IIT JEE 1994 Mathematics Question Paper with Answer and Solution

(C) Given that $\omega$ is an imaginary cube root of unity,we know that $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.
First,simplify the powers of $\omega$:
$\omega^{10} = (\omega^3)^3 \cdot \omega = 1^3 \cdot \omega = \omega$
$\omega^{23} = (\omega^3)^7 \cdot \omega^2 = 1^7 \cdot \omega^2 = \omega^2$
Substitute these into the expression:
$\sin \left[ (\omega + \omega^2)\pi - \frac{\pi}{4} \right]$
Since $1 + \omega + \omega^2 = 0$,we have $\omega + \omega^2 = -1$.
So,the expression becomes $\sin \left[ -1 \cdot \pi - \frac{\pi}{4} \right] = \sin \left( -\pi - \frac{\pi}{4} \right)$.
Using the property $\sin(-\theta) = -\sin(\theta)$:
$-\sin \left( \pi + \frac{\pi}{4} \right) = -(-\sin \frac{\pi}{4}) = \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$.

(A) The vertices of an equilateral triangle inscribed in a circle with center at the origin are given by rotating the first vertex by $120^\circ$ ($2\pi/3$ radians) and $240^\circ$ ($4\pi/3$ radians) about the origin.
Given $z_1 = 1 + i\sqrt{3} = 2e^{i\pi/3}$.
The other vertices are $z_2 = z_1 e^{i2\pi/3} = 2e^{i\pi/3} e^{i2\pi/3} = 2e^{i\pi} = -2$.
And $z_3 = z_1 e^{i4\pi/3} = 2e^{i\pi/3} e^{i4\pi/3} = 2e^{i5\pi/3} = 2(\cos(5\pi/3) + i\sin(5\pi/3)) = 2(1/2 - i\sqrt{3}/2) = 1 - i\sqrt{3}$.
Thus,the values of $z_3$ and $z_2$ are $1 - i\sqrt{3}$ and $-2$ respectively,or vice versa depending on the order of rotation. Comparing with the options,option $A$ provides the correct set of values.

(D) Given that $\ln(a + c)$,$\ln(c - a)$,and $\ln(a - 2b + c)$ are in $A.P.$
Therefore,$2\ln(c - a) = \ln(a + c) + \ln(a - 2b + c)$
Using the property $\ln(x) + \ln(y) = \ln(xy)$,we get:
$\ln((c - a)^2) = \ln((a + c)(a - 2b + c))$
Removing the logarithms:
$(c - a)^2 = (a + c)(a - 2b + c)$
$c^2 + a^2 - 2ac = a^2 - 2ab + ac + ac - 2bc + c^2$
$c^2 + a^2 - 2ac = a^2 + c^2 + 2ac - 2ab - 2bc$
$-2ac = 2ac - 2b(a + c)$
$2b(a + c) = 4ac$
$b = \frac{2ac}{a + c}$
This is the condition for $a, b, c$ to be in $H.P.$

(D) The roots of the equation $x^2 + x + 1 = 0$ are the complex cube roots of unity,$\omega$ and $\omega^2$.
Let $\alpha = \omega$ and $\beta = \omega^2$.
We need to find the equation whose roots are $\alpha^{19}$ and $\beta^7$.
$\alpha^{19} = \omega^{19} = (\omega^3)^6 \cdot \omega = 1^6 \cdot \omega = \omega$.
$\beta^7 = (\omega^2)^7 = \omega^{14} = (\omega^3)^4 \cdot \omega^2 = 1^4 \cdot \omega^2 = \omega^2$.
The roots of the new equation are $\omega$ and $\omega^2$,which are the same as the roots of the original equation $x^2 + x + 1 = 0$.

(C) For the quadratic equation $px^2 + qx + 1 = 0$ to have real roots,the discriminant $D$ must be greater than or equal to $0$.
$D = q^2 - 4(p)(1) \ge 0$
$q^2 \ge 4p$
Given $p, q \in \{1, 2, 3, 4\}$,we test the possible values:
If $p = 1$,$q^2 \ge 4 \Rightarrow q \in \{2, 3, 4\}$ ($3$ solutions).
If $p = 2$,$q^2 \ge 8 \Rightarrow q \in \{3, 4\}$ ($2$ solutions).
If $p = 3$,$q^2 \ge 12 \Rightarrow q \in \{4\}$ ($1$ solution).
If $p = 4$,$q^2 \ge 16 \Rightarrow q \in \{4\}$ ($1$ solution).
Total number of solutions = $3 + 2 + 1 + 1 = 7$.

(B) Given expression: $\sec 2x - \tan 2x = \frac{1 - \sin 2x}{\cos 2x}$
Using identities $\sin 2x = 2 \sin x \cos x$,$\cos 2x = \cos^2 x - \sin^2 x$,and $1 = \cos^2 x + \sin^2 x$:
$= \frac{\cos^2 x + \sin^2 x - 2 \sin x \cos x}{\cos^2 x - \sin^2 x} = \frac{(\cos x - \sin x)^2}{(\cos x - \sin x)(\cos x + \sin x)}$
$= \frac{\cos x - \sin x}{\cos x + \sin x}$
Dividing numerator and denominator by $\cos x$:
$= \frac{1 - \tan x}{1 + \tan x} = \frac{\tan \frac{\pi}{4} - \tan x}{1 + \tan \frac{\pi}{4} \tan x}$
Using the formula $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$:
$= \tan \left( \frac{\pi}{4} - x \right)$.

(C) Let the sides of the triangle be $a = 3, b = 5,$ and $c = 7.$
Since the largest side is $c = 7,$ the largest angle is $\angle C.$
Using the Law of Cosines: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$
Substituting the values: $\cos C = \frac{3^2 + 5^2 - 7^2}{2 \times 3 \times 5}$
$\cos C = \frac{9 + 25 - 49}{30} = \frac{-15}{30} = -\frac{1}{2}$
Since $\cos C = -\frac{1}{2},$ we have $\angle C = \arccos(-1/2) = \frac{2\pi}{3}.$

(C) In $\triangle ABC$,$AD = b \sin C = c \sin B$.
Given $AD = \frac{abc}{b^2 - c^2}$,we have $AD(b^2 - c^2) = abc$.
Substituting $AD = b \sin C$ and $AD = c \sin B$:
$b \sin C (b^2 - c^2) = abc \implies \sin C (b^2 - c^2) = ac$.
Using the sine rule $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$:
$\sin C (4R^2 \sin^2 B - 4R^2 \sin^2 C) = (2R \sin A)(2R \sin C)$.
$\sin^2 B - \sin^2 C = \sin A = \sin(180^\circ - (B+C)) = \sin(B+C)$.
$(\sin B - \sin C)(\sin B + \sin C) = \sin B \cos C + \cos B \sin C$.
Since $\sin B - \sin C = 2 \sin(\frac{B-C}{2}) \cos(\frac{B+C}{2})$ and $\sin B + \sin C = 2 \sin(\frac{B+C}{2}) \cos(\frac{B-C}{2})$,
$2 \sin(\frac{B-C}{2}) \cos(\frac{B+C}{2}) \cdot 2 \sin(\frac{B+C}{2}) \cos(\frac{B-C}{2}) = \sin(B+C)$.
$2 \sin(B-C) \cdot 2 \sin(B+C) \cdot \frac{1}{2} = \sin(B+C)$.
$2 \sin(B-C) = 1 \implies \sin(B-C) = \frac{1}{2} = \sin 30^\circ$.
$B - C = 30^\circ \implies B = 30^\circ + 23^\circ = 53^\circ$ (Not in options).
Re-evaluating the condition $AD = \frac{abc}{b^2-c^2}$:
$AD = \frac{abc}{b^2-c^2} \implies \frac{1}{AD} = \frac{b^2-c^2}{abc} = \frac{b}{ac} - \frac{c}{ab} = \frac{\sin B}{a \sin C} - \frac{\sin C}{a \sin B} = \frac{\sin^2 B - \sin^2 C}{a \sin B \sin C}$.
Since $AD = c \sin B$,$\frac{1}{c \sin B} = \frac{\sin(B-C)\sin(B+C)}{a \sin B \sin C}$.
Using $a = \frac{c \sin A}{\sin C} = \frac{c \sin(B+C)}{\sin C}$,we get $\sin(B-C) = 1$.
$B - C = 90^\circ \implies B = 90^\circ + 23^\circ = 113^\circ$.

(C) The given equations are $x^2 - 5x + 6 = 0$ and $y^2 - 6y + 5 = 0$.
Solving these,we get $x = 2, 3$ and $y = 1, 5$.
These represent the lines $x=2, x=3, y=1, y=5$,which form a rectangle with vertices $(2,1), (3,1), (3,5),$ and $(2,5)$.
The diagonal $d_1$ passes through $(2,1)$ and $(3,5)$. Its equation is $y - 1 = \frac{5 - 1}{3 - 2}(x - 2)$ $\Rightarrow y - 1 = 4(x - 2)$ $\Rightarrow y = 4x - 7$.
The diagonal $d_2$ passes through $(3,1)$ and $(2,5)$. Its equation is $y - 1 = \frac{5 - 1}{2 - 3}(x - 3)$ $\Rightarrow y - 1 = -4(x - 3)$ $\Rightarrow y - 1 = -4x + 12$ $\Rightarrow 4x + y = 13$.
Thus,the equations of the diagonals are $4x + y = 13$ and $y = 4x - 7$.

(C) Let $p$ be the altitude of the equilateral triangle. Then $p = a \sin 60^{\circ} = \frac{a\sqrt{3}}{2}$.
Since the triangle is equilateral,the centroid,orthocentre,circumcentre,and incentre all coincide.
Therefore,the radius $r$ of the inscribed circle is $\frac{1}{3}p = \frac{1}{3} \times \frac{a\sqrt{3}}{2} = \frac{a}{2\sqrt{3}}$.
The diameter of the circle is $D = 2r = \frac{a}{\sqrt{3}}$.
Let $x$ be the side length of the square inscribed in the circle. The diagonal of the square is equal to the diameter of the circle.
Thus,$x^2 + x^2 = D^2$,which gives $2x^2 = D^2$.
Substituting $D = \frac{a}{\sqrt{3}}$,we get $2x^2 = (\frac{a}{\sqrt{3}})^2 = \frac{a^2}{3}$.
Therefore,the area of the square is $x^2 = \frac{a^2}{6}$.

(A) Let the variable point be $P(x, y)$.
According to the definition of a conic section,the distance from a fixed point (focus) is $e$ times the distance from a fixed line (directrix).
Here,the focus is $S(-2, 0)$,the directrix is $x = -\frac{9}{2}$,and the eccentricity $e = \frac{2}{3}$.
Since $e < 1$,the locus is an ellipse.
Mathematically,$\sqrt{(x + 2)^2 + y^2} = \frac{2}{3} |x + \frac{9}{2}|$.
Squaring both sides: $(x + 2)^2 + y^2 = \frac{4}{9} (x + \frac{9}{2})^2$.
$(x^2 + 4x + 4) + y^2 = \frac{4}{9} (x^2 + 9x + \frac{81}{4})$.
$x^2 + 4x + 4 + y^2 = \frac{4}{9}x^2 + 4x + 9$.
$\frac{5}{9}x^2 + y^2 = 5$.
Dividing by $5$: $\frac{x^2}{9} + \frac{y^2}{5} = 1$.
This is the equation of an ellipse.

(D) The ellipse $E$ is given by $f(x, y) = \frac{x^2}{9} + \frac{y^2}{4} - 1 = 0$.
For point $P(1, 2)$,$f(1, 2) = \frac{1}{9} + \frac{4}{4} - 1 = \frac{1}{9} > 0$,so $P$ lies outside $E$.
For point $Q(2, 1)$,$f(2, 1) = \frac{4}{9} + \frac{1}{4} - 1 = \frac{16+9-36}{36} = -\frac{11}{36} < 0$,so $Q$ lies inside $E$.
The circle $C$ is given by $g(x, y) = x^2 + y^2 - 9 = 0$.
For point $P(1, 2)$,$g(1, 2) = 1^2 + 2^2 - 9 = 1 + 4 - 9 = -4 < 0$,so $P$ lies inside $C$.
For point $Q(2, 1)$,$g(2, 1) = 2^2 + 1^2 - 9 = 4 + 1 - 9 = -4 < 0$,so $Q$ lies inside $C$.
Thus,$P$ lies inside $C$ but outside $E$.

(C) Given equation: $2x^2 + 3y^2 - 8x - 18y + 35 - k = 0$.
Completing the square for $x$ and $y$ terms:
$2(x^2 - 4x) + 3(y^2 - 6y) + 35 - k = 0$
$2(x^2 - 4x + 4) + 3(y^2 - 6y + 9) + 35 - k - 8 - 27 = 0$
$2(x - 2)^2 + 3(y - 3)^2 = k$.
Case $1$: If $k = 0$,the equation becomes $2(x - 2)^2 + 3(y - 3)^2 = 0$,which represents the point $(2, 3)$.
Case $2$: If $k > 0$,the equation represents an ellipse.
Case $3$: If $k < 0$,the equation represents no real locus (imaginary ellipse).
Thus,the correct statement is that it represents a point if $k = 0$.

(A) Let $f(x) = 5x^2 + 2x + 3 - 2\sin x$.
We analyze the minimum value of $f(x) = 5(x^2 + \frac{2}{5}x) + 3 - 2\sin x = 5(x + \frac{1}{5})^2 + 3 - \frac{1}{5} - 2\sin x = 5(x + \frac{1}{5})^2 + 2.8 - 2\sin x$.
Since $-1 \le \sin x \le 1$,the term $-2\sin x$ has a minimum value of $-2$.
Thus,$f(x) \ge 5(x + \frac{1}{5})^2 + 2.8 - 2 = 5(x + \frac{1}{5})^2 + 0.8$.
Since $5(x + \frac{1}{5})^2 \ge 0$,it follows that $f(x) \ge 0.8 > 0$ for all real $x$.
Therefore,$f(x)$ is never equal to $0$,which means the curves never intersect.
The number of intersection points is $0$.

(B) For the parabola $y^2 = 4ax$,differentiating with respect to $x$ gives:
$2y \frac{dy}{dx} = 4a \implies \left(\frac{dy}{dx}\right)_1 = \frac{2a}{y} \dots (i)$
For the curve $y = e^{-x/2a}$,differentiating with respect to $x$ gives:
$\left(\frac{dy}{dx}\right)_2 = e^{-x/2a} \left(-\frac{1}{2a}\right) = -\frac{y}{2a} \dots (ii)$
Two curves intersect orthogonally if the product of their slopes at the point of intersection is $-1$:
$\left(\frac{dy}{dx}\right)_1 \times \left(\frac{dy}{dx}\right)_2 = \left(\frac{2a}{y}\right) \times \left(-\frac{y}{2a}\right) = -1$
Since the product is $-1$,the curve $y = e^{-x/2a}$ cuts the parabola $y^2 = 4ax$ at right angles.

(B) The coordinates of the foci are $F_1(-ae, 0)$ and $F_2(ae, 0)$.
Let the point $P$ be $(x, y)$. The base of the triangle $PF_1F_2$ is the distance between the foci,which is $F_1F_2 = 2ae$.
The height of the triangle is the perpendicular distance from $P$ to the $x$-axis,which is $|y|$.
The area $A$ of the triangle $PF_1F_2$ is given by $A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (2ae) \times |y| = ae|y|$.
For the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,the maximum value of $|y|$ is $b$ (at $x = 0$).
Therefore,the maximum area $A = ae \times b = abe$.

(B) Let $S = \sum_{k=0}^{99} \left[ \frac{1}{2} + \frac{k}{100} \right]$.
For $0 \le k \le 49$,we have $0 \le \frac{k}{100} \le 0.49$,so $\frac{1}{2} + \frac{k}{100} = 0.5 + \frac{k}{100} < 1$. Thus,$\left[ \frac{1}{2} + \frac{k}{100} \right] = 0$ for $k = 0, 1, \dots, 49$.
For $50 \le k \le 99$,we have $0.50 \le \frac{k}{100} \le 0.99$,so $1 \le \frac{1}{2} + \frac{k}{100} < 1.49$. Thus,$\left[ \frac{1}{2} + \frac{k}{100} \right] = 1$ for $k = 50, 51, \dots, 99$.
The number of terms from $k = 50$ to $k = 99$ is $99 - 50 + 1 = 50$.
Therefore,the sum is $0 \times 50 + 1 \times 50 = 50$.

(D) Given $2\sin^2 x + 3\sin x - 2 > 0$.
Factoring the quadratic: $2\sin^2 x + 4\sin x - \sin x - 2 > 0$.
$2\sin x(\sin x + 2) - 1(\sin x + 2) > 0$.
$(\sin x + 2)(2\sin x - 1) > 0$.
Since $\sin x + 2 > 0$ for all real $x$,we must have $2\sin x - 1 > 0$,which implies $\sin x > 1/2$.
For $x$ in a reasonable range,$\sin x > 1/2$ implies $x > \pi/6$.
Given $x^2 - x - 2 < 0$.
Factoring the quadratic: $(x - 2)(x + 1) < 0$.
This inequality holds for $x \in (-1, 2)$.
Combining the two conditions: $x > \pi/6$ and $-1 < x < 2$.
Since $\pi/6 \approx 0.523$,the intersection is $x \in (\pi/6, 2)$.

(D) Total committee size is $12$. Let $W$ be the number of women and $M$ be the number of men. We require $W + M = 12$ and $W \ge 5$.
$(i)$ Women are in the majority if $W > M$. Since $W + M = 12$,this implies $W > 6$. Possible values for $W$ are $7, 8, 9$.
Number of ways = $^9C_7 \times ^8C_5 + ^9C_8 \times ^8C_4 + ^9C_9 \times ^8C_3$
$= (36 \times 56) + (9 \times 70) + (1 \times 56) = 2016 + 630 + 56 = 2702$.
(ii) Men are in the majority if $M > W$. Since $W + M = 12$,this implies $M > 6$. Possible values for $W$ are $5, 6$ (since $W \ge 5$).
Number of ways = $^9C_5 \times ^8C_7 + ^9C_6 \times ^8C_6$
$= (126 \times 8) + (84 \times 28) = 1008 + 2352 = 3360$.
Wait,re-evaluating: The question asks for committees where $W \ge 5$. If $M > W$,then $W$ can be $5$ or $6$.
For $W=5, M=7$: $^9C_5 \times ^8C_7 = 126 \times 8 = 1008$.
For $W=6, M=6$: This is not a majority. Thus,only $W=5$ gives a male majority.
Correct answer is $2702, 1008$.

(A) In the expansion of $(1 + x)^n$,the coefficients of the second,third,and fourth terms are $^nC_1, ^nC_2$,and $^nC_3$ respectively.
Given that $^nC_1, ^nC_2, ^nC_3$ are in $A.P.$,we have:
$2(^nC_2) = ^nC_1 + ^nC_3$
$2 \times \frac{n(n - 1)}{2} = n + \frac{n(n - 1)(n - 2)}{6}$
Dividing by $n$ (since $n \neq 0$):
$n - 1 = 1 + \frac{(n - 1)(n - 2)}{6}$
$6(n - 1) = 6 + (n^2 - 3n + 2)$
$6n - 6 = n^2 - 3n + 8$
$n^2 - 9n + 14 = 0$
$(n - 7)(n - 2) = 0$
So,$n = 7$ or $n = 2$.
For $n = 2$,the expansion $(1 + x)^2$ has only three terms,so the fourth term does not exist. Thus,$n = 7$ is the only valid solution.

(C) For $n = 1$,we get,
${P^{1 + 1}} + {(P + 1)^{2(1) - 1}} = {P^2} + {(P + 1)^1} = {P^2} + P + 1$.
This expression is clearly divisible by ${P^2} + P + 1$.
Let us assume that the given result is true for $n = m \in N$,i.e.,${P^{m + 1}} + {(P + 1)^{2m - 1}} = k({P^2} + P + 1)$ for some $k \in N$ .....$(i)$
Now,for $n = m + 1$:
${P^{(m + 1) + 1}} + {(P + 1)^{2(m + 1) - 1}} = {P^{m + 2}} + {(P + 1)^{2m + 1}} = {P^{m + 2}} + {(P + 1)^2}{(P + 1)^{2m - 1}}$.
Using equation $(i)$,we substitute ${(P + 1)^{2m - 1}} = k({P^2} + P + 1) - {P^{m + 1}}$:
$= {P^{m + 2}} + {(P + 1)^2}[k({P^2} + P + 1) - {P^{m + 1}}]
= {P^{m + 2}} + {(P + 1)^2} \cdot k({P^2} + P + 1) - {(P + 1)^2}{P^{m + 1}}
= {P^{m + 1}}[P - {(P + 1)^2}] + {(P + 1)^2} \cdot k({P^2} + P + 1)
= {P^{m + 1}}[P - ({P^2} + 2P + 1)] + {(P + 1)^2} \cdot k({P^2} + P + 1)
= {P^{m + 1}}[-{P^2} - P - 1] + {(P + 1)^2} \cdot k({P^2} + P + 1)
= -{P^{m + 1}}({P^2} + P + 1) + {(P + 1)^2} \cdot k({P^2} + P + 1)
= ({P^2} + P + 1)[k{(P + 1)^2} - {P^{m + 1}}]$.
Since this is a multiple of ${P^2} + P + 1$,the result is true for $n = m + 1$.
By the principle of mathematical induction,the result is true for all $n \in N$.

(B) Given $\sin \frac{\pi }{2^n} + \cos \frac{\pi }{2^n} = \frac{\sqrt{n}}{2}$.
Multiply both sides by $\frac{1}{\sqrt{2}}$:
$\frac{1}{\sqrt{2}} \sin \frac{\pi }{2^n} + \frac{1}{\sqrt{2}} \cos \frac{\pi }{2^n} = \frac{\sqrt{n}}{2\sqrt{2}}$.
$\sin \left( \frac{\pi }{2^n} + \frac{\pi }{4} \right) = \frac{\sqrt{n}}{2\sqrt{2}}$.
Since the maximum value of $\sin \theta$ is $1$,we have $\frac{\sqrt{n}}{2\sqrt{2}} \le 1$ $\Rightarrow \sqrt{n} \le 2\sqrt{2}$ $\Rightarrow n \le 8$.
Also,for $n=1$,$\sin \frac{\pi}{2} + \cos \frac{\pi}{2} = 1 + 0 = 1$,while $\frac{\sqrt{1}}{2} = 0.5$. $1 \neq 0.5$.
For $n=2$,$\sin \frac{\pi}{4} + \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2} \approx 1.414$,while $\frac{\sqrt{2}}{2} \approx 0.707$.
For $n=3$,$\sin \frac{\pi}{8} + \cos \frac{\pi}{8} = \sqrt{1 + \sin \frac{\pi}{4}} = \sqrt{1 + \frac{1}{\sqrt{2}}} \approx 1.306$,while $\frac{\sqrt{3}}{2} \approx 0.866$.
For $n=4$,$\sin \frac{\pi}{16} + \cos \frac{\pi}{16} > 1$,while $\frac{\sqrt{4}}{2} = 1$.
Since the function $f(n) = \sin \frac{\pi}{2^n} + \cos \frac{\pi}{2^n}$ is decreasing for $n \ge 2$ and $g(n) = \frac{\sqrt{n}}{2}$ is increasing,the equality holds for $n > 4$ and $n \le 8$.

(C) Let $h$ be the height of the tower $AB$ and $x$ be the horizontal distance of $B$ from $A$. Then $x = h \tan \alpha$ and $h = h \sec \alpha$ (not needed here). Actually,$AB = h$ (slant height). The horizontal projection of $B$ is $B'$,where $AB' = h \sin \alpha$ and the vertical height of $B$ is $H = h \cos \alpha$.
Using the $m-n$ theorem in $\triangle B'AD$ with $C$ on $AD$ such that $AC = d$ and $CD = 2d$:
In $\triangle B'AC$,$\cot \beta = \frac{AC}{H} = \frac{d}{H} \Rightarrow H = d \tan \beta$.
In $\triangle B'AD$,$\cot \gamma = \frac{AD}{H} = \frac{3d}{H} \Rightarrow H = 3d \tan \gamma$.
Also,the horizontal distance of $B'$ from $A$ is $x = h \sin \alpha$. Since $B'$ is to the west,its coordinate is $-x$. $A$ is at $0$,$C$ is at $d$,$D$ is at $3d$.
Using the cotangent rule in $\triangle B'CD$:
$(d + 2d) \cot \beta = d \cot \gamma - 2d \cot(180^o - \alpha) = d \cot \gamma + 2d \cot \alpha$ is incorrect. The correct approach is:
$H \cot \beta = d + x$ and $H \cot \gamma = 3d + x$.
Subtracting: $H(\cot \gamma - \cot \beta) = 2d$.
From $H = d \tan \beta$,we have $d = H \cot \beta$.
So $H(\cot \gamma - \cot \beta) = 2H \cot \beta \Rightarrow \cot \gamma - \cot \beta = 2 \cot \beta$ (This assumes $x=0$).
Given the geometry,$x = H \tan \alpha$.
$H \cot \beta = d + H \tan \alpha \Rightarrow d = H(\cot \beta - \tan \alpha)$.
$H \cot \gamma = 3d + H \tan \alpha \Rightarrow 3d = H(\cot \gamma - \tan \alpha)$.
$3H(\cot \beta - \tan \alpha) = H(\cot \gamma - \tan \alpha)$.
$3 \cot \beta - 3 \tan \alpha = \cot \gamma - \tan \alpha$.
$3 \cot \beta - \cot \gamma = 2 \tan \alpha$.

(C) For the first circle $x^2 + y^2 - 10x + 16 = 0$,the center $C_1 = (5, 0)$ and radius $r_1 = \sqrt{5^2 - 16} = \sqrt{25 - 16} = \sqrt{9} = 3$.
For the second circle $x^2 + y^2 = r^2$,the center $C_2 = (0, 0)$ and radius $r_2 = r$.
The distance between the centers $d = \sqrt{(5-0)^2 + (0-0)^2} = 5$.
Two circles intersect at two distinct points if $|r_1 - r_2| < d < r_1 + r_2$.
Substituting the values,we get $|3 - r| < 5 < 3 + r$.
From $5 < 3 + r$,we get $r > 2$.
From $|3 - r| < 5$,we get $-5 < 3 - r < 5$,which implies $-8 < -r < 2$,or $r < 8$ and $r > -2$.
Combining these conditions,we get $2 < r < 8$.

(B) The equation of the tangent at point $(x_1, y_1)$ on the parabola $y^2 = 4ax$ is given by $yy_1 = 2a(x + x_1)$.
For the given parabola $y^2 = 4x$,we have $a = 1$.
The coordinates of the ends of the latus rectum are $L(1, 2)$ and $L'(1, -2)$.
The equation of the tangent at $L(1, 2)$ is $2y = 2(x + 1)$,which simplifies to $y = x + 1$.
The equation of the tangent at $L'(1, -2)$ is $-2y = 2(x + 1)$,which simplifies to $y = -(x + 1)$.
Solving these two equations simultaneously:
$x + 1 = -(x + 1)$
$2(x + 1) = 0$
$x = -1$
Substituting $x = -1$ into $y = x + 1$,we get $y = 0$.
Thus,the point of intersection is $(-1, 0)$.

(B) Let $P, Q,$ and $R$ be points with position vectors $\vec{p} = \alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}$,$\vec{q} = \beta \hat{i} + \gamma \hat{j} + \alpha \hat{k}$,and $\vec{r} = \gamma \hat{i} + \alpha \hat{j} + \beta \hat{k}$.
The distance between $P$ and $Q$ is $|\vec{q} - \vec{p}| = \sqrt{(\beta - \alpha)^2 + (\gamma - \beta)^2 + (\alpha - \gamma)^2}$.
The distance between $Q$ and $R$ is $|\vec{r} - \vec{q}| = \sqrt{(\gamma - \beta)^2 + (\alpha - \gamma)^2 + (\beta - \alpha)^2}$.
The distance between $R$ and $P$ is $|\vec{p} - \vec{r}| = \sqrt{(\alpha - \gamma)^2 + (\beta - \alpha)^2 + (\gamma - \beta)^2}$.
Since $|\vec{PQ}| = |\vec{QR}| = |\vec{RP}|$,the points form an equilateral triangle.

(D) Let $\vec{a} = \frac{1}{3}(2i - 2j + k)$.
$1$. Check if it is a unit vector:
$|\vec{a}| = \sqrt{(\frac{2}{3})^2 + (-\frac{2}{3})^2 + (\frac{1}{3})^2} = \sqrt{\frac{4}{9} + \frac{4}{9} + \frac{1}{9}} = \sqrt{\frac{9}{9}} = 1$.
So,it is a unit vector.
$2$. Check if it is perpendicular to $\vec{b} = 3i + 2j - 2k$:
$\vec{a} \cdot \vec{b} = \frac{1}{3}(2i - 2j + k) \cdot (3i + 2j - 2k) = \frac{1}{3}(2 \times 3 + (-2) \times 2 + 1 \times (-2)) = \frac{1}{3}(6 - 4 - 2) = \frac{1}{3}(0) = 0$.
Since the dot product is $0$,they are perpendicular.
$3$. Check if it is parallel to $\vec{c} = -i + j - \frac{1}{2}k$:
$\vec{a} = \frac{1}{3}(2i - 2j + k) = -\frac{2}{3}(-i + j - \frac{1}{2}k) = -\frac{2}{3}\vec{c}$.
Since $\vec{a} = k\vec{c}$ for some scalar $k$,they are parallel.
Since all conditions are satisfied,the correct option is $(d)$.

(B) Let the points be $P(1, -1, 2)$,$Q(2, 0, -1)$,and $R(0, 2, 1)$.
First,we find two vectors in the plane: $\overrightarrow{PQ} = (2-1)i + (0-(-1))j + (-1-2)k = i + j - 3k$ and $\overrightarrow{PR} = (0-1)i + (2-(-1))j + (1-2)k = -i + 3j - k$.
$A$ vector perpendicular to the plane is given by the cross product $\vec{n} = \overrightarrow{PQ} \times \overrightarrow{PR}$.
$\vec{n} = \begin{vmatrix} i & j & k \\ 1 & 1 & -3 \\ -1 & 3 & -1 \end{vmatrix} = i(-1 - (-9)) - j(-1 - 3) + k(3 - (-1)) = 8i + 4j + 4k$.
To simplify,we can take a parallel vector $\vec{v} = 2i + j + k$.
The magnitude of $\vec{v}$ is $|\vec{v}| = \sqrt{2^2 + 1^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$.
The unit vector perpendicular to the plane is $\pm \frac{\vec{v}}{|\vec{v}|} = \pm \frac{2i + j + k}{\sqrt{6}}$.
Comparing this with the given options,the correct option is $\frac{2i + j + k}{\sqrt{6}}$.

(A) Given $f(x) = \begin{cases} x \sin \frac{1}{x}, & x \ne 0 \\ 0, & x = 0 \end{cases}$.
Then $g(x) = x \cdot f(x) = \begin{cases} x^2 \sin \frac{1}{x}, & x \ne 0 \\ 0, & x = 0 \end{cases}$.
To check the differentiability of $g(x)$ at $x = 0$:
$g'(0) = \lim_{h \to 0} \frac{g(0+h) - g(0)}{h} = \lim_{h \to 0} \frac{h^2 \sin(1/h) - 0}{h} = \lim_{h \to 0} h \sin(1/h)$.
Since $\lim_{h \to 0} h = 0$ and $\sin(1/h)$ is bounded in $[-1, 1]$,by the squeeze theorem,$g'(0) = 0$.
Thus,$g(x)$ is differentiable at $x = 0$.
Now,for $x \ne 0$,$g'(x) = \frac{d}{dx} (x^2 \sin \frac{1}{x}) = 2x \sin \frac{1}{x} + x^2 \cos \frac{1}{x} \cdot (-\frac{1}{x^2}) = 2x \sin \frac{1}{x} - \cos \frac{1}{x}$.
As $x \to 0$,$2x \sin \frac{1}{x} \to 0$,but $\lim_{x \to 0} \cos \frac{1}{x}$ does not exist.
Therefore,$\lim_{x \to 0} g'(x)$ does not exist,which means $g'(x)$ is not continuous at $x = 0$.

(D) Given $f(x) = (x + 2)e^{-x}$.
To find the intervals of increase and decrease,we find the derivative $f'(x)$ using the product rule:
$f'(x) = (1)e^{-x} + (x + 2)(-e^{-x})$
$f'(x) = e^{-x}(1 - x - 2)$
$f'(x) = -e^{-x}(x + 1)$.
For the function to be increasing,$f'(x) > 0$:
$-e^{-x}(x + 1) > 0$
Since $e^{-x} > 0$ for all $x$,we must have $-(x + 1) > 0$,which implies $x + 1 < 0$,so $x < -1$.
Thus,the function is increasing in $(-\infty, -1)$.
For the function to be decreasing,$f'(x) < 0$:
$-e^{-x}(x + 1) < 0$
Since $e^{-x} > 0$ for all $x$,we must have $-(x + 1) < 0$,which implies $x + 1 > 0$,so $x > -1$.
Thus,the function is decreasing in $(-1, \infty)$.
Therefore,the function is increasing in $(-\infty, -1)$ and decreasing in $(-1, \infty)$.

(D) Let $I = \int \cos 2\theta \log \left( \frac{\cos \theta + \sin \theta }{\cos \theta - \sin \theta } \right) d\theta$.
First,simplify the logarithmic term: $\log \left( \frac{\cos \theta + \sin \theta }{\cos \theta - \sin \theta } \right) = \log \left( \frac{1 + \tan \theta }{1 - \tan \theta } \right) = \log \tan \left( \frac{\pi }{4} + \theta \right)$.
Let $u = \log \tan \left( \frac{\pi }{4} + \theta \right)$ and $dv = \cos 2\theta d\theta$.
Then $du = \frac{1}{\tan(\frac{\pi}{4} + \theta)} \cdot \sec^2(\frac{\pi}{4} + \theta) d\theta = \frac{\cos(\frac{\pi}{4} + \theta)}{\sin(\frac{\pi}{4} + \theta)} \cdot \frac{1}{\cos^2(\frac{\pi}{4} + \theta)} d\theta = \frac{1}{\sin(\frac{\pi}{4} + \theta)\cos(\frac{\pi}{4} + \theta)} d\theta = \frac{2}{\sin(\frac{\pi}{2} + 2\theta)} d\theta = 2 \sec 2\theta d\theta$.
Also,$v = \int \cos 2\theta d\theta = \frac{1}{2} \sin 2\theta$.
Using integration by parts $\int u dv = uv - \int v du$:
$I = \frac{1}{2} \sin 2\theta \log \tan \left( \frac{\pi }{4} + \theta \right) - \int \left( \frac{1}{2} \sin 2\theta \right) (2 \sec 2\theta) d\theta$.
$I = \frac{1}{2} \sin 2\theta \log \tan \left( \frac{\pi }{4} + \theta \right) - \int \tan 2\theta d\theta$.
$I = \frac{1}{2} \sin 2\theta \log \tan \left( \frac{\pi }{4} + \theta \right) - \frac{1}{2} \log |\sec 2\theta| + C$.

(D) Let $I = \int_2^3 {\frac{{\sqrt x }}{{\sqrt {5 - x} + \sqrt x }}} \,dx$ .....$(i)$
Using the property $\int_a^b f(x) \,dx = \int_a^b f(a + b - x) \,dx$,we replace $x$ with $(2 + 3 - x) = (5 - x)$.
$\therefore I = \int_2^3 {\frac{{\sqrt {5 - x} }}{{\sqrt {5 - (5 - x)} + \sqrt {5 - x} }}} \,dx$
$I = \int_2^3 {\frac{{\sqrt {5 - x} }}{{\sqrt x + \sqrt {5 - x} }}} \,dx$ .....$(ii)$
Adding $(i)$ and $(ii)$:
$2I = \int_2^3 {\frac{{\sqrt x + \sqrt {5 - x} }}{{\sqrt {5 - x} + \sqrt x }}} \,dx$
$2I = \int_2^3 1 \,dx$
$2I = [x]_2^3 = 3 - 2 = 1$
$I = \frac{1}{2}$.

(B) We know that the function $f(x) = |\sin x|$ is periodic with period $\pi$.
Thus,$\int_0^{n\pi} |\sin x| \, dx = n \int_0^{\pi} |\sin x| \, dx = n \int_0^{\pi} \sin x \, dx = n [-\cos x]_0^{\pi} = n(1 - (-1)) = 2n$.
Now,consider the integral $\int_{n\pi}^{n\pi + v} |\sin x| \, dx$. Let $x = n\pi + t$,then $dx = dt$. When $x = n\pi$,$t = 0$. When $x = n\pi + v$,$t = v$.
Since $|sin(n\pi + t)| = |\sin(n\pi) \cos t + \cos(n\pi) \sin t| = |(-1)^n \sin t| = |\sin t|$.
So,$\int_{n\pi}^{n\pi + v} |\sin x| \, dx = \int_0^v |\sin t| \, dt$.
Assuming $0 \le v \le \pi$,we have $\int_0^v \sin t \, dt = [-\cos t]_0^v = 1 - \cos v$.
Therefore,$\int_0^{n\pi + v} |\sin x| \, dx = 2n + 1 - \cos v$.

(B) The given curves are $y = |x - 1|$ and $y = 1$.
To find the intersection points,set $|x - 1| = 1$,which gives $x - 1 = 1$ or $x - 1 = -1$. Thus,$x = 2$ or $x = 0$.
The region is bounded between $x = 0$ and $x = 2$.
The area $A$ is given by $\int_{0}^{2} (1 - |x - 1|) dx$.
Since $|x - 1| = 1 - x$ for $x < 1$ and $|x - 1| = x - 1$ for $x \ge 1$,we split the integral:
$A = \int_{0}^{1} (1 - (1 - x)) dx + \int_{1}^{2} (1 - (x - 1)) dx$
$A = \int_{0}^{1} x dx + \int_{1}^{2} (2 - x) dx$
$A = \left[ \frac{x^2}{2} \right]_{0}^{1} + \left[ 2x - \frac{x^2}{2} \right]_{1}^{2}$
$A = \left( \frac{1}{2} - 0 \right) + \left( (4 - 2) - (2 - \frac{1}{2}) \right)$
$A = \frac{1}{2} + (2 - \frac{3}{2}) = \frac{1}{2} + \frac{1}{2} = 1$.
Thus,the area is $1$ square unit.

(A) If $A, B,$ and $C$ are mutually independent,then $P(A \cap B) = P(A)P(B), P(A \cap C) = P(A)P(C),$ and $P(A \cap B \cap C) = P(A)P(B)P(C)$.
For $S_1$: $P(A \cap (B \cup C)) = P((A \cap B) \cup (A \cap C)) = P(A \cap B) + P(A \cap C) - P(A \cap B \cap C) = P(A)P(B) + P(A)P(C) - P(A)P(B)P(C) = P(A)[P(B) + P(C) - P(B)P(C)] = P(A)P(B \cup C)$. Thus,$S_1$ is true.
For $S_2$: $P(A \cap (B \cap C)) = P(A \cap B \cap C) = P(A)P(B)P(C) = P(A)P(B \cap C)$. Thus,$S_2$ is true.

(B) The probability of getting a number greater than $4$ (i.e.,$5$ or $6$) in a single toss is $p = \frac{2}{6} = \frac{1}{3}$.
The probability of failure (getting $1, 2, 3,$ or $4$) is $q = 1 - p = 1 - \frac{1}{3} = \frac{2}{3}$.
We need the probability that the success occurs on an even number of tosses,which corresponds to the sequence of outcomes: (Failure,Success),(Failure,Failure,Failure,Success),(Failure,Failure,Failure,Failure,Failure,Success),and so on.
The required probability is $P = pq + q^3p + q^5p + \dots$
This is an infinite geometric series with the first term $a = pq$ and common ratio $r = q^2$.
The sum of an infinite geometric series is $S = \frac{a}{1 - r}$.
Substituting the values: $P = \frac{pq}{1 - q^2} = \frac{(\frac{1}{3})(\frac{2}{3})}{1 - (\frac{2}{3})^2} = \frac{\frac{2}{9}}{1 - \frac{4}{9}} = \frac{\frac{2}{9}}{\frac{5}{9}} = \frac{2}{5}$.

(C) Given $P(A^c) = 0.3$,so $P(A) = 1 - 0.3 = 0.7$.
Given $P(B) = 0.4$,so $P(B^c) = 1 - 0.4 = 0.6$.
Given $P(A \cap B^c) = 0.5$.
We know that $P(A) = P(A \cap B) + P(A \cap B^c)$,so $0.7 = P(A \cap B) + 0.5$,which implies $P(A \cap B) = 0.2$.
We need to find $P(B | A \cup B^c) = \frac{P(B \cap (A \cup B^c))}{P(A \cup B^c)}$.
Numerator: $P(B \cap (A \cup B^c)) = P((B \cap A) \cup (B \cap B^c)) = P(A \cap B) \cup \emptyset = P(A \cap B) = 0.2$.
Denominator: $P(A \cup B^c) = P(A) + P(B^c) - P(A \cap B^c) = 0.7 + 0.6 - 0.5 = 0.8$.
Therefore,$P(B | A \cup B^c) = \frac{0.2}{0.8} = \frac{1}{4}$.

(B) Let $H$ be the event of getting a head and $T$ be the event of getting a tail. $P(H) = P(T) = \frac{1}{2}$.
Case $1$: If $H$ occurs,two dice are rolled. The sum can be $7$ or $8$.
$P(7|H) = \frac{6}{36} = \frac{1}{6}$ and $P(8|H) = \frac{5}{36}$.
Case $2$: If $T$ occurs,a card is picked from $11$ cards numbered $2$ to $12$.
$P(7|T) = \frac{1}{11}$ and $P(8|T) = \frac{1}{11}$.
Total probability of getting $7$ is $P(7) = P(H)P(7|H) + P(T)P(7|T) = \frac{1}{2} \times \frac{1}{6} + \frac{1}{2} \times \frac{1}{11} = \frac{1}{2} \left( \frac{1}{6} + \frac{1}{11} \right) = \frac{1}{2} \left( \frac{17}{66} \right) = \frac{17}{132}$.
Total probability of getting $8$ is $P(8) = P(H)P(8|H) + P(T)P(8|T) = \frac{1}{2} \times \frac{5}{36} + \frac{1}{2} \times \frac{1}{11} = \frac{1}{2} \left( \frac{5}{36} + \frac{1}{11} \right) = \frac{1}{2} \left( \frac{55 + 36}{396} \right) = \frac{91}{792}$.
Required probability $P = P(7) + P(8) = \frac{17}{132} + \frac{91}{792} = \frac{102 + 91}{792} = \frac{193}{792} \approx 0.24368 \approx 0.244$.

(C) The probability of drawing a white ball in a single draw is $p = \frac{12}{24} = \frac{1}{2}$.
For the $4^{th}$ white ball to be drawn on the $7^{th}$ draw,there must be exactly $3$ white balls in the first $6$ draws,and the $7^{th}$ draw must be a white ball.
This follows a binomial distribution scenario.
The probability is given by: $P = \binom{6}{3} \times p^3 \times (1-p)^3 \times p$.
Substituting $p = \frac{1}{2}$:
$P = 20 \times (\frac{1}{2})^3 \times (\frac{1}{2})^3 \times \frac{1}{2} = 20 \times (\frac{1}{2})^7 = \frac{20}{128} = \frac{5}{32}$.

(A) The given determinant is $\Delta = \left| \begin{array}{ccc} \cos(A-P) & \cos(A-Q) & \cos(A-R) \\ \cos(B-P) & \cos(B-Q) & \cos(B-R) \\ \cos(C-P) & \cos(C-Q) & \cos(C-R) \end{array} \right|$.
Using the formula $\cos(x-y) = \cos x \cos y + \sin x \sin y$,we can write each element as a sum of two terms.
This allows us to express the determinant as the product of two matrices:
$\Delta = \left| \begin{array}{ccc} \cos A & \sin A & 0 \\ \cos B & \sin B & 0 \\ \cos C & \sin C & 0 \end{array} \right| \times \left| \begin{array}{ccc} \cos P & \cos Q & \cos R \\ \sin P & \sin Q & \sin R \\ 0 & 0 & 0 \end{array} \right|$.
Since the third column of the first matrix is all zeros and the third row of the second matrix is all zeros,the product of these two matrices is zero.
Alternatively,the determinant can be split into $8$ determinants,each of which has at least two identical columns (or rows),making the value of each determinant $0$.
Thus,the value of the determinant is $0$.

(D) Let $\alpha = \cos^{-1} \frac{1}{5\sqrt{2}}$. Then $\cos \alpha = \frac{1}{5\sqrt{2}}$.
Since $\tan^2 \alpha = \sec^2 \alpha - 1 = (5\sqrt{2})^2 - 1 = 50 - 1 = 49$,we have $\tan \alpha = 7$.
Thus,$\alpha = \tan^{-1} 7$.
Let $\beta = \sin^{-1} \frac{4}{\sqrt{17}}$. Then $\sin \beta = \frac{4}{\sqrt{17}}$.
Since $\cos^2 \beta = 1 - \sin^2 \beta = 1 - \frac{16}{17} = \frac{1}{17}$,we have $\cos \beta = \frac{1}{\sqrt{17}}$.
Then $\tan \beta = \frac{\sin \beta}{\cos \beta} = \frac{4/\sqrt{17}}{1/\sqrt{17}} = 4$.
Thus,$\beta = \tan^{-1} 4$.
The expression becomes $\tan(\alpha - \beta) = \tan(\tan^{-1} 7 - \tan^{-1} 4)$.
Using the formula $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$,we get:
$\tan(\tan^{-1} 7 - \tan^{-1} 4) = \frac{7 - 4}{1 + (7)(4)} = \frac{3}{1 + 28} = \frac{3}{29}$.

(A) The position vectors of $R$ and $S$ are $\frac{3p + 2q}{5}$ and $3p - 2q$ respectively.
Therefore,$\overrightarrow{OR} = \frac{3p + 2q}{5}$ and $\overrightarrow{OS} = 3p - 2q.$
Since $\overrightarrow{OR} \perp \overrightarrow{OS},$ their dot product is zero,so $\overrightarrow{OR} \cdot \overrightarrow{OS} = 0.$
$\left( \frac{3p + 2q}{5} \right) \cdot (3p - 2q) = 0$
$9|p|^2 - 6(p \cdot q) + 6(q \cdot p) - 4|q|^2 = 0$
$9|p|^2 - 4|q|^2 = 0$
$9|p|^2 = 4|q|^2$
Given $|p| = p$ and $|q| = q,$ we get $9p^2 = 4q^2.$

(D) Given $f(x) = \sin x$ and $g(x) = \ln |x|$.
For $fog(x) = f(g(x)) = \sin(\ln |x|)$.
Since the range of $\ln |x|$ is $(-\infty, \infty)$,the range of $\sin(\ln |x|)$ is $[-1, 1]$. Thus,$R_1 = [-1, 1]$.
For $gof(x) = g(f(x)) = \ln |\sin x|$.
Since the range of $|\sin x|$ is $(0, 1]$,the range of $\ln |\sin x|$ is $(-\infty, 0]$. Thus,$R_2 = (-\infty, 0]$.
Therefore,$R_1 = \{u: -1 \le u \le 1\}$ and $R_2 = \{v: -\infty < v \le 0\}$.

(B) For $f(x)$ to be continuous at $x = 0$,we must have $\lim_{x \to 0^-} f(x) = f(0) = \lim_{x \to 0^+} f(x)$.
First,calculate the left-hand limit $(LHL)$:
$\lim_{x \to 0^-} (1 + |\sin x|)^{a/|\sin x|} = e^{\lim_{x \to 0^-} |\sin x| \cdot \frac{a}{|\sin x|}} = e^a$.
Next,calculate the right-hand limit $(RHL)$:
$\lim_{x \to 0^+} e^{\tan 2x/\tan 3x} = e^{\lim_{x \to 0^+} \frac{\tan 2x}{\tan 3x}} = e^{\lim_{x \to 0^+} \frac{(\tan 2x / 2x) \cdot 2x}{(\tan 3x / 3x) \cdot 3x}} = e^{2/3}$.
Since $f(0) = b$,we equate the limits:
$e^a = b = e^{2/3}$.
Thus,$a = 2/3$ and $b = e^{2/3}$.

(A) Since the point $(2,3)$ lies on the curve $y^2=px^3+q$,we have:
$3^2 = p(2)^3 + q$
$9 = 8p + q$ ...$(i)$
Now,differentiate the curve equation $y^2=px^3+q$ with respect to $x$:
$2y \frac{dy}{dx} = 3px^2$
$\frac{dy}{dx} = \frac{3px^2}{2y}$
The slope of the tangent $y=4x-5$ is $4$. Therefore,the derivative at $(2,3)$ must be equal to $4$:
$\left. \frac{dy}{dx} \right|_{(2,3)} = \frac{3p(2)^2}{2(3)} = 4$
$\frac{12p}{6} = 4$
$2p = 4 \Rightarrow p = 2$ ...$(ii)$
Substitute $p=2$ into equation $(i)$:
$8(2) + q = 9$
$16 + q = 9$
$q = 9 - 16 = -7$
Thus,$p=2$ and $q=-7$.

(A) Given $y=(\sin x)^{\tan x}$.
Taking the natural logarithm on both sides,we get $\log y = \tan x \cdot \log(\sin x)$.
Differentiating both sides with respect to $x$ using the product rule:
$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(\tan x) \cdot \log(\sin x) + \tan x \cdot \frac{d}{dx}(\log(\sin x))$.
$\frac{1}{y} \frac{dy}{dx} = \sec^2 x \cdot \log(\sin x) + \tan x \cdot \frac{1}{\sin x} \cdot \cos x$.
Since $\frac{\cos x}{\sin x} = \cot x$,we have $\tan x \cdot \cot x = 1$.
Thus,$\frac{1}{y} \frac{dy}{dx} = \sec^2 x \log(\sin x) + 1$.
Multiplying by $y$,we get $\frac{dy}{dx} = y[1 + \sec^2 x \log(\sin x)]$.
Substituting $y = (\sin x)^{\tan x}$,we get $\frac{dy}{dx} = (\sin x)^{\tan x}[1 + \sec^2 x \log(\sin x)]$.