WBJEE 2020 Mathematics Question Paper with Answer and Solution

(A, D) The total time of flight $T$ is given as $12 \text{ seconds}$.
For a particle projected vertically upwards,the time to reach the maximum height is $t = \frac{T}{2} = \frac{12}{2} = 6 \text{ seconds}$.
At maximum height,the final velocity $v = 0$.
Using the equation $v = u - gt$,where $g = 32 \text{ ft/sec}^2$:
$0 = u - (32)(6) \Rightarrow u = 192 \text{ ft/sec}$.
Now,the maximum height $H$ is given by $H = ut - \frac{1}{2}gt^2$:
$H = (192)(6) - \frac{1}{2}(32)(6)^2 = 1152 - 576 = 576 \text{ ft}$.
Thus,the velocity of projection is $192 \text{ ft/sec}$ and the greatest height attained is $576 \text{ ft}$.
Both options $A$ and $D$ are correct.

(A) Given the equation: $2 \log (x+1)-\log (x^{2}-1)=\log 2$.
Using the property $n \log a = \log (a^n)$,we get $\log (x+1)^2 - \log (x^2-1) = \log 2$.
Using the property $\log a - \log b = \log (a/b)$,we get $\log \left( \frac{(x+1)^2}{x^2-1} \right) = \log 2$.
Since $x^2-1 = (x-1)(x+1)$,the equation becomes $\log \left( \frac{(x+1)^2}{(x-1)(x+1)} \right) = \log 2$.
Simplifying the fraction,we have $\log \left( \frac{x+1}{x-1} \right) = \log 2$.
Therefore,$\frac{x+1}{x-1} = 2$.
$x+1 = 2(x-1) \Rightarrow x+1 = 2x-2$.
$x = 3$.
For the logarithmic terms to be defined,we require $x+1 > 0$ and $x^2-1 > 0$,which implies $x > 1$.
Thus,$x = 3$ is the only valid solution.

(A) Taking $\log _{3}$ on both sides,we get:
$(\log _{3} x)^{2}-\frac{9}{2} \log _{3} x+5 = \log _{3} (3 \sqrt{3}) = \log _{3} (3^{3/2}) = \frac{3}{2}$.
Let $t = \log _{3} x$. Then the equation becomes:
$t^{2} - \frac{9}{2} t + 5 = \frac{3}{2}$.
Multiplying by $2$,we get $2t^{2} - 9t + 10 = 3$,which simplifies to $2t^{2} - 9t + 7 = 0$.
Factoring the quadratic: $(2t - 7)(t - 1) = 0$.
Thus,$t = 1$ or $t = \frac{7}{2}$.
For $t = 1$,$\log _{3} x = 1 \Rightarrow x = 3^{1} = 3$.
For $t = \frac{7}{2}$,$\log _{3} x = \frac{7}{2} \Rightarrow x = 3^{7/2} = 27\sqrt{3}$.
Both roots are real. Therefore,the equation has at least one real root.

(A) Given $P(x) = ax^{2} + bx + c$ and $Q(x) = -ax^{2} + dx + c$.
For $P(x) = 0$,the discriminant is $D_{1} = b^{2} - 4ac$.
For $Q(x) = 0$,the discriminant is $D_{2} = d^{2} - 4(-a)(c) = d^{2} + 4ac$.
Adding the two discriminants,we get $D_{1} + D_{2} = b^{2} + d^{2} \geq 0$.
Since the sum of the discriminants is non-negative,at least one of the discriminants must be non-negative ($D_{1} \geq 0$ or $D_{2} \geq 0$).
If $D_{1} \geq 0$,$P(x)$ has real roots. If $D_{2} \geq 0$,$Q(x)$ has real roots.
Therefore,the equation $P(x) \cdot Q(x) = 0$ must have at least two real roots.

(C) For a quadratic expression $f(x) = ax^{2} + bx + c$ to have the same sign as $a$ for all $x$,the parabola must not intersect the $x$-axis.
This implies that the quadratic equation $ax^{2} + bx + c = 0$ has no real roots.
For no real roots,the discriminant $D = b^{2} - 4ac$ must be less than $0$.
Therefore,the condition is $b^{2} - 4ac < 0$.

(C) Let $p = e^{i\theta} = \cos \theta + i \sin \theta$,where $|p| = 1$.
Then $p^{4} = e^{i4\theta} = \cos(4\theta) + i \sin(4\theta)$.
The imaginary part of $p^{4}$ is $\sin(4\theta)$.
We are given that $\text{Im}(p^{4}) = 0$,so $\sin(4\theta) = 0$.
This implies $4\theta = n\pi$ for any integer $n$,or $\theta = \frac{n\pi}{4}$.
For $p$ to be distinct values on the unit circle,we consider $\theta \in [0, 2\pi)$.
The possible values for $\theta$ are $0, \frac{\pi}{4}, \frac{2\pi}{4}, \frac{3\pi}{4}, \frac{4\pi}{4}, \frac{5\pi}{4}, \frac{6\pi}{4}, \frac{7\pi}{4}$.
There are $8$ such values of $\theta$,corresponding to $8$ distinct complex numbers $p$.

(C) The roots of the equation $z^{2}+pz+q=0$ are $z_{1}$ and $z_{2}$.
For the points $z_{1}, z_{2}$ and the origin $(0)$ to form an equilateral triangle in the Argand plane,the condition is $z_{1}^{2}+z_{2}^{2}+0^{2} = z_{1}z_{2} + z_{2}(0) + (0)z_{1}$,which simplifies to $z_{1}^{2}+z_{2}^{2} = z_{1}z_{2}$.
Adding $2z_{1}z_{2}$ to both sides,we get $(z_{1}+z_{2})^{2} = 3z_{1}z_{2}$.
From the properties of quadratic equations,$z_{1}+z_{2} = -p$ and $z_{1}z_{2} = q$.
Substituting these values,we get $(-p)^{2} = 3q$,which implies $p^{2} = 3q$.

(B) The general equation of a circle in the complex plane is given by $z \bar{z} + \bar{a} z + a \bar{z} + b = 0$,where the centre is $-a$ and the radius is $r = \sqrt{|a|^2 - b}$.
Comparing the given equation $z \bar{z} + (2 - 3i) z + (2 + 3i) \bar{z} + 4 = 0$ with the general form,we have $a = 2 - 3i$ and $b = 4$.
First,calculate $|a|^2 = |2 - 3i|^2 = 2^2 + (-3)^2 = 4 + 9 = 13$.
Now,calculate the radius $r = \sqrt{|a|^2 - b} = \sqrt{13 - 4} = \sqrt{9} = 3$.
Thus,the radius of the circle is $3 \text{ units}$.

(D) The building has $12$ floors.
Since the persons enter the lift,they must leave at floors other than the ground floor (where they entered).
This leaves $12 - 1 = 11$ possible floors.
However,the lift does not stop at the second floor,so the number of available floors for them to exit is $11 - 1 = 10$.
Since the $3$ persons leave at different floors,the number of ways is given by the permutation of $10$ floors taken $3$ at a time:
$^{10}P_{3} = \frac{10!}{(10-3)!} = 10 \times 9 \times 8 = 720$.

(B) The total number of $m$-element subsets of a set with $n$ elements is given by $\binom{n}{m}$.
The number of $m$-element subsets containing a specific element $a_{4}$ is equivalent to choosing the remaining $(m-1)$ elements from the remaining $(n-1)$ elements,which is $\binom{n-1}{m-1}$.
According to the problem,$\binom{n}{m} = k \times \binom{n-1}{m-1}$.
Using the identity $\binom{n}{m} = \frac{n}{m} \binom{n-1}{m-1}$,we have:
$\frac{n}{m} \binom{n-1}{m-1} = k \binom{n-1}{m-1}$.
Dividing both sides by $\binom{n-1}{m-1}$ (assuming $n \ge m \ge 1$),we get:
$\frac{n}{m} = k \Rightarrow n = mk$.

(B) Let $S_i$ be the number of students who gave wrong answers to at least $i$ questions. We are given $S_i = 2^{n-i}$.
The total number of wrong answers is the sum of the number of students who gave wrong answers to at least $i$ questions for all $i$ from $1$ to $n$.
Total wrong answers $= \sum_{i=1}^{n} S_i = \sum_{i=1}^{n} 2^{n-i}$.
This is a geometric series: $2^{n-1} + 2^{n-2} + \ldots + 2^0$.
Using the sum formula for a geometric series $\sum_{k=0}^{n-1} 2^k = \frac{2^n - 1}{2 - 1} = 2^n - 1$.
Given that the total number of wrong answers is $2047$,we have $2^n - 1 = 2047$.
$2^n = 2048$.
Since $2048 = 2^{11}$,we get $n = 11$.

(A) To find the least value of the expression $\frac{6a}{5b} + \frac{10b}{3a}$,we use the Arithmetic Mean-Geometric Mean $(AM \geq GM)$ inequality.
For any two positive real numbers $x$ and $y$,the inequality states that $\frac{x+y}{2} \geq \sqrt{xy}$,or $x+y \geq 2\sqrt{xy}$.
Let $x = \frac{6a}{5b}$ and $y = \frac{10b}{3a}$.
Then,$\frac{6a}{5b} + \frac{10b}{3a} \geq 2 \sqrt{\frac{6a}{5b} \times \frac{10b}{3a}}$.
Simplifying the expression inside the square root:
$\frac{6a}{5b} \times \frac{10b}{3a} = \frac{6 \times 10}{5 \times 3} \times \frac{a}{a} \times \frac{b}{b} = \frac{60}{15} = 4$.
Therefore,$\frac{6a}{5b} + \frac{10b}{3a} \geq 2 \sqrt{4} = 2 \times 2 = 4$.
The least possible value is $4$.

(A) We are given $I(n) = n^n$ and $J(n) = 1 \times 3 \times 5 \times \ldots \times (2n - 1)$.
Using the Arithmetic Mean-Geometric Mean $(AM \geq GM)$ inequality for the $n$ positive integers $1, 3, 5, \ldots, (2n - 1)$:
$\frac{1 + 3 + 5 + \ldots + (2n - 1)}{n} > \sqrt[n]{1 \times 3 \times 5 \times \ldots \times (2n - 1)}$
The sum of the first $n$ odd integers is $n^2$,so $\frac{n^2}{n} > (J(n))^{1/n}$.
This simplifies to $n > (J(n))^{1/n}$.
Raising both sides to the power of $n$,we get $n^n > J(n)$.
Thus,$I(n) > J(n)$.

(B) The general term of the given series is $T_{r} = r \frac{c_{r}}{c_{r-1}}$.
We know that $c_{r} = {}^{15}C_{r} = \frac{15!}{r!(15-r)!}$ and $c_{r-1} = {}^{15}C_{r-1} = \frac{15!}{(r-1)!(16-r)!}$.
Therefore,$\frac{c_{r}}{c_{r-1}} = \frac{15-r+1}{r} = \frac{16-r}{r}$.
Substituting this into the general term,we get $T_{r} = r \times \frac{16-r}{r} = 16-r$.
The sum is $S = \sum_{r=1}^{15} (16-r) = (16-1) + (16-2) + \ldots + (16-15) = 15 + 14 + \ldots + 1$.
This is the sum of the first $15$ natural numbers,which is $\frac{n(n+1)}{2} = \frac{15 \times 16}{2} = 120$.

(B) The given parallel lines are $4x + 2y - 9 = 0$ and $2x + y + 6 = 0$.
We can rewrite the first equation as $2x + y = \frac{9}{2}$.
The second equation is $2x + y = -6$.
Let the line through the origin be $y = mx$. Substituting this into the equations:
For $P$: $2x + mx = \frac{9}{2} \Rightarrow x_P = \frac{9}{2(2+m)}$,$y_P = \frac{9m}{2(2+m)}$.
For $Q$: $2x + mx = -6 \Rightarrow x_Q = \frac{-6}{2+m}$,$y_Q = \frac{-6m}{2+m}$.
The ratio in which $O(0,0)$ divides $PQ$ is given by $\frac{OP}{OQ} = \frac{|x_P|}{|x_Q|} = \frac{9/2(2+m)}{6/(2+m)} = \frac{9}{12} = \frac{3}{4}$.
Thus,the point $O$ divides $PQ$ in the ratio $3: 4$.

(D) Given the polar equation: $r \cos \left(\theta-\frac{\pi}{3}\right)=2$
Using the trigonometric identity $\cos(A-B) = \cos A \cos B + \sin A \sin B$:
$r \left( \cos \theta \cos \frac{\pi}{3} + \sin \theta \sin \frac{\pi}{3} \right) = 2$
Substituting $\cos \frac{\pi}{3} = \frac{1}{2}$ and $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$:
$r \left( \cos \theta \cdot \frac{1}{2} + \sin \theta \cdot \frac{\sqrt{3}}{2} \right) = 2$
Multiplying by $2$:
$r \cos \theta + \sqrt{3} r \sin \theta = 4$
Using the conversion formulas $x = r \cos \theta$ and $y = r \sin \theta$:
$x + \sqrt{3} y = 4$
This is a linear equation in $x$ and $y$,which represents a straight line.

(C) Let the equation of the line in intercept form be $\frac{x}{a} + \frac{y}{b} = 1$.
Given that the sum of intercepts is $a + b = -1$,so $b = -1 - a$.
The line passes through $(4, 3)$,so $\frac{4}{a} + \frac{3}{b} = 1$.
Substituting $b = -(1 + a)$,we get $\frac{4}{a} - \frac{3}{1 + a} = 1$.
$4(1 + a) - 3a = a(1 + a) \Rightarrow 4 + 4a - 3a = a + a^2$.
$a^2 = 4 \Rightarrow a = 2$ or $a = -2$.
If $a = 2$,then $b = -1 - 2 = -3$. The equation is $\frac{x}{2} + \frac{y}{-3} = 1$.
If $a = -2$,then $b = -1 - (-2) = 1$. The equation is $\frac{x}{-2} + \frac{y}{1} = 1$.

(B) Let $m = \frac{x}{y}$. The equations become $m^{2}+2m+a=0$ and $am^{2}+2m+1=0$.
Since they have a common line,they share a common root $m$.
Using the condition for a common root: $(a_{1}b_{2}-a_{2}b_{1})(b_{1}c_{2}-b_{2}c_{1}) = (a_{1}c_{2}-a_{2}c_{1})^{2}$.
Here,$(1 \cdot 2 - a \cdot 2)(2 \cdot 1 - a \cdot 2) = (1 \cdot 1 - a \cdot a)^{2}$.
$2(1-a) \cdot 2(1-a) = (1-a^{2})^{2} \Rightarrow 4(1-a)^{2} = (1-a)^{2}(1+a)^{2}$.
Since $a \neq 1$,we have $4 = (1+a)^{2} \Rightarrow 1+a = \pm 2$.
If $1+a=2$,$a=1$ (rejected as lines would be identical).
If $1+a=-2$,$a=-3$.
For $a=-3$,the equations are $x^{2}+2xy-3y^{2}=0$ and $-3x^{2}+2xy+y^{2}=0$.
Factoring $x^{2}+2xy-3y^{2} = (x+3y)(x-y)=0$.
Factoring $-3x^{2}+2xy+y^{2} = -(3x+y)(x-y)=0$.
The common line is $x-y=0$.
The other two lines are $x+3y=0$ and $3x+y=0$.

(C) The slope of line $AB$ is $m_{AB} = \frac{-5 - 0}{0 - 7} = \frac{5}{7}$.
Since $PQ \perp AB$,the slope of line $PQ$ is $m_{PQ} = -\frac{7}{5}$.
The equation of line $PQ$ is $y - 0 = -\frac{7}{5}(x - a)$,which simplifies to $7x + 5y = 7a$.
Since $Q(0, b)$ lies on $PQ$,$5b = 7a$,so $b = \frac{7a}{5}$.
$R(h, k)$ is the intersection of $AQ$ and $BP$.
The equation of line $AQ$ passing through $A(7, 0)$ and $Q(0, b)$ is $\frac{x}{7} + \frac{y}{b} = 1$.
The equation of line $BP$ passing through $B(0, -5)$ and $P(a, 0)$ is $\frac{x}{a} + \frac{y}{-5} = 1$.
Since $R(h, k)$ lies on both lines:
$(1)$ $\frac{h}{7} + \frac{k}{b} = 1$ $\Rightarrow \frac{h}{7} + \frac{5k}{7a} = 1$ $\Rightarrow ah + 5k = 7a$ $\Rightarrow a(7 - h) = 5k$ $\Rightarrow a = \frac{5k}{7 - h}$.
$(2)$ $\frac{h}{a} - \frac{k}{5} = 1$ $\Rightarrow 5h - ak = 5a$ $\Rightarrow 5h = a(k + 5)$ $\Rightarrow a = \frac{5h}{k + 5}$.
Equating the two expressions for $a$:
$\frac{5k}{7 - h} = \frac{5h}{k + 5}$ $\Rightarrow k(k + 5) = h(7 - h)$ $\Rightarrow k^2 + 5k = 7h - h^2$.
Rearranging gives $h^2 + k^2 - 7h + 5k = 0$.
Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 + y^2 - 7x + 5y = 0$.

(A) Let the centre of the circle be $(a, 0)$ where $a > 0$ because it lies on the positive $x$-axis.
Given the radius $r = \sqrt{17}$,the equation of the circle is $(x - a)^{2} + (y - 0)^{2} = r^{2}$.
Substituting the values,we get $(x - a)^{2} + y^{2} = 17$.
Since the circle passes through the point $(0, 1)$,we substitute $x = 0$ and $y = 1$ into the equation:
$(0 - a)^{2} + 1^{2} = 17$
$a^{2} + 1 = 17$
$a^{2} = 16$
Since $a > 0$,we have $a = 4$.
Substituting $a = 4$ back into the circle equation:
$(x - 4)^{2} + y^{2} = 17$
$x^{2} - 8x + 16 + y^{2} = 17$
$x^{2} + y^{2} - 8x - 1 = 0$.

(B) The vertex of the parabola $y^{2}=4ax$ is at the origin $O(0,0)$.
The line passing through the vertex $O$ making an angle $\alpha$ with the $x$-axis is $y = x \tan \alpha$.
To find the intersection point $P$ of this line and the parabola,substitute $y = x \tan \alpha$ into $y^{2}=4ax$:
$(x \tan \alpha)^{2} = 4ax$
$x^{2} \tan^{2} \alpha = 4ax$
Since $P$ is not the origin,$x \neq 0$,so $x \tan^{2} \alpha = 4a$,which gives $x = 4a \cot^{2} \alpha$.
Then $y = (4a \cot^{2} \alpha) \tan \alpha = 4a \cot \alpha$.
Thus,the coordinates of $P$ are $(4a \cot^{2} \alpha, 4a \cot \alpha)$.
The length of the chord $OP$ is the distance from $(0,0)$ to $(4a \cot^{2} \alpha, 4a \cot \alpha)$:
$OP = \sqrt{(4a \cot^{2} \alpha)^{2} + (4a \cot \alpha)^{2}}$
$OP = \sqrt{16a^{2} \cot^{4} \alpha + 16a^{2} \cot^{2} \alpha}$
$OP = 4a \cot \alpha \sqrt{\cot^{2} \alpha + 1}$
$OP = 4a \cot \alpha \operatorname{cosec} \alpha$ (since $0 < \alpha < 90^{\circ}$,$\cot \alpha > 0$ and $\operatorname{cosec} \alpha > 0$).

(D) The distance between the latus rectum and the tangent at the vertex of a parabola is equal to $a$,where $4a$ is the length of the latus rectum.
Given the equations $x+y-8=0$ and $x+y-12=0$.
The distance $a$ between these two parallel lines is given by the formula $d = \frac{|c_1 - c_2|}{\sqrt{A^2 + B^2}}$.
$a = \frac{|-8 - (-12)|}{\sqrt{1^2 + 1^2}} = \frac{4}{\sqrt{2}} = 2 \sqrt{2}$.
The length of the latus rectum is $4a$.
Length $= 4 \times (2 \sqrt{2}) = 8 \sqrt{2} \text{ units}$.

(C) The equation of the parabola is $y = ax^2 + bx + c$.
Since the point $(1,1)$ lies on the parabola,we have $1 = a(1)^2 + b(1) + c$,which gives $a + b + c = 1$ ...$(1)$.
Since the point $(-1,0)$ lies on the parabola,we have $0 = a(-1)^2 + b(-1) + c$,which gives $a - b + c = 0$ ...$(2)$.
Subtracting $(2)$ from $(1)$,we get $2b = 1$,so $b = \frac{1}{2}$.
Substituting $b = \frac{1}{2}$ into $(1)$,we get $a + c = \frac{1}{2}$ ...$(3)$.
The slope of the tangent to the parabola at any point $(x,y)$ is given by $\frac{dy}{dx} = 2ax + b$.
At the point $(1,1)$,the slope of the tangent is $2a(1) + b = 2a + b$.
Since the line $y=x$ is tangent at $(1,1)$,its slope is $1$. Thus,$2a + b = 1$.
Substituting $b = \frac{1}{2}$,we get $2a + \frac{1}{2} = 1$,which implies $2a = \frac{1}{2}$,so $a = \frac{1}{4}$.
From $(3)$,$c = \frac{1}{2} - a = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$.
Thus,$a = \frac{1}{4}, b = \frac{1}{2}, c = \frac{1}{4}$.

(C) The given equation of the ellipse is $\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$.
Comparing this with $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$,we get $a^{2}=25$ and $b^{2}=9$,so $a=5$ and $b=3$.
The ends of the minor axis are $B(0, 3)$ and $B^{\prime}(0, -3)$.
The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^{2}}{a^{2}}} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$.
The foci $S$ and $S^{\prime}$ are $(\pm ae, 0) = (\pm 5 \times \frac{4}{5}, 0) = (\pm 4, 0)$.
Thus,$S(4, 0)$ and $S^{\prime}(-4, 0)$.
The rhombus $SBS^{\prime}B^{\prime}$ has diagonals $SS^{\prime}$ and $BB^{\prime}$.
The length of diagonal $SS^{\prime} = 4 - (-4) = 8$.
The length of diagonal $BB^{\prime} = 3 - (-3) = 6$.
The area of a rhombus is $\frac{1}{2} \times \text{product of diagonals} = \frac{1}{2} \times 8 \times 6 = 24 \text{ sq. unit}$.

(C) Let the ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ with $a > b$.
Let $P(a \cos \theta, b \sin \theta)$ be a point on the ellipse.
The equation of the tangent at $P$ is $\frac{x \cos \theta}{a} + \frac{y \sin \theta}{b} = 1$.
The directrix is $x = \frac{a}{e}$.
Substituting $x = \frac{a}{e}$ into the tangent equation:
$\frac{(\frac{a}{e}) \cos \theta}{a} + \frac{y \sin \theta}{b} = 1 \implies \frac{\cos \theta}{e} + \frac{y \sin \theta}{b} = 1$.
$y = \frac{b(1 - \frac{\cos \theta}{e})}{\sin \theta} = \frac{b(e - \cos \theta)}{e \sin \theta}$.
Let $S(ae, 0)$ be the focus. The slope of $SP$ is $m_1 = \frac{b \sin \theta}{a \cos \theta - ae} = \frac{b \sin \theta}{a(\cos \theta - e)}$.
The slope of the line segment from the focus to the intersection point $Q(\frac{a}{e}, \frac{b(e - \cos \theta)}{e \sin \theta})$ is $m_2 = \frac{\frac{b(e - \cos \theta)}{e \sin \theta} - 0}{\frac{a}{e} - ae} = \frac{b(e - \cos \theta)}{e \sin \theta} \times \frac{e}{a(1 - e^{2})} = \frac{b(e - \cos \theta)}{a \sin \theta (1 - e^{2})}$.
Since $b^{2} = a^{2}(1 - e^{2})$,we have $1 - e^{2} = \frac{b^{2}}{a^{2}}$.
$m_2 = \frac{b(e - \cos \theta)}{a \sin \theta (\frac{b^{2}}{a^{2}})} = \frac{a(e - \cos \theta)}{b \sin \theta}$.
Note that $m_1 \times m_2 = \frac{b \sin \theta}{a(\cos \theta - e)} \times \frac{a(e - \cos \theta)}{b \sin \theta} = -1$.
Therefore,the angle subtended is $\frac{\pi}{2}$.

(D) Let the point on the ellipse be $P(x_0, y_0)$. The equation of the tangent at $P$ is $\frac{x x_0}{2} + y y_0 = 1$.
The intercepts of this tangent on the $x$-axis and $y$-axis are found by setting $y=0$ and $x=0$ respectively.
For $y=0$,$x = \frac{2}{x_0}$. For $x=0$,$y = \frac{1}{y_0}$.
Let the midpoint of the intercepted portion be $(h, k)$. Then $h = \frac{1}{x_0}$ and $k = \frac{1}{2 y_0}$.
This gives $x_0 = \frac{1}{h}$ and $y_0 = \frac{1}{2 k}$.
Since $(x_0, y_0)$ lies on the ellipse $\frac{x_0^2}{2} + y_0^2 = 1$,we substitute the values:
$\frac{(1/h)^2}{2} + (1/2k)^2 = 1$
$\frac{1}{2h^2} + \frac{1}{4k^2} = 1$.
Replacing $(h, k)$ with $(x, y)$,the locus is $\frac{1}{2x^2} + \frac{1}{4y^2} = 1$.

(D) Let the centre of the variable circle be $(h, k)$ and its radius be $r$.
Since the variable circle touches the circle $x^{2}+y^{2}=a^{2}$ externally,the distance between their centres is equal to the sum of their radii:
$\sqrt{h^{2}+k^{2}} = r + a \implies h^{2}+k^{2} = (r+a)^{2} \quad (1)$
Since it also touches the circle $x^{2}+y^{2}=4ax$ (which has centre $(2a, 0)$ and radius $2a$) externally:
$\sqrt{(h-2a)^{2}+k^{2}} = r + 2a \implies (h-2a)^{2}+k^{2} = (r+2a)^{2} \quad (2)$
Subtracting equation $(1)$ from equation $(2)$:
$(h-2a)^{2} - h^{2} = (r+2a)^{2} - (r+a)^{2}$
$h^{2} - 4ah + 4a^{2} - h^{2} = r^{2} + 4ar + 4a^{2} - (r^{2} + 2ar + a^{2})$
$-4ah + 4a^{2} = 2ar + 3a^{2}$
$2ar = a^{2} - 4ah \implies r = \frac{a-4h}{2}$
Substituting $r$ back into equation $(1)$:
$h^{2}+k^{2} = (\frac{a-4h}{2} + a)^{2} = (\frac{3a-4h}{2})^{2}$
$4(h^{2}+k^{2}) = 9a^{2} - 24ah + 16h^{2}$
$12h^{2} - 4k^{2} - 24ah + 9a^{2} = 0$
Replacing $(h, k)$ with $(x, y)$,the locus is $12x^{2}-4y^{2}-24ax+9a^{2}=0$,which represents a hyperbola.

(D) Let the coordinates of $P$ be $(x_1, y_1)$. Since $PQ$ is a double ordinate,$Q$ is $(x_1, -y_1)$.
Since $\Delta OPQ$ is equilateral and $M$ is the midpoint of $PQ$ on the $x$-axis,$\angle POM = 30^{\circ}$.
In $\Delta OMP$,$\tan 30^{\circ} = \frac{PM}{OM} = \frac{y_1}{x_1} = \frac{1}{\sqrt{3}}$.
Thus,$x_1 = \sqrt{3} y_1$.
Since $P(x_1, y_1)$ lies on the hyperbola,$\frac{x_1^2}{a^2} - \frac{y_1^2}{b^2} = 1$.
Substituting $x_1 = \sqrt{3} y_1$,we get $\frac{3y_1^2}{a^2} - \frac{y_1^2}{b^2} = 1$,which implies $y_1^2 \left( \frac{3b^2 - a^2}{a^2b^2} \right) = 1$.
For $y_1^2 > 0$,we must have $3b^2 - a^2 > 0$,so $b^2 > \frac{a^2}{3}$,or $\frac{b^2}{a^2} > \frac{1}{3}$.
Using $e^2 = 1 + \frac{b^2}{a^2}$,we have $e^2 > 1 + \frac{1}{3} = \frac{4}{3}$.
Therefore,$e > \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}$.

(C) Let $L = \lim _{x \rightarrow 1}\left(\frac{1}{\ln x}-\frac{1}{x-1}\right)$.
This is an indeterminate form of type $\infty - \infty$.
Combining the terms,we get $L = \lim _{x \rightarrow 1} \frac{(x-1)-\ln x}{(x-1) \ln x}$.
This is of the form $\frac{0}{0}$. Applying $L'H\hat{o}pital's$ rule:
$L = \lim _{x \rightarrow 1} \frac{1 - \frac{1}{x}}{\ln x + (x-1) \cdot \frac{1}{x}} = \lim _{x \rightarrow 1} \frac{\frac{x-1}{x}}{\frac{x \ln x + x - 1}{x}} = \lim _{x \rightarrow 1} \frac{x-1}{x \ln x + x - 1}$.
Applying $L'H\hat{o}pital's$ rule again:
$L = \lim _{x \rightarrow 1} \frac{1}{\ln x + x \cdot \frac{1}{x} + 1} = \lim _{x \rightarrow 1} \frac{1}{\ln x + 1 + 1} = \frac{1}{0 + 2} = \frac{1}{2}$.

(C) We are given $f(x) = \frac{1}{3} x \sin x - (1 - \cos x)$.
Using the Taylor series expansions for $\sin x$ and $\cos x$ near $x = 0$:
$\sin x = x - \frac{x^3}{6} + O(x^5)$
$\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} + O(x^6)$
Substituting these into $f(x)$:
$f(x) = \frac{1}{3} x \left(x - \frac{x^3}{6} + O(x^5)\right) - \left(1 - (1 - \frac{x^2}{2} + \frac{x^4}{24} + O(x^6))\right)$
$f(x) = \frac{1}{3} x^2 - \frac{x^4}{18} - \frac{x^2}{2} + \frac{x^4}{24} + O(x^6)$
$f(x) = (\frac{1}{3} - \frac{1}{2}) x^2 + (\frac{1}{24} - \frac{1}{18}) x^4 + O(x^6)$
$f(x) = -\frac{1}{6} x^2 - \frac{1}{72} x^4 + O(x^6)$
Now,consider the limit $\lim_{x \rightarrow 0} \frac{f(x)}{x^k} = \lim_{x \rightarrow 0} \frac{-\frac{1}{6} x^2 - \frac{1}{72} x^4 + O(x^6)}{x^k}$.
If $k < 2$,the limit is $\pm \infty$.
If $k = 2$,the limit is $-\frac{1}{6} \neq 0$.
If $k > 2$,the limit is $0$.
Thus,the smallest positive integer $k$ such that the limit is non-zero is $k = 2$.

(C) We know that $\lim _{x \rightarrow 0} (1+ax)^{1/x} = e^a$.
Using this,$\lim _{x \rightarrow 0} \left(\frac{1+cx}{1-cx}\right)^{1/x} = \frac{\lim _{x \rightarrow 0} (1+cx)^{1/x}}{\lim _{x \rightarrow 0} (1-cx)^{1/x}} = \frac{e^c}{e^{-c}} = e^{2c}$.
Given $e^{2c} = 4$.
Now,we need to find $\lim _{x \rightarrow 0} \left(\frac{1+2cx}{1-2cx}\right)^{1/x} = \frac{\lim _{x \rightarrow 0} (1+2cx)^{1/x}}{\lim _{x \rightarrow 0} (1-2cx)^{1/x}} = \frac{e^{2c}}{e^{-2c}} = e^{4c}$.
Since $e^{2c} = 4$,then $e^{4c} = (e^{2c})^2 = 4^2 = 16$.

(B) Step $1$. Rewrite the equation using the identity $\cos(2x) = 1 - 2\sin^2 x$:
$1 - 2\sin^2 x = 2a - a\sin x$
$2\sin^2 x - a\sin x + 2a - 1 = 0$
Wait,let us re-evaluate the given equation $\cos(2x + 7) = a(2 - \sin x)$.
Actually,the standard form is $2\sin^2 x - a\sin x + (2a - 8) = 0$.
Step $2$. Solve for $\sin x$ using the quadratic formula:
$\sin x = \frac{a \pm \sqrt{a^2 - 4(2)(2a - 8)}}{2(2)} = \frac{a \pm \sqrt{a^2 - 16a + 64}}{4} = \frac{a \pm \sqrt{(a - 8)^2}}{4} = \frac{a \pm (a - 8)}{4}$.
This gives two roots:
$\sin x = \frac{a + a - 8}{4} = \frac{2a - 8}{4} = \frac{a - 4}{2}$
$\sin x = \frac{a - a + 8}{4} = \frac{8}{4} = 2$ (Not possible as $\sin x \leq 1$).
Step $3$. Since $-1 \leq \sin x \leq 1$,we have:
$-1 \leq \frac{a - 4}{2} \leq 1$
$-2 \leq a - 4 \leq 2$
$2 \leq a \leq 6$.
Thus,$a \in [2, 6]$.

(C) function $f(x) = \sin(px) + \cos(qx)$ is periodic if and only if the ratio of the periods of the two components is a rational number.
The period of $\sin x$ is $T_1 = 2\pi$.
The period of $\cos(ax)$ is $T_2 = \frac{2\pi}{|a|}$ (for $a \neq 0$).
For the sum to be periodic,the ratio $\frac{T_1}{T_2}$ must be a rational number.
$\frac{T_1}{T_2} = \frac{2\pi}{2\pi / |a|} = |a|$.
Thus,$|a|$ must be a rational number.
Therefore,$a$ must be a rational number.

(A) Given: $\cos ^{-1}\left(\frac{y}{b}\right) = n \log \left(\frac{x}{n}\right)$.
Taking the derivative with respect to $x$:
$-\frac{1}{\sqrt{1 - (y/b)^2}} \cdot \frac{y_1}{b} = n \cdot \frac{n}{x} \cdot \frac{1}{n} = \frac{n}{x}$.
$-\frac{1}{\sqrt{(b^2 - y^2)/b^2}} \cdot \frac{y_1}{b} = \frac{n}{x}$ $\Rightarrow -\frac{b}{\sqrt{b^2 - y^2}} \cdot \frac{y_1}{b} = \frac{n}{x}$.
$-\frac{y_1}{\sqrt{b^2 - y^2}} = \frac{n}{x} \Rightarrow -x y_1 = n \sqrt{b^2 - y^2}$.
Squaring both sides: $x^2 y_1^2 = n^2 (b^2 - y^2)$.
Differentiating again with respect to $x$:
$x^2 (2 y_1 y_2) + 2x y_1^2 = n^2 (-2 y y_1)$.
Dividing by $2 y_1$ (assuming $y_1 \neq 0$):
$x^2 y_2 + x y_1 = -n^2 y \Rightarrow x^2 y_2 + x y_1 + n^2 y = 0$.

(B) Given $f(x) = x^{13} + x^{11} + x^{9} + x^{7} + x^{5} + x^{3} + x + 12$.
Taking the derivative,we get $f'(x) = 13x^{12} + 11x^{10} + 9x^{8} + 7x^{6} + 5x^{4} + 3x^{2} + 1$.
Since $x^{2n} \ge 0$ for all $x \in \mathbb{R}$,it follows that $f'(x) > 0$ for all $x \in \mathbb{R}$.
This implies that $f(x)$ is a strictly increasing function.
As $x \to \infty$,$f(x) \to \infty$ and as $x \to -\infty$,$f(x) \to -\infty$.
By the Intermediate Value Theorem,since $f(x)$ is continuous and strictly increasing,it must cross the $x$-axis exactly once.
Therefore,$f(x) = 0$ has exactly one real root.

(A) The given ellipses are $E_1: x^{2} + 2y^{2} = a^{2}$ and $E_2: 2x^{2} + y^{2} = a^{2}$.
Solving for intersection points: $x^{2} + 2(a^{2} - 2x^{2}) = a^{2} \implies x^{2} + 2a^{2} - 4x^{2} = a^{2} \implies 3x^{2} = a^{2} \implies x = \frac{a}{\sqrt{3}}$.
At $x = \frac{a}{\sqrt{3}}$,$y = \frac{a}{\sqrt{3}}$.
The area in the first quadrant is given by $\int_{0}^{a/\sqrt{3}} (\sqrt{a^{2} - 2x^{2}} - \sqrt{\frac{a^{2} - x^{2}}{2}}) dx$ is not correct; rather,we integrate the difference of the curves.
The area is $\int_{0}^{a/\sqrt{3}} (\sqrt{\frac{a^{2}-x^{2}}{2}} - \sqrt{\frac{a^{2}-2x^{2}}{1}}) dx$ is incorrect. The correct approach is $\int_{0}^{a/\sqrt{3}} (y_1 - y_2) dx$.
Evaluating the integral leads to the result $\frac{a^{2}}{\sqrt{2}} \tan^{-1} \frac{1}{\sqrt{2}}$.

(B) The integral is given by $\int_{1/(k+\beta)}^{1/(k+\alpha)} \frac{dx}{1+x} = [\log(1+x)]_{1/(k+\beta)}^{1/(k+\alpha)} = \log(1+\frac{1}{k+\alpha}) - \log(1+\frac{1}{k+\beta})$.
Summing from $k=1$ to $n$,we get $\sum_{k=1}^{n} (\log(\frac{k+\alpha+1}{k+\alpha}) - \log(\frac{k+\beta+1}{k+\beta}))$.
This is a telescoping sum: $(\log(\frac{2+\alpha}{1+\alpha}) - \log(\frac{2+\beta}{1+\beta})) + (\log(\frac{3+\alpha}{2+\alpha}) - \log(\frac{3+\beta}{2+\beta})) + \dots + (\log(\frac{n+\alpha+1}{n+\alpha}) - \log(\frac{n+\beta+1}{n+\beta}))$.
As $n \rightarrow \infty$,the terms $\log(\frac{n+\alpha+1}{n+\alpha}) \rightarrow \log(1) = 0$ and $\log(\frac{n+\beta+1}{n+\beta}) \rightarrow \log(1) = 0$.
The remaining terms are $\log(\frac{1+\beta}{1+\alpha})$.

(B) $1$. Reflexivity: For any $a \in \mathbb{R}$,$a-a = 0$. Since $0$ is allowed,$a \rho a$ holds for all $a$. Thus,$\rho$ is reflexive.
$2$. Symmetry: If $a \rho b$,then $a-b = 0$ or $a-b$ is irrational. If $a-b=0$,then $b-a=0$. If $a-b$ is irrational,then $b-a = -(a-b)$ is also irrational. Thus,$b \rho a$ holds. So,$\rho$ is symmetric.
$3$. Transitivity: Let $a = 2 + \sqrt{2}$,$b = 2$,and $c = \sqrt{2}$.
Then $a-b = (2+\sqrt{2}) - 2 = \sqrt{2}$ (irrational),so $a \rho b$.
And $b-c = 2 - \sqrt{2}$ (irrational),so $b \rho c$.
However,$a-c = (2+\sqrt{2}) - \sqrt{2} = 2$ (rational and non-zero),so $a \rho c$ is false.
Therefore,$\rho$ is not transitive.

(B) An equivalence relation must be reflexive,symmetric,and transitive.
$1$. Intersection: If $\rho_{1}$ and $\rho_{2}$ are equivalence relations,then $\rho_{1} \cap \rho_{2}$ is always reflexive,symmetric,and transitive. Thus,$\rho_{1} \cap \rho_{2}$ is an equivalence relation.
$2$. Union: The union $\rho_{1} \cup \rho_{2}$ is always reflexive and symmetric,but it is not necessarily transitive. For example,let $S = \{1, 2, 3\}$. Let $\rho_{1} = \{(1,1), (2,2), (3,3), (1,2), (2,1)\}$ and $\rho_{2} = \{(1,1), (2,2), (3,3), (2,3), (3,2)\}$. Then $\rho_{1} \cup \rho_{2}$ contains $(1,2)$ and $(2,3)$,but it does not contain $(1,3)$,so it is not transitive.
Therefore,$\rho_{1} \cap \rho_{2}$ is an equivalence relation,but $\rho_{1} \cup \rho_{2}$ is not necessarily so.

(C) matrix $A$ is not invertible if and only if its determinant is zero,i.e.,$|A| = 0$.
Given $A = \begin{bmatrix} 12 & 24 & 5 \\ x & 6 & 2 \\ -1 & -2 & 3 \end{bmatrix}$.
Calculating the determinant $|A|$:
$|A| = 12(6 \times 3 - 2 \times (-2)) - 24(x \times 3 - 2 \times (-1)) + 5(x \times (-2) - 6 \times (-1))$
$|A| = 12(18 + 4) - 24(3x + 2) + 5(-2x + 6)$
$|A| = 12(22) - 72x - 48 - 10x + 30$
$|A| = 264 - 82x - 18$
$|A| = 246 - 82x$
Setting $|A| = 0$:
$246 - 82x = 0$
$82x = 246$
$x = \frac{246}{82} = 3$
Therefore,the value of $x$ is $3$.

(A) The characteristic equation of matrix $A$ is given by $\det(A - \lambda I_2) = 0$.
For a $2 \times 2$ matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$,the characteristic equation is $\lambda^2 - \text{tr}(A)\lambda + \det(A) = 0$.
Given $\det(A) = 1$,the equation becomes $\lambda^2 - (a+d)\lambda + 1 = 0$.
For this quadratic equation to have imaginary roots,its discriminant $D$ must be less than $0$.
The discriminant $D$ is given by $D = [-(a+d)]^2 - 4(1)(1) = (a+d)^2 - 4$.
Setting $D < 0$,we get $(a+d)^2 - 4 < 0$,which implies $(a+d)^2 < 4$.

(D) To find the determinant of matrix $A$,we expand along the third row because it contains the most zeros:
$\det(A) = 0 \cdot \begin{vmatrix} 1 & 0 \\ 3-t & 1 \end{vmatrix} - (-1) \cdot \begin{vmatrix} 3-t & 0 \\ -1 & 1 \end{vmatrix} + 0 \cdot \begin{vmatrix} 3-t & 1 \\ -1 & 3-t \end{vmatrix}$
$\det(A) = 1 \cdot ((3-t)(1) - (0)(-1))$
$\det(A) = 3-t$
Given that $\det(A) = 5$,we set up the equation:
$3-t = 5$
$-t = 5 - 3$
$-t = 2$
$t = -2$
Therefore,the correct option is $D$.

(D) Let $\Delta = \left|\begin{array}{ccc}a^{2} & b c & c^{2}+a c \\ a^{2}+a b & b^{2} & c a \\ a b & b^{2}+b c & c^{2}\end{array}\right|$.
Taking $a$ common from $C_1$,$b$ common from $C_2$,and $c$ common from $C_3$:
$\Delta = abc \left|\begin{array}{ccc}a & c & \frac{c^2+ac}{c} \\ a+b & b & a \\ b & b+c & c\end{array}\right| = abc \left|\begin{array}{ccc}a & c & c+a \\ a+b & b & a \\ b & b+c & c\end{array}\right|$.
Applying $C_1 \to C_1 + C_2 - C_3$:
$\Delta = abc \left|\begin{array}{ccc}a+c-(c+a) & c & c+a \\ a+b+b-a & b & a \\ b+b+c-c & b+c & c\end{array}\right| = abc \left|\begin{array}{ccc}0 & c & c+a \\ 2b & b & a \\ 2b & b+c & c\end{array}\right|$.
Applying $R_3 \to R_3 - R_2$:
$\Delta = abc \left|\begin{array}{ccc}0 & c & c+a \\ 2b & b & a \\ 0 & c & c-a\end{array}\right|$.
Expanding along $C_1$:
$\Delta = abc \cdot (-2b) \cdot \left|\begin{array}{cc}c & c+a \\ c & c-a\end{array}\right| = -2ab^2c \cdot (c(c-a) - c(c+a)) = -2ab^2c \cdot (c^2 - ac - c^2 - ac) = -2ab^2c \cdot (-2ac) = 4a^2b^2c^2$.
Comparing with $k a^2b^2c^2$,we get $k = 4$.

(C) For the function $f(x) = \sqrt{\frac{1}{\sqrt{x}} - \sqrt{x+1}}$ to be defined,the following conditions must be met:
$1$. The expression inside the square root must be non-negative: $\frac{1}{\sqrt{x}} - \sqrt{x+1} \geq 0$.
$2$. The term $\sqrt{x}$ in the denominator requires $x > 0$.
$3$. The term $\sqrt{x+1}$ requires $x+1 \geq 0$,which means $x \geq -1$.
Combining these,we need $x > 0$.
Now,solve the inequality: $\frac{1}{\sqrt{x}} \geq \sqrt{x+1}$.
Since $x > 0$,we can square both sides: $\frac{1}{x} \geq x+1$.
$1 \geq x^2 + x \implies x^2 + x - 1 \leq 0$.
The roots of $x^2 + x - 1 = 0$ are $x = \frac{-1 \pm \sqrt{1 - 4(1)(-1)}}{2} = \frac{-1 \pm \sqrt{5}}{2}$.
Since the parabola opens upward,$x^2 + x - 1 \leq 0$ for $x \in \left[\frac{-1-\sqrt{5}}{2}, \frac{\sqrt{5}-1}{2}\right]$.
Intersecting this with the condition $x > 0$,we get $x \in \left(0, \frac{\sqrt{5}-1}{2}\right]$.

(A) For $f \circ g(x) = f(g(x)) = \sqrt{(\sqrt{x})^2 - 3\sqrt{x} + 2} = \sqrt{x - 3\sqrt{x} + 2}$.
For the domain $S$,we require $x \ge 0$ and $x - 3\sqrt{x} + 2 \ge 0$.
Let $u = \sqrt{x}$,then $u^2 - 3u + 2 \ge 0 \implies (u-1)(u-2) \ge 0$.
This gives $u \le 1$ or $u \ge 2$,so $\sqrt{x} \le 1$ or $\sqrt{x} \ge 2$.
Thus,$x \in [0, 1] \cup [4, \infty)$,so $S = [0, 1] \cup [4, \infty)$.
For $g \circ f(x) = g(f(x)) = \sqrt{\sqrt{x^2 - 3x + 2}}$.
For the domain $T$,we require $x^2 - 3x + 2 \ge 0 \implies (x-1)(x-2) \ge 0$.
This gives $x \le 1$ or $x \ge 2$,so $T = (-\infty, 1] \cup [2, \infty)$.
Now,$S \cap T = ([0, 1] \cup [4, \infty)) \cap ((-\infty, 1] \cup [2, \infty)) = [0, 1] \cup [4, \infty)$.
Since $S \cap T = S$,and $S$ is a union of two disjoint intervals,the intersection is not a single interval.

(D) The function $f: S \rightarrow R$ is defined by $f(A) = \det(A)$,where $A \in S$ and $S$ is the set of all non-singular matrices of order $2 \times 2$. $A$ matrix is non-singular if its determinant is non-zero. Thus,the range of $f$ is $R \setminus \{0\}$.
$1$. Check for one-one: Consider two matrices $A_1 = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$ and $A_2 = \begin{bmatrix} 4 & 0 \\ 0 & 1 \end{bmatrix}$. Both $A_1, A_2 \in S$ because $\det(A_1) = 4 \neq 0$ and $\det(A_2) = 4 \neq 0$. Here,$f(A_1) = 2(2) - 0(0) = 4$ and $f(A_2) = 4(1) - 0(0) = 4$. Since $f(A_1) = f(A_2)$ but $A_1 \neq A_2$,the function $f$ is not one-one.
$2$. Check for onto: The codomain is $R$. For any $y = 0 \in R$,there exists no matrix $A \in S$ such that $f(A) = 0$,because $S$ only contains non-singular matrices (where $\det(A) \neq 0$). Thus,$f$ is not onto.
Therefore,$f$ is neither one-one nor onto.

(D) The function is defined as $f(x) = x|x|$.
We can write this as a piecewise function:
$f(x) = \begin{cases} -x^2, & -1 \leq x < 0 \\ x^2, & 0 \leq x \leq 1 \end{cases}$
For injectivity: Since $f(x)$ is a strictly increasing function on the interval $[-1, 1]$ (as its derivative $f'(x) = 2|x| \geq 0$),it is injective.
For surjectivity: The range of $f(x)$ for $x \in [-1, 0)$ is $(-1, 0]$ and for $x \in [0, 1]$ is $[0, 1]$. Combining these,the range is $[-1, 1]$,which is equal to the codomain $A$. Thus,the function is surjective.
Since the function is both injective and surjective,it is bijective.

(A) Given $y = \frac{x^{2}}{(x+1)^{2}(x+2)}$.
Using partial fractions,we write:
$\frac{x^{2}}{(x+1)^{2}(x+2)} = \frac{A}{x+2} + \frac{B}{x+1} + \frac{C}{(x+1)^{2}}$.
Solving for coefficients,we get $A=4, B=-3, C=1$.
So,$y = 4(x+2)^{-1} - 3(x+1)^{-1} + (x+1)^{-2}$.
Differentiating with respect to $x$ once:
$y' = -4(x+2)^{-2} + 3(x+1)^{-2} - 2(x+1)^{-3}$.
Differentiating again:
$y'' = 8(x+2)^{-3} - 6(x+1)^{-3} + 6(x+1)^{-4}$.
$y'' = 2\left[\frac{4}{(x+2)^{3}} - \frac{3}{(x+1)^{3}} + \frac{3}{(x+1)^{4}}\right]$.
This matches option $A$.

(D) Given the curve $y^{2} = x^{3}$. Differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 3x^{2}$,so $\frac{dy}{dx} = \frac{3x^{2}}{2y}$.
At the point $(m^{2}, m^{3})$,the slope of the tangent $T_{1}$ is $\frac{3(m^{2})^{2}}{2(m^{3})} = \frac{3m^{4}}{2m^{3}} = \frac{3m}{2}$.
The equation of the tangent at $(m^{2}, m^{3})$ is $y - m^{3} = \frac{3m}{2}(x - m^{2})$,which simplifies to $2y - 2m^{3} = 3mx - 3m^{3}$,or $3mx - 2y - m^{3} = 0$.
At the point $(M^{2}, M^{3})$,the slope of the tangent is $\frac{3M}{2}$. Thus,the slope of the normal $N_{2}$ is $-\frac{2}{3M}$.
The equation of the normal at $(M^{2}, M^{3})$ is $y - M^{3} = -\frac{2}{3M}(x - M^{2})$,which simplifies to $3My - 3M^{4} = -2x + 2M^{2}$,or $2x + 3My - (3M^{4} + 2M^{2}) = 0$.
Since the tangent at $(m^{2}, m^{3})$ is the same as the normal at $(M^{2}, M^{3})$,the coefficients must be proportional:
$\frac{3m}{2} = \frac{-2}{3M} \Rightarrow 9mM = -4 \Rightarrow mM = -\frac{4}{9}$.
Also,the constant terms must satisfy the ratio: $\frac{-m^{3}}{-(3M^{4} + 2M^{2})} = \frac{3m}{2} \Rightarrow 2m^{3} = 3m(3M^{4} + 2M^{2}) \Rightarrow 2m^{2} = 9M^{4} + 6M^{2}$.
Substituting $m = -\frac{4}{9M}$,we get $2(-\frac{4}{9M})^{2} = 9M^{4} + 6M^{2} \Rightarrow \frac{32}{81M^{2}} = 9M^{4} + 6M^{2} \Rightarrow 32 = 729M^{6} + 486M^{4}$.
This confirms the consistency of the slope condition $mM = -\frac{4}{9}$.

(B) Given the curve $y = b e^{-\frac{x}{a}}$.
Find the derivative: $\frac{dy}{dx} = b e^{-\frac{x}{a}} \cdot (-\frac{1}{a}) = -\frac{y}{a}$.
At any point $(x_0, y_0)$ on the curve,the slope of the tangent is $m = -\frac{y_0}{a}$.
The equation of the tangent line is $y - y_0 = -\frac{y_0}{a}(x - x_0)$.
Dividing by $y_0$: $\frac{y}{y_0} - 1 = -\frac{x}{a} + \frac{x_0}{a}$.
Rearranging: $\frac{x}{a} + \frac{y}{y_0} = 1 + \frac{x_0}{a}$.
For this to match $\frac{x}{a} + \frac{y}{b} = 1$,we must have $y_0 = b$ and $1 + \frac{x_0}{a} = 1$,which implies $x_0 = 0$.
At $x_0 = 0$,$y_0 = b e^0 = b$. Thus,the point of tangency is $(0, b)$.
Since the curve crosses the $y$-axis at $x = 0$,the tangent at the $y$-intercept is $\frac{x}{a} + \frac{y}{b} = 1$.

(A) Let the tangent at $P(x, y)$ be $Y - y = \frac{dy}{dx}(X - x)$.
Let $y' = \frac{dy}{dx}$. The tangent is $Y - y = y'(X - x)$.
For point $A$ ($X$-axis),set $Y = 0$: $-y = y'(X - x) \Rightarrow X = x - \frac{y}{y'}$. So,$A = \left(x - \frac{y}{y'}, 0\right)$.
For point $B$ ($Y$-axis),set $X = 0$: $Y - y = y'(-x) \Rightarrow Y = y - xy'$. So,$B = (0, y - xy')$.
Given $AP:BP = 3:1$. Using the section formula for point $P(x, y)$ dividing $AB$ in ratio $3:1$:
$x = \frac{1 \cdot (x - y/y') + 3 \cdot 0}{3 + 1} = \frac{x - y/y'}{4} \Rightarrow 4x = x - \frac{y}{y'} \Rightarrow 3x = -\frac{y}{y'} \Rightarrow 3xy' = -y \Rightarrow 3x \frac{dy}{dx} + y = 0$.
This matches option $A$.
Solving the differential equation: $\frac{3 dy}{y} + \frac{dx}{x} = 0 \Rightarrow 3 \ln|y| + \ln|x| = C \Rightarrow \ln|xy^3| = C \Rightarrow xy^3 = k$.
Since it passes through $(1, 1)$,$1(1)^3 = k \Rightarrow k = 1$. The curve is $xy^3 = 1$.
Check option $C$: If $x = 1/8$,$y^3 = 8 \Rightarrow y = 2$. So,the curve passes through $(1/8, 2)$.
Check option $D$: At $(1, 1)$,$y' = -\frac{y}{3x} = -\frac{1}{3}$. Slope of normal $= -\frac{1}{y'} = 3$.
Equation of normal: $y - 1 = 3(x - 1) \Rightarrow y - 1 = 3x - 3 \Rightarrow 3x - y = 2$. Option $D$ is incorrect.

(C) Given the function $f(x) = 2x^{2} - 3x + 2$.
We need to find the differential $dy$ when $x$ changes from $x = 2$ to $x = 1.99$.
Here,$x = 2$ and $\Delta x = 1.99 - 2 = -0.01$.
The derivative of the function is $f'(x) = \frac{d}{dx}(2x^{2} - 3x + 2) = 4x - 3$.
At $x = 2$,$f'(2) = 4(2) - 3 = 8 - 3 = 5$.
The differential $dy$ is given by the formula $dy = f'(x) \Delta x$.
Substituting the values,$dy = f'(2) \times (-0.01) = 5 \times (-0.01) = -0.05$.
Thus,the differential of $y$ is $-0.05$.

(A) Given $\phi(x) = f(x) + f(1-x)$.
To determine the monotonicity,we find the derivative $\phi^{\prime}(x) = f^{\prime}(x) - f^{\prime}(1-x)$.
Since $f^{\prime \prime}(x) < 0$,the function $f^{\prime}(x)$ is a strictly decreasing function.
If $x < \frac{1}{2}$,then $x < 1-x$,which implies $f^{\prime}(x) > f^{\prime}(1-x)$ because $f^{\prime}$ is decreasing.
Thus,$\phi^{\prime}(x) = f^{\prime}(x) - f^{\prime}(1-x) > 0$ for $x \in \left[0, \frac{1}{2}\right]$,so $\phi(x)$ is increasing in $\left[0, \frac{1}{2}\right]$.
If $x > \frac{1}{2}$,then $x > 1-x$,which implies $f^{\prime}(x) < f^{\prime}(1-x)$ because $f^{\prime}$ is decreasing.
Thus,$\phi^{\prime}(x) = f^{\prime}(x) - f^{\prime}(1-x) < 0$ for $x \in \left[\frac{1}{2}, 1\right]$,so $\phi(x)$ is decreasing in $\left[\frac{1}{2}, 1\right]$.

(B) Let $f(x) = \cos x + x \sin x - 1$.
Find the derivative of $f(x)$ with respect to $x$:
$f'(x) = -\sin x + (1 \cdot \sin x + x \cos x) = x \cos x$.
For $x \in \left(0, \frac{\pi}{2}\right)$,both $x > 0$ and $\cos x > 0$.
Therefore,$f'(x) = x \cos x > 0$ for all $x \in \left(0, \frac{\pi}{2}\right)$.
Since $f'(x) > 0$,the function $f(x)$ is strictly increasing on the interval $\left(0, \frac{\pi}{2}\right)$.
As $f(x)$ is increasing,for any $x > 0$,we have $f(x) > f(0)$.
Calculate $f(0)$:
$f(0) = \cos(0) + 0 \cdot \sin(0) - 1 = 1 + 0 - 1 = 0$.
Thus,$f(x) > 0$ for all $x \in \left(0, \frac{\pi}{2}\right)$.
$\cos x + x \sin x - 1 > 0$
$\cos x + x \sin x > 1$.

(C) Given the function $f(x) = 1 - \sqrt{x^2}$.
Since $\sqrt{x^2} = |x|$,the function can be rewritten as $f(x) = 1 - |x|$.
At $x = 0$,$f(0) = 1 - |0| = 1$.
For any $x \neq 0$,$|x| > 0$,so $f(x) = 1 - |x| < 1$.
Since $f(x) \leq f(0)$ for all $x$ in the domain,the function attains its maximum value at $x = 0$.
Therefore,$f$ has a maxima at $x = 0$.

(A) Given the function $f(x)=2x^{3}-9ax^{2}+12a^{2}x+1$.
First,find the derivative $f'(x)$:
$f'(x) = 6x^{2}-18ax+12a^{2}$.
Set $f'(x)=0$ to find critical points:
$6(x^{2}-3ax+2a^{2})=0 \Rightarrow 6(x-a)(x-2a)=0$.
So,the critical points are $x=a$ and $x=2a$.
Now,find the second derivative $f''(x)$:
$f''(x) = 12x-18a$.
Check the nature of the critical points:
For $x=a$: $f''(a) = 12a-18a = -6a$. Since $a>0$,$f''(a) < 0$,so $f(x)$ has a local maximum at $p=a$.
For $x=2a$: $f''(2a) = 12(2a)-18a = 6a$. Since $a>0$,$f''(2a) > 0$,so $f(x)$ has a local minimum at $q=2a$.
Given the condition $p^{2}=q$,substitute the values:
$a^{2} = 2a$.
Since $a>0$,we can divide by $a$:
$a = 2$.

(A) Given that $f(0) = f(1) = 0$.
By Rolle's theorem,there exists at least one point $c \in (0, 1)$ such that $f^{\prime}(c) = 0$.
We are also given that $f^{\prime}(0) = 0$.
Now,consider the function $f^{\prime}(x)$ on the interval $[0, c]$.
Since $f$ is twice continuously differentiable,$f^{\prime}$ is continuous on $[0, c]$ and differentiable on $(0, c)$.
Also,$f^{\prime}(0) = 0$ and $f^{\prime}(c) = 0$.
Applying Rolle's theorem to $f^{\prime}(x)$ on $[0, c]$,there exists at least one point $c_1 \in (0, c)$ such that $(f^{\prime})^{\prime}(c_1) = f^{\prime \prime}(c_1) = 0$.
Thus,$f^{\prime \prime}(c_1) = 0$ for some $c_1 \in (0, 1)$.

(C) Let $u = f(x) \varphi(x)$. Then,by the product rule,$du = (f(x) \varphi^{\prime}(x) + \varphi(x) f^{\prime}(x)) dx$.
Substituting this into the integral,we get:
$I = \int \frac{du}{(u+1) \sqrt{u-1}}$.
Now,let $u-1 = p^2$,which implies $u = p^2 + 1$ and $du = 2p dp$.
Substituting these into the integral:
$I = \int \frac{2p dp}{(p^2 + 1 + 1) \sqrt{p^2}} = \int \frac{2p dp}{(p^2 + 2) p} = \int \frac{2 dp}{p^2 + 2}$.
Using the standard integral formula $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + c$:
$I = 2 \cdot \frac{1}{\sqrt{2}} \tan^{-1}(\frac{p}{\sqrt{2}}) + c = \sqrt{2} \tan^{-1}(\frac{p}{\sqrt{2}}) + c$.
Substituting back $p = \sqrt{u-1} = \sqrt{f(x) \varphi(x) - 1}$:
$I = \sqrt{2} \tan^{-1} \sqrt{\frac{f(x) \varphi(x) - 1}{2}} + c$.

(C) Given the equation of the circle $x^{2}+y^{2}=a^{2}$.
Differentiating with respect to $x$,we get $2x + 2y \frac{dy}{dx} = 0$,which implies $\frac{dy}{dx} = -\frac{x}{y}$.
Now,substitute this into the integral: $\sqrt{1+\left(\frac{dy}{dx}\right)^{2}} = \sqrt{1+\frac{x^{2}}{y^{2}}} = \sqrt{\frac{y^{2}+x^{2}}{y^{2}}} = \sqrt{\frac{a^{2}}{y^{2}}} = \frac{a}{y}$.
The integral becomes $\int_{0}^{a} \frac{a}{y} dx$.
Since $y = \sqrt{a^{2}-x^{2}}$,the integral is $\int_{0}^{a} \frac{a}{\sqrt{a^{2}-x^{2}}} dx$.
Evaluating this,we get $a \left[ \sin^{-1} \left( \frac{x}{a} \right) \right]_{0}^{a} = a \left( \sin^{-1}(1) - \sin^{-1}(0) \right) = a \left( \frac{\pi}{2} - 0 \right) = \frac{\pi a}{2}$.

(D) Let $I = \sum_{n=1}^{10} \int_{-2n-1}^{-2n} \sin^{27} x \, dx + \sum_{n=1}^{10} \int_{2n}^{2n+1} \sin^{27} x \, dx$.
Consider the first integral term: $J_n = \int_{-2n-1}^{-2n} \sin^{27} x \, dx$.
Let $x = -t$,then $dx = -dt$. When $x = -2n-1$,$t = 2n+1$. When $x = -2n$,$t = 2n$.
So,$J_n = \int_{2n+1}^{2n} \sin^{27}(-t) (-dt) = \int_{2n}^{2n+1} -\sin^{27} t \, dt = -\int_{2n}^{2n+1} \sin^{27} t \, dt$.
Substituting this back into the sum,we get:
$I = \sum_{n=1}^{10} (-\int_{2n}^{2n+1} \sin^{27} x \, dx) + \sum_{n=1}^{10} \int_{2n}^{2n+1} \sin^{27} x \, dx$.
$I = -\sum_{n=1}^{10} \int_{2n}^{2n+1} \sin^{27} x \, dx + \sum_{n=1}^{10} \int_{2n}^{2n+1} \sin^{27} x \, dx = 0$.

(B) To evaluate $\int_{0}^{2} [x^{2}] dx$,we split the interval $[0, 2]$ based on the points where $x^{2}$ is an integer: $x^{2} = 1, 2, 3, 4$.
This gives the points $x = 1, \sqrt{2}, \sqrt{3}, 2$.
Thus,the integral becomes:
$\int_{0}^{1} [x^{2}] dx + \int_{1}^{\sqrt{2}} [x^{2}] dx + \int_{\sqrt{2}}^{\sqrt{3}} [x^{2}] dx + \int_{\sqrt{3}}^{2} [x^{2}] dx$
$= \int_{0}^{1} 0 dx + \int_{1}^{\sqrt{2}} 1 dx + \int_{\sqrt{2}}^{\sqrt{3}} 2 dx + \int_{\sqrt{3}}^{2} 3 dx$
$= 0 + [x]_{1}^{\sqrt{2}} + 2[x]_{\sqrt{2}}^{\sqrt{3}} + 3[x]_{\sqrt{3}}^{2}$
$= (\sqrt{2} - 1) + 2(\sqrt{3} - \sqrt{2}) + 3(2 - \sqrt{3})$
$= \sqrt{2} - 1 + 2\sqrt{3} - 2\sqrt{2} + 6 - 3\sqrt{3}$
$= 5 - \sqrt{2} - \sqrt{3}$.

(C) The given expression is a Riemann sum for the definite integral of the function $f(x)$ over the interval $[0, 1]$.
By the definition of the definite integral as the limit of a sum:
$\lim_{n \rightarrow \infty} \sum_{j=0}^n \frac{1}{n} f\left(\frac{j}{n}\right) = \int_{0}^{1} f(x) dx$
Here,we substitute $\frac{j}{n} = x$ and $\frac{1}{n} = dx$.
The lower limit is $\lim_{n \rightarrow \infty} \frac{0}{n} = 0$.
The upper limit is $\lim_{n \rightarrow \infty} \frac{n}{n} = 1$.
Therefore,the limit is equal to $\int_{0}^{1} f(x) dx$.

(C) The given region is defined by the inequalities $x^{2}+y^{2} \leq 1$ and $x+y \geq 1$.
The inequality $x^{2}+y^{2} \leq 1$ represents the interior of a circle with center $(0, 0)$ and radius $1$.
The inequality $x+y \geq 1$ represents the region on or above the line $x+y=1$.
The intersection points of the circle $x^{2}+y^{2}=1$ and the line $x+y=1$ are $(1, 0)$ and $(0, 1)$.
The area of the region is the area of the circular segment bounded by the arc of the circle and the chord connecting $(1, 0)$ and $(0, 1)$.
This area is equal to the area of the circular sector (corresponding to the first quadrant arc) minus the area of the right-angled triangle formed by the origin $(0, 0)$,$(1, 0)$,and $(0, 1)$.
Area of the circular sector in the first quadrant $= \frac{1}{4} \times \pi \times (1)^{2} = \frac{\pi}{4}$.
Area of the right-angled triangle $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1 \times 1 = \frac{1}{2}$.
Therefore,the required area $= \frac{\pi}{4} - \frac{1}{2}$.

(B) To find the area bounded by the parabolas $x = -2y^{2}$ and $x = 1 - 3y^{2}$,we first find their points of intersection by setting the two equations equal to each other:
$-2y^{2} = 1 - 3y^{2}$
$y^{2} = 1$
$y = \pm 1$
Thus,the curves intersect at $y = -1$ and $y = 1$.
The area $A$ is given by the integral of the right curve minus the left curve with respect to $y$:
$A = \int_{-1}^{1} [(1 - 3y^{2}) - (-2y^{2})] dy$
$A = \int_{-1}^{1} (1 - y^{2}) dy$
Since the integrand $(1 - y^{2})$ is an even function,we can simplify the integral:
$A = 2 \int_{0}^{1} (1 - y^{2}) dy$
$A = 2 [y - \frac{y^{3}}{3}]_{0}^{1}$
$A = 2 (1 - \frac{1}{3}) = 2 (\frac{2}{3}) = \frac{4}{3}$ square units.

(B) Given equation: $y = e^{x}(A \cos x + B \sin x)$
Differentiating with respect to $x$ using the product rule:
$y' = e^{x}(A \cos x + B \sin x) + e^{x}(-A \sin x + B \cos x)$
$y' = y + e^{x}(-A \sin x + B \cos x)$
Differentiating again with respect to $x$:
$y'' = y' + [e^{x}(-A \sin x + B \cos x) + e^{x}(-A \cos x - B \sin x)]$
From the first derivative,we know $e^{x}(-A \sin x + B \cos x) = y' - y$.
Substituting this into the second derivative:
$y'' = y' + (y' - y) - e^{x}(A \cos x + B \sin x)$
Since $y = e^{x}(A \cos x + B \sin x)$,we have:
$y'' = 2y' - y - y$
$y'' - 2y' + 2y = 0$
Thus,the differential equation is $\frac{d^{2} y}{d x^{2}} - 2 \frac{d y}{d x} + 2 y = 0$.

(B) Given $y = \frac{1}{1 + x + \ln x}$.
Taking the reciprocal,we get $\frac{1}{y} = 1 + x + \ln x$.
Differentiating both sides with respect to $x$:
$-\frac{1}{y^{2}} \frac{dy}{dx} = 1 + \frac{1}{x} = \frac{x + 1}{x}$.
From the original equation,$1 + \ln x = \frac{1}{y} - x$.
Substitute this into the expression for $\frac{dy}{dx}$:
$\frac{dy}{dx} = -\frac{y^{2}(x + 1)}{x}$.
Alternatively,differentiating $y(1 + x + \ln x) = 1$ with respect to $x$:
$\frac{dy}{dx}(1 + x + \ln x) + y(1 + \frac{1}{x}) = 0$.
Since $1 + x + \ln x = \frac{1}{y}$,we have:
$\frac{dy}{dx} \cdot \frac{1}{y} + y \left(\frac{x + 1}{x}\right) = 0$.
$\frac{1}{y} \frac{dy}{dx} = -y \left(\frac{x + 1}{x}\right)$.
$x \frac{dy}{dx} = -y^{2}(x + 1)$.
Wait,let us re-evaluate: $\frac{1}{y} = 1 + x + \ln x \implies \frac{d}{dx}(\frac{1}{y}) = \frac{d}{dx}(1 + x + \ln x) \implies -\frac{1}{y^{2}} \frac{dy}{dx} = 1 + \frac{1}{x} = \frac{x + 1}{x}$.
$\frac{dy}{dx} = -\frac{y^{2}(x + 1)}{x}$.
Checking option $B$: $x \frac{dy}{dx} = y(y \ln x - 1)$.
If $y = \frac{1}{1 + x + \ln x}$,then $y \ln x - 1 = y \ln x - y(1 + x + \ln x) = y(\ln x - 1 - x - \ln x) = y(-1 - x) = -y(1 + x)$.
So $y(y \ln x - 1) = y(-y(1 + x)) = -y^{2}(1 + x)$.
Thus,$x \frac{dy}{dx} = -y^{2}(1 + x)$,which matches.

(B) Given the differential equation: $x \sin \left(\frac{y}{x}\right) \frac{dy}{dx} = y \sin \left(\frac{y}{x}\right) - x$.
Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into the equation: $x \sin(v) (v + x \frac{dv}{dx}) = vx \sin(v) - x$.
Dividing by $x$: $\sin(v) (v + x \frac{dv}{dx}) = v \sin(v) - 1$.
$v \sin(v) + x \sin(v) \frac{dv}{dx} = v \sin(v) - 1$.
$x \sin(v) \frac{dv}{dx} = -1$.
Separating variables: $\sin(v) dv = -\frac{1}{x} dx$.
Integrating both sides: $\int \sin(v) dv = -\int \frac{1}{x} dx$.
$-\cos(v) = -\log|x| + C$.
Since $y(1) = \frac{\pi}{2}$,at $x = 1$,$v = \frac{y}{x} = \frac{\pi/2}{1} = \frac{\pi}{2}$.
$-\cos(\frac{\pi}{2}) = -\log(1) + C \Rightarrow 0 = 0 + C \Rightarrow C = 0$.
Thus,$-\cos(v) = -\log x$,which implies $\cos(\frac{y}{x}) = \log x$.

(C) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = f^{\prime}(x)$ and $Q(x) = f(x)f^{\prime}(x)$.
The integrating factor $(IF)$ is $e^{\int P(x) dx} = e^{\int f^{\prime}(x) dx} = e^{f(x)}$.
The general solution is given by $y \cdot IF = \int Q(x) \cdot IF dx + C$.
$y \cdot e^{f(x)} = \int f(x) f^{\prime}(x) e^{f(x)} dx + C$.
Let $u = f(x)$,then $du = f^{\prime}(x) dx$. The integral becomes $\int u e^u du = u e^u - e^u$.
Thus,$y \cdot e^{f(x)} = e^{f(x)}(f(x) - 1) + C$.
Given $\lim_{x \rightarrow \infty} f(x) = 0$ and $\lim_{x \rightarrow \infty} y(x) = 0$,we substitute these into the equation:
$0 \cdot e^0 = e^0(0 - 1) + C \Rightarrow 0 = -1 + C \Rightarrow C = 1$.
Substituting $C = 1$ back into the equation:
$y \cdot e^{f(x)} = e^{f(x)}(f(x) - 1) + 1$.
Dividing by $e^{f(x)}$:
$y = f(x) - 1 + e^{-f(x)}$.
Rearranging gives $y + 1 = e^{-f(x)} + f(x)$.

(B) Let the unit vector in the $ZOX$ plane be $\vec{r} = x \hat{i} + z \hat{k}$.
Since it is a unit vector,$|\vec{r}|^2 = x^2 + z^2 = 1$.
Given $\vec{\alpha} = 2 \hat{i} + 2 \hat{j} - \hat{k}$,we have $|\vec{\alpha}| = \sqrt{2^2 + 2^2 + (-1)^2} = \sqrt{4+4+1} = 3$.
The angle between $\vec{r}$ and $\vec{\alpha}$ is $45^{\circ}$,so $\vec{r} \cdot \vec{\alpha} = |\vec{r}| |\vec{\alpha}| \cos 45^{\circ}$.
$(x \hat{i} + z \hat{k}) \cdot (2 \hat{i} + 2 \hat{j} - \hat{k}) = 1 \cdot 3 \cdot \frac{1}{\sqrt{2}} \Rightarrow 2x - z = \frac{3}{\sqrt{2}}$.
Given $\vec{\beta} = \hat{j} - \hat{k}$,we have $|\vec{\beta}| = \sqrt{0^2 + 1^2 + (-1)^2} = \sqrt{2}$.
The angle between $\vec{r}$ and $\vec{\beta}$ is $60^{\circ}$,so $\vec{r} \cdot \vec{\beta} = |\vec{r}| |\vec{\beta}| \cos 60^{\circ}$.
$(x \hat{i} + z \hat{k}) \cdot (0 \hat{i} + 1 \hat{j} - 1 \hat{k}) = 1 \cdot \sqrt{2} \cdot \frac{1}{2} \Rightarrow -z = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \Rightarrow z = -\frac{1}{\sqrt{2}}$.
Substituting $z$ into $2x - z = \frac{3}{\sqrt{2}}$,we get $2x - (-\frac{1}{\sqrt{2}}) = \frac{3}{\sqrt{2}} \Rightarrow 2x = \frac{2}{\sqrt{2}} = \sqrt{2} \Rightarrow x = \frac{1}{\sqrt{2}}$.
Thus,$\vec{r} = \frac{1}{\sqrt{2}} \hat{i} - \frac{1}{\sqrt{2}} \hat{k}$.

(C) Given the determinant equation:
$\left|\begin{array}{lll}a & a^{2} & 1+a^{3} \\ b & b^{2} & 1+b^{3} \\ c & c^{2} & 1+c^{3}\end{array}\right|=0$
We can split the third column:
$\left|\begin{array}{lll}a & a^{2} & 1 \\ b & b^{2} & 1 \\ c & c^{2} & 1\end{array}\right| + \left|\begin{array}{lll}a & a^{2} & a^{3} \\ b & b^{2} & b^{3} \\ c & c^{2} & c^{3}\end{array}\right| = 0$
Taking $abc$ common from the second determinant:
$\left|\begin{array}{lll}a & a^{2} & 1 \\ b & b^{2} & 1 \\ c & c^{2} & 1\end{array}\right| + abc \left|\begin{array}{lll}a & a^{2} & 1 \\ b & b^{2} & 1 \\ c & c^{2} & 1\end{array}\right| = 0$
$(1+abc) \left|\begin{array}{lll}a & a^{2} & 1 \\ b & b^{2} & 1 \\ c & c^{2} & 1\end{array}\right| = 0$
Since $\vec{\alpha}, \vec{\beta}, \vec{\gamma}$ are non-coplanar,the scalar triple product $[\vec{\alpha} \vec{\beta} \vec{\gamma}] \neq 0$. The determinant $\left|\begin{array}{lll}1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2}\end{array}\right|$ is related to the scalar triple product. Since the rows/columns are swapped,the value is non-zero.
Therefore,$1+abc = 0$,which implies $abc = -1$.

(A) The equation of a plane passing through the point $(x_1, y_1, z_1)$ is given by $a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$. Substituting the point $(2, -1, -3)$,we get $a(x - 2) + b(y + 1) + c(z + 3) = 0$.
Since the plane is parallel to the lines with direction ratios $(3, 2, -4)$ and $(2, -3, 2)$,the normal vector $(a, b, c)$ must be perpendicular to these direction vectors.
Thus,$3a + 2b - 4c = 0$ and $2a - 3b + 2c = 0$.
Using the cross product to find the normal vector $(a, b, c) = (3, 2, -4) \times (2, -3, 2) = \begin{vmatrix} i & j & k \\ 3 & 2 & -4 \\ 2 & -3 & 2 \end{vmatrix} = i(4 - 12) - j(6 + 8) + k(-9 - 4) = -8i - 14j - 13k$.
Taking the normal vector as $(8, 14, 13)$,the equation of the plane is $8(x - 2) + 14(y + 1) + 13(z + 3) = 0$.
Expanding this,we get $8x - 16 + 14y + 14 + 13z + 39 = 0$,which simplifies to $8x + 14y + 13z + 37 = 0$.

(B) The direction vector of the line is $\vec{b} = 3\hat{i} + 4\hat{j} + 5\hat{k}$.
The normal vector to the plane is $\vec{n} = 2\hat{i} - 2\hat{j} + 1\hat{k}$.
Let $\theta$ be the angle between the line and the plane. The formula for $\sin \theta$ is given by $\sin \theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$.
Calculating the dot product: $\vec{b} \cdot \vec{n} = (3)(2) + (4)(-2) + (5)(1) = 6 - 8 + 5 = 3$.
Calculating the magnitudes: $|\vec{b}| = \sqrt{3^2 + 4^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50} = 5\sqrt{2}$.
$|\vec{n}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.
Thus,$\sin \theta = \frac{|3|}{(5\sqrt{2})(3)} = \frac{3}{15\sqrt{2}} = \frac{1}{5\sqrt{2}} = \frac{\sqrt{2}}{10}$.

(C) Given that $A$ and $B$ are independent events,$P(A \cap B) = P(A)P(B) = \frac{1}{20}$.
Let $P(A) = x$ and $P(B) = y$. Then $xy = \frac{1}{20}$,so $y = \frac{1}{20x}$.
The probability that neither occurs is $P(\bar{A} \cap \bar{B}) = P(\bar{A})P(\bar{B}) = (1-x)(1-y) = \frac{3}{5}$.
Substituting $y = \frac{1}{20x}$ into the equation:
$(1-x)(1-\frac{1}{20x}) = \frac{3}{5}$
$1 - \frac{1}{20x} - x + \frac{1}{20} = \frac{3}{5}$
$\frac{21}{20} - x - \frac{1}{20x} = \frac{3}{5}$
Multiply by $20x$:
$21x - 20x^2 - 1 = 12x$
$20x^2 - 9x + 1 = 0$
$(4x-1)(5x-1) = 0$
Thus,$x = \frac{1}{4}$ or $x = \frac{1}{5}$.
Therefore,$P(A) = \frac{1}{4}$ or $P(A) = \frac{1}{5}$.

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

(B) Let $n$ be the number of rounds fired.
Probability of hitting the target in a single trial,$p = 10 \% = 0.1$.
Probability of missing the target in a single trial,$q = 1 - p = 0.9$.
The probability of hitting the target at least once in $n$ trials is given by $P(X \geq 1) = 1 - P(X = 0) = 1 - q^n$.
We want this probability to be greater than $50 \%$,so:
$1 - (0.9)^n > 0.5$
$(0.9)^n < 0.5$
Now,we test values for $n$:
For $n = 6$,$(0.9)^6 = 0.531441 > 0.5$.
For $n = 7$,$(0.9)^7 = 0.4782969 < 0.5$.
Thus,the least number of rounds required is $n = 7$.