WBJEE 2021 Mathematics Question Paper with Answer and Solution

(B) The equation of the incident ray is $x + \sqrt{3}y = \sqrt{3}$.
Find the point of incidence $A$ on the $x$-axis (where $y=0$): $x + \sqrt{3}(0) = \sqrt{3} \Rightarrow x = \sqrt{3}$. So,$A = (\sqrt{3}, 0)$.
Take a point $B$ on the incident ray,e.g.,$x=0$ $\Rightarrow \sqrt{3}y = \sqrt{3}$ $\Rightarrow y=1$. So,$B = (0, 1)$.
The reflected ray passes through $A(\sqrt{3}, 0)$ and the image of $B(0, 1)$ with respect to the $x$-axis,which is $B'(0, -1)$.
The slope of the reflected ray $AB'$ is $m = \frac{-1 - 0}{0 - \sqrt{3}} = \frac{-1}{-\sqrt{3}} = \frac{1}{\sqrt{3}}$.
The equation of the reflected ray is $y - 0 = \frac{1}{\sqrt{3}}(x - \sqrt{3})$.
$\sqrt{3}y = x - \sqrt{3}$.

(C) Given the quadratic equation $x^{2}-6x-2=0$.
Since $\alpha$ and $\beta$ are roots,they satisfy the equation:
$x^{n}-6x^{n-1}-2x^{n-2}=0$
$\Rightarrow x^{n}-2x^{n-2}=6x^{n-1}$
For $x = \alpha$ and $x = \beta$,we have:
$\alpha^{n}-2\alpha^{n-2}=6\alpha^{n-1}$
$\beta^{n}-2\beta^{n-2}=6\beta^{n-1}$
Subtracting the two equations:
$(\alpha^{n}-\beta^{n})-2(\alpha^{n-2}-\beta^{n-2})=6(\alpha^{n-1}-\beta^{n-1})$
Using the definition $a_{n}=\alpha^{n}-\beta^{n}$,we get:
$a_{n}-2a_{n-2}=6a_{n-1}$
For $n=10$:
$a_{10}-2a_{8}=6a_{9}$
Therefore,the value of $\frac{a_{10}-2a_{8}}{2a_{9}} = \frac{6a_{9}}{2a_{9}} = 3$.

(C) Given equation: $6^{x}+8^{x}=10^{x}$
Divide both sides by $10^{x}$:
$\left(\frac{6}{10}\right)^{x}+\left(\frac{8}{10}\right)^{x}=1$
$\left(\frac{3}{5}\right)^{x}+\left(\frac{4}{5}\right)^{x}=1$
Let $f(x) = \left(\frac{3}{5}\right)^{x}+\left(\frac{4}{5}\right)^{x}$.
Since $\frac{3}{5} < 1$ and $\frac{4}{5} < 1$,both functions $\left(\frac{3}{5}\right)^{x}$ and $\left(\frac{4}{5}\right)^{x}$ are strictly decreasing functions.
Therefore,their sum $f(x)$ is also a strictly decreasing function.
$A$ strictly decreasing function can take the value $1$ at most once.
By observation,for $x=2$,$\left(\frac{3}{5}\right)^{2}+\left(\frac{4}{5}\right)^{2} = \frac{9}{25}+\frac{16}{25} = \frac{25}{25} = 1$.
Thus,$x=2$ is the unique solution.
Hence,the equation has exactly one real root.

(C) Given $A = \{(z, w) \mid |z| = |w|\}$ and $B = \{(z, w) \mid z^2 = w^2\}$.
For any $(z, w) \in B$,we have $z^2 = w^2$,which implies $z^2 - w^2 = 0$,so $(z - w)(z + w) = 0$.
This means $z = w$ or $z = -w$.
If $z = w$,then $|z| = |w|$,so $(z, w) \in A$.
If $z = -w$,then $|z| = |-w| = |w|$,so $(z, w) \in A$.
Thus,every element of $B$ is also an element of $A$,which means $B \subseteq A$.
However,consider $(z, w) = (1, i)$. Here $|z| = |1| = 1$ and $|w| = |i| = 1$,so $|z| = |w|$,meaning $(1, i) \in A$.
But $z^2 = 1^2 = 1$ and $w^2 = i^2 = -1$,so $z^2 \neq w^2$,meaning $(1, i) \notin B$.
Since $A$ contains elements not in $B$,$B \subset A$ is the correct relation.

(D) Let $z = e^{i \theta}$,where $\theta \in \mathbb{R}$ and $\theta \neq n\pi$ for $n \in \mathbb{Z}$ (since $z \neq \pm 1$).
Let $w = \frac{z}{1-z^2}$.
Substituting $z = e^{i \theta}$,we get:
$w = \frac{e^{i \theta}}{1 - e^{i 2 \theta}}$
Divide numerator and denominator by $e^{i \theta}$:
$w = \frac{1}{e^{-i \theta} - e^{i \theta}}$
Using the identity $e^{i \theta} = \cos \theta + i \sin \theta$,we have $e^{-i \theta} - e^{i \theta} = -2i \sin \theta$.
Thus,$w = \frac{1}{-2i \sin \theta} = \frac{i}{2 \sin \theta}$.
Since $w$ is of the form $0 + i \left( \frac{1}{2 \sin \theta} \right)$,the real part is $0$.
Therefore,the locus of $w$ is the $y$-axis.

(B) Given $|z|-2=0 \implies |z|=2$. This represents a circle centered at the origin with radius $2$,so $x^2+y^2=4$.
Also,$|z+i|-|z-1|=0 \implies |z+i|=|z-1|$.
Let $z=x+iy$. Then $|x+i(y+1)|=|x-1+iy|$.
Squaring both sides: $x^2+(y+1)^2=(x-1)^2+y^2$.
$x^2+y^2+2y+1=x^2-2x+1+y^2$.
$2y=-2x \implies y=-x$.
Substituting $y=-x$ into $x^2+y^2=4$:
$x^2+(-x)^2=4 \implies 2x^2=4 \implies x^2=2 \implies x=\pm\sqrt{2}$.
If $x=\sqrt{2}$,then $y=-\sqrt{2}$,so $z=\sqrt{2}-i\sqrt{2}=\sqrt{2}(1-i)$.
If $x=-\sqrt{2}$,then $y=\sqrt{2}$,so $z=-\sqrt{2}+i\sqrt{2}=\sqrt{2}(-1+i)$.
Thus,the possible values for $z$ are $\sqrt{2}(1-i)$ and $\sqrt{2}(-1+i)$.

(A) We need to find the unit digit of the sum $S = 1! + 2! + 3! + 4! + 5! + \ldots + 99!$.
First,calculate the factorials:
$1! = 1$
$2! = 2 \times 1 = 2$
$3! = 3 \times 2 \times 1 = 6$
$4! = 4 \times 3 \times 2 \times 1 = 24$
$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$
For any $n \ge 5$,$n!$ contains the factors $5$ and $2$,so $n!$ ends in $0$.
Thus,for all $n \ge 5$,the unit digit of $n!$ is $0$.
The sum is $S = 1! + 2! + 3! + 4! + (5! + 6! + \ldots + 99!)$.
The unit digit of $S$ is the unit digit of $(1! + 2! + 3! + 4!) + (0 + 0 + \ldots + 0)$.
$1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33$.
The unit digit of $33$ is $3$.

(C) The word '$EQUATION$' has $8$ distinct letters: $5$ vowels $(E, U, A, I, O)$ and $3$ consonants $(Q, T, N)$.
We need to select $3$ vowels out of $5$ and $2$ consonants out of $3$.
Number of ways to select the letters = $^5C_3 \times ^3C_2 = 10 \times 3 = 30$.
Since all $3$ vowels must be together,we treat the $3$ vowels as a single block. Now we have $1$ block of vowels and $2$ consonants,totaling $3$ units to arrange.
Number of ways to arrange these $3$ units = $3! = 6$.
Within the vowel block,the $3$ vowels can be arranged in $3! = 6$ ways.
Total number of words = $30 \times 6 \times 6 = 1080$.

(C) Let $x_{1}, x_{2}, x_{3}, x_{4}$ be the marks assigned to the $4$ questions respectively.
We are given the equation: $x_{1} + x_{2} + x_{3} + x_{4} = 10$,where $x_{i} \geq 2$ for all $i \in \{1, 2, 3, 4\}$.
Let $y_{i} = x_{i} - 2$. Since $x_{i} \geq 2$,we have $y_{i} \geq 0$.
Substituting $x_{i} = y_{i} + 2$ into the equation:
$(y_{1} + 2) + (y_{2} + 2) + (y_{3} + 2) + (y_{4} + 2) = 10$
$y_{1} + y_{2} + y_{3} + y_{4} + 8 = 10$
$y_{1} + y_{2} + y_{3} + y_{4} = 2$.
The number of non-negative integer solutions is given by the formula $\binom{n+r-1}{r-1}$,where $n = 2$ and $r = 4$.
Number of ways $= \binom{2+4-1}{4-1} = \binom{5}{3} = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10$.

(A) Given the recurrence relation $h(p) = \frac{1}{2}\{h(p+1) + h(p-1)\}$,we can rewrite it as $2h(p) = h(p+1) + h(p-1)$,which implies $h(p+1) - h(p) = h(p) - h(p-1)$.
This indicates that the sequence $h(0), h(1), \ldots, h(100)$ forms an Arithmetic Progression ($A$.$P$.).
Let the common difference be $d$. Then $h(n) = h(0) + nd$.
Using $h(100) = 20$ and $h(0) = 5$,we have $20 = 5 + 100d$.
$100d = 15 \Rightarrow d = \frac{15}{100} = 0.15$.
Thus,$h(1) = h(0) + d = 5 + 0.15 = 5.15$.

(C) Since $\log \left(\frac{5 c}{2 a}\right), \log \left(\frac{7 b}{5 c}\right), \log \left(\frac{2 a}{7 b}\right)$ are in $A.P.$,we have: $\log \left(\frac{5 c}{2 a}\right) + \log \left(\frac{2 a}{7 b}\right) = 2 \log \left(\frac{7 b}{5 c}\right)$
$\log \left(\frac{5 c}{2 a} \times \frac{2 a}{7 b}\right) = \log \left(\frac{7 b}{5 c}\right)^2$
$\log \left(\frac{5 c}{7 b}\right) = \log \left(\frac{49 b^2}{25 c^2}\right)$
$\frac{5 c}{7 b} = \frac{49 b^2}{25 c^2}$ $\Rightarrow 125 c^3 = 343 b^3$ $\Rightarrow 5 c = 7 b$ $\Rightarrow c = \frac{7}{5} b$
Since $a, b, c$ are in $G.P.$,$b^2 = ac$. Substituting $c = \frac{7}{5} b$,we get $b^2 = a \left(\frac{7}{5} b\right) \Rightarrow a = \frac{5}{7} b$.
The sides are $\frac{5}{7} b, b, \frac{7}{5} b$.
Let $b = 35k$. Then the sides are $25k, 35k, 49k$.
Since $25k + 35k = 60k > 49k$,$25k + 49k = 74k > 35k$,and $35k + 49k = 84k > 25k$,these sides form a triangle.
Since all sides are unequal,it is a scalene triangle.

(D) Given the equation: $\frac{2}{3} \log _{b} a+\frac{3}{5} \log _{c} b+\frac{5}{2} \log _{a} c=3$.
Let $x = \frac{2}{3} \log _{b} a$,$y = \frac{3}{5} \log _{c} b$,and $z = \frac{5}{2} \log _{a} c$.
Note that $x \cdot y \cdot z = (\frac{2}{3} \log _{b} a) \cdot (\frac{3}{5} \log _{c} b) \cdot (\frac{5}{2} \log _{a} c) = (\frac{2}{3} \cdot \frac{3}{5} \cdot \frac{5}{2}) \cdot (\log _{b} a \cdot \log _{c} b \cdot \log _{a} c) = 1 \cdot 1 = 1$.
By the Arithmetic Mean-Geometric Mean inequality $(AM \ge GM)$,for positive real numbers $x, y, z$ with $x+y+z=3$ and $xyz=1$,the equality holds only when $x=y=z=1$.
Thus,$\frac{2}{3} \log _{b} a = 1 \Rightarrow \log _{b} a = \frac{3}{2}$.
Given $b=9$,we have $\log _{9} a = \frac{3}{2}$.
Therefore,$a = 9^{3/2} = (3^2)^{3/2} = 3^3 = 27$.

(D) The given expression is a geometric series with first term $A = (1+x)^{2016}$,common ratio $r = \frac{x}{1+x}$,and $n = 2017$ terms.
Sum $S = A \frac{1-r^n}{1-r} = (1+x)^{2016} \frac{1 - (\frac{x}{1+x})^{2017}}{1 - \frac{x}{1+x}} = (1+x)^{2016} \frac{\frac{(1+x)^{2017} - x^{2017}}{(1+x)^{2017}}}{\frac{1+x-x}{1+x}} = (1+x)^{2016} \frac{(1+x)^{2017} - x^{2017}}{(1+x)^{2017}} \cdot (1+x) = (1+x)^{2017} - x^{2017}$.
We are given $\sum_{i=0}^{2016} a_i x^i = (1+x)^{2017} - x^{2017}$.
Expanding $(1+x)^{2017}$ using the binomial theorem: $(1+x)^{2017} = \sum_{i=0}^{2017} {}^{2017}C_i x^i$.
Thus,$\sum_{i=0}^{2016} a_i x^i = \left( \sum_{i=0}^{2017} {}^{2017}C_i x^i \right) - x^{2017} = \sum_{i=0}^{2016} {}^{2017}C_i x^i$.
Comparing the coefficients of $x^{17}$,we get $a_{17} = {}^{2017}C_{17} = \frac{2017!}{17! (2017-17)!} = \frac{2017!}{17! 2000!}$.

(B) Let $x = 7^{7^{...7}}$ ($22$ times $7$). We want to find $x \pmod{48}$.
Note that $7^2 = 49 \equiv 1 \pmod{48}$.
Since the exponent of $7$ is $7^{7^{...7}}$ ($21$ times $7$),which is clearly an odd number,let the exponent be $2k+1$ for some integer $k$.
Then $7^{2k+1} = 7 \times (7^2)^k = 7 \times (49)^k$.
Since $49 \equiv 1 \pmod{48}$,we have $49^k \equiv 1^k \equiv 1 \pmod{48}$.
Therefore,$7 \times (49)^k \equiv 7 \times 1 \equiv 7 \pmod{48}$.
The remainder is $7$.

(C) The general term in the expansion of $(bc + ca + ab)^{6}$ is given by the multinomial theorem as $\frac{6!}{p! q! r!} (bc)^{p} (ca)^{q} (ab)^{r} = \frac{6!}{p! q! r!} a^{q+r} b^{p+r} c^{p+q}$.
We need the coefficient of $a^{3} b^{4} c^{5}$,so we set the exponents equal to the required powers:
$q + r = 3$
$p + r = 4$
$p + q = 5$
Adding these three equations: $2(p + q + r) = 3 + 4 + 5 = 12$,so $p + q + r = 6$.
Subtracting each equation from the sum:
$p = (p + q + r) - (q + r) = 6 - 3 = 3$
$q = (p + q + r) - (p + r) = 6 - 4 = 2$
$r = (p + q + r) - (p + q) = 6 - 5 = 1$
The coefficient is $\frac{6!}{3! 2! 1!} = \frac{720}{6 \times 2} = \frac{720}{12} = 60$.

(A) Let the vertices of the parallelogram be $O(0,0)$,$A(a \cos \theta, b \sin \theta)$,$B(a \cos \theta, -b \sin \theta)$,and $C(x, y)$.
Since $OABC$ is a parallelogram,the diagonals $OB$ and $AC$ bisect each other at the same midpoint.
Alternatively,the area of a parallelogram with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ and $(x_4, y_4)$ can be calculated using the cross product of vectors.
The area of $\triangle OAB = \frac{1}{2} |x_A y_B - x_B y_A| = \frac{1}{2} |(a \cos \theta)(-b \sin \theta) - (a \cos \theta)(b \sin \theta)| = \frac{1}{2} |-2ab \sin \theta \cos \theta| = ab |\sin \theta \cos \theta| = \frac{ab}{2} |\sin 2\theta|$.
The area of the parallelogram $OABC = 2 \times \text{Area}(\triangle OAB) = 2 \times \frac{ab}{2} |\sin 2\theta| = ab |\sin 2\theta|$.
The maximum value of $|\sin 2\theta|$ is $1$,which occurs when $2\theta = \pi / 2$,i.e.,$\theta = \pi / 4$.
Thus,the maximum area is $ab$ when $\theta = \pi / 4$.

(A) The circle is $x^{2}+y^{2}=4$,so its radius $OA = 2$ and the center $O$ is $(0,0)$.
Since $MA$ is a tangent,$\angle OAM = 90^{\circ}$.
In the right-angled triangle $\triangle OAM$,the hypotenuse $OM = 4$ and the side $OA = 2$.
Using the Pythagorean theorem,$MA = \sqrt{OM^{2} - OA^{2}} = \sqrt{4^{2} - 2^{2}} = \sqrt{16 - 4} = \sqrt{12} = 2\sqrt{3}$.
The quadrilateral $MAOB$ consists of two congruent right-angled triangles,$\triangle OAM$ and $\triangle OBM$.
Therefore,the area of quadrilateral $MAOB = 2 \times \text{Area}(\triangle OAM)$.
Area $(\triangle OAM) = \frac{1}{2} \times OA \times MA = \frac{1}{2} \times 2 \times 2\sqrt{3} = 2\sqrt{3}$.
Thus,the area of quadrilateral $MAOB = 2 \times 2\sqrt{3} = 4\sqrt{3}$ sq. units.

(A) Let $B = (2\alpha, 0)$.
Since $A = (0, 4)$,the midpoint $M$ of $AB$ is $(\alpha, 2)$.
The slope of $AB$ is $m_{AB} = \frac{0-4}{2\alpha-0} = -\frac{2}{\alpha}$.
The slope of the perpendicular bisector of $AB$ is $m_{MR} = -\frac{1}{m_{AB}} = \frac{\alpha}{2}$.
The equation of the line $MR$ passing through $M(\alpha, 2)$ with slope $\frac{\alpha}{2}$ is $y-2 = \frac{\alpha}{2}(x-\alpha)$.
To find $R$,set $x=0$: $y-2 = \frac{\alpha}{2}(0-\alpha) \Rightarrow y = 2 - \frac{\alpha^{2}}{2}$.
So,$R = (0, 2 - \frac{\alpha^{2}}{2})$.
Let $P(x, y)$ be the midpoint of $MR$. Then $x = \frac{\alpha+0}{2} = \frac{\alpha}{2}$ and $y = \frac{2 + (2 - \alpha^{2}/2)}{2} = 2 - \frac{\alpha^{2}}{4}$.
From $x = \frac{\alpha}{2}$,we have $\alpha = 2x$.
Substituting $\alpha = 2x$ into the equation for $y$: $y = 2 - \frac{(2x)^{2}}{4} = 2 - x^{2}$.
Thus,$y+x^{2}=2$.

(B) Let the mid-point of $AB$ be $(h, k)$.
Let $A = (\alpha, -\alpha)$ and $B = (\beta, \beta)$.
Then,the mid-point $(h, k) = \left(\frac{\alpha+\beta}{2}, \frac{\beta-\alpha}{2}\right)$.
So,$\alpha+\beta = 2h$ and $\beta-\alpha = 2k$.
The area of $\triangle AOB = \frac{1}{2} \times |OA| \times |OB| = \frac{1}{2} \sqrt{\alpha^{2} + (-\alpha)^{2}} \sqrt{\beta^{2} + \beta^{2}} = \frac{1}{2} \sqrt{2\alpha^{2}} \sqrt{2\beta^{2}} = |\alpha\beta| = C$.
Thus,$\alpha^{2}\beta^{2} = C^{2}$.
We know that $(\beta+\alpha)^{2} - (\beta-\alpha)^{2} = 4\alpha\beta$.
So,$4\alpha\beta = (2h)^{2} - (2k)^{2} = 4(h^{2}-k^{2})$,which implies $\alpha\beta = h^{2}-k^{2}$.
Substituting this into $\alpha^{2}\beta^{2} = C^{2}$,we get $(h^{2}-k^{2})^{2} = C^{2}$.
Replacing $(h, k)$ with $(x, y)$,the locus is $(x^{2}-y^{2})^{2} = C^{2}$.

(A) Let the circle $C$ have center $O(x_0, y_0)$ and radius $r$. The coordinates of any point $P$ on the circle can be represented as $P(x_0 + r \cos \theta, y_0 + r \sin \theta)$.
Let $Q$ be the fixed point $(a, b)$.
Let $R(h, k)$ be the midpoint of the line segment $PQ$.
Then,$h = \frac{x_0 + r \cos \theta + a}{2}$ and $k = \frac{y_0 + r \sin \theta + b}{2}$.
Rearranging these equations,we get:
$2h - (x_0 + a) = r \cos \theta$
$2k - (y_0 + b) = r \sin \theta$
Squaring and adding both equations:
$(2h - (x_0 + a))^2 + (2k - (y_0 + b))^2 = r^2(\cos^2 \theta + \sin^2 \theta)$
$4(h - \frac{x_0 + a}{2})^2 + 4(k - \frac{y_0 + b}{2})^2 = r^2$
$(h - \frac{x_0 + a}{2})^2 + (k - \frac{y_0 + b}{2})^2 = (\frac{r}{2})^2$
This is the equation of a circle with center $(\frac{x_0 + a}{2}, \frac{y_0 + b}{2})$ and radius $\frac{r}{2}$.

(B) The equation of the parabola is $y^{2} = x$,which is of the form $y^{2} = 4ax$,where $4a = 1$,so $a = \frac{1}{4}$.
For a parabola $y^{2} = 4ax$,the condition for three distinct normals to be drawn from a point $(h, k)$ is that $h > 2a$.
Here,the point is $(d, 0)$,so $h = d$.
Substituting the values,we get $d > 2 \times \frac{1}{4}$.
Therefore,$d > \frac{1}{2}$.

(A) The given equation is $6y = 2a^3x^2 + 3a^2x - 12a$.
Dividing by $2a^3$ (assuming $a \neq 0$),we get $x^2 + \frac{3}{2a}x = \frac{6y + 12a}{2a^3}$.
Completing the square for $x$: $(x + \frac{3}{4a})^2 = \frac{6y + 12a}{2a^3} + \frac{9}{16a^2} = \frac{48y + 96a + 9a}{16a^3} = \frac{48y + 105a}{16a^3}$.
Thus,$(x + \frac{3}{4a})^2 = \frac{3}{a^2}(y + \frac{35a}{16})$.
The vertex $(h, k)$ is given by $h = -\frac{3}{4a}$ and $k = -\frac{35a}{16}$.
From $h = -\frac{3}{4a}$,we have $a = -\frac{3}{4h}$.
Substituting $a$ into $k$: $k = -\frac{35}{16} \times (-\frac{3}{4h}) = \frac{105}{64h}$.
Therefore,$hk = \frac{105}{64}$.
The locus is $xy = \frac{105}{64}$.

(C) The equation of the family of curves passing through the intersection of the two ellipses is given by $S_1 + \lambda S_2 = 0$,where $S_1: x^{2}+2y^{2}-6x-12y+20=0$ and $S_2: 2x^{2}+y^{2}-10x-6y+15=0$.
$(x^{2}+2y^{2}-6x-12y+20) + \lambda(2x^{2}+y^{2}-10x-6y+15) = 0$
$(1+2\lambda)x^{2} + (2+\lambda)y^{2} - (6+10\lambda)x - (12+6\lambda)y + (20+15\lambda) = 0$
For this to represent a circle,the coefficients of $x^{2}$ and $y^{2}$ must be equal:
$1+2\lambda = 2+\lambda \Rightarrow \lambda = 1$
Substituting $\lambda = 1$ into the equation:
$3x^{2} + 3y^{2} - 16x - 18y + 35 = 0$
$x^{2} + y^{2} - \frac{16}{3}x - 6y + \frac{35}{3} = 0$
The centre of the circle is given by $\left(-\frac{g}{2}, -\frac{f}{2}\right)$,where $2g = -\frac{16}{3}$ and $2f = -6$.
Centre $= \left(\frac{16/3}{2}, \frac{6}{2}\right) = \left(\frac{8}{3}, 3\right)$.

(B) The given equation of the ellipse is $x^{2}+2y^{2}=4$.
Dividing by $4$,we get $\frac{x^{2}}{4}+\frac{y^{2}}{2}=1$.
Comparing this with the standard form $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$,we have $a^{2}=4$,so $a=2$.
The auxiliary circle of an ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ is given by $x^{2}+y^{2}=a^{2}$.
Here,the equation of the auxiliary circle is $x^{2}+y^{2}=4$.
The coordinates of a point on the auxiliary circle corresponding to an eccentric angle $\theta$ are given by $(a \cos \theta, a \sin \theta)$.
Given $\theta = 60^{\circ}$ and $a=2$,the coordinates are $(2 \cos 60^{\circ}, 2 \sin 60^{\circ})$.
Substituting the values,we get $(2 \times \frac{1}{2}, 2 \times \frac{\sqrt{3}}{2}) = (1, \sqrt{3})$.
Thus,the correct option is $B$.

(B) Let the center of the variable circle be $O(h, k)$ and its radius be $r$. Let the centers of the two given circles be $O_1$ and $O_2$ with radii $r_1$ and $r_2$ respectively.
Since the variable circle touches the two given circles externally,the distance between the centers is the sum of the radii:
$OO_1 = r + r_1$
$OO_2 = r + r_2$
Subtracting the two equations:
$OO_2 - OO_1 = (r + r_2) - (r + r_1) = r_2 - r_1$
Since $r_1$ and $r_2$ are constants,the difference of the distances of the point $O$ from two fixed points $O_1$ and $O_2$ is constant. By definition,this is the locus of a hyperbola.

(C) We evaluate the limit inside the sine function first:
$L = \lim_{x \rightarrow 0} \frac{e^{x}-x-1-\frac{x^{2}}{2}}{x^{2}}$
Using the Taylor series expansion for $e^{x} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \dots$,we get:
$L = \lim_{x \rightarrow 0} \frac{(1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{6} + \dots) - x - 1 - \frac{x^{2}}{2}}{x^{2}}$
$L = \lim_{x \rightarrow 0} \frac{\frac{x^{3}}{6} + \dots}{x^{2}} = \lim_{x \rightarrow 0} (\frac{x}{6} + \dots) = 0$
Since the function $\sin(u)$ is continuous at $u=0$,we have:
$I = \sin(L) = \sin(0) = 0$

(C) We evaluate the limit of the form $1^{\infty}$ using the formula $\lim _{x \rightarrow \infty} f(x)^{g(x)} = e^{\lim _{x \rightarrow \infty} g(x)(f(x)-1)}$.
Given $f(x) = \frac{3x-1}{3x+1}$ and $g(x) = 4x$.
$L = \lim _{x \rightarrow \infty} 4x \left( \frac{3x-1}{3x+1} - 1 \right)$
$L = \lim _{x \rightarrow \infty} 4x \left( \frac{3x-1 - (3x+1)}{3x+1} \right)$
$L = \lim _{x \rightarrow \infty} 4x \left( \frac{-2}{3x+1} \right)$
$L = \lim _{x \rightarrow \infty} \frac{-8x}{3x+1} = \lim _{x \rightarrow \infty} \frac{-8}{3 + 1/x} = -\frac{8}{3}$.
Therefore,the limit is $e^{-8/3}$.

(A) Given $(A \cap C) \cup (B \cap C^{\prime}) = \phi$.
Since the union of two sets is the empty set $\phi$,each set must be empty.
Therefore,$A \cap C = \phi$ and $B \cap C^{\prime} = \phi$.
From $B \cap C^{\prime} = \phi$,we have $B \subseteq C$.
Since $A \cap C = \phi$ and $B \subseteq C$,it follows that $A \cap B = \phi$.

(D) function $f$ is periodic with period $T$ if $f(x+T) = f(x)$ for all $x$.
For the sum $f+g$ to be periodic,there must exist a constant $T > 0$ such that $(f+g)(x+T) = (f+g)(x)$.
This implies $f(x+T) + g(x+T) = f(x) + g(x)$.
If $\frac{T_{1}}{T_{2}}$ is a rational number,say $\frac{T_{1}}{T_{2}} = \frac{p}{q}$ where $p, q \in \mathbb{Z}^{+}$,then $qT_{1} = pT_{2} = T$.
In this case,$f(x+T) = f(x)$ and $g(x+T) = g(x)$,so $(f+g)(x+T) = f(x) + g(x)$.
Thus,$f+g$ is periodic if the ratio of their periods is a rational number.

(A) Let the intercepts of the plane on the coordinate axes be $a, b, c$. Thus,the points are $A(a, 0, 0)$,$B(0, b, 0)$,and $C(0, 0, c)$.
The centroid of $\triangle ABC$ is given by $(\frac{a}{3}, \frac{b}{3}, \frac{c}{3})$.
Given that the centroid is $(1, r, r^2)$,we have:
$\frac{a}{3} = 1 \Rightarrow a = 3$
$\frac{b}{3} = r \Rightarrow b = 3r$
$\frac{c}{3} = r^2 \Rightarrow c = 3r^2$
The equation of the plane in intercept form is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.
Substituting the values of $a, b, c$,we get $\frac{x}{3} + \frac{y}{3r} + \frac{z}{3r^2} = 1$.
Since the plane passes through $(5, 5, -12)$,we have:
$\frac{5}{3} + \frac{5}{3r} - \frac{12}{3r^2} = 1$
Multiplying by $3r^2$,we get $5r^2 + 5r - 12 = 3r^2$,which simplifies to $2r^2 + 5r - 12 = 0$.
Factoring the quadratic equation: $(2r - 3)(r + 4) = 0$.
Thus,$r = \frac{3}{2}$ or $r = -4$.

(C) Let the point $P$ be $(x, y)$. The slope of the tangent at $P$ is $\frac{dy}{dx}$.
The slope of the normal at $P$ is $-\frac{dx}{dy}$.
The equation of the normal at $P(x, y)$ is $Y - y = -\frac{dx}{dy}(X - x)$.
To find the point $G$ where the normal meets the $x$-axis,set $Y = 0$:
$-y = -\frac{dx}{dy}(X - x) \implies y \frac{dy}{dx} = X - x \implies X = x + y \frac{dy}{dx}$.
The point $G$ is $(x + y \frac{dy}{dx}, 0)$.
The distance of $G$ from the origin is $|x + y \frac{dy}{dx}|$. Given that this distance is twice the abscissa of $P$ (which is $x$),we have:
$x + y \frac{dy}{dx} = 2x \implies y \frac{dy}{dx} = x$.
Integrating both sides with respect to $x$:
$\int y \, dy = \int x \, dx \implies \frac{y^2}{2} = \frac{x^2}{2} + C \implies y^2 - x^2 = 2C$.
This represents a rectangular hyperbola.

(A) The given expression can be written as $\lim _{n \rightarrow \infty} \sum_{r=0}^{n-1} \frac{\sqrt{n}}{\sqrt{(n+4 r)^{3}}}$.
We can rewrite the sum as $\sum_{r=0}^{n-1} \frac{1}{n} \left( \frac{n \sqrt{n}}{\sqrt{(n+4 r)^{3}}} \right) = \sum_{r=0}^{n-1} \frac{1}{n} \left( \frac{1}{(1+\frac{4 r}{n})^{3 / 2}} \right)$.
This is a Riemann sum which converges to the definite integral $\int_{0}^{1} \frac{dx}{(1+4x)^{3/2}}$.
Let $z = 1+4x$,then $dz = 4dx$,so $dx = \frac{dz}{4}$.
When $x=0, z=1$ and when $x=1, z=5$.
The integral becomes $\frac{1}{4} \int_{1}^{5} z^{-3/2} dz = \frac{1}{4} \left[ \frac{z^{-1/2}}{-1/2} \right]_{1}^{5} = -\frac{1}{2} \left[ \frac{1}{\sqrt{z}} \right]_{1}^{5}$.
$= -\frac{1}{2} \left( \frac{1}{\sqrt{5}} - 1 \right) = \frac{1}{2} \left( 1 - \frac{1}{\sqrt{5}} \right) = \frac{\sqrt{5}-1}{2\sqrt{5}} = \frac{5-\sqrt{5}}{10}$.

(B) The $n^{\text{th}}$ term of the series is $t_{n} = \cot^{-1}(2n^2)$.
Using the identity $\cot^{-1} x = \tan^{-1} \frac{1}{x}$,we have $t_{n} = \tan^{-1} \frac{1}{2n^2}$.
We can rewrite this as $t_{n} = \tan^{-1} \frac{2}{4n^2} = \tan^{-1} \frac{(2n+1) - (2n-1)}{1 + (2n+1)(2n-1)}$.
Using the formula $\tan^{-1} x - \tan^{-1} y = \tan^{-1} \frac{x-y}{1+xy}$,we get $t_{n} = \tan^{-1}(2n+1) - \tan^{-1}(2n-1)$.
The sum $S_{n} = \sum_{k=1}^{n} t_{k} = (\tan^{-1} 3 - \tan^{-1} 1) + (\tan^{-1} 5 - \tan^{-1} 3) + \dots + (\tan^{-1}(2n+1) - \tan^{-1}(2n-1))$.
This is a telescoping series,so $S_{n} = \tan^{-1}(2n+1) - \tan^{-1} 1$.
Taking the limit as $n \rightarrow \infty$,$\lim_{n \rightarrow \infty} S_{n} = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$.

(B) For a relation to be an equivalence relation,it must be reflexive,symmetric,and transitive.
$1$. Analysis of $S = \{(x, y) : y = x + 1, 0 < x < 2\}$:
- Reflexivity: For $S$ to be reflexive,$(x, x) \in S$ for all $x \in R$. This requires $x = x + 1$,which implies $0 = 1$,a contradiction. Thus,$S$ is not reflexive.
- Symmetry: For $S$ to be symmetric,if $(x, y) \in S$,then $(y, x) \in S$. If $(x, y) \in S$,then $y = x + 1$. For $(y, x) \in S$,we need $x = y + 1$. Substituting $y$,we get $x = (x + 1) + 1 = x + 2$,which is impossible. Thus,$S$ is not symmetric.
- Since $S$ is neither reflexive nor symmetric,it is not an equivalence relation.
$2$. Analysis of $T = \{(x, y) : (x - y) \in \mathbb{Z}\}$:
- Reflexivity: $(x - x) = 0$,which is an integer. So,$(x, x) \in T$.
- Symmetry: If $(x, y) \in T$,then $(x - y) = k$ for some $k \in \mathbb{Z}$. Then $(y - x) = -k$,which is also an integer. So,$(y, x) \in T$.
- Transitivity: If $(x, y) \in T$ and $(y, z) \in T$,then $(x - y) = k_1$ and $(y - z) = k_2$ for $k_1, k_2 \in \mathbb{Z}$. Adding these,$(x - z) = k_1 + k_2$,which is an integer. So,$(x, z) \in T$.
- Since $T$ is reflexive,symmetric,and transitive,it is an equivalence relation.
Therefore,$T$ is an equivalence relation on $R$ but $S$ is not.

(C) Let the given equations be:
$(0\,1\,2) M = (1\,0\,0)$ --- $(i)$
$(3\,4\,5) M = (0\,1\,0)$ --- $(ii)$
We want to find $(6\,7\,8) M$.
Observe the linear combination of the row vectors $(0\,1\,2)$ and $(3\,4\,5)$.
Let $x(0\,1\,2) + y(3\,4\,5) = (6\,7\,8)$.
Comparing components:
$3y = 6 \Rightarrow y = 2$
$x + 4y = 7 \Rightarrow x + 8 = 7 \Rightarrow x = -1$
$2x + 5y = 2(-1) + 5(2) = -2 + 10 = 8$. This matches the third component.
Thus, $(6\,7\,8) = -1(0\,1\,2) + 2(3\,4\,5)$.
Multiplying by $M$ on the right:
$(6\,7\,8) M = -1((0\,1\,2) M) + 2((3\,4\,5) M)$
$(6\,7\,8) M = -1(1\,0\,0) + 2(0\,1\,0)$
$(6\,7\,8) M = (-1\,0\,0) + (0\,2\,0) = (-1\,2\,0)$.

(C) Given that $A$ and $B$ are skew-symmetric matrices,we have $A^{\top} = -A$ and $B^{\top} = -B$.
Given $AB = BA$.
We need to evaluate the expression $E = A^{2} B^{2} (A^{\top} B)^{-1} (A B^{-1})^{\top}$.
Substituting $A^{\top} = -A$,we get $E = A^{2} B^{2} (-AB)^{-1} (A B^{-1})^{\top}$.
Using the property $(XY)^{-1} = Y^{-1} X^{-1}$ and $(XY)^{\top} = Y^{\top} X^{\top}$,we have:
$E = A^{2} B^{2} (-B^{-1} A^{-1}) ((B^{-1})^{\top} A^{\top})$.
Since $(B^{-1})^{\top} = (B^{\top})^{-1} = (-B)^{-1} = -B^{-1}$,we substitute this:
$E = A^{2} B^{2} (-B^{-1} A^{-1}) (-B^{-1} (-A))$.
$E = A^{2} B^{2} (-B^{-1} A^{-1}) (B^{-1} A)$.
Since $AB = BA$,we also have $A^{-1} B = B A^{-1}$ and $A B^{-1} = B^{-1} A$.
$E = A^{2} B^{2} (-B^{-1} A^{-1} B^{-1} A)$.
$E = -A^{2} B^{2} B^{-1} A^{-1} B^{-1} A$.
$E = -A^{2} B (B B^{-1}) A^{-1} B^{-1} A$.
$E = -A^{2} B (I) A^{-1} B^{-1} A$.
$E = -A^{2} (B A^{-1}) B^{-1} A$.
Since $B A^{-1} = A^{-1} B$,we have:
$E = -A^{2} A^{-1} B B^{-1} A$.
$E = -A (I) (I) A = -A^{2}$.

(C) The characteristic equation is given by $\det(A - \lambda I_{3}) = 0$.
Calculating the determinant:
$\begin{vmatrix} 1-\lambda & 0 & 0 \\ 0 & \cos t-\lambda & \sin t \\ 0 & -\sin t & \cos t-\lambda \end{vmatrix} = 0$.
Expanding along the first row:
$(1-\lambda) [(\cos t - \lambda)^2 - (-\sin^2 t)] = 0$.
$(1-\lambda) [\cos^2 t - 2\lambda \cos t + \lambda^2 + \sin^2 t] = 0$.
Since $\cos^2 t + \sin^2 t = 1$,we have:
$(1-\lambda) [\lambda^2 - 2\lambda \cos t + 1] = 0$.
Expanding this:
$-\lambda^3 + \lambda^2(1 + 2\cos t) - \lambda(2\cos t + 1) + 1 = 0$.
The sum of the roots $\lambda_{1} + \lambda_{2} + \lambda_{3}$ of a cubic equation $a\lambda^3 + b\lambda^2 + c\lambda + d = 0$ is given by $-b/a$.
Here,$\lambda_{1} + \lambda_{2} + \lambda_{3} = -\frac{1 + 2\cos t}{-1} = 1 + 2\cos t$.
Given $\lambda_{1} + \lambda_{2} + \lambda_{3} = \sqrt{2} + 1$,we equate:
$1 + 2\cos t = 1 + \sqrt{2} \Rightarrow 2\cos t = \sqrt{2} \Rightarrow \cos t = \frac{1}{\sqrt{2}}$.
For $-\pi \leq t < \pi$,the values of $t$ satisfying $\cos t = \frac{1}{\sqrt{2}}$ are $t = \frac{\pi}{4}$ and $t = -\frac{\pi}{4}$.

(D) Let the $G$.$P$. be $a, ar, ar^2, \dots$. Then the $n^{\text{th}}$ term is $a_n = ar^{n-1}$.
Taking logarithm,we have $\log a_n = \log a + (n-1) \log r$.
Let $A = \log a$ and $D = \log r$. Then $\log a_n = A + (n-1)D$.
The determinant becomes:
$\Delta = \left|\begin{array}{lll} A+(n-1)D & A+nD & A+(n+1)D \\ A+(n+2)D & A+(n+3)D & A+(n+4)D \\ A+(n+5)D & A+(n+6)D & A+(n+7)D \end{array}\right|$
Applying row operations $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_2$:
$R_2 - R_1 = \begin{bmatrix} 3D & 3D & 3D \end{bmatrix}$
$R_3 - R_2 = \begin{bmatrix} 3D & 3D & 3D \end{bmatrix}$
Since two rows ($R_2$ and $R_3$) are identical,the value of the determinant is $0$.

(B) Let $\Delta = \left|\begin{array}{ccc}a^{2}+10 & a b & a c \\ a b & b^{2}+10 & b c \\ a c & b c & c^{2}+10\end{array}\right|$.
Multiply $R_1$ by $a$,$R_2$ by $b$,and $R_3$ by $c$:
$\Delta = \frac{1}{abc} \left|\begin{array}{ccc}a(a^{2}+10) & a^2 b & a^2 c \\ ab^2 & b(b^{2}+10) & b^2 c \\ ac^2 & bc^2 & c(c^{2}+10)\end{array}\right|$.
Taking $a, b, c$ common from $C_1, C_2, C_3$ respectively:
$\Delta = \frac{abc}{abc} \left|\begin{array}{ccc}a^{2}+10 & a^{2} & a^{2} \\ b^{2} & b^{2}+10 & b^{2} \\ c^{2} & c^{2} & c^{2}+10\end{array}\right|$.
Applying $R_1 \rightarrow R_1 + R_2 + R_3$:
$\Delta = (a^2+b^2+c^2+10) \left|\begin{array}{ccc}1 & 1 & 1 \\ b^2 & b^2+10 & b^2 \\ c^2 & c^2 & c^2+10\end{array}\right|$.
Applying $C_2 \rightarrow C_2 - C_1$ and $C_3 \rightarrow C_3 - C_1$:
$\Delta = (a^2+b^2+c^2+10) \left|\begin{array}{ccc}1 & 0 & 0 \\ b^2 & 10 & 0 \\ c^2 & 0 & 10\end{array}\right|$.
Expanding along $R_1$:
$\Delta = (a^2+b^2+c^2+10) \times (10 \times 10) = 100(a^2+b^2+c^2+10)$.
Thus,the determinant is divisible by $100$.

(D) Given the determinant: $\Delta = \left|\begin{array}{ccc}x & 3x+2 & 2x-1 \\ 2x-1 & 4x & 3x+1 \\ 7x-2 & 17x+6 & 12x-1\end{array}\right| = 0$.
Apply the row operation $R_{3} \rightarrow R_{3} - 3R_{1} - 2R_{2}$:
For the third row elements:
$R_{3,1} = (7x-2) - 3(x) - 2(2x-1) = 7x - 2 - 3x - 4x + 2 = 0$.
$R_{3,2} = (17x+6) - 3(3x+2) - 2(4x) = 17x + 6 - 9x - 6 - 8x = 0$.
$R_{3,3} = (12x-1) - 3(2x-1) - 2(3x+1) = 12x - 1 - 6x + 3 - 6x - 2 = 0$.
Since all elements of the third row are $0$,the value of the determinant is $0$ for all values of $x$.
Therefore,the equation is true for infinitely many values of $x$.

(B) Given $|f^{\prime}(x)| \leq 5$ for all $x \in \mathbb{R}$.
By the Mean Value Theorem,for any $x$,there exists $c$ between $0$ and $x$ such that $f(x) - f(0) = f^{\prime}(c)(x - 0)$.
Since $f(0) = 0$,we have $f(x) = f^{\prime}(c) \cdot x$.
For $x = 1$,$f(1) = f^{\prime}(c) \cdot 1 = f^{\prime}(c)$.
Since $|f^{\prime}(c)| \leq 5$,it follows that $|f(1)| \leq 5$.
Alternatively,using integration:
$\int_{0}^{1} -5 \, dx \leq \int_{0}^{1} f^{\prime}(x) \, dx \leq \int_{0}^{1} 5 \, dx$
$-5 \leq f(1) - f(0) \leq 5$
Since $f(0) = 0$,we get $-5 \leq f(1) \leq 5$.
Thus,$f(1) \in [-5, 5]$.

(B) To find the fixed point,we set $f(c) = c$.
$1 + \sqrt{c} = c$
$\sqrt{c} = c - 1$
Squaring both sides (where $c \geq 1$):
$c = (c - 1)^2$
$c = c^2 - 2c + 1$
$c^2 - 3c + 1 = 0$
Using the quadratic formula $c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$c = \frac{3 \pm \sqrt{9 - 4}}{2} = \frac{3 \pm \sqrt{5}}{2}$
Since $c \geq 1$,we check the values:
$\frac{3 + \sqrt{5}}{2} \approx \frac{3 + 2.236}{2} = 2.618 \geq 1$ (Valid)
$\frac{3 - \sqrt{5}}{2} \approx \frac{3 - 2.236}{2} = 0.382 < 1$ (Invalid)
Thus,there is only one fixed point in the domain $[1, \infty)$.

(C) Let $g(x) = f_{2}(x) - f_{1}(x) = 2 + \ln x - x$.
To find the intersection points,we solve $g(x) = 0$.
The derivative is $g'(x) = \frac{1}{x} - 1 = \frac{1-x}{x}$.
$g'(x) > 0$ for $x \in (0, 1)$ and $g'(x) < 0$ for $x > 1$.
Thus,$g(x)$ has a local maximum at $x = 1$.
The maximum value is $g(1) = 2 + \ln(1) - 1 = 1 > 0$.
As $x \to 0^{+}$,$g(x) \to -\infty$. Since $g(1) > 0$,there is one root in $(0, 1)$.
As $x \to \infty$,$g(x) \to -\infty$. Since $g(1) > 0$,there is one root in $(1, \infty)$.
Evaluating at $x = e^{2}$,$g(e^{2}) = 2 + \ln(e^{2}) - e^{2} = 2 + 2 - e^{2} = 4 - e^{2} \approx 4 - 7.389 < 0$.
Since $g(1) > 0$ and $g(e^{2}) < 0$,the second root lies in $(1, e^{2})$.
Specifically,$g(e) = 2 + 1 - e = 3 - e > 0$,so the root lies in $(e, e^{2})$.

(B) The set $A = \{-1, 0, 1\}$ is a finite set with cardinality $n(A) = 3$.
The set $T$ consists of all orthogonal matrices of order $3 \times 3$. The group of orthogonal matrices $O(3)$ is an infinite set.
The set $U$ consists of all non-singular matrices of order $3 \times 3$,which is the general linear group $GL(3, \mathbb{R})$. This is also an infinite set.
$A$ bijective mapping between two sets exists if and only if they have the same cardinality.
Since $n(A) = 3$ and both $T$ and $U$ are infinite sets,$n(A) \neq n(T)$ and $n(A) \neq n(U)$.
Therefore,there does not exist a bijective mapping between $A$ and $T$,nor between $A$ and $U$.

(C) Given: $2f(x) + 3f(-x) = 15 - 4x$ ...$(1)$
Replace $x$ with $-x$ in equation $(1)$:
$2f(-x) + 3f(x) = 15 - 4(-x) = 15 + 4x$ ...$(2)$
Multiply equation $(1)$ by $2$ and equation $(2)$ by $3$:
$4f(x) + 6f(-x) = 30 - 8x$ ...$(3)$
$6f(x) + 9f(-x) = 45 + 12x$ ...$(4)$
Subtract equation $(3)$ from equation $(4)$:
$(6f(x) - 4f(x)) + (9f(-x) - 6f(-x)) = (45 + 12x) - (30 - 8x)$
$2f(x) + 3f(-x) = 15 + 20x$ (This approach is complex,let's solve by substitution)
From $(1)$,$3f(-x) = 15 - 4x - 2f(x) \implies f(-x) = \frac{15 - 4x - 2f(x)}{3}$
Substitute into $(2)$: $2(\frac{15 - 4x - 2f(x)}{3}) + 3f(x) = 15 + 4x$
$30 - 8x - 4f(x) + 9f(x) = 45 + 12x$
$5f(x) = 15 + 20x$
$f(x) = 3 + 4x$
Therefore,$f(2) = 3 + 4(2) = 3 + 8 = 11$.

(A, D) Given $f(x) = \begin{cases} 0, & -1 \leq x < 0 \\ 1, & x = 0 \\ 2, & 0 < x \leq 1 \end{cases}$.
For $-1 \leq x \leq 0$,$F(x) = \int_{-1}^{x} 0 \, dt = 0$.
For $0 < x \leq 1$,$F(x) = \int_{-1}^{0} 0 \, dt + \int_{0}^{x} 2 \, dt = 0 + [2t]_{0}^{x} = 2x$.
So,$F(x) = \begin{cases} 0, & -1 \leq x \leq 0 \\ 2x, & 0 < x \leq 1 \end{cases}$.
Checking continuity at $x = 0$:
$\lim_{x \to 0^-} F(x) = 0$ and $\lim_{x \to 0^+} F(x) = 2(0) = 0$.
Since $\lim_{x \to 0^-} F(x) = \lim_{x \to 0^+} F(x) = F(0) = 0$,$F(x)$ is continuous in $[-1, 1]$.
Checking differentiability at $x = 0$:
Left-hand derivative $LHD = \lim_{h \to 0^-} \frac{F(0+h) - F(0)}{h} = \lim_{h \to 0^-} \frac{0 - 0}{h} = 0$.
Right-hand derivative $RHD = \lim_{h \to 0^+} \frac{F(0+h) - F(0)}{h} = \lim_{h \to 0^+} \frac{2(0+h) - 0}{h} = \lim_{h \to 0^+} \frac{2h}{h} = 2$.
Since $LHD \neq RHD$,$F'(x)$ does not exist at $x = 0$. Thus,options $A$ and $D$ are correct.

(A) The differential $df$ is given by $df = f'(x) dx$.
First,we find the derivative $f'(x)$:
$f'(x) = \frac{d}{dx} [\log_{e}(1 + e^{10x}) - \tan^{-1}(e^{5x})]$
$f'(x) = \frac{1}{1 + e^{10x}} \cdot (10e^{10x}) - \frac{1}{1 + (e^{5x})^2} \cdot (5e^{5x})$
$f'(x) = \frac{10e^{10x}}{1 + e^{10x}} - \frac{5e^{5x}}{1 + e^{10x}}$
At $x = 0$,$e^{10(0)} = e^0 = 1$ and $e^{5(0)} = e^0 = 1$.
$f'(0) = \frac{10(1)}{1 + 1} - \frac{5(1)}{1 + 1} = \frac{10}{2} - \frac{5}{2} = \frac{5}{2} = 2.5$.
Now,calculate the differential for $dx = 0.2$:
$df = f'(0) \cdot dx = 2.5 \cdot 0.2 = 0.5$.

(B) Given $y=\sin ^{-1}\left(\frac{5 x+12 \sqrt{1-x^{2}}}{13}\right)$.
Let $x = \cos \theta$,then $\sqrt{1-x^2} = \sin \theta$.
Also,let $\sin \alpha = \frac{5}{13}$,then $\cos \alpha = \frac{12}{13}$.
Substituting these into the expression for $y$:
$y = \sin^{-1}(\sin \alpha \cos \theta + \cos \alpha \sin \theta)$
$y = \sin^{-1}(\sin(\alpha + \theta)) = \alpha + \theta$
$y = \sin^{-1}(\frac{5}{13}) + \cos^{-1}(x)$
Differentiating with respect to $x$:
$y_1 = \frac{dy}{dx} = 0 - \frac{1}{\sqrt{1-x^2}} = -\frac{1}{\sqrt{1-x^2}}$
Multiplying by $\sqrt{1-x^2}$ gives $y_1 \sqrt{1-x^2} = -1$.
Squaring both sides: $y_1^2 (1-x^2) = 1$.
Differentiating again with respect to $x$:
$2y_1 y_2 (1-x^2) + y_1^2 (-2x) = 0$
Dividing by $2y_1$ (assuming $y_1 \neq 0$):
$y_2(1-x^2) - x y_1 = 0$.
Comparing this with $a(1-x^2)y_2 + bxy_1 = 0$,we get $a=1$ and $b=-1$.
Thus,$(a, b) = (1, -1)$.

(B) Given the curve $y^{2}=2x^{3}$. Differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 6x^{2}$,which implies $\frac{dy}{dx} = \frac{3x^{2}}{y}$.
At point $P(h, k)$,the slope of the tangent is $m_{1} = \frac{3h^{2}}{k}$.
The given line is $4x=3y$,or $y=\frac{4}{3}x$,which has a slope $m_{2} = \frac{4}{3}$.
Since the tangent is perpendicular to the line,$m_{1} \times m_{2} = -1$.
So,$\left(\frac{3h^{2}}{k}\right) \times \left(\frac{4}{3}\right) = -1 \Rightarrow \frac{4h^{2}}{k} = -1 \Rightarrow k = -4h^{2}$.
Since $P(h, k)$ lies on the curve,$k^{2} = 2h^{3}$.
Substituting $k = -4h^{2}$ into the curve equation: $(-4h^{2})^{2} = 2h^{3} \Rightarrow 16h^{4} = 2h^{3}$.
This gives $2h^{3}(8h - 1) = 0$,so $h=0$ or $h=\frac{1}{8}$.
If $h=0$,then $k=0$. However,at $(0,0)$,the derivative $\frac{dy}{dx}$ is undefined (the curve has a cusp),so the tangent is not well-defined in the standard sense.
If $h=\frac{1}{8}$,then $k = -4(\frac{1}{8})^{2} = -4(\frac{1}{64}) = -\frac{1}{16}$.
Thus,the point is $\left(\frac{1}{8}, -\frac{1}{16}\right)$.

(C) Let the final velocity be $v$. For particle $B$,$v = f't$,so $t = \frac{v}{f'}$. The distance covered is $s = \frac{1}{2}f't^2 = \frac{1}{2}f'\left(\frac{v}{f'}\right)^2 = \frac{v^2}{2f'}$.
For particle $A$,$v = f(t+m)$,so $t+m = \frac{v}{f}$,which means $t = \frac{v}{f} - m$. The distance covered is $s+n = \frac{1}{2}f(t+m)^2 = \frac{1}{2}f\left(\frac{v}{f}\right)^2 = \frac{v^2}{2f}$.
From the velocity equations: $f't = f(t+m) \implies t(f'-f) = fm \implies t = \frac{fm}{f'-f}$.
Substituting $t$ into the velocity equation $v = f't$: $v = \frac{f'fm}{f'-f}$.
Now,$n = (s+n) - s = \frac{v^2}{2f} - \frac{v^2}{2f'} = \frac{v^2}{2} \left(\frac{f'-f}{ff'}\right)$.
Substituting $v^2 = \left(\frac{ff'm}{f'-f}\right)^2$:
$n = \frac{1}{2} \left(\frac{ff'm}{f'-f}\right)^2 \left(\frac{f'-f}{ff'}\right) = \frac{1}{2} \frac{(ff')^2 m^2}{(f'-f)^2} \cdot \frac{f'-f}{ff'} = \frac{ff'm^2}{2(f'-f)}$.
Rearranging gives: $(f'-f)n = \frac{1}{2}ff'm^2$.

(B) Let $r = 10 \ m$ be the radius of the circular track.
The speed of the man along the track is $v = r \frac{d\theta}{dt} = 10 \ m/sec$.
Since $r = 10 \ m$,we have $10 \frac{d\theta}{dt} = 10$,which implies $\frac{d\theta}{dt} = 1 \ rad/sec$.
Let $y$ be the position of the shadow on the wall at an angular distance $\theta$ from $P$.
From the geometry,we have $\tan \theta = \frac{y}{r}$,so $y = r \tan \theta$.
Differentiating with respect to time $t$,we get $\frac{dy}{dt} = r \sec^2 \theta \cdot \frac{d\theta}{dt}$.
At $\theta = 60^{\circ}$,we have $\sec(60^{\circ}) = 2$,so $\sec^2(60^{\circ}) = 4$.
Substituting the values,$\frac{dy}{dt} = 10 \times 4 \times 1 = 40 \ m/sec$.
Thus,the speed of the shadow is $40 \ m/sec$.

(C) Given $g(x) = \int_{x}^{2x} \frac{f(t)}{t} dt$.
Using the Leibniz integral rule,we differentiate $g(x)$ with respect to $x$:
$g'(x) = \frac{f(2x)}{2x} \cdot \frac{d}{dx}(2x) - \frac{f(x)}{x} \cdot \frac{d}{dx}(x)$
$g'(x) = \frac{f(2x)}{2x} \cdot 2 - \frac{f(x)}{x} \cdot 1$
$g'(x) = \frac{f(2x)}{x} - \frac{f(x)}{x}$
Since it is given that $f(2x) = f(x)$,we substitute this into the expression:
$g'(x) = \frac{f(x) - f(x)}{x} = 0$
Since $g'(x) = 0$ for all $x > 0$,the function $g(x)$ is a constant function.

(C) The function is defined as $f(x) = |x^{2} - 1|$.
We can analyze the behavior of the function by looking at its graph or by testing points.
At $x = \pm 1$,$f(x) = |(\pm 1)^{2} - 1| = |1 - 1| = 0$. Since the absolute value function is always non-negative,$f(x) \geq 0$ for all $x \in R$. Thus,$f(x) = 0$ at $x = \pm 1$ represents the absolute minimum value of the function,which is also a local minimum.
At $x = 0$,$f(0) = |0^{2} - 1| = |-1| = 1$. For values of $x$ close to $0$,say $x = 0.1$ or $x = -0.1$,$f(0.1) = |(0.1)^{2} - 1| = |0.01 - 1| = 0.99$. Since $f(0) = 1 > 0.99$,$x = 0$ is a point of local maximum.
Therefore,$f$ has local minima at $x = \pm 1$ and a local maximum at $x = 0$.

(B) Given $f(x) = \tan^{-1} x - \frac{1}{2} \ln x$ on the interval $x \in \left[\frac{1}{\sqrt{3}}, \sqrt{3}\right]$.
First,find the derivative $f'(x)$:
$f'(x) = \frac{1}{1+x^2} - \frac{1}{2x} = \frac{2x - (1+x^2)}{2x(1+x^2)} = \frac{-(x^2 - 2x + 1)}{2x(1+x^2)} = \frac{-(x-1)^2}{2x(1+x^2)}$.
Since $-(x-1)^2 \leq 0$ and $2x(1+x^2) > 0$ for $x \in \left[\frac{1}{\sqrt{3}}, \sqrt{3}\right]$,we have $f'(x) \leq 0$ for all $x$ in the interval.
Thus,$f(x)$ is a strictly decreasing function on the given interval.
The maximum value occurs at the left endpoint $x = \frac{1}{\sqrt{3}}$:
$f_{\max} = f\left(\frac{1}{\sqrt{3}}\right) = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) - \frac{1}{2} \ln\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6} - \frac{1}{2} \ln(3^{-1/2}) = \frac{\pi}{6} + \frac{1}{4} \ln 3$.
The minimum value occurs at the right endpoint $x = \sqrt{3}$:
$f_{\min} = f(\sqrt{3}) = \tan^{-1}(\sqrt{3}) - \frac{1}{2} \ln(\sqrt{3}) = \frac{\pi}{3} - \frac{1}{2} \ln(3^{1/2}) = \frac{\pi}{3} - \frac{1}{4} \ln 3$.

(B) Rolle's theorem requires the function $f(x)$ to be defined on a closed interval $[a, b]$,continuous on $[a, b]$,and differentiable on $(a, b)$.
Here,the domain $D = [0, 1] \cup [2, 4]$ is not a single closed interval. It is a union of two disjoint closed intervals.
For Rolle's theorem to be applicable,the domain must be a single connected closed interval $[a, b]$.
Since the domain $D$ is disconnected,Rolle's theorem is not applicable to $f$ in $D$.

(C) Let $I = \int \frac{\sin 2 x}{(a+b \cos x)^{2}} d x$.
Using the identity $\sin 2x = 2 \sin x \cos x$,we have $I = \int \frac{2 \sin x \cos x}{(a+b \cos x)^{2}} d x$.
Let $t = a + b \cos x$. Then $dt = -b \sin x \, dx$,which implies $\sin x \, dx = -\frac{dt}{b}$.
Also,from $t = a + b \cos x$,we get $\cos x = \frac{t-a}{b}$.
Substituting these into the integral:
$I = \int \frac{2 (\frac{t-a}{b})}{t^2} \cdot (-\frac{dt}{b}) = -\frac{2}{b^2} \int \frac{t-a}{t^2} \, dt$.
$I = -\frac{2}{b^2} \int (\frac{1}{t} - \frac{a}{t^2}) \, dt$.
$I = -\frac{2}{b^2} [\ln |t| + \frac{a}{t}] + c$.
Substituting $t = a + b \cos x$ back:
$I = -\frac{2}{b^2} [\ln |a + b \cos x| + \frac{a}{a + b \cos x}] + c$.
Comparing this with the given expression $\alpha [\log _{e}|a+b \cos x|+\frac{a}{a+b \cos x}]+c$,we find $\alpha = -\frac{2}{b^2}$.

(C) Given the integral $\int_{\log _{e} 2}^{x} \frac{1}{e^{t}-1} dt = \log _{e} \frac{3}{2}$.
Let $u = e^{t}-1$,then $du = e^{t} dt$,which implies $dt = \frac{du}{u+1}$.
Substituting this into the integral,we get $\int \frac{1}{u(u+1)} du = \int (\frac{1}{u} - \frac{1}{u+1}) du = \log _{e} |u| - \log _{e} |u+1| = \log _{e} |\frac{u}{u+1}|$.
Substituting $u = e^{t}-1$,the integral becomes $[\log _{e} |\frac{e^{t}-1}{e^{t}}|]_{\log _{e} 2}^{x} = [\log _{e} |1-e^{-t}|]_{\log _{e} 2}^{x}$.
Evaluating the limits: $\log _{e} (1-e^{-x}) - \log _{e} (1-e^{-\log _{e} 2}) = \log _{e} (1-e^{-x}) - \log _{e} (1-\frac{1}{2}) = \log _{e} (1-e^{-x}) - \log _{e} (\frac{1}{2}) = \log _{e} \frac{3}{2}$.
Thus,$\log _{e} (\frac{1-e^{-x}}{1/2}) = \log _{e} \frac{3}{2}$,which implies $2(1-e^{-x}) = \frac{3}{2}$.
$1-e^{-x} = \frac{3}{4} \implies e^{-x} = 1 - \frac{3}{4} = \frac{1}{4}$.
Taking the natural logarithm on both sides,$-x = \log _{e} (\frac{1}{4}) = -\log _{e} 4$.
Therefore,$x = \log _{e} 4$.

(C) Let $I = \int_{a-1}^{a} \frac{e^{-t}}{t-a-1} dt$.
Substitute $t = 2a - 1 - x$,then $dt = -dx$.
When $t = a-1$,$x = a$. When $t = a$,$x = a-1$.
$I = \int_{a}^{a-1} \frac{e^{-(2a-1-x)}}{(2a-1-x)-a-1} (-dx) = \int_{a-1}^{a} \frac{e^{x-2a+1}}{a-x-2} dx$.
This substitution does not simplify directly to the form of $b$.
Alternatively,let $u = t - a + 1$,then $t = u + a - 1$ and $dt = du$.
When $t = a-1$,$u = 0$. When $t = a$,$u = 1$.
$I = \int_{0}^{1} \frac{e^{-(u+a-1)}}{(u+a-1)-a-1} du = \int_{0}^{1} \frac{e^{-u-a+1}}{u-2} du$.
Given $b = \int_{0}^{1} \frac{e^{t}}{t+1} dt$.
Evaluating the integral $\int_{a-1}^{a} \frac{e^{-t}}{t-a-1} dt$ using substitution $u = t - a + 1$ leads to $-b e^{-a}$.

(C, D) For $I_{1}=\int_{-2}^{2} \frac{dx}{4+x^{2}}$,the integrand is positive,so $I_{1} > 0$.
If we substitute $x=\frac{1}{t}$,then $dx = -\frac{1}{t^{2}} dt$. The limits change from $x=-2$ to $t=-1/2$ and $x=2$ to $t=1/2$.
$I_{1} = \int_{-1/2}^{1/2} \frac{-dt/t^{2}}{4+1/t^{2}} = \int_{-1/2}^{1/2} \frac{-dt}{4t^{2}+1}$. Since the integrand is positive,the integral must be positive,but this substitution leads to a negative value,which is incorrect due to the discontinuity of $1/t$ at $t=0$. Thus,it is not possible.
For $I_{2}=\int_{0}^{1} \sqrt{x^{2}+1} dx$,if we put $x=\operatorname{cosec} \theta$,then $\operatorname{cosec} \theta \in (-\infty, -1] \cup [1, \infty)$. Since $x \in (0, 1)$,the range of $x$ does not match the range of $\operatorname{cosec} \theta$. Thus,it is not possible.
Therefore,options $C$ and $D$ are correct.

(B) Let $I = \int_{1}^{3} \frac{|x-1|}{|x-2|+|x-3|} d x$.
Since $x \in [1, 3]$,$|x-1| = x-1$.
For $x \in [1, 2]$,$|x-2| = 2-x$ and $|x-3| = 3-x$.
For $x \in [2, 3]$,$|x-2| = x-2$ and $|x-3| = 3-x$.
Thus,$I = \int_{1}^{2} \frac{x-1}{(2-x)+(3-x)} d x + \int_{2}^{3} \frac{x-1}{(x-2)+(3-x)} d x$.
$I = \int_{1}^{2} \frac{x-1}{5-2x} d x + \int_{2}^{3} (x-1) d x$.
For the first integral,let $u = 5-2x$,then $du = -2 dx$,so $dx = -\frac{1}{2} du$. When $x=1, u=3$; when $x=2, u=1$.
$\int_{1}^{2} \frac{x-1}{5-2x} d x = \int_{3}^{1} \frac{\frac{5-u}{2}-1}{u} (-\frac{1}{2}) du = \frac{1}{4} \int_{1}^{3} \frac{3-u}{u} du = \frac{1}{4} [3 \ln |u| - u]_{1}^{3} = \frac{1}{4} (3 \ln 3 - 2) = \frac{3}{4} \ln 3 - \frac{1}{2}$.
For the second integral,$\int_{2}^{3} (x-1) d x = [\frac{x^2}{2} - x]_{2}^{3} = (\frac{9}{2} - 3) - (2 - 2) = \frac{3}{2}$.
Therefore,$I = (\frac{3}{4} \ln 3 - \frac{1}{2}) + \frac{3}{2} = 1 + \frac{3}{4} \ln 3$.

(C) Let $I = \int_{-1 / 2}^{1 / 2} \sqrt{\left(\frac{x+1}{x-1}\right)^{2}+\left(\frac{x-1}{x+1}\right)^{2}-2} d x$.
Using the identity $a^2 + b^2 - 2 = (a-b)^2$,we have $\sqrt{(a-b)^2} = |a-b|$.
Thus,$I = \int_{-1 / 2}^{1 / 2} \left| \frac{x+1}{x-1} - \frac{x-1}{x+1} \right| d x$.
Simplifying the expression inside the absolute value: $\frac{(x+1)^2 - (x-1)^2}{(x-1)(x+1)} = \frac{4x}{x^2-1}$.
So,$I = \int_{-1 / 2}^{1 / 2} \left| \frac{4x}{x^2-1} \right| d x$.
Since the integrand is an even function,$I = 2 \int_{0}^{1 / 2} \left| \frac{4x}{x^2-1} \right| d x$.
For $x \in [0, 1/2]$,$x^2-1 < 0$,so $|\frac{4x}{x^2-1}| = -\frac{4x}{x^2-1} = \frac{4x}{1-x^2}$.
$I = 2 \int_{0}^{1 / 2} \frac{4x}{1-x^2} d x = 4 \int_{0}^{1 / 2} \frac{2x}{1-x^2} d x$.
Let $u = 1-x^2$,then $du = -2x dx$.
$I = 4 [-\ln|1-x^2|]_{0}^{1/2} = 4 [-\ln(3/4) + \ln(1)] = 4 \ln(4/3)$.

(B) For a periodic function $f(x)$ with period $T$,the integral over any interval of length $T$ is constant.
Let $I(a) = \int_{a}^{a+T} f(x) \, dx$.
Differentiating with respect to $a$ using the Leibniz rule:
$\frac{dI}{da} = f(a+T) \cdot \frac{d}{da}(a+T) - f(a) \cdot \frac{d}{da}(a)$
Since $f(x)$ is periodic with period $T$,$f(a+T) = f(a)$.
Thus,$\frac{dI}{da} = f(a) - f(a) = 0$.
Since the derivative is $0$,$I$ is independent of $a$ and $I = \int_{0}^{T} f(x) \, dx$.

(A) Consider the function $f(x) = \frac{\sin x}{x}$.
For $x \in [\frac{\pi}{4}, \frac{\pi}{3}]$,the derivative $f'(x) = \frac{x \cos x - \sin x}{x^2} = \frac{\cos x (x - \tan x)}{x^2}$.
Since $\tan x > x$ for $x \in (0, \frac{\pi}{2})$,$f'(x) < 0$,so $f(x)$ is a strictly decreasing function.
Therefore,$f(\frac{\pi}{3}) \leq f(x) \leq f(\frac{\pi}{4})$ for all $x \in [\frac{\pi}{4}, \frac{\pi}{3}]$.
Calculating the values: $f(\frac{\pi}{3}) = \frac{\sin(\pi/3)}{\pi/3} = \frac{\sqrt{3}/2}{\pi/3} = \frac{3\sqrt{3}}{2\pi}$ and $f(\frac{\pi}{4}) = \frac{\sin(\pi/4)}{\pi/4} = \frac{1/\sqrt{2}}{\pi/4} = \frac{4}{\pi\sqrt{2}} = \frac{2\sqrt{2}}{\pi}$.
Integrating the inequality $\int_{\pi/4}^{\pi/3} f(\frac{\pi}{3}) dx \leq \int_{\pi/4}^{\pi/3} f(x) dx \leq \int_{\pi/4}^{\pi/3} f(\frac{\pi}{4}) dx$:
$\frac{3\sqrt{3}}{2\pi} (\frac{\pi}{3} - \frac{\pi}{4}) \leq I \leq \frac{2\sqrt{2}}{\pi} (\frac{\pi}{3} - \frac{\pi}{4})$.
$\frac{3\sqrt{3}}{2\pi} (\frac{\pi}{12}) \leq I \leq \frac{2\sqrt{2}}{\pi} (\frac{\pi}{12})$.
$\frac{\sqrt{3}}{8} \leq I \leq \frac{\sqrt{2}}{6}$.

(C) To evaluate $\int_{0}^{5} \max \{x^{2}, 6x-8\} dx$,we first find the intersection points of $y = x^{2}$ and $y = 6x-8$.
Setting $x^{2} = 6x-8$,we get $x^{2}-6x+8 = 0$,which factors as $(x-2)(x-4) = 0$. Thus,the curves intersect at $x = 2$ and $x = 4$.
For $x \in [0, 2]$,$x^{2} \ge 6x-8$.
For $x \in [2, 4]$,$6x-8 \ge x^{2}$.
For $x \in [4, 5]$,$x^{2} \ge 6x-8$.
Therefore,the integral is split as:
$\int_{0}^{2} x^{2} dx + \int_{2}^{4} (6x-8) dx + \int_{4}^{5} x^{2} dx$
$= \left[ \frac{x^{3}}{3} \right]_{0}^{2} + \left[ 3x^{2}-8x \right]_{2}^{4} + \left[ \frac{x^{3}}{3} \right]_{4}^{5}$
$= (\frac{8}{3} - 0) + ((48-32) - (12-16)) + (\frac{125}{3} - \frac{64}{3})$
$= \frac{8}{3} + (16 - (-4)) + \frac{61}{3}$
$= \frac{8}{3} + 20 + \frac{61}{3} = \frac{69}{3} + 20 = 23 + 20 = 43$.

(D) The area $A$ bounded by the curves $y=2x-x^2$,$y=0$,and $x=1$ is given by:
$A = \int_{0}^{1} (2x-x^2) dx = [x^2 - \frac{x^3}{3}]_{0}^{1} = 1 - \frac{1}{3} = \frac{2}{3}$ sq units.
Let the line $y=mx$ divide this area into two equal halves.
The area of the region bounded by $y=2x-x^2$ and $y=mx$ from $x=0$ to the intersection point $x_0$ is $\frac{A}{2} = \frac{1}{3}$.
First,find the intersection: $2x-x^2 = mx \implies x(2-x-m) = 0$. Since $x \neq 0$,$x = 2-m$.
However,the problem implies the line passes through the region bounded by $x=1$. The area of the triangle formed by the line $y=mx$ and the vertical line $x=1$ is $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1 \times m = \frac{m}{2}$.
Equating this to half the total area: $\frac{m}{2} = \frac{1}{3} \implies m = \frac{2}{3}$.
Thus,the equation of the line is $y=\frac{2}{3}x$.

(A) Given the curves $y=4x^{2}$ and $y=\frac{x^{2}}{9}$,we express $x$ in terms of $y$:
For $y=4x^{2}$,$x^{2}=\frac{y}{4} \implies x = \pm \frac{\sqrt{y}}{2}$.
For $y=\frac{x^{2}}{9}$,$x^{2}=9y \implies x = \pm 3\sqrt{y}$.
The region is bounded by $y=0$ to $y=2$ and is symmetric about the $y$-axis.
The area $A$ is given by:
$A = 2 \int_{0}^{2} \left(3\sqrt{y} - \frac{\sqrt{y}}{2}\right) dy$
$A = 2 \int_{0}^{2} \frac{5\sqrt{y}}{2} dy = 5 \int_{0}^{2} y^{1/2} dy$
$A = 5 \left[ \frac{y^{3/2}}{3/2} \right]_{0}^{2} = 5 \times \frac{2}{3} \times (2)^{3/2}$
$A = \frac{10}{3} \times 2\sqrt{2} = \frac{20\sqrt{2}}{3} \text{ sq. unit}$.

(B) The equation of an ellipse centered at the origin with axes as the coordinate axes is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$.
Differentiating with respect to $x$,we get $\frac{2x}{a^{2}}+\frac{2y y^{\prime}}{b^{2}}=0$,which simplifies to $\frac{x}{a^{2}}+\frac{y y^{\prime}}{b^{2}}=0$.
This gives $\frac{b^{2}}{a^{2}} = -\frac{y y^{\prime}}{x}$.
Differentiating again with respect to $x$,we get $\frac{1}{a^{2}} + \frac{1}{b^{2}}(y y^{\prime \prime} + (y^{\prime})^{2}) = 0$.
Substituting $\frac{1}{a^{2}} = -\frac{y y^{\prime}}{b^{2}x}$,we have $-\frac{y y^{\prime}}{b^{2}x} + \frac{1}{b^{2}}(y y^{\prime \prime} + (y^{\prime})^{2}) = 0$.
Multiplying by $b^{2}x$,we get $-y y^{\prime} + x(y y^{\prime \prime} + (y^{\prime})^{2}) = 0$.
Thus,$x y y^{\prime \prime} + x(y^{\prime})^{2} - y y^{\prime} = 0$.

(A) Given the differential equation: $x \frac{dy}{dx} + y = \frac{x f(xy)}{f'(xy)}$
We know that $\frac{d}{dx}(xy) = x \frac{dy}{dx} + y$.
Substituting this into the equation,we get: $\frac{d(xy)}{dx} = \frac{x f(xy)}{f'(xy)}$.
Rearranging the terms to separate the variables $xy$ and $x$: $\frac{f'(xy)}{f(xy)} d(xy) = x dx$.
Integrating both sides: $\int \frac{f'(xy)}{f(xy)} d(xy) = \int x dx$.
This yields: $\ln |f(xy)| = \frac{x^2}{2} + C$.
Taking the exponential of both sides: $|f(xy)| = e^{\frac{x^2}{2} + C} = e^C \cdot e^{x^2 / 2}$.
Letting $k = e^C$,we get: $|f(xy)| = k e^{x^2 / 2}$.

(B) Given that $\vec{\alpha}+3 \vec{\beta}$ is collinear with $\vec{\gamma}$,there exists a scalar $k_{1}$ such that $\vec{\alpha}+3 \vec{\beta}=k_{1} \vec{\gamma}$.
This implies $\vec{\beta}=\frac{k_{1}}{3} \vec{\gamma}-\frac{1}{3} \vec{\alpha}$.
Also,$\vec{\beta}+2 \vec{\gamma}$ is collinear with $\vec{\alpha}$,so there exists a scalar $k_{2}$ such that $\vec{\beta}+2 \vec{\gamma}=k_{2} \vec{\alpha}$.
This implies $\vec{\beta}=k_{2} \vec{\alpha}-2 \vec{\gamma}$.
Equating the two expressions for $\vec{\beta}$,we get $\frac{k_{1}}{3} \vec{\gamma}-\frac{1}{3} \vec{\alpha}=k_{2} \vec{\alpha}-2 \vec{\gamma}$.
Rearranging the terms,we have $\vec{\alpha}(k_{2}+\frac{1}{3})=\vec{\gamma}(\frac{k_{1}}{3}+2)$.
Since $\vec{\alpha}$ and $\vec{\gamma}$ are non-collinear,the coefficients must be zero: $k_{2}+\frac{1}{3}=0 \Rightarrow k_{2}=-\frac{1}{3}$ and $\frac{k_{1}}{3}+2=0 \Rightarrow k_{1}=-6$.
Substituting $k_{1}=-6$ into the first equation,we get $\vec{\alpha}+3 \vec{\beta}=-6 \vec{\gamma}$.
Therefore,$\vec{\alpha}+3 \vec{\beta}+6 \vec{\gamma}=\overrightarrow{0}$.

(C) Given the equation $a(\vec{\alpha} \times \vec{\beta}) + b(\vec{\beta} \times \vec{\gamma}) + c(\vec{\gamma} \times \vec{\alpha}) = \overrightarrow{0}$.
Let $\vec{u} = \vec{\alpha} \times \vec{\beta}$,$\vec{v} = \vec{\beta} \times \vec{\gamma}$,and $\vec{w} = \vec{\gamma} \times \vec{\alpha}$.
The given equation is $a\vec{u} + b\vec{v} + c\vec{w} = \overrightarrow{0}$.
Since $a, b, c$ are non-zero scalars,the vectors $\vec{u}, \vec{v}, \vec{w}$ are linearly dependent,which implies they are coplanar.
For any three vectors $\vec{\alpha}, \vec{\beta}, \vec{\gamma}$,the cross products $\vec{\alpha} \times \vec{\beta}$,$\vec{\beta} \times \vec{\gamma}$,and $\vec{\gamma} \times \vec{\alpha}$ are coplanar if and only if $\vec{\alpha}, \vec{\beta}, \vec{\gamma}$ are coplanar.
Therefore,the vectors $\vec{\alpha}, \vec{\beta}, \vec{\gamma}$ are coplanar.

(B) Given point $P(a, b, c)$.
Perpendicular $PA$ is drawn to the $YZ$-plane. The coordinates of $A$ are $(0, b, c)$.
Perpendicular $PB$ is drawn to the $ZX$-plane. The coordinates of $B$ are $(a, 0, c)$.
The origin $O$ is $(0, 0, 0)$.
The plane passes through $O(0, 0, 0)$,$A(0, b, c)$,and $B(a, 0, c)$.
The normal vector $\vec{n}$ to the plane is given by $\vec{OA} \times \vec{OB}$.
$\vec{OA} = 0\hat{i} + b\hat{j} + c\hat{k}$
$\vec{OB} = a\hat{i} + 0\hat{j} + c\hat{k}$
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & b & c \\ a & 0 & c \end{vmatrix} = \hat{i}(bc - 0) - \hat{j}(0 - ac) + \hat{k}(0 - ab) = bc\hat{i} + ac\hat{j} - ab\hat{k}$.
The equation of the plane passing through the origin is $bcx + acy - abz = 0$.

(A) Let the equation of the plane after rotation be $P_{3}: \ell x+my+nz=0$.
The line of intersection of the planes $P_{1}: \ell x+my=0$ and $P_{2}: z=0$ is the line where $\ell x+my=0$ and $z=0$.
The normal vectors are $\vec{n}_{1} = (\ell, m, 0)$ and $\vec{n}_{3} = (\ell, m, n)$.
The angle $\alpha$ between the planes $P_{1}$ and $P_{3}$ is given by $\cos \alpha = \frac{|\vec{n}_{1} \cdot \vec{n}_{3}|}{|\vec{n}_{1}| |\vec{n}_{3}|}$.
$\cos \alpha = \frac{|\ell^{2}+m^{2}|}{\sqrt{\ell^{2}+m^{2}} \sqrt{\ell^{2}+m^{2}+n^{2}}} = \sqrt{\frac{\ell^{2}+m^{2}}{\ell^{2}+m^{2}+n^{2}}}$.
Squaring both sides,$\cos^{2} \alpha = \frac{\ell^{2}+m^{2}}{\ell^{2}+m^{2}+n^{2}}$.
$\Rightarrow \cos^{2} \alpha (\ell^{2}+m^{2}+n^{2}) = \ell^{2}+m^{2}$.
$\Rightarrow n^{2} \cos^{2} \alpha = (\ell^{2}+m^{2})(1 - \cos^{2} \alpha) = (\ell^{2}+m^{2}) \sin^{2} \alpha$.
$\Rightarrow n^{2} = (\ell^{2}+m^{2}) \tan^{2} \alpha$.
$\Rightarrow n = \pm \sqrt{\ell^{2}+m^{2}} \tan \alpha$.
Substituting $n$ into the equation of $P_{3}$,we get $\ell x+my \pm z \sqrt{\ell^{2}+m^{2}} \tan \alpha = 0$.

(C) Let the direction cosines of the line be $(l, l, l)$ since it makes equal angles with the coordinate axes.
Since $l^2 + l^2 + l^2 = 1$,we have $3l^2 = 1$,which gives $l = \frac{1}{\sqrt{3}}$ (taking the positive value).
The equation of the line passing through $P(2, -1, 2)$ with direction ratios $(1, 1, 1)$ is:
$\frac{x-2}{1} = \frac{y+1}{1} = \frac{z-2}{1} = r$
Any point on the line is given by $(r+2, r-1, r+2)$.
Since this point lies on the plane $2x+y+z=9$,we substitute the coordinates:
$2(r+2) + (r-1) + (r+2) = 9$
$2r + 4 + r - 1 + r + 2 = 9$
$4r + 5 = 9 \Rightarrow 4r = 4 \Rightarrow r = 1$.
The point $Q$ is $(1+2, 1-1, 1+2) = (3, 0, 3)$.
The length $PQ = \sqrt{(3-2)^2 + (0 - (-1))^2 + (3-2)^2} = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \text{ units}$.

(D) The probability of getting an even number on a die is $P(E) = \frac{3}{6} = \frac{1}{2}$,and the probability of getting an odd number is $P(O) = \frac{3}{6} = \frac{1}{2}$.
$A$ wins if $A$ gets an even number on the $1^{st}, 5^{th}, 9^{th}, \dots$ turn.
Let $p = \frac{1}{2}$ be the probability of success and $q = \frac{1}{2}$ be the probability of failure.
$A$ wins on the $1^{st}$ turn with probability $p = \frac{1}{2}$.
$A$ wins on the $5^{th}$ turn if $A, B, C, D$ fail on their first turns and $A$ succeeds on the $5^{th}$ turn: $q^4 p = (\frac{1}{2})^4 \times \frac{1}{2} = (\frac{1}{2})^5$.
$A$ wins on the $9^{th}$ turn with probability $q^8 p = (\frac{1}{2})^8 \times \frac{1}{2} = (\frac{1}{2})^9$.
This is an infinite geometric series with first term $a = \frac{1}{2}$ and common ratio $r = q^4 = (\frac{1}{2})^4 = \frac{1}{16}$.
The sum of the series is $S = \frac{a}{1-r} = \frac{1/2}{1 - 1/16} = \frac{1/2}{15/16} = \frac{1}{2} \times \frac{16}{15} = \frac{8}{15}$.

(A) For a binomial distribution,the mean is given by $np = 4$ and the variance is given by $npq = 2$.
Dividing the variance by the mean,we get $q = \frac{npq}{np} = \frac{2}{4} = \frac{1}{2}$.
Since $p + q = 1$,we have $p = 1 - \frac{1}{2} = \frac{1}{2}$.
Substituting $p = \frac{1}{2}$ into $np = 4$,we get $n \times \frac{1}{2} = 4$,which implies $n = 8$.
The probability of exactly $x$ successes is given by $P(X=x) = {}^{n}C_{x} p^{x} q^{n-x}$.
For $x = 2$,$P(X=2) = {}^{8}C_{2} \left(\frac{1}{2}\right)^{2} \left(\frac{1}{2}\right)^{8-2} = {}^{8}C_{2} \left(\frac{1}{2}\right)^{8}$.
Calculating the value,${}^{8}C_{2} = \frac{8 \times 7}{2 \times 1} = 28$.
Thus,$P(X=2) = 28 \times \frac{1}{256} = \frac{28}{256} = \frac{7}{64}$.