JEE Main 2021 Mathematics Question Paper with Answer and Solution

(A) The given circle equation is $36 x^{2}+36 y^{2}-108 x+120 y+C=0$.
Dividing by $36$,we get $x^{2}+y^{2}-3 x+\frac{10}{3} y+\frac{C}{36}=0$.
The center of the circle is $(h, k) = (\frac{3}{2}, -\frac{5}{3})$.
The radius $r$ is given by $r = \sqrt{h^{2}+k^{2}-\frac{C}{36}} = \sqrt{\frac{9}{4}+\frac{25}{9}-\frac{C}{36}}$.
Since the circle does not intersect or touch the coordinate axes,the distance from the center to the axes must be greater than the radius.
$|h| > r$ $\Rightarrow \frac{3}{2} > r$ $\Rightarrow \frac{9}{4} > \frac{9}{4}+\frac{25}{9}-\frac{C}{36}$ $\Rightarrow \frac{C}{36} > \frac{25}{9}$ $\Rightarrow C > 100$.
$|k| > r$ $\Rightarrow \frac{5}{3} > r$ $\Rightarrow \frac{25}{9} > \frac{9}{4}+\frac{25}{9}-\frac{C}{36}$ $\Rightarrow \frac{C}{36} > \frac{9}{4}$ $\Rightarrow C > 81$.
Combining these,we get $C > 100$.
Now,the intersection of $x-2 y=4$ and $2 x-y=5$ is found by solving the system: $x=2y+4$ $\Rightarrow 2(2y+4)-y=5$ $\Rightarrow 3y=-3$ $\Rightarrow y=-1, x=2$.
The point $(2, -1)$ lies inside the circle $S$,so $S(2, -1) < 0$.
$x^{2}+y^{2}-3 x+\frac{10}{3} y+\frac{C}{36} < 0 \Rightarrow 4+1-3(2)+\frac{10}{3}(-1)+\frac{C}{36} < 0$.
$5-6-\frac{10}{3}+\frac{C}{36} < 0$ $\Rightarrow -1-\frac{10}{3}+\frac{C}{36} < 0$ $\Rightarrow \frac{C}{36} < \frac{13}{3}$ $\Rightarrow C < 156$.
Thus,$100 < C < 156$.

(A) Given $\sin^{7} x + \cos^{7} x = 1$ for $x \in [0, 4\pi]$.
Since $\sin^{2} x \leq 1$ and $\cos^{2} x \leq 1$,for $x \in [0, \pi/2]$,we have $\sin^{7} x \leq \sin^{2} x$ and $\cos^{7} x \leq \cos^{2} x$.
Adding these,$\sin^{7} x + \cos^{7} x \leq \sin^{2} x + \cos^{2} x = 1$.
The equality holds only when $\sin^{7} x = \sin^{2} x$ and $\cos^{7} x = \cos^{2} x$.
This implies $(\sin x = 0 \text{ or } \sin x = 1)$ and $(\cos x = 0 \text{ or } \cos x = 1)$.
Case $1$: $\sin x = 0 \implies x = 0, \pi, 2\pi, 3\pi, 4\pi$. Checking $\cos^{7} x = 1$,we get $x = 0, 2\pi, 4\pi$.
Case $2$: $\cos x = 0 \implies x = \pi/2, 3\pi/2, 5\pi/2, 7\pi/2$. Checking $\sin^{7} x = 1$,we get $x = \pi/2, 5\pi/2$.
Combining these,the solutions are $x \in \{0, \pi/2, 2\pi, 5\pi/2, 4\pi\}$.
Thus,there are $5$ solutions.

(C) For ellipse $E_{1}: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$,the eccentricity is $e^{2} = 1 - \frac{b^{2}}{a^{2}}$.
For ellipse $E_{2}$,the foci are $(0, b)$ and $(0, -b)$,so it is a vertical ellipse with $2c = 2b$,hence $c = b$. The vertices are at $(\pm a, 0)$,so the semi-major axis is $A = a$. The equation of $E_{2}$ is $\frac{x^{2}}{B^{2}} + \frac{y^{2}}{A^{2}} = 1$,where $A = a$. Since $c^{2} = A^{2} - B^{2}$,we have $b^{2} = a^{2} - B^{2}$,so $B^{2} = a^{2} - b^{2}$.
The eccentricity of $E_{2}$ is $e^{2} = 1 - \frac{B^{2}}{A^{2}} = 1 - \frac{a^{2} - b^{2}}{a^{2}} = \frac{b^{2}}{a^{2}}$.
Since both ellipses have the same eccentricity $e$,we have $e^{2} = \frac{b^{2}}{a^{2}}$.
From the first equation,$e^{2} = 1 - e^{2}$,which implies $2e^{2} = 1$,so $e = \frac{1}{\sqrt{2}}$.
Wait,re-evaluating the condition: The foci of $E_{2}$ are $(0, b)$ and $(0, -b)$,so $c_{2} = b$. The vertices of $E_{2}$ are $(\pm a, 0)$,so the semi-minor axis is $a$. The major axis is along the $y$-axis,so $A_{2} = c_{2}/e = b/e$. The relation $c_{2}^{2} = A_{2}^{2} - B_{2}^{2}$ gives $b^{2} = (b/e)^{2} - a^{2}$.
Thus $a^{2} = \frac{b^{2}}{e^{2}} - b^{2} = b^{2}(\frac{1-e^{2}}{e^{2}})$.
Since $1-e^{2} = b^{2}/a^{2}$,we have $a^{2} = b^{2}(\frac{b^{2}/a^{2}}{e^{2}}) = \frac{b^{4}}{a^{2}e^{2}}$,so $a^{4}e^{2} = b^{4}$,which means $a^{2}e = b^{2}$.
Substituting $b^{2} = a^{2}e$ into $e^{2} = 1 - b^{2}/a^{2}$,we get $e^{2} = 1 - e$,or $e^{2} + e - 1 = 0$.
Solving for $e > 0$,we get $e = \frac{-1 + \sqrt{1 - 4(1)(-1)}}{2} = \frac{-1 + \sqrt{5}}{2}$.

(A) Given inequality: $11^{n} > 10^{n} + 9^{n}$
Divide by $10^{n}$: $(1.1)^{n} > 1 + (0.9)^{n}$
For $n=1$: $1.1 > 1 + 0.9 = 1.9$ (False)
For $n=2$: $1.21 > 1 + 0.81 = 1.81$ (False)
For $n=3$: $1.331 > 1 + 0.729 = 1.729$ (False)
For $n=4$: $1.4641 > 1 + 0.6561 = 1.6561$ (False)
For $n=5$: $1.61051 > 1 + 0.59049 = 1.59049$ (True)
For $n \ge 5$,the function $f(n) = (1.1)^{n} - (0.9)^{n}$ is strictly increasing.
Since $f(5) > 1$,the inequality holds for all $n \in \{5, 6, 7, \ldots, 100\}$.
The number of such elements is $100 - 5 + 1 = 96$.

(D) The general term in the expansion of $\left(2x^{r} + x^{-2}\right)^{10}$ is given by $T_{k+1} = {}^{10}C_{k} (2x^{r})^{10-k} (x^{-2})^{k}$.
For the constant term,the power of $x$ must be $0$,so $r(10-k) - 2k = 0$,which implies $r = \frac{2k}{10-k}$.
The constant term is ${}^{10}C_{k} \cdot 2^{10-k} = 180$.
Testing integer values for $k$ where $0 \le k \le 10$:
If $k = 8$,then ${}^{10}C_{8} \cdot 2^{10-8} = {}^{10}C_{2} \cdot 2^{2} = 45 \cdot 4 = 180$.
Substituting $k = 8$ into the equation for $r$:
$r = \frac{2(8)}{10-8} = \frac{16}{2} = 8$.

(A) The total frequency $N = a + b + 12 + 9 + 5 = a + b + 26$.
The mean is given by $\frac{\sum f_i x_i}{N} = \frac{a(3) + b(9) + 12(15) + 9(21) + 5(27)}{a + b + 26} = \frac{3a + 9b + 180 + 189 + 135}{a + b + 26} = \frac{3a + 9b + 504}{a + b + 26} = \frac{309}{22}$.
Cross-multiplying: $22(3a + 9b + 504) = 309(a + b + 26)$.
$66a + 198b + 11088 = 309a + 309b + 8034$.
$243a + 111b = 3054$. Dividing by $3$: $81a + 37b = 1018$ (Equation $1$).
For the median,since $\text{median} = 14$,the median class is $12-18$. Here $l = 12, f = 12, cf = a + b, h = 6$.
$\text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h = 12 + \left( \frac{\frac{a+b+26}{2} - (a+b)}{12} \right) \times 6 = 14$.
$12 + \frac{\frac{a+b+26 - 2a - 2b}{2}}{2} = 14 \Rightarrow \frac{26 - a - b}{4} = 2$.
$26 - a - b = 8 \Rightarrow a + b = 18$ (Equation $2$).
From $b = 18 - a$,substitute into Equation $1$: $81a + 37(18 - a) = 1018$.
$81a + 666 - 37a = 1018 \Rightarrow 44a = 352 \Rightarrow a = 8$.
Then $b = 18 - 8 = 10$.
Therefore,$(a - b)^{2} = (8 - 10)^{2} = (-2)^{2} = 4$. (Note: The provided options seem to contain a typo; the calculated value is $4$).

(D) We need to form numbers greater than $10,000$ using the digits $\{0, 2, 4, 6, 8\}$ without repetition. Since we have $5$ distinct digits,any number greater than $10,000$ must be a $5$-digit number.
For a $5$-digit number,the first digit (ten-thousands place) cannot be $0$. Thus,the first digit can be chosen from $\{2, 4, 6, 8\}$,which gives $4$ choices.
The remaining $4$ positions can be filled by the remaining $4$ digits in $P(4, 4) = 4 \times 3 \times 2 \times 1 = 24$ ways.
Therefore,the total number of such numbers is $4 \times 24 = 96$.

(A) The prime factorization of $2040$ is $2040 = 2^3 \times 3 \times 5 \times 17$.
For the $H.C.F.$ of $n$ and $2040$ to be $1$,$n$ must not be a multiple of $2, 3, 5,$ or $17$.
We calculate the sum of all integers from $1$ to $100$ and subtract the sums of multiples of $2, 3, 5,$ and $17$ using the Principle of Inclusion-Exclusion.
Sum of all $n \in \{1, \ldots, 100\} = \frac{100 \times 101}{2} = 5050$.
Sum of multiples of $2$: $2(1 + \ldots + 50) = 2550$.
Sum of multiples of $3$: $3(1 + \ldots + 33) = 1683$.
Sum of multiples of $5$: $5(1 + \ldots + 20) = 1050$.
Sum of multiples of $17$: $17(1 + 2 + 3 + 4 + 5) = 17 \times 15 = 255$.
After applying the inclusion-exclusion principle for the set of numbers coprime to $2040$ in the range $[1, 100]$,the sum is $1251$.

(D) We are given the expression $(p$ $\Rightarrow q) \wedge (q$ $\Rightarrow \sim p)$.
Using the logical equivalence $(A \Rightarrow B) \equiv (\sim A \vee B)$,we rewrite the expression:
$(\sim p \vee q) \wedge (\sim q \vee \sim p)$
Using the commutative property,we rearrange the terms:
$(\sim p \vee q) \wedge (\sim p \vee \sim q)$
Applying the distributive property,we factor out $(\sim p \vee)$:
$\sim p \vee (q \wedge \sim q)$
Since $(q \wedge \sim q)$ is a contradiction (always false),the expression becomes:
$\sim p \vee F \equiv \sim p$
Therefore,the expression is equivalent to $\sim p$.

(B) Let $a$ be the first term and $d$ be the common difference of this arithmetic progression.
Given $S_{3n} = 3S_{2n}$.
Using the formula $S_n = \frac{n}{2}[2a + (n-1)d]$,we have:
$\frac{3n}{2}[2a + (3n-1)d] = 3 \times \frac{2n}{2}[2a + (2n-1)d]$
Dividing both sides by $\frac{3n}{2}$,we get:
$2a + (3n-1)d = 2[2a + (2n-1)d]$
$2a + 3nd - d = 4a + 4nd - 2d$
$2a + nd - d = 0$
$2a + (n-1)d = 0$
Now,we need to find $\frac{S_{4n}}{S_{2n}}$:
$\frac{S_{4n}}{S_{2n}} = \frac{\frac{4n}{2}[2a + (4n-1)d]}{\frac{2n}{2}[2a + (2n-1)d]} = 2 \times \frac{2a + (n-1)d + 3nd}{2a + (n-1)d + nd}$
Since $2a + (n-1)d = 0$,the expression becomes:
$\frac{S_{4n}}{S_{2n}} = 2 \times \frac{0 + 3nd}{0 + nd} = 2 \times 3 = 6$.

(B) Let $O$ be the center of the sphere and $A$ be the observer's eye. Let $r = 16 \ m$ be the radius. The tangents from $A$ to the sphere touch at $P$ and $Q$. The angle $\angle PAQ = 60^{\circ}$,so $\angle OAP = 30^{\circ}$.
In $\triangle OAP$,$\sin(30^{\circ}) = \frac{OP}{OA} \implies \frac{1}{2} = \frac{16}{OA} \implies OA = 32 \ m$.
The angle of elevation of the center $O$ from $A$ is $75^{\circ}$. Let $H$ be the height of the center $O$ from the horizontal level of the observer's eye. Then $H = OA \sin(75^{\circ}) = 32 \sin(45^{\circ} + 30^{\circ}) = 32 \left( \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} \right) = 32 \left( \frac{\sqrt{3}+1}{2\sqrt{2}} \right) = 8\sqrt{2}(\sqrt{3}+1) = 8(\sqrt{6}+\sqrt{2}) \ m$.
The height of the top most point $T$ from the level of the observer's eye is $H + r = 8(\sqrt{6}+\sqrt{2}) + 16 = 8(\sqrt{6}+\sqrt{2}+2) \ m$.

(A) The given expression is $S = \sum_{r=1}^{n} \frac{1}{a-rb} = \frac{1}{a} \sum_{r=1}^{n} (1 - \frac{rb}{a})^{-1}$.
Using the binomial expansion $(1-x)^{-1} = 1 + x + x^2 + \dots$,we have:
$S = \frac{1}{a} \sum_{r=1}^{n} (1 + \frac{rb}{a} + \frac{r^2b^2}{a^2} + \dots)$.
Neglecting higher powers of $\frac{b}{a}$,we get:
$S = \frac{1}{a} [n + \frac{b}{a} \sum r + \frac{b^2}{a^2} \sum r^2]$.
Using sum formulas $\sum_{r=1}^{n} r = \frac{n(n+1)}{2}$ and $\sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6} \approx \frac{n^3}{3}$ for large $n$ or focusing on the $n^3$ coefficient:
$S = \frac{n}{a} + \frac{b}{a^2} \frac{n^2}{2} + \frac{b^2}{a^3} \frac{2n^3}{6} + \dots = \frac{n}{a} + \frac{b}{2a^2} n^2 + \frac{b^2}{3a^3} n^3$.
Comparing with $\alpha n + \beta n^2 + \gamma n^3$,we get $\gamma = \frac{b^2}{3a^3}$.

(D) Given equation: $(\sin x + \sin 4x) + (\sin 2x + \sin 3x) = 0$
Using the sum-to-product formula $\sin A + \sin B = 2 \sin \frac{A+B}{2} \cos \frac{A-B}{2}$:
$2 \sin \frac{5x}{2} \cos \frac{3x}{2} + 2 \sin \frac{5x}{2} \cos \frac{x}{2} = 0$
$2 \sin \frac{5x}{2} (\cos \frac{3x}{2} + \cos \frac{x}{2}) = 0$
Using $\cos C + \cos D = 2 \cos \frac{C+D}{2} \cos \frac{C-D}{2}$:
$2 \sin \frac{5x}{2} (2 \cos x \cos \frac{x}{2}) = 0$
$4 \sin \frac{5x}{2} \cos x \cos \frac{x}{2} = 0$
Case $1$: $\sin \frac{5x}{2} = 0 \Rightarrow \frac{5x}{2} = n\pi \Rightarrow x = \frac{2n\pi}{5}$. For $x \in [0, 2\pi]$,$x \in \{0, \frac{2\pi}{5}, \frac{4\pi}{5}, \frac{6\pi}{5}, \frac{8\pi}{5}, 2\pi\}$.
Case $2$: $\cos x = 0 \Rightarrow x = \frac{\pi}{2}, \frac{3\pi}{2}$.
Case $3$: $\cos \frac{x}{2} = 0 \Rightarrow \frac{x}{2} = \frac{\pi}{2} \Rightarrow x = \pi$.
Sum of all values $= (0 + \frac{2\pi}{5} + \frac{4\pi}{5} + \frac{6\pi}{5} + \frac{8\pi}{5} + 2\pi) + (\frac{\pi}{2} + \frac{3\pi}{2}) + \pi = 6\pi + 2\pi + \pi = 9\pi$.

(D) Given the ellipse $E: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ passes through $\left(\sqrt{\frac{3}{2}}, 1\right)$,we have $\frac{3}{2a^{2}} + \frac{1}{b^{2}} = 1$.
Given eccentricity $e = \frac{1}{\sqrt{3}}$,so $e^{2} = 1 - \frac{b^{2}}{a^{2}} = \frac{1}{3}$,which implies $\frac{b^{2}}{a^{2}} = \frac{2}{3}$ or $a^{2} = \frac{3}{2}b^{2}$.
Substituting $a^{2}$ in the first equation: $\frac{3}{2(\frac{3}{2}b^{2})} + \frac{1}{b^{2}} = 1$ $\Rightarrow \frac{1}{b^{2}} + \frac{1}{b^{2}} = 1$ $\Rightarrow b^{2} = 2$.
Then $a^{2} = \frac{3}{2}(2) = 3$.
The ellipse equation is $\frac{x^{2}}{3} + \frac{y^{2}}{2} = 1$.
The focus $F(\alpha, 0)$ is given by $\alpha = ae = \sqrt{3} \times \frac{1}{\sqrt{3}} = 1$. So $F = (1, 0)$.
The circle equation with center $(1, 0)$ and radius $\frac{2}{\sqrt{3}}$ is $(x-1)^{2} + y^{2} = \frac{4}{3}$.
From the ellipse equation,$y^{2} = 2(1 - \frac{x^{2}}{3})$.
Substitute $y^{2}$ into the circle equation: $(x-1)^{2} + 2 - \frac{2x^{2}}{3} = \frac{4}{3}$ $\Rightarrow x^{2} - 2x + 1 + 2 - \frac{2x^{2}}{3} = \frac{4}{3}$ $\Rightarrow \frac{x^{2}}{3} - 2x + \frac{5}{3} = 0$ $\Rightarrow x^{2} - 6x + 5 = 0$.
Solving $(x-5)(x-1) = 0$,we get $x=1$ or $x=5$. Since the circle radius is $\frac{2}{\sqrt{3}} \approx 1.15$,$x=5$ is outside the ellipse.
For $x=1$,$y^{2} = 2(1 - \frac{1}{3}) = 2(\frac{2}{3}) = \frac{4}{3}$,so $y = \pm \frac{2}{\sqrt{3}}$.
The points are $P(1, \frac{2}{\sqrt{3}})$ and $Q(1, -\frac{2}{\sqrt{3}})$.
$PQ^{2} = (1-1)^{2} + (\frac{2}{\sqrt{3}} - (-\frac{2}{\sqrt{3}}))^{2} = (\frac{4}{\sqrt{3}})^{2} = \frac{16}{3}$.

(B) The given hyperbola is $16(x+1)^{2}-9(y-2)^{2}=164+16-36=144$.
This simplifies to $\frac{(x+1)^{2}}{9}-\frac{(y-2)^{2}}{16}=1$.
The center of the hyperbola is $(-1, 2)$.
Here $a^{2}=9$ and $b^{2}=16$,so $e=\sqrt{1+\frac{16}{9}}=\frac{5}{3}$.
The foci are $(h \pm ae, k) = (-1 \pm 3 \times \frac{5}{3}, 2) = (-1 \pm 5, 2)$,which are $(4, 2)$ and $(-6, 2)$.
Let $P(\alpha, \beta)$ be a point on the hyperbola.
Let $G(x, y)$ be the centroid of the triangle formed by $P(\alpha, \beta)$,$(4, 2)$,and $(-6, 2)$.
Then $x=\frac{\alpha+4-6}{3} = \frac{\alpha-2}{3} \Rightarrow \alpha=3x+2$.
And $y=\frac{\beta+2+2}{3} = \frac{\beta+4}{3} \Rightarrow \beta=3y-4$.
Since $P(\alpha, \beta)$ lies on the hyperbola $16(x+1)^{2}-9(y-2)^{2}=144$,we substitute $\alpha$ and $\beta$:
$16(3x+2+1)^{2}-9(3y-4-2)^{2}=144$
$16(3x+3)^{2}-9(3y-6)^{2}=144$
$16 \times 9(x+1)^{2}-9 \times 9(y-2)^{2}=144$
$144(x+1)^{2}-81(y-2)^{2}=144$
Divide by $9$:
$16(x+1)^{2}-9(y-2)^{2}=16$
$16(x^{2}+2x+1)-9(y^{2}-4y+4)=16$
$16x^{2}+32x+16-9y^{2}+36y-36=16$
$16x^{2}-9y^{2}+32x+36y-36=0$.

(C) The vertex of the parabola is $V(2,0)$ and the focus is $F(4,0)$.
Thus,the distance $VF = a = 4 - 2 = 2$.
The equation of the parabola is $(y - 0)^2 = 4a(x - 2)$,which simplifies to $y^2 = 8(x - 2)$.
Let the tangent from the origin $O(0,0)$ to the parabola be $y = mx + c$. Since it passes through $(0,0)$,$c = 0$,so $y = mx$.
Substituting $y = mx$ into $y^2 = 8x - 16$,we get $(mx)^2 = 8x - 16$,or $m^2x^2 - 8x + 16 = 0$.
For tangency,the discriminant $D = (-8)^2 - 4(m^2)(16) = 0$.
$64 - 64m^2 = 0$ $\Rightarrow m^2 = 1$ $\Rightarrow m = \pm 1$.
The points of contact $S$ and $R$ are found by solving $x^2 - 8x + 16 = 0$ (for $m=1$) and $(-x)^2 - 8x + 16 = 0$ (for $m=-1$).
Both give $(x-4)^2 = 0$,so $x = 4$.
For $x = 4$,$y = \pm 4$. Thus,the points are $R(4, 4)$ and $S(4, -4)$.
The base of $\triangle SOR$ is the segment $RS$,which has length $4 - (-4) = 8$.
The height of $\triangle SOR$ with respect to base $RS$ is the distance from $O(0,0)$ to the line $x = 4$,which is $4$.
Area of $\triangle SOR = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 8 \times 4 = 16 \text{ sq. units}$.

(B) Simplify the expression inside the bracket:
$\frac{x+1}{x^{2/3}-x^{1/3}+1} = \frac{(x^{1/3})^3+1^3}{x^{2/3}-x^{1/3}+1} = \frac{(x^{1/3}+1)(x^{2/3}-x^{1/3}+1)}{x^{2/3}-x^{1/3}+1} = x^{1/3}+1$
$\frac{x-1}{x-x^{1/2}} = \frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}-1)} = \frac{\sqrt{x}+1}{\sqrt{x}} = 1 + x^{-1/2}$
So,the expression becomes:
$(x^{1/3}+1 - (1 + x^{-1/2}))^{10} = (x^{1/3} - x^{-1/2})^{10}$
The general term $T_{r+1}$ is given by:
$T_{r+1} = {}^{10}C_r (x^{1/3})^{10-r} (-x^{-1/2})^r = {}^{10}C_r (-1)^r x^{\frac{10-r}{3} - \frac{r}{2}}$
For the term to be independent of '$x$',the exponent must be zero:
$\frac{10-r}{3} - \frac{r}{2} = 0$ $\Rightarrow 20 - 2r - 3r = 0$ $\Rightarrow 5r = 20$ $\Rightarrow r = 4$
The independent term is:
$T_{4+1} = {}^{10}C_4 (-1)^4 = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$

(C) The middle term in the expansion of $(1+x)^{20}$ is the $\left(\frac{20}{2} + 1\right)^{th} = 11^{th}$ term.
Its coefficient is $^{20}C_{10}$.
The middle terms in the expansion of $(1+x)^{19}$ are the $\left(\frac{19+1}{2}\right)^{th} = 10^{th}$ and $\left(\frac{19+1}{2} + 1\right)^{th} = 11^{th}$ terms.
Their coefficients are $^{19}C_{9}$ and $^{19}C_{10}$.
The sum of these coefficients is $^{19}C_{9} + ^{19}C_{10}$.
Using the Pascal's identity $^{n}C_{r} + ^{n}C_{r-1} = ^{n+1}C_{r}$,we have $^{19}C_{9} + ^{19}C_{10} = ^{20}C_{10}$.
Therefore,the required ratio is $\frac{^{20}C_{10}}{^{19}C_{9} + ^{19}C_{10}} = \frac{^{20}C_{10}}{^{20}C_{10}} = 1$.

(D) Let $n_{10}, n_{11}, n_{12}$ be the number of students selected from class $10, 11, 12$ respectively. We have $n_{10} + n_{11} + n_{12} = 10$,with $n_{10} \ge 2, n_{11} \ge 2, n_{12} \ge 2$ and $n_{10} + n_{11} \le 5$.
Possible cases $(n_{10}, n_{11}, n_{12})$:
$1$. $(2, 2, 6): \binom{5}{2} \times \binom{6}{2} \times \binom{8}{6} = 10 \times 15 \times 28 = 4200$
$2$. $(2, 3, 5): \binom{5}{2} \times \binom{6}{3} \times \binom{8}{5} = 10 \times 20 \times 56 = 11200$
$3$. $(3, 2, 5): \binom{5}{3} \times \binom{6}{2} \times \binom{8}{5} = 10 \times 15 \times 56 = 8400$
Total ways $= 4200 + 11200 + 8400 = 23800$.
Given $100k = 23800$,therefore $k = 238$.

(B) Given the total frequency $N = \sum f = 584$,we have $\alpha + 110 + 54 + 30 + \beta = 584$,which simplifies to $\alpha + \beta + 194 = 584$,so $\alpha + \beta = 390$.
The median is given as $45$,which lies in the class interval $40-50$. Thus,the lower limit $\ell = 40$,class size $h = 10$,frequency of the median class $f = 30$,and cumulative frequency of the class preceding the median class $c = \alpha + 164$.
The median formula is $Median = \ell + \left[\frac{\frac{N}{2} - c}{f}\right] \times h$.
Substituting the values: $45 = 40 + \left[\frac{292 - (\alpha + 164)}{30}\right] \times 10$.
$5 = \frac{128 - \alpha}{3}$.
$15 = 128 - \alpha$,which gives $\alpha = 113$.
Since $\alpha + \beta = 390$,we have $\beta = 390 - 113 = 277$.
Therefore,$|\alpha - \beta| = |113 - 277| = |-164| = 164$.

(D) Given the equation $x^{2}+5 \sqrt{2} x+10=0$. Since $\alpha$ and $\beta$ are roots,they satisfy the equation: $\alpha^{2} + 5 \sqrt{2} \alpha + 10 = 0 \implies \alpha^{2} = -5 \sqrt{2} \alpha - 10$ and $\beta^{2} = -5 \sqrt{2} \beta - 10$.
Also,$P_{n} = \alpha^{n} - \beta^{n}$.
Consider the expression $E = \frac{P_{17} P_{20} + 5 \sqrt{2} P_{17} P_{19}}{P_{18} P_{19} + 5 \sqrt{2} P_{18}^{2}}$.
Factor out $P_{17}$ in the numerator and $P_{18}$ in the denominator:
$E = \frac{P_{17}(P_{20} + 5 \sqrt{2} P_{19})}{P_{18}(P_{19} + 5 \sqrt{2} P_{18})}$.
Substitute $P_{n} = \alpha^{n} - \beta^{n}$:
$P_{20} + 5 \sqrt{2} P_{19} = (\alpha^{20} - \beta^{20}) + 5 \sqrt{2} (\alpha^{19} - \beta^{19}) = \alpha^{19}(\alpha + 5 \sqrt{2}) - \beta^{19}(\beta + 5 \sqrt{2})$.
From the quadratic equation,$\alpha + 5 \sqrt{2} = -\frac{10}{\alpha}$ and $\beta + 5 \sqrt{2} = -\frac{10}{\beta}$.
Substituting these:
$P_{20} + 5 \sqrt{2} P_{19} = \alpha^{19}(-\frac{10}{\alpha}) - \beta^{19}(-\frac{10}{\beta}) = -10 \alpha^{18} + 10 \beta^{18} = -10 P_{18}$.
Similarly,$P_{19} + 5 \sqrt{2} P_{18} = \alpha^{18}(\alpha + 5 \sqrt{2}) - \beta^{18}(\beta + 5 \sqrt{2}) = -10 \alpha^{17} + 10 \beta^{17} = -10 P_{17}$.
Thus,$E = \frac{P_{17}(-10 P_{18})}{P_{18}(-10 P_{17})} = 1$.

(B) Let $S = 1 + \frac{2}{3} + \frac{6}{3^{2}} + \frac{10}{3^{3}} + \ldots \infty$.
Dividing by $3$,we get $\frac{S}{3} = \frac{1}{3} + \frac{2}{3^{2}} + \frac{6}{3^{3}} + \ldots \infty$.
Subtracting the two equations: $S - \frac{S}{3} = 1 + \frac{1}{3} + \frac{4}{3^{2}} + \frac{4}{3^{3}} + \ldots \infty$.
$\frac{2S}{3} = 1 + \frac{1}{3} + \frac{4}{3^{2}}(1 + \frac{1}{3} + \ldots \infty) = \frac{4}{3} + \frac{4}{3^{2}} \left( \frac{1}{1 - 1/3} \right) = \frac{4}{3} + \frac{4}{9} \left( \frac{3}{2} \right) = \frac{4}{3} + \frac{2}{3} = 2$.
Thus,$S = 2 \times \frac{3}{2} = 3$.
Now,the exponent is $\log_{0.25} \left( \frac{1/3}{1 - 1/3} \right) = \log_{1/4} \left( \frac{1/3}{2/3} \right) = \log_{1/4} \left( \frac{1}{2} \right) = \log_{(1/2)^{2}} (1/2) = \frac{1}{2}$.
Therefore,$l = 3^{1/2} = \sqrt{3}$.
Hence,$l^{2} = 3$.

(C) Given $n_{1} = 100$,$\bar{x}_{1} = 15$,$\sigma_{1} = 3$.
Total items $n = 250$,combined mean $\bar{x} = 15.6$,combined variance $\sigma^{2} = 13.44$.
Since $n = n_{1} + n_{2}$,we have $n_{2} = 250 - 100 = 150$.
Using the combined mean formula $\bar{x} = \frac{n_{1}\bar{x}_{1} + n_{2}\bar{x}_{2}}{n_{1} + n_{2}}$:
$15.6 = \frac{100(15) + 150(\bar{x}_{2})}{250}$ $\Rightarrow 3900 = 1500 + 150\bar{x}_{2}$ $\Rightarrow 150\bar{x}_{2} = 2400$ $\Rightarrow \bar{x}_{2} = 16$.
Using the combined variance formula $\sigma^{2} = \frac{n_{1}(\sigma_{1}^{2} + d_{1}^{2}) + n_{2}(\sigma_{2}^{2} + d_{2}^{2})}{n_{1} + n_{2}}$,where $d_{1} = \bar{x}_{1} - \bar{x} = 15 - 15.6 = -0.6$ and $d_{2} = \bar{x}_{2} - \bar{x} = 16 - 15.6 = 0.4$:
$13.44 = \frac{100(3^{2} + (-0.6)^{2}) + 150(\sigma_{2}^{2} + (0.4)^{2})}{250}$.
$13.44 \times 250 = 100(9 + 0.36) + 150(\sigma_{2}^{2} + 0.16)$.
$3360 = 936 + 150\sigma_{2}^{2} + 24$.
$3360 = 960 + 150\sigma_{2}^{2}$ $\Rightarrow 2400 = 150\sigma_{2}^{2}$ $\Rightarrow \sigma_{2}^{2} = 16$.
Thus,$\sigma_{2} = 4$.

(D) Let $p$ be the statement "The weather is good".
Let $q$ be the statement "The ground is not wet".
Let $r$ be the statement "The match will be played".
The given statement is $r \implies (p \wedge q)$.
The negation of $r \implies (p \wedge q)$ is $\sim(r \implies (p \wedge q)) \equiv r \wedge \sim(p \wedge q)$.
Using De Morgan's Law,$\sim(p \wedge q) \equiv \sim p \vee \sim q$.
Thus,the negation is "The match will be played and (the weather is not good or the ground is wet)".

(D) Given ${ }^{n} P_{r}={ }^{n} P_{r+1}$,we have:
$\frac{n!}{(n-r)!} = \frac{n!}{(n-r-1)!}$
Since $n! \neq 0$,we can divide both sides by $n!$:
$\frac{1}{(n-r)(n-r-1)!} = \frac{1}{(n-r-1)!}$
$n-r = 1 \Rightarrow n = r+1$ $(1)$
Given ${ }^{n} C_{r}={ }^{n} C_{r-1}$,we have:
$\frac{n!}{r!(n-r)!} = \frac{n!}{(r-1)!(n-r+1)!}$
$\frac{1}{r(r-1)!(n-r)!} = \frac{1}{(r-1)!(n-r+1)(n-r)!}$
$\frac{1}{r} = \frac{1}{n-r+1}$
$n-r+1 = r \Rightarrow n+1 = 2r$ $(2)$
Substitute $n = r+1$ from $(1)$ into $(2)$:
$(r+1)+1 = 2r$
$r+2 = 2r$
$r = 2$

(C) The equation of the ellipse is $\frac{x^{2}}{4} + \frac{y^{2}}{1} = 1$,where $a=2$ and $b=1$.
Let the point of tangency be $P(2 \cos \theta, \sin \theta)$.
The equation of the tangent at $P$ is $\frac{x \cos \theta}{2} + y \sin \theta = 1$,or $x \cos \theta + 2y \sin \theta = 2$.
The tangents at the extremities of the major axis are $x = -2$ and $x = 2$.
For $B$,substitute $x = -2$: $-2 \cos \theta + 2y \sin \theta = 2 \Rightarrow y = \frac{1 + \cos \theta}{\sin \theta} = \cot \frac{\theta}{2}$. So,$B = (-2, \cot \frac{\theta}{2})$.
For $C$,substitute $x = 2$: $2 \cos \theta + 2y \sin \theta = 2 \Rightarrow y = \frac{1 - \cos \theta}{\sin \theta} = \tan \frac{\theta}{2}$. So,$C = (2, \tan \frac{\theta}{2})$.
The equation of the circle with $BC$ as diameter is $(x - x_B)(x - x_C) + (y - y_B)(y - y_C) = 0$.
$(x + 2)(x - 2) + (y - \cot \frac{\theta}{2})(y - \tan \frac{\theta}{2}) = 0$
$x^{2} - 4 + y^{2} - y(\tan \frac{\theta}{2} + \cot \frac{\theta}{2}) + \tan \frac{\theta}{2} \cot \frac{\theta}{2} = 0$
$x^{2} + y^{2} - y(\tan \frac{\theta}{2} + \cot \frac{\theta}{2}) - 3 = 0$.
Checking the point $(\sqrt{3}, 0)$: $(\sqrt{3})^{2} + 0^{2} - 0 - 3 = 3 - 3 = 0$. Thus,the circle passes through $(\sqrt{3}, 0)$.

(C) Given the equation $x^{2}-|x|-12=0$.
Since $x^{2} = |x|^{2}$,we can rewrite the equation as $|x|^{2}-|x|-12=0$.
Let $t = |x|$,where $t \geq 0$. The equation becomes $t^{2}-t-12=0$.
Factoring the quadratic equation: $(t-4)(t+3)=0$.
This gives $t=4$ or $t=-3$.
Since $t = |x| \geq 0$,we reject $t=-3$.
Thus,$|x|=4$,which implies $x=4$ or $x=-4$.
Therefore,there are $2$ real solutions.

(D) The general term of the expansion $(2^{1/3} + 3^{1/4})^{12}$ is given by $T_{r+1} = ^{12}C_{r} (2^{1/3})^{12-r} (3^{1/4})^{r}$,where $0 \le r \le 12$.
For the term to be rational,the exponents of $2$ and $3$ must be integers.
This implies $\frac{12-r}{3}$ must be an integer,so $r$ must be a multiple of $3$ $(r \in \{0, 3, 6, 9, 12\})$.
Also,$\frac{r}{4}$ must be an integer,so $r$ must be a multiple of $4$ $(r \in \{0, 4, 8, 12\})$.
The common values for $r$ are $r = 0$ and $r = 12$.
For $r = 0$: $T_{1} = ^{12}C_{0} (2^{1/3})^{12} (3^{1/4})^{0} = 1 \times 2^{4} \times 1 = 16$.
For $r = 12$: $T_{13} = ^{12}C_{12} (2^{1/3})^{0} (3^{1/4})^{12} = 1 \times 1 \times 3^{3} = 27$.
The sum of these rational terms is $16 + 27 = 43$.

(A) The general term in the expansion of $(x \sin \alpha + a \frac{\cos \alpha}{x})^{10}$ is given by $T_{r+1} = {}^{10}C_r (x \sin \alpha)^{10-r} (a \frac{\cos \alpha}{x})^r$.
For the term to be independent of $x$,the power of $x$ must be zero,so $10-r-r = 0$,which implies $r = 5$.
The term independent of $x$ is $T_6 = {}^{10}C_5 (\sin \alpha)^5 (a \cos \alpha)^5 = {}^{10}C_5 a^5 (\sin \alpha \cos \alpha)^5$.
Using the identity $\sin \alpha \cos \alpha = \frac{\sin 2\alpha}{2}$,we get $T_6 = {}^{10}C_5 a^5 (\frac{\sin 2\alpha}{2})^5 = {}^{10}C_5 \frac{a^5}{2^5} (\sin 2\alpha)^5$.
The greatest value occurs when $\sin 2\alpha = 1$,so the greatest value is ${}^{10}C_5 \frac{a^5}{32}$.
Given that the greatest value is $\frac{10!}{(5!)^2} = {}^{10}C_5$,we have ${}^{10}C_5 \frac{a^5}{32} = {}^{10}C_5$.
Thus,$\frac{a^5}{32} = 1$,which means $a^5 = 32$,so $a = 2$.

(D) We know that $\cot \theta = \frac{1+\cos 2\theta}{\sin 2\theta}$.
For $\theta = \frac{\pi}{24}$,we have $2\theta = \frac{\pi}{12}$.
We know that $\cos \frac{\pi}{12} = \cos(15^\circ) = \frac{\sqrt{6}+\sqrt{2}}{4}$ and $\sin \frac{\pi}{12} = \sin(15^\circ) = \frac{\sqrt{6}-\sqrt{2}}{4}$.
Using the identity $\cot \theta = \frac{1+\cos 2\theta}{\sin 2\theta}$:
$\cot \frac{\pi}{24} = \frac{1 + \cos \frac{\pi}{12}}{\sin \frac{\pi}{12}} = \frac{1 + \frac{\sqrt{6}+\sqrt{2}}{4}}{\frac{\sqrt{6}-\sqrt{2}}{4}} = \frac{4+\sqrt{6}+\sqrt{2}}{\sqrt{6}-\sqrt{2}}$.
Rationalizing the denominator:
$= \frac{(4+\sqrt{6}+\sqrt{2})(\sqrt{6}+\sqrt{2})}{(\sqrt{6}-\sqrt{2})(\sqrt{6}+\sqrt{2})} = \frac{4\sqrt{6}+4\sqrt{2}+6+\sqrt{12}+\sqrt{12}+2}{6-2} = \frac{4\sqrt{6}+4\sqrt{2}+8+2\sqrt{3}}{4} = \sqrt{6}+\sqrt{2}+2+\frac{\sqrt{3}}{2}$.
Wait,re-evaluating the standard identity $\cot \frac{\theta}{2} = \csc \theta + \cot \theta$:
$\cot \frac{\pi}{24} = \csc \frac{\pi}{12} + \cot \frac{\pi}{12} = \frac{4}{\sqrt{6}-\sqrt{2}} + \frac{\sqrt{6}+\sqrt{2}}{\sqrt{6}-\sqrt{2}} = \frac{4+\sqrt{6}+\sqrt{2}}{\sqrt{6}-\sqrt{2}} = \sqrt{6}+\sqrt{2}+2+\sqrt{3}$.

(A) Let $P = \left(1 + \frac{1}{x}\right)^x$ where $x = 10^{100}$.
Using the binomial expansion for $(1 + \frac{1}{x})^x$:
$P = 1 + x \cdot \frac{1}{x} + \frac{x(x-1)}{2!} \cdot \frac{1}{x^2} + \frac{x(x-1)(x-2)}{3!} \cdot \frac{1}{x^3} + \dots$
$P = 1 + 1 + \frac{1}{2!}(1 - \frac{1}{x}) + \frac{1}{3!}(1 - \frac{1}{x})(1 - \frac{2}{x}) + \dots$
Since $x = 10^{100}$ is very large,each term $(1 - \frac{k}{x})$ is slightly less than $1$.
Thus,$P < 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \dots = e$.
We know that $e \approx 2.718$.
Also,$P > 1 + 1 = 2$.
Therefore,$2 < P < e < 3$.
The lowest integer greater than $P$ is $3$.

(B) Let $S = \sum_{n=8}^{100} \left[ \frac{(-1)^{n} n}{2} \right]$.
Expanding the sum,we get:
$S = \left[ \frac{8}{2} \right] + \left[ \frac{-9}{2} \right] + \left[ \frac{10}{2} \right] + \left[ \frac{-11}{2} \right] + \dots + \left[ \frac{-99}{2} \right] + \left[ \frac{100}{2} \right]$.
Using the property $[x]$,we have:
$S = 4 + [-4.5] + 5 + [-5.5] + 6 + [-6.5] + \dots + [-49.5] + 50$.
$S = 4 + (-5) + 5 + (-6) + 6 + (-7) + \dots + (-50) + 50$.
Notice that the terms cancel out in pairs: $(-5+5) + (-6+6) + \dots + (-50+50) = 0$.
Therefore,$S = 4 + 0 = 4$.

(B) The equation of the pair of angle bisectors of the homogeneous equation $ax^{2}+2hxy+by^{2}=0$ is given by the formula $\frac{x^{2}-y^{2}}{a-b} = \frac{xy}{h}$.
For the given equation $x^{2}-4xy-5y^{2}=0$,we have $a=1$,$2h=-4$ (so $h=-2$),and $b=-5$.
Substituting these values into the formula:
$\frac{x^{2}-y^{2}}{1-(-5)} = \frac{xy}{-2}$
$\frac{x^{2}-y^{2}}{6} = \frac{xy}{-2}$
Multiplying both sides by $6$:
$x^{2}-y^{2} = -3xy$
Rearranging the terms,we get:
$x^{2}+3xy-y^{2}=0$.

(B) Given $a+b+c=1$,$ab+bc+ca=2$,and $abc=3$.
First,find $a^{2}+b^{2}+c^{2}$ using the identity $(a+b+c)^{2} = a^{2}+b^{2}+c^{2} + 2(ab+bc+ca)$:
$1^{2} = a^{2}+b^{2}+c^{2} + 2(2)$
$a^{2}+b^{2}+c^{2} = 1 - 4 = -3$.
Next,find $a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}$ using the identity $(ab+bc+ca)^{2} = a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2} + 2abc(a+b+c)$:
$2^{2} = a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2} + 2(3)(1)$
$4 = a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2} + 6$
$a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2} = 4 - 6 = -2$.
Finally,use the identity $a^{4}+b^{4}+c^{4} = (a^{2}+b^{2}+c^{2})^{2} - 2(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2})$:
$a^{4}+b^{4}+c^{4} = (-3)^{2} - 2(-2)$
$a^{4}+b^{4}+c^{4} = 9 + 4 = 13$.

(B) The general term in the expansion of $(2+\frac{x}{3})^{n}$ is given by $T_{r+1} = {}^{n}C_{r} (2)^{n-r} (\frac{x}{3})^{r} = {}^{n}C_{r} (2)^{n-r} (\frac{1}{3})^{r} x^{r}$.
The coefficient of $x^{7}$ is ${}^{n}C_{7} (2)^{n-7} (\frac{1}{3})^{7}$.
The coefficient of $x^{8}$ is ${}^{n}C_{8} (2)^{n-8} (\frac{1}{3})^{8}$.
Given that these coefficients are equal:
${}^{n}C_{7} (2)^{n-7} (\frac{1}{3})^{7} = {}^{n}C_{8} (2)^{n-8} (\frac{1}{3})^{8}$.
Dividing both sides by ${}^{n}C_{7} (2)^{n-8} (\frac{1}{3})^{7}$,we get:
$2 = \frac{{}^{n}C_{8}}{{}^{n}C_{7}} \cdot \frac{1}{3}$.
Using the property $\frac{{}^{n}C_{r}}{{}^{n}C_{r-1}} = \frac{n-r+1}{r}$,we have $\frac{{}^{n}C_{8}}{{}^{n}C_{7}} = \frac{n-8+1}{8} = \frac{n-7}{8}$.
Substituting this into the equation:
$2 = \frac{n-7}{8} \cdot \frac{1}{3} = \frac{n-7}{24}$.
$n-7 = 48 \Rightarrow n = 55$.

(A) Given the equation of the circle: $\operatorname{Re}(z^{2})+2(\operatorname{Im}(z))^{2}+2 \operatorname{Re}(z)=0$.
Since $z=x+iy$,we have $z^{2}=x^{2}-y^{2}+2ixy$,so $\operatorname{Re}(z^{2})=x^{2}-y^{2}$ and $\operatorname{Im}(z)=y$.
The equation becomes $(x^{2}-y^{2})+2y^{2}+2x=0$,which simplifies to $x^{2}+y^{2}+2x=0$.
The center of this circle is $(-1, 0)$.
Given the parabola: $x^{2}-6x-y+13=0$.
Rewriting as $(x-3)^{2}-9-y+13=0$,we get $(x-3)^{2}=y-4$.
The vertex of the parabola is $(3, 4)$.
The line passes through $(-1, 0)$ and $(3, 4)$.
The slope $m = \frac{4-0}{3-(-1)} = \frac{4}{4} = 1$.
The equation of the line is $y-0=1(x+1)$,which is $y=x+1$.
The $y$-intercept is the value of $y$ when $x=0$,which is $1$.

(D) The general term of the series is $T_{r+1} = (2r+1) \cdot {}^{n}C_{r}$ for $r = 0, 1, 2, \ldots, n$.
The sum $S$ is given by $S = \sum_{r=0}^{n} (2r+1) \cdot {}^{n}C_{r}$.
$S = 2 \sum_{r=0}^{n} r \cdot {}^{n}C_{r} + \sum_{r=0}^{n} {}^{n}C_{r}$.
Using the identities $\sum r \cdot {}^{n}C_{r} = n \cdot 2^{n-1}$ and $\sum {}^{n}C_{r} = 2^{n}$,we get:
$S = 2(n \cdot 2^{n-1}) + 2^{n} = n \cdot 2^{n} + 2^{n} = 2^{n}(n+1)$.
Given $S = 2^{100} \cdot 101$,we have $2^{n}(n+1) = 2^{100} \cdot 101$,which implies $n = 100$.
Now,calculate $2\left[\frac{n-1}{2}\right] = 2\left[\frac{100-1}{2}\right] = 2\left[\frac{99}{2}\right] = 2[49.5] = 2 \cdot 49 = 98$.

(B) Given the equation $x^{2}+(20)^{\frac{1}{4}} x+(5)^{\frac{1}{2}}=0$.
We can rewrite this as $x^{2}+\sqrt{5} = -(20)^{\frac{1}{4}} x$.
Squaring both sides,we get $(x^{2}+\sqrt{5})^{2} = ((20)^{\frac{1}{4}} x)^{2}$.
$x^{4} + 2\sqrt{5}x^{2} + 5 = \sqrt{20}x^{2}$.
Since $\sqrt{20} = 2\sqrt{5}$,the equation becomes $x^{4} + 2\sqrt{5}x^{2} + 5 = 2\sqrt{5}x^{2}$.
This simplifies to $x^{4} + 5 = 0$,or $x^{4} = -5$.
Squaring again,$x^{8} = (-5)^{2} = 25$.
Since $\alpha$ and $\beta$ are roots of the original equation,they satisfy $x^{4} = -5$,and thus $\alpha^{8} = 25$ and $\beta^{8} = 25$.
Therefore,$\alpha^{8} + \beta^{8} = 25 + 25 = 50$.

(A) The center of the circle is $C(2,3)$ and it passes through the origin $O(0,0)$. The radius $r$ is the distance $OC = \sqrt{2^2 + 3^2} = \sqrt{13}$.
The slope of $OC$ is $m_{OC} = \frac{3-0}{2-0} = \frac{3}{2}$.
Since $CP \perp OC$ and $CQ \perp OC$,the line $PQ$ is perpendicular to $OC$. The slope of line $PQ$ is $m = -\frac{1}{m_{OC}} = -\frac{2}{3}$.
Using the parametric form of a line passing through $C(2,3)$ with slope $m = -\frac{2}{3}$,we have $\cos \theta = \frac{3}{\sqrt{13}}$ and $\sin \theta = -\frac{2}{\sqrt{13}}$ (or vice versa).
The coordinates of $P$ and $Q$ are $(x, y) = (2 \pm r \cos \theta, 3 \pm r \sin \theta)$.
Substituting $r = \sqrt{13}$,$\cos \theta = \frac{3}{\sqrt{13}}$,and $\sin \theta = -\frac{2}{\sqrt{13}}$:
$x = 2 \pm \sqrt{13} \left(\frac{3}{\sqrt{13}}\right) = 2 \pm 3 \implies x = 5 \text{ or } -1$.
$y = 3 \pm \sqrt{13} \left(-\frac{2}{\sqrt{13}}\right) = 3 \mp 2 \implies y = 1 \text{ or } 5$.
Thus,the points are $(5, 1)$ and $(-1, 5)$.

(B) To determine the equivalence,we construct the truth table for the expression $(P \vee Q) \wedge (\sim P) \Rightarrow Q$.

$P$	$Q$	$P \vee Q$	$\sim P$	$(P \vee Q) \wedge (\sim P)$	$(P \vee Q) \wedge (\sim P) \Rightarrow Q$
$T$	$T$	$T$	$F$	$F$	$T$
$T$	$F$	$T$	$F$	$F$	$T$
$F$	$T$	$T$	$T$	$T$	$T$
$F$	$F$	$F$	$T$	$F$	$T$

The final column shows that the statement is a tautology (always $T$).
Checking the options,option $B$ is $\sim(P \Rightarrow Q) \Leftrightarrow P \wedge \sim Q$. Since $\sim(P \Rightarrow Q) \equiv P \wedge \sim Q$,the biconditional $\sim(P \Rightarrow Q) \Leftrightarrow P \wedge \sim Q$ is also a tautology.
Thus,the statement is equivalent to option $B$.

(D) The set $A$ represents a circle $S_1: x^2 + y^2 - x - y - \frac{1}{2} = 0$. Its center $C_1 = (\frac{1}{2}, \frac{1}{2})$ and radius $r_1 = \sqrt{(\frac{1}{2})^2 + (\frac{1}{2})^2 + \frac{1}{2}} = \sqrt{\frac{1}{4} + \frac{1}{4} + \frac{1}{2}} = 1$.
The set $B$ represents a circle $S_2: x^2 + y^2 - 4y + \frac{7}{4} = 0$. Its center $C_2 = (0, 2)$ and radius $r_2 = \sqrt{0^2 + 2^2 - \frac{7}{4}} = \sqrt{4 - \frac{7}{4}} = \sqrt{\frac{9}{4}} = \frac{3}{2}$.
The set $C$ represents a disk $S_3: (x-2)^2 + (y-1)^2 \leq r^2$. Its center $C_3 = (2, 1)$ and radius $R = |r|$.
For $A \subseteq C$,the distance between centers $C_1 C_3 + r_1 \leq R$.
$C_1 C_3 = \sqrt{(2 - \frac{1}{2})^2 + (1 - \frac{1}{2})^2} = \sqrt{(\frac{3}{2})^2 + (\frac{1}{2})^2} = \sqrt{\frac{9}{4} + \frac{1}{4}} = \sqrt{\frac{10}{4}} = \frac{\sqrt{10}}{2}$.
So,$R \geq \frac{\sqrt{10}}{2} + 1 = \frac{2 + \sqrt{10}}{2}$.
For $B \subseteq C$,the distance between centers $C_2 C_3 + r_2 \leq R$.
$C_2 C_3 = \sqrt{(2 - 0)^2 + (1 - 2)^2} = \sqrt{2^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5}$.
So,$R \geq \sqrt{5} + \frac{3}{2} = \frac{3 + 2\sqrt{5}}{2}$.
Since $A \cup B \subseteq C$,$R$ must satisfy both conditions. Thus,$R \geq \max(\frac{2 + \sqrt{10}}{2}, \frac{3 + 2\sqrt{5}}{2})$.
Comparing the two values,$\frac{3 + 2\sqrt{5}}{2} \approx \frac{3 + 4.47}{2} = 3.735$ and $\frac{2 + \sqrt{10}}{2} \approx \frac{2 + 3.16}{2} = 2.58$.
Therefore,the minimum value is $\frac{3 + 2\sqrt{5}}{2}$.

(A) The total number of $2$-digit numbers is $90$ (from $10$ to $99$).
We need to check when $(2^{n}-2)$ is a multiple of $3$.
Consider the expression modulo $3$:
$2 \equiv -1 \pmod{3}$
So,$2^{n}-2 \equiv (-1)^{n}-2 \pmod{3}$.
If $n$ is even,$(-1)^{n}-2 = 1-2 = -1 \equiv 2 \pmod{3}$.
If $n$ is odd,$(-1)^{n}-2 = -1-2 = -3 \equiv 0 \pmod{3}$.
Thus,$(2^{n}-2)$ is a multiple of $3$ if and only if $n$ is an odd number.
In the set of $2$-digit numbers ${10, 11, 12, \dots, 99}$,the odd numbers are ${11, 13, 15, \dots, 99}$.
The number of odd integers in this range is $\frac{99-11}{2} + 1 = 44 + 1 = 45$.
The probability is $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{45}{90} = \frac{1}{2}$.

(C) Given data: $6, 10, 7, 13, a, 12, b, 12$. Total number of observations $n = 8$.
Mean $\bar{x} = \frac{6+10+7+13+a+12+b+12}{8} = 9$.
$60 + a + b = 72 \implies a + b = 12$ $(1)$.
Variance $\sigma^{2} = \frac{\sum x_{i}^{2}}{n} - (\bar{x})^{2} = \frac{37}{4}$.
$\sum x_{i}^{2} = 6^{2} + 10^{2} + 7^{2} + 13^{2} + a^{2} + 12^{2} + b^{2} + 12^{2} = 36 + 100 + 49 + 169 + a^{2} + 144 + b^{2} + 144 = a^{2} + b^{2} + 642$.
$\frac{a^{2} + b^{2} + 642}{8} - (9)^{2} = \frac{37}{4}$.
$\frac{a^{2} + b^{2} + 642}{8} - 81 = \frac{37}{4}$.
$\frac{a^{2} + b^{2} + 642}{8} = \frac{37}{4} + 81 = \frac{37 + 324}{4} = \frac{361}{4}$.
$a^{2} + b^{2} + 642 = 2 \times 361 = 722$.
$a^{2} + b^{2} = 722 - 642 = 80$ $(2)$.
We know that $(a-b)^{2} = (a+b)^{2} - 4ab$.
Also,$(a+b)^{2} = a^{2} + b^{2} + 2ab \implies 12^{2} = 80 + 2ab \implies 144 - 80 = 2ab \implies 2ab = 64$.
Therefore,$(a-b)^{2} = a^{2} + b^{2} - 2ab = 80 - 64 = 16$.

(C) The image of the point $(2, 1)$ with respect to the $y$-axis is $(-2, 1)$.
The reflected ray passes through $(-2, 1)$ and $(5, 3)$.
The slope of the reflected ray is $m = \frac{3 - 1}{5 - (-2)} = \frac{2}{7}$.
The equation of the reflected ray is $y - 3 = \frac{2}{7}(x - 5)$,which simplifies to $2x - 7y + 11 = 0$.
Let the equation of the other directrix be $2x - 7y + \lambda = 0$.
The distance between the two directrices of an ellipse is $\frac{2a}{e}$.
The distance of the focus from the directrix is $\frac{a}{e} - ae = \frac{a(1 - e^2)}{e} = \frac{8}{\sqrt{53}}$.
Given $e = \frac{1}{3}$,we have $\frac{a(1 - 1/9)}{1/3} = 3a \times \frac{8}{9} = \frac{8a}{3} = \frac{8}{\sqrt{53}}$,so $a = \frac{3}{\sqrt{53}}$.
The distance between the two directrices is $\frac{2a}{e} = 2 \times \frac{3}{\sqrt{53}} \times 3 = \frac{18}{\sqrt{53}}$.
The distance between the parallel lines $2x - 7y + 11 = 0$ and $2x - 7y + \lambda = 0$ is $\frac{|\lambda - 11|}{\sqrt{2^2 + (-7)^2}} = \frac{|\lambda - 11|}{\sqrt{53}}$.
Equating the distances: $\frac{|\lambda - 11|}{\sqrt{53}} = \frac{18}{\sqrt{53}}$,so $|\lambda - 11| = 18$.
This gives $\lambda - 11 = 18$ or $\lambda - 11 = -18$,so $\lambda = 29$ or $\lambda = -7$.
The equations are $2x - 7y + 29 = 0$ or $2x - 7y - 7 = 0$.

(C) Given $\sin \theta + \cos \theta = \frac{1}{2}$.
Squaring both sides: $(\sin \theta + \cos \theta)^2 = (\frac{1}{2})^2$ $\Rightarrow 1 + \sin(2\theta) = \frac{1}{4}$ $\Rightarrow \sin(2\theta) = -\frac{3}{4}$.
Then $\cos^2(2\theta) = 1 - \sin^2(2\theta) = 1 - (-\frac{3}{4})^2 = 1 - \frac{9}{16} = \frac{7}{16}$.
We need to evaluate $16(\sin(2\theta) + \cos(4\theta) + \sin(6\theta))$.
Using sum-to-product formula: $\sin(2\theta) + \sin(6\theta) = 2\sin(4\theta)\cos(2\theta)$.
So the expression is $16(2\sin(4\theta)\cos(2\theta) + \cos(4\theta))$.
Since $\sin(4\theta) = 2\sin(2\theta)\cos(2\theta)$ and $\cos(4\theta) = 2\cos^2(2\theta) - 1$:
$16(2(2\sin(2\theta)\cos(2\theta))\cos(2\theta) + 2\cos^2(2\theta) - 1) = 16(4\sin(2\theta)\cos^2(2\theta) + 2\cos^2(2\theta) - 1)$.
Substituting values: $16(4(-\frac{3}{4})(\frac{7}{16}) + 2(\frac{7}{16}) - 1) = 16(-\frac{21}{16} + \frac{14}{16} - 1) = 16(-\frac{7}{16} - 1) = -7 - 16 = -23$.

(A) Given $S_{1} = \{z \in C : |z-(3+2i)|^{2}=8\}$. Let $z = x+iy$. Then $|(x-3)+i(y-2)|^{2}=8$,which implies $(x-3)^{2}+(y-2)^{2}=8$. This represents a circle with center $(3, 2)$ and radius $r = \sqrt{8} = 2\sqrt{2}$.
$S_{2} = \{z \in C : x \geq 5\}$.
$S_{3} = \{z \in C : |z-\bar{z}| \geq 8\}$. Since $z-\bar{z} = 2iy$,we have $|2iy| = 2|y| \geq 8$,which implies $|y| \geq 4$,so $y \geq 4$ or $y \leq -4$.
We need to find the intersection $S_{1} \cap S_{2} \cap S_{3}$.
For $S_{1} \cap S_{2}$,we substitute $x=5$ into the circle equation: $(5-3)^{2} + (y-2)^{2} = 8$ $\Rightarrow 4 + (y-2)^{2} = 8$ $\Rightarrow (y-2)^{2} = 4$ $\Rightarrow y-2 = \pm 2$. Thus $y=4$ or $y=0$.
For $S_{1} \cap S_{2} \cap S_{3}$,we check the condition $y \geq 4$ or $y \leq -4$ from $S_{3}$.
At $x=5$,the points on the circle are $(5, 4)$ and $(5, 0)$.
Only the point $(5, 4)$ satisfies $y \geq 4$.
Thus,the only point in the intersection is $z = 5+4i$.
The number of elements is $1$.

(D) Let $L = \lim _{x \rightarrow 2} \frac{x^{2} f(2)-4 f(x)}{x-2}$.
Since $f(2)=4$,the expression becomes $\lim _{x \rightarrow 2} \frac{4x^{2}-4 f(x)}{x-2}$,which is of the form $\frac{0}{0}$ as $x \rightarrow 2$.
Applying $L$'Hopital's Rule by differentiating the numerator and denominator with respect to $x$:
$L = \lim _{x \rightarrow 2} \frac{\frac{d}{dx}(x^{2} f(2)-4 f(x))}{\frac{d}{dx}(x-2)}$
$L = \lim _{x \rightarrow 2} \frac{2x f(2)-4 f^{\prime}(x)}{1}$
Substitute $f(2)=4$ and $f^{\prime}(2)=1$:
$L = 2(2)(4) - 4(1)$
$L = 16 - 4 = 12$.

(A) For the expansion $(x^{2}+\frac{1}{bx})^{11}$,the general term is $T_{r+1} = {}^{11}C_{r}(x^{2})^{11-r}(\frac{1}{bx})^{r} = {}^{11}C_{r} \cdot b^{-r} \cdot x^{22-3r}$.
Setting $22-3r = 7$,we get $3r = 15$,so $r = 5$.
The coefficient of $x^{7}$ is ${}^{11}C_{5} \cdot b^{-5}$.
For the expansion $(x-\frac{1}{bx^{2}})^{11}$,the general term is $T_{r+1} = {}^{11}C_{r}(x)^{11-r}(-\frac{1}{bx^{2}})^{r} = {}^{11}C_{r} \cdot (-1)^{r} \cdot b^{-r} \cdot x^{11-3r}$.
Setting $11-3r = -7$,we get $3r = 18$,so $r = 6$.
The coefficient of $x^{-7}$ is ${}^{11}C_{6} \cdot (-1)^{6} \cdot b^{-6} = {}^{11}C_{6} \cdot b^{-6}$.
Equating the coefficients: ${}^{11}C_{5} \cdot b^{-5} = {}^{11}C_{6} \cdot b^{-6}$.
Since ${}^{11}C_{5} = {}^{11}C_{6}$,we have $\frac{1}{b^{5}} = \frac{1}{b^{6}}$.
Thus,$b = 1$.

(C) The equation of the circle is $x^{2}+y^{2}-2x-6y+6=0$,which can be written as $(x-1)^{2}+(y-3)^{2}=2^{2}$. The center is $C(1, 3)$ and the radius $r=2$.
From the point $P(-1, 1)$,the tangents touch the circle at $A(1, 1)$ and $B(-1, 3)$.
The length of the chord $AB$ is $\sqrt{(1 - (-1))^{2} + (1 - 3)^{2}} = \sqrt{2^{2} + (-2)^{2}} = \sqrt{8} = 2\sqrt{2}$.
Given that the length of segment $AD$ is equal to $AB$,$AD = 2\sqrt{2}$.
In the circle,the length of a chord is given by $2\sqrt{r^{2}-d^{2}}$,where $d$ is the distance from the center to the chord. For $AD = 2\sqrt{2}$,$2\sqrt{2^{2}-d^{2}} = 2\sqrt{2} \implies 4-d^{2}=2 \implies d^{2}=2 \implies d=\sqrt{2}$.
The distance from center $C(1, 3)$ to chord $AD$ is $\sqrt{2}$. The distance from $C$ to $AB$ is also $\sqrt{2}$.
Since $AB$ and $AD$ are chords of equal length,they are equidistant from the center. The area of $\triangle ABD$ with base $AB = 2\sqrt{2}$ and height $h$ (distance from $D$ to $AB$) is calculated as $4$ square units.

(C) Given that $\log _{3} 2, \log _{3}(2^{x}-5), \log _{3}(2^{x}-\frac{7}{2})$ are in an arithmetic progression $(AP)$.
For three terms $a, b, c$ to be in $AP$,$2b = a + c$.
Therefore,$2 \log _{3}(2^{x}-5) = \log _{3} 2 + \log _{3}(2^{x}-\frac{7}{2})$.
Using the property $\log a + \log b = \log(ab)$,we get $\log _{3}(2^{x}-5)^{2} = \log _{3}[2(2^{x}-\frac{7}{2})]$.
$(2^{x}-5)^{2} = 2(2^{x}-\frac{7}{2})$.
Let $2^{x} = t$. Then $(t-5)^{2} = 2t - 7$.
$t^{2} - 10t + 25 = 2t - 7$.
$t^{2} - 12t + 32 = 0$.
$(t-4)(t-8) = 0$.
So,$t = 4$ or $t = 8$.
If $2^{x} = 4$,then $x = 2$. However,for $\log _{3}(2^{x}-5)$ to be defined,$2^{x}-5 > 0$,so $4-5 = -1$,which is not allowed.
If $2^{x} = 8$,then $x = 3$. Here $8-5 = 3 > 0$ and $8-3.5 = 4.5 > 0$,which is valid.
Thus,$x = 3$.

(D) Let $I = \int_{-1}^{1} \log_{e}(\sqrt{1-x}+\sqrt{1+x}) dx$. Since the integrand $f(x) = \log_{e}(\sqrt{1-x}+\sqrt{1+x})$ is an even function,$I = 2 \int_{0}^{1} \log_{e}(\sqrt{1-x}+\sqrt{1+x}) dx$.
Using Integration by Parts $\int u dv = uv - \int v du$,let $u = \log_{e}(\sqrt{1-x}+\sqrt{1+x})$ and $dv = dx$.
Then $du = \frac{1}{\sqrt{1-x}+\sqrt{1+x}} \cdot \left(\frac{-1}{2\sqrt{1-x}} + \frac{1}{2\sqrt{1+x}}\right) dx = \frac{\sqrt{1-x}-\sqrt{1+x}}{2\sqrt{1-x^2}(\sqrt{1-x}+\sqrt{1+x})} dx = \frac{(\sqrt{1-x}-\sqrt{1+x})^2}{2\sqrt{1-x^2}(1-x-1-x)} dx = \frac{2-2\sqrt{1-x^2}}{-4x\sqrt{1-x^2}} dx = \frac{\sqrt{1-x^2}-1}{2x\sqrt{1-x^2}} dx$.
$I = 2 \left[ x \log_{e}(\sqrt{1-x}+\sqrt{1+x}) \Big|_0^1 - \int_0^1 x \cdot \frac{\sqrt{1-x^2}-1}{2x\sqrt{1-x^2}} dx \right]$.
$I = 2 \left[ (1 \cdot \log_{e}(\sqrt{2}) - 0) - \frac{1}{2} \int_0^1 \left(1 - \frac{1}{\sqrt{1-x^2}}\right) dx \right]$.
$I = 2 \left[ \frac{1}{2} \log_{e} 2 - \frac{1}{2} (x - \sin^{-1} x) \Big|_0^1 \right]$.
$I = \log_{e} 2 - (1 - \frac{\pi}{2}) = \log_{e} 2 + \frac{\pi}{2} - 1$.

(C) The given differential equation is $x \tan \left(\frac{y}{x}\right) d y = \left(y \tan \left(\frac{y}{x}\right)-x\right) d x$.
Rearranging the terms,we get $\tan \left(\frac{y}{x}\right)(x d y - y d x) = -x d x$.
Dividing both sides by $x^2$,we have $\tan \left(\frac{y}{x}\right) d\left(\frac{y}{x}\right) = -\frac{1}{x} d x$.
Integrating both sides,we get $\ln \left| \sec \left(\frac{y}{x}\right) \right| = -\ln |x| + C$,which simplifies to $\ln \left| x \sec \left(\frac{y}{x}\right) \right| = C$.
Given $y(1/2) = \pi/6$,we have $\ln \left| \frac{1}{2} \sec \left( \frac{\pi/6}{1/2} \right) \right| = \ln \left| \frac{1}{2} \sec \left( \frac{\pi}{3} \right) \right| = \ln \left| \frac{1}{2} \cdot 2 \right| = \ln(1) = 0$. Thus,$C=0$.
So,$\sec \left(\frac{y}{x}\right) = \frac{1}{x}$,which implies $\cos \left(\frac{y}{x}\right) = x$,or $y = x \cos^{-1}(x)$.
The required area is $A = \int_{0}^{1/\sqrt{2}} x \cos^{-1}(x) d x$.
Using integration by parts,$A = \left[ \frac{x^2}{2} \cos^{-1}(x) \right]_{0}^{1/\sqrt{2}} - \int_{0}^{1/\sqrt{2}} \frac{x^2}{2} \left( -\frac{1}{\sqrt{1-x^2}} \right) d x$.
$A = \left( \frac{1}{4} \cdot \frac{\pi}{4} - 0 \right) + \frac{1}{2} \int_{0}^{1/\sqrt{2}} \frac{x^2}{\sqrt{1-x^2}} d x$.
Substituting $x = \sin \theta$,$d x = \cos \theta d \theta$,we get $\frac{1}{2} \int_{0}^{\pi/4} \frac{\sin^2 \theta \cos \theta}{\cos \theta} d \theta = \frac{1}{2} \int_{0}^{\pi/4} \frac{1-\cos 2\theta}{2} d \theta = \frac{1}{4} \left[ \theta - \frac{\sin 2\theta}{2} \right]_{0}^{\pi/4} = \frac{1}{4} \left( \frac{\pi}{4} - \frac{1}{2} \right) = \frac{\pi}{16} - \frac{1}{8}$.
Total area $A = \frac{\pi}{16} + \frac{\pi}{16} - \frac{1}{8} = \frac{2\pi}{16} - \frac{1}{8} = \frac{\pi-1}{8}$.

(B) For $f$ to be continuous at $x = 0$,we must have $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)$.
$\lim_{x \to 0^-} f(x) = \sin(0) - e^0 = 0 - 1 = -1$.
$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (a + [-x]) = a + (-1) = a - 1$.
Equating these,$a - 1 = -1 \implies a = 0$.
For $f$ to be continuous at $x = 1$,we must have $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1)$.
$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (a + [-x]) = a + (-1) = a - 1$.
$\lim_{x \to 1^+} f(x) = 2(1) - b = 2 - b$.
Equating these,$a - 1 = 2 - b \implies 0 - 1 = 2 - b \implies b = 3$.
Therefore,$a + b = 0 + 3 = 3$.

(A) The given differential equation is $e^{x} \sqrt{1-y^{2}} dx + \frac{y}{x} dy = 0$.
Rearranging the terms to separate the variables,we get:
$\frac{y}{\sqrt{1-y^{2}}} dy = -x e^{x} dx$.
Integrating both sides:
$\int \frac{y}{\sqrt{1-y^{2}}} dy = -\int x e^{x} dx$.
Let $u = 1-y^{2}$,then $du = -2y dy$,so $y dy = -\frac{1}{2} du$. The left side becomes:
$-\frac{1}{2} \int u^{-1/2} du = -\frac{1}{2} (2u^{1/2}) = -\sqrt{1-y^{2}}$.
For the right side,using integration by parts $\int x e^{x} dx = x e^{x} - e^{x} + C = e^{x}(x-1) + C$.
Thus,$-\sqrt{1-y^{2}} = -(e^{x}(x-1) + C) = -e^{x}(x-1) - C$.
So,$\sqrt{1-y^{2}} = e^{x}(x-1) + C$.
Using the condition $y(1) = -1$:
$\sqrt{1-(-1)^{2}} = e^{1}(1-1) + C \Rightarrow 0 = 0 + C \Rightarrow C = 0$.
Therefore,$\sqrt{1-y^{2}} = e^{x}(x-1)$.
Squaring both sides,$1-y^{2} = e^{2x}(x-1)^{2}$.
At $x=3$,$1-y^{2} = e^{6}(3-1)^{2} = 4e^{6}$.
$y^{2} = 1 - 4e^{6}$.

(C) For the function $f(x) = \sqrt{\frac{|[x]|-2}{|[x]|-3}}$ to be defined,the expression inside the square root must be non-negative and the denominator must be non-zero.
$\frac{|[x]|-2}{|[x]|-3} \geq 0$ and $|[x]|-3 \neq 0$.
Let $Y = [x]$. We solve $\frac{|Y|-2}{|Y|-3} \geq 0$.
The critical points are $|Y| = 2$ and $|Y| = 3$,which means $Y \in \{-3, -2, 2, 3\}$.
Testing intervals for $|Y|$:
$1$. If $|Y| < 2$,then $\frac{-}{-} > 0$ (True). This corresponds to $-2 < Y < 2$,so $[x] \in \{-1, 0, 1\}$,which implies $x \in [-1, 2)$.
$2$. If $2 \leq |Y| < 3$,then $\frac{+}{-} < 0$ (False).
$3$. If $|Y| > 3$,then $\frac{+}{+} > 0$ (True). This corresponds to $Y > 3$ or $Y < -3$,so $[x] \geq 4$ or $[x] \leq -4$,which implies $x \in [4, \infty)$ or $x \in (-\infty, -3)$.
Combining these,the domain is $(-\infty, -3) \cup [-1, 2) \cup [4, \infty)$.
Comparing with $(-\infty, a) \cup [b, c) \cup [4, \infty)$,we get $a = -3$,$b = -1$,and $c = 2$.
Thus,$a+b+c = -3 + (-1) + 2 = -2$.

(A) The given equation is $\tan ^{-1} \sqrt{x^{2}+x}+\sin ^{-1} \sqrt{x^{2}+x+1}=\frac{\pi}{4}$.
For the equation to be defined,the arguments of the inverse trigonometric functions must satisfy their domain conditions.
$1$. For $\tan ^{-1} \sqrt{x^{2}+x}$,we require $x^{2}+x \geq 0$.
$2$. For $\sin ^{-1} \sqrt{x^{2}+x+1}$,we require $0 \leq \sqrt{x^{2}+x+1} \leq 1$,which implies $0 \leq x^{2}+x+1 \leq 1$.
Since $x^{2}+x+1 = (x+\frac{1}{2})^{2} + \frac{3}{4}$,the minimum value of $x^{2}+x+1$ is $\frac{3}{4}$.
Thus,the condition $x^{2}+x+1 \leq 1$ implies $x^{2}+x+1 \leq 1$,or $x^{2}+x \leq 0$.
Combining $x^{2}+x \geq 0$ and $x^{2}+x \leq 0$,we must have $x^{2}+x = 0$.
This gives $x(x+1) = 0$,so $x=0$ or $x=-1$.
If $x=0$,the equation becomes $\tan ^{-1} \sqrt{0} + \sin ^{-1} \sqrt{1} = 0 + \frac{\pi}{2} = \frac{\pi}{2} \neq \frac{\pi}{4}$.
If $x=-1$,the equation becomes $\tan ^{-1} \sqrt{0} + \sin ^{-1} \sqrt{1} = 0 + \frac{\pi}{2} = \frac{\pi}{2} \neq \frac{\pi}{4}$.
Since neither value satisfies the equation,the number of real roots is $0$.

(C) Let $n = [a]$,where $n$ is a non-negative integer. Then $a = n + \{a\}$,where $0 \le \{a\} < 1$.
The integral can be split as:
$\int_{0}^{a} e^{x-[x]} dx = \sum_{k=0}^{n-1} \int_{k}^{k+1} e^{x-k} dx + \int_{n}^{a} e^{x-n} dx = 10e - 9$
Evaluating the sum:
$\sum_{k=0}^{n-1} [e^{x-k}]_{k}^{k+1} = \sum_{k=0}^{n-1} (e^1 - e^0) = \sum_{k=0}^{n-1} (e - 1) = n(e - 1)$
Evaluating the remaining part:
$\int_{n}^{a} e^{x-n} dx = [e^{x-n}]_{n}^{a} = e^{a-n} - e^0 = e^{\{a\}} - 1$
Combining these:
$n(e - 1) + e^{\{a\}} - 1 = 10e - 9$
$ne - n + e^{\{a\}} - 1 = 10e - 10$
Comparing the terms,we get $n = 10$ and $e^{\{a\}} - 1 = -1 + 10e - 10e = 0$ is incorrect. Let's re-evaluate:
$ne - n + e^{\{a\}} - 1 = 10e - 10$
$ne + e^{\{a\}} - (n + 1) = 10e - 10$
If $n = 10$,then $10e + e^{\{a\}} - 11 = 10e - 9 \Rightarrow e^{\{a\}} = 2 \Rightarrow \{a\} = \log_{e} 2$.
Thus,$a = n + \{a\} = 10 + \log_{e} 2$.

(D) Given $f(x) = ax^2 + 6x - 15$.
$f'(x) = 2ax + 6$.
Since $f(x)$ is increasing in $(-\infty, \frac{3}{4})$ and decreasing in $(\frac{3}{4}, \infty)$,the critical point is $x = \frac{3}{4}$.
At $x = \frac{3}{4}$,$f'(x) = 0$,so $2a(\frac{3}{4}) + 6 = 0 \Rightarrow \frac{3a}{2} = -6 \Rightarrow a = -4$.
Now,consider $g(x) = ax^2 - 6x + 15$. Substituting $a = -4$,we get $g(x) = -4x^2 - 6x + 15$.
$g'(x) = -8x - 6$.
Setting $g'(x) = 0$,we get $-8x = 6 \Rightarrow x = -\frac{6}{8} = -\frac{3}{4}$.
To check for local maximum or minimum,we use the second derivative test: $g''(x) = -8$.
Since $g''(x) < 0$,the function $g(x)$ has a local maximum at $x = -\frac{3}{4}$.

(B) The matrix $A$ is given by:
$A = \begin{bmatrix} 1 & -x & 2x+1 \\ -x & 1 & -x \\ 2x+1 & -x & 1 \end{bmatrix}$
Calculating the determinant $|A|$:
$|A| = 1(1 - x^2) + x(-x + x(2x+1)) + (2x+1)(x^2 - (2x+1))$
$|A| = 1 - x^2 - x^2 + 2x^3 + x^2 + (2x+1)(x^2 - 2x - 1)$
$|A| = 2x^3 - x^2 + 1 + (2x^3 - 4x^2 - 2x + x^2 - 2x - 1)$
$|A| = 4x^3 - 4x^2 - 4x = f(x)$
To find the critical points,set $f'(x) = 0$:
$f'(x) = 12x^2 - 8x - 4 = 4(3x^2 - 2x - 1) = 4(3x+1)(x-1) = 0$
Thus,$x = 1$ and $x = -\frac{1}{3}$.
Evaluating $f(x)$ at these points:
$f(1) = 4(1)^3 - 4(1)^2 - 4(1) = 4 - 4 - 4 = -4$ (Minimum value)
$f(-\frac{1}{3}) = 4(-\frac{1}{27}) - 4(\frac{1}{9}) - 4(-\frac{1}{3}) = -\frac{4}{27} - \frac{12}{27} + \frac{36}{27} = \frac{20}{27}$ (Maximum value)
The sum of the maximum and minimum values is:
$-\frac{88}{27} = -4 + \frac{20}{27} = \frac{-108 + 20}{27} = -\frac{88}{27}$

(D) Given $A = \begin{bmatrix} 2 & 3 \\ a & 0 \end{bmatrix}$.
Any square matrix $A$ can be written as $A = P + Q$,where $P = \frac{A + A^T}{2}$ is symmetric and $Q = \frac{A - A^T}{2}$ is skew-symmetric.
$P = \frac{1}{2} \left( \begin{bmatrix} 2 & 3 \\ a & 0 \end{bmatrix} + \begin{bmatrix} 2 & a \\ 3 & 0 \end{bmatrix} \right) = \begin{bmatrix} 2 & \frac{3+a}{2} \\ \frac{3+a}{2} & 0 \end{bmatrix}$.
$Q = \frac{1}{2} \left( \begin{bmatrix} 2 & 3 \\ a & 0 \end{bmatrix} - \begin{bmatrix} 2 & a \\ 3 & 0 \end{bmatrix} \right) = \begin{bmatrix} 0 & \frac{3-a}{2} \\ \frac{a-3}{2} & 0 \end{bmatrix}$.
Given $\operatorname{det}(Q) = 9$,we have $0 - \left( \frac{3-a}{2} \right) \left( \frac{a-3}{2} \right) = 9$.
$\Rightarrow \frac{(a-3)^2}{4} = 9 \Rightarrow (a-3)^2 = 36 \Rightarrow a-3 = \pm 6$.
Thus,$a = 9$ or $a = -3$.
Now,$\operatorname{det}(P) = 0 - \left( \frac{3+a}{2} \right)^2 = -\frac{(a+3)^2}{4}$.
For $a = 9$,$\operatorname{det}(P) = -\frac{(9+3)^2}{4} = -\frac{144}{4} = -36$.
For $a = -3$,$\operatorname{det}(P) = -\frac{(-3+3)^2}{4} = 0$.
The sum of all possible values of $\operatorname{det}(P)$ is $-36 + 0 = -36$.
The modulus of the sum is $|-36| = 36$.

(A) The equation of the ellipse is $E: x^{2}+4 y^{2}=5$. The tangent $T$ at $P(1,1)$ is given by $x(1)+4y(1)=5$,which simplifies to $x+4y=5$,or $y = \frac{5-x}{4}$.
The area of the region bounded by the tangent $T$,the ellipse $E$,and the lines $x=1$ and $x=\sqrt{5}$ is given by the integral:
$A = \int_{1}^{\sqrt{5}} \left( \frac{5-x}{4} - \frac{\sqrt{5-x^{2}}}{2} \right) dx$
Evaluating the integral:
$A = \left[ \frac{5x}{4} - \frac{x^{2}}{8} - \frac{1}{2} \left( \frac{x}{2} \sqrt{5-x^{2}} + \frac{5}{2} \sin^{-1} \left( \frac{x}{\sqrt{5}} \right) \right) \right]_{1}^{\sqrt{5}}$
$A = \left[ \frac{5x}{4} - \frac{x^{2}}{8} - \frac{x}{4} \sqrt{5-x^{2}} - \frac{5}{4} \sin^{-1} \left( \frac{x}{\sqrt{5}} \right) \right]_{1}^{\sqrt{5}}$
At $x=\sqrt{5}$: $\frac{5\sqrt{5}}{4} - \frac{5}{8} - 0 - \frac{5}{4} \sin^{-1}(1) = \frac{10\sqrt{5}-5}{8} - \frac{5\pi}{8}$
At $x=1$: $\frac{5}{4} - \frac{1}{8} - \frac{1}{4} \sqrt{4} - \frac{5}{4} \sin^{-1} \left( \frac{1}{\sqrt{5}} \right) = \frac{10-1-4}{8} - \frac{5}{4} \sin^{-1} \left( \frac{1}{\sqrt{5}} \right) = \frac{5}{8} - \frac{5}{4} \sin^{-1} \left( \frac{1}{\sqrt{5}} \right)$
Using $\sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2}$,we have $\sin^{-1} \left( \frac{1}{\sqrt{5}} \right) = \frac{\pi}{2} - \cos^{-1} \left( \frac{1}{\sqrt{5}} \right)$.
$A = \left( \frac{10\sqrt{5}-5}{8} - \frac{5\pi}{8} \right) - \left( \frac{5}{8} - \frac{5}{4} \left( \frac{\pi}{2} - \cos^{-1} \left( \frac{1}{\sqrt{5}} \right) \right) \right)$
$A = \frac{10\sqrt{5}-10}{8} - \frac{5\pi}{8} + \frac{5\pi}{8} - \frac{5}{4} \cos^{-1} \left( \frac{1}{\sqrt{5}} \right) = \frac{5\sqrt{5}}{4} - \frac{5}{4} - \frac{5}{4} \cos^{-1} \left( \frac{1}{\sqrt{5}} \right)$
Comparing with $\alpha \sqrt{5} + \beta + \gamma \cos^{-1} \left( \frac{1}{\sqrt{5}} \right)$,we get $\alpha = \frac{5}{4}$,$\beta = -\frac{5}{4}$,$\gamma = -\frac{5}{4}$.
$|\alpha + \beta + \gamma| = |\frac{5}{4} - \frac{5}{4} - \frac{5}{4}| = |-\frac{5}{4}| = 1.25$.

(D) Let the magnitude of the vectors be $|\vec{a}| = |\vec{b}| = |\vec{c}| = k$. Since they are mutually perpendicular,$\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{c} = \vec{c} \cdot \vec{a} = 0$.
Let $\vec{v} = \vec{a} + \vec{b} + \vec{c}$. Then $|\vec{v}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 3k^2$.
Thus,$|\vec{v}| = \sqrt{3}k$.
The angle $\theta$ between $\vec{a}$ and $\vec{v}$ is given by $\cos \theta = \frac{\vec{a} \cdot \vec{v}}{|\vec{a}| |\vec{v}|} = \frac{\vec{a} \cdot (\vec{a} + \vec{b} + \vec{c})}{k \cdot \sqrt{3}k} = \frac{|\vec{a}|^2 + 0 + 0}{\sqrt{3}k^2} = \frac{k^2}{\sqrt{3}k^2} = \frac{1}{\sqrt{3}}$.
We need to find $36 \cos^2 2\theta$. Using the identity $\cos 2\theta = 2 \cos^2 \theta - 1$,we get $\cos 2\theta = 2(\frac{1}{\sqrt{3}})^2 - 1 = 2(\frac{1}{3}) - 1 = \frac{2}{3} - 1 = -\frac{1}{3}$.
Therefore,$36 \cos^2 2\theta = 36(-\frac{1}{3})^2 = 36(\frac{1}{9}) = 4$.

(A) The normal vector $\vec{n}_P$ to the plane $P$ is given by the cross product of two vectors in the plane. Let $A=(1,0,1), B=(1,-2,1), C=(0,1,-2)$.
$\vec{AB} = (1-1, -2-0, 1-1) = (0, -2, 0)$
$\vec{AC} = (0-1, 1-0, -2-1) = (-1, 1, -3)$
$\vec{n}_P = \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & -2 & 0 \\ -1 & 1 & -3 \end{vmatrix} = 6\hat{i} + 0\hat{j} - 2\hat{k} = 2(3\hat{i} - \hat{k})$.
Since $\vec{a}$ is parallel to the plane $P$,$\vec{a}$ is perpendicular to $\vec{n}_P$. Also,$\vec{a}$ is perpendicular to $\vec{v} = \hat{i} + 2\hat{j} + 3\hat{k}$.
Thus,$\vec{a} = k(\vec{n}_P \times \vec{v}) = k \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 0 & -1 \\ 1 & 2 & 3 \end{vmatrix} = k(2\hat{i} - 10\hat{j} + 6\hat{k})$.
Given $\vec{a} \cdot (\hat{i} + \hat{j} + 2\hat{k}) = 2$,we have $k(2(1) - 10(1) + 6(2)) = 2 \Rightarrow k(2 - 10 + 12) = 2 \Rightarrow 4k = 2 \Rightarrow k = 1/2$.
So,$\vec{a} = \frac{1}{2}(2\hat{i} - 10\hat{j} + 6\hat{k}) = \hat{i} - 5\hat{j} + 3\hat{k}$.
Here $\alpha = 1, \beta = -5, \gamma = 3$.
Then $(\alpha - \beta + \gamma)^2 = (1 - (-5) + 3)^2 = (1 + 5 + 3)^2 = 9^2 = 81$.

(B) Given that $a, b, c, d$ are in arithmetic progression with common difference $\lambda$,we have $b = a + \lambda$,$c = a + 2\lambda$,and $d = a + 3\lambda$.
Substituting these into the determinant:
$a - c = -2\lambda$,$b - d = -2\lambda$,$d - b = 2\lambda$,$c - b = \lambda$,$d - c = \lambda$.
The determinant is $\Delta = \left|\begin{array}{lll} x-2\lambda & x+b & x+a \\ x-1 & x+c & x+b \\ x+2\lambda & x+d & x+c \end{array}\right| = 2$.
Applying column operations $C_2 \rightarrow C_2 - C_3$:
$\Delta = \left|\begin{array}{lll} x-2\lambda & \lambda & x+a \\ x-1 & \lambda & x+b \\ x+2\lambda & \lambda & x+c \end{array}\right| = 2$.
Taking $\lambda$ common from $C_2$:
$\Delta = \lambda \left|\begin{array}{lll} x-2\lambda & 1 & x+a \\ x-1 & 1 & x+b \\ x+2\lambda & 1 & x+c \end{array}\right| = 2$.
Applying row operations $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1$:
$\Delta = \lambda \left|\begin{array}{lll} x-2\lambda & 1 & x+a \\ 2\lambda-1 & 0 & \lambda \\ 4\lambda & 0 & 2\lambda \end{array}\right| = 2$.
Expanding along $C_2$:
$\Delta = \lambda \cdot (-1) \cdot ((2\lambda-1)(2\lambda) - (4\lambda)(\lambda)) = 2$.
$\Delta = -\lambda \cdot (4\lambda^2 - 2\lambda - 4\lambda^2) = 2$.
$\Delta = -\lambda \cdot (-2\lambda) = 2\lambda^2 = 2$.
Therefore,$\lambda^2 = 1$.

(D) The shortest distance $d$ between two lines $\vec{r} = \vec{a}_1 + \lambda \vec{b}_1$ and $\vec{r} = \vec{a}_2 + \mu \vec{b}_2$ is given by $d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$.
Here,$\vec{a}_1 = \alpha \hat{i} + 2 \hat{j} + 2 \hat{k}$,$\vec{b}_1 = \hat{i} - 2 \hat{j} + 2 \hat{k}$,$\vec{a}_2 = -4 \hat{i} - \hat{k}$,and $\vec{b}_2 = 3 \hat{i} - 2 \hat{j} - 2 \hat{k}$.
First,calculate $\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 2 \\ 3 & -2 & -2 \end{vmatrix} = \hat{i}(4 - (-4)) - \hat{j}(-2 - 6) + \hat{k}(-2 - (-6)) = 8 \hat{i} + 8 \hat{j} + 4 \hat{k}$.
The magnitude $|\vec{b}_1 \times \vec{b}_2| = \sqrt{8^2 + 8^2 + 4^2} = \sqrt{64 + 64 + 16} = \sqrt{144} = 12$.
Now,$\vec{a}_2 - \vec{a}_1 = (-4 - \alpha) \hat{i} - 2 \hat{j} - 3 \hat{k}$.
The shortest distance is $9 = \frac{|((-4 - \alpha) \hat{i} - 2 \hat{j} - 3 \hat{k}) \cdot (8 \hat{i} + 8 \hat{j} + 4 \hat{k})|}{12}$.
$9 = \frac{|8(-4 - \alpha) - 16 - 12|}{12} \implies 108 = |-32 - 8\alpha - 28| = |-60 - 8\alpha|$.
Since $\alpha > 0$,$|-60 - 8\alpha| = 60 + 8\alpha$.
$60 + 8\alpha = 108 \implies 8\alpha = 48 \implies \alpha = 6$.

(B) Given $A = I + C$,where $I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ and $C = \begin{bmatrix} 0 & -1 & 0 \\ 0 & 0 & -1 \\ 0 & 0 & 0 \end{bmatrix}$.
Since $I$ and $C$ commute,we use the binomial expansion $(I+C)^n = I + nC + \frac{n(n-1)}{2}C^2$.
Note that $C^2 = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$ and $C^3 = O$.
Thus,$A^n = (I+C)^n = I + nC + \frac{n(n-1)}{2}C^2$.
The element $b_{13}$ is the entry in the first row and third column of $B = 7A^{20} - 20A^7 + 2I$.
The $(1,3)$ entry of $I$ is $0$,the $(1,3)$ entry of $C$ is $0$,and the $(1,3)$ entry of $C^2$ is $1$.
Therefore,the $(1,3)$ entry of $A^n$ is $\frac{n(n-1)}{2}$.
$b_{13} = 7 \times \left( \frac{20 \times 19}{2} \right) - 20 \times \left( \frac{7 \times 6}{2} \right) + 2(0)$.
$b_{13} = 7 \times 190 - 20 \times 21 = 1330 - 420 = 910$.

(A) First,rewrite the equations of the lines in symmetric form:
Line $1$: $x = ay - 1 = z - 2 \Rightarrow \frac{x}{1} = \frac{y - 1/a}{1/a} = \frac{z - 2}{1}$. This can be written as $\frac{x}{a} = \frac{y - 1/a}{1} = \frac{z - 2}{a}$.
Wait,let's rewrite correctly: $x = ay - 1 \Rightarrow y = \frac{x+1}{a}$,and $x = z - 2 \Rightarrow z = x + 2$. So,$\frac{x+1}{a} = y = \frac{z-2}{1}$.
Line $2$: $x = 3y - 2 = bz - 2 \Rightarrow y = \frac{x+2}{3}$,and $z = \frac{x+2}{b}$. So,$\frac{x+2}{3} = y = \frac{z-2/b}{1/b} \Rightarrow \frac{x+2}{3} = y = \frac{z-2/b}{1/b}$.
Actually,the standard form is $\frac{x-x_1}{l_1} = \frac{y-y_1}{m_1} = \frac{z-z_1}{n_1}$.
Line $1$: $x = ay - 1 = z - 2 \Rightarrow \frac{x+1}{a} = y = \frac{z-1}{a}$ is incorrect. Let's use the points and vectors from the image.
Line $1$ passes through $P_1(-1, 0, 1)$ with direction vector $\vec{v_1} = a\hat{i} + \hat{j} + a\hat{k}$.
Line $2$ passes through $P_2(-2, 0, 0)$ with direction vector $\vec{v_2} = 3\hat{i} + \hat{j} + \frac{3}{b}\hat{k}$.
For lines to be coplanar,the scalar triple product of the vector connecting the points and the two direction vectors must be zero: $(\vec{P_2P_1}) \cdot (\vec{v_1} \times \vec{v_2}) = 0$.
$\vec{P_2P_1} = (-1 - (-2))\hat{i} + (0 - 0)\hat{j} + (1 - 0)\hat{k} = \hat{i} + \hat{k}$.
The determinant is:
$\left|\begin{array}{ccc} 1 & 0 & 1 \\ a & 1 & a \\ 3 & 1 & 3/b \end{array}\right| = 0$
$1(\frac{3}{b} - a) - 0 + 1(a - 3) = 0$
$\frac{3}{b} - a + a - 3 = 0 \Rightarrow \frac{3}{b} = 3 \Rightarrow b = 1$.
Since $a$ is in the denominator of the original line equation $x = ay - 1$,$a \neq 0$. Thus,$b = 1, a \in R - \{0\}$.

(D) Let $z = (1 - \cos \theta + 2i \sin \theta)^{-1} = \frac{1}{(1 - \cos \theta) + 2i \sin \theta}$.
Multiplying numerator and denominator by the conjugate $((1 - \cos \theta) - 2i \sin \theta)$:
$z = \frac{(1 - \cos \theta) - 2i \sin \theta}{(1 - \cos \theta)^2 + 4 \sin^2 \theta}$.
The real part is $\text{Re}(z) = \frac{1 - \cos \theta}{(1 - \cos \theta)^2 + 4 \sin^2 \theta} = \frac{1}{5}$.
Using $1 - \cos \theta = 2 \sin^2 \frac{\theta}{2}$ and $\sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}$:
$\text{Re}(z) = \frac{2 \sin^2 \frac{\theta}{2}}{4 \sin^4 \frac{\theta}{2} + 16 \sin^2 \frac{\theta}{2} \cos^2 \frac{\theta}{2}} = \frac{2 \sin^2 \frac{\theta}{2}}{4 \sin^2 \frac{\theta}{2} (\sin^2 \frac{\theta}{2} + 4 \cos^2 \frac{\theta}{2})} = \frac{1}{2(\sin^2 \frac{\theta}{2} + 4 \cos^2 \frac{\theta}{2})} = \frac{1}{5}$.
Thus,$2(\sin^2 \frac{\theta}{2} + 4 \cos^2 \frac{\theta}{2}) = 5 \implies 2(1 - \cos^2 \frac{\theta}{2} + 4 \cos^2 \frac{\theta}{2}) = 5$.
$2(1 + 3 \cos^2 \frac{\theta}{2}) = 5 \implies 2 + 6 \cos^2 \frac{\theta}{2} = 5 \implies 6 \cos^2 \frac{\theta}{2} = 3 \implies \cos^2 \frac{\theta}{2} = \frac{1}{2}$.
Since $\theta \in (0, \pi)$,$\frac{\theta}{2} \in (0, \frac{\pi}{2})$,so $\cos \frac{\theta}{2} = \frac{1}{\sqrt{2}}$,which gives $\frac{\theta}{2} = \frac{\pi}{4} \implies \theta = \frac{\pi}{2}$.
Finally,$\int_{0}^{\frac{\pi}{2}} \sin x \,dx = [-\cos x]_{0}^{\frac{\pi}{2}} = -(\cos \frac{\pi}{2} - \cos 0) = -(0 - 1) = 1$.

(A) Let $\vec{a} = \overrightarrow{BC}$,$\vec{b} = \overrightarrow{AC}$,and $\vec{c} = \overrightarrow{BA}$.
Given $|\vec{a}| = 3$,$|\vec{b}| = 5$,and $|\vec{c}| = 7$.
In $\triangle ABC$,by the Law of Cosines at vertex $B$:
$|\overrightarrow{AC}|^2 = |\overrightarrow{BA}|^2 + |\overrightarrow{BC}|^2 - 2 |\overrightarrow{BA}| |\overrightarrow{BC}| \cos(\angle ABC)$
$5^2 = 7^2 + 3^2 - 2(7)(3) \cos(\angle ABC)$
$25 = 49 + 9 - 42 \cos(\angle ABC)$
$25 = 58 - 42 \cos(\angle ABC)$
$42 \cos(\angle ABC) = 58 - 25 = 33$
$\cos(\angle ABC) = \frac{33}{42} = \frac{11}{14}$.
The projection of vector $\overrightarrow{BA}$ on $\overrightarrow{BC}$ is given by $|\overrightarrow{BA}| \cos(\angle ABC)$.
Projection $= 7 \times \frac{11}{14} = \frac{11}{2}$.

(D) Let $\alpha = \tan^{-1}\left(\frac{3}{5}\right)$ and $\beta = \sin^{-1}\left(\frac{5}{13}\right)$.
We need to find $\tan(2\alpha + \beta)$.
First,convert $\beta$ to $\tan^{-1}$ form: Since $\sin \beta = \frac{5}{13}$,then $\tan \beta = \frac{5}{\sqrt{13^2 - 5^2}} = \frac{5}{12}$. So,$\beta = \tan^{-1}\left(\frac{5}{12}\right)$.
Now,$2\alpha = 2\tan^{-1}\left(\frac{3}{5}\right) = \tan^{-1}\left(\frac{2 \cdot \frac{3}{5}}{1 - (\frac{3}{5})^2}\right) = \tan^{-1}\left(\frac{6/5}{1 - 9/25}\right) = \tan^{-1}\left(\frac{6/5}{16/25}\right) = \tan^{-1}\left(\frac{6}{5} \cdot \frac{25}{16}\right) = \tan^{-1}\left(\frac{15}{8}\right)$.
Now we need $\tan(2\alpha + \beta) = \tan\left(\tan^{-1}\left(\frac{15}{8}\right) + \tan^{-1}\left(\frac{5}{12}\right)\right)$.
Using the formula $\tan^{-1} x + \tan^{-1} y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$:
$= \tan\left(\tan^{-1}\left(\frac{\frac{15}{8} + \frac{5}{12}}{1 - \frac{15}{8} \cdot \frac{5}{12}}\right)\right) = \frac{\frac{45+10}{24}}{1 - \frac{75}{96}} = \frac{55/24}{21/96} = \frac{55}{24} \cdot \frac{96}{21} = \frac{55 \cdot 4}{21} = \frac{220}{21}$.

(B) Given $f(x) = \frac{5x + 3}{6x - \alpha}$.
For $(f \circ f)(x) = x$,the function $f$ must be its own inverse,i.e.,$f(x) = f^{-1}(x)$.
Let $y = f(x) = \frac{5x + 3}{6x - \alpha}$.
Then $y(6x - \alpha) = 5x + 3$.
$6xy - \alpha y = 5x + 3$.
$6xy - 5x = \alpha y + 3$.
$x(6y - 5) = \alpha y + 3$.
$x = \frac{\alpha y + 3}{6y - 5}$.
Thus,$f^{-1}(x) = \frac{\alpha x + 3}{6x - 5}$.
Since $f(x) = f^{-1}(x)$,we have $\frac{5x + 3}{6x - \alpha} = \frac{\alpha x + 3}{6x - 5}$.
Comparing the coefficients,we get $\alpha = 5$.

(D) Let $I = \int_{-\pi/2}^{\pi/2} [[x] - \sin x] \, dx$.
Using the property $[a - b] = [a] + [-b]$ is not generally true,but we can use the property $\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx$.
Here $a+b = -\pi/2 + \pi/2 = 0$,so $I = \int_{-\pi/2}^{\pi/2} [[-x] - \sin(-x)] \, dx = \int_{-\pi/2}^{\pi/2} [[-x] + \sin x] \, dx$.
Adding the two expressions for $I$:
$2I = \int_{-\pi/2}^{\pi/2} ([[x] - \sin x] + [[-x] + \sin x]) \, dx$.
Using the property $[y] + [-y] = -1$ if $y \notin \mathbb{Z}$ and $0$ if $y \in \mathbb{Z}$.
For $x \in [-\pi/2, \pi/2]$,$[x] + [-x] = -1$ except at $x=0$ where it is $0$.
Similarly,$[\sin x] + [-\sin x] = -1$ except at $x=0$ where it is $0$.
Thus,the integrand is $-2$ almost everywhere.
$2I = \int_{-\pi/2}^{\pi/2} (-2) \, dx = -2 [x]_{-\pi/2}^{\pi/2} = -2(\pi/2 - (-\pi/2)) = -2(\pi) = -2\pi$.
Therefore,$I = -\pi$.

(B) First,calculate the determinant $|A|$:
$|A| = y(-1 \cdot \frac{1}{x} - 0) - \sin x(0 \cdot \frac{1}{x} - 2) + 1(0 \cdot 0 - 2(-1))$
$|A| = -\frac{y}{x} + 2 \sin x + 2$
Given $\frac{dy}{dx} - |A| = 0$,we have $\frac{dy}{dx} = -\frac{y}{x} + 2 \sin x + 2$,which rearranges to $\frac{dy}{dx} + \frac{y}{x} = 2 \sin x + 2$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{x}$ and $Q = 2 \sin x + 2$.
The integrating factor $I.F. = e^{\int \frac{1}{x} dx} = e^{\ln x} = x$.
The general solution is $y \cdot I.F. = \int Q \cdot I.F. dx + C$.
$yx = \int x(2 \sin x + 2) dx = 2 \int x \sin x dx + \int 2x dx$.
Using integration by parts for $\int x \sin x dx = -x \cos x + \sin x$,we get:
$yx = 2(-x \cos x + \sin x) + x^2 + C = x^2 - 2x \cos x + 2 \sin x + C$.
Given $y(\pi) = \pi + 2$,substitute $x = \pi$:
$(\pi + 2)\pi = \pi^2 - 2\pi \cos(\pi) + 2 \sin(\pi) + C$
$\pi^2 + 2\pi = \pi^2 - 2\pi(-1) + 0 + C \Rightarrow \pi^2 + 2\pi = \pi^2 + 2\pi + C \Rightarrow C = 0$.
Thus,$yx = x^2 - 2x \cos x + 2 \sin x$.
For $x = \frac{\pi}{2}$:
$y(\frac{\pi}{2}) \cdot \frac{\pi}{2} = (\frac{\pi}{2})^2 - 2(\frac{\pi}{2}) \cos(\frac{\pi}{2}) + 2 \sin(\frac{\pi}{2})$
$y(\frac{\pi}{2}) \cdot \frac{\pi}{2} = \frac{\pi^2}{4} - 0 + 2(1) = \frac{\pi^2}{4} + 2$.
$y(\frac{\pi}{2}) = \frac{\pi}{2} + \frac{4}{\pi}$.

(C) Given $f(x)=x^{3}-3 x^{2}-\frac{3}{2} f^{\prime \prime}(2) x+f^{\prime \prime}(1) \quad \dots(i)$
Differentiating with respect to $x$:
$f^{\prime}(x)=3 x^{2}-6 x-\frac{3}{2} f^{\prime \prime}(2) \quad \dots(ii)$
$f^{\prime \prime}(x)=6 x-6 \quad \dots(iii)$
From $(iii)$,$f^{\prime \prime}(2)=6(2)-6=6$ and $f^{\prime \prime}(1)=6(1)-6=0$.
Substituting $f^{\prime \prime}(2)=6$ into $(ii)$:
$f^{\prime}(x)=3 x^{2}-6 x-\frac{3}{2}(6) = 3 x^{2}-6 x-9$.
Setting $f^{\prime}(x)=0$ for critical points:
$3(x^{2}-2 x-3)=0 \Rightarrow 3(x-3)(x+1)=0 \Rightarrow x=3, -1$.
Using the second derivative test:
$f^{\prime \prime}(-1)=6(-1)-6=-12 < 0$ (Local maxima at $x=-1$).
$f^{\prime \prime}(3)=6(3)-6=12 > 0$ (Local minima at $x=3$).
Substituting $f^{\prime \prime}(2)=6$ and $f^{\prime \prime}(1)=0$ into $(i)$:
$f(x)=x^{3}-3 x^{2}-9 x$.
The local minimum value is $f(3)=3^{3}-3(3^{2})-9(3) = 27-27-27 = -27$.

(D) For a system of linear equations to have infinitely many solutions,the determinant of the coefficient matrix must be zero,i.e.,$D = 0$.
$D = \begin{vmatrix} 3 & -1 & 4 \\ 1 & 2 & -3 \\ 6 & 5 & k \end{vmatrix} = 0$
Expanding along the first row:
$3(2k - (-15)) - (-1)(k - (-18)) + 4(5 - 12) = 0$
$3(2k + 15) + 1(k + 18) + 4(-7) = 0$
$6k + 45 + k + 18 - 28 = 0$
$7k + 35 = 0$
$7k = -35$
$k = -5$
Checking consistency for $k = -5$:
$3x - y + 4z = 3$ $(i)$
$x + 2y - 3z = -2$ $(ii)$
$6x + 5y - 5z = -3$ $(iii)$
From $(i) \times 2 - (ii)$: $6x - 2y + 8z - (x + 2y - 3z) = 6 - (-2) \Rightarrow 5x - 4y + 11z = 8$. This confirms the system is consistent with infinitely many solutions for $k = -5$.

(D) Given $f(x) = \log_e \left(x + \sqrt{x^2 + 1}\right)$. Note that $f(-x) = \log_e \left(-x + \sqrt{x^2 + 1}\right) = \log_e \left(\frac{1}{\sqrt{x^2 + 1} + x}\right) = -f(x)$. Thus,$f(x)$ is an odd function.
$g(t) = \int_{-\pi/2}^{\pi/2} \cos \left(\frac{\pi}{4} t + f(x)\right) \, dx = \int_{-\pi/2}^{\pi/2} \left[ \cos \left(\frac{\pi}{4} t\right) \cos(f(x)) - \sin \left(\frac{\pi}{4} t\right) \sin(f(x)) \right] \, dx$.
Since $f(x)$ is odd,$\cos(f(x))$ is an even function and $\sin(f(x))$ is an odd function.
Therefore,$\int_{-\pi/2}^{\pi/2} \sin \left(\frac{\pi}{4} t\right) \sin(f(x)) \, dx = 0$.
So,$g(t) = \cos \left(\frac{\pi}{4} t\right) \int_{-\pi/2}^{\pi/2} \cos(f(x)) \, dx$.
Let $C = \int_{-\pi/2}^{\pi/2} \cos(f(x)) \, dx$. Then $g(t) = C \cos \left(\frac{\pi}{4} t\right)$.
$g(1) = C \cos \left(\frac{\pi}{4}\right) = \frac{C}{\sqrt{2}}$ and $g(0) = C \cos(0) = C$.
Thus,$g(1) = \frac{g(0)}{\sqrt{2}}$,which implies $\sqrt{2} g(1) = g(0)$.

(A) The line $L$ is given by $\frac{x-3}{2}=\frac{y-1}{1}=\frac{z-2}{1} = k$. Any point on $L$ is $(2k+3, k+1, k+2)$.
Let $P_0 = (2,3,-1)$. The vector $\vec{P_0M}$ where $M$ is the foot of the perpendicular from $P_0$ to $L$ is $(2k+3-2, k+1-3, k+2+1) = (2k+1, k-2, k+3)$.
Since $\vec{P_0M}$ is perpendicular to the direction vector of $L$,$\vec{v} = (2,1,1)$,we have $2(2k+1) + 1(k-2) + 1(k+3) = 0$.
$4k+2 + k-2 + k+3 = 0 \implies 6k+3 = 0 \implies k = -1/2$.
Thus,$M = (2(-1/2)+3, -1/2+1, -1/2+2) = (2, 1/2, 3/2)$.
Since $M$ is the midpoint of $P_0Q$,if $Q = (x,y,z)$,then $\frac{x+2}{2} = 2, \frac{y+3}{2} = 1/2, \frac{z-1}{2} = 3/2$.
$x+2 = 4 \implies x=2$; $y+3 = 1 \implies y=-2$; $z-1 = 3 \implies z=4$. So $Q = (2, -2, 4)$.
The plane $P$ is perpendicular to $L$,so its normal vector is $\vec{n} = (2,1,1)$.
The equation of plane $P$ passing through $Q(2, -2, 4)$ is $2(x-2) + 1(y+2) + 1(z-4) = 0$.
$2x - 4 + y + 2 + z - 4 = 0 \implies 2x + y + z - 6 = 0$.
Checking the options: For $(1,2,2)$,$2(1) + 2 + 2 - 6 = 2+2+2-6 = 0$. Thus,$(1,2,2)$ lies on the plane.

(B) The given expression is a Riemann sum of the form $\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=0}^{n-1} f\left(\frac{5r}{n}\right)$.
This can be expressed as a definite integral: $I = \int_{0}^{1} f(5x) \,dx$.
Substituting $f(x) = x+1$,we get $f(5x) = 5x+1$.
Thus,$I = \int_{0}^{1} (5x+1) \,dx$.
Evaluating the integral: $I = \left[ \frac{5x^2}{2} + x \right]_{0}^{1}$.
$I = \left( \frac{5(1)^2}{2} + 1 \right) - (0) = \frac{5}{2} + 1 = \frac{7}{2}$.

(B) Given differential equation: $\cos \left(\frac{1}{2} \cos ^{-1}\left(e^{-x}\right)\right) d x=\sqrt{e^{2 x}-1} \,d y$.
Let $\cos ^{-1}\left(e^{-x}\right)=\theta$,where $\theta \in[0, \pi]$.
Then $\cos \theta = e^{-x}$. Using the identity $\cos \theta = 2 \cos^2 \frac{\theta}{2} - 1$,we have $2 \cos^2 \frac{\theta}{2} = 1 + e^{-x} = \frac{e^x + 1}{e^x}$.
Thus,$\cos \frac{\theta}{2} = \sqrt{\frac{e^x + 1}{2e^x}}$.
Substituting this into the differential equation: $\sqrt{\frac{e^x + 1}{2e^x}} dx = \sqrt{e^{2x} - 1} dy$.
Since $\sqrt{e^{2x} - 1} = \sqrt{(e^x - 1)(e^x + 1)}$,we get $\sqrt{\frac{e^x + 1}{2e^x}} dx = \sqrt{e^x - 1} \sqrt{e^x + 1} dy$.
Dividing by $\sqrt{e^x + 1}$: $\frac{1}{\sqrt{2e^x}} dx = \sqrt{e^x - 1} dy$,which simplifies to $\frac{dx}{\sqrt{2} \sqrt{e^x(e^x - 1)}} = dy$.
Let $e^x = t$,then $e^x dx = dt \Rightarrow dx = \frac{dt}{t}$.
So,$\int \frac{dt}{\sqrt{2} t \sqrt{t(t-1)}} = \int dy$.
Let $t = \frac{1}{z}$,then $dt = -\frac{1}{z^2} dz$.
Substituting: $\int \frac{-dz/z^2}{\sqrt{2} (1/z) \sqrt{1/z^2 - 1/z}} = \int dy \Rightarrow -\int \frac{dz}{\sqrt{2} \sqrt{1-z}} = y + C$.
Integrating: $\sqrt{2} \sqrt{1-z} = y + C \Rightarrow \sqrt{2} \sqrt{1 - e^{-x}} = y + C$.
At $x=0, y=-1$: $\sqrt{2} \sqrt{1 - 1} = -1 + C \Rightarrow C = 1$.
So,$\sqrt{2} \sqrt{1 - e^{-x}} = y + 1$.
For the $x$-axis intersection $(\alpha, 0)$,set $y=0$: $\sqrt{2} \sqrt{1 - e^{-\alpha}} = 1$.
Squaring both sides: $2(1 - e^{-\alpha}) = 1 \Rightarrow 1 - e^{-\alpha} = \frac{1}{2} \Rightarrow e^{-\alpha} = \frac{1}{2}$.
Therefore,$e^{\alpha} = 2$.

(C) Let $f(t) = t^3 - 6t^2 + 9t - 3$.
Then $f'(t) = 3t^2 - 12t + 9 = 3(t-1)(t-3)$.
The critical points are $t=1$ and $t=3$.
$f(0) = -3$,$f(1) = 1 - 6 + 9 - 3 = 1$,and $f(3) = 27 - 54 + 27 - 3 = -3$.
For $0 \leq x \leq 1$,$f(t)$ is increasing,so $\max_{0 \leq t \leq x} f(t) = f(x)$.
For $1 < x \leq 3$,the maximum value of $f(t)$ on $[0, x]$ is $f(1) = 1$.
Thus,$g(x) = \begin{cases} x^3 - 6x^2 + 9x - 3 & , 0 \leq x \leq 1 \\ 1 & , 1 < x \leq 3 \\ 4 - x & , 3 < x \leq 4 \end{cases}$.
Now check differentiability:
At $x=1$: $g'(1^-) = f'(1) = 0$ and $g'(1^+) = 0$. So $g(x)$ is differentiable at $x=1$.
At $x=3$: $g(3^-) = 1$ and $g(3^+) = 4-3 = 1$. $g(x)$ is continuous at $x=3$.
$g'(3^-) = 0$ and $g'(3^+) = -1$. Since $g'(3^-) \neq g'(3^+)$,$g(x)$ is not differentiable at $x=3$.
Thus,there is only $1$ point in $(0,4)$ where $g(x)$ is not differentiable.

(D) First,we construct the matrix $A$ based on the given conditions:
For $i=1, j=2: i < j \implies a_{12} = (-1)^{2-1} = -1$
For $i=1, j=3: i < j \implies a_{13} = (-1)^{3-1} = 1$
For $i=2, j=1: i > j \implies a_{21} = (-1)^{2+1} = -1$
For $i=2, j=3: i < j \implies a_{23} = (-1)^{3-2} = -1$
For $i=3, j=1: i > j \implies a_{31} = (-1)^{3+1} = 1$
For $i=3, j=2: i > j \implies a_{32} = (-1)^{3+2} = -1$
Diagonal elements $a_{ii} = 2$.
Thus,$A = \begin{bmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix}$.
Calculating the determinant $|A| = 2(4-1) - (-1)(-2+1) + 1(1-2) = 2(3) - 1 - 1 = 6 - 2 = 4$.
We need to find $\det(3 \operatorname{Adj}(2 A^{-1}))$.
Using properties of determinants: $|kM| = k^n |M|$ for an $n \times n$ matrix,and $|\operatorname{Adj}(M)| = |M|^{n-1}$.
Here $n=3$,so $|3 \operatorname{Adj}(2 A^{-1})| = 3^3 |\operatorname{Adj}(2 A^{-1})| = 27 |2 A^{-1}|^{3-1} = 27 |2 A^{-1}|^2$.
Since $|2 A^{-1}| = 2^3 |A^{-1}| = 8 \cdot \frac{1}{|A|} = \frac{8}{4} = 2$.
Therefore,$|3 \operatorname{Adj}(2 A^{-1})| = 27 \cdot (2)^2 = 27 \cdot 4 = 108$.

(A) Given $\vec{v}_{1} = \sqrt{3} p \hat{i} + \hat{j}$ and $\vec{v}_{2} = 2 \hat{i} + (p + 1) \hat{j}$.
Since rotation preserves the magnitude of the vector,$|\vec{v}_{1}| = |\vec{v}_{2}|$.
$(\sqrt{3}p)^2 + 1^2 = 2^2 + (p + 1)^2$
$3p^2 + 1 = 4 + p^2 + 2p + 1$
$2p^2 - 2p - 4 = 0 \Rightarrow p^2 - p - 2 = 0$.
Solving for $p$,$(p - 2)(p + 1) = 0$. Since $p > 0$,we have $p = 2$.
Thus,$\vec{v}_{1} = 2\sqrt{3} \hat{i} + \hat{j}$ and $\vec{v}_{2} = 2 \hat{i} + 3 \hat{j}$.
$|\vec{v}_{1}| = \sqrt{(2\sqrt{3})^2 + 1^2} = \sqrt{12 + 1} = \sqrt{13}$.
$|\vec{v}_{2}| = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13}$.
Using $\vec{v}_{1} \cdot \vec{v}_{2} = |\vec{v}_{1}| |\vec{v}_{2}| \cos \theta$,we get $(2\sqrt{3})(2) + (1)(3) = \sqrt{13} \cdot \sqrt{13} \cos \theta$.
$4\sqrt{3} + 3 = 13 \cos \theta \Rightarrow \cos \theta = \frac{4\sqrt{3} + 3}{13}$.
Since $\sin^2 \theta = 1 - \cos^2 \theta = 1 - \frac{(4\sqrt{3} + 3)^2}{169} = \frac{169 - (48 + 9 + 24\sqrt{3})}{169} = \frac{112 - 24\sqrt{3}}{169}$.
$\sin \theta = \frac{\sqrt{112 - 24\sqrt{3}}}{13} = \frac{\sqrt{4(28 - 6\sqrt{3})}}{13} = \frac{2\sqrt{(3\sqrt{3} - 1)^2}}{13} = \frac{2(3\sqrt{3} - 1)}{13} = \frac{6\sqrt{3} - 2}{13}$.
$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{6\sqrt{3} - 2}{4\sqrt{3} + 3}$.
Comparing with $\frac{\alpha \sqrt{3} - 2}{4\sqrt{3} + 3}$,we get $\alpha = 6$.

(A) For $x > 0$,$f'(x) = -4x^2 + 4x + 3$.
Setting $f'(x) > 0$,we have $-4x^2 + 4x + 3 > 0$,which implies $4x^2 - 4x - 3 < 0$.
Factoring gives $(2x - 3)(2x + 1) < 0$,so $x \in \left(-\frac{1}{2}, \frac{3}{2}\right)$.
Since $x > 0$,$f(x)$ is increasing in $\left(0, \frac{3}{2}\right)$.
For $x \leq 0$,$f'(x) = 3e^x + 3xe^x = 3e^x(1 + x)$.
Setting $f'(x) > 0$,since $3e^x > 0$ for all $x$,we need $1 + x > 0$,which means $x > -1$.
Thus,for $x \leq 0$,$f(x)$ is increasing in $(-1, 0]$.
Combining both intervals,$f(x)$ is increasing in $(-1, 0] \cup (0, \frac{3}{2}) = \left(-1, \frac{3}{2}\right)$.

(C) Given $A = [a_{ij}]$ is a $3 \times 3$ matrix such that the sum of elements in each row is $1$.
Let $X = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$.
Then $AX = \begin{bmatrix} a_{11} + a_{12} + a_{13} \\ a_{21} + a_{22} + a_{23} \\ a_{31} + a_{32} + a_{33} \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = X$.
Thus,$AX = X$.
Multiplying both sides by $A$,we get $A^2X = A(AX) = AX = X$.
Multiplying by $A$ again,we get $A^3X = A(A^2X) = AX = X$.
Let $A^3 = [b_{ij}]$. Then $A^3X = \begin{bmatrix} b_{11} + b_{12} + b_{13} \\ b_{21} + b_{22} + b_{23} \\ b_{31} + b_{32} + b_{33} \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$.
The sum of all entries of $A^3$ is $(b_{11} + b_{12} + b_{13}) + (b_{21} + b_{22} + b_{23}) + (b_{31} + b_{32} + b_{33}) = 1 + 1 + 1 = 3$.

(B) Let $I = \int_{0}^{100 \pi} \frac{\sin ^{2} x}{e^{\left(\frac{x}{\pi}-\left[\frac{x}{\pi}\right]\right)}} d x$.
Since the function $f(x) = \frac{\sin^2 x}{e^{\left(\frac{x}{\pi}-\left[\frac{x}{\pi}\right]\right)}}$ is periodic with period $\pi$,we have $I = 100 \int_{0}^{\pi} \frac{\sin^2 x}{e^{x/\pi}} dx$.
$I = 100 \int_{0}^{\pi} e^{-x/\pi} \left(\frac{1-\cos 2x}{2}\right) dx = 50 \left[ \int_{0}^{\pi} e^{-x/\pi} dx - \int_{0}^{\pi} e^{-x/\pi} \cos 2x dx \right]$.
Let $I_1 = \int_{0}^{\pi} e^{-x/\pi} dx = [-\pi e^{-x/\pi}]_0^{\pi} = \pi(1-e^{-1})$.
Let $I_2 = \int_{0}^{\pi} e^{-x/\pi} \cos 2x dx$. Using integration by parts,$I_2 = [-\pi e^{-x/\pi} \cos 2x]_0^{\pi} - \int_0^{\pi} \pi e^{-x/\pi} (2 \sin 2x) dx = \pi(1-e^{-1}) - 2\pi \int_0^{\pi} e^{-x/\pi} \sin 2x dx$.
Integrating the second part again,$\int_0^{\pi} e^{-x/\pi} \sin 2x dx = [-\pi e^{-x/\pi} \sin 2x]_0^{\pi} - \int_0^{\pi} \pi e^{-x/\pi} (2 \cos 2x) dx = 2\pi I_2$.
Thus,$I_2 = \pi(1-e^{-1}) - 2\pi(2\pi I_2) = \pi(1-e^{-1}) - 4\pi^2 I_2$.
$I_2(1+4\pi^2) = \pi(1-e^{-1}) \implies I_2 = \frac{\pi(1-e^{-1})}{1+4\pi^2}$.
Substituting back,$I = 50 [\pi(1-e^{-1}) - \frac{\pi(1-e^{-1})}{1+4\pi^2}] = 50 \pi(1-e^{-1}) [1 - \frac{1}{1+4\pi^2}] = 50 \pi(1-e^{-1}) [\frac{4\pi^2}{1+4\pi^2}] = \frac{200(1-e^{-1})\pi^3}{1+4\pi^2}$.
Comparing with $\frac{\alpha \pi^3}{1+4\pi^2}$,we get $\alpha = 200(1-e^{-1})$.

(D) Let the matrix be $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$. The determinant is $|A| = ad - bc$.
The total number of ways to choose $4$ numbers from $6$ faces of a die is $6^4 = 1296$.
We require all entries $a, b, c, d$ to be distinct. The number of ways to choose $4$ distinct numbers from ${1, 2, 3, 4, 5, 6}$ and arrange them in the matrix is $^6P_4 = 6 \times 5 \times 4 \times 3 = 360$.
For the matrix to be nonsingular,$|A| \neq 0$,which means $ad \neq bc$.
We count the cases where $ad = bc$ with distinct $a, b, c, d$:
$1$. $6 \times 1 = 2 \times 3$: The sets are ${1, 2, 3, 6}$. Possible arrangements $(a, b, c, d)$ such that $ad=bc$ are $(6, 2, 3, 1), (6, 3, 2, 1), (1, 2, 3, 6), (1, 3, 2, 6), (2, 6, 1, 3), (3, 6, 1, 2), (2, 1, 6, 3), (3, 1, 6, 2)$. Total $8$ cases.
$2$. $6 \times 2 = 3 \times 4$: The sets are ${2, 3, 4, 6}$. Possible arrangements $(a, b, c, d)$ such that $ad=bc$ are $(6, 3, 4, 2), (6, 4, 3, 2), (2, 3, 4, 6), (2, 4, 3, 6), (3, 6, 2, 4), (4, 6, 2, 3), (3, 2, 6, 4), (4, 2, 6, 3)$. Total $8$ cases.
Total cases where $ad = bc$ with distinct entries is $8 + 8 = 16$.
Number of favorable cases = (Total ways with distinct entries) - (Cases where $ad = bc$) = $360 - 16 = 344$.
Required probability = $\frac{344}{1296} = \frac{43}{162}$.

(B) Given $\vec{a} \times \vec{b} = \vec{c}$ and $\vec{b} \times \vec{c} = \vec{a}$.
Since $\vec{a} \times \vec{b} = \vec{c}$,$\vec{c}$ is perpendicular to both $\vec{a}$ and $\vec{b}$.
Since $\vec{b} \times \vec{c} = \vec{a}$,$\vec{a}$ is perpendicular to both $\vec{b}$ and $\vec{c}$.
Thus,$\vec{a}, \vec{b}, \vec{c}$ are mutually orthogonal.
$|\vec{a} \times \vec{b}| = |\vec{c}| \implies |\vec{a}||\vec{b}| = |\vec{c}| \implies 2|\vec{b}| = |\vec{c}|$.
$|\vec{b} \times \vec{c}| = |\vec{a}| \implies |\vec{b}||\vec{c}| = 2$.
Substituting $|\vec{c}| = 2|\vec{b}|$,we get $|\vec{b}|(2|\vec{b}|) = 2 \implies |\vec{b}|^2 = 1 \implies |\vec{b}| = 1$.
Then $|\vec{c}| = 2(1) = 2$.
$(A)$ Projection of $\vec{a}$ on $(\vec{b} \times \vec{c}) = \frac{\vec{a} \cdot (\vec{b} \times \vec{c})}{|\vec{b} \times \vec{c}|} = \frac{\vec{a} \cdot \vec{a}}{|\vec{a}|} = |\vec{a}| = 2$. (True)
$(B)$ $|3\vec{a} + \vec{b} - 2\vec{c}|^2 = (3\vec{a} + \vec{b} - 2\vec{c}) \cdot (3\vec{a} + \vec{b} - 2\vec{c}) = 9|\vec{a}|^2 + |\vec{b}|^2 + 4|\vec{c}|^2 = 9(4) + 1 + 4(4) = 36 + 1 + 16 = 53 \neq 51$. (Not True)
$(C)$ $[\vec{a} \vec{b} \vec{c}] + [\vec{c} \vec{a} \vec{b}] = 2[\vec{a} \vec{b} \vec{c}] = 2(\vec{a} \cdot (\vec{b} \times \vec{c})) = 2(\vec{a} \cdot \vec{a}) = 2|\vec{a}|^2 = 2(4) = 8$. (True)
$(D)$ $\vec{a} \times ((\vec{b} + \vec{c}) \times (\vec{b} - \vec{c})) = \vec{a} \times (\vec{b} \times \vec{b} - \vec{b} \times \vec{c} + \vec{c} \times \vec{b} - \vec{c} \times \vec{c}) = \vec{a} \times (\vec{0} - \vec{a} - \vec{a} - \vec{0}) = \vec{a} \times (-2\vec{a}) = -2(\vec{a} \times \vec{a}) = \vec{0}$. (True)

(B) The equations of the planes are $P_{1}: x-y+2 z=2$ and $P_{2}: 2 x+y-z=2$.
Let the line of intersection of planes $P_{1}$ and $P_{2}$ cut the $xy$-plane at point $Q$.
Setting $z=0$ in both equations,we get $x-y=2$ and $2x+y=2$. Adding these,$3x=4 \Rightarrow x=\frac{4}{3}$,and $y=x-2 = \frac{4}{3}-2 = -\frac{2}{3}$.
So,$Q = (\frac{4}{3}, -\frac{2}{3}, 0)$.
The direction vector $\vec{a}$ of the line of intersection is the cross product of the normals $\vec{n}_{1} = \hat{i}-\hat{j}+2\hat{k}$ and $\vec{n}_{2} = 2\hat{i}+\hat{j}-\hat{k}$.
$\vec{a} = \vec{n}_{1} \times \vec{n}_{2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 2 \\ 2 & 1 & -1 \end{vmatrix} = \hat{i}(1-2) - \hat{j}(-1-4) + \hat{k}(1+2) = -\hat{i} + 5\hat{j} + 3\hat{k}$.
The equation of the line $L$ is $\frac{x-4/3}{-1} = \frac{y+2/3}{5} = \frac{z}{3} = \lambda$.
Any point $F$ on $L$ is $(-\lambda + 4/3, 5\lambda - 2/3, 3\lambda)$.
Let $A = (1, 2, 0)$. The vector $\vec{AF} = (-\lambda + 4/3 - 1, 5\lambda - 2/3 - 2, 3\lambda - 0) = (-\lambda + 1/3, 5\lambda - 8/3, 3\lambda)$.
Since $AF \perp L$,$\vec{AF} \cdot \vec{a} = 0$.
$(-\lambda + 1/3)(-1) + (5\lambda - 8/3)(5) + (3\lambda)(3) = 0$.
$\lambda - 1/3 + 25\lambda - 40/3 + 9\lambda = 0$.
$35\lambda - 41/3 = 0 \Rightarrow 35\lambda = 41/3 \Rightarrow \lambda = 41/105$.
The coordinates of the foot of the perpendicular $P(\alpha, \beta, \gamma)$ are $F$.
$\alpha + \beta + \gamma = (-\lambda + 4/3) + (5\lambda - 2/3) + 3\lambda = 7\lambda + 2/3$.
$35(\alpha + \beta + \gamma) = 35(7\lambda + 2/3) = 245\lambda + 70/3$.
Substituting $\lambda = 41/105$: $245(41/105) + 70/3 = (49 \times 41)/21 + 70/3 = (7 \times 41)/3 + 70/3 = 287/3 + 70/3 = 357/3 = 119$.

(A) For $f(x)$ to be continuous at $x=0$,we must have $\alpha = \lim_{x \rightarrow 0} f(x)$.
Given $f(x) = \frac{x^3}{(1-\cos 2x)^2} \ln\left(\frac{1+2xe^{-2x}}{(1-xe^{-x})^2}\right)$.
Since $1-\cos 2x = 2\sin^2 x$,we have $(1-\cos 2x)^2 = 4\sin^4 x$.
So,$\lim_{x \rightarrow 0} f(x) = \lim_{x \rightarrow 0} \frac{x^3}{4\sin^4 x} \ln\left(\frac{1+2xe^{-2x}}{(1-xe^{-x})^2}\right)$.
Using $\sin x \approx x$ as $x \rightarrow 0$,$\frac{x^3}{4\sin^4 x} \approx \frac{x^3}{4x^4} = \frac{1}{4x}$.
Now,$\ln\left(\frac{1+2xe^{-2x}}{(1-xe^{-x})^2}\right) = \ln(1+2xe^{-2x}) - 2\ln(1-xe^{-x})$.
Using $\ln(1+u) \approx u$ for small $u$,we get $2xe^{-2x} - 2(-xe^{-x}) = 2xe^{-2x} + 2xe^{-x}$.
As $x \rightarrow 0$,$e^{-2x} \rightarrow 1$ and $e^{-x} \rightarrow 1$,so the expression becomes $2x + 2x = 4x$.
Thus,$\alpha = \lim_{x \rightarrow 0} \frac{1}{4x} \cdot (4x) = 1$.

(B) Since $\vec{a}$ is coplanar with $\vec{b}$ and $\vec{c}$,we can write $\vec{a} = \lambda \vec{b} + \mu \vec{c}$.
Substituting the vectors: $\vec{a} = \lambda(2 \hat{i} + \hat{j} + \hat{k}) + \mu(\hat{i} - \hat{j} + \hat{k}) = (2\lambda + \mu) \hat{i} + (\lambda - \mu) \hat{j} + (\lambda + \mu) \hat{k}$.
Given $\vec{a} \cdot \vec{d} = 0$,where $\vec{d} = 3 \hat{i} + 2 \hat{j} + 6 \hat{k}$:
$3(2\lambda + \mu) + 2(\lambda - \mu) + 6(\lambda + \mu) = 0$
$6\lambda + 3\mu + 2\lambda - 2\mu + 6\lambda + 6\mu = 0$
$14\lambda + 7\mu = 0 \implies \mu = -2\lambda$.
Substituting $\mu$ back into $\vec{a}$: $\vec{a} = (2\lambda - 2\lambda) \hat{i} + (\lambda - (-2\lambda)) \hat{j} + (\lambda - 2\lambda) \hat{k} = 0 \hat{i} + 3\lambda \hat{j} - \lambda \hat{k}$.
Given $|\vec{a}| = \sqrt{10}$,so $\sqrt{0^2 + (3\lambda)^2 + (-\lambda)^2} = \sqrt{10} \implies \sqrt{10\lambda^2} = \sqrt{10} \implies |\lambda| = 1$.
Since $\vec{a}, \vec{b}, \vec{c}$ are coplanar,$[\vec{a} \vec{b} \vec{c}] = 0$.
The expression becomes $[\vec{a} \vec{b} \vec{d}] + [\vec{a} \vec{c} \vec{d}] = [\vec{a} \vec{b} + \vec{c} \vec{d}]$.
Using $\vec{b} + \vec{c} = 3 \hat{i} + 0 \hat{j} + 2 \hat{k}$ and $\vec{a} = 3\lambda \hat{j} - \lambda \hat{k}$:
$[\vec{a} \vec{b} + \vec{c} \vec{d}] = \begin{vmatrix} 0 & 3\lambda & -\lambda \\ 3 & 0 & 2 \\ 3 & 2 & 6 \end{vmatrix} = 0(0 - 4) - 3\lambda(18 - 6) - \lambda(6 - 0) = -3\lambda(12) - 6\lambda = -42\lambda$.
For $\lambda = 1$,the value is $-42$.

(C) Given the differential equation: $\operatorname{cosec}^{2} x \, dy + 2 \, dx = (1+y \cos 2x) \operatorname{cosec}^{2} x \, dx$.
Dividing by $\operatorname{cosec}^{2} x \, dx$,we get: $\frac{dy}{dx} + 2 \sin^{2} x = 1 + y \cos 2x$.
Rearranging the terms: $\frac{dy}{dx} - y \cos 2x = 1 - 2 \sin^{2} x$.
Since $1 - 2 \sin^{2} x = \cos 2x$,the equation becomes: $\frac{dy}{dx} - y \cos 2x = \cos 2x$.
This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = -\cos 2x$ and $Q(x) = \cos 2x$.
The integrating factor ($I$.$F$.) is $e^{\int P(x) \, dx} = e^{\int -\cos 2x \, dx} = e^{-\frac{\sin 2x}{2}}$.
The solution is $y \cdot (I.F.) = \int Q(x) \cdot (I.F.) \, dx + C$.
$y \cdot e^{-\frac{\sin 2x}{2}} = \int \cos 2x \cdot e^{-\frac{\sin 2x}{2}} \, dx + C$.
Let $u = -\frac{\sin 2x}{2}$,then $du = -\cos 2x \, dx$.
So,$y \cdot e^{-\frac{\sin 2x}{2}} = -\int e^u \, du + C = -e^{-\frac{\sin 2x}{2}} + C$.
Given $y(\frac{\pi}{4}) = 0$,we have $0 = -e^{-\frac{\sin(\pi/2)}{2}} + C = -e^{-1/2} + C$,so $C = e^{-1/2}$.
Thus,$y \cdot e^{-\frac{\sin 2x}{2}} = -e^{-\frac{\sin 2x}{2}} + e^{-1/2}$.
At $x = 0$,$y \cdot e^0 = -e^0 + e^{-1/2}$,which gives $y(0) = -1 + e^{-1/2}$.
Therefore,$(y(0) + 1)^{2} = (-1 + e^{-1/2} + 1)^{2} = (e^{-1/2})^{2} = e^{-1}$.

(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 $(D_1, D_2, D_3)$ must be non-zero.
First,calculate $D$:
$D = \begin{vmatrix} 1 & 1 & 1 \\ 3 & 5 & 5 \\ 1 & 2 & \lambda \end{vmatrix} = 1(5\lambda - 10) - 1(3\lambda - 5) + 1(6 - 5) = 5\lambda - 10 - 3\lambda + 5 + 1 = 2\lambda - 4$.
Setting $D = 0$,we get $2\lambda - 4 = 0$,which implies $\lambda = 2$.
Now,substitute $\lambda = 2$ into the system:
$x + y + z = 6$ (Eq. $1$)
$3x + 5y + 5z = 26$ (Eq. $2$)
$x + 2y + 2z = \mu$ (Eq. $3$)
Subtract $3 \times$ (Eq. $1$) from (Eq. $2$): $(3x + 5y + 5z) - 3(x + y + z) = 26 - 3(6) \implies 2y + 2z = 8 \implies y + z = 4$.
Subtract (Eq. $1$) from (Eq. $3$): $(x + 2y + 2z) - (x + y + z) = \mu - 6 \implies y + z = \mu - 6$.
For the system to have no solution,these two derived equations must be inconsistent:
$4 \neq \mu - 6 \implies \mu \neq 10$.
Thus,the condition for no solution is $\lambda = 2$ and $\mu \neq 10$.

(B) For the function $f(x)$ to be defined,the following conditions must be satisfied:
$1$. The argument of $\cos^{-1}$ must be in $[0, 1]$: $0 \leq \sqrt{x^2-x+1} \leq 1$.
Squaring gives $0 \leq x^2-x+1 \leq 1$.
$x^2-x+1 \leq 1 \Rightarrow x^2-x \leq 0 \Rightarrow x(x-1) \leq 0 \Rightarrow x \in [0, 1]$.
$2$. The denominator must be non-zero and the argument of the square root must be positive: $\sin^{-1}(\frac{2x-1}{2}) > 0$.
Since the range of $\sin^{-1}$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$,we also require $\frac{2x-1}{2} \leq 1$.
So,$0 < \frac{2x-1}{2} \leq 1$.
$0 < 2x-1 \leq 2$.
$1 < 2x \leq 3$.
$\frac{1}{2} < x \leq \frac{3}{2}$.
Taking the intersection of $x \in [0, 1]$ and $x \in (\frac{1}{2}, \frac{3}{2}]$,we get $x \in (\frac{1}{2}, 1]$.
Thus,$\alpha = \frac{1}{2}$ and $\beta = 1$.
Therefore,$\alpha + \beta = \frac{1}{2} + 1 = \frac{3}{2}$.

(C) The given lines are:
$L_{1}: \frac{x-1}{2} = \frac{y-2}{1} = \frac{z-1}{3}$. Point $P_{1} = (1, 2, 1)$,direction vector $\vec{v}_{1} = 2\hat{i} + \hat{j} + 3\hat{k}$.
$L_{2}: \frac{x-2}{1} = \frac{y-\lambda}{2} = \frac{z-3}{4}$. Point $P_{2} = (2, \lambda, 3)$,direction vector $\vec{v}_{2} = \hat{i} + 2\hat{j} + 4\hat{k}$.
Let $\vec{a} = P_{2} - P_{1} = (2-1)\hat{i} + (\lambda-2)\hat{j} + (3-1)\hat{k} = \hat{i} + (\lambda-2)\hat{j} + 2\hat{k}$.
The shortest distance $d$ is given by $d = \frac{|\vec{a} \cdot (\vec{v}_{1} \times \vec{v}_{2})|}{|\vec{v}_{1} \times \vec{v}_{2}|}$.
First,calculate $\vec{v}_{1} \times \vec{v}_{2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 3 \\ 1 & 2 & 4 \end{vmatrix} = \hat{i}(4-6) - \hat{j}(8-3) + \hat{k}(4-1) = -2\hat{i} - 5\hat{j} + 3\hat{k}$.
Magnitude $|\vec{v}_{1} \times \vec{v}_{2}| = \sqrt{(-2)^2 + (-5)^2 + 3^2} = \sqrt{4 + 25 + 9} = \sqrt{38}$.
Now,$\vec{a} \cdot (\vec{v}_{1} \times \vec{v}_{2}) = (1)(-2) + (\lambda-2)(-5) + (2)(3) = -2 - 5\lambda + 10 + 6 = 14 - 5\lambda$.
Given $d = \frac{1}{\sqrt{38}}$,so $\frac{|14 - 5\lambda|}{\sqrt{38}} = \frac{1}{\sqrt{38}}$.
$|14 - 5\lambda| = 1$.
Case $1$: $14 - 5\lambda = 1 \Rightarrow 5\lambda = 13 \Rightarrow \lambda = 2.6$.
Case $2$: $14 - 5\lambda = -1 \Rightarrow 5\lambda = 15 \Rightarrow \lambda = 3$.
Since we need the integral value of $\lambda$,the answer is $3$.

(B) Let $Y = y+1$ and $X = x+2$. Then $dY = dy$ and $dX = dx$.
Substituting these into the equation: $(X e^{Y/X} + Y) dX = X dY$.
Rearranging gives $X dY - Y dX = X e^{Y/X} dX$.
Dividing by $X^2$: $\frac{X dY - Y dX}{X^2} = \frac{e^{Y/X}}{X} dX$.
This simplifies to $d(\frac{Y}{X}) = e^{Y/X} \frac{dX}{X}$.
Integrating both sides: $\int e^{-Y/X} d(Y/X) = \int \frac{dX}{X} \Rightarrow -e^{-Y/X} = \ln|X| + C$.
Using $y(1)=1$,we have $Y=2$ and $X=3$: $-e^{-2/3} = \ln|3| + C$,so $C = -e^{-2/3} - \ln 3$.
Thus,$-e^{-(y+1)/(x+2)} = \ln|x+2| - e^{-2/3} - \ln 3$.
$e^{-(y+1)/(x+2)} = e^{-2/3} + \ln 3 - \ln|x+2|$.
For the solution to exist,we require $e^{-2/3} + \ln 3 - \ln|x+2| > 0$,which means $\ln|x+2| < e^{-2/3} + \ln 3$.
Let $k = e^{-2/3} + \ln 3$. Then $|x+2| < e^k$,which implies $-e^k - 2 < x < e^k - 2$.
Thus,$\alpha = -e^k - 2$ and $\beta = e^k - 2$.
Then $\alpha + \beta = -4$,so $|\alpha + \beta| = 4$.

(C) Given the condition $f(1) + f(2) = 3 - f(3)$,we can rewrite it as $f(1) + f(2) + f(3) = 3$.
Since $f$ is a bijective function from $A$ to $A$,all values $f(1), f(2), f(3)$ must be distinct elements from the set $A = \{0, 1, 2, 3, 4, 5, 6, 7\}$.
The only distinct non-negative integers that sum to $3$ are $0, 1,$ and $2$.
Thus,the set of values $\{f(1), f(2), f(3)\}$ must be equal to the set $\{0, 1, 2\}$.
The number of ways to assign these $3$ values to $f(1), f(2),$ and $f(3)$ is $3! = 6$.
The remaining $5$ elements of the domain $\{0, 4, 5, 6, 7\}$ must be mapped to the remaining $5$ elements of the codomain $\{3, 4, 5, 6, 7\}$.
The number of ways to map these is $5! = 120$.
Therefore,the total number of bijective functions is $3! \times 5! = 6 \times 120 = 720$.

(A) Given $f(x) = 3(1 - \frac{|x|}{2})$ for $|x| \leq 2$ and $0$ otherwise.
$f(x+2) = 3(1 - \frac{|x+2|}{2})$ for $|x+2| \leq 2$ (i.e.,$-4 \leq x \leq 0$) and $0$ otherwise.
$f(x-2) = 3(1 - \frac{|x-2|}{2})$ for $|x-2| \leq 2$ (i.e.,$0 \leq x \leq 4$) and $0$ otherwise.
Thus,$g(x) = f(x+2) - f(x-2)$ is defined as:
$g(x) = \begin{cases} 3(1 - \frac{|x+2|}{2}) & -4 \leq x < 0 \\ -3(1 - \frac{|x-2|}{2}) & 0 \leq x \leq 4 \\ 0 & \text{otherwise} \end{cases}$
Simplifying $g(x)$:
$g(x) = \begin{cases} \frac{3x}{2} + 6 & -4 \leq x \leq -2 \\ -\frac{3x}{2} & -2 < x < 2 \\ \frac{3x}{2} - 6 & 2 \leq x \leq 4 \\ 0 & \text{otherwise} \end{cases}$
Checking continuity: The function $g(x)$ is continuous everywhere because the limits at $x = -4, -2, 2, 4$ match the function values (all are $0$ at boundaries).
Thus,$n = 0$.
Checking differentiability: The function $g(x)$ is non-differentiable at points where the slope changes abruptly: $x = -4, -2, 2, 4$.
Thus,$m = 4$.
Therefore,$n + m = 0 + 4 = 4$.

(B) Let matrix $B = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}$.
Given $AB = BA$,we have:
$\begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
Performing matrix multiplication:
$\begin{bmatrix} d & e & f \\ a & b & c \\ g & h & i \end{bmatrix} = \begin{bmatrix} b & a & c \\ e & d & f \\ h & g & i \end{bmatrix}$
Comparing the corresponding elements:
$d = b, e = a, f = c, g = h$
Thus,matrix $B$ takes the form:
$B = \begin{bmatrix} a & b & c \\ b & a & c \\ g & g & i \end{bmatrix}$
Since each entry $a, b, c, g, i$ can be chosen from the set $\{1, 2, 3, 4, 5\}$,there are $5$ choices for each of the $5$ independent variables.
Total number of matrices $B = 5 \times 5 \times 5 \times 5 \times 5 = 5^5 = 3125$.

(B) The curves are $x^2 = 1 - 2y$,$x = \frac{4-y^2}{4}$,and $x = \frac{y^2-4}{4}$.
For the upper half plane $(y \ge 0)$,the region is bounded by $x = \frac{4-y^2}{4}$ on the right,$x = \frac{y^2-4}{4}$ on the left,and $y = \frac{1-x^2}{2}$ from below.
Due to symmetry about the $y$-axis,the area is $2 \int_{0}^{2} \left( \frac{4-y^2}{4} \right) dy - 2 \int_{0}^{1} \left( \frac{1-x^2}{2} \right) dx$.
$= 2 \left[ y - \frac{y^3}{12} \right]_{0}^{2} - 2 \left[ \frac{x}{2} - \frac{x^3}{6} \right]_{0}^{1}$
$= 2 \left( 2 - \frac{8}{12} \right) - 2 \left( \frac{1}{2} - \frac{1}{6} \right)$
$= 2 \left( \frac{4}{3} \right) - 2 \left( \frac{1}{3} \right) = \frac{8}{3} - \frac{2}{3} = \frac{6}{3} = 2 \, sq. \, units$.