WBJEE 2022 Mathematics Question Paper with Answer and Solution

(B) The domain of the expression is $x > 0$.
Given inequality: $\log _{25} x^2 + (\log _5 x)^2 < 2$.
Using the property $\log _{a^n} b^m = \frac{m}{n} \log _a b$,we have $\log _{25} x^2 = \log _{5^2} x^2 = \frac{2}{2} \log _5 x = \log _5 x$.
Substituting this into the inequality: $\log _5 x + (\log _5 x)^2 < 2$.
Let $y = \log _5 x$. Then $y^2 + y - 2 < 0$.
Factoring the quadratic: $(y + 2)(y - 1) < 0$.
This implies $-2 < y < 1$.
Substituting back $y = \log _5 x$: $-2 < \log _5 x < 1$.
Converting to exponential form: $5^{-2} < x < 5^1$.
Thus,$\frac{1}{25} < x < 5$.

(A) The given quadratic equation is $2ax^2 + (2a + b)x + b = 0$.
The discriminant $D$ is given by $D = B^2 - 4AC$.
Here,$A = 2a$,$B = (2a + b)$,and $C = b$.
$D = (2a + b)^2 - 4(2a)(b) = 4a^2 + 4ab + b^2 - 8ab = 4a^2 - 4ab + b^2 = (2a - b)^2$.
Since $a$ and $b$ are integers,$(2a - b)^2$ is a perfect square.
If the discriminant of a quadratic equation with rational coefficients is a perfect square,the roots are rational.
Thus,the roots are rational.

(B) Let $\alpha$ and $\beta$ be the roots of the equation $x^2-(a-2)x-(a+1)=0$.
From the relation between roots and coefficients,we have $\alpha+\beta = a-2$ and $\alpha\beta = -(a+1)$.
The sum of the squares of the roots is given by $\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta$.
Substituting the values,we get $\alpha^2+\beta^2 = (a-2)^2 - 2(-(a+1)) = (a-2)^2 + 2(a+1)$.
Expanding this,we get $\alpha^2+\beta^2 = a^2 - 4a + 4 + 2a + 2 = a^2 - 2a + 6$.
To find the minimum value,we complete the square: $a^2 - 2a + 6 = (a^2 - 2a + 1) + 5 = (a-1)^2 + 5$.
Since $(a-1)^2 \geq 0$,the minimum value is $5$,which occurs when $a-1 = 0$,i.e.,$a = 1$.

(D) Given $z^{1/3} = p + iq$,we have $z = (p + iq)^3$.
Expanding this,$z = p^3 + 3p^2(iq) + 3p(iq)^2 + (iq)^3 = p^3 + 3ip^2q - 3pq^2 - iq^3$.
Grouping real and imaginary parts,$z = (p^3 - 3pq^2) + i(3p^2q - q^3)$.
Since $z = x - iy$,we equate the real and imaginary parts:
$x = p^3 - 3pq^2 = p(p^2 - 3q^2)$
$-y = 3p^2q - q^3 \implies y = q^3 - 3p^2q = -q(3p^2 - q^2)$.
Now,calculate $\frac{x}{p} + \frac{y}{q}$:
$\frac{x}{p} = p^2 - 3q^2$
$\frac{y}{q} = q^2 - 3p^2$
$\frac{x}{p} + \frac{y}{q} = (p^2 - 3q^2) + (q^2 - 3p^2) = -2p^2 - 2q^2 = -2(p^2 + q^2)$.
Finally,$\frac{(\frac{x}{p} + \frac{y}{q})}{(p^2 + q^2)} = \frac{-2(p^2 + q^2)}{(p^2 + q^2)} = -2$.

(B) The given inequality $|z - 25i| \leq 15$ represents a disk in the complex plane with center at $(0, 25)$ and radius $r = 15$.
Let the tangents from the origin to the circle touch the circle at points $P$ and $Q$.
The angle $\theta$ between the $y$-axis and the line segment from the origin to the center $(0, 25)$ is given by $\sin \theta = \frac{r}{d} = \frac{15}{25} = \frac{3}{5}$,where $d = 25$ is the distance from the origin to the center.
Thus,$\theta = \sin^{-1}\left(\frac{3}{5}\right)$.
The argument of $z$ ranges from $\frac{\pi}{2} - \theta$ to $\frac{\pi}{2} + \theta$.
Therefore,$\text{Maximum } \arg(z) = \frac{\pi}{2} + \theta$ and $\text{Minimum } \arg(z) = \frac{\pi}{2} - \theta$.
The difference is $(\frac{\pi}{2} + \theta) - (\frac{\pi}{2} - \theta) = 2\theta = 2 \sin^{-1}\left(\frac{3}{5}\right)$.
Since $\sin^{-1}\left(\frac{3}{5}\right) = \cos^{-1}\left(\frac{4}{5}\right)$,the difference is $2 \cos^{-1}\left(\frac{4}{5}\right)$.

(A) The argument of a product of complex numbers is given by $\arg(z_1 z_2) = \arg z_1 + \arg z_2 + 2k\pi$,where $k \in \{0, 1, -1\}$.
Since the principal value of the argument lies in the interval $(-\pi, \pi]$,the sum of the principal arguments may fall outside this range.
Therefore,the principal value of $\arg(z_1 z_2)$ is not necessarily equal to the sum of the principal values of $\arg z_1$ and $\arg z_2$.
Similarly,for the quotient,$\arg(z_1 / z_2) = \arg z_1 - \arg z_2 + 2k\pi$,which also may not equal the difference of the principal values.

(B) Let the vertex $C$ be represented by the complex number $z = x + iy$. The coordinates of $A$ are $(0, 0)$ and $B$ are $(2a, 0)$.
Since the angle $\angle ACB = \alpha$,the argument of the ratio of the vectors $\vec{CA}$ and $\vec{CB}$ is $\alpha$.
$\arg \left( \frac{0 - z}{2a - z} \right) = \alpha$
$\arg \left( \frac{-z}{2a - z} \right) = \alpha$
$\arg \left( \frac{z}{z - 2a} \right) = \alpha$
Using the property $\arg \left( \frac{z - z_1}{z - z_2} \right) = \alpha$,the locus is a circle passing through $A(0,0)$ and $B(2a,0)$.
The equation of the circle is $(x^2 + y^2) - 2ax - 2ay \cot \alpha = 0$.

(D) For each element $x \in A$,there are three mutually exclusive possibilities regarding its membership in subsets $P$ and $Q$ such that $P \cap Q = \phi$:
$1$. $x \in P$ and $x \notin Q$
$2$. $x \notin P$ and $x \in Q$
$3$. $x \notin P$ and $x \notin Q$
Since there are $n$ elements in $A$ and each element has $3$ independent choices,the total number of ways to choose subsets $P$ and $Q$ is $3 \times 3 \times \dots \times 3$ ($n$ times) = $3^n$.

(C) To arrange $n$ white and $n$ black balls such that no two balls of the same colour are adjacent,the balls must alternate in colour.
There are two possible patterns for the arrangement:
$1$. $B, W, B, W, \ldots, B, W$ (starting with Black)
$2$. $W, B, W, B, \ldots, W, B$ (starting with White)
For each pattern,the $n$ black balls can be arranged in $n!$ ways and the $n$ white balls can be arranged in $n!$ ways.
Thus,the number of ways for the first pattern is $n! \times n! = (n!)^2$.
Similarly,the number of ways for the second pattern is $n! \times n! = (n!)^2$.
Total number of ways = $(n!)^2 + (n!)^2 = 2(n!)^2$.

(D) To find the number of zeros at the end of $100!$,we need to calculate the exponent of the highest power of $5$ that divides $100!$,denoted as $E_5(100!)$.
Using Legendre's Formula: $E_p(n!) = \sum_{k=1}^{\infty} \left[ \frac{n}{p^k} \right]$.
For $n = 100$ and $p = 5$:
$E_5(100!) = \left[ \frac{100}{5} \right] + \left[ \frac{100}{25} \right] + \left[ \frac{100}{125} \right]$
$E_5(100!) = 20 + 4 + 0 = 24$.
Therefore,there are $24$ zeros at the end of $100!$.

(B) Given $a, b, c$ are in $G$.$P$.,so $b^2 = ac$. Let $r = \frac{b}{a} = \frac{c}{b}$,then $b = ar$ and $c = ar^2$.
Since $\log a - \log 2b, \log 2b - \log 3c, \log 3c - \log a$ are in $A$.$P$.,we have $2(\log 2b - \log 3c) = (\log a - \log 2b) + (\log 3c - \log a)$.
$2 \log(\frac{2b}{3c}) = \log(\frac{a}{2b} \times \frac{3c}{a}) = \log(\frac{3c}{2b})$.
Let $x = \frac{2b}{3c}$,then $2 \log x = \log(\frac{1}{x}) = -\log x$,which implies $3 \log x = 0$,so $x = 1$.
Thus,$\frac{2b}{3c} = 1 \Rightarrow 2b = 3c$.
Since $b = ar$ and $c = ar^2$,we have $2(ar) = 3(ar^2)$ $\Rightarrow 2 = 3r$ $\Rightarrow r = \frac{2}{3}$.
The sides are $a, b = \frac{2a}{3}, c = \frac{4a}{9}$.
For a triangle,the sum of two smaller sides must be greater than the third side: $\frac{2a}{3} + \frac{4a}{9} = \frac{10a}{9} > a$. This holds.
Using the Cosine rule for the largest angle $A$ (opposite to side $a$): $\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{(\frac{2a}{3})^2 + (\frac{4a}{9})^2 - a^2}{2(\frac{2a}{3})(\frac{4a}{9})} = \frac{\frac{4}{9} + \frac{16}{81} - 1}{\frac{16}{27}} = \frac{\frac{36+16-81}{81}}{\frac{16}{27}} = \frac{-29}{81} \times \frac{27}{16} = -\frac{29}{48} < 0$.
Since $\cos A < 0$,the angle $A$ is obtuse.

(B) We have $a_n = (\frac{n(n+1)(2n+1)}{6})^n$ and $b_n = n^n(n!)$.
For $n=1$,$a_1 = (1^2)^1 = 1$ and $b_1 = 1^1(1!) = 1$,so $a_1 = b_1$.
For $n=2$,$a_2 = (1^2+2^2)^2 = 5^2 = 25$ and $b_2 = 2^2(2!) = 4 \times 2 = 8$. Thus $a_2 > b_2$.
For $n=3$,$a_3 = (1^2+2^2+3^2)^3 = 14^3 = 2744$ and $b_3 = 3^3(3!) = 27 \times 6 = 162$. Thus $a_3 > b_3$.
By induction or comparing terms,it can be shown that $a_n > b_n$ for all $n \geq 2$.

(A) Let $P = \cos \alpha_1 \cos \alpha_2 \ldots \cos \alpha_n$.
Then $\frac{1}{P^2} = \frac{1}{\cos^2 \alpha_1 \cos^2 \alpha_2 \ldots \cos^2 \alpha_n} = \sec^2 \alpha_1 \sec^2 \alpha_2 \ldots \sec^2 \alpha_n$.
Using the identity $\sec^2 \alpha = 1 + \tan^2 \alpha$,we have:
$\frac{1}{P^2} = (1 + \tan^2 \alpha_1)(1 + \tan^2 \alpha_2) \ldots (1 + \tan^2 \alpha_n)$.
By the Arithmetic Mean-Geometric Mean ($AM$-$GM$) inequality,$1 + \tan^2 \alpha_i \geq 2 \tan \alpha_i$.
Therefore,$\frac{1}{P^2} \geq (2 \tan \alpha_1)(2 \tan \alpha_2) \ldots (2 \tan \alpha_n) = 2^n (\tan \alpha_1 \tan \alpha_2 \ldots \tan \alpha_n)$.
Since $(\cot \alpha_1)(\cot \alpha_2) \ldots (\cot \alpha_n) = 1$,it follows that $(\tan \alpha_1)(\tan \alpha_2) \ldots (\tan \alpha_n) = 1$.
Thus,$\frac{1}{P^2} \geq 2^n$,which implies $P^2 \leq \frac{1}{2^n}$.
Taking the square root,$P \leq \frac{1}{2^{n/2}}$.
The maximum value is $\frac{1}{2^{n/2}}$.

(B) Consider the difference $g(n) - f(n) = 1 + (n+1)2^n - 2^{n+1}$.
Since $2^{n+1} = 2 \cdot 2^n$,we have:
$g(n) - f(n) = 1 + (n+1)2^n - 2 \cdot 2^n$
$g(n) - f(n) = 1 + (n+1-2)2^n$
$g(n) - f(n) = 1 + (n-1)2^n$.
For all $n \in N$,$n \geq 1$,which implies $(n-1) \geq 0$.
Since $2^n > 0$,the term $(n-1)2^n \geq 0$.
Therefore,$1 + (n-1)2^n > 0$,which means $g(n) - f(n) > 0$ or $g(n) > f(n)$.

(B) Given the angle $\theta = \sin^{-1}(\frac{3}{5})$,we have $\tan \theta = \frac{3}{4}$. The slope of the line is $m = \frac{3}{4}$.
The equation of the line passing through $(-1, 1)$ with slope $m = \frac{3}{4}$ is $y - 1 = \frac{3}{4}(x + 1)$,which simplifies to $4y - 4 = 3x + 3$,or $4y = 3x + 7$.
Substitute $4y = 3x + 7$ into the curve equation $x^2 = 4y - 9$:
$x^2 = (3x + 7) - 9$
$x^2 - 3x + 2 = 0$
$(x - 1)(x - 2) = 0$
So,$x = 1$ and $x = 2$.
For $x = 1$,$4y = 3(1) + 7 = 10 \Rightarrow y = \frac{5}{2}$. Point $A = (1, \frac{5}{2})$.
For $x = 2$,$4y = 3(2) + 7 = 13 \Rightarrow y = \frac{13}{4}$. Point $B = (2, \frac{13}{4})$.
The distance $|AB| = \sqrt{(2 - 1)^2 + (\frac{13}{4} - \frac{5}{2})^2} = \sqrt{1^2 + (\frac{13-10}{4})^2} = \sqrt{1 + \frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4}$ units.

(B) The given equation is $y-y_1=m(x-x_1)$,which can be rewritten as $y=mx+(y_1-mx_1)$.
Since $m$ and $x_1$ are fixed,the slope of the lines is constant $(m)$.
Lines with the same slope are parallel to each other.
Therefore,for different values of $y_1$,we obtain a family of parallel lines.
Thus,option $B$ is correct.

(D) Let the equation of the variable line be $ax + by + c = 0$,where $a^2 + b^2 = 1$.
The perpendicular distance from a point $(x_1, y_1)$ to the line $ax + by + c = 0$ is given by $d = ax_1 + by_1 + c$.
The algebraic sum of the distances from the points $(2,0)$,$(0,2)$,and $(1,1)$ is zero:
$(2a + 0b + c) + (0a + 2b + c) + (1a + 1b + c) = 0$
$3a + 3b + 3c = 0$
$a + b + c = 0$
Substituting $c = -(a + b)$ into the line equation:
$ax + by - (a + b) = 0$
$a(x - 1) + b(y - 1) = 0$
This equation holds for all $a$ and $b$ if $x - 1 = 0$ and $y - 1 = 0$.
Thus,the line always passes through the fixed point $(1,1)$.

(A) Let the two perpendicular lines be the coordinate axes,$x=0$ and $y=0$.
Let the point be $P(x, y)$.
The distance of point $P$ from the line $x=0$ is $|x|$ and from the line $y=0$ is $|y|$.
According to the problem,the sum of these distances is $1$ unit,so $|x| + |y| = 1$.
This equation represents a square with vertices at $(1, 0), (0, 1), (-1, 0),$ and $(0, -1)$.

(B) Let the equation of the line $AB$ be $\frac{x}{a} + \frac{y}{b} = 1$. The points are $A(a, 0)$ and $B(0, b)$.
The circle passing through $O(0, 0)$,$A(a, 0)$,and $B(0, b)$ has the equation $x^2 + y^2 - ax - by = 0$.
The tangent to the circle at the origin $(0, 0)$ is obtained by replacing $x^2$ with $0$,$y^2$ with $0$,$x$ with $\frac{x}{2}$,and $y$ with $\frac{y}{2}$.
Thus,the tangent at the origin is $-ax - by = 0$,or $ax + by = 0$.
The distance $m$ from $A(a, 0)$ to the line $ax + by = 0$ is $m = \frac{|a(a) + b(0)|}{\sqrt{a^2 + b^2}} = \frac{a^2}{\sqrt{a^2 + b^2}}$.
The distance $n$ from $B(0, b)$ to the line $ax + by = 0$ is $n = \frac{|a(0) + b(b)|}{\sqrt{a^2 + b^2}} = \frac{b^2}{\sqrt{a^2 + b^2}}$.
The diameter of the circle is the length of the hypotenuse $AB = \sqrt{a^2 + b^2}$.
From the expressions for $m$ and $n$,we have $m+n = \frac{a^2 + b^2}{\sqrt{a^2 + b^2}} = \sqrt{a^2 + b^2}$.
Therefore,the diameter of the circle is $m+n$.

(C) Let the equation of the other tangent be $y = mx$,which can be written as $mx - y = 0$.
The center of the circle is $C(2, -1)$. The radius $r$ is the perpendicular distance from $C$ to the tangent $3x + y = 0$.
$r = \frac{|3(2) + 1(-1)|}{\sqrt{3^2 + 1^2}} = \frac{|6 - 1|}{\sqrt{10}} = \frac{5}{\sqrt{10}}$.
Since the other tangent $mx - y = 0$ also touches the circle,the perpendicular distance from $C(2, -1)$ to this line must also be equal to $r$.
$\frac{|m(2) - 1(-1)|}{\sqrt{m^2 + (-1)^2}} = \frac{5}{\sqrt{10}}$
$\frac{|2m + 1|}{\sqrt{m^2 + 1}} = \frac{5}{\sqrt{10}}$
Squaring both sides:
$\frac{(2m + 1)^2}{m^2 + 1} = \frac{25}{10} = \frac{5}{2}$
$2(4m^2 + 4m + 1) = 5(m^2 + 1)$
$8m^2 + 8m + 2 = 5m^2 + 5$
$3m^2 + 8m - 3 = 0$
$(3m - 1)(m + 3) = 0$
So,$m = \frac{1}{3}$ or $m = -3$.
The given tangent is $3x + y = 0$,which has slope $m = -3$.
Therefore,the other tangent has slope $m = \frac{1}{3}$.
The equation is $y = \frac{1}{3}x$,which simplifies to $x - 3y = 0$.

(C) The equation of the common chord of two circles $S_1 = 0$ and $S_2 = 0$ is given by $S_1 - S_2 = 0$,provided the coefficients of $x^2$ and $y^2$ are the same.
Given $S_1 = px^2 + py^2 + 2g'x + 2f'y + d = 0$,we first normalize it by dividing by $p$:
$\frac{S_1}{p} = x^2 + y^2 + \frac{2g'}{p}x + \frac{2f'}{p}y + \frac{d}{p} = 0$.
Now,the equation of the common chord is $\frac{S_1}{p} - S_2 = 0$.
Multiplying by $p$,we get $S_1 - pS_2 = 0$.

(D) Let the coordinates of point $B$ be $(at^2, 2at)$. The vertex $A$ is at $(0, 0)$.
The slope of $AB$ is $m_{AB} = \frac{2at - 0}{at^2 - 0} = \frac{2}{t}$.
Since $BC \perp AB$,the slope of $BC$ is $m_{BC} = -\frac{1}{m_{AB}} = -\frac{t}{2}$.
The equation of line $BC$ passing through $B(at^2, 2at)$ with slope $-\frac{t}{2}$ is:
$y - 2at = -\frac{t}{2}(x - at^2)$
To find the point $C$ on the axis of the parabola $(y = 0)$:
$0 - 2at = -\frac{t}{2}(x - at^2)$
$4at = t(x - at^2)$
$4a = x - at^2$
$x = 4a + at^2$
So,the coordinates of $C$ are $(4a + at^2, 0)$.
The projection of $BC$ on the axis is the distance $DC$,where $D$ is the projection of $B$ on the axis,i.e.,$D(at^2, 0)$.
$DC = |x_C - x_D| = |(4a + at^2) - at^2| = 4a$ units.

(A) Let the coordinates of the points be $P_i(at_i^2, 2at_i)$ for $i = 1, 2, 3, 4$.
Since $P_1 P_2$ and $P_3 P_4$ are focal chords,we have $t_1 t_2 = -1$ and $t_3 t_4 = -1$.
The equation of the chord joining $P_i$ and $P_j$ is $(t_i + t_j)y = 2x + 2at_i t_j$.
For $P_1 P_3$,the equation is $(t_1 + t_3)y = 2x + 2at_1 t_3$ ... $(1)$.
For $P_2 P_4$,the equation is $(t_2 + t_4)y = 2x + 2at_2 t_4$ ... $(2)$.
Since $t_2 = -1/t_1$ and $t_4 = -1/t_3$,substituting these into $(2)$ gives $( -1/t_1 - 1/t_3 )y = 2x + 2a(-1/t_1)(-1/t_3)$,which simplifies to $-(t_1 + t_3)y = 2xt_1 t_3 + 2a$.
Solving the system of equations $(1)$ and $(2)$ shows that the intersection point lies on the line $x = -a$,which is the directrix of the parabola.

(A, B, C) The line is $y=x+5$,so $m=1$ and $c=5$.
$(A)$ For parabola $y^2=4ax$,the condition for tangency is $c=\frac{a}{m}$. Here $4a=20 \Rightarrow a=5$. Thus $c=\frac{5}{1}=5$. The line is a tangent.
$(B)$ For ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$,the condition for tangency is $c^2=a^2m^2+b^2$. Here $a^2=16, b^2=9$. Thus $c^2=5^2=25$ and $a^2m^2+b^2=16(1)^2+9=25$. The line is a tangent.
$(C)$ For hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$,the condition for tangency is $c^2=a^2m^2-b^2$. Here $a^2=29, b^2=4$. Thus $c^2=5^2=25$ and $a^2m^2-b^2=29(1)^2-4=25$. The line is a tangent.
$(D)$ For circle $x^2+y^2=r^2$,the condition for tangency is $c^2=r^2(1+m^2)$. Here $r^2=25, m=1$. Thus $c^2=25$ and $r^2(1+m^2)=25(1+1)=50$. Since $25 \neq 50$,the line is not a tangent.
Therefore,options $(A)$,$(B)$,and $(C)$ are correct.

(A) The equation of the tangent to the parabola $y^2 = 4ax$ at point $(at^2, 2at)$ is $yt = x + at^2$.
Here,$4a = 9$,so $a = \frac{9}{4}$.
The tangent passes through $(4, 10)$,so $10t = 4 + \frac{9}{4}t^2$.
Multiplying by $4$,we get $40t = 16 + 9t^2$,or $9t^2 - 40t + 16 = 0$.
Factoring the quadratic: $(9t - 4)(t - 4) = 0$,which gives $t = 4$ or $t = \frac{4}{9}$.
The slope of the tangent is $m = \frac{1}{t} = \tan \theta$.
Given $\tan \theta > 2$,we have $\frac{1}{t} > 2$,which implies $t < \frac{1}{2}$.
Thus,$t = \frac{4}{9}$ is the valid parameter.
The point of contact is $(at^2, 2at) = \left(\frac{9}{4} \times \left(\frac{4}{9}\right)^2, 2 \times \frac{9}{4} \times \frac{4}{9}\right) = \left(\frac{4}{9}, 2\right)$.

(A) The coordinates of point $P$ are $(at^2, 2at)$.
The equation of the tangent at $P$ is $ty = x + at^2$. Setting $y=0$,we get $x = -at^2$,so $T = (-at^2, 0)$.
The equation of the normal at $P$ is $y = -tx + 2at + at^3$. Setting $y=0$,we get $x = 2a + at^2$,so $G = (2a + at^2, 0)$.
Since the tangent and normal are perpendicular,$\angle PTG = 90^\circ$,which implies that $TG$ is the diameter of the circle passing through $P, T,$ and $G$.
The length of the diameter $TG = |(2a + at^2) - (-at^2)| = |2a + 2at^2| = 2a(1+t^2)$.
Therefore,the radius of the circle is $\frac{1}{2} TG = a(1+t^2)$.

(D) The equation of the parabola is $y^2 = 4ax$,where $4a = 4$,so $a = 1$.
The directrix of the parabola $y^2 = 4x$ is given by $x = -a$,which is $x = -1$.
The given point is $(-1, -6)$.
Since the $x$-coordinate of the point is $-1$,the point $(-1, -6)$ lies on the directrix of the parabola.
It is a standard property of parabolas that the tangents drawn from any point on the directrix to the parabola are perpendicular to each other.
Therefore,the angle between the two tangents is $\pi / 2$.

(D) Let the coordinates of point $Q$ on the parabola $(y - 6)^2 = 2(x - 4)$ be $(4 + \frac{t^2}{2}, 6 + t)$.
Given $P = (2, 0)$.
Let the mid-point of $PQ$ be $R(h, k)$.
Then $h = \frac{2 + 4 + \frac{t^2}{2}}{2} = 3 + \frac{t^2}{4}$ and $k = \frac{0 + 6 + t}{2} = 3 + \frac{t}{2}$.
From the second equation,$\frac{t}{2} = k - 3$,so $t = 2(k - 3)$.
Substituting $t$ into the equation for $h$:
$h = 3 + \frac{(2(k - 3))^2}{4} = 3 + \frac{4(k - 3)^2}{4} = 3 + (k - 3)^2$.
$h - 3 = (k - 3)^2$.
Replacing $(h, k)$ with $(x, y)$,we get $(y - 3)^2 = x - 3$.
$y^2 - 6y + 9 = x - 3$.
$y^2 - 6y - x + 12 = 0$.

(A) Let the equation of the chord $AB$ be $lx + my = 1$.
Homogenizing the equation of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with the help of the chord equation $lx + my = 1$,we get:
$\frac{x^2}{a^2} + \frac{y^2}{b^2} = (lx + my)^2$
$\frac{x^2}{a^2} + \frac{y^2}{b^2} = l^2x^2 + m^2y^2 + 2lmxy$
$x^2(\frac{1}{a^2} - l^2) + y^2(\frac{1}{b^2} - m^2) - 2lmxy = 0$
Since $AB$ subtends a right angle at the origin,the sum of the coefficients of $x^2$ and $y^2$ must be zero.
$(\frac{1}{a^2} - l^2) + (\frac{1}{b^2} - m^2) = 0$
$l^2 + m^2 = \frac{1}{a^2} + \frac{1}{b^2}$
Now,the distance of the chord $lx + my = 1$ from the origin is $p = \frac{1}{\sqrt{l^2 + m^2}}$.
If $A$ and $B$ are points on the ellipse,the coordinates can be represented such that $OA^2$ and $OB^2$ relate to the chord properties. For a chord subtending a right angle at the origin,the expression $\frac{1}{OA^2} + \frac{1}{OB^2}$ is constant and equal to $\frac{1}{a^2} + \frac{1}{b^2}$.

(C) Let the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. The positive end of the minor axis is $P(0, b)$.
Let $(h, k)$ be the midpoint of a chord passing through $(0, b)$ and some point $(x_1, y_1)$ on the ellipse.
Then,$h = \frac{x_1 + 0}{2} \Rightarrow x_1 = 2h$ and $k = \frac{y_1 + b}{2} \Rightarrow y_1 = 2k - b$.
Since $(x_1, y_1)$ lies on the ellipse,we have $\frac{x_1^2}{a^2} + \frac{y_1^2}{b^2} = 1$.
Substituting the values of $x_1$ and $y_1$,we get $\frac{(2h)^2}{a^2} + \frac{(2k - b)^2}{b^2} = 1$.
This simplifies to $\frac{4h^2}{a^2} + \frac{4(k - b/2)^2}{b^2} = 1$,which can be rewritten as $\frac{h^2}{(a/2)^2} + \frac{(k - b/2)^2}{(b/2)^2} = 1$.
Replacing $(h, k)$ with $(x, y)$,the locus is $\frac{x^2}{(a/2)^2} + \frac{(y - b/2)^2}{(b/2)^2} = 1$,which represents an ellipse.

(D) Let the coordinates of $P$ be $(a \sec \theta, b \tan \theta)$. Since $PQ$ is a double ordinate,the coordinates of $Q$ are $(a \sec \theta, -b \tan \theta)$.
Given that $\triangle OPQ$ is an equilateral triangle,the angle $\angle POD = 30^{\circ}$,where $D$ is the point $(a \sec \theta, 0)$.
In $\triangle OPD$,we have $\tan 30^{\circ} = \frac{PD}{OD} = \frac{b \tan \theta}{a \sec \theta}$.
$\frac{1}{\sqrt{3}} = \frac{b}{a} \sin \theta \implies \frac{b}{a} = \frac{1}{\sqrt{3} \sin \theta}$.
We know that $e^2 = 1 + \frac{b^2}{a^2} = 1 + \frac{1}{3 \sin^2 \theta}$.
Since $0 < \sin^2 \theta < 1$,we have $\sin^2 \theta < 1$,which implies $\frac{1}{\sin^2 \theta} > 1$.
Therefore,$e^2 = 1 + \frac{1}{3 \sin^2 \theta} > 1 + \frac{1}{3} = \frac{4}{3}$.
Thus,$e > \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}$.

(B) The equation of the normal to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ at point $(a \sec \theta, b \tan \theta)$ is given by $ax \cos \theta + by \cot \theta = a^2 + b^2$.
For the hyperbola $\frac{x^2}{9} - \frac{y^2}{4} = 1$,we have $a^2 = 9$ and $b^2 = 4$,so $a = 3$ and $b = 2$.
The normal at $P(3 \sec \theta, 2 \tan \theta)$ is $3x \cos \theta + 2y \cot \theta = 9 + 4 = 13$ ... $(1)$.
The normal at $Q(3 \sec \phi, 2 \tan \phi)$ is $3x \cos \phi + 2y \cot \phi = 13$.
Given $\phi = \frac{\pi}{2} - \theta$,we have $\cos \phi = \sin \theta$ and $\cot \phi = \tan \theta$.
So,the normal at $Q$ is $3x \sin \theta + 2y \tan \theta = 13$ ... $(2)$.
From $(1)$,$3x \cos \theta = 13 - 2y \cot \theta \Rightarrow 3x = \frac{13 - 2y \cot \theta}{\cos \theta}$.
From $(2)$,$3x \sin \theta = 13 - 2y \tan \theta \Rightarrow 3x = \frac{13 - 2y \tan \theta}{\sin \theta}$.
Equating the two expressions for $3x$:
$\frac{13 - 2y \cot \theta}{\cos \theta} = \frac{13 - 2y \tan \theta}{\sin \theta}$
$(13 - 2y \cot \theta) \sin \theta = (13 - 2y \tan \theta) \cos \theta$
$13 \sin \theta - 2y \cos \theta = 13 \cos \theta - 2y \sin \theta$
$2y \sin \theta - 2y \cos \theta = 13 \cos \theta - 13 \sin \theta$
$2y(\sin \theta - \cos \theta) = -13(\sin \theta - \cos \theta)$
Since $\theta \neq \frac{\pi}{4}$ (as $0 < \theta, \phi < \frac{\pi}{2}$ and $\theta + \phi = \frac{\pi}{2}$),$\sin \theta - \cos \theta \neq 0$.
Thus,$2y = -13 \Rightarrow y = -\frac{13}{2}$.

(B) Given $\lim _{x \rightarrow \infty} \left(\frac{x^2+1}{x+1}-ax-b\right) = 0$.
Taking the common denominator:
$\lim _{x \rightarrow \infty} \frac{x^2+1-ax(x+1)-b(x+1)}{x+1} = 0$
$\lim _{x \rightarrow \infty} \frac{x^2+1-ax^2-ax-bx-b}{x+1} = 0$
$\lim _{x \rightarrow \infty} \frac{(1-a)x^2 - (a+b)x + (1-b)}{x+1} = 0$
For the limit to be $0$ as $x \rightarrow \infty$,the coefficient of the highest power of $x$ in the numerator must be $0$ and the degree of the numerator must be less than the degree of the denominator.
Thus,$1-a = 0 \implies a = 1$.
Substituting $a=1$ into the coefficient of $x$: $-(a+b) = 0 \implies -(1+b) = 0 \implies b = -1$.
Therefore,$a=1$ and $b=-1$.

(C) Let $L = \lim _{x \rightarrow 0} \frac{1}{x} \ln \sqrt{\frac{1+x}{1-x}}$.
Using the property $\ln(a^b) = b \ln(a)$,we can rewrite the expression as:
$L = \lim _{x \rightarrow 0} \frac{1}{x} \cdot \frac{1}{2} \ln \left( \frac{1+x}{1-x} \right)$.
$L = \frac{1}{2} \lim _{x \rightarrow 0} \frac{\ln(1+x) - \ln(1-x)}{x}$.
Using the standard limit $\lim _{x \rightarrow 0} \frac{\ln(1+ax)}{x} = a$,we have:
$L = \frac{1}{2} \left( \lim _{x \rightarrow 0} \frac{\ln(1+x)}{x} - \lim _{x \rightarrow 0} \frac{\ln(1-x)}{x} \right)$.
$L = \frac{1}{2} (1 - (-1)) = \frac{1}{2} (2) = 1$.

(B) An equivalence relation must be reflexive,symmetric,and transitive.
$1$. Reflexivity: Since $R$ and $S$ are equivalence relations,$(a, a) \in R$ and $(a, a) \in S$ for all $a \in A$. Thus,$(a, a) \in R \cap S$.
$2$. Symmetry: If $(a, b) \in R \cap S$,then $(a, b) \in R$ and $(a, b) \in S$. Since $R$ and $S$ are symmetric,$(b, a) \in R$ and $(b, a) \in S$,so $(b, a) \in R \cap S$.
$3$. Transitivity: If $(a, b) \in R \cap S$ and $(b, c) \in R \cap S$,then $(a, b), (b, c) \in R$ and $(a, b), (b, c) \in S$. Since $R$ and $S$ are transitive,$(a, c) \in R$ and $(a, c) \in S$,so $(a, c) \in R \cap S$.
Therefore,$R \cap S$ is an equivalence relation.

(B) Given $A = \begin{bmatrix} 1 & 1 \\ 0 & i \end{bmatrix}$.
Calculate the powers of $A$:
$A^2 = \begin{bmatrix} 1 & 1 \\ 0 & i \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & i \end{bmatrix} = \begin{bmatrix} 1 & 1+i \\ 0 & i^2 \end{bmatrix} = \begin{bmatrix} 1 & 1+i \\ 0 & -1 \end{bmatrix}$.
$A^3 = A^2 \times A = \begin{bmatrix} 1 & 1+i \\ 0 & -1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & i \end{bmatrix} = \begin{bmatrix} 1 & 1+i+i \\ 0 & -i \end{bmatrix} = \begin{bmatrix} 1 & 1+2i \\ 0 & -i \end{bmatrix}$.
Wait,let us re-evaluate $A^4$:
$A^4 = A^2 \times A^2 = \begin{bmatrix} 1 & 1+i \\ 0 & -1 \end{bmatrix} \begin{bmatrix} 1 & 1+i \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} 1 & 1+i-1-i \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$.
Since $A^4 = I$,we have $A^{2018} = A^{2016} \times A^2 = (A^4)^{504} \times A^2 = I^{504} \times A^2 = A^2$.
Thus,$A^{2018} = \begin{bmatrix} 1 & 1+i \\ 0 & -1 \end{bmatrix}$.
Comparing with $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$,we get $a=1$ and $d=-1$.
Therefore,$a+d = 1 + (-1) = 0$.

(B) We know that for a $3 \times 3$ matrix $A$,the determinant of its adjoint matrix is given by $\det(\text{adj}(A)) = (\det(A))^{n-1}$,where $n$ is the order of the matrix.
Here,$n = 3$,so $\det(P) = (\det(A))^{3-1} = (\det(A))^2$.
Given $\det(A) = 4$,we have $\det(P) = 4^2 = 16$.
Now,calculate the determinant of matrix $P = \begin{bmatrix} 1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4 \end{bmatrix}$:
$\det(P) = 1(3 \times 4 - 3 \times 4) - \alpha(1 \times 4 - 3 \times 2) + 3(1 \times 4 - 3 \times 2)$
$\det(P) = 1(12 - 12) - \alpha(4 - 6) + 3(4 - 6)$
$\det(P) = 0 - \alpha(-2) + 3(-2) = 2\alpha - 6$.
Equating the two values: $2\alpha - 6 = 16$.
$2\alpha = 22 \Rightarrow \alpha = 11$.

(D) The characteristic equation is given by $\operatorname{det}(A-\lambda I_2)=0$.
$\left|\begin{array}{cc} 2-\lambda & 3 \\ x & y-\lambda \end{array}\right|=0$
$(2-\lambda)(y-\lambda)-3x=0$
$\lambda^2 - (y+2)\lambda + 2y - 3x = 0$
Since the roots are $4$ and $8$,the sum of roots is $4+8=12$ and the product of roots is $4 \times 8 = 32$.
Comparing with $\lambda^2 - (y+2)\lambda + (2y-3x) = 0$:
$y+2 = 12 \Rightarrow y=10$.
$2y-3x = 32 \Rightarrow 2(10)-3x=32 \Rightarrow 20-3x=32 \Rightarrow -3x=12 \Rightarrow x=-4$.
Thus,$x=-4$ and $y=10$.

(B) To find the coefficient of $x$ in $\Delta(x)$,we use the property that the coefficient of $x$ is given by $\Delta'(0)$.
Given $\Delta(x) = \begin{vmatrix} x-2 & (x-1)^2 & x^3 \\ x-1 & x^2 & (x+1)^3 \\ x & (x+1)^2 & (x+2)^3 \end{vmatrix}$.
Using the property of differentiation of a determinant,$\Delta'(x) = \Delta_1(x) + \Delta_2(x) + \Delta_3(x)$,where $\Delta_i$ is the determinant with the $i$-th row differentiated.
Evaluating $\Delta'(0)$ involves calculating the derivative of each row at $x=0$.
After performing the differentiation and substituting $x=0$,we obtain $\Delta'(0) = -2$.
Thus,the coefficient of $x$ is $-2$.

(B, D) Expanding the determinant $\Delta$ along the third column:
$\Delta = \cos \theta \begin{vmatrix} \cos \theta \cos \phi & \cos \theta \sin \phi \\ -\sin \theta \sin \phi & \sin \theta \cos \phi \end{vmatrix} - (- \sin \theta) \begin{vmatrix} \sin \theta \cos \phi & \sin \theta \sin \phi \\ -\sin \theta \sin \phi & \sin \theta \cos \phi \end{vmatrix} + 0$
$\Delta = \cos \theta [(\cos \theta \cos \phi)(\sin \theta \cos \phi) - (\cos \theta \sin \phi)(-\sin \theta \sin \phi)] + \sin \theta [(\sin \theta \cos \phi)(\sin \theta \cos \phi) - (\sin \theta \sin \phi)(-\sin \theta \sin \phi)]$
$\Delta = \cos \theta [\sin \theta \cos \theta \cos^2 \phi + \sin \theta \cos \theta \sin^2 \phi] + \sin \theta [\sin^2 \theta \cos^2 \phi + \sin^2 \theta \sin^2 \phi]$
$\Delta = \cos \theta [\sin \theta \cos \theta (\cos^2 \phi + \sin^2 \phi)] + \sin \theta [\sin^2 \theta (\cos^2 \phi + \sin^2 \phi)]$
$\Delta = \cos \theta (\sin \theta \cos \theta) + \sin \theta (\sin^2 \theta) = \sin \theta \cos^2 \theta + \sin^3 \theta = \sin \theta (\cos^2 \theta + \sin^2 \theta) = \sin \theta$
Since $\Delta = \sin \theta$,it is independent of $\phi$.
Also,$\frac{d \Delta}{d \theta} = \cos \theta$.
At $\theta = \frac{\pi}{2}$,$\frac{d \Delta}{d \theta} = \cos \frac{\pi}{2} = 0$.
Thus,both options $B$ and $D$ are correct.

(D) system of linear equations $AX = B$ has a unique solution if and only if the determinant of the coefficient matrix $A$ is non-zero,i.e.,$|A| \neq 0$.
Given the matrix $A = \begin{bmatrix} 1 & 2 & 4 \\ 2 & 1 & 2 \\ 1 & 2 & a-4 \end{bmatrix}$.
Calculating the determinant $|A|$:
$|A| = 1((1)(a-4) - (2)(2)) - 2((2)(a-4) - (2)(1)) + 4((2)(2) - (1)(1))$
$|A| = 1(a-4-4) - 2(2a-8-2) + 4(4-1)$
$|A| = (a-8) - 2(2a-10) + 4(3)$
$|A| = a - 8 - 4a + 20 + 12$
$|A| = -3a + 24$
For a unique solution,$|A| \neq 0$.
$-3a + 24 \neq 0 \Rightarrow -3a \neq -24 \Rightarrow a \neq 8$.

(C) For the function $y = \sqrt{\log _{10} \frac{3x - x^2}{2}}$ to be defined,the expression inside the square root must be non-negative:
$\log _{10} \left( \frac{3x - x^2}{2} \right) \geq 0$
Since $\log _{10} 1 = 0$,we have:
$\frac{3x - x^2}{2} \geq 1$
$3x - x^2 \geq 2$
$x^2 - 3x + 2 \leq 0$
$(x - 1)(x - 2) \leq 0$
This inequality holds for $x \in [1, 2]$.
Additionally,the argument of the logarithm must be positive:
$\frac{3x - x^2}{2} > 0$
$x(3 - x) > 0$
$x(x - 3) < 0$
This holds for $x \in (0, 3)$.
The intersection of $x \in [1, 2]$ and $x \in (0, 3)$ is $x \in [1, 2]$.

(C) Let $x_1, x_2 \in S$ such that $f(x_1) = f(x_2)$.
Applying $g$ to both sides,we get $g(f(x_1)) = g(f(x_2))$.
This is equivalent to $(g \circ f)(x_1) = (g \circ f)(x_2)$.
Since $g \circ f$ is given as an injective mapping,$(g \circ f)(x_1) = (g \circ f)(x_2)$ implies $x_1 = x_2$.
Since $f(x_1) = f(x_2)$ leads to $x_1 = x_2$,the function $f$ must be injective.
Therefore,$f$ is necessarily injective.

(D) To check for one-one: Let $f(x_1) = f(x_2)$.
$\frac{2x_1}{x_1-1} = \frac{2x_2}{x_2-1}$
$x_1(x_2-1) = x_2(x_1-1)$
$x_1x_2 - x_1 = x_1x_2 - x_2$
$-x_1 = -x_2 \Rightarrow x_1 = x_2$.
Thus,$f$ is one-one.
To check for onto: Let $y = \frac{2x}{x-1}$.
$y(x-1) = 2x \Rightarrow yx - y = 2x \Rightarrow x(y-2) = y \Rightarrow x = \frac{y}{y-2}$.
Since $y \in R-\{2\}$,$x$ is always defined and $x \neq 1$. Thus,for every $y$ in the codomain,there exists an $x$ in the domain.
Therefore,$f$ is onto.
Hence,$f$ is both one-one and onto.

(C) Let $t = 2x-1$. Since $0 < x < 1$,we have $-1 < 2x-1 < 1$,so $-1 < t < 1$.
The function becomes $f(t) = \frac{t}{1-|t|}$ for $t \in (-1, 1)$.
We can write this as:
$f(t) = \begin{cases} \frac{t}{1+t}, & -1 < t \leq 0 \\ \frac{t}{1-t}, & 0 < t < 1 \end{cases}$
Since $f$ is continuous and $\lim_{t \to -1^+} f(t) = -\infty$ and $\lim_{t \to 1^-} f(t) = +\infty$,the range of $f$ is $(-\infty, \infty) = R$. Thus,$f$ is surjective.
Now,check for injectivity by finding the derivative:
$f'(t) = \begin{cases} \frac{1}{(1+t)^2}, & -1 < t < 0 \\ \frac{1}{(1-t)^2}, & 0 < t < 1 \end{cases}$
Since $f'(t) > 0$ for all $t \in (-1, 1)$,the function is strictly increasing.
Therefore,$f$ is injective.
Since $f$ is both injective and surjective,it is bijective.

(D) For the function to be continuous at $x = 0$,we must have $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)$.
First,calculate the left-hand limit $(LHL)$:
$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{\sin(a+1)x + \sin x}{x} = \lim_{x \to 0^-} \left( \frac{\sin(a+1)x}{x} + \frac{\sin x}{x} \right) = (a+1) + 1 = a+2$.
Next,calculate the right-hand limit $(RHL)$:
$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{\sqrt{x+bx^2} - \sqrt{x}}{bx^{1/2}} = \lim_{x \to 0^+} \frac{\sqrt{x}(\sqrt{1+bx} - 1)}{bx^{1/2}} = \lim_{x \to 0^+} \frac{\sqrt{1+bx} - 1}{b}$.
Using the binomial expansion $(1+bx)^{1/2} \approx 1 + \frac{1}{2}bx$,we get:
$\lim_{x \to 0^+} \frac{1 + \frac{1}{2}bx - 1}{b} = \lim_{x \to 0^+} \frac{\frac{1}{2}bx}{b} = \lim_{x \to 0^+} \frac{x}{2} = 0$.
Since $f(0) = c$,for continuity we require $a+2 = 0$ and $c = 0$.
Thus,$a = -2$ and $c = 0$. Since the denominator in the $RHL$ expression contains $b$,we must have $b \neq 0$ for the function to be defined for $x > 0$.
Therefore,$a = -2, b \in \mathbb{R} - \{0\}, c = 0$.

(C) Given $f(x) = a_0 + a_1|x| + a_2|x|^2 + a_3|x|^3$.
Since $|x|^2 = x^2$ and $|x|^3 = |x| \cdot x^2$,we can write $f(x) = a_0 + a_1|x| + a_2x^2 + a_3|x|x^2$.
For $f(x)$ to be differentiable at $x=0$,the left-hand derivative and right-hand derivative must be equal.
Right-hand derivative at $x=0$: $f'(0^+) = \lim_{h \to 0^+} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^+} \frac{a_0 + a_1h + a_2h^2 + a_3h^3 - a_0}{h} = \lim_{h \to 0^+} (a_1 + a_2h + a_3h^2) = a_1$.
Left-hand derivative at $x=0$: $f'(0^-) = \lim_{h \to 0^-} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^-} \frac{a_0 + a_1(-h) + a_2h^2 + a_3(-h^3) - a_0}{h} = \lim_{h \to 0^-} (-a_1 + a_2h - a_3h^2) = -a_1$.
For differentiability,$f'(0^+) = f'(0^-) \implies a_1 = -a_1 \implies 2a_1 = 0 \implies a_1 = 0$.
Thus,$f(x)$ is differentiable at $x=0$ only if $a_1=0$.

(A) Given $y=e^{\tan ^{-1} x}$.
Differentiating with respect to $x$,we get:
$\frac{dy}{dx} = e^{\tan ^{-1} x} \times \frac{d}{dx}(\tan ^{-1} x) = e^{\tan ^{-1} x} \times \frac{1}{1+x^2}$.
Since $y = e^{\tan ^{-1} x}$,we can write:
$\frac{dy}{dx} = \frac{y}{1+x^2}$.
Multiplying both sides by $(1+x^2)$,we get:
$(1+x^2) y_1 = y$.
Differentiating again with respect to $x$ using the product rule on the left side:
$\frac{d}{dx}[(1+x^2) y_1] = \frac{d}{dx}(y)$.
$(1+x^2) y_2 + y_1(2x) = y_1$.
Rearranging the terms:
$(1+x^2) y_2 + (2x - 1) y_1 = 0$.

(B) Given the transformation $z = \log \tan \frac{x}{2}$.
First,find $\frac{d z}{d x}$:
$\frac{d z}{d x} = \frac{1}{\tan \frac{x}{2}} \cdot \sec^2 \frac{x}{2} \cdot \frac{1}{2} = \frac{\cos \frac{x}{2}}{\sin \frac{x}{2}} \cdot \frac{1}{\cos^2 \frac{x}{2}} \cdot \frac{1}{2} = \frac{1}{2 \sin \frac{x}{2} \cos \frac{x}{2}} = \frac{1}{\sin x} = \operatorname{cosec} x$.
Now,express $\frac{d y}{d z}$ in terms of $x$:
$\frac{d y}{d z} = \frac{d y}{d x} \cdot \frac{d x}{d z} = \frac{d y}{d x} \cdot \sin x$.
Next,find $\frac{d^2 y}{d z^2}$:
$\frac{d^2 y}{d z^2} = \frac{d}{d z} \left( \sin x \frac{d y}{d x} \right) = \frac{d}{d x} \left( \sin x \frac{d y}{d x} \right) \cdot \frac{d x}{d z} = \left( \cos x \frac{d y}{d x} + \sin x \frac{d^2 y}{d x^2} \right) \cdot \sin x = \sin x \cos x \frac{d y}{d x} + \sin^2 x \frac{d^2 y}{d x^2}$.
Substitute this into the target equation $\frac{d^2 y}{d z^2} + k y = 0$:
$\sin^2 x \frac{d^2 y}{d x^2} + \sin x \cos x \frac{d y}{d x} + k y = 0$.
Divide by $\sin^2 x$:
$\frac{d^2 y}{d x^2} + \cot x \frac{d y}{d x} + k \operatorname{cosec}^2 x \cdot y = 0$.
Comparing this with the given equation $\frac{d^2 y}{d x^2} + \cot x \frac{d y}{d x} + 4 \operatorname{cosec}^2 x \cdot y = 0$,we get $k = 4$.

(C) Let the point of contact be $(x_1, y_1)$. The equation of the tangent at $(x_1, y_1)$ is $y - y_1 = m(x - x_1)$,where $m = \frac{dy}{dx}$.
Since the segment of the tangent between the coordinate axes is bisected at $(x_1, y_1)$,the points where the tangent meets the axes are $(2x_1, 0)$ and $(0, 2y_1)$.
The slope of the tangent is $m = \frac{2y_1 - 0}{0 - 2x_1} = -\frac{y_1}{x_1}$.
Thus,$\frac{dy}{dx} = -\frac{y}{x}$.
Separating variables,we get $\frac{dy}{y} = -\frac{dx}{x}$.
Integrating both sides,$\ln y = -\ln x + C_0$,which implies $\ln(xy) = C_0$,or $xy = C$.
Since the curve passes through $(3, 2)$,we have $3 \times 2 = C$,so $C = 6$.
Therefore,the equation of the curve is $xy = 6$.

(A) Given acceleration $v'(t) = a - kt^2$. Since the particle starts from rest,$v(0) = 0$.
Integrating with respect to $t$,we get $v(t) = \int (a - kt^2) dt = at - \frac{k}{3}t^3 + C$.
Using $v(0) = 0$,we find $C = 0$,so $v(t) = at - \frac{k}{3}t^3$.
To find the maximum velocity,we set the acceleration to zero: $a - kt^2 = 0$,which gives $t^2 = \frac{a}{k}$,or $t = \sqrt{\frac{a}{k}}$ (since $t > 0$).
Substituting this value of $t$ into the velocity equation:
$v_{\max} = a \sqrt{\frac{a}{k}} - \frac{k}{3} \left(\sqrt{\frac{a}{k}}\right)^3$
$v_{\max} = a \sqrt{\frac{a}{k}} - \frac{k}{3} \cdot \frac{a}{k} \sqrt{\frac{a}{k}}$
$v_{\max} = a \sqrt{\frac{a}{k}} - \frac{a}{3} \sqrt{\frac{a}{k}} = \frac{2a}{3} \sqrt{\frac{a}{k}} = \frac{2}{3} \sqrt{\frac{a^3}{k}}$.

(C) Let the height of the balloon when the stone is dropped be $h$ and its upward velocity be $v$.
When the stone is dropped,its initial velocity is $u = v$ (upwards).
Using the equation of motion for the stone: $s = ut + \frac{1}{2}at^2$.
Here,$s = -h$ (displacement is downwards),$u = v$,$a = -g = -32 \ ft/sec^2$,and $t = 4 \ sec$.
$-h = v(4) + \frac{1}{2}(-32)(4)^2$.
$-h = 4v - 16(16)$.
$-h = 4v - 256$.
$h = 256 - 4v$.
In $4 \ sec$,the balloon rises further by a distance $d = v \times t = v \times 4 = 4v$.
The total height of the balloon from the ground when the stone hits the ground is $H = h + d$.
$H = (256 - 4v) + 4v = 256 \ ft$.

(A) Given $f(x) = x^2 + x \sin x - \cos x$.
We find the derivative: $f'(x) = 2x + (\sin x + x \cos x) + \sin x = 2x + 2 \sin x + x \cos x = x(2 + \cos x) + 2 \sin x$.
For $x > 0$,$f'(x) = x(2 + \cos x) + 2 \sin x$. Since $2 + \cos x > 0$ and $x > 0$,and $2 \sin x$ is not always negative,we analyze the behavior. Alternatively,note that $f(0) = -1$.
As $x \to \infty$,$f(x) \to \infty$. Since $f(0) = -1 < 0$ and $f(x)$ is continuous,by the Intermediate Value Theorem,there exists at least one root in $(0, \infty)$.
Similarly,as $x \to -\infty$,$f(x) \to \infty$. Since $f(0) = -1 < 0$,there exists at least one root in $(-\infty, 0)$.
Thus,$f(x) = 0$ has at least two real roots (one positive and one negative),which implies it has at least one real root.

(C) Given $f(x)=(x-2)^{17}(x+5)^{24}$.
To find the critical points,we find the derivative $f'(x)$:
$f'(x) = 17(x-2)^{16}(x+5)^{24} + 24(x-2)^{17}(x+5)^{23}$
$f'(x) = (x-2)^{16}(x+5)^{23} [17(x+5) + 24(x-2)]$
$f'(x) = (x-2)^{16}(x+5)^{23} [17x + 85 + 24x - 48]$
$f'(x) = (x-2)^{16}(x+5)^{23} (41x + 37)$
The critical points are $x=2, x=-5, x=-\frac{37}{41}$.
Now,examine the sign of $f'(x)$ around $x=2$:
Since $(x-2)^{16}$ is always non-negative,the sign of $f'(x)$ depends on $(x+5)^{23}(41x+37)$.
For $x$ slightly less than $2$,$(x+5)^{23} > 0$ and $(41x+37) > 0$,so $f'(x) > 0$.
For $x$ slightly greater than $2$,$(x+5)^{23} > 0$ and $(41x+37) > 0$,so $f'(x) > 0$.
Since the sign of $f'(x)$ does not change as $x$ passes through $2$,$f(x)$ has neither a maximum nor a minimum at $x=2$.

(C) Given $f(x) = e^{\sin x} + e^{\cos x}$.
To find the maximum value,we use the Arithmetic Mean-Geometric Mean ($AM$-$GM$) inequality or derivative analysis.
Let $u = \sin x$ and $v = \cos x$. We know that $u^2 + v^2 = 1$.
Using the property that for $f(x) = e^{\sin x} + e^{\cos x}$,the function attains its maximum when $\sin x = \cos x$.
Setting $\sin x = \cos x$,we get $\tan x = 1$,which implies $x = \frac{\pi}{4} + n\pi$.
For $x = \frac{\pi}{4}$,$\sin x = \frac{1}{\sqrt{2}}$ and $\cos x = \frac{1}{\sqrt{2}}$.
Substituting these values into $f(x)$:
$f\left(\frac{\pi}{4}\right) = e^{1/\sqrt{2}} + e^{1/\sqrt{2}} = 2e^{1/\sqrt{2}}$.
Thus,the maximum value is $2e^{1/\sqrt{2}}$.

(C) Let the radius of the circular sector be $r$ and the arc length be $\ell$.
Given that the total length of the wire is $20 \ m$,the perimeter of the sector is $2r + \ell = 20$.
Thus,$\ell = 20 - 2r$.
The area $A$ of a circular sector is given by $A = \frac{1}{2} r \ell$.
Substituting the value of $\ell$,we get $A = \frac{1}{2} r (20 - 2r) = 10r - r^2$.
To find the maximum area,we differentiate $A$ with respect to $r$ and set it to zero:
$\frac{dA}{dr} = 10 - 2r = 0$.
Solving for $r$,we get $2r = 10$,which implies $r = 5 \ m$.
To verify that this is a maximum,we check the second derivative: $\frac{d^2A}{dr^2} = -2$,which is less than $0$,confirming that the area is maximum at $r = 5 \ m$.

(A) Define a function $g(x) = e^{-x} f(x)$.
Since $f(x)$ is continuous on $[a, b]$ and differentiable on $(a, b)$,$g(x)$ is also continuous on $[a, b]$ and differentiable on $(a, b)$.
We have $g(a) = e^{-a} f(a) = e^{-a} \cdot 0 = 0$ and $g(b) = e^{-b} f(b) = e^{-b} \cdot 0 = 0$.
Since $g(a) = g(b) = 0$,by Rolle's Theorem,there exists at least one point $c \in (a, b)$ such that $g^{\prime}(c) = 0$.
Calculating the derivative: $g^{\prime}(x) = -e^{-x} f(x) + e^{-x} f^{\prime}(x) = e^{-x} (f^{\prime}(x) - f(x))$.
Setting $g^{\prime}(c) = 0$,we get $e^{-c} (f^{\prime}(c) - f(c)) = 0$.
Since $e^{-c} \neq 0$ for any real $c$,we must have $f^{\prime}(c) - f(c) = 0$,which implies $f^{\prime}(c) = f(c)$ for at least one $c \in (a, b)$.

(C) Given that $p^{\prime}(x) > 0$ for all $x \in \mathbb{R}$,the polynomial $p(x)$ is a strictly increasing function on the entire real line.
Since $p(x)$ is strictly increasing,it can have at most one real root.
We are given $p(0) = 1$.
Since $p(x)$ is strictly increasing and $p(0) = 1 > 0$,for any $x > 0$,$p(x) > p(0) = 1$,so $p(x)$ cannot have any positive real roots.
As $x \to -\infty$,$p(x) \to -\infty$ (assuming the degree is odd) or $p(x) \to c$ (if degree is even,but $p^{\prime}(x) > 0$ implies the degree must be odd for the limit to be $-\infty$).
Since $p(0) = 1$ and the function is continuous and strictly increasing,by the Intermediate Value Theorem,there must exist some $x_0 < 0$ such that $p(x_0) = 0$.
Thus,$p(x)$ has exactly one negative real root.

(B) We are given the integral $I = \int \frac{x^{1/2}}{\sqrt{1-x^3}} dx$.
Let $t = x^{3/2}$.
Then $dt = \frac{3}{2} x^{1/2} dx$,which implies $x^{1/2} dx = \frac{2}{3} dt$.
Substituting these into the integral,we get:
$I = \int \frac{\frac{2}{3} dt}{\sqrt{1-t^2}} = \frac{2}{3} \int \frac{dt}{\sqrt{1-t^2}} = \frac{2}{3} \sin^{-1}(t) + c$.
Substituting $t = x^{3/2}$ back,we get:
$I = \frac{2}{3} \sin^{-1}(x^{3/2}) + c$.
Comparing this with the given form $\frac{2}{3} g(f(x)) + c$,we identify $f(x) = x^{3/2}$ and $g(x) = \sin^{-1}(x)$.

(A) Let $I = \int \cos(\ln x) \, dx$.
Using integration by parts,let $u = \cos(\ln x)$ and $dv = dx$. Then $du = -\sin(\ln x) \cdot \frac{1}{x} \, dx$ and $v = x$.
$I = x \cos(\ln x) - \int x \cdot (-\sin(\ln x)) \cdot \frac{1}{x} \, dx$
$I = x \cos(\ln x) + \int \sin(\ln x) \, dx$.
Now,integrate $\int \sin(\ln x) \, dx$ by parts again,with $u = \sin(\ln x)$ and $dv = dx$. Then $du = \cos(\ln x) \cdot \frac{1}{x} \, dx$ and $v = x$.
$\int \sin(\ln x) \, dx = x \sin(\ln x) - \int x \cdot \cos(\ln x) \cdot \frac{1}{x} \, dx = x \sin(\ln x) - I$.
Substituting this back into the equation for $I$:
$I = x \cos(\ln x) + x \sin(\ln x) - I$
$2I = x \{\cos(\ln x) + \sin(\ln x)\}$
$I = \frac{x}{2} \{\cos(\ln x) + \sin(\ln x)\} + c$.

(A) Let $I = \int_0^{\pi / 2} \frac{(\cos x)^{\sin x}}{(\cos x)^{\sin x} + (\sin x)^{\cos x}} dx$ --- $(1)$
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get:
$I = \int_0^{\pi / 2} \frac{(\cos(\pi/2 - x))^{\sin(\pi/2 - x)}}{(\cos(\pi/2 - x))^{\sin(\pi/2 - x)} + (\sin(\pi/2 - x))^{\cos(\pi/2 - x)}} dx$
$I = \int_0^{\pi / 2} \frac{(\sin x)^{\cos x}}{(\sin x)^{\cos x} + (\cos x)^{\sin x}} dx$ --- $(2)$
Adding $(1)$ and $(2)$:
$2I = \int_0^{\pi / 2} \frac{(\cos x)^{\sin x} + (\sin x)^{\cos x}}{(\cos x)^{\sin x} + (\sin x)^{\cos x}} dx$
$2I = \int_0^{\pi / 2} 1 dx$
$2I = [x]_0^{\pi / 2} = \pi / 2$
$I = \pi / 4$

(A) Define a function $g(x) = (1-x) \int_0^x f(t) dt$.
Since $f$ is derivable on $[0, 1]$,it is continuous on $[0, 1]$,and thus the integral $\int_0^x f(t) dt$ is differentiable.
We observe that $g(0) = (1-0) \int_0^0 f(t) dt = 1 \times 0 = 0$.
Also,$g(1) = (1-1) \int_0^1 f(t) dt = 0 \times \int_0^1 f(t) dt = 0$.
Since $g(x)$ is continuous on $[0, 1]$ and differentiable on $(0, 1)$ with $g(0) = g(1) = 0$,by Rolle's Theorem,there exists at least one $c \in (0, 1)$ such that $g'(c) = 0$.
Calculating the derivative: $g'(x) = -1 \int_0^x f(t) dt + (1-x) f(x)$.
Setting $g'(c) = 0$,we get $-(1-c) f(c) + \int_0^c f(t) dt = 0$,which implies $\int_0^c f(t) dt = (1-c) f(c)$.

(B) Let $g(x) = \lim _{c \rightarrow 0} \int_c^x \frac{b t \cos 4 t - a \sin 4 t}{t^2} d t = \frac{a \sin 4 x}{x} - 1$.
Applying the limit as $x \rightarrow 0$,we have $g(0) = \lim _{x \rightarrow 0} (\frac{a \sin 4 x}{x} - 1) = 4a - 1$.
Since the integral from $c$ to $c$ is $0$,we must have $g(0) = 0$,so $4a - 1 = 0$,which gives $a = 1/4$.
Now,differentiate both sides with respect to $x$ using the Fundamental Theorem of Calculus:
$g'(x) = \frac{b x \cos 4 x - a \sin 4 x}{x^2}$.
Also,differentiating the right side:
$g'(x) = \frac{d}{dx} (\frac{a \sin 4 x}{x} - 1) = \frac{4ax \cos 4 x - a \sin 4 x}{x^2}$.
Comparing the two expressions for $g'(x)$,we get $b = 4a$.
Since $a = 1/4$,we have $b = 4(1/4) = 1$.
Thus,$a = 1/4$ and $b = 1$.

(B) Using the Leibniz integral rule,the derivative of $f(x) = \int_{g(x)}^{h(x)} F(t) dt$ is given by $f^{\prime}(x) = F(h(x)) \cdot h^{\prime}(x) - F(g(x)) \cdot g^{\prime}(x)$.
Here,$F(t) = e^{-t^2}$,$h(x) = \cos x$,and $g(x) = \sin x$.
So,$f^{\prime}(x) = e^{-(\cos x)^2} \cdot (-\sin x) - e^{-(\sin x)^2} \cdot (\cos x)$.
At $x = \frac{\pi}{4}$,we have $\sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$ and $\cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$.
Substituting these values:
$f^{\prime}\left(\frac{\pi}{4}\right) = -e^{-1/2} \cdot \frac{1}{\sqrt{2}} - e^{-1/2} \cdot \frac{1}{\sqrt{2}}$
$f^{\prime}\left(\frac{\pi}{4}\right) = -\frac{2}{\sqrt{2} \cdot \sqrt{e}} = -\frac{\sqrt{2}}{\sqrt{e}} = -\sqrt{\frac{2}{e}}$.

(C) Given $\int_0^x \sqrt{1-\left(f^{\prime}(t)\right)^2} dt = \int_0^x f(t) dt$.
Applying the Leibniz rule by differentiating both sides with respect to $x$,we get $\sqrt{1-\left(f^{\prime}(x)\right)^2} = f(x)$.
Squaring both sides,$1 - (f^{\prime}(x))^2 = f^2(x)$,which implies $(f^{\prime}(x))^2 = 1 - f^2(x)$.
Thus,$f^{\prime}(x) = \pm \sqrt{1 - f^2(x)}$.
Separating variables,$\int \frac{df}{\sqrt{1-f^2}} = \pm \int dx$,which gives $\sin^{-1}(f(x)) = \pm x + C$.
Since $f(0) = 0$,we have $\sin^{-1}(0) = 0 + C$,so $C = 0$.
Thus,$f(x) = \sin(x)$ or $f(x) = -\sin(x)$.
Since $f$ is non-negative on $[0, \pi/2]$,we must have $f(x) = \sin(x)$.
We know that for $x > 0$,$\sin(x) < x$.
Therefore,$f(4/3) = \sin(4/3) < 4/3$ and $f(2/3) = \sin(2/3) < 2/3$.

(D) For $0 < x < 1$,we have $x^2 < x$ and $0 \le \cos^2 x \le 1$.
Comparing the integrands:
$e^{-x} \cos^2 x < e^{-x^2} \cos^2 x < e^{-x^2} < e^{-x^2/2}$.
Since the integrand of $I_4$ is the largest for all $x \in (0, 1)$,the integral $I_4$ will be the greatest.
Therefore,$I = I_4$.

(B) The given equations of the parabolas are:
$y^2 = -8(x-2) \quad \dots(1)$
$y^2 = 24(x+2) \quad \dots(2)$
To find the intersection points,equate the expressions for $y^2$:
$-8(x-2) = 24(x+2)$
$-8x + 16 = 24x + 48$
$-32x = 32 \implies x = -1$
At $x = -1$,$y^2 = 24(-1+2) = 24$,so $y = \pm \sqrt{24} = \pm 2\sqrt{6}$.
The area is symmetric about the $x$-axis,so the total area is twice the area above the $x$-axis:
$\text{Area} = 2 \left[ \int_{-2}^{-1} \sqrt{24(x+2)} \, dx + \int_{-1}^{2} \sqrt{-8(x-2)} \, dx \right]$
$\text{Area} = 2 \left[ 2\sqrt{6} \int_{-2}^{-1} \sqrt{x+2} \, dx + 2\sqrt{2} \int_{-1}^{2} \sqrt{2-x} \, dx \right]$
$\text{Area} = 4 \left[ \sqrt{6} \left( \frac{2}{3}(x+2)^{3/2} \right)_{-2}^{-1} + \sqrt{2} \left( -\frac{2}{3}(2-x)^{3/2} \right)_{-1}^{2} \right]$
$\text{Area} = 4 \left[ \sqrt{6} \left( \frac{2}{3}(1) - 0 \right) + \sqrt{2} \left( 0 - (-\frac{2}{3}(3)^{3/2}) \right) \right]$
$\text{Area} = 4 \left[ \frac{2\sqrt{6}}{3} + \frac{2\sqrt{2}}{3} \cdot 3\sqrt{3} \right] = 4 \left[ \frac{2\sqrt{6}}{3} + 2\sqrt{6} \right] = 4 \left[ \frac{8\sqrt{6}}{3} \right] = \frac{32}{3} \sqrt{6} \text{ sq. unit}$.

(D) Given the differential equation: $\cos y \frac{dy}{dx} = e^x e^{\sin y} + x^2 e^{\sin y}$.
Divide both sides by $e^{\sin y}$: $e^{-\sin y} \cos y \frac{dy}{dx} = e^x + x^2$.
Let $u = \sin y$,then $\frac{du}{dx} = \cos y \frac{dy}{dx}$.
The equation becomes $e^{-u} \frac{du}{dx} = e^x + x^2$.
Integrating both sides with respect to $x$: $\int e^{-u} du = \int (e^x + x^2) dx$.
$-e^{-u} = e^x + \frac{x^3}{3} + C_1$.
Substitute $u = \sin y$ back: $-e^{-\sin y} = e^x + \frac{x^3}{3} + C_1$.
Rearranging to the form $f(x) + e^{-\sin y} = C$: $e^x + \frac{x^3}{3} + e^{-\sin y} = C$.
Comparing this with $f(x) + e^{-\sin y} = C$,we get $f(x) = e^x + \frac{x^3}{3}$.

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

(B) Given $\vec{a} = \hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$.
Since $\vec{c}$ is coplanar with $\vec{a}$ and $\vec{b}$ and perpendicular to $\vec{a}$,$\vec{c}$ lies in the plane of $\vec{a}$ and $\vec{b}$.
The vector $\vec{d}$ is perpendicular to both $\vec{a}$ and $\vec{c}$. Since $\vec{c}$ is in the plane of $\vec{a}$ and $\vec{b}$,the vector $\vec{d}$ must be perpendicular to the plane containing $\vec{a}$ and $\vec{b}$.
Thus,$\vec{d}$ is parallel to $\vec{a} \times \vec{b}$.
Calculate $\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -1 \\ 1 & -1 & 1 \end{vmatrix} = \hat{i}(1 - 1) - \hat{j}(1 - (-1)) + \hat{k}(-1 - 1) = 0\hat{i} - 2\hat{j} - 2\hat{k} = -2(\hat{j} + \hat{k})$.
The unit vector $\vec{d}$ is $\pm \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} = \pm \frac{-2(\hat{j} + \hat{k})}{\sqrt{(-2)^2 + (-2)^2}} = \pm \frac{-2(\hat{j} + \hat{k})}{\sqrt{8}} = \pm \frac{-2(\hat{j} + \hat{k})}{2\sqrt{2}} = \pm \frac{1}{\sqrt{2}}(\hat{j} + \hat{k})$.

(D) The scalar triple product is defined as $[\vec{\alpha} \vec{\beta} \vec{\gamma}] = \vec{\alpha} \cdot (\vec{\beta} \times \vec{\gamma})$.
First,calculate the cross product $\vec{\beta} \times \vec{\gamma}$:
$\vec{\beta} \times \vec{\gamma} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -1 \\ 1 & 0 & 1 \end{vmatrix} = \hat{i}(1 - 0) - \hat{j}(1 - (-1)) + \hat{k}(0 - 1) = \hat{i} - 2\hat{j} - \hat{k}$.
Thus,$[\vec{\alpha} \vec{\beta} \vec{\gamma}] = \vec{\alpha} \cdot (\hat{i} - 2\hat{j} - \hat{k})$.
Since $\vec{\alpha}$ is a unit vector,the dot product $\vec{\alpha} \cdot \vec{v}$ is maximized when $\vec{\alpha}$ is in the same direction as $\vec{v}$,and the maximum value is equal to the magnitude $|\vec{v}|$.
Here,$\vec{v} = \hat{i} - 2\hat{j} - \hat{k}$.
$|\vec{v}| = \sqrt{(1)^2 + (-2)^2 + (-1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6}$.
Therefore,the maximum value is $\sqrt{6}$.

(C) The equation of any plane passing through the intersection of the planes $P_1: x+y+z-1=0$ and $P_2: 2x+3y-z+4=0$ is given by $P_1 + \lambda P_2 = 0$.
$(x+y+z-1) + \lambda(2x+3y-z+4) = 0$
$(1+2\lambda)x + (1+3\lambda)y + (1-\lambda)z + (4\lambda-1) = 0$.
Since the plane is parallel to the $x$-axis,its normal vector $\vec{n} = (1+2\lambda, 1+3\lambda, 1-\lambda)$ must be perpendicular to the $x$-axis vector $\hat{i} = (1, 0, 0)$.
Therefore,the dot product is zero: $(1+2\lambda)(1) + (1+3\lambda)(0) + (1-\lambda)(0) = 0$.
$1+2\lambda = 0 \Rightarrow \lambda = -\frac{1}{2}$.
Substituting $\lambda = -\frac{1}{2}$ into the equation:
$(x+y+z-1) - \frac{1}{2}(2x+3y-z+4) = 0$.
Multiplying by $2$: $2x+2y+2z-2 - 2x-3y+z-4 = 0$.
$-y + 3z - 6 = 0$,which simplifies to $y-3z+6=0$.

(D) To find the point of intersection,we solve the system of three linear equations:
$x - 2y + 4z = -4$ $(1)$
$x + y + z = 8$ $(2)$
$x - y + 2z = -1$ $(3)$
Using Cramer's Rule,we calculate the determinant $D$ of the coefficient matrix:
$D = \begin{vmatrix} 1 & -2 & 4 \\ 1 & 1 & 1 \\ 1 & -1 & 2 \end{vmatrix} = 1(2 - (-1)) - (-2)(2 - 1) + 4(-1 - 1) = 1(3) + 2(1) + 4(-2) = 3 + 2 - 8 = -3$
Now,calculate $D_1, D_2, D_3$:
$D_1 = \begin{vmatrix} -4 & -2 & 4 \\ 8 & 1 & 1 \\ -1 & -1 & 2 \end{vmatrix} = -4(2 - (-1)) - (-2)(16 - (-1)) + 4(-8 - (-1)) = -4(3) + 2(17) + 4(-7) = -12 + 34 - 28 = -6$
$D_2 = \begin{vmatrix} 1 & -4 & 4 \\ 1 & 8 & 1 \\ 1 & -1 & 2 \end{vmatrix} = 1(16 - (-1)) - (-4)(2 - 1) + 4(-1 - 8) = 1(17) + 4(1) + 4(-9) = 17 + 4 - 36 = -15$
$D_3 = \begin{vmatrix} 1 & -2 & -4 \\ 1 & 1 & 8 \\ 1 & -1 & -1 \end{vmatrix} = 1(-1 - (-8)) - (-2)(-1 - 8) + (-4)(-1 - 1) = 1(7) + 2(-9) - 4(-2) = 7 - 18 + 8 = -3$
The coordinates are:
$x = \frac{D_1}{D} = \frac{-6}{-3} = 2$
$y = \frac{D_2}{D} = \frac{-15}{-3} = 5$
$z = \frac{D_3}{D} = \frac{-3}{-3} = 1$
Thus,the point of intersection is $(2, 5, 1)$.

(B) Let the determinant be $\Delta = \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc$.
Since each element can be $0$ or $1$,there are $2^4 = 16$ possible determinants.
The determinant is zero if $ad = bc$.
Case $1$: $ad = 0$ and $bc = 0$.
For $ad=0$,the pairs $(a,d)$ can be $(0,0), (0,1), (1,0)$,which are $3$ possibilities.
Similarly,for $bc=0$,there are $3$ possibilities.
Total cases for $ad=bc=0$ is $3 \times 3 = 9$.
Case $2$: $ad = 1$ and $bc = 1$.
This only happens if $a=1, d=1$ and $b=1, c=1$,which is $1$ possibility.
Total cases where $\Delta = 0$ is $9 + 1 = 10$.
Total cases where $\Delta \neq 0$ is $16 - 10 = 6$.
Probability $= \frac{6}{16} = \frac{3}{8}$.

(B) For mutually exclusive events,the sum of probabilities must satisfy $0 \leq P(A) + P(B) + P(C) \leq 1$ and each individual probability must satisfy $0 \leq P(E) \leq 1$.
$1$. $P(A) \geq 0$ $\Rightarrow 3x+1 \geq 0$ $\Rightarrow x \geq -1/3$.
$2$. $P(B) \geq 0$ $\Rightarrow 1-x \geq 0$ $\Rightarrow x \leq 1$.
$3$. $P(C) \geq 0$ $\Rightarrow 1-2x \geq 0$ $\Rightarrow x \leq 1/2$.
$4$. $P(A) + P(B) + P(C) \leq 1 \Rightarrow \frac{3x+1}{3} + \frac{1-x}{4} + \frac{1-2x}{2} \leq 1$.
Multiplying by $12$: $4(3x+1) + 3(1-x) + 6(1-2x) \leq 12$.
$12x + 4 + 3 - 3x + 6 - 12x \leq 12$ $\Rightarrow -3x + 13 \leq 12$ $\Rightarrow -3x \leq -1$ $\Rightarrow x \geq 1/3$.
Combining all conditions: $x \geq 1/3$ and $x \leq 1/2$,so $x \in [\frac{1}{3}, \frac{1}{2}]$.