JEE Main 2021 Mathematics Question Paper with Answer and Solution

(A) Given the equation: $81^{\sin^2 x} + 81^{\cos^2 x} = 30$.
Let $u = 81^{\sin^2 x}$. Since $\cos^2 x = 1 - \sin^2 x$,the equation becomes $u + 81^{1 - \sin^2 x} = 30$.
This simplifies to $u + \frac{81}{u} = 30$.
Multiplying by $u$,we get $u^2 - 30u + 81 = 0$.
Factoring the quadratic: $(u - 27)(u - 3) = 0$.
So,$u = 27$ or $u = 3$.
Case $1$: $81^{\sin^2 x} = 27 \implies (3^4)^{\sin^2 x} = 3^3 \implies 4 \sin^2 x = 3 \implies \sin^2 x = 3/4 \implies \sin x = \pm \sqrt{3}/2$.
For $0 \le x \le \pi$,$x = \pi/3$ or $x = 2\pi/3$.
Case $2$: $81^{\sin^2 x} = 3 \implies (3^4)^{\sin^2 x} = 3^1 \implies 4 \sin^2 x = 1 \implies \sin^2 x = 1/4 \implies \sin x = \pm 1/2$.
For $0 \le x \le \pi$,$x = \pi/6$ or $x = 5\pi/6$.
Checking the options,$\pi/6$ is the correct value.

(B) To find the number of elements in the set,we solve the equation $(|x|-3)|x+4|=6$.
Let $f(x) = |x|-3$ and $g(x) = \frac{6}{|x+4|}$.
We look for the number of intersection points of the graphs of $y = |x|-3$ and $y = \frac{6}{|x+4|}$.
The graph of $y = |x|-3$ is a $V$-shaped curve with its vertex at $(0, -3)$ and x-intercepts at $x = 3$ and $x = -3$.
The graph of $y = \frac{6}{|x+4|}$ is a hyperbola-like curve with a vertical asymptote at $x = -4$ and it is always positive.
By observing the graphs,the curve $y = |x|-3$ intersects $y = \frac{6}{|x+4|}$ at two distinct points: one in the interval $(3, \infty)$ and one in the interval $(-\infty, -4)$.
Therefore,the number of elements in the set is $2$.

(B) Given $x \in \left(0, \frac{\pi}{2}\right)$.
From the first equation: $\log_{10} \sin x + \log_{10} \cos x = -1$
$\Rightarrow \log_{10}(\sin x \cos x) = -1$
$\Rightarrow \sin x \cos x = 10^{-1} = \frac{1}{10} \quad ....(1)$
From the second equation: $\log_{10}(\sin x + \cos x) = \frac{1}{2}(\log_{10} n - \log_{10} 10) = \frac{1}{2} \log_{10} \left(\frac{n}{10}\right) = \log_{10} \sqrt{\frac{n}{10}}$
$\Rightarrow \sin x + \cos x = \sqrt{\frac{n}{10}}$
Squaring both sides:
$(\sin x + \cos x)^2 = \frac{n}{10}$
$\sin^2 x + \cos^2 x + 2 \sin x \cos x = \frac{n}{10}$
$1 + 2 \left(\frac{1}{10}\right) = \frac{n}{10}$
$1 + \frac{1}{5} = \frac{n}{10}$
$\frac{6}{5} = \frac{n}{10}$
$n = \frac{6 \times 10}{5} = 12$.

(D) The equation of the parabola is $y^{2} = 2x$. Comparing this with $y^{2} = 4Ax$,we get $4A = 2$,so $A = 1/2$.
For a parabola $y^{2} = 4Ax$,the normal at any point $(x_{1}, y_{1})$ passes through $(a, 0)$ if $a > 2A$.
Substituting the value of $A$:
$a > 2 \times (1/2)$
$a > 1$.
Thus,for three distinct normals to pass through the point $(a, 0)$,$a$ must be greater than $1$.

(D) tautology is a statement that is true for all possible truth values of its components. We evaluate the truth table for the expression $(p \wedge q)$ $\rightarrow (p$ $\rightarrow q)$:

$p$	$q$	$p \wedge q$	$p \rightarrow q$	$(p \wedge q)$ $\rightarrow (p$ $\rightarrow q)$
$T$	$T$	$T$	$T$	$T$
$T$	$F$	$F$	$F$	$T$
$F$	$T$	$F$	$T$	$T$
$F$	$F$	$F$	$T$	$T$

Since the final column contains only $T$ (True),the expression $(p \wedge q)$ $\rightarrow (p$ $\rightarrow q)$ is a tautology.

(B) Given the inequality: $\log_{\frac{1}{\sqrt{2}}} \left( \frac{|z|+11}{(|z|-1)^2} \right) \leq 2$.
Since the base $\frac{1}{\sqrt{2}} < 1$,the inequality sign reverses when removing the logarithm:
$\frac{|z|+11}{(|z|-1)^2} \geq \left( \frac{1}{\sqrt{2}} \right)^2 = \frac{1}{2}$.
Cross-multiplying (since $(|z|-1)^2 > 0$ for $|z| \neq 1$):
$2(|z|+11) \geq (|z|-1)^2$.
$2|z| + 22 \geq |z|^2 - 2|z| + 1$.
Rearranging the terms:
$|z|^2 - 4|z| - 21 \leq 0$.
Factoring the quadratic expression:
$(|z|-7)(|z|+3) \leq 0$.
Since $|z| \geq 0$,we have $|z|+3 > 0$,so $|z|-7 \leq 0$,which implies $|z| \leq 7$.
Given $|z| \neq 1$,the largest value of $|z|$ is $7$.

(A) The general term in the expansion of $(3^{1/4} + 5^{1/8})^{60}$ is given by $T_{r+1} = {}^{60}C_r (3^{1/4})^{60-r} (5^{1/8})^r = {}^{60}C_r (3)^{(60-r)/4} (5)^{r/8}$.
For the term to be rational,the exponents of $3$ and $5$ must be integers.
Thus,$r$ must be a multiple of $8$ such that $0 \leq r \leq 60$.
The possible values for $r$ are $0, 8, 16, 24, 32, 40, 48, 56$.
There are $8$ rational terms.
The total number of terms in the expansion is $60 + 1 = 61$.
Therefore,the number of irrational terms $n = 61 - 8 = 53$.
We need to check divisibility for $(n - 1) = 53 - 1 = 52$.
Since $52 = 26 \times 2$,$(n - 1)$ is divisible by $26$.

(D) Let the midpoint of the chord be $(h, k)$.
The equation of the chord of the circle $x^{2}+y^{2}=25$ with midpoint $(h, k)$ is given by $T=S_1$,which is $xh+yk=h^{2}+k^{2}$.
This can be rewritten as $y = -\frac{h}{k}x + \frac{h^{2}+k^{2}}{k}$.
This line is tangent to the hyperbola $\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$.
The condition for the line $y=mx+c$ to be a tangent to the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ is $c^{2}=a^{2}m^{2}-b^{2}$.
Here,$m=-\frac{h}{k}$ and $c=\frac{h^{2}+k^{2}}{k}$,$a^{2}=9$,$b^{2}=16$.
Substituting these values into the condition: $\left(\frac{h^{2}+k^{2}}{k}\right)^{2} = 9\left(-\frac{h}{k}\right)^{2} - 16$.
$\frac{(h^{2}+k^{2})^{2}}{k^{2}} = \frac{9h^{2}}{k^{2}} - 16$.
$(h^{2}+k^{2})^{2} = 9h^{2} - 16k^{2}$.
Replacing $(h, k)$ with $(x, y)$,the locus is $(x^{2}+y^{2})^{2} = 9x^{2}-16y^{2}$,or $(x^{2}+y^{2})^{2}-9x^{2}+16y^{2}=0$.

(B) Given equation: $(81)^{\sin ^{2} x} + (81)^{\cos ^{2} x} = 30$.
Since $\cos ^{2} x = 1 - \sin ^{2} x$,we have $(81)^{\sin ^{2} x} + (81)^{1 - \sin ^{2} x} = 30$.
$(81)^{\sin ^{2} x} + \frac{81}{(81)^{\sin ^{2} x}} = 30$.
Let $t = (81)^{\sin ^{2} x}$. Then $t + \frac{81}{t} = 30$,which implies $t^{2} - 30t + 81 = 0$.
$(t - 3)(t - 27) = 0$,so $t = 3$ or $t = 27$.
Case $1$: $(81)^{\sin ^{2} x} = 3 \implies 3^{4 \sin ^{2} x} = 3^{1} \implies 4 \sin ^{2} x = 1 \implies \sin ^{2} x = \frac{1}{4}$.
In $[0, \pi]$,$\sin x = \frac{1}{2}$ or $\sin x = -\frac{1}{2}$. Since $\sin x \ge 0$ in $[0, \pi]$,$\sin x = \frac{1}{2}$ gives $x = \frac{\pi}{6}, \frac{5\pi}{6}$ ($2$ solutions).
Case $2$: $(81)^{\sin ^{2} x} = 27 \implies 3^{4 \sin ^{2} x} = 3^{3} \implies 4 \sin ^{2} x = 3 \implies \sin ^{2} x = \frac{3}{4}$.
In $[0, \pi]$,$\sin x = \frac{\sqrt{3}}{2}$ or $\sin x = -\frac{\sqrt{3}}{2}$. Since $\sin x \ge 0$ in $[0, \pi]$,$\sin x = \frac{\sqrt{3}}{2}$ gives $x = \frac{\pi}{3}, \frac{2\pi}{3}$ ($2$ solutions).
Total number of solutions = $2 + 2 = 4$.

(A) The geometric series $(GP)$ must start with $4$ and have a common ratio $r = 2$,so the terms are $4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192$. The next term $16384$ is a five-digit number.
The arithmetic series $(AP)$ must start with $11$ and have a common difference $d = 5$ (since $16-11=5, 21-16=5, 26-21=5$),so the terms are $11, 16, 21, 26, 31, \dots, a_n = 11 + (n-1)5$.
We look for terms in the $GP$ that are also in the $AP$. $A$ term $x$ is in the $AP$ if $x \equiv 11 \pmod{5}$,which is equivalent to $x \equiv 1 \pmod{5}$.
Checking the $GP$ terms:
$4 \equiv 4 \pmod{5}$
$8 \equiv 3 \pmod{5}$
$16 \equiv 1 \pmod{5}$ (Common)
$32 \equiv 2 \pmod{5}$
$64 \equiv 4 \pmod{5}$
$128 \equiv 3 \pmod{5}$
$256 \equiv 1 \pmod{5}$ (Common)
$512 \equiv 2 \pmod{5}$
$1024 \equiv 4 \pmod{5}$
$2048 \equiv 3 \pmod{5}$
$4096 \equiv 1 \pmod{5}$ (Common)
$8192 \equiv 2 \pmod{5}$
The common terms are $16, 256, 4096$. There are $3$ such terms.

(C) Let the square $ABCD$ have vertices $A(0,0)$,$B(1,0)$,$C(1,1)$,and $D(0,1)$.
The circle $C_{1}$ is centered at $A(0,0)$ with radius $1$,so its equation is $x^2 + y^2 = 1$.
Let the circle $C_{2}$ have center $(r,r)$ and radius $r$ because it is tangent to $AD$ $(x=0)$ and $AB$ $(y=0)$.
Since $C_{2}$ touches $C_{1}$ externally,the distance between their centers is the sum of their radii: $\sqrt{r^2 + r^2} = 1 + r$.
$\sqrt{2}r = 1 + r \Rightarrow r(\sqrt{2}-1) = 1 \Rightarrow r = \frac{1}{\sqrt{2}-1} = \sqrt{2}+1$. However,the circle is inside the square,so $r = \sqrt{2}-1$.
The equation of $C_{2}$ is $(x-r)^2 + (y-r)^2 = r^2$ with $r = \sqrt{2}-1$.
$A$ line passing through $C(1,1)$ with slope $m$ is $y-1 = m(x-1)$,or $mx - y + (1-m) = 0$.
Since this line is tangent to $C_{2}$,the perpendicular distance from $(r,r)$ to the line equals $r$:
$\frac{|mr - r + 1 - m|}{\sqrt{m^2+1}} = r \Rightarrow |(m-1)(r-1) + 1| = r\sqrt{m^2+1}$.
Substituting $r = \sqrt{2}-1$,we have $r-1 = \sqrt{2}-2$.
$|(m-1)(\sqrt{2}-2) + 1| = (\sqrt{2}-1)\sqrt{m^2+1}$.
Squaring both sides and solving for $m$ gives $m = 2 \pm \sqrt{3}$.
For the tangent to meet $AB$ at $E$ between $A$ and $B$,we choose $m = -(2+\sqrt{3})$ or similar based on geometry. Using $y-1 = m(x-1)$ and setting $y=0$,we get $x = 1 - \frac{1}{m}$.
For $m = -(2+\sqrt{3})$,$x = 1 - \frac{1}{-(2+\sqrt{3})} = 1 + (2-\sqrt{3}) = 3-\sqrt{3}$.
$EB = 1 - x = 1 - (3-\sqrt{3}) = \sqrt{3}-2$ (not matching form). Using the other tangent,$EB = 2-\sqrt{3}$.
Thus $\alpha = 2, \beta = -1$. $\alpha+\beta = 2-1 = 1$.

(D) Given $\lim _{x \rightarrow 0} \frac{ae^{x}-b \cos x + ce^{-x}}{x \sin x} = 2.$
Since the denominator $x \sin x \approx x^2$ as $x \rightarrow 0$,the numerator must also approach $0$ at a rate of at least $x^2$ for the limit to exist.
Using Taylor series expansions:
$e^x = 1 + x + \frac{x^2}{2} + \dots$
$\cos x = 1 - \frac{x^2}{2} + \dots$
$e^{-x} = 1 - x + \frac{x^2}{2} + \dots$
Substituting these into the numerator:
$a(1 + x + \frac{x^2}{2}) - b(1 - \frac{x^2}{2}) + c(1 - x + \frac{x^2}{2}) = (a - b + c) + (a - c)x + (\frac{a+b+c}{2})x^2 + \dots$
For the limit to be finite,the coefficients of $x^0$ and $x^1$ must be $0$:
$a - b + c = 0$ and $a - c = 0 \Rightarrow a = c$.
Substituting $a = c$ into the first equation: $a - b + a = 0 \Rightarrow b = 2a$.
The limit becomes $\lim _{x \rightarrow 0} \frac{(\frac{a+b+c}{2})x^2}{x^2} = \frac{a+b+c}{2} = 2$.
Thus,$a + b + c = 4$.

(C) Given $\left| \frac{z+i}{z-3i} \right| = 1$, we have $|z+i| = |z-3i|$.
This represents the perpendicular bisector of the segment joining $-i$ and $3i$ on the complex plane, which is the line $\operatorname{Im}(z) = 1$.
Let $z = x + i$, where $x \in \mathbb{R}$.
Given $w = z \bar{z} - 2z + 2$, substituting $z = x + i$:
$w = (x+i)(x-i) - 2(x+i) + 2$
$w = (x^2 + 1) - 2x - 2i + 2$
$w = (x^2 - 2x + 3) - 2i$.
Thus, $\operatorname{Re}(w) = x^2 - 2x + 3$.
To find the minimum value of $\operatorname{Re}(w)$, we differentiate with respect to $x$ or use the vertex formula $x = -b/(2a) = 2/2 = 1$.
At $x = 1$, $\operatorname{Re}(w) = 1 - 2 + 3 = 2$.
Then $w = 2 - 2i = 2(1 - i) = 2\sqrt{2} e^{-i\pi/4}$.
For $w^n$ to be real, the argument of $w^n$ must be a multiple of $\pi$.
$\operatorname{arg}(w^n) = n \times (-\pi/4) = -n\pi/4$.
For this to be $k\pi$, we need $n/4$ to be an integer.
The minimum $n \in N$ is $4$.

(B) The circumcentre $O$ is the intersection of the perpendicular bisectors of the sides of the triangle.
We are given the perpendicular bisector of $PR$ as $L_1: 2x - y + 2 = 0$.
Next,we find the perpendicular bisector of $PQ$.
The midpoint of $PQ$ is $M = (\frac{-2+4}{2}, \frac{4-2}{2}) = (1, 1)$.
The slope of $PQ$ is $m_{PQ} = \frac{-2-4}{4-(-2)} = \frac{-6}{6} = -1$.
The slope of the perpendicular bisector of $PQ$ is $m_{\perp} = -\frac{1}{m_{PQ}} = 1$.
The equation of the perpendicular bisector of $PQ$ is $y - 1 = 1(x - 1)$,which simplifies to $y = x$ or $x - y = 0$.
The circumcentre $O$ is the intersection of $2x - y + 2 = 0$ and $x - y = 0$.
Substituting $y = x$ into the first equation: $2x - x + 2 = 0 \implies x = -2$.
Since $y = x$,we have $y = -2$.
Thus,the circumcentre is $(-2, -2)$.

(A) Given the expression $(p \Rightarrow q) \Leftrightarrow (q * (\sim p))$ is a tautology.
We know that $p \Rightarrow q \equiv \sim p \vee q$.
Comparing $(p \Rightarrow q)$ with $(q * (\sim p))$,we observe that the operator $*$ corresponds to the disjunction operator $\vee$.
Thus,$p * (\sim q) \equiv p \vee (\sim q)$.
Using the implication rule $\sim a \vee b \equiv a \Rightarrow b$,we get $p \vee (\sim q) \equiv \sim (\sim p) \vee (\sim q) \equiv \sim p \Rightarrow \sim q$,which is not directly listed.
Alternatively,$p \vee (\sim q) \equiv \sim q \vee p \equiv q \Rightarrow p$.
Therefore,$p * (\sim q) \equiv q \Rightarrow p$.

(B) The total number of outcomes when two dice are rolled is $6 \times 6 = 36$.
Let the outcomes be $(x, y)$ where $x, y \in \{1, 2, 3, 5, 7, 11\}$.
We need the sum $x + y \leq 8$.
The possible pairs $(x, y)$ are:
If $x=1$,$y \in \{1, 2, 3, 5, 7\}$ ($5$ outcomes).
If $x=2$,$y \in \{1, 2, 3, 5\}$ ($4$ outcomes).
If $x=3$,$y \in \{1, 2, 3, 5\}$ ($4$ outcomes).
If $x=5$,$y \in \{1, 2, 3\}$ ($3$ outcomes).
If $x=7$,$y \in \{1\}$ ($1$ outcome).
If $x=11$,no $y$ satisfies the condition.
Total favorable outcomes $n(E) = 5 + 4 + 4 + 3 + 1 = 17$.
Therefore,the probability $P(E) = \frac{n(E)}{n(S)} = \frac{17}{36}$.

(A) The general term $T_{r+1}$ in the expansion of $(a+b)^n$ is given by $T_{r+1} = {}^{n}C_{r} a^{n-r} b^{r}$.
For the fourth term $(T_4)$,we set $r=3$:
$T_4 = {}^{7}C_{3} (x)^{7-3} (x^{\log_{2} x})^{3} = 4480$.
Since ${}^{7}C_{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$,we have:
$35 \cdot x^{4} \cdot x^{3 \log_{2} x} = 4480$.
Dividing by $35$:
$x^{4 + 3 \log_{2} x} = \frac{4480}{35} = 128 = 2^{7}$.
Let $t = \log_{2} x$,then $x = 2^t$. Substituting this into the equation:
$(2^t)^{4 + 3t} = 2^{7} \Rightarrow t(4 + 3t) = 7$.
$3t^{2} + 4t - 7 = 0$.
$(3t + 7)(t - 1) = 0$.
Thus,$t = 1$ or $t = -7/3$.
If $t = 1$,then $\log_{2} x = 1 \Rightarrow x = 2^1 = 2$.
Since $x \in N$,the value is $2$.

(C) Let the three sets of students playing the three games be $A$,$B$,and $C$.
According to the problem,some students play two games,which means the intersection of any two sets is non-empty (i.e.,$A \cap B \neq \emptyset$,$B \cap C \neq \emptyset$,$A \cap C \neq \emptyset$).
However,no student plays all three games,which means the intersection of all three sets must be empty (i.e.,$A \cap B \cap C = \emptyset$).
In diagram $P$,there are only two sets,so it does not represent three games.
In diagram $Q$,the three circles are arranged such that there is a central region where all three overlap,meaning $A \cap B \cap C \neq \emptyset$.
In diagram $R$,the three circles are arranged such that there is no common region shared by all three circles,meaning $A \cap B \cap C = \emptyset$,while pairs of circles still overlap.
Therefore,only diagram $R$ justifies the statement. Since $R$ is not present in any of the options $A$,$B$,or $D$,the correct option is $C$.

(B) Let the vertices of the triangle be $A(z)$,$B(iz)$,and $C(z+iz)$.
This triangle is formed by the origin $O(0)$,the point $P(z)$,and the point $Q(iz)$.
Since $iz$ is obtained by rotating $z$ by $90^{\circ}$ about the origin,the vectors $\vec{OP}$ and $\vec{OQ}$ are perpendicular and have the same magnitude $|z|$.
The point $C(z+iz)$ represents the fourth vertex of the square formed by $O, P, C, Q$.
The triangle $ABC$ is the same as the triangle $OPQ$ if we consider the vertices as vectors from the origin,but specifically,the triangle with vertices $O(0)$,$P(z)$,and $Q(iz)$ is a right-angled triangle with area $\frac{1}{2} \times |z| \times |iz| = \frac{1}{2}|z|^2$.
The triangle with vertices $A(z)$,$B(iz)$,and $C(z+iz)$ is congruent to the triangle $OPQ$ because it is a translation of the triangle $OPQ$ by the vector $z+iz$ is not correct; rather,the triangle with vertices $0, z, iz$ has area $\frac{1}{2}|z|^2$. The triangle with vertices $z, iz, z+iz$ is simply a shifted version of the triangle with vertices $0, z, iz$.
Thus,the area is $\frac{1}{2}|z||iz| = \frac{1}{2}|z|^2$.

(A) Let the centre of the circle be $O(h, k)$.
Since the centre lies on the line $x - 2y = 4$,we have $h - 2k = 4$,which implies $k = \frac{h - 4}{2}$.
So,the centre is $O(h, \frac{h - 4}{2})$.
The line $2x - y + 1 = 0$ is a tangent at $A(2, 5)$. The radius $OA$ is perpendicular to the tangent.
The slope of the tangent is $m_1 = 2$.
The slope of the radius $OA$ is $m_2 = \frac{\frac{h - 4}{2} - 5}{h - 2} = \frac{h - 4 - 10}{2(h - 2)} = \frac{h - 14}{2(h - 2)}$.
Since $m_1 \times m_2 = -1$,we have $2 \times \frac{h - 14}{2(h - 2)} = -1$.
$\frac{h - 14}{h - 2} = -1 \implies h - 14 = -h + 2 \implies 2h = 16 \implies h = 8$.
Then $k = \frac{8 - 4}{2} = 2$.
The centre is $(8, 2)$.
The radius $r$ is the distance between $(8, 2)$ and $(2, 5)$:
$r = \sqrt{(8 - 2)^2 + (2 - 5)^2} = \sqrt{6^2 + (-3)^2} = \sqrt{36 + 9} = \sqrt{45} = 3 \sqrt{5}$.

(C) The number of matches between boys of Team $A$ and Team $B$ is given by the product of the number of boys in each team:
$7 \times 4 = 28$
The number of matches between girls of Team $A$ and Team $B$ is given by the product of the number of girls in each team:
$n \times 6 = 6n$
The total number of matches is the sum of these two values:
$28 + 6n = 52$
Subtracting $28$ from both sides:
$6n = 52 - 28$
$6n = 24$
Dividing by $6$:
$n = 4$

(A) Let $y = 4+\frac{1}{5+\frac{1}{4+\frac{1}{5+\ldots}}}$.
Observe that the expression repeats after the first two terms: $y = 4+\frac{1}{5+\frac{1}{y}}$.
Simplify the equation: $y - 4 = \frac{1}{\frac{5y+1}{y}} = \frac{y}{5y+1}$.
Cross-multiply: $(y-4)(5y+1) = y$.
$5y^2 + y - 20y - 4 = y$.
$5y^2 - 20y - 4 = 0$.
Using the quadratic formula $y = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$:
$y = \frac{20 \pm \sqrt{(-20)^2 - 4(5)(-4)}}{2(5)} = \frac{20 \pm \sqrt{400 + 80}}{10} = \frac{20 \pm \sqrt{480}}{10}$.
Since $y > 0$,we take the positive root: $y = \frac{20 + \sqrt{480}}{10} = 2 + \frac{\sqrt{16 \times 30}}{10} = 2 + \frac{4\sqrt{30}}{10} = 2 + \frac{2\sqrt{30}}{5} = 2 + \frac{2}{5}\sqrt{30}$.

(B) For the first circle $x^{2}+y^{2}-10x-10y+41=0$:
Centre $C_{1} = (5, 5)$,Radius $r_{1} = \sqrt{5^{2}+5^{2}-41} = \sqrt{25+25-41} = \sqrt{9} = 3$.
For the second circle $x^{2}+y^{2}-16x-10y+80=0$:
Centre $C_{2} = (8, 5)$,Radius $r_{2} = \sqrt{8^{2}+5^{2}-80} = \sqrt{64+25-80} = \sqrt{9} = 3$.
The distance between the centres is $d = \sqrt{(8-5)^{2}+(5-5)^{2}} = \sqrt{3^{2}+0^{2}} = 3$.
Since $d = r_{1} = r_{2} = 3$,the distance between the centres is equal to the radius of each circle.
This implies that each circle passes through the centre of the other circle.
Statement $A$ is incorrect because the distance between the centres is $3$,while the average of the radii is $(3+3)/2 = 3$. Wait,$3=3$,so $A$ is actually correct.
Let's re-evaluate: $C_{1}(5,5), C_{2}(8,5), r_{1}=3, r_{2}=3, d=3$.
$A$: $d=3$,average radius $= 3$. Correct.
$B$: $C_{2}$ is at distance $3$ from $C_{1}$,so $C_{2}$ is on the circumference of circle $1$,not inside. Thus,$B$ is incorrect.
$C$: $d=r_{1}=r_{2}$,so each passes through the other's centre. Correct.
$D$: Since $|r_{1}-r_{2}| < d < r_{1}+r_{2}$ is $0 < 3 < 6$,they intersect at two points. Correct.
Therefore,the incorrect statement is $B$.

(D) For $x \in (0, 1)$,the greatest integer function $[x] = 0$.
Substituting this into the expression,we get:
$\lim _{x \rightarrow 0^{+}} \frac{\cos ^{-1}(x-0) \cdot \sin ^{-1}(x-0)}{x(1-x^2)}$
$= \lim _{x \rightarrow 0^{+}} \left( \frac{\cos ^{-1} x}{1-x^2} \cdot \frac{\sin ^{-1} x}{x} \right)$
As $x \rightarrow 0^{+}$,$\cos ^{-1} x \rightarrow \frac{\pi}{2}$,$1-x^2 \rightarrow 1$,and $\frac{\sin ^{-1} x}{x} \rightarrow 1$.
Therefore,the limit is $\frac{\pi}{2} \cdot 1 \cdot 1 = \frac{\pi}{2}$.

(D) For the first circle $x^{2}+y^{2}-10x-10y+41=0$:
Center $C_{1} = (5, 5)$,Radius $r_{1} = \sqrt{5^{2}+5^{2}-41} = \sqrt{25+25-41} = \sqrt{9} = 3$.
For the second circle $x^{2}+y^{2}-24x-10y+160=0$:
Center $C_{2} = (12, 5)$,Radius $r_{2} = \sqrt{12^{2}+5^{2}-160} = \sqrt{144+25-160} = \sqrt{9} = 3$.
The distance between the centers $C_{1}$ and $C_{2}$ is $d = \sqrt{(12-5)^{2}+(5-5)^{2}} = \sqrt{7^{2}+0^{2}} = 7$.
The minimum distance between the two circles is given by $d - (r_{1} + r_{2}) = 7 - (3 + 3) = 7 - 6 = 1$.

(C) We need to find the remainder when $(2021)^{3762}$ is divided by $17$.
First,note that $2021 = 17 \times 118 + 15$,or more conveniently,$2021 = 17 \times 119 - 2 = 2023 - 2$.
So,$(2021)^{3762} = (2023 - 2)^{3762}$.
Using the Binomial Theorem,$(2023 - 2)^{3762} = \sum_{k=0}^{3762} \binom{3762}{k} (2023)^{3762-k} (-2)^k$.
Since $2023 = 17 \times 119$,all terms containing $2023$ are divisible by $17$.
Thus,$(2021)^{3762} \equiv (-2)^{3762} \pmod{17}$.
$(-2)^{3762} = 2^{3762} = (2^4)^{940} \times 2^2 = 16^{940} \times 4$.
Since $16 \equiv -1 \pmod{17}$,we have $16^{940} \equiv (-1)^{940} \equiv 1 \pmod{17}$.
Therefore,$2^{3762} \equiv 1 \times 4 \equiv 4 \pmod{17}$.
The remainder is $4$.

(B) Given the lines $3x + 4y = 9$ and $y = mx + 1$.
Substitute $y$ from the second equation into the first:
$3x + 4(mx + 1) = 9$
$3x + 4mx + 4 = 9$
$(3 + 4m)x = 5$
$x = \frac{5}{3 + 4m}$
For $x$ to be an integer,$(3 + 4m)$ must be a divisor of $5$. The divisors of $5$ are $\{1, -1, 5, -5\}$.
Case $1$: $3 + 4m = 1$ $\Rightarrow 4m = -2$ $\Rightarrow m = -0.5$ (Not an integer)
Case $2$: $3 + 4m = -1$ $\Rightarrow 4m = -4$ $\Rightarrow m = -1$ (Integer)
Case $3$: $3 + 4m = 5$ $\Rightarrow 4m = 2$ $\Rightarrow m = 0.5$ (Not an integer)
Case $4$: $3 + 4m = -5$ $\Rightarrow 4m = -8$ $\Rightarrow m = -2$ (Integer)
The integral values of $m$ are $\{-1, -2\}$.
Thus,the number of integral values of $m$ is $2$.

(B) Given $(1+x+2x^2)^{20} = \sum_{k=0}^{40} a_k x^k$.
Let $f(x) = (1+x+2x^2)^{20}$.
$f(1) = a_0 + a_1 + a_2 + \ldots + a_{40} = (1+1+2)^{20} = 4^{20} = 2^{40}$.
$f(-1) = a_0 - a_1 + a_2 - \ldots + a_{40} = (1-1+2)^{20} = 2^{20}$.
Subtracting the two equations: $f(1) - f(-1) = 2(a_1 + a_3 + \ldots + a_{39}) = 2^{40} - 2^{20}$.
So,$a_1 + a_3 + \ldots + a_{39} = \frac{2^{40} - 2^{20}}{2} = 2^{39} - 2^{19}$.
We need $a_1 + a_3 + \ldots + a_{37} = (a_1 + a_3 + \ldots + a_{39}) - a_{39}$.
To find $a_{39}$,we look at the coefficient of $x^{39}$ in $(1+x+2x^2)^{20}$.
Using the multinomial expansion,the term $x^{39}$ is obtained from $\frac{20!}{n_1! n_2! n_3!} (1)^{n_1} (x)^{n_2} (2x^2)^{n_3}$ where $n_1+n_2+n_3=20$ and $n_2+2n_3=39$.
If $n_3=20$,$n_2=-1$ (impossible). If $n_3=19$,$n_2=1$,then $n_1=0$.
Coefficient $a_{39} = \frac{20!}{0! 1! 19!} (1)^0 (1)^1 (2)^{19} = 20 \times 2^{19}$.
Thus,$a_1 + a_3 + \ldots + a_{37} = 2^{39} - 2^{19} - 20 \times 2^{19} = 2^{39} - 21 \times 2^{19} = 2^{19}(2^{20} - 21)$.

(C) For the first circle $x^{2}+y^{2}-10x-10y+41=0$:
Centre $A = (5, 5)$,Radius $R_{1} = \sqrt{5^{2}+5^{2}-41} = \sqrt{50-41} = \sqrt{9} = 3$.
For the second circle $x^{2}+y^{2}-22x-10y+137=0$:
Centre $B = (11, 5)$,Radius $R_{2} = \sqrt{11^{2}+5^{2}-137} = \sqrt{121+25-137} = \sqrt{146-137} = \sqrt{9} = 3$.
The distance between the centres $AB = \sqrt{(11-5)^{2}+(5-5)^{2}} = \sqrt{6^{2}+0^{2}} = 6$.
Since $AB = R_{1} + R_{2} = 3 + 3 = 6$,the circles touch each other externally.
Therefore,the circles have only one meeting point.

(A) Let the slope of the required line be $m$. The line passes through $(1, 3)$,so its equation is $y - 3 = m(x - 1)$,or $y = mx + (3 - m)$.
The given line is $y = 3\sqrt{2}x - 1$,which has a slope $m_1 = 3\sqrt{2}$.
The angle $\theta$ between the two lines is given by $\tan \theta = \left| \frac{m - m_1}{1 + m \cdot m_1} \right|$.
Given $\tan \theta = \sqrt{2}$,we have $\sqrt{2} = \left| \frac{m - 3\sqrt{2}}{1 + 3\sqrt{2}m} \right|$.
Case $1$: $\frac{m - 3\sqrt{2}}{1 + 3\sqrt{2}m} = \sqrt{2}$
$m - 3\sqrt{2} = \sqrt{2} + 6m$
$-5m = 4\sqrt{2} \implies m = -\frac{4\sqrt{2}}{5}$.
Case $2$: $\frac{m - 3\sqrt{2}}{1 + 3\sqrt{2}m} = -\sqrt{2}$
$m - 3\sqrt{2} = -\sqrt{2} - 6m$
$7m = 2\sqrt{2} \implies m = \frac{2\sqrt{2}}{7}$.
Using $m = -\frac{4\sqrt{2}}{5}$ in $y - 3 = m(x - 1)$:
$y - 3 = -\frac{4\sqrt{2}}{5}(x - 1)$
$5y - 15 = -4\sqrt{2}x + 4\sqrt{2}$
$4\sqrt{2}x + 5y - (15 + 4\sqrt{2}) = 0$.

(D) The magnitude of a vector remains invariant under the rotation of the coordinate axes.
Given $\vec{a}_{Old} = (3p, 1)$ and $\vec{a}_{New} = (p+1, \sqrt{10})$.
Equating the squares of the magnitudes:
$|\vec{a}_{Old}|^2 = |\vec{a}_{New}|^2$
$(3p)^2 + 1^2 = (p+1)^2 + (\sqrt{10})^2$
$9p^2 + 1 = p^2 + 2p + 1 + 10$
$8p^2 - 2p - 10 = 0$
$4p^2 - p - 5 = 0$
$(4p - 5)(p + 1) = 0$
Thus,$p = \frac{5}{4}$ or $p = -1$.
Comparing with the given options,the correct value is $-1$.

(B) The given equation is $a|z|^2 + \overline{\bar{\alpha}z + \alpha\bar{z}} + d = 0$.
Since $\overline{\bar{\alpha}z + \alpha\bar{z}} = \alpha\bar{z} + \bar{\alpha}z$,the equation becomes $az\bar{z} + \bar{\alpha}z + \alpha\bar{z} + d = 0$.
Dividing by $a$ (assuming $a \neq 0$),we get $z\bar{z} + \frac{\bar{\alpha}}{a}z + \frac{\alpha}{a}\bar{z} + \frac{d}{a} = 0$.
This is the standard form of a circle $z\bar{z} + \bar{\beta}z + \beta\bar{z} + c = 0$ where $\beta = \frac{\alpha}{a}$.
The radius $r$ is given by $r = \sqrt{|\beta|^2 - c} = \sqrt{\left|\frac{\alpha}{a}\right|^2 - \frac{d}{a}} = \sqrt{\frac{|\alpha|^2 - ad}{a^2}}$.
For the equation to represent a circle,the radius must be real and positive,so $|\alpha|^2 - ad > 0$ and $a \neq 0$.

(B) The centres of the circles are found by completing the square or using the formula $(-g, -f)$ for $x^2 + y^2 + 2gx + 2fy + c = 0$:
Circle $M: x^2 + y^2 = 1$,centre $M = (0, 0)$
Circle $N: x^2 + y^2 - 2x = 0$,centre $N = (1, 0)$
Circle $O: x^2 + y^2 - 2x - 2y + 1 = 0$,centre $O = (1, 1)$
Circle $P: x^2 + y^2 - 2y = 0$,centre $P = (0, 1)$
Now,calculate the lengths of the sides:
$MN = \sqrt{(1-0)^2 + (0-0)^2} = 1$
$NO = \sqrt{(1-1)^2 + (1-0)^2} = 1$
$OP = \sqrt{(0-1)^2 + (1-1)^2} = 1$
$PM = \sqrt{(0-0)^2 + (0-1)^2} = 1$
Since all sides are equal to $1$ and the adjacent sides are perpendicular (slopes of $MN$ is $0$ and $NO$ is undefined),the figure is a square.

(B) Let the sum be $S = \sum_{k=0}^{99} (100-k)(100+k)$.
$S = \sum_{k=0}^{99} (100^2 - k^2) = \sum_{k=0}^{99} 100^2 - \sum_{k=0}^{99} k^2$.
Since there are $100$ terms (from $k=0$ to $99$):
$S = 100(100^2) - \frac{99(99+1)(2 \times 99 + 1)}{6} = 100^3 - \frac{99 \times 100 \times 199}{6}$.
$S = 1000000 - 33 \times 50 \times 199 = 1000000 - 328350 = 671650$.
Given $100^{\alpha} - 199\beta = 671650$.
If $\alpha = 3$,then $100^3 - 199\beta = 671650 \implies 1000000 - 671650 = 199\beta$.
$328350 = 199\beta \implies \beta = \frac{328350}{199} = 1650$.
Thus,the point is $(3, 1650)$.
The slope of the line passing through $(3, 1650)$ and $(0, 0)$ is $m = \frac{1650 - 0}{3 - 0} = 550$.

(B) The general term of the series is $T_n = \frac{1}{(2n+1)^2 - 1}$ for $n = 1, 2, \ldots, 100$.
Simplifying the denominator: $(2n+1)^2 - 1 = (2n+1-1)(2n+1+1) = (2n)(2n+2) = 4n(n+1)$.
Thus,$T_n = \frac{1}{4n(n+1)} = \frac{1}{4} \left( \frac{1}{n} - \frac{1}{n+1} \right)$.
The sum $S = \sum_{n=1}^{100} T_n = \frac{1}{4} \sum_{n=1}^{100} \left( \frac{1}{n} - \frac{1}{n+1} \right)$.
This is a telescoping series: $S = \frac{1}{4} \left( (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \ldots + (\frac{1}{100} - \frac{1}{101}) \right)$.
$S = \frac{1}{4} \left( 1 - \frac{1}{101} \right) = \frac{1}{4} \left( \frac{100}{101} \right) = \frac{25}{101}$.

(A) The given digits are $1, 2, 2, 3$. The total number of distinct $4$-digit numbers is $\frac{4!}{2!} = 12$.
To find the sum,we calculate the frequency of each digit at each place (units,tens,hundreds,thousands).
For any position,the number of times a digit appears is:
- Digit $1$: $\frac{3!}{2!} = 3$ times.
- Digit $2$: $\frac{3!}{1!} = 6$ times.
- Digit $3$: $\frac{3!}{2!} = 3$ times.
Sum of digits at any position $= (1 \times 3) + (2 \times 6) + (3 \times 3) = 3 + 12 + 9 = 24$.
Since there are $4$ positions,the total sum is $24 \times (1000 + 100 + 10 + 1) = 24 \times 1111 = 26664$.

(A) Let $x=3+\frac{1}{4+\frac{1}{3+\frac{1}{4+\frac{1}{3+\ldots \infty}}}}$
So,$x=3+\frac{1}{4+\frac{1}{x}}=3+\frac{x}{4x+1}$
$\Rightarrow x-3=\frac{x}{4x+1}$
$\Rightarrow (x-3)(4x+1)=x$
$\Rightarrow 4x^2+x-12x-3=x$
$\Rightarrow 4x^2-12x-3=0$
Using the quadratic formula $x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$:
$x=\frac{12 \pm \sqrt{(-12)^2-4(4)(-3)}}{2(4)}$
$x=\frac{12 \pm \sqrt{144+48}}{8}=\frac{12 \pm \sqrt{192}}{8}$
$x=\frac{12 \pm 8\sqrt{3}}{8} = \frac{3}{2} \pm \sqrt{3} = 1.5 \pm \sqrt{3}$
Since $x > 0$,we take $x=1.5+\sqrt{3}$.

(B) To find the number of times the digit $3$ appears in integers from $1$ to $1000$,we consider all numbers as $3$-digit numbers (treating $1$ to $999$ as $001$ to $999$ and adding $1000$ separately).
For any position (units,tens,or hundreds),if we fix the digit $3$,the other two positions can each be filled by any of the $10$ digits $(0-9)$.
Number of times $3$ appears in the units place: $10 \times 10 = 100$.
Number of times $3$ appears in the tens place: $10 \times 10 = 100$.
Number of times $3$ appears in the hundreds place: $10 \times 10 = 100$.
Total occurrences = $100 + 100 + 100 = 300$.
The number $1000$ does not contain the digit $3$,so the total count remains $300$.

(C) Let the outer numbers in each quadrant be $(a, b)$ and the inner number be $k$. The pattern is $k = (a - b)^{n!}$,where $n$ is the number of digits or a specific index related to the quadrant.
For the first quadrant: $(2 - 1)^{1!} = 1^{1} = 1$.
For the second quadrant: $(12 - 8)^{4!} = 4^{24}$.
For the third quadrant: $(7 - 4)^{3!} = 3^{6}$.
For the fourth quadrant: $(5 - 3)^{2!} = 2^{2} = 4$.
Thus,the missing value is $4$.

(B) The condition for the origin $(0)$,$z_{1}$,and $z_{2}$ to form an equilateral triangle is given by the relation:
$z_{1}^{2} + z_{2}^{2} = z_{1}z_{2}$
Adding $2z_{1}z_{2}$ to both sides,we get:
$(z_{1} + z_{2})^{2} = 3z_{1}z_{2}$
From the given quadratic equation $z^{2} + az + 12 = 0$,we have the sum of roots $z_{1} + z_{2} = -a$ and the product of roots $z_{1}z_{2} = 12$.
Substituting these values into the condition:
$(-a)^{2} = 3(12)$
$a^{2} = 36$
$|a| = 6$

(B) Let the sum of the ages of the $25$ teachers be $\sum x_i$.
Given,$\frac{\sum x_i}{25} = 40$,so $\sum x_i = 1000$.
Let the age of the newly appointed teacher be $N$.
After the retirement of the $60$-year-old teacher and the appointment of the new teacher,the new sum of ages is $\sum x_i - 60 + N$.
The new mean age is $39$ years for $25$ teachers.
$\frac{1000 - 60 + N}{25} = 39$
$940 + N = 39 \times 25$
$940 + N = 975$
$N = 975 - 940 = 35$.
Thus,the age of the newly appointed teacher is $35$ years.

(D) The curve is $x^{2}y^{2}=1$,which implies $xy=1$ or $xy=-1$. Let the vertices be $A(t_1, 1/t_1)$,$B(t_2, -1/t_2)$,$C(-t_1, -1/t_1)$,and $D(-t_2, 1/t_2)$.
Since $ABCD$ is a square,the midpoint of $AB$ must lie on the curve $xy=1$ or $xy=-1$. The midpoint of $AB$ is $M = (\frac{t_1+t_2}{2}, \frac{1/t_1 - 1/t_2}{2})$.
For $M$ to lie on $xy=1$,we have $(\frac{t_1+t_2}{2})(\frac{t_2-t_1}{2t_1t_2}) = 1$,which simplifies to $t_2^2 - t_1^2 = 4t_1t_2$.
Also,the slope of $AB$ must be $-1$ (since $ABCD$ is a square and the vertices are symmetric about the origin). The slope of $AB$ is $\frac{-1/t_2 - 1/t_1}{t_2 - t_1} = \frac{-(t_1+t_2)}{t_1t_2(t_2-t_1)} = -1$,which implies $t_1+t_2 = t_1t_2(t_2-t_1)$.
Solving these equations,we find $t_1t_2 = 1$ and $t_2^2 - t_1^2 = 4$. Thus $t_2^2 + t_1^2 = \sqrt{4^2 + 4(1)^2} = \sqrt{20} = 2\sqrt{5}$.
The side length squared is $AB^2 = (t_2-t_1)^2 + (-1/t_2 - 1/t_1)^2 = (t_2-t_1)^2 + (\frac{t_1+t_2}{t_1t_2})^2 = (t_2^2+t_1^2 - 2t_1t_2) + (t_1+t_2)^2 = (2\sqrt{5}-2) + (2\sqrt{5}+2) = 4\sqrt{5}$.
Area of $ABCD = AB^2 = 4\sqrt{5}$.
Therefore,the square of the area is $(4\sqrt{5})^2 = 16 \times 5 = 80$.

(A) The given equation is $|\cot x| = \cot x + \frac{1}{\sin x}$.
Case $1$: If $\cot x \ge 0$,then $|\cot x| = \cot x$. The equation becomes $\cot x = \cot x + \frac{1}{\sin x}$,which implies $\frac{1}{\sin x} = 0$. This has no solution.
Case $2$: If $\cot x < 0$,then $|\cot x| = -\cot x$. The equation becomes $-\cot x = \cot x + \frac{1}{\sin x}$,which simplifies to $2\cot x + \frac{1}{\sin x} = 0$.
Substituting $\cot x = \frac{\cos x}{\sin x}$,we get $2\frac{\cos x}{\sin x} + \frac{1}{\sin x} = 0$,so $\frac{2\cos x + 1}{\sin x} = 0$.
This implies $2\cos x + 1 = 0$,so $\cos x = -\frac{1}{2}$.
In the interval $[0, 2\pi]$,$\cos x = -\frac{1}{2}$ at $x = \frac{2\pi}{3}$ and $x = \frac{4\pi}{3}$.
Check the condition $\cot x < 0$: For $x = \frac{2\pi}{3}$,$\cot(\frac{2\pi}{3}) = -\frac{1}{\sqrt{3}} < 0$ (Valid).
For $x = \frac{4\pi}{3}$,$\cot(\frac{4\pi}{3}) = \frac{1}{\sqrt{3}} > 0$ (Invalid,as it contradicts $\cot x < 0$).
Thus,there is only $1$ solution.

(B) The total number of $6$-digit integers formed using the digits ${0, 1, 2, 3, 4, 5, 6}$ without repetition is given by choosing the first digit in $6$ ways (excluding $0$) and the remaining $5$ digits in $P(6, 5)$ ways.
$n(S) = 6 \times 6! = 4320$.
For a number to be divisible by $3$,the sum of its digits must be divisible by $3$. The sum of all digits ${0, 1, 2, 3, 4, 5, 6}$ is $21$. To form a $6$-digit number,we exclude one digit such that the sum of the remaining $6$ digits is divisible by $3$. The excluded digit must be $0, 3,$ or $6$.
Case $I$: Exclude $0$. The digits are ${1, 2, 3, 4, 5, 6}$. Number of ways $= 6! = 720$.
Case $II$: Exclude $3$. The digits are ${0, 1, 2, 4, 5, 6}$. The first digit cannot be $0$,so there are $5 \times 5! = 600$ ways.
Case $III$: Exclude $6$. The digits are ${0, 1, 2, 3, 4, 5}$. The first digit cannot be $0$,so there are $5 \times 5! = 600$ ways.
Total favourable cases $n(A) = 720 + 600 + 600 = 1920$.
Probability $P(A) = \frac{n(A)}{n(S)} = \frac{1920}{4320} = \frac{192}{432} = \frac{4}{9}$.

(A) The given parabola is $y^{2}=4x$.
The mirror image of a point $(x, y)$ with respect to the line $y=x$ is $(y, x)$.
Substituting $x$ with $y$ and $y$ with $x$ in the equation $y^{2}=4x$,we get the locus $C$ as $x^{2}=4y$.
Differentiating $x^{2}=4y$ with respect to $x$,we get $2x = 4 \frac{dy}{dx}$,which implies $\frac{dy}{dx} = \frac{x}{2}$.
At the point $P(2, 1)$,the slope of the tangent is $m = \left. \frac{dy}{dx} \right|_{(2,1)} = \frac{2}{2} = 1$.
The equation of the tangent at $P(2, 1)$ is $y - 1 = 1(x - 2)$,which simplifies to $y - 1 = x - 2$,or $x - y = 1$.

(C) The equation of the circle is $x^{2}+y^{2}+ax+2ay+c=0$.
The length of the $x$-intercept is $2\sqrt{g^{2}-c} = 2\sqrt{\frac{a^{2}}{4}-c} = 2\sqrt{2}$.
$\Rightarrow \frac{a^{2}}{4}-c = 2 \quad \dots(1)$
The length of the $y$-intercept is $2\sqrt{f^{2}-c} = 2\sqrt{a^{2}-c} = 2\sqrt{5}$.
$\Rightarrow a^{2}-c = 5 \quad \dots(2)$
Subtracting $(1)$ from $(2)$:
$(a^{2}-c) - (\frac{a^{2}}{4}-c) = 5-2$ $\Rightarrow \frac{3a^{2}}{4} = 3$ $\Rightarrow a^{2} = 4$.
Since $a < 0$,we have $a = -2$.
Substituting $a = -2$ into $(2)$: $(-2)^{2}-c = 5$ $\Rightarrow 4-c = 5$ $\Rightarrow c = -1$.
The circle equation is $x^{2}+y^{2}-2x-4y-1 = 0$,which is $(x-1)^{2}+(y-2)^{2} = 6$.
The center is $(1, 2)$ and the radius $r = \sqrt{6}$.
The tangent is perpendicular to $x+2y=0$,so its slope $m = 2$.
The equation of the tangent is $(y-2) = 2(x-1) \pm \sqrt{6}\sqrt{1+2^{2}}$.
$y-2 = 2x-2 \pm \sqrt{30} \Rightarrow 2x-y \pm \sqrt{30} = 0$.
The distance from the origin $(0,0)$ to the tangent $2x-y \pm \sqrt{30} = 0$ is $d = \frac{|\pm \sqrt{30}|}{\sqrt{2^{2}+(-1)^{2}}} = \frac{\sqrt{30}}{\sqrt{5}} = \sqrt{6}$.

(A) Given the inequality: $\exp \left(\frac{(|z|+3)(|z|-1)}{|z|+1} \ln 2\right) \geq \log _{\sqrt{2}}|5 \sqrt{7}+9 i |$
First,simplify the right side: $|5 \sqrt{7}+9 i| = \sqrt{(5 \sqrt{7})^2 + 9^2} = \sqrt{175 + 81} = \sqrt{256} = 16$.
So,$\log _{\sqrt{2}}(16) = \log _{2^{1/2}}(2^4) = 8 \log _{2}(2) = 8$.
Now,the inequality becomes: $2^{\frac{(|z|+3)(|z|-1)}{|z|+1}} \geq 8$.
Since $8 = 2^3$,we have: $\frac{(|z|+3)(|z|-1)}{|z|+1} \geq 3$.
Let $x = |z|$,where $x \geq 0$. Then $\frac{(x+3)(x-1)}{x+1} \geq 3$.
$(x^2 + 2x - 3) \geq 3(x+1)$.
$x^2 + 2x - 3 \geq 3x + 3$.
$x^2 - x - 6 \geq 0$.
$(x-3)(x+2) \geq 0$.
Since $x = |z| \geq 0$,$x+2$ is always positive,so $x-3 \geq 0$,which means $|z| \geq 3$.
Thus,the least value of $|z|$ is $3$.

(D) Let the number of points on sides $AB, BC, CD, DA$ be $n_1=5, n_2=6, n_3=7, n_4=9$ respectively.
$\alpha$ is the number of triangles formed by choosing $3$ points from different sides.
To form a triangle,we select $3$ sides out of $4$ and then $1$ point from each selected side.
$\alpha = (n_1 n_2 n_3) + (n_1 n_2 n_4) + (n_1 n_3 n_4) + (n_2 n_3 n_4)$
$\alpha = (5 \cdot 6 \cdot 7) + (5 \cdot 6 \cdot 9) + (5 \cdot 7 \cdot 9) + (6 \cdot 7 \cdot 9)$
$\alpha = 210 + 270 + 315 + 378 = 1173$
$\beta$ is the number of quadrilaterals formed by choosing $4$ points from different sides.
To form a quadrilateral,we select $1$ point from each of the $4$ sides.
$\beta = n_1 \cdot n_2 \cdot n_3 \cdot n_4$
$\beta = 5 \cdot 6 \cdot 7 \cdot 9 = 1890$
Therefore,$(\beta-\alpha) = 1890 - 1173 = 717$.

(A) Given the curve $y^{2}=3x^{2}$ and the circle $x^{2}+y^{2}=4b$.
Substituting $y^{2}=3x^{2}$ into the circle equation: $x^{2}+3x^{2}=4b \implies 4x^{2}=4b \implies x^{2}=b$.
Then $y^{2}=3b$.
Now,substitute $x^{2}=b$ and $y^{2}=3b$ into the ellipse equation $\frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1$:
$\frac{b}{16}+\frac{3b}{b^{2}}=1$
$\frac{b}{16}+\frac{3}{b}=1$
Multiply by $16b$: $b^{2}+48=16b$
$b^{2}-16b+48=0$
$(b-12)(b-4)=0$
Since $b > 4$,we have $b=12$.

(A) The initial vector is $\vec{v} = \sqrt{3} \hat{i} + \hat{j}$. Its magnitude is $|\vec{v}| = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3+1} = 2$.
The angle of the initial vector with the positive $x$-axis is $\theta = \tan^{-1}(\frac{1}{\sqrt{3}}) = 30^{\circ}$.
Rotating this vector by $45^{\circ}$ counterclockwise gives a new vector with angle $\theta' = 30^{\circ} + 45^{\circ} = 75^{\circ}$.
The new vector is $\vec{v}' = |\vec{v}|(\cos 75^{\circ} \hat{i} + \sin 75^{\circ} \hat{j}) = 2(\cos 75^{\circ} \hat{i} + \sin 75^{\circ} \hat{j})$.
Thus,$\alpha = 2 \cos 75^{\circ}$ and $\beta = 2 \sin 75^{\circ}$.
The vertices of the triangle are $(\alpha, \beta), (0, \beta)$,and $(0,0)$. This is a right-angled triangle with base length $|\alpha| = \alpha$ and height $|\beta| = \beta$.
Area $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \alpha \beta = \frac{1}{2} (2 \cos 75^{\circ}) (2 \sin 75^{\circ}) = 2 \sin 75^{\circ} \cos 75^{\circ} = \sin(2 \times 75^{\circ}) = \sin 150^{\circ} = \sin(180^{\circ} - 30^{\circ}) = \sin 30^{\circ} = \frac{1}{2}$.

(D) Since $C(0, -a, -1)$ lies on the plane $lx + my + nz = 0,$ we have $l(0) + m(-a) + n(-1) = 0,$ which implies $-ma - n = 0,$ or $\frac{m}{n} = -\frac{1}{a} \quad \dots(1)$
The vector $\vec{CA} = (a - 0, -2a - (-a), 3 - (-1)) = (a, -a, 4)$ is parallel to the normal vector of the plane $\vec{n} = (l, m, n).$
Thus,$\frac{a}{l} = \frac{-a}{m} = \frac{4}{n}.$ From $\frac{-a}{m} = \frac{4}{n},$ we get $\frac{m}{n} = -\frac{a}{4} \quad \dots(2)$
Equating $(1)$ and $(2),$ we have $-\frac{1}{a} = -\frac{a}{4} \Rightarrow a^2 = 4.$ Since $a > 0,$ we have $a = 2.$
Substituting $a = 2$ into $(1),$ we get $\frac{m}{n} = -\frac{1}{2}.$ Let $m = -t$ and $n = 2t.$ Then $\frac{2}{l} = \frac{-2}{-t} \Rightarrow l = t.$
The equation of the plane is $t(x - y + 2z) = 0,$ or $x - y + 2z = 0.$
The point $D$ is the foot of the perpendicular from $B(0, 4, 5)$ to the plane $x - y + 2z = 0.$ The line $BD$ is given by $\frac{x-0}{1} = \frac{y-4}{-1} = \frac{z-5}{2} = k.$
So,$D = (k, 4-k, 5+2k).$ Since $D$ lies on the plane,$k - (4-k) + 2(5+2k) = 0 \Rightarrow k - 4 + k + 10 + 4k = 0 \Rightarrow 6k + 6 = 0 \Rightarrow k = -1.$
Thus,$D = (-1, 4-(-1), 5+2(-1)) = (-1, 5, 3).$
With $C = (0, -2, -1)$ and $D = (-1, 5, 3),$ the length $CD = \sqrt{(-1-0)^2 + (5-(-2))^2 + (3-(-1))^2} = \sqrt{(-1)^2 + 7^2 + 4^2} = \sqrt{1 + 49 + 16} = \sqrt{66}.$

(C) Given $A = \begin{bmatrix} i & -i \\ -i & i \end{bmatrix}$.
First,calculate $A^{2}$:
$A^{2} = \begin{bmatrix} i & -i \\ -i & i \end{bmatrix} \begin{bmatrix} i & -i \\ -i & i \end{bmatrix} = \begin{bmatrix} i^{2} + (-i)(-i) & -i^{2} - i^{2} \\ -i^{2} - i^{2} & (-i)(-i) + i^{2} \end{bmatrix} = \begin{bmatrix} -1 - 1 & 1 + 1 \\ 1 + 1 & -1 - 1 \end{bmatrix} = \begin{bmatrix} -2 & 2 \\ 2 & -2 \end{bmatrix} = 2 \begin{bmatrix} -1 & 1 \\ 1 & -1 \end{bmatrix}$.
Next,calculate $A^{4} = (A^{2})^{2}$:
$A^{4} = \left( 2 \begin{bmatrix} -1 & 1 \\ 1 & -1 \end{bmatrix} \right)^{2} = 4 \begin{bmatrix} -1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} -1 & 1 \\ 1 & -1 \end{bmatrix} = 4 \begin{bmatrix} 1+1 & -1-1 \\ -1-1 & 1+1 \end{bmatrix} = 4 \begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix} = 8 \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}$.
Then,calculate $A^{8} = (A^{4})^{2}$:
$A^{8} = \left( 8 \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \right)^{2} = 64 \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} = 64 \begin{bmatrix} 1+1 & -1-1 \\ -1-1 & 1+1 \end{bmatrix} = 64 \begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix} = 128 \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}$.
The system is $128 \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 8 \\ 64 \end{bmatrix}$.
This implies $128(x - y) = 8 \Rightarrow x - y = \frac{8}{128} = \frac{1}{16}$ and $128(-x + y) = 64 \Rightarrow -x + y = \frac{64}{128} = \frac{1}{2}$.
From the first equation,$x - y = \frac{1}{16}$,and from the second,$x - y = -\frac{1}{2}$.
Since $\frac{1}{16} \neq -\frac{1}{2}$,the system has no solution.

(B) Given points $P(3, -1, 2)$ and $Q(1, 2, -4)$.
Direction vectors of lines $PR$ and $QS$ are $\vec{v_1} = 4\hat{i} - \hat{j} + 2\hat{k}$ and $\vec{v_2} = -2\hat{i} + \hat{j} - 2\hat{k}$.
The normal vector $\vec{n}$ to the plane containing $P, T, Q$ is $\vec{v_1} \times \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -1 & 2 \\ -2 & 1 & -2 \end{vmatrix} = \hat{i}(2-2) - \hat{j}(-8+4) + \hat{k}(4-2) = 0\hat{i} + 4\hat{j} + 2\hat{k}$.
Unit normal vector $\hat{n} = \frac{4\hat{j} + 2\hat{k}}{\sqrt{0^2 + 4^2 + 2^2}} = \frac{4\hat{j} + 2\hat{k}}{\sqrt{20}} = \frac{2\hat{j} + \hat{k}}{\sqrt{5}}$.
For intersection point $T$,line $PT: \vec{r} = (3\hat{i} - \hat{j} + 2\hat{k}) + \lambda(4\hat{i} - \hat{j} + 2\hat{k})$ and line $QT: \vec{r} = (\hat{i} + 2\hat{j} - 4\hat{k}) + \mu(-2\hat{i} + \hat{j} - 2\hat{k})$.
Equating coordinates: $3+4\lambda = 1-2\mu \Rightarrow 4\lambda + 2\mu = -2 \Rightarrow 2\lambda + \mu = -1$.
$-1-\lambda = 2+\mu \Rightarrow \lambda + \mu = -3$.
Subtracting equations: $\lambda = 2$,then $\mu = -5$.
Point $T = (3+4(2), -1-(2), 2+2(2)) = (11, -3, 6)$.
Vector $\vec{OA} = \vec{OT} \pm |\vec{TA}|\hat{n} = (11\hat{i} - 3\hat{j} + 6\hat{k}) \pm \sqrt{5} \left(\frac{2\hat{j} + \hat{k}}{\sqrt{5}}\right) = (11\hat{i} - 3\hat{j} + 6\hat{k}) \pm (2\hat{j} + \hat{k})$.
Case $1$: $\vec{OA} = 11\hat{i} - \hat{j} + 7\hat{k} \Rightarrow |\vec{OA}| = \sqrt{121 + 1 + 49} = \sqrt{171}$.
Case $2$: $\vec{OA} = 11\hat{i} - 5\hat{j} + 5\hat{k} \Rightarrow |\vec{OA}| = \sqrt{121 + 25 + 25} = \sqrt{171}$.

(B) First,we find the composite function $(f \circ g)(x) = f(g(x))$.
Given $g(x) < 0$ when $x < 0$ (since $x^3 < 0$ for $x < 0$),and $g(x) \geq 0$ when $x \geq 0$ (since $x^3 \geq 0$ for $x \in [0, 1)$ and $3x-2 \geq 1$ for $x \geq 1$).
Thus,$(f \circ g)(x) = \begin{cases} g(x)+2, & g(x) < 0 \\ (g(x))^2, & g(x) \geq 0 \end{cases} = \begin{cases} x^3+2, & x < 0 \\ (x^3)^2, & 0 \leq x < 1 \\ (3x-2)^2, & x \geq 1 \end{cases} = \begin{cases} x^3+2, & x < 0 \\ x^6, & 0 \leq x < 1 \\ (3x-2)^2, & x \geq 1 \end{cases}$.
Now,check for continuity and differentiability at the transition points $x=0$ and $x=1$.
At $x=0$: $\lim_{x \to 0^-} (x^3+2) = 2$ and $\lim_{x \to 0^+} (x^6) = 0$. Since $2 \neq 0$,the function is discontinuous at $x=0$,hence not differentiable at $x=0$.
At $x=1$: $\lim_{x \to 1^-} (x^6) = 1$ and $\lim_{x \to 1^+} (3x-2)^2 = (3(1)-2)^2 = 1$. The function is continuous at $x=1$.
Check derivatives at $x=1$: $LHD = \frac{d}{dx}(x^6)|_{x=1} = 6(1)^5 = 6$. $RHD = \frac{d}{dx}(3x-2)^2|_{x=1} = 2(3x-2) \cdot 3|_{x=1} = 6(3-2) = 6$.
Since $LHD = RHD$,the function is differentiable at $x=1$.
Therefore,the only point where $(f \circ g)(x)$ is not differentiable is $x=0$. The number of such points is $1$.

(C) The line is given by $\frac{x-1}{-1} = \frac{y+4}{2} = \frac{z+2}{3}$.
Since the plane $lx + my + nz = 0$ contains this line,it passes through the point $(1, -4, -2)$ and its normal vector $(l, m, n)$ is perpendicular to the direction vector $(-1, 2, 3)$ of the line.
Thus,$l(1) + m(-4) + n(-2) = 0 \implies l - 4m - 2n = 0$ (Equation $1$).
And $-l + 2m + 3n = 0$ (Equation $2$).
Adding $(1)$ and $(2)$,we get $-2m + n = 0 \implies n = 2m$.
Substituting $n = 2m$ into $(2)$,we get $-l + 2m + 3(2m) = 0 \implies l = 8m$.
Thus,the plane equation is $8x + y + 2z = 0$.
The point $C$ dividing $AB$ in ratio $k:1$ is $\left(\frac{2k-3}{k+1}, \frac{4k-6}{k+1}, \frac{-3k+1}{k+1}\right)$.
Since $C$ lies on the plane,$8\left(\frac{2k-3}{k+1}\right) + \left(\frac{4k-6}{k+1}\right) + 2\left(\frac{-3k+1}{k+1}\right) = 0$.
$16k - 24 + 4k - 6 - 6k + 2 = 0$.
$14k - 28 = 0 \implies k = 2$.

(B) Given $f(x)=(4 a-3)\left(x+\log _{e} 5\right)+2(a-7) \cot \left(\frac{x}{2}\right) \sin ^{2}\left(\frac{x}{2}\right)$.
Since $\cot \left(\frac{x}{2}\right) \sin ^{2}\left(\frac{x}{2}\right) = \frac{\cos(x/2)}{\sin(x/2)} \cdot \sin^2(x/2) = \sin(x/2)\cos(x/2) = \frac{1}{2} \sin x$,the function simplifies to:
$f(x)=(4 a-3)\left(x+\log _{e} 5\right)+(a-7) \sin x$.
For critical points,we set $f'(x) = 0$:
$f'(x)=(4 a-3)(1)+(a-7) \cos x = 0$.
This implies $\cos x = \frac{3-4 a}{a-7}$.
Since $-1 \leq \cos x \leq 1$ and $x \neq 2n\pi$ (which means $\cos x \neq 1$),we have:
$-1 \leq \frac{3-4 a}{a-7} < 1$.
Case $1$: $\frac{3-4 a}{a-7} \geq -1 \Rightarrow \frac{3-4 a + a - 7}{a-7} \geq 0 \Rightarrow \frac{-3 a - 4}{a-7} \geq 0 \Rightarrow \frac{3 a + 4}{a-7} \leq 0$.
This gives $a \in [-\frac{4}{3}, 7)$.
Case $2$: $\frac{3-4 a}{a-7} < 1 \Rightarrow \frac{3-4 a - a + 7}{a-7} < 0 \Rightarrow \frac{-5 a + 10}{a-7} < 0 \Rightarrow \frac{5(a-2)}{a-7} > 0$.
This gives $a \in (-\infty, 2) \cup (7, \infty)$.
Taking the intersection of Case $1$ and Case $2$,we get $a \in [-\frac{4}{3}, 2)$.
However,checking the boundary $a=2$,$\cos x = \frac{3-8}{2-7} = \frac{-5}{-5} = 1$,which is excluded as $x \neq 2n\pi$. Thus,the range is $[-\frac{4}{3}, 2)$.

(C) Let $E_1$ be the event that the missing card is a spade,and $E_2$ be the event that the missing card is not a spade.
$P(E_1) = \frac{13}{52} = \frac{1}{4}$ and $P(E_2) = \frac{39}{52} = \frac{3}{4}$.
Let $A$ be the event that two cards drawn from the remaining $51$ cards are spades.
If $E_1$ occurs,there are $12$ spades left in $51$ cards. Thus,$P(A|E_1) = \frac{^{12}C_2}{^{51}C_2} = \frac{12 \times 11}{51 \times 50} = \frac{132}{2550}$.
If $E_2$ occurs,there are $13$ spades left in $51$ cards. Thus,$P(A|E_2) = \frac{^{13}C_2}{^{51}C_2} = \frac{13 \times 12}{51 \times 50} = \frac{156}{2550}$.
We need to find $P(E_2|A)$. By Bayes' Theorem:
$P(E_2|A) = \frac{P(E_2)P(A|E_2)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)}$
$P(E_2|A) = \frac{\frac{3}{4} \times \frac{156}{2550}}{\frac{1}{4} \times \frac{132}{2550} + \frac{3}{4} \times \frac{156}{2550}} = \frac{3 \times 156}{132 + 3 \times 156} = \frac{468}{132 + 468} = \frac{468}{600} = \frac{39}{50}$.

(D) The given differential equation is $\frac{dy}{dx} + 2y \tan x = \sin x$. This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = 2 \tan x$ and $Q = \sin x$.
The integrating factor $(I.F.)$ is given by $I.F. = e^{\int P dx} = e^{\int 2 \tan x dx} = e^{2 \ln |\sec x|} = \sec^2 x$.
The general solution is $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + C$.
$y \sec^2 x = \int \sin x \cdot \sec^2 x dx = \int \sec x \tan x dx = \sec x + C$.
Given $y(\frac{\pi}{3}) = 0$,we substitute $x = \frac{\pi}{3}$ and $y = 0$:
$0 \cdot \sec^2(\frac{\pi}{3}) = \sec(\frac{\pi}{3}) + C \implies 0 = 2 + C \implies C = -2$.
Thus,$y \sec^2 x = \sec x - 2$,which simplifies to $y = \frac{\sec x - 2}{\sec^2 x} = \cos x - 2 \cos^2 x$.
Let $t = \cos x$. Since $-1 \le \cos x \le 1$,we have $y = t - 2t^2$. To find the maximum,we differentiate with respect to $t$:
$\frac{dy}{dt} = 1 - 4t = 0 \implies t = \frac{1}{4}$.
Since $t = \frac{1}{4}$ is within the range $[-1, 1]$,the maximum value is $y = \frac{1}{4} - 2(\frac{1}{4})^2 = \frac{1}{4} - \frac{2}{16} = \frac{1}{4} - \frac{1}{8} = \frac{1}{8}$.

(C) We have $S_{k} = \sum_{r=1}^{k} \tan^{-1}\left(\frac{6^{r}}{2 \cdot 2^{2r} + 3 \cdot 3^{2r}}\right)$.
Dividing the numerator and denominator by $3^{2r}$,we get:
$S_{k} = \sum_{r=1}^{k} \tan^{-1}\left(\frac{\frac{2^r}{3^r} \cdot 3^r}{2 \cdot \frac{2^{2r}}{3^{2r}} + 3}\right) = \sum_{r=1}^{k} \tan^{-1}\left(\frac{3^r \cdot (2/3)^r}{2 \cdot (2/3)^{2r} + 3}\right)$.
Actually,rewrite the term inside $\tan^{-1}$ as:
$\frac{6^r}{2 \cdot 4^r + 3 \cdot 9^r} = \frac{6^r / 9^r}{2 \cdot (4/9)^r + 3} = \frac{(2/3)^r}{2 \cdot (2/3)^{2r} + 3}$.
This can be written as $\frac{(2/3)^r / 3}{1 + \frac{2}{3} (2/3)^{2r}} = \frac{(2/3)^r - (2/3)^{r+1}}{1 + (2/3)^r \cdot (2/3)^{r+1}}$.
Thus,the sum is a telescoping series:
$S_{k} = \sum_{r=1}^{k} \left( \tan^{-1}((2/3)^r) - \tan^{-1}((2/3)^{r+1}) \right)$.
$S_{k} = \left( \tan^{-1}(2/3) - \tan^{-1}((2/3)^2) \right) + \left( \tan^{-1}((2/3)^2) - \tan^{-1}((2/3)^3) \right) + \dots + \left( \tan^{-1}((2/3)^k) - \tan^{-1}((2/3)^{k+1}) \right)$.
$S_{k} = \tan^{-1}(2/3) - \tan^{-1}((2/3)^{k+1})$.
Taking the limit as $k \rightarrow \infty$,$(2/3)^{k+1} \rightarrow 0$.
$S_{\infty} = \tan^{-1}(2/3) - \tan^{-1}(0) = \tan^{-1}(2/3) = \cot^{-1}(3/2)$.

(B) The given limit is $E = 2 \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} f\left(\frac{r}{n}\right)$.
This is a Riemann sum,which can be written as the definite integral $E = 2 \int_{0}^{1} f(x) dx$.
Substituting $f(x) = \log_{2}\left(1+\tan\left(\frac{\pi x}{4}\right)\right) = \frac{\ln(1+\tan(\pi x/4))}{\ln 2}$,we get:
$E = \frac{2}{\ln 2} \int_{0}^{1} \ln\left(1+\tan\frac{\pi x}{4}\right) dx \quad \dots(i)$
Using the property $\int_{0}^{a} g(x) dx = \int_{0}^{a} g(a-x) dx$,we replace $x$ with $1-x$:
$E = \frac{2}{\ln 2} \int_{0}^{1} \ln\left(1+\tan\left(\frac{\pi}{4}(1-x)\right)\right) dx$
$E = \frac{2}{\ln 2} \int_{0}^{1} \ln\left(1+\frac{1-\tan(\pi x/4)}{1+\tan(\pi x/4)}\right) dx$
$E = \frac{2}{\ln 2} \int_{0}^{1} \ln\left(\frac{2}{1+\tan(\pi x/4)}\right) dx$
$E = \frac{2}{\ln 2} \int_{0}^{1} (\ln 2 - \ln(1+\tan(\pi x/4))) dx \quad \dots(ii)$
Adding $(i)$ and $(ii)$:
$2E = \frac{2}{\ln 2} \int_{0}^{1} \ln 2 dx = \frac{2}{\ln 2} \cdot \ln 2 = 2$
Therefore,$E = 1$.

(D) Let $A = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}$.
The diagonal entries of $AA^{T}$ are given by the sum of squares of the elements in each row of $A$.
Specifically,the diagonal entries are $(a^{2}+b^{2}+c^{2})$,$(d^{2}+e^{2}+f^{2})$,and $(g^{2}+h^{2}+i^{2})$.
The sum of all diagonal entries is $a^{2}+b^{2}+c^{2}+d^{2}+e^{2}+f^{2}+g^{2}+h^{2}+i^{2} = 9$,where $a, b, c, d, e, f, g, h, i \in \{0, 1, 2, 3\}$.
We need to find the number of ways to choose these $9$ elements such that their squares sum to $9$.

Case	Description	Number of Matrices
$1$. All $1$s	$9$ ones,$0$ zeros	$\frac{9!}{9!} = 1$
$2$. One $3$	One $3$,$8$ zeros	$\frac{9!}{1! \times 8!} = 9$
$3$. One $2$,one $1$	One $2$,one $1$,$7$ zeros	$\frac{9!}{1! \times 1! \times 7!} = 72$
$4$. Two $2$s,one $1$	Two $2$s,one $1$,$6$ zeros	$\frac{9!}{2! \times 1! \times 6!} = 252$
$5$. Three $1$s,two $2$s	Wait,let us re-evaluate combinations: $x_1^2 + ... + x_9^2 = 9$. Possible sets of squares: $\{1,1,1,1,1,1,1,1,1\}$,$\{9,0,0,0,0,0,0,0,0\}$,$\{4,4,1,0,0,0,0,0,0\}$,$\{4,1,1,1,1,1,0,0,0\}$.

Recalculating based on partitions of $9$ into $9$ squares from $\{0, 1, 4, 9\}$:
$1$. Nine $1$s: $\frac{9!}{9!} = 1$
$2$. One $3$ $(3^2=9)$: $\frac{9!}{1!8!} = 9$
$3$. One $2$ $(2^2=4)$ and five $1$s $(1^2=1)$: $\frac{9!}{1!5!3!} = 504$
$4$. Two $2$s $(2^2=4)$ and one $1$ $(1^2=1)$: $\frac{9!}{2!1!6!} = 252$
Total $= 1 + 9 + 504 + 252 = 766$.

(C) Let $M = (P^{-1}AP - I)^{2}$.
We know that $\det(P^{-1}AP - I) = \det(P^{-1}(A - I)P) = \det(P^{-1}) \det(A - I) \det(P) = \det(A - I)$.
Therefore,$\det(M) = (\det(A - I))^{2}$.
Now,calculate $A - I$:
$A - I = \begin{bmatrix} 2-1 & 7 & \omega^{2} \\ -1 & -\omega-1 & 1 \\ 0 & -\omega & -\omega+1-1 \end{bmatrix} = \begin{bmatrix} 1 & 7 & \omega^{2} \\ -1 & -\omega-1 & 1 \\ 0 & -\omega & -\omega \end{bmatrix}$.
Using the property $1 + \omega + \omega^{2} = 0$,we have $-\omega - 1 = \omega^{2}$.
$A - I = \begin{bmatrix} 1 & 7 & \omega^{2} \\ -1 & \omega^{2} & 1 \\ 0 & -\omega & -\omega \end{bmatrix}$.
Calculating the determinant:
$\det(A - I) = 1(\omega^{2}(-\omega) - (1)(-\omega)) - 7((-1)(-\omega) - (1)(0)) + \omega^{2}((-1)(-\omega) - (\omega^{2})(0))$
$= 1(-\omega^{3} + \omega) - 7(\omega) + \omega^{2}(\omega)$
$= -1 + \omega - 7\omega + \omega^{3} = -1 - 6\omega + 1 = -6\omega$.
Thus,$\det(M) = (-6\omega)^{2} = 36\omega^{2}$.
Given $\det(M) = \alpha \omega^{2}$,we get $\alpha = 36$.

(B) Given the curve $y(x)=\int_{0}^{x}(2t^{2}-15t+10)dt$.
By the Fundamental Theorem of Calculus,the slope of the tangent at $x=a$ is $y'(a) = 2a^{2}-15a+10$.
The slope of the normal at $(a, b)$ is $-\frac{1}{y'(a)} = -\frac{1}{2a^{2}-15a+10}$.
The given line is $x+3y=-5$,which can be written as $y=-\frac{1}{3}x-\frac{5}{3}$. The slope of this line is $-\frac{1}{3}$.
Since the normal is parallel to the line,their slopes are equal:
$-\frac{1}{2a^{2}-15a+10} = -\frac{1}{3} \Rightarrow 2a^{2}-15a+10 = 3$.
$2a^{2}-15a+7 = 0 \Rightarrow (2a-1)(a-7) = 0$.
Since $a>1$,we have $a=7$.
Now,calculate $b = y(7) = \int_{0}^{7}(2t^{2}-15t+10)dt = [\frac{2}{3}t^{3} - \frac{15}{2}t^{2} + 10t]_{0}^{7}$.
$b = \frac{2}{3}(343) - \frac{15}{2}(49) + 10(7) = \frac{686}{3} - \frac{735}{2} + 70 = \frac{1372 - 2205 + 420}{6} = -\frac{413}{6}$.
Thus,$6b = -413$.
$|a+6b| = |7 - 413| = |-406| = 406$.

(A) Given the differential equation $\frac{dy}{dx} = 2(x + 1)$.
Integrating both sides with respect to $x$,we get:
$y(x) = \int 2(x + 1) dx = (x + 1)^2 + C = x^2 + 2x + 1 + C$.
Let $K = 1 + C$,so $y(x) = (x + 1)^2 + C$.
The curve is a parabola opening upwards with vertex at $(-1, C)$. For the curve to bound an area with the $x-$axis,it must lie below the $x-$axis,so $C < 0$. Let $C = -k^2$ where $k > 0$.
The roots of $y(x) = 0$ are $(x + 1)^2 = -C$,so $x = -1 \pm \sqrt{-C}$.
The area bounded by the curve and the $x-$axis is given by:
$A = \int_{-1-\sqrt{-C}}^{-1+\sqrt{-C}} (0 - ((x + 1)^2 + C)) dx = -\int_{-1-\sqrt{-C}}^{-1+\sqrt{-C}} ((x + 1)^2 + C) dx$.
Let $u = x + 1$,then $du = dx$. The limits change to $-\sqrt{-C}$ to $\sqrt{-C}$.
$A = -\int_{-\sqrt{-C}}^{\sqrt{-C}} (u^2 + C) du = -[\frac{u^3}{3} + Cu]_{-\sqrt{-C}}^{\sqrt{-C}} = -[(\frac{(-C)^{3/2}}{3} + C\sqrt{-C}) - (\frac{-(-C)^{3/2}}{3} - C\sqrt{-C})]$.
$A = -[\frac{2}{3}(-C)^{3/2} + 2C\sqrt{-C}] = -[\frac{2}{3}(-C)\sqrt{-C} - 2(-C)\sqrt{-C}] = -[-\frac{4}{3}(-C)^{3/2}] = \frac{4}{3}(-C)^{3/2}$.
Given $A = \frac{4\sqrt{8}}{3} = \frac{4(2\sqrt{2})}{3} = \frac{8\sqrt{2}}{3}$.
So,$\frac{4}{3}(-C)^{3/2} = \frac{8\sqrt{2}}{3} \Rightarrow (-C)^{3/2} = 2\sqrt{2} = (\sqrt{2})^3$.
Thus,$-C = 2$,which means $C = -2$.
The function is $y(x) = (x + 1)^2 - 2 = x^2 + 2x - 1$.
Therefore,$y(1) = (1)^2 + 2(1) - 1 = 1 + 2 - 1 = 2$.

(B) Given $f(x)+f(x+1)=2$.
Integrating both sides from $0$ to $1$,we get $\int_{0}^{1} f(x) d x + \int_{0}^{1} f(x+1) d x = \int_{0}^{1} 2 d x$.
Using the substitution $t=x+1$ in the second integral,we have $\int_{0}^{1} f(x) d x + \int_{1}^{2} f(t) d t = 2$,which implies $\int_{0}^{2} f(x) d x = 2$.
Since $f(x)+f(x+1)=2$,$f(x+2) = 2-f(x+1) = 2-(2-f(x)) = f(x)$,so $f(x)$ is periodic with period $T=2$.
$I_{1} = \int_{0}^{8} f(x) d x = \frac{8}{2} \int_{0}^{2} f(x) d x = 4 \times 2 = 8$.
$I_{2} = \int_{-1}^{3} f(x) d x$. Since $f(x)$ has period $2$,$\int_{a}^{a+T} f(x) d x$ is independent of $a$.
Thus,$I_{2} = \int_{0}^{4} f(x) d x = \frac{4}{2} \int_{0}^{2} f(x) d x = 2 \times 2 = 4$.
Therefore,$I_{1}+2 I_{2} = 8 + 2(4) = 8 + 8 = 16$.

(B) Given the function $y = 5^{\log x}$.
To find the inverse,we interchange $x$ and $y$:
$x = 5^{\log y}$.
Using the property of logarithms $a^{\log b} = b^{\log a}$,we can rewrite the expression:
$x = y^{\log 5}$.
Thus,the inverse function is $f^{-1}(y) = y^{\log 5}$.

(A) Given $\overrightarrow{r} \times \overrightarrow{a} = \overrightarrow{r} \times \overrightarrow{b}$,we have $\overrightarrow{r} \times (\overrightarrow{a} - \overrightarrow{b}) = 0$.
This implies $\overrightarrow{r}$ is parallel to $(\overrightarrow{a} - \overrightarrow{b})$.
Calculate $\overrightarrow{a} - \overrightarrow{b} = (2-7)\hat{i} + (-3-1)\hat{j} + (4 - (-6))\hat{k} = -5\hat{i} - 4\hat{j} + 10\hat{k}$.
So,$\overrightarrow{r} = \lambda(-5\hat{i} - 4\hat{j} + 10\hat{k})$.
Given $\overrightarrow{r} \cdot (\hat{i} + 2\hat{j} + \hat{k}) = -3$,substitute $\overrightarrow{r}$:
$\lambda(-5\hat{i} - 4\hat{j} + 10\hat{k}) \cdot (\hat{i} + 2\hat{j} + \hat{k}) = -3$.
$\lambda(-5 - 8 + 10) = -3 \Rightarrow -3\lambda = -3 \Rightarrow \lambda = 1$.
Thus,$\overrightarrow{r} = -5\hat{i} - 4\hat{j} + 10\hat{k}$.
Finally,calculate $\overrightarrow{r} \cdot (2\hat{i} - 3\hat{j} + \hat{k}) = (-5)(2) + (-4)(-3) + (10)(1) = -10 + 12 + 10 = 12$.

(D) The given system of equations is:
$kx + y + z = 1$
$x + ky + z = k$
$x + y + kz = k^2$
First,we calculate the determinant $\Delta$ of the coefficient matrix:
$\Delta = \begin{vmatrix} k & 1 & 1 \\ 1 & k & 1 \\ 1 & 1 & k \end{vmatrix}$
$= k(k^2 - 1) - 1(k - 1) + 1(1 - k)$
$= k(k - 1)(k + 1) - (k - 1) - (k - 1)$
$= (k - 1)[k(k + 1) - 1 - 1]$
$= (k - 1)(k^2 + k - 2)$
$= (k - 1)(k + 2)(k - 1) = (k - 1)^2(k + 2)$
For the system to have no solution,we require $\Delta = 0$ and at least one of the Cramer's determinants $(\Delta_1, \Delta_2, \Delta_3)$ to be non-zero.
If $k = 1$,the equations become $x + y + z = 1$,$x + y + z = 1$,and $x + y + z = 1$. This represents the same plane,so there are infinitely many solutions.
If $k = -2$,$\Delta = 0$. Let us check $\Delta_1$:
$\Delta_1 = \begin{vmatrix} 1 & 1 & 1 \\ -2 & -2 & 1 \\ 4 & 1 & -2 \end{vmatrix} = 1(4 - 1) - 1(4 - 4) + 1(-2 + 8) = 3 - 0 + 6 = 9 \neq 0$.
Since $\Delta = 0$ and $\Delta_1 \neq 0$,the system has no solution when $k = -2$.

(A) The given series is $\sum_{n=1}^{100} \cot^{-1}(2n^2)$.
Using the identity $\cot^{-1}(x) = \tan^{-1}(\frac{1}{x})$,we have $\sum_{n=1}^{100} \tan^{-1}(\frac{1}{2n^2})$.
Multiplying numerator and denominator by $2$,we get $\sum_{n=1}^{100} \tan^{-1}(\frac{2}{4n^2})$.
This can be written as $\sum_{n=1}^{100} \tan^{-1}(\frac{(2n+1)-(2n-1)}{1+(2n+1)(2n-1)})$.
Using the formula $\tan^{-1}(x) - \tan^{-1}(y) = \tan^{-1}(\frac{x-y}{1+xy})$,the sum becomes $\sum_{n=1}^{100} (\tan^{-1}(2n+1) - \tan^{-1}(2n-1))$.
This is a telescoping series: $(\tan^{-1} 3 - \tan^{-1} 1) + (\tan^{-1} 5 - \tan^{-1} 3) + \dots + (\tan^{-1} 201 - \tan^{-1} 199)$.
After cancellation,we are left with $\tan^{-1} 201 - \tan^{-1} 1$.
Using the formula again,$\tan^{-1}(\frac{201-1}{1+201 \times 1}) = \tan^{-1}(\frac{200}{202}) = \tan^{-1}(\frac{100}{101})$.
Since $\cot^{-1}(\alpha) = \tan^{-1}(\frac{101}{100})$,we have $\alpha = \frac{101}{100} = 1.01$.

(D) The equation of a plane passing through the origin $(0, 0, 0)$ and containing the $y$-axis (which has direction vector $\hat{j}$) and passing through the point $P(1, 2, 3)$ can be found using the normal vector $\vec{n}$.
The plane contains the vector $\vec{v} = \hat{j} = (0, 1, 0)$ and the vector $\vec{OP} = \hat{i} + 2\hat{j} + 3\hat{k} = (1, 2, 3)$.
The normal vector $\vec{n}$ is given by the cross product:
$\vec{n} = \vec{v} \times \vec{OP} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 0 \\ 1 & 2 & 3 \end{vmatrix}$
$\vec{n} = \hat{i}(3 - 0) - \hat{j}(0 - 0) + \hat{k}(0 - 1) = 3\hat{i} - \hat{k}$.
The equation of the plane passing through $(0, 0, 0)$ with normal $\vec{n} = (3, 0, -1)$ is:
$3(x - 0) + 0(y - 0) - 1(z - 0) = 0$
$3x - z = 0$.

(C) Given $A = \begin{bmatrix} 0 & \sin \alpha \\ \sin \alpha & 0 \end{bmatrix}$.
First,calculate $A^{2}$:
$A^{2} = \begin{bmatrix} 0 & \sin \alpha \\ \sin \alpha & 0 \end{bmatrix} \begin{bmatrix} 0 & \sin \alpha \\ \sin \alpha & 0 \end{bmatrix} = \begin{bmatrix} \sin^{2} \alpha & 0 \\ 0 & \sin^{2} \alpha \end{bmatrix} = \sin^{2} \alpha I$.
Now,substitute this into the determinant equation:
$\det\left(A^{2} - \frac{1}{2} I\right) = \det\left(\sin^{2} \alpha I - \frac{1}{2} I\right) = \det\left(\left(\sin^{2} \alpha - \frac{1}{2}\right) I\right) = 0$.
Since $I$ is a $2 \times 2$ identity matrix,$\det(kI) = k^{2} \det(I) = k^{2}$.
Thus,$\left(\sin^{2} \alpha - \frac{1}{2}\right)^{2} = 0$.
This implies $\sin^{2} \alpha = \frac{1}{2}$,so $\sin \alpha = \pm \frac{1}{\sqrt{2}}$.
For $\sin \alpha = \frac{1}{\sqrt{2}}$,a possible value is $\alpha = \frac{\pi}{4}$.

(B) Given equation: $\tan ^{-1}(x+1)+\cot ^{-1}\left(\frac{1}{x-1}\right)=\tan ^{-1}\left(\frac{8}{31}\right)$.
Using the property $\cot ^{-1}(u) = \tan ^{-1}\left(\frac{1}{u}\right)$ for $u > 0$,we have $\cot ^{-1}\left(\frac{1}{x-1}\right) = \tan ^{-1}(x-1)$ if $x-1 > 0$,i.e.,$x > 1$.
If $x > 1$,then $\tan ^{-1}(x+1) + \tan ^{-1}(x-1) = \tan ^{-1}\left(\frac{8}{31}\right)$.
Applying $\tan(A+B)$ formula: $\frac{(x+1)+(x-1)}{1-(x+1)(x-1)} = \frac{8}{31} \Rightarrow \frac{2x}{1-(x^2-1)} = \frac{8}{31} \Rightarrow \frac{2x}{2-x^2} = \frac{8}{31}$.
$62x = 16 - 8x^2 \Rightarrow 8x^2 + 62x - 16 = 0 \Rightarrow 4x^2 + 31x - 8 = 0$.
Solving for $x$: $(4x-1)(x+8) = 0$,so $x = \frac{1}{4}$ or $x = -8$. Both values are $\leq 1$,so the assumption $x > 1$ is invalid.
If $x < 1$,then $\cot ^{-1}\left(\frac{1}{x-1}\right) = \pi + \tan ^{-1}(x-1)$.
Then $\tan ^{-1}(x+1) + \pi + \tan ^{-1}(x-1) = \tan ^{-1}\left(\frac{8}{31}\right)$.
$\tan ^{-1}\left(\frac{2x}{2-x^2}\right) = \tan ^{-1}\left(\frac{8}{31}\right) - \pi$.
Taking $\tan$ on both sides: $\frac{2x}{2-x^2} = \tan(\tan ^{-1}(\frac{8}{31}) - \pi) = \frac{8}{31}$.
This leads to $4x^2 + 31x - 8 = 0$,giving $x = \frac{1}{4}$ or $x = -8$. Since $x < 1$,both are candidates. Checking $x = \frac{1}{4}$: $\tan^{-1}(1.25) + \cot^{-1}(1/(-0.75)) = \tan^{-1}(1.25) + \cot^{-1}(-1.33) \approx 0.89 + 2.21 > \pi/2$. Thus $x = 1/4$ is rejected.
For $x = -8$: $\tan^{-1}(-7) + \cot^{-1}(-1/9) = -1.42 + 1.68 = 0.26 = \tan^{-1}(8/31)$.
Thus,the only solution is $x = -8$.

(A) Let $I = \int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$.
Here,$a = \frac{\pi}{6}$ and $b = \frac{\pi}{3}$,so $a+b = \frac{\pi}{2}$.
$g(\alpha) = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin^{\alpha} x}{\sin^{\alpha} x + \cos^{\alpha} x} dx$ ... $(i)$
Using the property,$g(\alpha) = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin^{\alpha}(\frac{\pi}{2}-x)}{\sin^{\alpha}(\frac{\pi}{2}-x) + \cos^{\alpha}(\frac{\pi}{2}-x)} dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\cos^{\alpha} x}{\cos^{\alpha} x + \sin^{\alpha} x} dx$ ... $(ii)$
Adding $(i)$ and $(ii)$:
$2g(\alpha) = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin^{\alpha} x + \cos^{\alpha} x}{\sin^{\alpha} x + \cos^{\alpha} x} dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 1 dx = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$.
Thus,$g(\alpha) = \frac{\pi}{12}$,which is a constant function.
$A$ constant function is neither strictly increasing nor strictly decreasing.
Therefore,statements $A$,$B$,and $C$ are all incorrect.

(A) Given the differential equation: $\frac{dy}{dx} = xy - 1 + x - y$.
Rearranging the terms: $\frac{dy}{dx} = x(y + 1) - 1(y + 1) = (x - 1)(y + 1)$.
This is a variable separable differential equation: $\frac{dy}{y + 1} = (x - 1) dx$.
Integrating both sides: $\int \frac{dy}{y + 1} = \int (x - 1) dx$.
$\ln|y + 1| = \frac{x^2}{2} - x + C$.
Using the initial condition $y(0) = 0$: $\ln|0 + 1| = \frac{0^2}{2} - 0 + C \Rightarrow \ln(1) = C \Rightarrow C = 0$.
Thus,$\ln|y + 1| = \frac{x^2}{2} - x$,which implies $y + 1 = e^{\frac{x^2}{2} - x}$.
So,$y(x) = e^{\frac{x^2}{2} - x} - 1$.
To find $y(1)$,substitute $x = 1$: $y(1) = e^{\frac{1^2}{2} - 1} - 1 = e^{\frac{1}{2} - 1} - 1 = e^{-\frac{1}{2}} - 1$.

(A) The objective function is $z = 6xy + y^2$. The constraints are $3x + 4y \leq 100$,$4x + 3y \leq 75$,$x \geq 0$,and $y \geq 0$.
From the constraints,the feasible region is a polygon with vertices $(0, 0)$,$(18.75, 0)$,and $(0, 25)$.
Evaluating $z$ at the vertices:
At $(0, 0)$,$z = 6(0)(0) + 0^2 = 0$.
At $(18.75, 0)$,$z = 6(18.75)(0) + 0^2 = 0$.
At $(0, 25)$,$z = 6(0)(25) + 25^2 = 625$.
However,since $z$ is not a linear function,we check the boundary $4x + 3y = 75$,which implies $x = \frac{75-3y}{4}$.
Substituting this into $z$: $z = 6y(\frac{75-3y}{4}) + y^2 = \frac{3y(75-3y)}{2} + y^2 = \frac{225y - 9y^2 + 2y^2}{2} = \frac{225y - 7y^2}{2}$.
To find the maximum,take the derivative with respect to $y$ and set to $0$: $\frac{dz}{dy} = \frac{225 - 14y}{2} = 0 \implies y = \frac{225}{14} \approx 16.07$.
Then $x = \frac{75 - 3(225/14)}{4} = \frac{1050 - 675}{56} = \frac{375}{56} \approx 6.7$.
Substituting these values into $z$: $z = \frac{225(225/14) - 7(225/14)^2}{2} = \frac{50625/14 - 354375/196}{2} = \frac{708750 - 354375}{392} = \frac{354375}{392} \approx 904.0178$.
The maximum value is approximately $904$.

(A) For the function to be continuous at $x = 0$,we must have $f(0) = \lim_{x \to 0} f(x)$.
Using the formula $\cos A - \cos B = -2 \sin(\frac{A+B}{2}) \sin(\frac{A-B}{2})$,we get:
$\lim_{x \to 0} \frac{-2 \sin(\frac{\sin x + x}{2}) \sin(\frac{\sin x - x}{2})}{x^4} = \frac{1}{k}$
$\lim_{x \to 0} \frac{2 \sin(\frac{\sin x + x}{2}) \sin(\frac{x - \sin x}{2})}{x^4} = \frac{1}{k}$
Using $\sin \theta \approx \theta$ as $\theta \to 0$ and the Taylor series expansion $\sin x = x - \frac{x^3}{6} + \dots$,we have:
$\lim_{x \to 0} 2 \left( \frac{\sin x + x}{2x} \right) \left( \frac{x - \sin x}{2x^3} \right) = \frac{1}{k}$
$\lim_{x \to 0} 2 \left( \frac{x + x}{2x} \right) \left( \frac{x - (x - \frac{x^3}{6})}{2x^3} \right) = \frac{1}{k}$
$2 \times (1) \times \frac{x^3/6}{2x^3} = \frac{1}{k}$
$2 \times 1 \times \frac{1}{12} = \frac{1}{k}$
$\frac{1}{6} = \frac{1}{k} \Rightarrow k = 6$.

(B) Given $f(x) = \sin \left(\cos ^{-1}\left(\frac{1-2^{2x}}{1+2^{2x}}\right)\right)$.
Let $2^x = t$. Then $f(x) = \sin(\cos^{-1}(\frac{1-t^2}{1+t^2}))$.
We know that $\cos^{-1}(\frac{1-t^2}{1+t^2}) = 2\tan^{-1}(t)$ for $t \ge 0$.
So,$f(x) = \sin(2\tan^{-1}(2^x))$.
Using the identity $\sin(2\theta) = \frac{2\tan\theta}{1+\tan^2\theta}$,where $\theta = \tan^{-1}(2^x)$,we get $\tan\theta = 2^x$.
Thus,$f(x) = \frac{2(2^x)}{1+(2^x)^2} = \frac{2 \cdot 2^x}{1+2^{2x}}$.
Now,differentiate $f(x)$ with respect to $x$ using the quotient rule:
$f'(x) = \frac{(1+2^{2x}) \cdot \frac{d}{dx}(2 \cdot 2^x) - (2 \cdot 2^x) \cdot \frac{d}{dx}(1+2^{2x})}{(1+2^{2x})^2}$.
$f'(x) = \frac{(1+2^{2x})(2 \cdot 2^x \ln 2) - (2 \cdot 2^x)(2^{2x} \ln 2 \cdot 2)}{(1+2^{2x})^2}$.
At $x=1$,$2^x = 2$ and $2^{2x} = 4$.
$f'(1) = \frac{(1+4)(2 \cdot 2 \ln 2) - (2 \cdot 2)(4 \ln 2 \cdot 2)}{(1+4)^2} = \frac{5(4 \ln 2) - 4(8 \ln 2)}{25} = \frac{20 \ln 2 - 32 \ln 2}{25} = -\frac{12}{25} \ln 2$.
Comparing with $-\frac{b}{a} \ln 2$,we get $b=12$ and $a=25$.
Therefore,$|a^2 - b^2| = |25^2 - 12^2| = |625 - 144| = 481$.

(B) Let $P(E_{1}) = P_{1}$,$P(E_{2}) = P_{2}$,and $P(E_{3}) = P_{3}$.
Since the events are independent,the probability that only $E_{1}$ occurs is $\alpha = P_{1}(1 - P_{2})(1 - P_{3})$.
The probability that only $E_{2}$ occurs is $\beta = (1 - P_{1})P_{2}(1 - P_{3})$.
The probability that only $E_{3}$ occurs is $\gamma = (1 - P_{1})(1 - P_{2})P_{3}$.
The probability that none of the events occur is $p = (1 - P_{1})(1 - P_{2})(1 - P_{3})$.
Given $(\alpha - 2\beta)p = \alpha\beta$,we substitute the expressions:
$\left(P_{1}(1 - P_{2})(1 - P_{3}) - 2(1 - P_{1})P_{2}(1 - P_{3})\right)p = P_{1}(1 - P_{2})(1 - P_{3}) \cdot (1 - P_{1})P_{2}(1 - P_{3})$.
Dividing both sides by $(1 - P_{1})(1 - P_{2})(1 - P_{3})^2$,we get:
$\frac{P_{1}}{1 - P_{1}} - \frac{2P_{2}}{1 - P_{2}} = \frac{P_{1}P_{2}}{(1 - P_{1})(1 - P_{2})}$.
This simplifies to $P_{1} = 2P_{2}$.
Similarly,from $(\beta - 3\gamma)p = 2\beta\gamma$,we get $P_{2} = 3P_{3}$.
Therefore,$P_{1} = 2(3P_{3}) = 6P_{3}$.
Thus,$\frac{P_{1}}{P_{3}} = 6$.

(C) Given $\vec{a} \cdot \vec{b} = 1 \Rightarrow (\alpha \hat{i} + \beta \hat{j} + 3 \hat{k}) \cdot (-\beta \hat{i} - \alpha \hat{j} - \hat{k}) = 1$.
This simplifies to $-\alpha \beta - \alpha \beta - 3 = 1 \Rightarrow -2 \alpha \beta = 4 \Rightarrow \alpha \beta = -2$ $(1)$.
Given $\vec{b} \cdot \vec{c} = -3 \Rightarrow (-\beta \hat{i} - \alpha \hat{j} - \hat{k}) \cdot (\hat{i} - 2 \hat{j} - \hat{k}) = -3$.
This simplifies to $-\beta + 2 \alpha + 1 = -3 \Rightarrow 2 \alpha - \beta = -4$ $(2)$.
From $(1)$,$\beta = -2/\alpha$. Substituting into $(2)$: $2 \alpha - (-2/\alpha) = -4 \Rightarrow 2 \alpha^2 + 2 = -4 \alpha \Rightarrow \alpha^2 + 2 \alpha + 1 = 0 \Rightarrow (\alpha + 1)^2 = 0 \Rightarrow \alpha = -1$.
Then $\beta = -2/(-1) = 2$.
Now,the scalar triple product is $\frac{1}{3}[\vec{a} \vec{b} \vec{c}] = \frac{1}{3} \begin{vmatrix} \alpha & \beta & 3 \\ -\beta & -\alpha & -1 \\ 1 & -2 & -1 \end{vmatrix} = \frac{1}{3} \begin{vmatrix} -1 & 2 & 3 \\ -2 & 1 & -1 \\ 1 & -2 & -1 \end{vmatrix}$.
Performing row operations $R_1 \to R_1 + R_3$: $\frac{1}{3} \begin{vmatrix} 0 & 0 & 2 \\ -2 & 1 & -1 \\ 1 & -2 & -1 \end{vmatrix} = \frac{1}{3} [2(4 - 1)] = \frac{1}{3} \times 6 = 2$.

(C) Given $A = \begin{bmatrix} 2 & 3 \\ 0 & -1 \end{bmatrix}$. The determinant $|A| = (2)(-1) - (3)(0) = -2$.
First,we calculate $\det(A^4) = |A|^4 = (-2)^4 = 16$.
Next,consider $\operatorname{adj}(2A)$. Since $2A = \begin{bmatrix} 4 & 6 \\ 0 & -2 \end{bmatrix}$,its determinant is $|2A| = 2^2 |A| = 4(-2) = -8$.
The adjoint of a $2 \times 2$ matrix $M = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is $\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.
Thus,$\operatorname{adj}(2A) = \begin{bmatrix} -2 & -6 \\ 0 & 4 \end{bmatrix}$.
Note that $\operatorname{adj}(2A) = -2 \begin{bmatrix} 1 & 3 \\ 0 & -2 \end{bmatrix}$.
Alternatively,using the property $\operatorname{adj}(kA) = k^{n-1} \operatorname{adj}(A)$,for $n=2$,$\operatorname{adj}(2A) = 2 \operatorname{adj}(A) = 2 \begin{bmatrix} -1 & -3 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} -2 & -6 \\ 0 & 4 \end{bmatrix}$.
The eigenvalues of $A$ are $\lambda_1 = 2$ and $\lambda_2 = -1$. The eigenvalues of $A^{10}$ are $2^{10}$ and $(-1)^{10} = 1$.
The eigenvalues of $\operatorname{adj}(2A)$ are $2 \times (-1) = -2$ and $2 \times 2 = 4$. Thus,the eigenvalues of $(\operatorname{adj}(2A))^{10}$ are $(-2)^{10} = 2^{10}$ and $4^{10} = 2^{20}$.
Since $A^{10}$ and $(\operatorname{adj}(2A))^{10}$ share the eigenvalue $2^{10}$,the matrix $A^{10} - (\operatorname{adj}(2A))^{10}$ has at least one eigenvalue equal to $2^{10} - 2^{10} = 0$.
Therefore,$\det(A^{10} - (\operatorname{adj}(2A))^{10}) = 0$.
Finally,the required value is $16 + 0 = 16$.

(C) Let $I = \int_{0}^{\sqrt{\pi / 2}} ([x^2] + [-\cos x]) dx$.
Since $0 \le x \le \sqrt{\pi / 2} \approx 1.25$,we have $0 \le x^2 \le \pi / 2 \approx 1.57$.
Thus,$[x^2] = 0$ for $0 \le x < 1$ and $[x^2] = 1$ for $1 \le x \le \sqrt{\pi / 2}$.
For $0 \le x \le \sqrt{\pi / 2}$,we have $0 \le \cos x \le 1$.
Specifically,for $x \in (0, \sqrt{\pi / 2}]$,$0 \le \cos x < 1$,so $-\cos x \in [-1, 0)$.
Thus,$[-\cos x] = -1$ for $x \in (0, \sqrt{\pi / 2}]$ and $[-\cos x] = 0$ at $x = 0$.
$I = \int_{0}^{1} (0 - 1) dx + \int_{1}^{\sqrt{\pi / 2}} (1 - 1) dx = \int_{0}^{1} (-1) dx + 0 = -1$.

(C) The equation of a plane passing through the intersection of two planes $P_1 = 0$ and $P_2 = 0$ is given by $P_1 + \lambda P_2 = 0$.
Here,$(2x - 7y + 4z - 3) + \lambda(3x - 5y + 4z + 11) = 0$.
Grouping the terms,we get $(2 + 3\lambda)x - (7 + 5\lambda)y + (4 + 4\lambda)z + (-3 + 11\lambda) = 0$.
Since the plane passes through the point $(-2, 1, 3)$,we substitute these coordinates into the equation:
$(2 + 3\lambda)(-2) - (7 + 5\lambda)(1) + (4 + 4\lambda)(3) - 3 + 11\lambda = 0$.
$-4 - 6\lambda - 7 - 5\lambda + 12 + 12\lambda - 3 + 11\lambda = 0$.
Combining the $\lambda$ terms and constants: $(-6 - 5 + 12 + 11)\lambda + (-4 - 7 + 12 - 3) = 0$.
$12\lambda - 2 = 0$,which gives $\lambda = \frac{1}{6}$.
Substituting $\lambda = \frac{1}{6}$ back into the plane equation:
$(2 + 3(\frac{1}{6}))x - (7 + 5(\frac{1}{6}))y + (4 + 4(\frac{1}{6}))z + (-3 + 11(\frac{1}{6})) = 0$.
$(\frac{15}{6})x - (\frac{47}{6})y + (\frac{28}{6})z - (\frac{7}{6}) = 0$.
Multiplying by $6$,we get $15x - 47y + 28z - 7 = 0$.
Comparing this with $ax + by + cz - 7 = 0$,we have $a = 15, b = -47, c = 28$.
Now,calculate $2a + b + c - 7 = 2(15) + (-47) + 28 - 7 = 30 - 47 + 28 - 7 = 4$.

(C) Given the equation of the parabola: $y^{2} = 4ax + 4a^{2}$.
Differentiating both sides with respect to $x$:
$2y \frac{dy}{dx} = 4a$
$\Rightarrow a = \frac{y}{2} \frac{dy}{dx}$.
Substitute the value of $a$ back into the original equation:
$y^{2} = 4\left(\frac{y}{2} \frac{dy}{dx}\right)x + 4\left(\frac{y}{2} \frac{dy}{dx}\right)^{2}$.
Simplify the equation:
$y^{2} = 2xy \frac{dy}{dx} + 4 \cdot \frac{y^{2}}{4} \left(\frac{dy}{dx}\right)^{2}$.
$y^{2} = 2xy \frac{dy}{dx} + y^{2} \left(\frac{dy}{dx}\right)^{2}$.
Dividing by $y$ (assuming $y \neq 0$):
$y = 2x \frac{dy}{dx} + y \left(\frac{dy}{dx}\right)^{2}$.
Rearranging the terms,we get:
$y \left(\frac{dy}{dx}\right)^{2} + 2x \frac{dy}{dx} - y = 0$.

(D) Given the determinant equation: $\left|\begin{array}{ccc}1+\sin ^{2} x & \sin ^{2} x & \sin ^{2} x \\ \cos ^{2} x & 1+\cos ^{2} x & \cos ^{2} x \\ 4 \sin 2 x & 4 \sin 2 x & 1+4 \sin 2 x\end{array}\right|=0$.
Apply the row operation $R_{1} \rightarrow R_{1} + R_{2} + R_{3}$:
Since $\sin^{2} x + \cos^{2} x = 1$,the sum of each column in the first row becomes $1 + \sin^{2} x + \cos^{2} x + 4 \sin 2x = 2 + 4 \sin 2x$.
Factoring out $(2 + 4 \sin 2x)$,we get:
$(2 + 4 \sin 2x) \left|\begin{array}{ccc}1 & 1 & 1 \\ \cos ^{2} x & 1+\cos ^{2} x & \cos ^{2} x \\ 4 \sin 2 x & 4 \sin 2 x & 1+4 \sin 2 x \end{array}\right| = 0$.
Applying $C_{2} \rightarrow C_{2} - C_{1}$ and $C_{3} \rightarrow C_{3} - C_{1}$:
$(2 + 4 \sin 2x) \left|\begin{array}{ccc}1 & 0 & 0 \\ \cos ^{2} x & 1 & 0 \\ 4 \sin 2 x & 0 & 1 \end{array}\right| = 0$.
This simplifies to $(2 + 4 \sin 2x)(1) = 0$,so $2 + 4 \sin 2x = 0$,which means $\sin 2x = -\frac{1}{2}$.
For $0 < x < \pi$,we have $0 < 2x < 2\pi$.
The values of $2x$ where $\sin 2x = -\frac{1}{2}$ in the interval $(0, 2\pi)$ are $2x = \pi + \frac{\pi}{6} = \frac{7\pi}{6}$ and $2x = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}$.
Thus,$x = \frac{7\pi}{12}$ and $x = \frac{11\pi}{12}$.

(B) For the system of linear equations to have a non-trivial solution,the determinant of the coefficient matrix must be zero:
$\begin{vmatrix} \alpha & \beta & \gamma \\ \beta & \gamma & \alpha \\ \gamma & \alpha & \beta \end{vmatrix} = 0$
Expanding the determinant:
$\alpha(\gamma\beta - \alpha^{2}) - \beta(\beta^{2} - \alpha\gamma) + \gamma(\beta\alpha - \gamma^{2}) = 0$
$\alpha\beta\gamma - \alpha^{3} - \beta^{3} + \alpha\beta\gamma + \alpha\beta\gamma - \gamma^{3} = 0$
$3\alpha\beta\gamma - (\alpha^{3} + \beta^{3} + \gamma^{3}) = 0$
Using the identity $\alpha^{3} + \beta^{3} + \gamma^{3} - 3\alpha\beta\gamma = (\alpha + \beta + \gamma)(\alpha^{2} + \beta^{2} + \gamma^{2} - (\alpha\beta + \beta\gamma + \gamma\alpha))$,we have:
$-(\alpha + \beta + \gamma)(\alpha^{2} + \beta^{2} + \gamma^{2} - (\alpha\beta + \beta\gamma + \gamma\alpha)) = 0$
From Vieta's formulas,$\alpha + \beta + \gamma = -a$ and $\alpha\beta + \beta\gamma + \gamma\alpha = b$. Also,$\alpha^{2} + \beta^{2} + \gamma^{2} = (\alpha + \beta + \gamma)^{2} - 2(\alpha\beta + \beta\gamma + \gamma\alpha) = a^{2} - 2b$.
Substituting these values:
$-(-a)(a^{2} - 2b - b) = 0$
$a(a^{2} - 3b) = 0$
Since $a \neq 0$,we must have $a^{2} - 3b = 0$,which implies $a^{2} = 3b$.
Therefore,$\frac{a^{2}}{b} = 3$.

(A) Let $I = \int \frac{(2 x-1) \cos \sqrt{(2 x-1)^{2}+5}}{\sqrt{4 x^{2}-4 x+6}} d x$.
Note that $4x^2 - 4x + 6 = (2x-1)^2 + 5$.
So,$I = \int \frac{(2 x-1) \cos \sqrt{(2 x-1)^{2}+5}}{\sqrt{(2 x-1)^{2}+5}} d x$.
Let $u = \sqrt{(2 x-1)^{2}+5}$.
Then $u^2 = (2x-1)^2 + 5$.
Differentiating both sides with respect to $x$,we get $2u \frac{du}{dx} = 2(2x-1) \cdot 2 = 4(2x-1)$.
Thus,$(2x-1) dx = \frac{u}{2} du$.
Substituting these into the integral:
$I = \int \frac{\cos u}{u} \cdot \frac{u}{2} du = \frac{1}{2} \int \cos u du$.
$I = \frac{1}{2} \sin u + c$.
Substituting back $u = \sqrt{(2 x-1)^{2}+5}$,we get $I = \frac{1}{2} \sin \sqrt{(2 x-1)^{2}+5} + c$.

(D) We use the Taylor series expansions for $\sin^{-1} x$ and $\tan^{-1} x$ near $x = 0$:
$\sin^{-1} x = x + \frac{x^3}{6} + O(x^5)$
$\tan^{-1} x = x - \frac{x^3}{3} + O(x^5)$
Substituting these into the limit expression:
$L = \lim_{x \rightarrow 0} \frac{(x + \frac{x^3}{6}) - (x - \frac{x^3}{3})}{3x^3}$
$L = \lim_{x \rightarrow 0} \frac{\frac{x^3}{6} + \frac{x^3}{3}}{3x^3} = \lim_{x \rightarrow 0} \frac{\frac{1}{6} + \frac{1}{3}}{3} = \frac{\frac{1+2}{6}}{3} = \frac{3/6}{3} = \frac{1/2}{3} = \frac{1}{6}$
Given $L = 1/6$,we calculate $6L + 1$:
$6(1/6) + 1 = 1 + 1 = 2$

(B) For the function $f(x) = \frac{\operatorname{cosec}^{-1} x}{\sqrt{x - [x]}}$ to be defined,two conditions must be satisfied:
$1$. The domain of $\operatorname{cosec}^{-1} x$ is $(-\infty, -1] \cup [1, \infty)$.
$2$. The denominator $\sqrt{x - [x]}$ must be non-zero and real. Since $x - [x] = \{x\}$ (the fractional part of $x$),we require $\{x\} > 0$,which means $x$ cannot be an integer.
Combining these conditions,$x$ must be in $(-\infty, -1] \cup [1, \infty)$ and $x \notin \mathbb{Z}$.
This corresponds to all non-integers in the set $(-\infty, -1] \cup [1, \infty)$,which is equivalent to all non-integers except the interval $(-1, 1)$.

(C) For $f(x) = \sqrt{x}$,the domain is $[0, \infty)$.
For $g(x) = \sqrt{1-x}$,the domain is $(-\infty, 1]$.
The domain of $f+g, f-g,$ and $g-f$ is the intersection of the domains of $f$ and $g$,which is $[0, 1]$.
For $f/g$,we require $g(x) \neq 0$,so $1-x \neq 0 \implies x \neq 1$. The domain is $[0, 1)$.
For $g/f$,we require $f(x) \neq 0$,so $x \neq 0$. The domain is $(0, 1]$.
The common domain for all these functions is the intersection of $[0, 1], [0, 1),$ and $(0, 1]$,which is $(0, 1)$.

(D) Given $f(x) = \begin{cases} \frac{1}{|x|} & ; |x| \geq 1 \\ ax^2 + b & ; |x| < 1 \end{cases}$.
For $x \geq 1$,$f(x) = \frac{1}{x}$. For $x \leq -1$,$f(x) = -\frac{1}{x}$.
Since $f(x)$ is differentiable at every point,it must be continuous at $x = 1$.
$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) \implies a(1)^2 + b = \frac{1}{1} \implies a + b = 1 \quad \dots(1)$.
Also,$f(x)$ must be differentiable at $x = 1$.
$f'(x) = \begin{cases} -\frac{1}{x^2} & ; x > 1 \\ 2ax & ; -1 < x < 1 \end{cases}$.
Equating the derivatives at $x = 1$:
$\lim_{x \to 1^-} f'(x) = \lim_{x \to 1^+} f'(x) \implies 2a(1) = -\frac{1}{(1)^2} \implies 2a = -1 \implies a = -\frac{1}{2}$.
Substituting $a = -\frac{1}{2}$ into equation $(1)$:
$-\frac{1}{2} + b = 1 \implies b = 1 + \frac{1}{2} = \frac{3}{2}$.
Thus,the values are $a = -\frac{1}{2}$ and $b = \frac{3}{2}$.

(B) Given equations are:
$A + 2B = \begin{bmatrix} 1 & 2 & 0 \\ 6 & -3 & 3 \\ -5 & 3 & 1 \end{bmatrix} \quad \dots(1)$
$2A - B = \begin{bmatrix} 2 & -1 & 5 \\ 2 & -1 & 6 \\ 0 & 1 & 2 \end{bmatrix} \quad \dots(2)$
Multiply equation $(2)$ by $2$:
$4A - 2B = \begin{bmatrix} 4 & -2 & 10 \\ 4 & -2 & 12 \\ 0 & 2 & 4 \end{bmatrix} \quad \dots(3)$
Adding $(1)$ and $(3)$:
$5A = \begin{bmatrix} 1+4 & 2-2 & 0+10 \\ 6+4 & -3-2 & 3+12 \\ -5+0 & 3+2 & 1+4 \end{bmatrix} = \begin{bmatrix} 5 & 0 & 10 \\ 10 & -5 & 15 \\ -5 & 5 & 5 \end{bmatrix}$
$A = \begin{bmatrix} 1 & 0 & 2 \\ 2 & -1 & 3 \\ -1 & 1 & 1 \end{bmatrix}$
Now,from $(2)$,$B = 2A - \begin{bmatrix} 2 & -1 & 5 \\ 2 & -1 & 6 \\ 0 & 1 & 2 \end{bmatrix}$:
$B = \begin{bmatrix} 2 & 0 & 4 \\ 4 & -2 & 6 \\ -2 & 2 & 2 \end{bmatrix} - \begin{bmatrix} 2 & -1 & 5 \\ 2 & -1 & 6 \\ 0 & 1 & 2 \end{bmatrix} = \begin{bmatrix} 0 & 1 & -1 \\ 2 & -1 & 0 \\ -2 & 1 & 0 \end{bmatrix}$
$\operatorname{Tr}(A) = 1 + (-1) + 1 = 1$
$\operatorname{Tr}(B) = 0 + (-1) + 0 = -1$
$\operatorname{Tr}(A) - \operatorname{Tr}(B) = 1 - (-1) = 2$

(C) The midpoint $M$ of the line segment $PQ$ is given by $M = \left(\frac{4+2}{2}, \frac{-3+3}{2}, \frac{1-5}{2}\right) = (3, 0, -2)$.
Since the plane bisects the line segment $PQ$ at right angles,the normal vector to the plane is parallel to the vector $\vec{PQ} = (2-4, 3-(-3), -5-1) = (-2, 6, -6)$.
We can take the normal vector as $\vec{n} = (1, -3, 3)$ by dividing by $-2$.
The equation of the plane passing through $M(3, 0, -2)$ with normal vector $\vec{n} = (1, -3, 3)$ is:
$1(x-3) - 3(y-0) + 3(z+2) = 0$
$x - 3y + 3z - 3 + 6 = 0$
$x - 3y + 3z + 3 = 0$
Comparing this with $ax+by+cz+d=0$,we get $a=1, b=-3, c=3, d=3$.
The value of $(a^{2}+b^{2}+c^{2}+d^{2})$ is $1^{2} + (-3)^{2} + 3^{2} + 3^{2} = 1 + 9 + 9 + 9 = 28$.
Since $a, b, c, d$ are integers and the plane equation is unique up to a scalar multiple,the minimum value for integer coefficients is $28$.

(D) Given $f(x^{2}) + g(4-x) = 4x^{3}$ and $g(4-x) = -g(x)$.
Substituting $g(4-x) = -g(x)$ into the first equation,we get $f(x^{2}) - g(x) = 4x^{3}$,so $f(x^{2}) = 4x^{3} + g(x)$.
We want to evaluate $I = \int_{-4}^{4} f(x) dx$. Let $x = t^{2}$,then $dx = 2t dt$. However,since $f(x)$ is defined via $f(x^{2})$,we consider the integral $\int_{-4}^{4} f(x^{2}) dx$.
Since $f(x^{2}) = 4x^{3} + g(x)$,we have $\int_{-4}^{4} f(x^{2}) dx = \int_{-4}^{4} (4x^{3} + g(x)) dx$.
$= \int_{-4}^{4} 4x^{3} dx + \int_{-4}^{4} g(x) dx$.
Since $4x^{3}$ is an odd function,$\int_{-4}^{4} 4x^{3} dx = 0$.
Given $g(4-x) = -g(x)$,let $u = 4-x$,then $du = -dx$. The integral $\int_{-4}^{4} g(x) dx$ becomes $\int_{8}^{0} g(4-u) (-du) = \int_{0}^{8} g(4-u) du = \int_{0}^{8} -g(u) du$. This implies the integral of $g(x)$ over symmetric intervals is $0$.
Thus,the integral evaluates to $0$.

(A) The equation of any plane parallel to $x - 2y + 2z - 3 = 0$ is of the form $x - 2y + 2z + \lambda = 0$.
The distance from the point $(1, 2, 3)$ to the plane $x - 2y + 2z + \lambda = 0$ is given by $\frac{|1 - 2(2) + 2(3) + \lambda|}{\sqrt{1^2 + (-2)^2 + 2^2}} = 1$.
Simplifying the expression: $\frac{|1 - 4 + 6 + \lambda|}{\sqrt{9}} = 1 \Rightarrow \frac{|\lambda + 3|}{3} = 1$.
This gives $|\lambda + 3| = 3$,so $\lambda + 3 = 3$ or $\lambda + 3 = -3$.
Thus,$\lambda = 0$ or $\lambda = -6$.
The two possible planes are $x - 2y + 2z = 0$ and $x - 2y + 2z - 6 = 0$.
Case $1$: $a=1, b=-2, c=2, d=0$. Then $(b - d) = -2 - 0 = -2$ and $(c - a) = 2 - 1 = 1$. So,$-2 = K(1) \Rightarrow K = -2$.
Case $2$: $a=1, b=-2, c=2, d=-6$. Then $(b - d) = -2 - (-6) = 4$ and $(c - a) = 2 - 1 = 1$. So,$4 = K(1) \Rightarrow K = 4$.
The positive value of $K$ is $4$.

(C) Given $f(x) = \int \frac{5x^{8} + 7x^{6}}{(x^{2} + 1 + 2x^{7})^{2}} dx$.
Divide the numerator and denominator by $x^{14}$ inside the integral:
$f(x) = \int \frac{5x^{-6} + 7x^{-8}}{(x^{-5} + x^{-7} + 2)^{2}} dx$.
Let $t = x^{-5} + x^{-7} + 2$.
Then $dt = (-5x^{-6} - 7x^{-8}) dx$,which implies $-(5x^{-6} + 7x^{-8}) dx = dt$.
Substituting this into the integral:
$f(x) = \int -\frac{dt}{t^{2}} = \frac{1}{t} + C = \frac{1}{x^{-5} + x^{-7} + 2} + C = \frac{x^{7}}{1 + x^{2} + 2x^{7}} + C$.
Given $f(0) = 0$,we find $C = 0$.
Thus,$f(x) = \frac{x^{7}}{x^{2} + 1 + 2x^{7}}$.
Evaluating at $x = 1$:
$f(1) = \frac{1^{7}}{1^{2} + 1 + 2(1)^{7}} = \frac{1}{1 + 1 + 2} = \frac{1}{4}$.
Since $f(1) = \frac{1}{K}$,we have $K = 4$.

(C) Given $f(x) = \left|\begin{array}{ccc} \sin^{2} x & 1+\cos^{2} x & \cos 2x \\ 1+\sin^{2} x & \cos^{2} x & \cos 2x \\ \sin^{2} x & \cos^{2} x & \sin 2x \end{array}\right|$.
Apply the column operation $C_{1} \rightarrow C_{1} + C_{2}$:
$f(x) = \left|\begin{array}{ccc} \sin^{2} x + 1 + \cos^{2} x & 1+\cos^{2} x & \cos 2x \\ 1+\sin^{2} x + \cos^{2} x & \cos^{2} x & \cos 2x \\ \sin^{2} x + \cos^{2} x & \cos^{2} x & \sin 2x \end{array}\right| = \left|\begin{array}{ccc} 2 & 1+\cos^{2} x & \cos 2x \\ 2 & \cos^{2} x & \cos 2x \\ 1 & \cos^{2} x & \sin 2x \end{array}\right|$.
Apply the row operation $R_{1} \rightarrow R_{1} - R_{2}$:
$f(x) = \left|\begin{array}{ccc} 0 & 1 & 0 \\ 2 & \cos^{2} x & \cos 2x \\ 1 & \cos^{2} x & \sin 2x \end{array}\right|$.
Expanding along $R_{1}$:
$f(x) = -1 \times (2 \sin 2x - \cos 2x) = \cos 2x - 2 \sin 2x$.
The maximum value of $a \cos \theta + b \sin \theta$ is $\sqrt{a^{2} + b^{2}}$.
Here $a = 1$ and $b = -2$.
So,$f(x)_{\max} = \sqrt{1^{2} + (-2)^{2}} = \sqrt{1+4} = \sqrt{5}$.

(C) For the function to be continuous at $x=0$,we must have $\operatorname{Lim}_{x \rightarrow 0^{+}} f(x) = \operatorname{Lim}_{x \rightarrow 0^{-}} f(x) = f(0) = \alpha$.
First,consider the Right Hand Limit $(RHL)$ as $x \rightarrow 0^{+}$:
$\operatorname{Lim}_{x \rightarrow 0^{+}} \frac{\cos^{-1}(1-x^2) \sin^{-1}(1-x)}{x(1-x)(1+x)} = \operatorname{Lim}_{x \rightarrow 0^{+}} \frac{\cos^{-1}(1-x^2)}{x} \cdot \frac{\sin^{-1}(1-x)}{1-x^2} = \frac{\pi}{2} \cdot \operatorname{Lim}_{x \rightarrow 0^{+}} \frac{\cos^{-1}(1-x^2)}{x}$.
Let $1-x^2 = \cos \theta$,then as $x \rightarrow 0^{+}$,$\theta \rightarrow 0^{+}$.
$\operatorname{Lim}_{\theta \rightarrow 0^{+}} \frac{\theta}{\sqrt{1-\cos \theta}} = \operatorname{Lim}_{\theta \rightarrow 0^{+}} \frac{\theta}{\sqrt{2} \sin(\theta/2)} = \frac{1}{\sqrt{2}} \cdot 2 = \sqrt{2}$.
So,$RHL = \frac{\pi}{2} \cdot \sqrt{2} = \frac{\pi}{\sqrt{2}}$.
Now,consider the Left Hand Limit $(LHL)$ as $x \rightarrow 0^{-}$:
For $x \in (-1, 0)$,$\{x\} = x+1$.
$\operatorname{Lim}_{x \rightarrow 0^{-}} \frac{\cos^{-1}(1-(x+1)^2) \sin^{-1}(1-(x+1))}{(x+1)-(x+1)^3} = \operatorname{Lim}_{x \rightarrow 0^{-}} \frac{\cos^{-1}(1-(x+1)^2) \sin^{-1}(-x)}{(x+1)(1-(x+1)^2)} = \operatorname{Lim}_{x \rightarrow 0^{-}} \frac{\cos^{-1}(1-(x+1)^2) \cdot (-x)}{(x+1)(-x)(2+x)} = \frac{\cos^{-1}(0)}{1 \cdot 2} = \frac{\pi/2}{2} = \frac{\pi}{4}$.
Since $RHL = \frac{\pi}{\sqrt{2}}$ and $LHL = \frac{\pi}{4}$,the limits are not equal.
Therefore,the function is not continuous at $x=0$ for any value of $\alpha$.

(B) The equation of a plane passing through $(42, 0, 0)$,$(0, 42, 0)$,and $(0, 0, 42)$ in intercept form is $\frac{x}{42} + \frac{y}{42} + \frac{z}{42} = 1$,which simplifies to $x + y + z = 42$.
We can rewrite this as $(x-11) + (y-19) + (z-12) = 42 - 11 - 19 - 12 = 0$.
Let $a = x-11$,$b = y-19$,and $c = z-12$. Then $a + b + c = 0$.
The given expression is $3 + \frac{a}{b^2 c^2} + \frac{b}{a^2 c^2} + \frac{c}{a^2 b^2} - \frac{42}{14abc}$.
Since $a+b+c=0$,we have $x+y+z=42$. Substituting this into the last term,we get $\frac{42}{14abc} = \frac{3}{abc}$.
The expression becomes $3 + \frac{a^3 + b^3 + c^3}{a^2 b^2 c^2} - \frac{3}{abc} = 3 + \frac{a^3 + b^3 + c^3 - 3abc}{a^2 b^2 c^2}$.
Since $a+b+c=0$,the identity $a^3 + b^3 + c^3 = 3abc$ holds.
Therefore,the expression simplifies to $3 + \frac{3abc - 3abc}{a^2 b^2 c^2} = 3 + 0 = 3$.

(C) The integral is given by $I = \int_{0}^{10} [x] e^{[x]-x+1} dx$.
Since $[x] = n$ for $x \in [n, n+1)$,we can split the integral as:
$I = \sum_{n=0}^{9} \int_{n}^{n+1} n \cdot e^{n-x+1} dx$.
Evaluating the integral for each term:
$I = \sum_{n=0}^{9} n \left[ -e^{n-x+1} \right]_{n}^{n+1} = \sum_{n=0}^{9} n \left( -e^{0} + e^{1} \right) = (e-1) \sum_{n=0}^{9} n$.
Using the sum formula $\sum_{n=0}^{9} n = \frac{9 \times 10}{2} = 45$,we get:
$I = 45(e-1)$.