IIT JEE 1992 Mathematics Question Paper with Answer and Solution

(C) Let $x - 1 = y$,so $x = y + 1$.
Then,$\frac{x^2}{(x - 1)^3(x - 2)} = \frac{(1 + y)^2}{y^3(y - 1)} = \frac{1 + 2y + y^2}{y^3(y - 1)} = -\frac{1 + 2y + y^2}{y^3(1 - y)}$.
Dividing $(1 + 2y + y^2)$ by $(y - 1)$ gives $(y^2 + 2y + 1) = (y - 1)(y + 3) + 4$.
Thus,$\frac{y^2 + 2y + 1}{y^3(y - 1)} = \frac{(y - 1)(y + 3) + 4}{y^3(y - 1)} = \frac{y + 3}{y^3} + \frac{4}{y^3(y - 1)}$.
Using partial fractions for $\frac{4}{y^3(y - 1)} = \frac{A}{y} + \frac{B}{y^2} + \frac{C}{y^3} + \frac{D}{y - 1}$,we find $D = 4$,$C = -4$,$B = -4$,$A = -4$.
Substituting back,the expression becomes $\frac{-1}{(x - 1)^3} + \frac{-3}{(x - 1)^2} + \frac{-4}{(x - 1)} + \frac{4}{(x - 2)}$.

(D) We have $H.M. = \frac{2ab}{a + b}$ and $G.M. = \sqrt{ab}$.
Given $\frac{H.M.}{G.M.} = \frac{4}{5}$.
$\Rightarrow \frac{2ab/(a + b)}{\sqrt{ab}} = \frac{4}{5}$ $\Rightarrow \frac{2\sqrt{ab}}{a + b} = \frac{4}{5}$.
$\Rightarrow \frac{a + b}{2\sqrt{ab}} = \frac{5}{4}$.
Applying componendo and dividendo:
$\frac{a + b + 2\sqrt{ab}}{a + b - 2\sqrt{ab}} = \frac{5 + 4}{5 - 4}$ $\Rightarrow \frac{(\sqrt{a} + \sqrt{b})^2}{(\sqrt{a} - \sqrt{b})^2} = \frac{9}{1}$.
Taking square root on both sides:
$\frac{\sqrt{a} + \sqrt{b}}{\sqrt{a} - \sqrt{b}} = \frac{3}{1}$.
Applying componendo and dividendo again:
$\frac{(\sqrt{a} + \sqrt{b}) + (\sqrt{a} - \sqrt{b})}{(\sqrt{a} + \sqrt{b}) - (\sqrt{a} - \sqrt{b})} = \frac{3 + 1}{3 - 1}$ $\Rightarrow \frac{2\sqrt{a}}{2\sqrt{b}} = \frac{4}{2} = 2$.
Squaring both sides:
$\frac{a}{b} = 2^2 = 4$.
Thus,$a:b = 4:1$ or $b:a = 1:4$.

(C) Given that $\alpha$ and $\beta$ are the roots of $(x - a)(x - b) - c = 0$.
Expanding this,we get $x^2 - (a + b)x + ab - c = 0$.
Comparing this with the standard quadratic equation $x^2 - (\alpha + \beta)x + \alpha\beta = 0$,we have:
$\alpha + \beta = a + b$ and $\alpha\beta = ab - c$.
Now,consider the equation $(x - \alpha)(x - \beta) + c = 0$.
Expanding this,we get $x^2 - (\alpha + \beta)x + \alpha\beta + c = 0$.
Substituting the values of $(\alpha + \beta)$ and $\alpha\beta$ into this equation:
$x^2 - (a + b)x + (ab - c) + c = 0$
$x^2 - (a + b)x + ab = 0$
$(x - a)(x - b) = 0$.
Thus,the roots of the equation are $a$ and $b$.

(C) This is a problem of derangements where $n = 4$ items are to be placed in $n$ boxes such that no item goes into its correct box.
The number of derangements $D_n$ is given by the formula:
$D_n = n! \left( 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \dots + \frac{(-1)^n}{n!} \right)$
For $n = 4$:
$D_4 = 4! \left( \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} \right)$
$D_4 = 24 \left( \frac{1}{2} - \frac{1}{6} + \frac{1}{24} \right)$
$D_4 = 12 - 4 + 1 = 9$.
Thus,there are $9$ ways.

(A) Let $f(x) = \sin \left( x + \frac{\pi}{6} \right) + \cos \left( x + \frac{\pi}{6} \right)$.
We can rewrite this as $f(x) = \sqrt{2} \left( \frac{1}{\sqrt{2}} \sin \left( x + \frac{\pi}{6} \right) + \frac{1}{\sqrt{2}} \cos \left( x + \frac{\pi}{6} \right) \right)$.
Using the identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$,we get $f(x) = \sqrt{2} \sin \left( x + \frac{\pi}{6} + \frac{\pi}{4} \right) = \sqrt{2} \sin \left( x + \frac{5\pi}{12} \right)$.
The maximum value of $\sin \theta$ is $1$,which occurs when $\theta = \frac{\pi}{2}$.
Setting $x + \frac{5\pi}{12} = \frac{\pi}{2}$,we get $x = \frac{\pi}{2} - \frac{5\pi}{12} = \frac{6\pi - 5\pi}{12} = \frac{\pi}{12}$.
Since $\frac{\pi}{12}$ lies in the interval $\left( 0, \frac{\pi}{2} \right)$,the maximum value is attained at $x = \frac{\pi}{12}$.

(A) Let $y = \frac{\tan x}{\tan 3x}$.
Using the formula $\tan 3x = \frac{3\tan x - \tan^3 x}{1 - 3\tan^2 x}$,we get:
$y = \frac{\tan x}{\frac{3\tan x - \tan^3 x}{1 - 3\tan^2 x}} = \frac{1 - 3\tan^2 x}{3 - \tan^2 x}$.
Let $t = \tan^2 x$,where $t \ge 0$. Then $y = \frac{1 - 3t}{3 - t}$.
To find the range,let $y = \frac{3(1 - 3t)}{3(3 - t)} = \frac{3 - 9t}{3(3 - t)} = \frac{(3 - t) - 8t}{3(3 - t)} = \frac{1}{3} - \frac{8t}{3(3 - t)}$.
Alternatively,solving for $t$: $y(3 - t) = 1 - 3t \implies 3y - yt = 1 - 3t \implies t(3 - y) = 1 - 3y \implies t = \frac{1 - 3y}{3 - y}$.
Since $t = \tan^2 x \ge 0$,we have $\frac{1 - 3y}{3 - y} \ge 0$.
This inequality holds when $y \in [1/3, 3)$.
Thus,the value of $y$ never lies in the interval $(1/3, 3)$.

(D) Given equation: $\cos^2 \theta + \sin \theta + 1 = 0$
Using the identity $\cos^2 \theta = 1 - \sin^2 \theta$,we get:
$1 - \sin^2 \theta + \sin \theta + 1 = 0$
$-\sin^2 \theta + \sin \theta + 2 = 0$
Multiplying by $-1$:
$\sin^2 \theta - \sin \theta - 2 = 0$
Factoring the quadratic equation:
$(\sin \theta - 2)(\sin \theta + 1) = 0$
This gives $\sin \theta = 2$ or $\sin \theta = -1$.
Since the range of $\sin \theta$ is $[-1, 1]$,$\sin \theta = 2$ is impossible.
Thus,$\sin \theta = -1$.
The general solution is $\theta = 2n\pi - \frac{\pi}{2}$.
For $n=1$,$\theta = 2\pi - \frac{\pi}{2} = \frac{3\pi}{2}$.
Since $\frac{5\pi}{4} < \frac{3\pi}{2} < \frac{7\pi}{4}$,the solution lies in the interval $\left( \frac{5\pi}{4}, \frac{7\pi}{4} \right)$.

(A) We know that for any real $x$,by the $A.M. \ge G.M.$ inequality,$\frac{5^x + 5^{-x}}{2} \ge \sqrt{5^x \cdot 5^{-x}} = 1$.
This implies $5^x + 5^{-x} \ge 2$.
Also,we know that the range of the cosine function is $[-1, 1]$,so $\cos(e^x) \le 1$.
This implies $2 \cos(e^x) \le 2$.
For the equation $2 \cos(e^x) = 5^x + 5^{-x}$ to hold,both sides must be equal to $2$.
This requires $5^x + 5^{-x} = 2$,which occurs only at $x = 0$.
Substituting $x = 0$ into the left side,we get $2 \cos(e^0) = 2 \cos(1)$.
Since $\cos(1) \approx 0.54$,$2 \cos(1) \approx 1.08 \neq 2$.
Therefore,there is no value of $x$ that satisfies the equation.

(A) The perimeter of $\Delta ABC$ is $a + b + c$. The arithmetic mean of the sines of its angles is $\frac{\sin A + \sin B + \sin C}{3}$.
Given: $a + b + c = 6 \times \frac{\sin A + \sin B + \sin C}{3} = 2(\sin A + \sin B + \sin C)$.
Using the Sine Rule,$a = k \sin A$,$b = k \sin B$,and $c = k \sin C$,where $k = 2R$ (the diameter of the circumcircle).
Substituting these into the equation: $k(\sin A + \sin B + \sin C) = 2(\sin A + \sin B + \sin C)$.
Since $\sin A + \sin B + \sin C \neq 0$ for a triangle,we have $k = 2$.
Given $a = 1$,we use $a = k \sin A$:
$1 = 2 \sin A \implies \sin A = \frac{1}{2}$.
Since $A$ is an angle of a triangle,$A = \frac{\pi}{6}$ or $A = \frac{5\pi}{6}$. Given the options,$A = \frac{\pi}{6}$.

(D) Let $R$ be the circumradius of $\Delta PQR$. Given $R = PQ = PR$.
By the sine rule,$R = \frac{PQ}{2 \sin R} = \frac{PR}{2 \sin Q} = \frac{QR}{2 \sin P}$.
Since $R = PQ$,we have $PQ = \frac{PQ}{2 \sin R}$,which implies $\sin R = \frac{1}{2}$.
Thus,$\angle R = \frac{\pi}{6}$.
Since $PQ = PR$,the triangle is isosceles,so $\angle Q = \angle R = \frac{\pi}{6}$.
The sum of angles in a triangle is $\pi$. Therefore,$\angle P = \pi - (\angle Q + \angle R) = \pi - (\frac{\pi}{6} + \frac{\pi}{6}) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$.

(D) The given line is $3x + y = 3$,which can be written as $y = -3x + 3$. The slope of this line is $m_1 = -3$.
The slope of a line perpendicular to this line is $m_2 = -1/m_1 = -1/(-3) = 1/3$.
The equation of the line passing through $(2, 2)$ with slope $m_2 = 1/3$ is given by $y - y_1 = m_2(x - x_1)$.
Substituting the values,we get $y - 2 = \frac{1}{3}(x - 2)$.
Multiplying by $3$,we get $3y - 6 = x - 2$,which simplifies to $x - 3y + 4 = 0$.
To find the $y$-intercept,we set $x = 0$ in the equation: $0 - 3y + 4 = 0$,which gives $3y = 4$ or $y = 4/3$.
Thus,the $y$-intercept is $4/3$.

(A) Let the two perpendicular lines be the coordinate axes,$X$-axis and $Y$-axis. Let the point be $P(x, y)$.
The distance of point $P$ from the $X$-axis is $|y|$ and from the $Y$-axis is $|x|$.
According to the problem,the sum of these distances is $1$,so $|x| + |y| = 1$.
If the point lies in the first quadrant,$x > 0$ and $y > 0$,so $x + y = 1$.
If the point lies in the second quadrant,$x < 0$ and $y > 0$,so $-x + y = 1$.
If the point lies in the third quadrant,$x < 0$ and $y < 0$,so $-x - y = 1$.
If the point lies in the fourth quadrant,$x > 0$ and $y < 0$,so $x - y = 1$.
These four equations represent the sides of a square with vertices at $(1, 0), (0, 1), (-1, 0),$ and $(0, -1)$.
Thus,the locus is a square.

(D) Let the centre of the circle be $(h, k)$. Since the circle passes through $(0, 0)$ and $(1, 0)$,the perpendicular bisector of the chord joining $(0, 0)$ and $(1, 0)$ must contain the centre. The midpoint is $(1/2, 0)$ and the line is $x = 1/2$. Thus,$h = 1/2$.
Since the circle passes through $(0, 0)$,its radius $r$ is the distance from $(1/2, k)$ to $(0, 0)$,so $r^2 = (1/2)^2 + k^2 = 1/4 + k^2$.
The circle touches $x^2 + y^2 = 9$,which has centre $(0, 0)$ and radius $R = 3$. The distance between the centres is $\sqrt{(1/2)^2 + k^2} = \sqrt{1/4 + k^2} = r$.
For the circles to touch,the distance between centres must be $|R \pm r|$. Here,the distance between centres is $r$,so $r = |3 \pm r|$.
Case $1$: $r = 3 - r$ $\Rightarrow 2r = 3$ $\Rightarrow r = 3/2$.
Then $r^2 = 9/4$,so $1/4 + k^2 = 9/4$ $\Rightarrow k^2 = 8/4 = 2$ $\Rightarrow k = \pm \sqrt{2}$.
Thus,the centre is $\left( \frac{1}{2}, \pm \sqrt{2} \right)$.

(B) Let $\tan^{-1} 2x = \theta$.
Then $2x = \tan \theta$,which implies $x = \frac{1}{2} \tan \theta$.
As $x \to 0$,$\theta \to 0$.
Substituting these into the limit:
$\mathop {\lim }\limits_{x \to 0} \frac{x}{\tan^{-1} 2x} = \mathop {\lim }\limits_{\theta \to 0} \frac{\frac{1}{2} \tan \theta}{\theta} = \frac{1}{2} \mathop {\lim }\limits_{\theta \to 0} \frac{\tan \theta}{\theta}$.
Since $\mathop {\lim }\limits_{\theta \to 0} \frac{\tan \theta}{\theta} = 1$,the result is $\frac{1}{2} \times 1 = \frac{1}{2}$.

(B) Using $L$'$H$ôpital's Rule repeatedly,we differentiate the numerator and denominator $n$ times:
$\mathop {\lim }\limits_{x \to \infty } \frac{{{x^n}}}{{{e^x}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{n \cdot x^{n-1}}}{{{e^x}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{n(n-1) \cdot x^{n-2}}}{{{e^x}}} = \dots = \mathop {\lim }\limits_{x \to \infty } \frac{{n!}}{{{e^x}}}$.
Since $\mathop {\lim }\limits_{x \to \infty } e^x = \infty$,we have $\frac{n!}{\infty} = 0$.
This holds true for any whole number $n \ge 0$.

(C) Let $y = 3^{40}$.
Taking $\log_{10}$ on both sides,we get $\log_{10} y = \log_{10} (3^{40})$.
Using the property $\log(a^b) = b \log a$,we have $\log_{10} y = 40 \times \log_{10} 3$.
Given $\log_{10} 3 = 0.477$,so $\log_{10} y = 40 \times 0.477 = 19.08$.
The number of digits in $3^{40}$ is given by $\lfloor \log_{10} y \rfloor + 1$.
Here,$\lfloor 19.08 \rfloor + 1 = 19 + 1 = 20$.
Therefore,the number of digits is $20$.

(C) Let $z = \frac{\beta - \alpha}{1 - \overline{\alpha}\beta}$.
We calculate $|z|^2 = z \cdot \overline{z} = \left( \frac{\beta - \alpha}{1 - \overline{\alpha}\beta} \right) \left( \frac{\overline{\beta} - \overline{\alpha}}{1 - \alpha\overline{\beta}} \right)$.
Expanding the numerator: $(\beta - \alpha)(\overline{\beta} - \overline{\alpha}) = \beta\overline{\beta} - \beta\overline{\alpha} - \alpha\overline{\beta} + \alpha\overline{\alpha} = |\beta|^2 - \beta\overline{\alpha} - \alpha\overline{\beta} + |\alpha|^2$.
Expanding the denominator: $(1 - \overline{\alpha}\beta)(1 - \alpha\overline{\beta}) = 1 - \alpha\overline{\beta} - \overline{\alpha}\beta + \alpha\overline{\alpha}\beta\overline{\beta} = 1 - \alpha\overline{\beta} - \overline{\alpha}\beta + |\alpha|^2|\beta|^2$.
Since $|\beta| = 1$,we have $|\beta|^2 = 1$.
Substituting $|\beta|^2 = 1$ into the expressions:
Numerator: $1 - \beta\overline{\alpha} - \alpha\overline{\beta} + |\alpha|^2$.
Denominator: $1 - \alpha\overline{\beta} - \overline{\alpha}\beta + |\alpha|^2(1) = 1 - \alpha\overline{\beta} - \overline{\alpha}\beta + |\alpha|^2$.
Since the numerator and denominator are equal,$|z|^2 = 1$,which implies $|z| = 1$.

(A) Let the $G$.$P$. be $a, ar, ar^2, \dots$ where $a > 0$ and $r > 0$.
Given $\alpha = \sum\limits_{n = 1}^{100} {{a_{2n}}} = a_2 + a_4 + \dots + a_{200}$.
This is a $G$.$P$. with first term $ar$ and common ratio $r^2$ having $100$ terms.
So,$\alpha = ar(1 + r^2 + r^4 + \dots + r^{198})$.
Given $\beta = \sum\limits_{n = 1}^{100} {{a_{2n - 1}}} = a_1 + a_3 + \dots + a_{199}$.
This is a $G$.$P$. with first term $a$ and common ratio $r^2$ having $100$ terms.
So,$\beta = a(1 + r^2 + r^4 + \dots + r^{198})$.
Dividing $\alpha$ by $\beta$:
$\frac{\alpha}{\beta} = \frac{ar(1 + r^2 + r^4 + \dots + r^{198})}{a(1 + r^2 + r^4 + \dots + r^{198})} = r$.
Thus,the common ratio is $\frac{\alpha}{\beta}$.

(C) Let $y = 3^{40}$.
Taking $\log_{10}$ on both sides:
$\log_{10} y = \log_{10} (3^{40})$
$\log_{10} y = 40 \times \log_{10} 3$
Given $\log_{10} 3 = 0.477$,so:
$\log_{10} y = 40 \times 0.477 = 19.08$
The number of digits in $y$ is given by $\lfloor \log_{10} y \rfloor + 1$.
Number of digits $= 19 + 1 = 20$.

(C) Given $A = \log_2 \log_2 \log_4 256 + 2 \log_{\sqrt{2}} 2$.
First,evaluate $\log_4 256$:
Since $256 = 4^4$,$\log_4 256 = 4$.
Next,evaluate $\log_2 \log_2 4$:
Since $\log_2 4 = 2$,we have $\log_2 2 = 1$.
Now,evaluate $2 \log_{\sqrt{2}} 2$:
Since $\sqrt{2} = 2^{1/2}$,$\log_{2^{1/2}} 2 = \frac{1}{1/2} \log_2 2 = 2 \times 1 = 2$.
Thus,$2 \log_{\sqrt{2}} 2 = 2 \times 2 = 4$.
Finally,$A = 1 + 4 = 5$.

(C) Let $(x - 1) = y$,then $x = y + 1$.
Substituting this into the expression:
$\frac{x^2}{(x - 1)^3(x - 2)} = \frac{(1 + y)^2}{y^3(y - 1)} = \frac{1 + 2y + y^2}{y^3(y - 1)} = \frac{1 + 2y + y^2}{y^3(y - 1)}$.
Performing division of $(y^2 + 2y + 1)$ by $(y - 1)$:
$\frac{y^2 + 2y + 1}{y - 1} = (y + 3) + \frac{4}{y - 1}$.
Thus,$\frac{1 + 2y + y^2}{y^3(y - 1)} = \frac{1}{y^3} \left( \frac{y^2 + 2y + 1}{y - 1} \right) = \frac{1}{y^3} \left( y + 3 + \frac{4}{y - 1} \right) = \frac{1}{y^2} + \frac{3}{y^3} + \frac{4}{y^3(y - 1)}$.
Alternatively,using partial fraction decomposition for $\frac{x^2}{(x - 1)^3(x - 2)} = \frac{A}{x - 1} + \frac{B}{(x - 1)^2} + \frac{C}{(x - 1)^3} + \frac{D}{x - 2}$.
Solving for coefficients gives $A = -4, B = -3, C = -1, D = 4$.
Therefore,the expression is $\frac{-4}{x - 1} - \frac{3}{(x - 1)^2} - \frac{1}{(x - 1)^3} + \frac{4}{x - 2}$.

(A) Given $f(x + y) = f(x)f(y)$ for all $x, y \in N$.
For $x = 1$,$f(2) = f(1+1) = f(1)f(1) = 3^2 = 9$.
For $x = 3$,$f(3) = f(2+1) = f(2)f(1) = 3^2 \times 3 = 3^3 = 27$.
By induction,$f(x) = 3^x$.
We are given $\sum_{x=1}^n f(x) = 120$.
Substituting $f(x) = 3^x$,we get $\sum_{x=1}^n 3^x = 120$.
This is a geometric progression with first term $a = 3$,common ratio $r = 3$,and $n$ terms.
The sum is $S_n = \frac{a(r^n - 1)}{r - 1} = \frac{3(3^n - 1)}{3 - 1} = 120$.
$\frac{3(3^n - 1)}{2} = 120$.
$3(3^n - 1) = 240$.
$3^n - 1 = 80$.
$3^n = 81$.
Since $81 = 3^4$,we have $n = 4$.

(A) We know the identity for inverse trigonometric functions: $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$ for all $x \in [-1, 1]$.
Given that $\sin^{-1} x = \frac{\pi}{5}$.
Substituting this value into the identity,we get: $\frac{\pi}{5} + \cos^{-1} x = \frac{\pi}{2}$.
Therefore,$\cos^{-1} x = \frac{\pi}{2} - \frac{\pi}{5}$.
Taking the least common multiple $(LCM)$ of $2$ and $5$,which is $10$,we get: $\cos^{-1} x = \frac{5\pi - 2\pi}{10} = \frac{3\pi}{10}$.

(B) Let the required vector be $\vec{v} = ai + bj + ck$.
For $\vec{v}$ to be coplanar with $\vec{u_1} = i + j + 2k$ and $\vec{u_2} = i + 2j + k$,it must be a linear combination of $\vec{u_1}$ and $\vec{u_2}$.
So,$\vec{v} = p(i + j + 2k) + r(i + 2j + k) = (p+r)i + (p+2r)j + (2p+r)k$.
Comparing components,we get $a = p+r$,$b = p+2r$,and $c = 2p+r$.
Since $\vec{v}$ is perpendicular to $\vec{w} = i + j + k$,their dot product is zero:
$(ai + bj + ck) \cdot (i + j + k) = a + b + c = 0$.
Substituting the expressions for $a, b, c$:
$(p+r) + (p+2r) + (2p+r) = 4p + 4r = 0$,which implies $p = -r$.
Substituting $p = -r$ into the expressions for $a, b, c$:
$a = -r + r = 0$,
$b = -r + 2r = r$,
$c = 2(-r) + r = -r$.
Thus,$\vec{v} = r(j - k)$.
For $\vec{v}$ to be a unit vector,$|\vec{v}| = 1$:
$\sqrt{0^2 + r^2 + (-r)^2} = 1 \Rightarrow \sqrt{2r^2} = 1 \Rightarrow |r|\sqrt{2} = 1 \Rightarrow r = \pm \frac{1}{\sqrt{2}}$.
Therefore,the required unit vector is $\pm \frac{1}{\sqrt{2}}(j - k)$.

(A) Given $y = \cot^{-1}(\sqrt{\cos 2x})$.
Using the chain rule,$\frac{dy}{dx} = -\frac{1}{1 + (\sqrt{\cos 2x})^2} \cdot \frac{d}{dx}(\sqrt{\cos 2x})$.
$\frac{dy}{dx} = -\frac{1}{1 + \cos 2x} \cdot \frac{1}{2\sqrt{\cos 2x}} \cdot (-\sin 2x \cdot 2)$.
$\frac{dy}{dx} = \frac{\sin 2x}{(1 + \cos 2x)\sqrt{\cos 2x}}$.
Using the identities $\sin 2x = 2\sin x \cos x$ and $1 + \cos 2x = 2\cos^2 x$,we get:
$\frac{dy}{dx} = \frac{2\sin x \cos x}{2\cos^2 x \sqrt{\cos 2x}} = \frac{\tan x}{\sqrt{\cos 2x}}$.
At $x = \frac{\pi}{6}$,$\tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}$ and $\cos 2(\frac{\pi}{6}) = \cos(\frac{\pi}{3}) = \frac{1}{2}$.
Substituting these values: $\frac{dy}{dx} = \frac{1/\sqrt{3}}{\sqrt{1/2}} = \frac{1}{\sqrt{3}} \cdot \sqrt{2} = \sqrt{\frac{2}{3}} = \left(\frac{2}{3}\right)^{1/2}$.

(D) Let the given integral be $I = \int_{-2}^{2} (px^2 + qx + s) \, dx$.
We can split this into three integrals:
$I = p \int_{-2}^{2} x^2 \, dx + q \int_{-2}^{2} x \, dx + s \int_{-2}^{2} 1 \, dx$.
Since $f(x) = x$ is an odd function,$\int_{-2}^{2} x \, dx = 0$.
Since $f(x) = x^2$ and $f(x) = 1$ are even functions,we can use the property $\int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx$.
Thus,$I = 2p \int_{0}^{2} x^2 \, dx + 0 + 2s \int_{0}^{2} 1 \, dx$.
$I = 2p \left[ \frac{x^3}{3} \right]_{0}^{2} + 2s [x]_{0}^{2}$.
$I = 2p \left( \frac{8}{3} \right) + 2s(2) = \frac{16p}{3} + 4s$.
Therefore,to find the numerical value of $I$,it is necessary to know the values of $p$ and $s$.

(D) Let $I = \int_{-\pi}^{\pi} (\cos^2 px + \sin^2 qx - 2 \sin qx \cos px) dx$.
Using the property $\int_{-a}^{a} f(x) dx = 2 \int_{0}^{a} f(x) dx$ if $f(x)$ is even and $0$ if $f(x)$ is odd:
$I = \int_{-\pi}^{\pi} \cos^2 px dx + \int_{-\pi}^{\pi} \sin^2 qx dx - 2 \int_{-\pi}^{\pi} \sin qx \cos px dx$.
Since $\sin qx \cos px$ is an odd function,its integral over $[-\pi, \pi]$ is $0$.
$I = \int_{-\pi}^{\pi} \frac{1 + \cos 2px}{2} dx + \int_{-\pi}^{\pi} \frac{1 - \cos 2qx}{2} dx$.
$I = \frac{1}{2} [x + \frac{\sin 2px}{2p}]_{-\pi}^{\pi} + \frac{1}{2} [x - \frac{\sin 2qx}{2q}]_{-\pi}^{\pi}$.
$I = \frac{1}{2} [(\pi - (-\pi)) + 0] + \frac{1}{2} [(\pi - (-\pi)) - 0] = \frac{1}{2} (2\pi) + \frac{1}{2} (2\pi) = \pi + \pi = 2\pi$.

(B) India plays a total of $4$ matches. The maximum points in any single match is $2$.
Therefore,the maximum total points in $4$ matches is $8$.
To get at least $7$ points,India must get either $7$ points or $8$ points.
Let $X_i$ be the points in the $i$-th match,where $P(X_i=0)=0.45, P(X_i=1)=0.05, P(X_i=2)=0.50$.
For a total of $8$ points,India must score $2$ points in all $4$ matches:
$P(8) = (0.50)^4 = 0.0625$.
For a total of $7$ points,India must score $2$ points in $3$ matches and $1$ point in $1$ match:
$P(7) = \binom{4}{1} \times (0.50)^3 \times (0.05)^1 = 4 \times 0.125 \times 0.05 = 0.0250$.
The probability of getting at least $7$ points is $P(7) + P(8) = 0.0250 + 0.0625 = 0.0875$.