JEE Main 2021 Mathematics Question Paper with Answer and Solution

(C) Let the statement $P$ be: "For all $M > 0$,there exists $x \in S$ such that $x \geq M$."
The negation of a statement involving quantifiers follows these rules:
$1$. The negation of "for all" $(\forall)$ is "there exists" $(\exists)$.
$2$. The negation of "there exists" $(\exists)$ is "for all" $(\forall)$.
$3$. The negation of $x \geq M$ is $x < M$.
Applying these rules to $P$:
$\sim P$: "There exists $M > 0$ such that for all $x \in S$,$x < M$."
Thus,the correct option is $C$.

(A) The reflection of point $P(a, b)$ about the line $y=x$ gives the point $(b, a)$.
Translating this point by $2$ units along the positive $x$-axis results in the point $(b+2, a)$.
Applying a rotation of $\frac{\pi}{4}$ anti-clockwise about the origin,the new coordinates $(x', y')$ are given by:
$x' = (b+2)\cos\frac{\pi}{4} - a\sin\frac{\pi}{4} = \frac{b+2-a}{\sqrt{2}}$
$y' = (b+2)\sin\frac{\pi}{4} + a\cos\frac{\pi}{4} = \frac{b+2+a}{\sqrt{2}}$
Given the final position is $\left(-\frac{1}{\sqrt{2}}, \frac{7}{\sqrt{2}}\right)$,we have:
$\frac{b+2-a}{\sqrt{2}} = -\frac{1}{\sqrt{2}}$ $\Rightarrow b-a+2 = -1$ $\Rightarrow b-a = -3$ (Equation $1$)
$\frac{b+2+a}{\sqrt{2}} = \frac{7}{\sqrt{2}}$ $\Rightarrow b+a+2 = 7$ $\Rightarrow b+a = 5$ (Equation $2$)
Adding Equation $1$ and Equation $2$:
$2b = 2 \Rightarrow b = 1$
Substituting $b=1$ into Equation $2$:
$1+a = 5 \Rightarrow a = 4$
Therefore,$2a+b = 2(4)+1 = 9$.

(B) Given $\alpha = \max \{8^{2 \sin 3x} \cdot 4^{4 \cos 3x}\}$ and $\beta = \min \{8^{2 \sin 3x} \cdot 4^{4 \cos 3x}\}$.
We can rewrite the expression as $2^{6 \sin 3x} \cdot 2^{8 \cos 3x} = 2^{6 \sin 3x + 8 \cos 3x}$.
The range of $6 \sin 3x + 8 \cos 3x$ is $[-\sqrt{6^2 + 8^2}, \sqrt{6^2 + 8^2}] = [-10, 10]$.
Thus,$\alpha = 2^{10}$ and $\beta = 2^{-10}$.
Then $\alpha^{1/5} = (2^{10})^{1/5} = 2^2 = 4$ and $\beta^{1/5} = (2^{-10})^{1/5} = 2^{-2} = 1/4$.
The roots of $8x^2 + bx + c = 0$ are $4$ and $1/4$.
Sum of roots: $4 + 1/4 = 17/4 = -b/8 \Rightarrow b = -34$.
Product of roots: $4 \cdot 1/4 = 1 = c/8 \Rightarrow c = 8$.
Therefore,$c - b = 8 - (-34) = 42$.

(D) Given $S_{1}: |z-2| \leq 1$,which represents a disk centered at $(2, 0)$ with radius $1$.
Given $S_{2}: z(1+i)+\overline{z}(1-i) \geq 4$. Let $z = x+iy$. Then $\overline{z} = x-iy$.
Substituting these into the inequality:
$(x+iy)(1+i) + (x-iy)(1-i) \geq 4$
$(x - y + i(x+y)) + (x - y - i(x+y)) \geq 4$
$2x - 2y \geq 4 \implies y \leq x-2$.
We need the maximum value of $|z - 2.5|^2$ for $z$ in the intersection of the disk $|z-2| \leq 1$ and the half-plane $y \leq x-2$.
The point $z = 2.5$ lies on the boundary of the disk. The maximum distance from $2.5$ in the disk occurs at the point on the boundary furthest from $2.5$. The line $y = x-2$ passes through the center $(2,0)$ of the circle.
The intersection points of the line $y = x-2$ and the circle $(x-2)^2 + y^2 = 1$ are found by substituting $y = x-2$ into the circle equation:
$(x-2)^2 + (x-2)^2 = 1 \implies 2(x-2)^2 = 1 \implies x-2 = \pm \frac{1}{\sqrt{2}}$.
So $x = 2 \pm \frac{1}{\sqrt{2}}$. The corresponding $y$ values are $y = x-2 = \pm \frac{1}{\sqrt{2}}$.
The intersection points are $(2 + \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$ and $(2 - \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}})$.
The point $P = (2 - \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}})$ is the point in the intersection region furthest from $2.5 + 0i$.
$|z - 2.5|^2 = |(2 - \frac{1}{\sqrt{2}} - 2.5) - i\frac{1}{\sqrt{2}}|^2 = |-\frac{1}{2} - \frac{1}{\sqrt{2}} - i\frac{1}{\sqrt{2}}|^2$
$= (-\frac{1}{2} - \frac{1}{\sqrt{2}})^2 + (-\frac{1}{\sqrt{2}})^2 = (\frac{1}{4} + \frac{1}{2} + \frac{1}{\sqrt{2}}) + \frac{1}{2} = \frac{5}{4} + \frac{1}{\sqrt{2}} = \frac{5 + 2\sqrt{2}}{4}$.

(C) Let the parallelogram be $ABCD$. The lines $4x + 5y = 0$ and $7x + 2y = 0$ intersect at the origin $A(0, 0)$.
Let the diagonal $BD$ lie on the line $11x + 7y = 9$.
Point $D$ is the intersection of $4x + 5y = 0$ and $11x + 7y = 9$. Solving these,we get $D = (5/3, -4/3)$.
Point $B$ is the intersection of $7x + 2y = 0$ and $11x + 7y = 9$. Solving these,we get $B = (-2/3, 7/3)$.
The diagonals of a parallelogram bisect each other. Let $M$ be the midpoint of $BD$.
$M = (\frac{5/3 - 2/3}{2}, \frac{-4/3 + 7/3}{2}) = (1/2, 1/2)$.
The other diagonal $AC$ passes through $A(0, 0)$ and $M(1/2, 1/2)$.
The equation of line $AC$ is $y - 0 = \frac{1/2 - 0}{1/2 - 0}(x - 0)$,which simplifies to $y = x$.
Checking the options,the point $(2, 2)$ satisfies $y = x$.

(C) The total frequency is $N = 4 + 4 + \alpha + \beta = 8 + \alpha + \beta$.
The mean is given by $\bar{x} = \frac{\sum f_i x_i}{N} = 6$.
$\frac{4(2) + 4(6) + \alpha(8) + \beta(9)}{8 + \alpha + \beta} = 6$
$8 + 24 + 8\alpha + 9\beta = 48 + 6\alpha + 6\beta$
$2\alpha + 3\beta = 16 \quad \dots (i)$
The variance is $\sigma^2 = \frac{\sum f_i x_i^2}{N} - (\bar{x})^2 = 6.8$.
$\frac{4(4) + 4(36) + \alpha(64) + \beta(81)}{8 + \alpha + \beta} - 36 = 6.8$
$\frac{16 + 144 + 64\alpha + 81\beta}{8 + \alpha + \beta} = 42.8$
$160 + 64\alpha + 81\beta = 42.8(8 + \alpha + \beta) = 342.4 + 42.8\alpha + 42.8\beta$
$21.2\alpha + 38.2\beta = 182.4$
Multiplying by $10$: $212\alpha + 382\beta = 1824 \Rightarrow 106\alpha + 191\beta = 912 \quad \dots (ii)$
Solving $(i)$ and $(ii)$: From $(i)$,$\alpha = \frac{16 - 3\beta}{2} = 8 - 1.5\beta$.
Substituting in $(ii)$: $106(8 - 1.5\beta) + 191\beta = 912$
$848 - 159\beta + 191\beta = 912 \Rightarrow 32\beta = 64 \Rightarrow \beta = 2$.
Then $\alpha = 8 - 1.5(2) = 5$.
Total frequency $N = 8 + 5 + 2 = 15$.
New sum of observations when $x_3$ changes from $8$ to $7$:
Original sum $= 6 \times 15 = 90$.
New sum $= 90 - (8 \times 5) + (7 \times 5) = 90 - 40 + 35 = 85$.
New mean $= \frac{85}{15} = \frac{17}{3}$.

(B) Let $a = 3^{\log _{3} \sqrt{25^{x-1}+7}} = \sqrt{25^{x-1}+7}$ and $b = 3^{\left(-\frac{1}{8}\right) \log _{3}\left(5^{x-1}+1\right)} = (5^{x-1}+1)^{-1/8}$.
The expansion is $(a+b)^{10}$. The ninth term $T_9$ is given by $T_9 = {}^{10}C_8 a^{10-8} b^8 = {}^{10}C_8 a^2 b^8$.
Substituting the values: ${}^{10}C_8 = 45$,$a^2 = 25^{x-1}+7$,and $b^8 = (5^{x-1}+1)^{-1}$.
So,$45 \times \frac{25^{x-1}+7}{5^{x-1}+1} = 180$.
Dividing by $45$,we get $\frac{25^{x-1}+7}{5^{x-1}+1} = 4$.
Let $t = 5^{x-1}$. Then $\frac{t^2+7}{t+1} = 4$.
$t^2+7 = 4t+4 \Rightarrow t^2-4t+3 = 0$.
$(t-1)(t-3) = 0$,so $t=1$ or $t=3$.
If $5^{x-1} = 1$,then $x-1 = 0 \Rightarrow x = 1$.
If $5^{x-1} = 3$,then $x-1 = \log_5 3 \Rightarrow x = 1 + \log_5 3$.

(A) Given that $\tan \left(\frac{\pi}{9}\right), x, \tan \left(\frac{7 \pi}{18}\right)$ are in arithmetic progression,we have $2x = \tan \left(\frac{\pi}{9}\right) + \tan \left(\frac{7 \pi}{18}\right)$,so $x = \frac{1}{2} \left(\tan \frac{\pi}{9} + \tan \frac{7 \pi}{18}\right)$.
Given that $\tan \left(\frac{\pi}{9}\right), y, \tan \left(\frac{5 \pi}{18}\right)$ are in arithmetic progression,we have $2y = \tan \left(\frac{\pi}{9}\right) + \tan \left(\frac{5 \pi}{18}\right)$.
We need to find $|x - 2y|$.
Substituting the expressions:
$x - 2y = \frac{1}{2} \left(\tan \frac{\pi}{9} + \tan \frac{7 \pi}{18}\right) - \left(\tan \frac{\pi}{9} + \tan \frac{5 \pi}{18}\right)$.
Using the identities $\tan \left(\frac{7 \pi}{18}\right) = \cot \left(\frac{\pi}{9}\right)$ and $\tan \left(\frac{5 \pi}{18}\right) = \cot \left(\frac{2 \pi}{9}\right)$:
$x - 2y = \frac{1}{2} \tan \frac{\pi}{9} + \frac{1}{2} \cot \frac{\pi}{9} - \tan \frac{\pi}{9} - \cot \frac{2 \pi}{9} = \frac{1}{2} \left(\cot \frac{\pi}{9} - \tan \frac{\pi}{9}\right) - \cot \frac{2 \pi}{9}$.
Since $\cot \theta - \tan \theta = 2 \cot(2\theta)$:
$x - 2y = \frac{1}{2} \left(2 \cot \frac{2 \pi}{9}\right) - \cot \frac{2 \pi}{9} = \cot \frac{2 \pi}{9} - \cot \frac{2 \pi}{9} = 0$.
Therefore,$|x - 2y| = 0$.

(B) Let $L = \lim _{x \rightarrow 0} \frac{x}{(1-\sin x)^{1/8}-(1+\sin x)^{1/8}}$.
Using the binomial expansion $(1+u)^n \approx 1+nu$ for small $u$,where $n = 1/8$:
$(1-\sin x)^{1/8} \approx 1 - \frac{1}{8} \sin x$.
$(1+\sin x)^{1/8} \approx 1 + \frac{1}{8} \sin x$.
Substituting these into the limit expression:
$L = \lim _{x \rightarrow 0} \frac{x}{(1 - \frac{1}{8} \sin x) - (1 + \frac{1}{8} \sin x)}$.
$L = \lim _{x \rightarrow 0} \frac{x}{-\frac{2}{8} \sin x} = \lim _{x \rightarrow 0} \frac{x}{-\frac{1}{4} \sin x}$.
Since $\lim _{x \rightarrow 0} \frac{\sin x}{x} = 1$,we have $\lim _{x \rightarrow 0} \frac{x}{\sin x} = 1$.
Therefore,$L = -4 \times 1 = -4$.

(B) Let the center of the circle be $(h, k)$. Since the circle touches the $y$-axis at $(0, 6)$,the $y$-coordinate of the center is $k = 6$ and the radius $r = |h|$.
Thus,the equation of the circle is $(x - h)^{2} + (y - 6)^{2} = h^{2}$.
This circle cuts an intercept of $6 \sqrt{5}$ on the $x$-axis. Setting $y = 0$ in the equation,we get $(x - h)^{2} + (0 - 6)^{2} = h^{2}$,which simplifies to $(x - h)^{2} + 36 = h^{2}$,or $(x - h)^{2} = h^{2} - 36$.
Taking the square root,$x - h = \pm \sqrt{h^{2} - 36}$,so $x = h \pm \sqrt{h^{2} - 36}$.
The length of the intercept on the $x$-axis is the difference between these two $x$-values: $(h + \sqrt{h^{2} - 36}) - (h - \sqrt{h^{2} - 36}) = 2 \sqrt{h^{2} - 36}$.
Given that the intercept is $6 \sqrt{5}$,we have $2 \sqrt{h^{2} - 36} = 6 \sqrt{5}$,so $\sqrt{h^{2} - 36} = 3 \sqrt{5}$.
Squaring both sides,$h^{2} - 36 = 9 \times 5 = 45$,so $h^{2} = 81$,which means $h = \pm 9$.
Since the radius $r = |h|$,the radius of the circle is $r = 9$.

(C) The center of the ellipse is $C(3, -4)$.
The focus is $S(4, -4)$ and the vertex is $A(5, -4)$.
Since the $y$-coordinates are the same,the major axis is horizontal.
The distance from the center to the vertex is $a = |5 - 3| = 2$.
The distance from the center to the focus is $ae = |4 - 3| = 1$.
Thus,$e = \frac{1}{2}$.
Using $b^{2} = a^{2}(1 - e^{2})$,we get $b^{2} = 4(1 - \frac{1}{4}) = 4(\frac{3}{4}) = 3$.
The equation of the ellipse is $\frac{(x - 3)^{2}}{4} + \frac{(y + 4)^{2}}{3} = 1$.
The condition for the line $y = mx - 4$ to be a tangent to the ellipse $\frac{(x - h)^{2}}{a^{2}} + \frac{(y - k)^{2}}{b^{2}} = 1$ is $(y - k) = m(x - h) \pm \sqrt{a^{2}m^{2} + b^{2}}$.
Here,$h = 3, k = -4, a^{2} = 4, b^{2} = 3$.
The line is $y + 4 = mx - 3m$,so $y - (-4) = m(x - 3) - 3m$.
Comparing with the tangent form,the constant term is $-3m = \pm \sqrt{4m^{2} + 3}$.
Squaring both sides,$9m^{2} = 4m^{2} + 3$.
$5m^{2} = 3$.

(A) Given $z = \frac{3 + 2i \cos \theta}{1 - 3i \cos \theta}$.
To find the real part, multiply the numerator and denominator by the conjugate of the denominator $(1 + 3i \cos \theta)$:
$z = \frac{(3 + 2i \cos \theta)(1 + 3i \cos \theta)}{(1 - 3i \cos \theta)(1 + 3i \cos \theta)}$
$z = \frac{3 + 9i \cos \theta + 2i \cos \theta + 6i^2 \cos^2 \theta}{1 + 9 \cos^2 \theta}$
Since $i^2 = -1$, we have:
$z = \frac{3 - 6 \cos^2 \theta + 11i \cos \theta}{1 + 9 \cos^2 \theta}$
The real part is $\operatorname{Re}(z) = \frac{3 - 6 \cos^2 \theta}{1 + 9 \cos^2 \theta}$.
Given $\operatorname{Re}(z) = 0$, we have $3 - 6 \cos^2 \theta = 0$, which implies $\cos^2 \theta = \frac{1}{2}$.
Since $\theta \in (0, \frac{\pi}{2})$, $\cos \theta = \frac{1}{\sqrt{2}}$, so $\theta = \frac{\pi}{4}$.
Now, calculate $\sin^2 3\theta + \cos^2 \theta$:
$\sin^2 3(\frac{\pi}{4}) + \cos^2(\frac{\pi}{4}) = \sin^2(\frac{3\pi}{4}) + (\frac{1}{\sqrt{2}})^2$
$= (\frac{1}{\sqrt{2}})^2 + \frac{1}{2} = \frac{1}{2} + \frac{1}{2} = 1$.

(B) Given equation: $e^{4x} - e^{3x} - 4e^{2x} - e^{x} + 1 = 0$.
Divide by $e^{2x}$ (since $e^{2x} > 0$):
$e^{2x} - e^{x} - 4 - e^{-x} + e^{-2x} = 0$.
Group terms: $(e^{2x} + e^{-2x}) - (e^{x} + e^{-x}) - 4 = 0$.
Let $t = e^{x} > 0$. Then $e^{x} + e^{-x} = t + \frac{1}{t} = \alpha$,where $\alpha \geq 2$.
Note that $e^{2x} + e^{-2x} = (t + \frac{1}{t})^2 - 2 = \alpha^2 - 2$.
Substituting into the equation: $(\alpha^2 - 2) - \alpha - 4 = 0$.
$\alpha^2 - \alpha - 6 = 0$.
$(\alpha - 3)(\alpha + 2) = 0$.
Since $\alpha \geq 2$,we have $\alpha = 3$.
Now,$t + \frac{1}{t} = 3 \Rightarrow t^2 - 3t + 1 = 0$.
The discriminant $D = (-3)^2 - 4(1)(1) = 5 > 0$.
Since $t = e^{x} > 0$ and the product of roots is $1$ and sum is $3$,both roots for $t$ are positive.
Thus,there are $2$ real roots for $x$.

(A) The given number is $N = (10)^{10} \cdot (11)^{11} \cdot (13)^{13} = 2^{10} \cdot 5^{10} \cdot 11^{11} \cdot 13^{13}$.
$A$ divisor $d$ of $N$ is of the form $2^a \cdot 5^b \cdot 11^c \cdot 13^d$,where $0 \le a \le 10, 0 \le b \le 10, 0 \le c \le 11, 0 \le d \le 13$.
For $d$ to be of the form $4n+1$,$d$ must be odd,so $a=0$.
Thus,$d = 5^b \cdot 11^c \cdot 13^d$.
Modulo $4$,we have $5 \equiv 1 \pmod{4}$,$11 \equiv -1 \equiv 3 \pmod{4}$,and $13 \equiv 1 \pmod{4}$.
So,$d \equiv 1^b \cdot (-1)^c \cdot 1^d \equiv (-1)^c \pmod{4}$.
For $d \equiv 1 \pmod{4}$,we must have $(-1)^c = 1$,which means $c$ must be even.
Possible values for $c \in \{0, 2, 4, 6, 8, 10\}$,which gives $6$ choices.
Possible values for $b \in \{0, 1, 2, \dots, 10\}$,which gives $11$ choices.
Possible values for $d \in \{0, 1, 2, \dots, 13\}$,which gives $14$ choices.
The total number of such divisors is $11 \times 6 \times 14 = 924$.

(A) First,find the set $A$: $n^{2} - n \leq 10,000$. Solving $n(n-1) \leq 10,000$,we find $n \approx 100.5$,so $A = \{1, 2, \ldots, 100\}$.
Next,find $B - C$: $B$ contains numbers of the form $3k+1$ (i.e.,$4, 7, 10, 13, 16, 19, \ldots$). $C$ contains even numbers. Thus,$B - C$ contains odd numbers of the form $3k+1$,which are $7, 13, 19, \ldots$.
The general term of $B - C$ is $6m + 1$ for $m \geq 1$.
We need $A \cap (B - C) = \{n \in \{1, \ldots, 100\} \mid n = 6m + 1\}$.
For $m=1, n=7$; for $m=16, n=97$. The number of terms is $16$.
The sum is an arithmetic progression: $S = \frac{16}{2}(7 + 97) = 8 \times 104 = 832$.

(C) The mean of $a, b, c$ is $\bar{x} = \frac{a+b+c}{3}$.
Since $b = a + c$,we have $\bar{x} = \frac{b+b}{3} = \frac{2b}{3}$.
The standard deviation of $a+2, b+2, c+2$ is the same as the standard deviation of $a, b, c$,which is $d$.
Thus,$d^2 = \frac{a^2+b^2+c^2}{3} - (\bar{x})^2$.
$d^2 = \frac{a^2+b^2+c^2}{3} - \left(\frac{2b}{3}\right)^2$.
$d^2 = \frac{a^2+b^2+c^2}{3} - \frac{4b^2}{9}$.
$d^2 = \frac{3(a^2+b^2+c^2) - 4b^2}{9}$.
$9d^2 = 3(a^2+c^2+b^2) - 4b^2$.
$9d^2 = 3(a^2+c^2) + 3b^2 - 4b^2$.
$9d^2 = 3(a^2+c^2) - b^2$.
Therefore,$b^2 = 3(a^2+c^2) - 9d^2$.

(C) Given $f(x) = \int_{0}^{x} [y] \, dy$.
For $x \in [n, n+1)$,where $n \in \mathbb{N}_0$,we have $[y] = n$ for $y \in [n, x)$.
Thus,$f(x) = \int_{0}^{1} 0 \, dy + \int_{1}^{2} 1 \, dy + \dots + \int_{n-1}^{n} (n-1) \, dy + \int_{n}^{x} n \, dy$.
$f(x) = 0 + 1 + 2 + \dots + (n-1) + n(x-n) = \frac{(n-1)n}{2} + nx - n^2$.
Since $n = [x]$,we have $f(x) = \frac{[x]([x]-1)}{2} + [x](x-[x])$.
At any integer $x=n$,the left-hand limit is $\lim_{x \to n^-} f(x) = \sum_{k=0}^{n-1} k = \frac{(n-1)n}{2}$ and the right-hand limit is $\lim_{x \to n^+} f(x) = \frac{(n-1)n}{2} + n(n-n) = \frac{n(n-1)}{2}$.
Since the limits are equal,$f(x)$ is continuous for all $x \geq 0$.
However,the derivative $f'(x) = [x]$ is discontinuous at integer points because $\lim_{x \to n^-} f'(x) = n-1$ and $\lim_{x \to n^+} f'(x) = n$.
Therefore,$f$ is continuous everywhere but not differentiable at integer points.

(B) Given $g(3n+1)=3n+2$,$g(3n+2)=3n+3$,and $g(3n+3)=3n+1$ for $n \geq 0$.
First,we calculate $g \circ g \circ g(x)$:
For $x = 3n+1$,$g(g(g(3n+1))) = g(g(3n+2)) = g(3n+3) = 3n+1$.
Similarly,$g(g(g(3n+2))) = 3n+2$ and $g(g(g(3n+3))) = 3n+3$.
Thus,$g \circ g \circ g(x) = x$ for all $x \in N$.
Now,consider the condition $f(g(x)) = f(x)$.
If $f$ is a constant function,say $f(x) = c$ for all $x \in N$,then $f(g(x)) = c = f(x)$. However,a constant function is not onto on $N$.
If we define $f$ such that it maps all elements in the same cycle of $g$ to the same value,$f$ will satisfy $f(g(x)) = f(x)$.
For example,define $f(3n+1) = f(3n+2) = f(3n+3) = n+1$. This function $f$ is onto because for any $y \in N$,we can choose $n = y-1$ such that $f(3(y-1)+1) = y$.
Thus,there exists an onto function $f$ such that $f \circ g = f$.

(C) The total number of ways to distribute $9$ distinct balls into $4$ boxes is $4^{9}$.
The number of ways to choose $3$ balls for box $B_{3}$ is ${}^{9}C_{3}$.
The remaining $6$ balls can be distributed among the other $3$ boxes $(B_{1}, B_{2}, B_{4})$ in $3^{6}$ ways.
Thus,the probability that $B_{3}$ contains exactly $3$ balls is $P = \frac{{}^{9}C_{3} \cdot 3^{6}}{4^{9}}$.
We can rewrite this as $P = \frac{{}^{9}C_{3} \cdot 3^{6}}{4^{9}} = \frac{84 \cdot 3^{6}}{4^{9}}$.
Since we want the form $k \left(\frac{3}{4}\right)^{9}$,we write $P = k \cdot \frac{3^{9}}{4^{9}}$.
Equating the two expressions: $k \cdot \frac{3^{9}}{4^{9}} = \frac{84 \cdot 3^{6}}{4^{9}}$.
$k = \frac{84 \cdot 3^{6}}{3^{9}} = \frac{84}{3^{3}} = \frac{84}{27} = \frac{28}{9} \approx 3.11$.
Checking the options for $k = \frac{28}{9} \approx 3.11$:
$|x-3| < 1 \Rightarrow 2 < x < 4$. Since $3.11$ lies in this interval,option $C$ is correct.

(D) The line $L$ is given by $\frac{x}{1}=\frac{y}{0}=\frac{z}{-1} = \lambda$. So,any point on $L$ is $N(\lambda, 0, -\lambda)$.
Since $PN \perp L$,the vector $\vec{PN} = (\lambda-1, -2, -\lambda+1)$ is perpendicular to the direction vector of $L$,which is $\vec{v} = (1, 0, -1)$.
Thus,$(\lambda-1)(1) + (-2)(0) + (-\lambda+1)(-1) = 0 \Rightarrow \lambda-1 + \lambda-1 = 0 \Rightarrow 2\lambda = 2 \Rightarrow \lambda = 1$.
So,$N = (1, 0, -1)$ and $\vec{PN} = (0, -2, 0)$.
Now,let the line through $P$ parallel to the plane $x+y+2z=0$ meet $L$ at $Q(\mu, 0, -\mu)$.
The vector $\vec{PQ} = (\mu-1, -2, -\mu+1)$. Since this line is parallel to the plane,$\vec{PQ}$ is perpendicular to the normal of the plane $\vec{n} = (1, 1, 2)$.
Thus,$(\mu-1)(1) + (-2)(1) + (-\mu+1)(2) = 0 \Rightarrow \mu-1 - 2 - 2\mu + 2 = 0 \Rightarrow -\mu - 1 = 0 \Rightarrow \mu = -1$.
So,$Q = (-1, 0, 1)$ and $\vec{PQ} = (-2, -2, 2)$.
The angle $\alpha$ between $\vec{PN} = (0, -2, 0)$ and $\vec{PQ} = (-2, -2, 2)$ is given by $\cos \alpha = \frac{|\vec{PN} \cdot \vec{PQ}|}{|\vec{PN}| |\vec{PQ}|}$.
$|\vec{PN}| = \sqrt{0^2 + (-2)^2 + 0^2} = 2$.
$|\vec{PQ}| = \sqrt{(-2)^2 + (-2)^2 + 2^2} = \sqrt{4+4+4} = \sqrt{12} = 2\sqrt{3}$.
$\vec{PN} \cdot \vec{PQ} = (0)(-2) + (-2)(-2) + (0)(2) = 4$.
$\cos \alpha = \frac{4}{2 \times 2\sqrt{3}} = \frac{4}{4\sqrt{3}} = \frac{1}{\sqrt{3}}$.

(C) Given $f(x)=3 \sin ^{4} x+10 \sin ^{3} x+6 \sin ^{2} x-3$.
Differentiating with respect to $x$:
$f'(x) = 12 \sin^3 x \cos x + 30 \sin^2 x \cos x + 12 \sin x \cos x$
$f'(x) = 6 \sin x \cos x (2 \sin^2 x + 5 \sin x + 2)$
Factoring the quadratic expression:
$f'(x) = 6 \sin x \cos x (2 \sin x + 1)(\sin x + 2)$
For $x \in \left[-\frac{\pi}{6}, \frac{\pi}{2}\right]$,we have:
$1. \cos x > 0$ for all $x \in \left[-\frac{\pi}{6}, \frac{\pi}{2}\right]$.
$2. (\sin x + 2) > 0$ for all $x \in \left[-\frac{\pi}{6}, \frac{\pi}{2}\right]$.
$3. (2 \sin x + 1) \ge 0$ for all $x \in \left[-\frac{\pi}{6}, \frac{\pi}{2}\right]$.
Thus,the sign of $f'(x)$ depends on $\sin x$:
- If $x \in \left(-\frac{\pi}{6}, 0\right)$,then $\sin x < 0$,so $f'(x) < 0$. Thus,$f$ is decreasing in $\left(-\frac{\pi}{6}, 0\right)$.
- If $x \in \left(0, \frac{\pi}{2}\right)$,then $\sin x > 0$,so $f'(x) > 0$. Thus,$f$ is increasing in $\left(0, \frac{\pi}{2}\right)$.

(B) For the vectors to be co-planar,their scalar triple product must be zero:
$\begin{vmatrix} a+b+2 & a+2 b+c & -(b+c) \\ b+1 & 2 b & -b \\ b+2 & 2 b & 1-b \end{vmatrix} = 0$
Applying row operations $R_3 \rightarrow R_3 - R_2$ and $R_1 \rightarrow R_1 - R_2$:
$\begin{vmatrix} a+1 & a+c & -c \\ b+1 & 2 b & -b \\ 1 & 0 & 1 \end{vmatrix} = 0$
Expanding along the third row:
$1((a+c)(-b) - (2b)(-c)) + 1((a+1)(2b) - (b+1)(a+c)) = 0$
$(-ab - bc + 2bc) + (2ab + 2b - (ab + ac + b + c)) = 0$
$(-ab + bc) + (ab - ac + b - c) = 0$
$bc - ac + b - c = 0$
$c(b-a) + 1(b-c) = 0$
Wait,let us re-evaluate the determinant expansion:
$(a+1)(2b) - (a+c)(b+1+b) + (-c)(-2b) = 0$
$(a+1)(2b) - (a+c)(2b+1) + 2bc = 0$
$2ab + 2b - (2ab + a + 2bc + c) + 2bc = 0$
$2ab + 2b - 2ab - a - 2bc - c + 2bc = 0$
$2b - a - c = 0$
Therefore,$2b = a+c$.

(C) For $f$ to be continuous at $x = 2$,we must have $\lim_{x \rightarrow 2^{-}} f(x) = \lim_{x \rightarrow 2^{+}} f(x) = f(2) = \mu$.
First,calculate the right-hand limit:
$\lim_{x \rightarrow 2^{+}} f(x) = \lim_{x \rightarrow 2^{+}} e^{\frac{\tan(x-2)}{x-[x]}}$. Since $x > 2$,$[x] = 2$,so $\lim_{x \rightarrow 2^{+}} e^{\frac{\tan(x-2)}{x-2}} = e^{1} = e$.
Thus,$\mu = e$.
Next,calculate the left-hand limit:
$\lim_{x \rightarrow 2^{-}} f(x) = \lim_{x \rightarrow 2^{-}} \frac{\lambda|x^{2}-5x+6|}{\mu(5x-x^{2}-6)}$.
Note that $x^{2}-5x+6 = (x-2)(x-3)$. For $x < 2$,$(x-2) < 0$ and $(x-3) < 0$,so $(x-2)(x-3) > 0$. Thus,$|x^{2}-5x+6| = (x-2)(x-3)$.
Also,$5x-x^{2}-6 = -(x^{2}-5x+6) = -(x-2)(x-3)$.
So,$\lim_{x \rightarrow 2^{-}} f(x) = \frac{\lambda(x-2)(x-3)}{\mu(-(x-2)(x-3))} = -\frac{\lambda}{\mu}$.
Equating the limits: $-\frac{\lambda}{\mu} = e \Rightarrow \lambda = -\mu e = -e^{2}$.
Therefore,$\lambda + \mu = -e^{2} + e = e(-e+1)$.

(D) Given equation: $e^{6x} - e^{4x} - 2e^{3x} - 12e^{2x} + e^{x} + 1 = 0$
Divide the entire equation by $e^{3x}$ (since $e^{3x} \neq 0$):
$e^{3x} - e^{x} - 2 - 12e^{-x} + e^{-2x} + e^{-3x} = 0$
Rearranging terms:
$(e^{3x} + e^{-3x}) - (e^{x} - e^{-2x}) - (e^{-x} - e^{x}) - 2 = 0$
Alternatively,divide by $e^{3x}$ and group:
$e^{3x} - 12 - e^{-3x} = e^{x} + 2 - e^{-x}$
Let $f(x) = e^{3x} - 12 - e^{-3x}$ and $g(x) = e^{x} + 2 - e^{-x}$.
$f'(x) = 3e^{3x} + 3e^{-3x} > 0$ (strictly increasing).
$g'(x) = e^{x} + e^{-x} > 0$ (strictly increasing).
By observing the intersection of the graphs of $f(x)$ and $g(x)$,or by analyzing the behavior of the function $h(x) = e^{6x} - e^{4x} - 2e^{3x} - 12e^{2x} + e^{x} + 1$,we find that the function crosses the $x$-axis at exactly $2$ points.
Thus,the number of real roots is $2$.

(A) The region is bounded by the curves $y = 2x^2$ and $y = 4-2x$ for $x \geq 0$.
To find the intersection points,set $2x^2 = 4-2x$,which simplifies to $x^2 + x - 2 = 0$.
Factoring gives $(x+2)(x-1) = 0$,so $x = 1$ or $x = -2$.
Since $x \geq 0$,we consider the interval $[0, 1]$.
The area is given by the integral $\int_{0}^{1} ((4-2x) - 2x^2) dx$.
$= \int_{0}^{1} (4 - 2x - 2x^2) dx$
$= [4x - x^2 - \frac{2x^3}{3}]_{0}^{1}$
$= (4(1) - (1)^2 - \frac{2(1)^3}{3}) - (0)$
$= 4 - 1 - \frac{2}{3} = 3 - \frac{2}{3} = \frac{7}{3} \, sq. \, units$.

(D) For a system of linear equations to have no solution,the determinant of the coefficient matrix $D$ must be $0$,and at least one of the Cramer's rule determinants $(D_x, D_y, D_z)$ must be non-zero.
First,calculate the determinant of the coefficient matrix $D$:
$D = \begin{vmatrix} 2 & 3 & 6 \\ 1 & 2 & a \\ 3 & 5 & 9 \end{vmatrix} = 2(18 - 5a) - 3(9 - 3a) + 6(5 - 6) = 36 - 10a - 27 + 9a - 6 = 3 - a$.
For $D = 0$,we must have $a = 3$.
Next,calculate the determinant $D_z$ (or check the consistency of the augmented matrix):
$D_z = \begin{vmatrix} 2 & 3 & 8 \\ 1 & 2 & 5 \\ 3 & 5 & b \end{vmatrix} = 2(2b - 25) - 3(b - 15) + 8(5 - 6) = 4b - 50 - 3b + 45 - 8 = b - 13$.
For the system to have no solution when $D = 0$,we require $D_z \neq 0$,which implies $b - 13 \neq 0$,or $b \neq 13$.
Thus,the system has no solution when $a = 3$ and $b \neq 13$.

(D) Let $I = \int_{\pi / 24}^{5 \pi / 24} \frac{1}{1 + (\tan 2x)^{1/3}} dx = \int_{\pi / 24}^{5 \pi / 24} \frac{(\cos 2x)^{1/3}}{(\cos 2x)^{1/3} + (\sin 2x)^{1/3}} dx \dots (i)$
Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,where $a+b = \frac{\pi}{24} + \frac{5\pi}{24} = \frac{6\pi}{24} = \frac{\pi}{4}$.
Then $I = \int_{\pi / 24}^{5 \pi / 24} \frac{(\cos(2(\frac{\pi}{4} - x)))^{1/3}}{(\cos(2(\frac{\pi}{4} - x)))^{1/3} + (\sin(2(\frac{\pi}{4} - x)))^{1/3}} dx$
$I = \int_{\pi / 24}^{5 \pi / 24} \frac{(\cos(\frac{\pi}{2} - 2x))^{1/3}}{(\cos(\frac{\pi}{2} - 2x))^{1/3} + (\sin(\frac{\pi}{2} - 2x))^{1/3}} dx$
$I = \int_{\pi / 24}^{5 \pi / 24} \frac{(\sin 2x)^{1/3}}{(\sin 2x)^{1/3} + (\cos 2x)^{1/3}} dx \dots (ii)$
Adding $(i)$ and $(ii)$:
$2I = \int_{\pi / 24}^{5 \pi / 24} \frac{(\cos 2x)^{1/3} + (\sin 2x)^{1/3}}{(\cos 2x)^{1/3} + (\sin 2x)^{1/3}} dx = \int_{\pi / 24}^{5 \pi / 24} 1 dx$
$2I = [x]_{\pi / 24}^{5 \pi / 24} = \frac{5\pi}{24} - \frac{\pi}{24} = \frac{4\pi}{24} = \frac{\pi}{6}$
$I = \frac{\pi}{12}$

(A) Given the differential equation $\frac{dy}{dx} = 1 + xe^{y-x}$.
Rearranging the terms: $\frac{dy}{dx} - 1 = xe^{y-x}$.
Since $y-x = u$,then $\frac{du}{dx} = \frac{dy}{dx} - 1$. Substituting this,we get $\frac{du}{dx} = xe^u$.
Separating variables: $e^{-u} du = x dx$.
Integrating both sides: $\int e^{-u} du = \int x dx \Rightarrow -e^{-u} = \frac{x^2}{2} + C$.
Substituting $u = y-x$: $-e^{-(y-x)} = \frac{x^2}{2} + C \Rightarrow -e^{x-y} = \frac{x^2}{2} + C$.
Given $y(0)=0$,at $x=0, y=0$: $-e^0 = 0 + C \Rightarrow C = -1$.
So,$-e^{x-y} = \frac{x^2}{2} - 1 \Rightarrow e^{x-y} = 1 - \frac{x^2}{2} = \frac{2-x^2}{2}$.
Taking natural log: $x-y = \ln\left(\frac{2-x^2}{2}\right) \Rightarrow y = x - \ln\left(\frac{2-x^2}{2}\right)$.
To find the minimum,set $\frac{dy}{dx} = 0$: $1 + xe^{y-x} = 0$. From the original equation,$\frac{dy}{dx} = 1 + x\left(\frac{2}{2-x^2}\right) = \frac{2-x^2+2x}{2-x^2} = 0$.
Solving $-x^2+2x+2=0 \Rightarrow x^2-2x-2=0$. Roots are $x = \frac{2 \pm \sqrt{4+8}}{2} = 1 \pm \sqrt{3}$.
Since $x \in(-\sqrt{2}, \sqrt{2})$,we take $x = 1-\sqrt{3}$.
Substituting $x = 1-\sqrt{3}$ into $y(x)$: $y = (1-\sqrt{3}) - \ln\left(\frac{2-(1-\sqrt{3})^2}{2}\right) = (1-\sqrt{3}) - \ln\left(\frac{2-(1+3-2\sqrt{3})}{2}\right) = (1-\sqrt{3}) - \ln\left(\frac{2-4+2\sqrt{3}}{2}\right) = (1-\sqrt{3}) - \ln(\sqrt{3}-1)$.

(A) Given vectors are $\vec{p}=2 \hat{i}+3 \hat{j}+\hat{k}$ and $\vec{q}=\hat{i}+2 \hat{j}+\hat{k}$.
First,calculate the sum and difference of the vectors:
$\vec{p}+\vec{q} = (2+1)\hat{i} + (3+2)\hat{j} + (1+1)\hat{k} = 3\hat{i} + 5\hat{j} + 2\hat{k}$
$\vec{p}-\vec{q} = (2-1)\hat{i} + (3-2)\hat{j} + (1-1)\hat{k} = \hat{i} + \hat{j} + 0\hat{k}$
Since $\vec{r}$ is perpendicular to both $(\vec{p}+\vec{q})$ and $(\vec{p}-\vec{q})$,$\vec{r}$ must be parallel to their cross product:
$(\vec{p}+\vec{q}) \times (\vec{p}-\vec{q}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 5 & 2 \\ 1 & 1 & 0 \end{vmatrix} = \hat{i}(0-2) - \hat{j}(0-2) + \hat{k}(3-5) = -2\hat{i} + 2\hat{j} - 2\hat{k}$.
Let $\vec{v} = -2\hat{i} + 2\hat{j} - 2\hat{k}$. The magnitude is $|\vec{v}| = \sqrt{(-2)^2 + 2^2 + (-2)^2} = \sqrt{4+4+4} = \sqrt{12} = 2\sqrt{3}$.
The vector $\vec{r}$ is given by $\vec{r} = \pm |\vec{r}| \frac{\vec{v}}{|\vec{v}|} = \pm \sqrt{3} \frac{-2\hat{i} + 2\hat{j} - 2\hat{k}}{2\sqrt{3}} = \pm (-\hat{i} + \hat{j} - \hat{k})$.
Thus,$\vec{r} = \mp \hat{i} \pm \hat{j} \mp \hat{k}$.
Comparing with $\vec{r} = \alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}$,we get $|\alpha|=1, |\beta|=1, |\gamma|=1$.
Therefore,$|\alpha|+|\beta|+|\gamma| = 1+1+1 = 3$.

(A) We are given $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ where $a, b, c, d \in \{\pm 3, \pm 2, \pm 1, 0\}$.
We need to find the number of matrices such that $\det(A) = ad - bc = 15$.
Since the maximum value of $ad$ is $3 \times 3 = 9$ and the minimum value of $bc$ is $-3 \times 3 = -9$,the maximum value of $ad - bc$ is $9 - (-9) = 18$.
Possible values for $ad$ and $bc$ such that $ad - bc = 15$:
Case $I$: $ad = 9$ and $bc = -6$.
For $ad = 9$,the pairs $(a, d)$ can be $(3, 3)$ or $(-3, -3)$ ($2$ pairs).
For $bc = -6$,the pairs $(b, c)$ can be $(3, -2), (-3, 2), (-2, 3), (2, -3)$ ($4$ pairs).
Total matrices for Case $I = 2 \times 4 = 8$.
Case $II$: $ad = 6$ and $bc = -9$.
For $ad = 6$,the pairs $(a, d)$ can be $(3, 2), (2, 3), (-3, -2), (-2, -3)$ ($4$ pairs).
For $bc = -9$,the pairs $(b, c)$ can be $(3, -3), (-3, 3)$ ($2$ pairs).
Total matrices for Case $II = 4 \times 2 = 8$.
Total number of such matrices $= 8 + 8 = 16$.

(C) Given the differential equation: $e^{y} \frac{d y}{d x}-2 e^{y} \sin x=-\sin x \cos ^{2} x$.
Let $e^{y}=t$,then $e^{y} \frac{d y}{d x}=\frac{d t}{d x}$.
The equation becomes $\frac{d t}{d x}-2 \sin x \cdot t=-\sin x \cos ^{2} x$.
This is a linear differential equation of the form $\frac{d t}{d x}+P(x)t=Q(x)$,where $P(x)=-2 \sin x$ and $Q(x)=-\sin x \cos ^{2} x$.
The integrating factor $I.F. = e^{\int -2 \sin x dx} = e^{2 \cos x}$.
The solution is $t \cdot e^{2 \cos x} = \int -\sin x \cos ^{2} x \cdot e^{2 \cos x} dx + C$.
Let $u = \cos x$,then $du = -\sin x dx$. The integral becomes $\int u^{2} e^{2u} du$.
Using integration by parts: $\int u^{2} e^{2u} du = u^{2} \frac{e^{2u}}{2} - \int 2u \frac{e^{2u}}{2} du = \frac{u^{2} e^{2u}}{2} - (u \frac{e^{2u}}{2} - \int \frac{e^{2u}}{2} du) = \frac{u^{2} e^{2u}}{2} - \frac{u e^{2u}}{2} + \frac{e^{2u}}{4}$.
So,$e^{y} e^{2 \cos x} = e^{2 \cos x} (\frac{\cos^{2} x}{2} - \frac{\cos x}{2} + \frac{1}{4}) + C$.
At $x = \frac{\pi}{2}$,$y = 0$,so $e^{0} e^{0} = e^{0} (0 - 0 + \frac{1}{4}) + C \Rightarrow 1 = \frac{1}{4} + C \Rightarrow C = \frac{3}{4}$.
Thus,$e^{y} = \frac{\cos^{2} x}{2} - \frac{\cos x}{2} + \frac{1}{4} + \frac{3}{4} e^{-2 \cos x}$.
At $x = 0$,$e^{y(0)} = \frac{1}{2} - \frac{1}{2} + \frac{1}{4} + \frac{3}{4} e^{-2} = \frac{1}{4} + \frac{3}{4} e^{-2}$.
Comparing with $\alpha + \beta e^{-2}$,we get $\alpha = \frac{1}{4}$ and $\beta = \frac{3}{4}$.
Therefore,$4(\alpha + \beta) = 4(\frac{1}{4} + \frac{3}{4}) = 4(1) = 4$.

(A) Let $A = \begin{bmatrix} 0 & i \\ 1 & 0 \end{bmatrix}$ and $X = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$.
The given condition is $A^n X = X$ for all $a, b, c, d \in R$.
Since $X$ can be any $2 \times 2$ matrix,we can choose $X = I$ (the identity matrix),which implies $A^n = I$.
Now,calculate powers of $A$:
$A^2 = \begin{bmatrix} 0 & i \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & i \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} i & 0 \\ 0 & i \end{bmatrix} = iI$.
$A^4 = (A^2)^2 = (iI)^2 = i^2 I = -I$.
$A^8 = (A^4)^2 = (-I)^2 = I$.
Thus,$A^n = I$ if and only if $n$ is a multiple of $8$.
We need to find the number of $2$-digit numbers $n$ such that $n$ is a multiple of $8$.
The $2$-digit multiples of $8$ are $16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96$.
Counting these,we have $11$ such numbers.

(B) Given the differential equation: $x dy = (y + x^3 \cos x) dx$
Divide by $x^2$ (assuming $x \neq 0$):
$\frac{x dy - y dx}{x^2} = x \cos x dx$
This is the derivative of the quotient:
$d(\frac{y}{x}) = x \cos x dx$
Integrating both sides:
$\int d(\frac{y}{x}) = \int x \cos x dx$
Using integration by parts for the right side:
$\frac{y}{x} = x \sin x - \int \sin x dx$
$\frac{y}{x} = x \sin x + \cos x + C$
Given $y(\pi) = 0$,substitute $x = \pi$ and $y = 0$:
$0 = \pi \sin(\pi) + \cos(\pi) + C$
$0 = 0 - 1 + C \implies C = 1$
So,$\frac{y}{x} = x \sin x + \cos x + 1$
$y = x^2 \sin x + x \cos x + x$
Now,calculate $y(\frac{\pi}{2})$:
$y(\frac{\pi}{2}) = (\frac{\pi}{2})^2 \sin(\frac{\pi}{2}) + \frac{\pi}{2} \cos(\frac{\pi}{2}) + \frac{\pi}{2}$
$y(\frac{\pi}{2}) = \frac{\pi^2}{4}(1) + \frac{\pi}{2}(0) + \frac{\pi}{2} = \frac{\pi^2}{4} + \frac{\pi}{2}$

(A) Given the determinant equation: $\left|\begin{array}{lll} \sin x & \cos x & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x \end{array}\right|=0$.
Applying row operations $R_{1} \rightarrow R_{1}-R_{2}$ and $R_{2} \rightarrow R_{2}-R_{3}$,we get:
$\left|\begin{array}{ccc} \sin x-\cos x & \cos x-\sin x & 0 \\ 0 & \sin x-\cos x & \cos x-\sin x \\ \cos x & \cos x & \sin x \end{array}\right|=0$.
Taking $(\sin x-\cos x)$ common from $R_{1}$ and $R_{2}$:
$(\sin x-\cos x)^{2} \left|\begin{array}{ccc} 1 & -1 & 0 \\ 0 & 1 & -1 \\ \cos x & \cos x & \sin x \end{array}\right|=0$.
Expanding the determinant:
$(\sin x-\cos x)^{2} [1(\sin x + \cos x) + 1(0 + \cos x)] = 0$.
$(\sin x-\cos x)^{2} (\sin x + 2 \cos x) = 0$.
This implies $\sin x = \cos x$ or $\sin x = -2 \cos x$.
Case $1$: $\tan x = 1 \implies x = \frac{\pi}{4}$. This value lies in the interval $[-\frac{\pi}{4}, \frac{\pi}{4}]$.
Case $2$: $\tan x = -2$. In the interval $[-\frac{\pi}{4}, \frac{\pi}{4}]$,$\tan x$ ranges from $-1$ to $1$. Since $-2$ is outside this range,there is no solution for $\tan x = -2$ in this interval.
Thus,there is only $1$ distinct real root,which is $x = \frac{\pi}{4}$.

(C) Given that $(g \circ f)^{-1}$ exists,the composite function $g \circ f: A \rightarrow C$ must be a bijection (both one-one and onto).
$1$. For $g \circ f$ to be one-one,$f$ must be one-one. If $f$ were not one-one,there would exist $x_1, x_2 \in A$ such that $f(x_1) = f(x_2)$,implying $g(f(x_1)) = g(f(x_2))$,which contradicts the one-one property of $g \circ f$.
$2$. For $g \circ f$ to be onto,$g$ must be onto. If $g$ were not onto,there would exist some $z \in C$ such that no $y \in B$ satisfies $g(y) = z$,meaning no $x \in A$ could satisfy $g(f(x)) = z$,contradicting the onto property of $g \circ f$.
Therefore,$f$ must be one-one and $g$ must be onto.

(B) Let $I = \int_{-1}^{1} f(x) \, dx$,where $f(x) = \log \left(x+\sqrt{x^{2}+1}\right)$.
Check if $f(x)$ is an odd function by evaluating $f(-x)$:
$f(-x) = \log \left(-x+\sqrt{(-x)^{2}+1}\right) = \log \left(\sqrt{x^{2}+1}-x\right)$.
Multiply and divide by $(\sqrt{x^{2}+1}+x)$:
$f(-x) = \log \left(\frac{(\sqrt{x^{2}+1}-x)(\sqrt{x^{2}+1}+x)}{\sqrt{x^{2}+1}+x}\right) = \log \left(\frac{x^{2}+1-x^{2}}{\sqrt{x^{2}+1}+x}\right) = \log \left(\frac{1}{\sqrt{x^{2}+1}+x}\right)$.
Using the property $\log(1/a) = -\log(a)$:
$f(-x) = -\log \left(x+\sqrt{x^{2}+1}\right) = -f(x)$.
Since $f(-x) = -f(x)$,the function is an odd function.
By the property of definite integrals,if $f(x)$ is an odd function,then $\int_{-a}^{a} f(x) \, dx = 0$.
Therefore,$\int_{-1}^{1} \log \left(x+\sqrt{x^{2}+1}\right) \, dx = 0$.

(B) Given $P = \begin{bmatrix} 1 & 0 \\ 1/2 & 1 \end{bmatrix}$.
Calculate $P^2 = P \times P = \begin{bmatrix} 1 & 0 \\ 1/2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 1/2 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 1/2 + 1/2 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$.
Calculate $P^3 = P^2 \times P = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 1/2 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 1 + 1/2 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 3/2 & 1 \end{bmatrix}$.
Calculate $P^4 = P^2 \times P^2 = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 1 + 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}$.
Observing the pattern,$P^n = \begin{bmatrix} 1 & 0 \\ n/2 & 1 \end{bmatrix}$.
Therefore,for $n = 50$,$P^{50} = \begin{bmatrix} 1 & 0 \\ 50/2 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 25 & 1 \end{bmatrix}$.

(A) Three vectors $\vec{u}, \vec{v}, \vec{w}$ are coplanar if their scalar triple product is zero,i.e.,$[\vec{u} \vec{v} \vec{w}] = 0$.
The given vectors are $\vec{u} = a \hat{i} + a \hat{j} + c \hat{k}$,$\vec{v} = 1 \hat{i} + 0 \hat{j} + 1 \hat{k}$,and $\vec{w} = c \hat{i} + c \hat{j} + b \hat{k}$.
The condition for coplanarity is given by the determinant:
$\left|\begin{array}{lll}a & a & c \\ 1 & 0 & 1 \\ c & c & b\end{array}\right| = 0$
Expanding the determinant along the second row:
$-1 \cdot \left|\begin{array}{ll}a & c \\ c & b\end{array}\right| + 0 \cdot \left|\begin{array}{ll}a & c \\ c & b\end{array}\right| - 1 \cdot \left|\begin{array}{ll}a & a \\ c & c\end{array}\right| = 0$
$-1(ab - c^2) - 1(ac - ac) = 0$
$-(ab - c^2) - 0 = 0$
$c^2 - ab = 0$
$c^2 = ab$
Since $a, b, c$ are positive numbers,$c = \sqrt{ab}$.

(D) The mean $E(X)$ is given by $\sum_{j=0}^{\infty} j \cdot P(X=j)$.
Since $P(X=0) = 0 \cdot \frac{1}{2} = 0$,we have $E(X) = \sum_{j=1}^{\infty} j \cdot \frac{1}{3^j}$.
This is an arithmetico-geometric series of the form $\sum_{j=1}^{\infty} j r^j$ where $r = \frac{1}{3}$.
The sum is given by $\frac{r}{(1-r)^2} = \frac{1/3}{(1-1/3)^2} = \frac{1/3}{4/9} = \frac{3}{4}$.
For $P(X \text{ is positive and even})$,we sum $P(X=j)$ for $j \in \{2, 4, 6, \ldots\}$.
$P(X \text{ is positive and even}) = \sum_{k=1}^{\infty} P(X=2k) = \sum_{k=1}^{\infty} \frac{1}{3^{2k}} = \sum_{k=1}^{\infty} \left(\frac{1}{9}\right)^k$.
This is a geometric series with first term $a = \frac{1}{9}$ and common ratio $r = \frac{1}{9}$.
The sum is $\frac{a}{1-r} = \frac{1/9}{1-1/9} = \frac{1/9}{8/9} = \frac{1}{8}$.
Thus,the mean is $\frac{3}{4}$ and the probability is $\frac{1}{8}$.

(C) Given: $|\vec{a}|=2$ and $|\vec{b}|=5$.
We know that $|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}| \sin \theta = 8$.
Substituting the values: $2 \times 5 \times \sin \theta = 8 \implies 10 \sin \theta = 8 \implies \sin \theta = \frac{8}{10} = \frac{4}{5}$.
Since $\sin^2 \theta + \cos^2 \theta = 1$,we have $\cos^2 \theta = 1 - \sin^2 \theta = 1 - (\frac{4}{5})^2 = 1 - \frac{16}{25} = \frac{9}{25}$.
Thus,$|\cos \theta| = \sqrt{\frac{9}{25}} = \frac{3}{5}$.
Now,$|\vec{a} \cdot \vec{b}| = |\vec{a}||\vec{b}| |\cos \theta|$.
$|\vec{a} \cdot \vec{b}| = 2 \times 5 \times \frac{3}{5} = 6$.

(B) For $x > 2$,$f(x) = \int_{0}^{1} (5 + (1-t)) \, dt + \int_{1}^{x} (5 + (t-1)) \, dt$.
Evaluating the integrals:
$f(x) = \int_{0}^{1} (6-t) \, dt + \int_{1}^{x} (4+t) \, dt = [6t - \frac{t^2}{2}]_{0}^{1} + [4t + \frac{t^2}{2}]_{1}^{x}$.
$f(x) = (6 - \frac{1}{2}) + (4x + \frac{x^2}{2} - 4 - \frac{1}{2}) = \frac{11}{2} + 4x + \frac{x^2}{2} - \frac{9}{2} = \frac{x^2}{2} + 4x + 1$.
Check continuity at $x=2$:
$f(2^-) = 5(2) + 1 = 11$.
$f(2^+) = \frac{2^2}{2} + 4(2) + 1 = 2 + 8 + 1 = 11$.
Since $f(2^-) = f(2^+) = 11$,$f(x)$ is continuous at $x=2$.
Check differentiability at $x=2$:
$Lf'(2) = \frac{d}{dx}(5x+1)|_{x=2} = 5$.
$Rf'(2) = \frac{d}{dx}(\frac{x^2}{2} + 4x + 1)|_{x=2} = (x+4)|_{x=2} = 2+4 = 6$.
Since $Lf'(2) \neq Rf'(2)$,$f(x)$ is not differentiable at $x=2$.

(C) Since $P''(x)$ is a constant,$P(x)$ must be a polynomial of degree $2$. Let $P(x) = ax^2 + bx + c$.
Given that $f(x)$ is continuous at $x = 2$,we have $\lim_{x \to 2} f(x) = f(2) = 7$.
This implies $\lim_{x \to 2} \frac{P(x)}{\sin(x-2)} = 7$.
For the limit to exist and be finite,$P(2)$ must be $0$ because $\sin(x-2) \to 0$ as $x \to 2$. Thus,$(x-2)$ is a factor of $P(x)$.
Let $P(x) = (x-2)(ax + k)$.
Then $\lim_{x \to 2} \frac{(x-2)(ax+k)}{\sin(x-2)} = \lim_{x \to 2} \frac{x-2}{\sin(x-2)} \cdot (ax+k) = 1 \cdot (2a+k) = 7$.
So,$2a + k = 7$.
We are given $P(3) = 9$. Substituting $x=3$ into $P(x) = (x-2)(ax+k)$ gives $(3-2)(3a+k) = 9$,so $3a + k = 9$.
Subtracting the two equations: $(3a+k) - (2a+k) = 9 - 7$,which gives $a = 2$.
Substituting $a=2$ into $2a+k=7$ gives $4+k=7$,so $k=3$.
Thus,$P(x) = (x-2)(2x+3)$.
Finally,$P(5) = (5-2)(2(5)+3) = 3(13) = 39$.

(C) Let the side length of the equilateral triangle be $a = 2 \sqrt{2}$.
Let the length of the rectangle be $\ell$ and its breadth be $b$.
From the similar triangles formed at the top,the ratio of the height of the small triangle to its base is equal to the ratio of the height of the large triangle to its base.
The height of the equilateral triangle is $H = \frac{\sqrt{3}}{2} a = \frac{\sqrt{3}}{2} (2 \sqrt{2}) = \sqrt{6}$.
By similar triangles,$\frac{H-b}{H} = \frac{\ell}{a}$.
$\frac{\sqrt{6}-b}{\sqrt{6}} = \frac{\ell}{2 \sqrt{2}}$.
$b = \sqrt{6} (1 - \frac{\ell}{2 \sqrt{2}}) = \sqrt{6} - \frac{\sqrt{6} \ell}{2 \sqrt{2}} = \sqrt{6} - \frac{\sqrt{3}}{2} \ell$.
The area $A = \ell \times b = \ell (\sqrt{6} - \frac{\sqrt{3}}{2} \ell) = \sqrt{6} \ell - \frac{\sqrt{3}}{2} \ell^2$.
To maximize $A$,set $\frac{dA}{d\ell} = 0$.
$\sqrt{6} - \sqrt{3} \ell = 0 \Rightarrow \ell = \frac{\sqrt{6}}{\sqrt{3}} = \sqrt{2}$.
The maximum area $A = \sqrt{2} (\sqrt{6} - \frac{\sqrt{3}}{2} \sqrt{2}) = \sqrt{12} - \frac{\sqrt{6}}{2} \sqrt{2} = 2 \sqrt{3} - \sqrt{3} = \sqrt{3}$.
The square of the largest area is $A^2 = (\sqrt{3})^2 = 3$.

(A) Given the differential equation: $\frac{dy}{dx} = \frac{2y}{x \ln x}$.
Separating the variables,we get: $\frac{dy}{y} = \frac{2 dx}{x \ln x}$.
Integrating both sides: $\int \frac{dy}{y} = \int \frac{2}{x \ln x} dx$.
Let $u = \ln x$,then $du = \frac{1}{x} dx$. The integral becomes: $\ln |y| = 2 \int \frac{du}{u} = 2 \ln |u| + C = 2 \ln |\ln x| + C$.
So,$\ln |y| = \ln |(\ln x)^2| + C$,which implies $y = k(\ln x)^2$ for some constant $k$.
Given that the curve passes through $(2, (\ln 2)^2)$,we substitute $x=2$ and $y=(\ln 2)^2$:
$(\ln 2)^2 = k(\ln 2)^2 \Rightarrow k = 1$.
Thus,the function is $f(x) = (\ln x)^2$.
To find $f(e)$,we substitute $x=e$: $f(e) = (\ln e)^2 = (1)^2 = 1$.

(B) The probability of getting a head in a single toss is $P(H) = \frac{1}{2}$.
The probability of getting no heads in $n$ tosses is $P(X=0) = \left(\frac{1}{2}\right)^n$.
The probability of getting at least one head is $P(X \geq 1) = 1 - P(X=0) = 1 - \left(\frac{1}{2}\right)^n$.
Given that $P(X \geq 1) \geq 0.9$,we have:
$1 - \left(\frac{1}{2}\right)^n \geq 0.9$
Rearranging the inequality:
$1 - 0.9 \geq \left(\frac{1}{2}\right)^n$
$0.1 \geq \left(\frac{1}{2}\right)^n$
$\frac{1}{10} \geq \frac{1}{2^n}$
$2^n \geq 10$
Testing values for $n$:
For $n=3$,$2^3 = 8 < 10$.
For $n=4$,$2^4 = 16 \geq 10$.
Thus,the minimum value of $n$ is $4$.

(C) Two lines $\frac{x-x_1}{a_1} = \frac{y-y_1}{b_1} = \frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2} = \frac{y-y_2}{b_2} = \frac{z-z_2}{c_2}$ are co-planar if the determinant of the vector connecting a point on each line and the direction vectors is zero:
$\left|\begin{array}{ccc} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{array}\right| = 0$
Here,$(x_1, y_1, z_1) = (k, 2, 3)$ and $(x_2, y_2, z_2) = (-1, -2, -3)$.
The direction vectors are $(a_1, b_1, c_1) = (1, 2, 3)$ and $(a_2, b_2, c_2) = (3, 2, 1)$.
Substituting these values into the determinant:
$\left|\begin{array}{ccc} -1-k & -2-2 & -3-3 \\ 1 & 2 & 3 \\ 3 & 2 & 1 \end{array}\right| = 0$
$\left|\begin{array}{ccc} -(k+1) & -4 & -6 \\ 1 & 2 & 3 \\ 3 & 2 & 1 \end{array}\right| = 0$
Expanding the determinant:
$-(k+1)(2-6) - (-4)(1-9) + (-6)(2-6) = 0$
$-(k+1)(-4) + 4(-8) - 6(-4) = 0$
$4(k+1) - 32 + 24 = 0$
$4k + 4 - 8 = 0$
$4k - 4 = 0$
$4k = 4$
$k = 1$

(B) Given $(\vec{a}+3 \vec{b}) \perp(7 \vec{a}-5 \vec{b})$,their dot product is zero:
$(\vec{a}+3 \vec{b}) \cdot(7 \vec{a}-5 \vec{b}) = 7|\vec{a}|^2 - 15|\vec{b}|^2 + 16(\vec{a} \cdot \vec{b}) = 0 \quad \dots(1)$
Given $(\vec{a}-4 \vec{b}) \perp(7 \vec{a}-2 \vec{b})$,their dot product is zero:
$(\vec{a}-4 \vec{b}) \cdot(7 \vec{a}-2 \vec{b}) = 7|\vec{a}|^2 + 8|\vec{b}|^2 - 30(\vec{a} \cdot \vec{b}) = 0 \quad \dots(2)$
Subtracting equation $(2)$ from equation $(1)$:
$(7|\vec{a}|^2 - 15|\vec{b}|^2 + 16(\vec{a} \cdot \vec{b})) - (7|\vec{a}|^2 + 8|\vec{b}|^2 - 30(\vec{a} \cdot \vec{b})) = 0$
$-23|\vec{b}|^2 + 46(\vec{a} \cdot \vec{b}) = 0 \implies 46(\vec{a} \cdot \vec{b}) = 23|\vec{b}|^2 \implies \vec{a} \cdot \vec{b} = \frac{1}{2}|\vec{b}|^2$
Substitute $\vec{a} \cdot \vec{b} = \frac{1}{2}|\vec{b}|^2$ into equation $(1)$:
$7|\vec{a}|^2 - 15|\vec{b}|^2 + 16(\frac{1}{2}|\vec{b}|^2) = 0$
$7|\vec{a}|^2 - 15|\vec{b}|^2 + 8|\vec{b}|^2 = 0 \implies 7|\vec{a}|^2 = 7|\vec{b}|^2 \implies |\vec{a}| = |\vec{b}|$
Now,$\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \frac{\frac{1}{2}|\vec{b}|^2}{|\vec{b}||\vec{b}|} = \frac{1}{2}$
$\theta = \cos^{-1}(\frac{1}{2}) = 60^{\circ}$

(D) The region $R$ is defined by $\frac{1}{2} \leq x \leq 2$ and $0 \leq y \leq 2^x$ for $x \in [\frac{1}{2}, 1]$,and $\log_e x \leq y \leq 2^x$ for $x \in [1, 2]$.
Area $A = \int_{1/2}^{1} 2^x \, dx + \int_{1}^{2} (2^x - \log_e x) \, dx$
$A = \int_{1/2}^{2} 2^x \, dx - \int_{1}^{2} \log_e x \, dx$
$A = \left[ \frac{2^x}{\log_e 2} \right]_{1/2}^{2} - [x \log_e x - x]_{1}^{2}$
$A = \frac{2^2 - 2^{1/2}}{\log_e 2} - ((2 \log_e 2 - 2) - (1 \log_e 1 - 1))$
$A = \frac{4 - \sqrt{2}}{\log_e 2} - (2 \log_e 2 - 2 + 1)$
$A = (4 - \sqrt{2})(\log_e 2)^{-1} - 2(\log_e 2) + 1$
Comparing with $\alpha(\log_e 2)^{-1} + \beta(\log_e 2) + \gamma$,we get $\alpha = 4 - \sqrt{2}$,$\beta = -2$,$\gamma = 1$.
Now,$(\alpha + \beta - 2\gamma)^2 = (4 - \sqrt{2} - 2 - 2(1))^2 = (4 - \sqrt{2} - 2 - 2)^2 = (-\sqrt{2})^2 = 2$.

(B) Given the differential equation $\log _{e}\left(\frac{d y}{d x}\right)=3 x+4 y$,we can write it as $\frac{d y}{d x}=e^{3 x+4 y}=e^{3 x} \cdot e^{4 y}$.
Separating the variables,we get $e^{-4 y} d y=e^{3 x} d x$.
Integrating both sides,$\int e^{-4 y} d y=\int e^{3 x} d x$,which gives $-\frac{1}{4} e^{-4 y}=\frac{1}{3} e^{3 x}+C$.
Using the initial condition $y(0)=0$,we substitute $x=0$ and $y=0$: $-\frac{1}{4} e^{0}=\frac{1}{3} e^{0}+C \Rightarrow -\frac{1}{4}=\frac{1}{3}+C \Rightarrow C=-\frac{1}{4}-\frac{1}{3}=-\frac{7}{12}$.
Thus,$-\frac{1}{4} e^{-4 y}=\frac{1}{3} e^{3 x}-\frac{7}{12}$.
Multiplying by $-12$,we get $3 e^{-4 y} = 7 - 4 e^{3 x}$,so $e^{-4 y} = \frac{7 - 4 e^{3 x}}{3}$.
Taking the reciprocal,$e^{4 y} = \frac{3}{7 - 4 e^{3 x}}$,so $4 y = \log _{e} \left(\frac{3}{7 - 4 e^{3 x}}\right)$.
For $x = -\frac{2}{3} \log _{e} 2$,we have $e^{3 x} = e^{3 \left(-\frac{2}{3} \log _{e} 2\right)} = e^{-2 \log _{e} 2} = e^{\log _{e} (2^{-2})} = 2^{-2} = \frac{1}{4}$.
Substituting this into the equation for $4y$: $4 y = \log _{e} \left(\frac{3}{7 - 4(1/4)}\right) = \log _{e} \left(\frac{3}{7 - 1}\right) = \log _{e} \left(\frac{3}{6}\right) = \log _{e} \left(\frac{1}{2}\right) = -\log _{e} 2$.
Therefore,$y = -\frac{1}{4} \log _{e} 2$. Comparing this with $y = \alpha \log _{e} 2$,we get $\alpha = -\frac{1}{4}$.

(B) Given $A = \begin{bmatrix} 1 & 2 \\ -1 & 4 \end{bmatrix}$.
First,we find the determinant of $A$: $|A| = (1)(4) - (2)(-1) = 4 + 2 = 6$.
Next,we find the adjoint of $A$: $\text{adj}(A) = \begin{bmatrix} 4 & -2 \\ 1 & 1 \end{bmatrix}$.
Thus,$A^{-1} = \frac{1}{|A|} \text{adj}(A) = \frac{1}{6} \begin{bmatrix} 4 & -2 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} \frac{2}{3} & -\frac{1}{3} \\ \frac{1}{6} & \frac{1}{6} \end{bmatrix}$.
We are given $A^{-1} = \alpha I + \beta A$,so:
$\begin{bmatrix} \frac{2}{3} & -\frac{1}{3} \\ \frac{1}{6} & \frac{1}{6} \end{bmatrix} = \alpha \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \beta \begin{bmatrix} 1 & 2 \\ -1 & 4 \end{bmatrix} = \begin{bmatrix} \alpha + \beta & 2\beta \\ -\beta & \alpha + 4\beta \end{bmatrix}$.
Comparing the elements,we get:
$2\beta = -\frac{1}{3} \implies \beta = -\frac{1}{6}$.
$\alpha + \beta = \frac{2}{3} \implies \alpha - \frac{1}{6} = \frac{2}{3} \implies \alpha = \frac{2}{3} + \frac{1}{6} = \frac{5}{6}$.
Finally,calculate $4(\alpha - \beta) = 4(\frac{5}{6} - (-\frac{1}{6})) = 4(\frac{5}{6} + \frac{1}{6}) = 4(1) = 4$.

(B) For $f(x)$ to be continuous at $x=0$,we must have $\lim_{x \rightarrow 0^-} f(x) = \lim_{x \rightarrow 0^+} f(x) = f(0) = b$.
First,calculate the right-hand limit:
$\lim_{x \rightarrow 0^+} f(x) = \lim_{x \rightarrow 0^+} e^{\frac{\cot 4x}{\cot 2x}} = \lim_{x \rightarrow 0^+} e^{\frac{\tan 2x}{\tan 4x}} = \lim_{x \rightarrow 0^+} e^{\frac{\tan 2x}{2\tan 2x(1-\tan^2 2x)}} = e^{1/2}$.
Thus,$b = e^{1/2}$.
Next,calculate the left-hand limit:
$\lim_{x \rightarrow 0^-} f(x) = \lim_{x \rightarrow 0^-} (1+|\sin x|)^{\frac{3a}{|\sin x|}} = \lim_{t \rightarrow 0^+} (1+t)^{\frac{3a}{t}} = e^{3a}$ (where $t = |\sin x|$).
Equating the limits,$e^{3a} = e^{1/2}$,which implies $3a = 1/2$,so $a = 1/6$.
Therefore,$6a = 1$.
Finally,$6a + b^2 = 1 + (e^{1/2})^2 = 1 + e$.

(D) Given $\vec{a}=\hat{i}+\hat{j}+2 \hat{k}$ and $\vec{b}=-\hat{i}+2 \hat{j}+3 \hat{k}$.
First,calculate $\vec{a}+\vec{b} = (1-1)\hat{i} + (1+2)\hat{j} + (2+3)\hat{k} = 3\hat{j} + 5\hat{k}$.
Next,simplify the expression $E = ((\vec{a} \times((\vec{a}-\vec{b}) \times \vec{b})) \times \vec{b})$.
Using the property $(\vec{a}-\vec{b}) \times \vec{b} = \vec{a} \times \vec{b} - \vec{b} \times \vec{b} = \vec{a} \times \vec{b}$.
So,$E = ((\vec{a} \times (\vec{a} \times \vec{b})) \times \vec{b})$.
Using the vector triple product formula $\vec{u} \times (\vec{v} \times \vec{w}) = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{u} \cdot \vec{v})\vec{w}$,we have:
$\vec{a} \times (\vec{a} \times \vec{b}) = (\vec{a} \cdot \vec{b})\vec{a} - (\vec{a} \cdot \vec{a})\vec{b}$.
Then $E = ((\vec{a} \cdot \vec{b})\vec{a} - (\vec{a} \cdot \vec{a})\vec{b}) \times \vec{b} = (\vec{a} \cdot \vec{b})(\vec{a} \times \vec{b}) - (\vec{a} \cdot \vec{a})(\vec{b} \times \vec{b}) = (\vec{a} \cdot \vec{b})(\vec{a} \times \vec{b})$.
Calculate $\vec{a} \cdot \vec{b} = (1)(-1) + (1)(2) + (2)(3) = -1 + 2 + 6 = 7$.
Calculate $\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 2 \\ -1 & 2 & 3 \end{vmatrix} = \hat{i}(3-4) - \hat{j}(3+2) + \hat{k}(2+1) = -\hat{i} - 5\hat{j} + 3\hat{k}$.
Thus,$E = 7(-\hat{i} - 5\hat{j} + 3\hat{k})$.
Finally,calculate $(\vec{a}+\vec{b}) \times E = (3\hat{j} + 5\hat{k}) \times 7(-\hat{i} - 5\hat{j} + 3\hat{k}) = 7 \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 3 & 5 \\ -1 & -5 & 3 \end{vmatrix}$.
$= 7 [\hat{i}(9 - (-25)) - \hat{j}(0 - (-5)) + \hat{k}(0 - (-3))] = 7(34\hat{i} - 5\hat{j} + 3\hat{k})$.

(C) Let $I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{d x}{\left(1+e^{x \cos x}\right)\left(\sin ^{4} x+\cos ^{4} x\right)} \cdots(1)$
Using the property $\int_{a}^{b} f(x) d x = \int_{a}^{b} f(a+b-x) d x$,where $a = -\frac{\pi}{4}$ and $b = \frac{\pi}{4}$,we get $a+b = 0$. Thus,$f(x)$ becomes $f(-x)$:
$I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{d x}{\left(1+e^{-x \cos(-x)}\right)\left(\sin ^{4}(-x)+\cos ^{4}(-x)\right)} = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{d x}{\left(1+e^{-x \cos x}\right)\left(\sin ^{4} x+\cos ^{4} x\right)} \cdots(2)$
Adding $(1)$ and $(2)$:
$2I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \left( \frac{1}{1+e^{x \cos x}} + \frac{1}{1+e^{-x \cos x}} \right) \frac{d x}{\sin ^{4} x+\cos ^{4} x}$
Since $\frac{1}{1+e^{x \cos x}} + \frac{1}{1+e^{-x \cos x}} = \frac{1}{1+e^{x \cos x}} + \frac{e^{x \cos x}}{e^{x \cos x}+1} = 1$,we have:
$2I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{d x}{\sin ^{4} x+\cos ^{4} x} = 2 \int_{0}^{\frac{\pi}{4}} \frac{d x}{\sin ^{4} x+\cos ^{4} x}$
$I = \int_{0}^{\frac{\pi}{4}} \frac{d x}{\sin ^{4} x+\cos ^{4} x} = \int_{0}^{\frac{\pi}{4}} \frac{\sec^4 x}{\tan^4 x + 1} dx = \int_{0}^{\frac{\pi}{4}} \frac{(1+\tan^2 x) \sec^2 x}{\tan^4 x + 1} dx$
Let $\tan x = u$,then $\sec^2 x dx = du$. As $x \to 0, u \to 0$ and $x \to \frac{\pi}{4}, u \to 1$:
$I = \int_{0}^{1} \frac{1+u^2}{u^4+1} du = \int_{0}^{1} \frac{1+\frac{1}{u^2}}{u^2+\frac{1}{u^2}} du = \int_{0}^{1} \frac{1+\frac{1}{u^2}}{(u-\frac{1}{u})^2 + 2} du$
Let $u-\frac{1}{u} = t$,then $(1+\frac{1}{u^2}) du = dt$. As $u \to 0, t \to -\infty$ and $u \to 1, t \to 0$:
$I = \int_{-\infty}^{0} \frac{dt}{t^2+2} = \left[ \frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{t}{\sqrt{2}} \right) \right]_{-\infty}^{0} = \frac{1}{\sqrt{2}} (0 - (-\frac{\pi}{2})) = \frac{\pi}{2\sqrt{2}}$

(B) The normal vector $\vec{n}$ of the required plane is perpendicular to the normals of the given planes $\vec{n}_1 = 2\hat{i} + \hat{j} - \hat{k}$ and $\vec{n}_2 = \hat{i} - \hat{j} - \hat{k}$.
Thus,$\vec{n} = \vec{n}_1 \times \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -1 \\ 1 & -1 & -1 \end{vmatrix} = \hat{i}(-1 - 1) - \hat{j}(-2 + 1) + \hat{k}(-2 - 1) = -2\hat{i} + \hat{j} - 3\hat{k}$.
The equation of the plane passing through $(-1, 0, -2)$ with normal vector $\vec{n} = -2\hat{i} + \hat{j} - 3\hat{k}$ is:
$-2(x + 1) + 1(y - 0) - 3(z + 2) = 0$
$-2x - 2 + y - 3z - 6 = 0$
$-2x + y - 3z - 8 = 0$
Multiplying by $-1$,we get $2x - y + 3z + 8 = 0$.
Comparing this with $ax + by + cz + 8 = 0$,we have $a = 2$,$b = -1$,and $c = 3$.
Therefore,$a + b + c = 2 - 1 + 3 = 4$.

(C) We are given the limit $L = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{j=1}^{n} \frac{(2 j-1)+8 n}{(2 j-1)+4 n}$.
Dividing the numerator and denominator of the term inside the sum by $n$,we get:
$L = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{j=1}^{n} \frac{2(\frac{j}{n}) - \frac{1}{n} + 8}{2(\frac{j}{n}) - \frac{1}{n} + 4}$.
Using the definition of the definite integral as the limit of a Riemann sum,where $\frac{j}{n} \rightarrow x$ and $\frac{1}{n} \rightarrow dx$ as $n \rightarrow \infty$,the expression becomes:
$L = \int_{0}^{1} \frac{2x + 8}{2x + 4} dx$.
We can simplify the integrand as follows:
$\frac{2x + 8}{2x + 4} = \frac{(2x + 4) + 4}{2x + 4} = 1 + \frac{4}{2x + 4} = 1 + \frac{2}{x + 2}$.
Now,integrate with respect to $x$ from $0$ to $1$:
$L = \int_{0}^{1} (1 + \frac{2}{x + 2}) dx = [x + 2 \ln|x + 2|]_{0}^{1}$.
Evaluating the definite integral:
$L = (1 + 2 \ln(3)) - (0 + 2 \ln(2)) = 1 + 2(\ln(3) - \ln(2)) = 1 + 2 \ln(\frac{3}{2})$.
Thus,the correct option is $C$.

(A) Given the determinant $f(x) = \left| \begin{array}{ccc} \sin^2 x & -2 + \cos^2 x & \cos 2x \\ 2 + \sin^2 x & \cos^2 x & \cos 2x \\ \sin^2 x & \cos^2 x & 1 + \cos 2x \end{array} \right|$.
Applying row operations $R_1 \rightarrow R_1 - R_2$ and $R_2 \rightarrow R_2 - R_3$:
$f(x) = \left| \begin{array}{ccc} -2 & -2 & 0 \\ 2 & 0 & -1 \\ \sin^2 x & \cos^2 x & 1 + \cos 2x \end{array} \right|$.
Expanding along the first row:
$f(x) = -2(0 - (-\cos^2 x)) - (-2)(2(1 + \cos 2x) - (-\sin^2 x)) + 0$
$f(x) = -2 \cos^2 x + 2(2 + 2 \cos 2x + \sin^2 x)$
$f(x) = -2 \cos^2 x + 4 + 4 \cos 2x + 2 \sin^2 x$
$f(x) = 4 + 4 \cos 2x - 2(\cos^2 x - \sin^2 x)$
Since $\cos 2x = \cos^2 x - \sin^2 x$,we have:
$f(x) = 4 + 4 \cos 2x - 2 \cos 2x = 4 + 2 \cos 2x$.
For $x \in [0, \pi]$,the maximum value of $\cos 2x$ is $1$.
Therefore,$f(x)_{\text{max}} = 4 + 2(1) = 6$.

(B) Given $F(x) = e^{-x} \int_{3}^{x} (3t^{2}+2t+4F^{\prime}(t)) \,dt$. Note that $F(3) = 0$.
Multiply by $e^{x}$: $e^{x}F(x) = \int_{3}^{x} (3t^{2}+2t+4F^{\prime}(t)) \,dt$.
Differentiating both sides with respect to $x$ using the Leibniz rule:
$e^{x}F(x) + e^{x}F^{\prime}(x) = 3x^{2}+2x+4F^{\prime}(x)$.
Rearranging terms: $(e^{x}-4)F^{\prime}(x) + e^{x}F(x) = 3x^{2}+2x$.
This is a linear differential equation of the form $\frac{d}{dx} [F(x)(e^{x}-4)] = 3x^{2}+2x$.
Integrating both sides with respect to $x$: $F(x)(e^{x}-4) = \int (3x^{2}+2x) \,dx = x^{3}+x^{2}+C$.
Since $F(3) = 0$,we have $0 = 3^{3}+3^{2}+C \Rightarrow C = -36$.
Thus,$F(x) = \frac{x^{3}+x^{2}-36}{e^{x}-4}$.
Now,$F^{\prime}(x) = \frac{(3x^{2}+2x)(e^{x}-4) - (x^{3}+x^{2}-36)e^{x}}{(e^{x}-4)^{2}}$.
At $x=4$: $F^{\prime}(4) = \frac{(3(16)+2(4))(e^{4}-4) - (64+16-36)e^{4}}{(e^{4}-4)^{2}} = \frac{56(e^{4}-4) - 44e^{4}}{(e^{4}-4)^{2}} = \frac{12e^{4}-224}{(e^{4}-4)^{2}}$.
Comparing with $\frac{\alpha e^{\beta}-224}{(e^{\beta}-4)^{2}}$,we get $\alpha=12$ and $\beta=4$.
Therefore,$\alpha+\beta = 12+4 = 16$.

(C) The line passes through point $A(2, 3, -2)$ and has direction vector $\vec{v} = -3\hat{i} + 2\hat{j} + \hat{k}$.
The plane passes through point $B(3, 7, -7)$ and point $A(2, 3, -2)$.
The vector $\vec{AB} = (3-2)\hat{i} + (7-3)\hat{j} + (-7 - (-2))\hat{k} = \hat{i} + 4\hat{j} - 5\hat{k}$.
The normal vector $\vec{n}$ to the plane is given by the cross product of $\vec{v}$ and $\vec{AB}$:
$\vec{n} = \vec{v} \times \vec{AB} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3 & 2 & 1 \\ 1 & 4 & -5 \end{vmatrix} = \hat{i}(-10-4) - \hat{j}(15-1) + \hat{k}(-12-2) = -14\hat{i} - 14\hat{j} - 14\hat{k}$.
We can take the normal vector as $\vec{n} = \hat{i} + \hat{j} + \hat{k}$.
The equation of the plane is $1(x-2) + 1(y-3) + 1(z+2) = 0$,which simplifies to $x + y + z - 3 = 0$.
The distance $d$ from the origin $(0, 0, 0)$ to the plane $x + y + z - 3 = 0$ is $d = \frac{|0+0+0-3|}{\sqrt{1^2+1^2+1^2}} = \frac{3}{\sqrt{3}} = \sqrt{3}$.
Therefore,$d^{2} = (\sqrt{3})^{2} = 3$.

(D) Given $f(m \cdot n) = f(m) \cdot f(n)$ for all $m, n \in S$ such that $m \cdot n \in S$.
$1$. For $m=1$,$f(n) = f(1) \cdot f(n)$,which implies $f(1) = 1$.
$2$. The values of $f(2), f(3), f(5), f(7)$ determine the function because $f(4) = f(2 \cdot 2) = f(2)^2$ and $f(6) = f(2 \cdot 3) = f(2) \cdot f(3)$.
$3$. Case $1$: $f(2) = 1$. Then $f(4) = 1^2 = 1$ and $f(6) = 1 \cdot f(3) = f(3)$.
$f(3)$ can be any value in $S$ ($7$ choices),$f(5)$ can be any value in $S$ ($7$ choices),and $f(7)$ can be any value in $S$ ($7$ choices).
Number of functions = $1 \times 1 \times 7 \times 1 \times 7 \times 7 = 343$.
$4$. Case $2$: $f(2) = 2$. Then $f(4) = 2^2 = 4 \in S$ (valid) and $f(6) = 2 \cdot f(3)$.
For $f(6) \in S$,$2 \cdot f(3) \in \{1, 2, 3, 4, 5, 6, 7\}$.
Possible values for $f(3)$ are $1, 2, 3$ (since $2 \cdot 1=2, 2 \cdot 2=4, 2 \cdot 3=6$,all $\in S$).
$f(5)$ can be any value in $S$ ($7$ choices),$f(7)$ can be any value in $S$ ($7$ choices).
Number of functions = $1 \times 3 \times 7 \times 7 = 147$.
$5$. Case $3$: $f(2) = 3$. Then $f(4) = 3^2 = 9 \notin S$ (invalid).
Total number of functions = $343 + 147 = 490$.

(A) Given $f(x) = \min \{x-[x], 1+[x]-x\}$. Let ${x} = x-[x]$. Then $f(x) = \min \{\{x\}, 1-\{x\}\}$.
For $x \in [0, 3)$,the function ${x}$ is periodic with period $1$. The graph of $f(x)$ consists of triangular waves between $0$ and $3$.
Specifically,$f(x) = \{x\}$ when $0 \le \{x\} \le 1/2$ and $f(x) = 1-\{x\}$ when $1/2 < \{x\} < 1$.
$1$. Continuity: The function $f(x)$ is continuous everywhere on $[0, 3]$ because ${x}$ is continuous except at integers,but the $\min$ function makes it continuous at integers as well. Thus,$P = \emptyset$ and the number of elements in $P$ is $0$.
$2$. Differentiability: The function $f(x)$ is not differentiable at points where the two expressions are equal,i.e.,${x} = 1/2$,and at points where ${x}$ is discontinuous,i.e.,$x \in \{1, 2\}$.
In the interval $(0, 3)$,${x} = 1/2$ occurs at $x \in \{1/2, 3/2, 5/2\}$.
The points where $f$ is not differentiable are $Q = \{1/2, 1, 3/2, 2, 5/2\}$.
The number of elements in $Q$ is $5$.
Sum of elements in $P$ and $Q = 0 + 5 = 5$.

(B) For a system of linear equations to have infinite solutions,the determinant of the coefficient matrix $\Delta$ must be $0$,and the determinants $\Delta_1, \Delta_2, \Delta_3$ must also be $0$.
First,calculate $\Delta$:
$\Delta = \begin{vmatrix} 1 & 1 & -1 \\ 1 & 2 & \alpha \\ 2 & -1 & 1 \end{vmatrix} = 1(2 + \alpha) - 1(1 - 2\alpha) - 1(-1 - 4) = 2 + \alpha - 1 + 2\alpha + 5 = 3\alpha + 6$.
Setting $\Delta = 0$,we get $3\alpha + 6 = 0$,which implies $\alpha = -2$.
Now,substitute $\alpha = -2$ into the system:
$x + y - z = 2$
$x + 2y - 2z = 1$
$2x - y + z = \beta$
For infinite solutions,the augmented matrix must have a rank less than $3$. We can use Cramer's rule or row reduction. Using $\Delta_2 = 0$:
$\Delta_2 = \begin{vmatrix} 1 & 2 & -1 \\ 1 & 1 & -2 \\ 2 & \beta & 1 \end{vmatrix} = 1(1 + 2\beta) - 2(1 + 4) - 1(\beta - 2) = 1 + 2\beta - 10 - \beta + 2 = \beta - 7$.
Setting $\Delta_2 = 0$,we get $\beta = 7$.
Thus,$\alpha + \beta = -2 + 7 = 5$.

(C) For the function to be defined,the arguments of the logarithms must be positive:
$\log_{5}(\log_{3}(18x - x^{2} - 77)) > 0 \implies \log_{3}(18x - x^{2} - 77) > 1 \implies 18x - x^{2} - 77 > 3$
$x^{2} - 18x + 80 < 0 \implies (x - 8)(x - 10) < 0 \implies x \in (8, 10)$.
Thus,$a = 8$ and $b = 10$.
Let $I = \int_{a}^{b} \frac{\sin^{3} x}{\sin^{3} x + \sin^{3}(a + b - x)} dx$.
Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a + b - x) dx$,we get:
$I = \int_{a}^{b} \frac{\sin^{3}(a + b - x)}{\sin^{3}(a + b - x) + \sin^{3} x} dx$.
Adding the two expressions for $I$:
$2I = \int_{a}^{b} \frac{\sin^{3} x + \sin^{3}(a + b - x)}{\sin^{3} x + \sin^{3}(a + b - x)} dx = \int_{a}^{b} 1 dx = b - a$.
$I = \frac{b - a}{2} = \frac{10 - 8}{2} = 1$.

(B) Given the differential equation: $\sec y \frac{dy}{dx} - (\sin(x+y) + \sin(x-y)) = 0$.
Using the identity $\sin(A+B) + \sin(A-B) = 2 \sin A \cos B$,we get:
$\sec y \frac{dy}{dx} - 2 \sin x \cos y = 0$.
$\sec y \frac{dy}{dx} = 2 \sin x \cos y$.
Dividing by $\cos y$ (since $\cos y \neq 0$ for $y \in [0, \frac{\pi}{2})$):
$\sec^2 y \frac{dy}{dx} = 2 \sin x$.
Integrating both sides with respect to $x$:
$\int \sec^2 y dy = \int 2 \sin x dx$.
$\tan y = -2 \cos x + C$.
Given $y(0) = 0$,we substitute $x=0$ and $y=0$:
$\tan(0) = -2 \cos(0) + C \Rightarrow 0 = -2(1) + C \Rightarrow C = 2$.
So,$\tan y = 2 - 2 \cos x$.
Differentiating with respect to $x$:
$\sec^2 y \frac{dy}{dx} = 2 \sin x$.
At $x = \frac{\pi}{2}$,$\tan y = 2 - 2 \cos(\frac{\pi}{2}) = 2 - 0 = 2$.
Since $\tan y = 2$,$\sec^2 y = 1 + \tan^2 y = 1 + 2^2 = 5$.
Substituting into the derivative equation:
$5 \frac{dy}{dx} = 2 \sin(\frac{\pi}{2}) = 2(1) = 2$.
Thus,$5y'(\frac{\pi}{2}) = 2$.

(C) Given $\vec{a} = \hat{i} + \hat{j} + \hat{k}$,$\vec{c} = \hat{j} - \hat{k}$,$\vec{a} \times \vec{b} = \vec{c}$,and $\vec{a} \cdot \vec{b} = 1$.
First,calculate $\vec{a} \times \vec{c} = (\hat{i} + \hat{j} + \hat{k}) \times (\hat{j} - \hat{k}) = \hat{i}(-1-1) - \hat{j}(-1-0) + \hat{k}(1-0) = -2\hat{i} + \hat{j} + \hat{k}$.
The magnitude $|\vec{a} \times \vec{c}| = \sqrt{(-2)^{2} + 1^{2} + 1^{2}} = \sqrt{4 + 1 + 1} = \sqrt{6}$.
The projection length $l$ of $\vec{b}$ on $\vec{a} \times \vec{c}$ is given by $l = \frac{|\vec{b} \cdot (\vec{a} \times \vec{c})|}{|\vec{a} \times \vec{c}|}$.
Using the scalar triple product property,$\vec{b} \cdot (\vec{a} \times \vec{c}) = -\vec{a} \cdot (\vec{b} \times \vec{c})$. Alternatively,since $\vec{a} \times \vec{b} = \vec{c}$,we have $\vec{b} \cdot (\vec{a} \times \vec{c}) = \vec{b} \cdot (\vec{a} \times (\vec{a} \times \vec{b})) = \vec{b} \cdot ((\vec{a} \cdot \vec{b})\vec{a} - (\vec{a} \cdot \vec{a})\vec{b}) = (\vec{a} \cdot \vec{b})(\vec{b} \cdot \vec{a}) - |\vec{a}|^{2}|\vec{b}|^{2}$.
Easier approach: $(\vec{a} \times \vec{b}) \cdot \vec{c} = \vec{c} \cdot \vec{c} = |\vec{c}|^{2} = 0^{2} + 1^{2} + (-1)^{2} = 2$.
Since $\vec{b} \cdot (\vec{a} \times \vec{c}) = -\vec{c} \cdot (\vec{a} \times \vec{b}) = -|\vec{c}|^{2} = -2$,we have $l = \frac{|-2|}{\sqrt{6}} = \frac{2}{\sqrt{6}}$.
Thus,$l^{2} = \frac{4}{6} = \frac{2}{3}$.
Therefore,$3l^{2} = 3 \times \frac{2}{3} = 2$.

(C) Let the first line be $\frac{x-\alpha}{1}=\frac{y-1}{2}=\frac{z-1}{3} = \phi$. Then any point on this line is $(\phi+\alpha, 2\phi+1, 3\phi+1)$.
Let the second line be $\frac{x-4}{\beta}=\frac{y-6}{3}=\frac{z-7}{3} = q$. Then any point on this line is $(q\beta+4, 3q+6, 3q+7)$.
For the lines to intersect,there must exist $\phi$ and $q$ such that:
$\phi+\alpha = q\beta+4$ $(i)$
$2\phi+1 = 3q+6$ (ii)
$3\phi+1 = 3q+7$ (iii)
Subtracting (ii) from (iii),we get $\phi = 1$. Substituting $\phi=1$ into (ii),we get $2(1)+1 = 3q+6$,which implies $3q = -3$,so $q = -1$.
Substituting $\phi=1$ and $q=-1$ into $(i)$,we get $1+\alpha = -\beta+4$,which simplifies to $\alpha+\beta = 3$.
The point of intersection is $(\phi+\alpha, 2\phi+1, 3\phi+1) = (1+\alpha, 3, 4)$.
Since this point lies on the plane $x+2y-z=8$,we have $(1+\alpha) + 2(3) - 4 = 8$.
$1+\alpha+6-4 = 8 \implies \alpha+3 = 8 \implies \alpha = 5$.
Since $\alpha+\beta = 3$,we have $5+\beta = 3 \implies \beta = -2$.
Thus,$\alpha-\beta = 5 - (-2) = 7$.

(A) The given functional equation is $f(x+y)+f(x-y)=2 f(x) f(y)$.
This is a well-known Cauchy-type functional equation whose solution is $f(x) = \cos(ax)$.
Given $f\left(\frac{1}{2}\right) = -1$,we have $\cos\left(\frac{a}{2}\right) = -1$,which implies $\frac{a}{2} = (2n+1)\pi$,so $a = 2(2n+1)\pi$.
For any integer $k$,$f(k) = \cos(2(2n+1)\pi k) = \cos(2m\pi) = 1$ where $m$ is an integer.
Thus,the expression becomes $\sum_{k=1}^{20} \frac{1}{\sin(k) \sin(k+1)}$.
Using the identity $\frac{1}{\sin(k) \sin(k+1)} = \frac{1}{\sin(1)} \left( \frac{\sin((k+1)-k)}{\sin(k) \sin(k+1)} \right) = \frac{1}{\sin(1)} (\cot(k) - \cot(k+1))$.
Summing from $k=1$ to $20$,we get $\frac{1}{\sin(1)} (\cot(1) - \cot(21))$.
$= \frac{1}{\sin(1)} \left( \frac{\cos(1)}{\sin(1)} - \frac{\cos(21)}{\sin(21)} \right) = \frac{1}{\sin(1)} \left( \frac{\cos(1)\sin(21) - \sin(1)\cos(21)}{\sin(1)\sin(21)} \right)$.
$= \frac{\sin(21-1)}{\sin^2(1) \sin(21)} = \frac{\sin(20)}{\sin^2(1) \sin(21)} = \operatorname{cosec}^2(1) \operatorname{cosec}(21) \sin(20)$.