + c \Rightarrow c = 2$.
The equation of the curve is $x^2 = \frac{x}{y} + 2$.
For $x = 1$,$1^2 = \frac{1}{y} + 2 \Rightarrow 1 = \frac{1}{y} + 2 \Rightarrow \frac{1}{y} = -1 \Rightarrow y = -1$.
Thus,$|y(1)| = |-1| = 1$.

(A) We need to evaluate the integral $I = \int_{-2}^{2} |3x^2 - 3x - 6| dx$.
First,factor the expression inside the absolute value: $3x^2 - 3x - 6 = 3(x^2 - x - 2) = 3(x - 2)(x + 1)$.
The roots of the quadratic are $x = -1$ and $x = 2$.
In the interval $[-2, -1]$,$3(x^2 - x - 2) \ge 0$.
In the interval $[-1, 2]$,$3(x^2 - x - 2) \le 0$.
Thus,$I = 3 \left[ \int_{-2}^{-1} (x^2 - x - 2) dx + \int_{-1}^{2} -(x^2 - x - 2) dx \right]$.
Evaluating the first integral: $\int_{-2}^{-1} (x^2 - x - 2) dx = [\frac{x^3}{3} - \frac{x^2}{2} - 2x]_{-2}^{-1} = (-\frac{1}{3} - \frac{1}{2} + 2) - (-\frac{8}{3} - 2 + 4) = \frac{7}{6} - (-\frac{2}{3}) = \frac{7}{6} + \frac{4}{6} = \frac{11}{6}$.
Evaluating the second integral: $-\int_{-1}^{2} (x^2 - x - 2) dx = -[\frac{x^3}{3} - \frac{x^2}{2} - 2x]_{-1}^{2} = -[(\frac{8}{3} - 2 - 4) - (-\frac{1}{3} - \frac{1}{2} + 2)] = -[-\frac{10}{3} - \frac{7}{6}] = -[-\frac{27}{6}] = \frac{9}{2} = \frac{27}{6}$.
Total value $I = 3 \times (\frac{11}{6} + \frac{27}{6}) = 3 \times \frac{38}{6} = 3 \times \frac{19}{3} = 19$.

(C) The direction ratios of $l_{1}$ are $\vec{v}_{1} = (1, 2, 2)$ and for $l_{2}$ are $\vec{v}_{2} = (2, 2, 1)$.
Since line $l$ is perpendicular to both $l_{1}$ and $l_{2}$,its direction vector is $\vec{v} = \vec{v}_{1} \times \vec{v}_{2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 2 \\ 2 & 2 & 1 \end{vmatrix} = \hat{i}(2-4) - \hat{j}(1-4) + \hat{k}(2-4) = -2\hat{i} + 3\hat{j} - 2\hat{k}$.
Since $l$ passes through the origin,its equation is $\vec{r} = \mu(-2\hat{i} + 3\hat{j} - 2\hat{k})$.
For the intersection of $l$ and $l_{1}$:
$3+t = -2\mu$,$-1+2t = 3\mu$,$4+2t = -2\mu$.
Solving these,we get $t = -1$ and $\mu = -1$. The point of intersection $P$ is $(2, -3, 2)$.
Let $Q$ be a point on $l_{2}$,$Q = (3+2s, 3+2s, 2+s)$.
Given $PQ = \sqrt{17}$,so $PQ^{2} = 17$.
$(3+2s-2)^{2} + (3+2s+3)^{2} + (2+s-2)^{2} = 17$.
$(2s+1)^{2} + (2s+6)^{2} + s^{2} = 17$.
$4s^{2} + 4s + 1 + 4s^{2} + 24s + 36 + s^{2} = 17$.
$9s^{2} + 28s + 20 = 0$.
$(9s+10)(s+2) = 0$,so $s = -2$ or $s = -10/9$.
For $s = -10/9$,$Q = (3-20/9, 3-20/9, 2-10/9) = (7/9, 7/9, 8/9)$,which is in the first octant.
Thus,$a=7/9, b=7/9, c=8/9$.
$18(a+b+c) = 18(7/9 + 7/9 + 8/9) = 18(22/9) = 44$.

(D) Since $\overrightarrow{a}_{1}$ and $\overrightarrow{a}_{2}$ are collinear,their components must be proportional:
$\frac{x}{1} = \frac{-1}{y} = \frac{1}{z} = k$ (where $k$ is a constant).
From this,we get $x = k$,$y = -\frac{1}{k}$,and $z = \frac{1}{k}$.
Substituting these into the vector $x \hat{i} + y \hat{j} + z \hat{k}$,we get $k \hat{i} - \frac{1}{k} \hat{j} + \frac{1}{k} \hat{k}$.
For the vectors to be collinear,the ratio must hold. If we consider the simplest case where $k=1$,then $x=1, y=-1, z=1$.
The vector becomes $\hat{i} - \hat{j} + \hat{k}$.
The magnitude of this vector is $\sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}$.
The unit vector is $\frac{1}{\sqrt{3}}(\hat{i} - \hat{j} + \hat{k})$.

(A) Given $f(k) = \begin{cases} k+1, & \text{if } k \text{ is odd} \\ k, & \text{if } k \text{ is even} \end{cases}$.
We are given the condition $g(f(k)) = f(k)$ for all $k \in A$.
If $k$ is even,$f(k) = k$. Thus,$g(f(k)) = g(k) = f(k) = k$. So,$g(k) = k$ for all even $k \in \{2, 4, 6, 8, 10\}$.
If $k$ is odd,$f(k) = k+1$. Thus,$g(f(k)) = g(k+1) = f(k) = k+1$. Since $k+1$ is always even,this confirms that $g$ must map even numbers to themselves,which we already established.
For odd $k \in \{1, 3, 5, 7, 9\}$,the condition $g(f(k)) = f(k)$ does not impose any restriction on the value of $g(k)$.
Since $g: A \rightarrow A$,for each of the $5$ odd values of $k$,$g(k)$ can be any of the $10$ elements in $A$.
Therefore,the number of such functions $g$ is $10 \times 10 \times 10 \times 10 \times 10 = 10^{5}$.

(B) $f(x)$ is continuous on $R$,so it must be continuous at $x = 1$ and $x = -1$.
For continuity at $x = 1$:
$\lim_{x \rightarrow 1^-} f(x) = f(1) = \lim_{x \rightarrow 1^+} f(x)$
$|a(1)^2 + 1 + b| = \lim_{x \rightarrow 1^+} \sin(\pi x)$
$|a + 1 + b| = \sin(\pi) = 0$
$a + b + 1 = 0 \Rightarrow a + b = -1$ ... $(1)$
For continuity at $x = -1$:
$\lim_{x \rightarrow -1^-} f(x) = f(-1) = \lim_{x \rightarrow -1^+} f(x)$
$\lim_{x \rightarrow -1^-} 2 \sin \left(-\frac{\pi x}{2}\right) = |a(-1)^2 + (-1) + b|$
$2 \sin \left(\frac{\pi}{2}\right) = |a - 1 + b|$
$2(1) = |a + b - 1|$
$|a + b - 1| = 2$
This implies $a + b - 1 = 2$ or $a + b - 1 = -2$.
$a + b = 3$ or $a + b = -1$.
Since we already established $a + b = -1$ from the continuity at $x = 1$,the value of $a + b$ is $-1$.

(C) Given $f(x) = \int_{1}^{x} \frac{\ln t}{1+t} dt.$
Consider $f\left(\frac{1}{x}\right) = \int_{1}^{1/x} \frac{\ln t}{1+t} dt.$
Let $t = \frac{1}{u},$ then $dt = -\frac{1}{u^2} du.$
When $t=1, u=1$ and when $t=1/x, u=x.$
So,$f\left(\frac{1}{x}\right) = \int_{1}^{x} \frac{\ln(1/u)}{1+(1/u)} \left(-\frac{1}{u^2}\right) du = \int_{1}^{x} \frac{-\ln u}{\frac{u+1}{u}} \left(\frac{1}{u^2}\right) du = \int_{1}^{x} \frac{-\ln u}{u(1+u)} du.$
Now,$f(x) + f\left(\frac{1}{x}\right) = \int_{1}^{x} \frac{\ln t}{1+t} dt - \int_{1}^{x} \frac{\ln t}{t(1+t)} dt = \int_{1}^{x} \left( \frac{\ln t}{1+t} - \frac{\ln t}{t(1+t)} \right) dt.$
$= \int_{1}^{x} \frac{t \ln t - \ln t}{t(1+t)} dt = \int_{1}^{x} \frac{(t-1) \ln t}{t(1+t)} dt.$
Wait,let's re-evaluate: $f(x) + f(1/x) = \int_{1}^{x} \frac{\ln t}{1+t} dt + \int_{1}^{x} \frac{-\ln t}{t(1+t)} dt = \int_{1}^{x} \ln t \left( \frac{1}{1+t} - \frac{1}{t(1+t)} \right) dt = \int_{1}^{x} \ln t \left( \frac{t-1}{t(1+t)} \right) dt.$
Actually,for $x=e$: $f(e) = \int_{1}^{e} \frac{\ln t}{1+t} dt$ and $f(1/e) = \int_{1}^{1/e} \frac{\ln t}{1+t} dt = \int_{1/e}^{1} \frac{-\ln t}{1+t} dt.$
Using the property $\int_{1}^{x} \frac{\ln t}{1+t} dt + \int_{1}^{1/x} \frac{\ln t}{1+t} dt = \int_{1}^{x} \frac{\ln t}{1+t} dt - \int_{1/x}^{1} \frac{\ln t}{1+t} dt = \int_{1}^{x} \frac{\ln t}{1+t} dt - \int_{x}^{1} \frac{\ln(1/u)}{1+(1/u)} \frac{1}{u^2} du = \int_{1}^{x} \frac{\ln t}{1+t} dt - \int_{1}^{x} \frac{\ln u}{u(1+u)} du = \int_{1}^{x} \frac{\ln t}{1+t} (1 - 1/t) dt = \int_{1}^{x} \frac{\ln t (t-1)}{t(1+t)} dt.$
Evaluating at $x=e$: $\int_{1}^{e} \frac{\ln t}{1+t} dt + \int_{1}^{1/e} \frac{\ln t}{1+t} dt = \int_{1}^{e} \frac{\ln t}{1+t} dt - \int_{1/e}^{1} \frac{\ln t}{1+t} dt = \int_{1}^{e} \frac{\ln t}{1+t} dt - \int_{e}^{1} \frac{\ln(1/u)}{1+1/u} \frac{1}{u^2} du = \int_{1}^{e} \frac{\ln t}{1+t} dt - \int_{1}^{e} \frac{\ln u}{u(1+u)} du = \int_{1}^{e} \ln t \left( \frac{1}{1+t} - \frac{1}{t(1+t)} \right) dt = \int_{1}^{e} \frac{\ln t (t-1)}{t(1+t)} dt.$
This evaluates to $\frac{1}{2}$.

(C) The domain of $(f \circ g)(x) = \sin^{-1}(g(x))$ requires $|g(x)| \leq 1$.
First,we find $g(2) = \lim_{x \to 2} \frac{(x+1)(x-2)}{(2x+3)(x-2)} = \lim_{x \to 2} \frac{x+1}{2x+3} = \frac{3}{7}$.
Thus,we solve $|\frac{x+1}{2x+3}| \leq 1$ for $x \neq 2$.
This implies $-1 \leq \frac{x+1}{2x+3} \leq 1$.
Case $1$: $\frac{x+1}{2x+3} \leq 1 \Rightarrow \frac{x+1 - (2x+3)}{2x+3} \leq 0 \Rightarrow \frac{-x-2}{2x+3} \leq 0 \Rightarrow \frac{x+2}{2x+3} \geq 0$.
The solution is $x \in (-\infty, -2] \cup (-\frac{3}{2}, \infty)$.
Case $2$: $\frac{x+1}{2x+3} \geq -1 \Rightarrow \frac{x+1 + 2x+3}{2x+3} \geq 0 \Rightarrow \frac{3x+4}{2x+3} \geq 0$.
The solution is $x \in (-\infty, -\frac{4}{3}] \cup (-\frac{3}{2}, \infty)$.
Taking the intersection of both cases and excluding $x=2$ (since $g(2)$ is defined as the limit),we get $x \in (-\infty, -2] \cup [-\frac{4}{3}, \infty)$.

(C) Let the triangle be $\Delta ABC$ inscribed in a circle of radius $r$ with center $O$.
Let $\theta$ be the angle $\angle OBP$,where $P$ is the midpoint of $BC$.
The height of the triangle is $h = r \sin \theta + r$.
The base of the triangle is $BC = 2r \cos \theta$.
The area of the triangle $\Delta$ is given by $\Delta = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} (2r \cos \theta) (r \sin \theta + r) = r^2 \cos \theta (1 + \sin \theta)$.
To find the maximum area,differentiate $\Delta$ with respect to $\theta$:
$\frac{d\Delta}{d\theta} = r^2 [-\sin \theta (1 + \sin \theta) + \cos \theta (\cos \theta)] = r^2 [-\sin \theta - \sin^2 \theta + \cos^2 \theta] = r^2 [-\sin \theta - \sin^2 \theta + (1 - \sin^2 \theta)] = r^2 [1 - \sin \theta - 2 \sin^2 \theta]$.
Setting $\frac{d\Delta}{d\theta} = 0$,we get $2 \sin^2 \theta + \sin \theta - 1 = 0$,which factors as $(2 \sin \theta - 1)(\sin \theta + 1) = 0$.
Since $\theta \in [0, \pi/2)$,$\sin \theta = 1/2$,so $\theta = \pi/6$.
At $\theta = \pi/6$,the side length of the triangle is $s = 2r \cos(\pi/6) = 2r (\sqrt{3}/2) = \sqrt{3}r$.
Thus,the triangle of maximum area is an equilateral triangle with side length $\sqrt{3}r$.

(D) The equations of the planes are $x+2y+z=6$ and $y+2z=4$.
To find the line of intersection,we express $x$ and $y$ in terms of $z$.
From $y+2z=4$,we get $y=4-2z$.
Substituting this into the first plane equation: $x+2(4-2z)+z=6 \Rightarrow x+8-4z+z=6 \Rightarrow x-3z=-2 \Rightarrow x=3z-2$.
Thus,the line $L$ can be written in symmetric form as $\frac{x+2}{3} = \frac{y-4}{-2} = z = \lambda$.
Any point $P$ on the line $L$ is given by $(3\lambda-2, -2\lambda+4, \lambda)$.
The direction vector of the line is $\vec{v} = 3\hat{i}-2\hat{j}+\hat{k}$.
Let $A = (3,2,1)$. The vector $\vec{AP} = (3\lambda-2-3, -2\lambda+4-2, \lambda-1) = (3\lambda-5, -2\lambda+2, \lambda-1)$.
Since $\vec{AP} \perp \vec{v}$,their dot product is zero: $(3\lambda-5)(3) + (-2\lambda+2)(-2) + (\lambda-1)(1) = 0$.
$9\lambda - 15 + 4\lambda - 4 + \lambda - 1 = 0 \Rightarrow 14\lambda - 20 = 0 \Rightarrow \lambda = \frac{10}{7}$.
Substituting $\lambda$ back,$P = (3(\frac{10}{7})-2, -2(\frac{10}{7})+4, \frac{10}{7}) = (\frac{16}{7}, \frac{8}{7}, \frac{10}{7})$.
Thus,$\alpha+\beta+\gamma = \frac{16+8+10}{7} = \frac{34}{7}$.
The value of $21(\alpha+\beta+\gamma) = 21 \times \frac{34}{7} = 3 \times 34 = 102$.

(C) The differential equation is given by $\frac{dy}{dx} = \frac{xy^2 + y}{x} = y^2 + \frac{y}{x}$.
Rearranging the terms,we get $\frac{dy}{dx} - \frac{y}{x} = y^2$.
Dividing by $y^2$,we have $y^{-2} \frac{dy}{dx} - \frac{1}{x} y^{-1} = 1$.
Let $v = y^{-1}$,then $\frac{dv}{dx} = -y^{-2} \frac{dy}{dx}$,which implies $y^{-2} \frac{dy}{dx} = -\frac{dv}{dx}$.
Substituting this into the equation,we get $-\frac{dv}{dx} - \frac{1}{x} v = 1$,or $\frac{dv}{dx} + \frac{1}{x} v = -1$.
This is a linear differential equation with integrating factor $IF = e^{\int \frac{1}{x} dx} = e^{\ln x} = x$.
The solution is $v \cdot x = \int (-1) \cdot x dx + C = -\frac{x^2}{2} + C$.
Since $v = \frac{1}{y}$,we have $\frac{x}{y} = -\frac{x^2}{2} + C$.
Given the curve intersects $x + 2y = 4$ at $x = -2$,we find $y$ by substituting $x = -2$: $-2 + 2y = 4 \Rightarrow 2y = 6 \Rightarrow y = 3$.
The curve passes through $(-2, 3)$,so $\frac{-2}{3} = -\frac{(-2)^2}{2} + C \Rightarrow -\frac{2}{3} = -2 + C \Rightarrow C = 2 - \frac{2}{3} = \frac{4}{3}$.
Thus,$\frac{x}{y} = -\frac{x^2}{2} + \frac{4}{3}$.
For the point $(3, y)$,we have $\frac{3}{y} = -\frac{3^2}{2} + \frac{4}{3} = -\frac{9}{2} + \frac{4}{3} = \frac{-27 + 8}{6} = -\frac{19}{6}$.
Therefore,$y = 3 \cdot (-\frac{6}{19}) = -\frac{18}{19}$.

(B) Let the given equations be:
$P_{1}: x+2y-3z=a$
$P_{2}: 2x+6y-11z=b$
$P_{3}: x-2y+7z=c$
First,we calculate the determinant of the coefficient matrix $D$:
$D = \begin{vmatrix} 1 & 2 & -3 \\ 2 & 6 & -11 \\ 1 & -2 & 7 \end{vmatrix} = 1(42-22) - 2(14+11) - 3(-4-6) = 20 - 50 + 30 = 0$.
Since $D=0$,the system does not have a unique solution.
Now,observe the linear combination of the equations:
$2P_{1} + P_{3} = 2(x+2y-3z) + (x-2y+7z) = 2x+4y-6z + x-2y+7z = 3x+2y+z$ (This does not match $P_{2}$ directly).
Let us check $5P_{1} = 2P_{2} + P_{3}$:
$2P_{2} + P_{3} = 2(2x+6y-11z) + (x-2y+7z) = 4x+12y-22z + x-2y+7z = 5x+10y-15z = 5(x+2y-3z) = 5P_{1}$.
Thus,if $5a = 2b + c$,the third equation is a linear combination of the first two,meaning the planes are consistent and share a common line of intersection.
Therefore,the system has an infinite number of solutions when $5a = 2b + c$.

(B) Given $f^{\prime}(a) = 2$ and $f(a) = 4$.
We need to evaluate $L = \lim_{x \rightarrow a} \frac{x f(a) - a f(x)}{x - a}$.
Since the limit is of the form $\frac{0}{0}$,we apply $L$'$H$ôpital's rule by differentiating the numerator and denominator with respect to $x$:
$L = \lim_{x \rightarrow a} \frac{\frac{d}{dx}(x f(a) - a f(x))}{\frac{d}{dx}(x - a)}$
$L = \lim_{x \rightarrow a} \frac{f(a) - a f^{\prime}(x)}{1}$
Substituting $x = a$:
$L = f(a) - a f^{\prime}(a)$
$L = 4 - a(2) = 4 - 2a$.

(A) Let $P = (1, 3, 5)$ and $Q = (\alpha, \beta, \gamma)$ be the image of $P$ with respect to the plane $4x - 5y + 2z = 8$.
The midpoint $M$ of $PQ$ is given by $M = \left(\frac{1+\alpha}{2}, \frac{3+\beta}{2}, \frac{5+\gamma}{2}\right)$.
Since $M$ lies on the plane,we have:
$4\left(\frac{1+\alpha}{2}\right) - 5\left(\frac{3+\beta}{2}\right) + 2\left(\frac{5+\gamma}{2}\right) = 8$
$2(1+\alpha) - 2.5(3+\beta) + (5+\gamma) = 8$
$2 + 2\alpha - 7.5 - 2.5\beta + 5 + \gamma = 8$
$2\alpha - 2.5\beta + \gamma = 8.5$ ... $(1)$
The line $PQ$ is perpendicular to the plane,so its direction ratios are proportional to the normal vector $(4, -5, 2)$:
$\frac{\alpha-1}{4} = \frac{\beta-3}{-5} = \frac{\gamma-5}{2} = k$
$\alpha = 1 + 4k, \beta = 3 - 5k, \gamma = 5 + 2k$ ... $(2)$
Substituting $(2)$ into the plane equation formula $\frac{\alpha-x_1}{a} = \frac{\beta-y_1}{b} = \frac{\gamma-z_1}{c} = -2 \frac{ax_1+by_1+cz_1-d}{a^2+b^2+c^2}$:
$k = -2 \frac{4(1) - 5(3) + 2(5) - 8}{4^2 + (-5)^2 + 2^2} = -2 \frac{4 - 15 + 10 - 8}{16 + 25 + 4} = -2 \frac{-9}{45} = \frac{18}{45} = \frac{2}{5}$.
Using $k = \frac{2}{5}$ in $(2)$:
$\alpha = 1 + 4(\frac{2}{5}) = \frac{13}{5}$
$\beta = 3 - 5(\frac{2}{5}) = 1$
$\gamma = 5 + 2(\frac{2}{5}) = \frac{29}{5}$
Therefore,$5(\alpha + \beta + \gamma) = 5(\frac{13}{5} + 1 + \frac{29}{5}) = 13 + 5 + 29 = 47$.

(A) Given $f(x) = \int_{0}^{x} e^{t} f(t) dt + e^{x}$.
At $x=0$,$f(0) = \int_{0}^{0} e^{t} f(t) dt + e^{0} = 0 + 1 = 1$.
Differentiating both sides with respect to $x$ using the Leibniz rule:
$f'(x) = e^{x} f(x) + e^{x} = e^{x}(f(x) + 1)$.
Rearranging the terms,we get $\frac{f'(x)}{f(x) + 1} = e^{x}$.
Integrating both sides with respect to $x$ from $0$ to $x$:
$\int_{0}^{x} \frac{f'(t)}{f(t) + 1} dt = \int_{0}^{x} e^{t} dt$.
$[\ln(f(t) + 1)]_{0}^{x} = [e^{t}]_{0}^{x}$.
$\ln(f(x) + 1) - \ln(f(0) + 1) = e^{x} - e^{0}$.
Since $f(0) = 1$,we have $\ln(f(x) + 1) - \ln(2) = e^{x} - 1$.
$\ln\left(\frac{f(x) + 1}{2}\right) = e^{x} - 1$.
Taking the exponential of both sides:
$\frac{f(x) + 1}{2} = e^{(e^{x} - 1)}$.
$f(x) = 2 e^{(e^{x} - 1)} - 1$.

(A) The curves $y = \sin x$ and $y = \cos x$ intersect at $x = \frac{\pi}{4}$ in the first quadrant.
$A_{1}$ is the area bounded by $y = \cos x$,$y = \sin x$ and the $y$-axis from $x = 0$ to $x = \frac{\pi}{4}$:
$A_{1} = \int_{0}^{\pi/4} (\cos x - \sin x) dx = [\sin x + \cos x]_{0}^{\pi/4} = (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) - (0 + 1) = \sqrt{2} - 1$.
$A_{2}$ is the area bounded by $y = \sin x$ from $x = 0$ to $x = \frac{\pi}{4}$ and $y = \cos x$ from $x = \frac{\pi}{4}$ to $x = \frac{\pi}{2}$:
$A_{2} = \int_{0}^{\pi/4} \sin x dx + \int_{\pi/4}^{\pi/2} \cos x dx = [-\cos x]_{0}^{\pi/4} + [\sin x]_{\pi/4}^{\pi/2}$
$A_{2} = (-\frac{1}{\sqrt{2}} - (-1)) + (1 - \frac{1}{\sqrt{2}}) = 1 - \frac{1}{\sqrt{2}} + 1 - \frac{1}{\sqrt{2}} = 2 - \frac{2}{\sqrt{2}} = 2 - \sqrt{2} = \sqrt{2}(\sqrt{2} - 1)$.
Now,$A_{1} : A_{2} = (\sqrt{2} - 1) : \sqrt{2}(\sqrt{2} - 1) = 1 : \sqrt{2}$.
And $A_{1} + A_{2} = (\sqrt{2} - 1) + (2 - \sqrt{2}) = 1$.

(C) We are given $I_{m, n} = \int_{0}^{1} x^{m-1}(1-x)^{n-1} dx$.
By substituting $x = \frac{1}{1+y}$,we have $dx = -\frac{1}{(1+y)^2} dy$.
When $x=0, y \to \infty$ and when $x=1, y=0$.
Thus,$I_{m, n} = \int_{\infty}^{0} (\frac{1}{1+y})^{m-1} (1 - \frac{1}{1+y})^{n-1} (-\frac{1}{(1+y)^2}) dy = \int_{0}^{\infty} \frac{y^{n-1}}{(1+y)^{m+n}} dy$.
Similarly,$I_{m, n} = \int_{0}^{\infty} \frac{y^{m-1}}{(1+y)^{m+n}} dy$.
Adding these,$2I_{m, n} = \int_{0}^{\infty} \frac{y^{m-1} + y^{n-1}}{(1+y)^{m+n}} dy$.
Splitting the integral at $y=1$:
$2I_{m, n} = \int_{0}^{1} \frac{y^{m-1} + y^{n-1}}{(1+y)^{m+n}} dy + \int_{1}^{\infty} \frac{y^{m-1} + y^{n-1}}{(1+y)^{m+n}} dy$.
In the second integral,substitute $y = \frac{1}{t}$,so $dy = -\frac{1}{t^2} dt$.
$\int_{1}^{\infty} \frac{y^{m-1} + y^{n-1}}{(1+y)^{m+n}} dy = \int_{1}^{0} \frac{(1/t)^{m-1} + (1/t)^{n-1}}{(1 + 1/t)^{m+n}} (-1/t^2) dt = \int_{0}^{1} \frac{t^{n-1} + t^{m-1}}{(t+1)^{m+n}} dt$.
Thus,$2I_{m, n} = \int_{0}^{1} \frac{y^{m-1} + y^{n-1}}{(1+y)^{m+n}} dy + \int_{0}^{1} \frac{y^{m-1} + y^{n-1}}{(1+y)^{m+n}} dy = 2 \int_{0}^{1} \frac{y^{m-1} + y^{n-1}}{(1+y)^{m+n}} dy$.
Therefore,$\int_{0}^{1} \frac{x^{m-1} + x^{n-1}}{(1+x)^{m+n}} dx = I_{m, n}$,which implies $\alpha = 1$.

(C) Given $A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 3 & 0 & -1 \end{bmatrix}$.
Calculating powers of $A$:
$A^2 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 1 \end{bmatrix}$,$A^3 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 8 & 0 \\ 3 & 0 & -1 \end{bmatrix}$,$A^4 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 16 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
By induction,for even $n$,$A^n = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2^n & 0 \\ 0 & 0 & 1 \end{bmatrix}$ and for odd $n$,$A^n = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2^n & 0 \\ 3 & 0 & -1 \end{bmatrix}$.
Thus,$A^{20} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2^{20} & 0 \\ 0 & 0 & 1 \end{bmatrix}$ and $A^{19} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2^{19} & 0 \\ 3 & 0 & -1 \end{bmatrix}$.
Substituting into the equation $A^{20} + \alpha A^{19} + \beta A = I_3$ (where $I_3$ is the identity matrix):
$\begin{bmatrix} 1+\alpha+\beta & 0 & 0 \\ 0 & 2^{20}+\alpha 2^{19}+2\beta & 0 \\ 3\alpha+3\beta & 0 & 1-\alpha-\beta \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
Comparing elements: $1+\alpha+\beta = 1 \Rightarrow \alpha+\beta = 0 \Rightarrow \beta = -\alpha$.
Also,$3\alpha+3\beta = 0$ (consistent with $\alpha+\beta=0$) and $1-\alpha-\beta = 1 \Rightarrow \alpha+\beta = 0$.
Finally,$2^{20} + \alpha 2^{19} + 2\beta = 4$. Substituting $\beta = -\alpha$:
$2^{20} + \alpha 2^{19} - 2\alpha = 4 \Rightarrow \alpha(2^{19}-2) = 4 - 2^{20}$.
$\alpha = \frac{4 - 2^{20}}{2^{19}-2} = \frac{2(2 - 2^{19})}{2^{19}-2} = -2$.
Since $\beta = -\alpha$,$\beta = 2$.
Therefore,$\beta - \alpha = 2 - (-2) = 4$.

(B) The given differential equation is $(y+1) \tan ^{2} x \,dx+\tan x \,dy+y \,dx=0$.
Rearranging the terms,we get $\tan x \,dy + (y \tan^2 x + y + \tan^2 x) \,dx = 0$.
Dividing by $\tan x \,dx$,we get $\frac{dy}{dx} + \frac{y(\tan^2 x + 1) + \tan^2 x}{\tan x} = 0$.
Since $1+\tan^2 x = \sec^2 x$,we have $\frac{dy}{dx} + y \frac{\sec^2 x}{\tan x} = -\tan x$.
This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{\sec^2 x}{\tan x}$ and $Q(x) = -\tan x$.
The integrating factor $IF = e^{\int \frac{\sec^2 x}{\tan x} \,dx} = e^{\ln(\tan x)} = \tan x$.
The general solution is $y \cdot IF = \int Q(x) \cdot IF \,dx + C$.
$y \tan x = \int -\tan^2 x \,dx + C = \int (1 - \sec^2 x) \,dx + C = x - \tan x + C$.
Thus,$y = \frac{x}{\tan x} - 1 + \frac{C}{\tan x}$.
Given $\lim_{x \to 0+} x y(x) = 1$,we have $\lim_{x \to 0+} x \left( \frac{x}{\tan x} - 1 + \frac{C}{\tan x} \right) = 1$.
Since $\lim_{x \to 0} \frac{x}{\tan x} = 1$,the limit becomes $1(1) - 0 + C(1) = 1$,which implies $1 + C = 1$,so $C = 0$.
Thus,$y(x) = \frac{x}{\tan x} - 1$.
Evaluating at $x = \frac{\pi}{4}$,$y\left(\frac{\pi}{4}\right) = \frac{\pi/4}{1} - 1 = \frac{\pi}{4} - 1$.

(D) The probability that exactly one of $A$ or $B$ occurs is given by $P(A \cap \overline{B}) + P(\overline{A} \cap B) = \frac{5}{9}$.
Since $A$ and $B$ are independent, this becomes $P(A)P(\overline{B}) + P(\overline{A})P(B) = \frac{5}{9}$.
Substituting the given values $P(A) = p$ and $P(B) = 2p$, we have $p(1 - 2p) + (1 - p)(2p) = \frac{5}{9}$.
Expanding this, we get $p - 2p^2 + 2p - 2p^2 = \frac{5}{9}$, which simplifies to $3p - 4p^2 = \frac{5}{9}$.
Rearranging into a quadratic equation: $36p^2 - 27p + 5 = 0$.
Factoring the quadratic: $(12p - 5)(3p - 1) = 0$.
Thus, $p = \frac{5}{12}$ or $p = \frac{1}{3}$.
The largest value of $p$ is $\frac{5}{12}$.

(B) For a system of homogeneous linear equations to have a non-trivial solution,the determinant of the coefficient matrix must be zero.
Let $D = \begin{vmatrix} 1+\cos^2 \theta & \sin^2 \theta & 4\sin 3\theta \\ \cos^2 \theta & 1+\sin^2 \theta & 4\sin 3\theta \\ \cos^2 \theta & \sin^2 \theta & 1+4\sin 3\theta \end{vmatrix} = 0$.
Applying $C_1 \rightarrow C_1 + C_2$,we get:
$D = \begin{vmatrix} 1+\cos^2 \theta + \sin^2 \theta & \sin^2 \theta & 4\sin 3\theta \\ \cos^2 \theta + 1+\sin^2 \theta & 1+\sin^2 \theta & 4\sin 3\theta \\ \cos^2 \theta + \sin^2 \theta & \sin^2 \theta & 1+4\sin 3\theta \end{vmatrix} = \begin{vmatrix} 2 & \sin^2 \theta & 4\sin 3\theta \\ 2 & 1+\sin^2 \theta & 4\sin 3\theta \\ 1 & \sin^2 \theta & 1+4\sin 3\theta \end{vmatrix} = 0$.
Applying $R_1 \rightarrow R_1 - R_2$ and $R_2 \rightarrow R_2 - R_3$:
$D = \begin{vmatrix} 0 & -1 & 0 \\ 1 & 1 & -1 \\ 1 & \sin^2 \theta & 1+4\sin 3\theta \end{vmatrix} = 0$.
Expanding along the first row:
$-(-1) \begin{vmatrix} 1 & -1 \\ 1 & 1+4\sin 3\theta \end{vmatrix} = 0 \implies 1(1+4\sin 3\theta) - (-1)(1) = 0$.
$1 + 4\sin 3\theta + 1 = 0 \implies 4\sin 3\theta = -2 \implies \sin 3\theta = -\frac{1}{2}$.
Since $\theta \in \left(0, \frac{\pi}{2}\right)$,$3\theta \in (0, \frac{3\pi}{2})$.
For $\sin 3\theta = -\frac{1}{2}$,$3\theta = \pi + \frac{\pi}{6} = \frac{7\pi}{6}$.
$\theta = \frac{7\pi}{18}$.

(C) Given $f(x) = \cos \left(2 \tan ^{-1} \sin \left(\cot ^{-1} \sqrt{\frac{1-x}{x}}\right)\right)$.
Let $\theta = \cot ^{-1} \sqrt{\frac{1-x}{x}}$. Then $\cot \theta = \sqrt{\frac{1-x}{x}}$,which implies $\tan \theta = \sqrt{\frac{x}{1-x}}$.
Using the identity $\sin \theta = \frac{\tan \theta}{\sqrt{1+\tan^2 \theta}} = \frac{\sqrt{x/(1-x)}}{\sqrt{1+x/(1-x)}} = \sqrt{x}$.
So,$\cot ^{-1} \sqrt{\frac{1-x}{x}} = \sin ^{-1} \sqrt{x}$.
Substituting this into $f(x)$:
$f(x) = \cos \left(2 \tan ^{-1} \sqrt{x}\right)$.
Using the identity $2 \tan ^{-1} u = \cos ^{-1} \left(\frac{1-u^2}{1+u^2}\right)$:
$f(x) = \cos \left(\cos ^{-1} \left(\frac{1-x}{1+x}\right)\right) = \frac{1-x}{1+x}$.
Now,differentiate $f(x)$ with respect to $x$ using the quotient rule:
$f^{\prime}(x) = \frac{(1+x)(-1) - (1-x)(1)}{(1+x)^2} = \frac{-1-x-1+x}{(1+x)^2} = \frac{-2}{(1+x)^2}$.
We need to check the options. Consider $(1-x)^2 f^{\prime}(x) + 2(f(x))^2$:
$(1-x)^2 \left(\frac{-2}{(1+x)^2}\right) + 2 \left(\frac{1-x}{1+x}\right)^2 = \frac{-2(1-x)^2}{(1+x)^2} + \frac{2(1-x)^2}{(1+x)^2} = 0$.
Thus,$(1-x)^{2} f^{\prime}(x)+2(f(x))^{2}=0$ is correct.

(A) Given $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = \hat{j} - \hat{k}$.
We are given $\vec{a} \times \vec{c} = \vec{b}$ and $\vec{a} \cdot \vec{c} = 3$.
We need to find $\vec{a} \cdot (\vec{b} \times \vec{c})$.
Using the scalar triple product property,$\vec{a} \cdot (\vec{b} \times \vec{c}) = (\vec{a} \times \vec{b}) \cdot \vec{c}$.
First,calculate $\vec{a} \times \vec{b}$:
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 0 & 1 & -1 \end{vmatrix} = \hat{i}(-1-1) - \hat{j}(-1-0) + \hat{k}(1-0) = -2\hat{i} + \hat{j} + \hat{k}$.
Now,use the vector triple product identity $\vec{a} \times (\vec{a} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{a} - (\vec{a} \cdot \vec{a})\vec{c}$.
Since $\vec{a} \times \vec{c} = \vec{b}$,we have $\vec{a} \times \vec{b} = (\vec{a} \cdot \vec{c})\vec{a} - |\vec{a}|^2 \vec{c}$.
Given $|\vec{a}|^2 = 1^2 + 1^2 + 1^2 = 3$ and $\vec{a} \cdot \vec{c} = 3$,we get:
$-2\hat{i} + \hat{j} + \hat{k} = 3(\hat{i} + \hat{j} + \hat{k}) - 3\vec{c}$.
$3\vec{c} = 3\hat{i} + 3\hat{j} + 3\hat{k} - (-2\hat{i} + \hat{j} + \hat{k}) = 5\hat{i} + 2\hat{j} + 2\hat{k}$.
$\vec{c} = \frac{5}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k}$.
Finally,$\vec{a} \cdot (\vec{b} \times \vec{c}) = (\vec{a} \times \vec{b}) \cdot \vec{c} = (-2\hat{i} + \hat{j} + \hat{k}) \cdot (\frac{5}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k}) = -\frac{10}{3} + \frac{2}{3} + \frac{2}{3} = -\frac{6}{3} = -2$.

(B) Let $I = \int_{-1 / \sqrt{2}}^{1 / \sqrt{2}} \sqrt{\left(\frac{x+1}{x-1}\right)^{2} + \left(\frac{x-1}{x+1}\right)^{2} - 2} \, dx$.
We know that $a^2 + b^2 - 2 = (a-b)^2$. Here $a = \frac{x+1}{x-1}$ and $b = \frac{x-1}{x+1}$.
So,the integrand becomes $\sqrt{(\frac{x+1}{x-1} - \frac{x-1}{x+1})^2} = |\frac{x+1}{x-1} - \frac{x-1}{x+1}|$.
Simplifying the expression inside the absolute value: $\frac{(x+1)^2 - (x-1)^2}{x^2-1} = \frac{4x}{x^2-1}$.
Thus,$I = \int_{-1 / \sqrt{2}}^{1 / \sqrt{2}} |\frac{4x}{x^2-1}| \, dx$.
Since the integrand is an even function,$I = 2 \int_{0}^{1 / \sqrt{2}} |\frac{4x}{x^2-1}| \, dx = 8 \int_{0}^{1 / \sqrt{2}} |\frac{x}{x^2-1}| \, dx$.
For $x \in [0, 1/\sqrt{2}]$,$x^2-1 < 0$,so $|\frac{x}{x^2-1}| = -\frac{x}{x^2-1}$.
$I = -8 \int_{0}^{1 / \sqrt{2}} \frac{x}{x^2-1} \, dx = -4 \int_{0}^{1 / \sqrt{2}} \frac{2x}{x^2-1} \, dx$.
$I = -4 [\ln |x^2-1|]_{0}^{1 / \sqrt{2}} = -4 (\ln |1/2 - 1| - \ln |0-1|) = -4 (\ln(1/2) - 0) = -4 \ln(1/2) = 4 \ln 2 = \ln 16$.