IIT JEE 1966 Mathematics Question Paper with Answer and Solution

(B) Given that $x, y, z$ are in $G.P.$,we have $y^2 = xz$.
Let $a^x = b^y = c^z = m$.
Taking logarithm on both sides,we get $x \log a = y \log b = z \log c = \log m$.
This implies $x = \frac{\log m}{\log a} = \log_a m$,$y = \log_b m$,and $z = \log_c m$.
Since $x, y, z$ are in $G.P.$,we have $\frac{y}{x} = \frac{z}{y}$.
Substituting the values,we get $\frac{\log_b m}{\log_a m} = \frac{\log_c m}{\log_b m}$.
Using the change of base formula $\frac{\log_b m}{\log_a m} = \log_b a$,we get $\log_b a = \log_c b$.

(D) Let the $n$ terms of the $G.P.$ be $a, ar, ar^2, \dots, ar^{n-1}$.
The sum $S = a + ar + \dots + ar^{n-1} = \frac{a(1-r^n)}{1-r}$.
The product $P = a \cdot ar \cdot ar^2 \dots ar^{n-1} = a^n r^{0+1+2+\dots+(n-1)} = a^n r^{\frac{n(n-1)}{2}}$.
Thus,$P^2 = a^{2n} r^{n(n-1)}$.
The sum of reciprocals $R = \frac{1}{a} + \frac{1}{ar} + \dots + \frac{1}{ar^{n-1}} = \frac{1}{a} \left( 1 + \frac{1}{r} + \dots + \frac{1}{r^{n-1}} \right) = \frac{1}{a} \left( \frac{1 - (1/r)^n}{1 - 1/r} \right) = \frac{1}{a} \left( \frac{(r^n-1)/r^n}{(r-1)/r} \right) = \frac{r^n-1}{a r^{n-1} (r-1)} = \frac{1-r^n}{a r^{n-1} (1-r)}$.
Now,$\frac{S}{R} = \frac{a(1-r^n)}{1-r} \div \frac{1-r^n}{a r^{n-1} (1-r)} = a^2 r^{n-1}$.
Therefore,$(\frac{S}{R})^n = (a^2 r^{n-1})^n = a^{2n} r^{n(n-1)} = P^2$.

(B) Given $(1 + x)^{15} = C_0 + C_1x + C_2x^2 + .... + C_{15}x^{15}$.
Subtracting $C_0 = 1$ and dividing by $x$,we get:
$\frac{(1 + x)^{15} - 1}{x} = C_1 + C_2x + C_3x^2 + .... + C_{15}x^{14}$.
Differentiating both sides with respect to $x$:
$\frac{d}{dx} \left( \frac{(1 + x)^{15} - 1}{x} \right) = C_2 + 2C_3x + 3C_4x^2 + .... + 14C_{15}x^{13}$.
Using the quotient rule on the left side:
$\frac{x \cdot 15(1 + x)^{14} - ((1 + x)^{15} - 1)}{x^2} = C_2 + 2C_3x + 3C_4x^2 + .... + 14C_{15}x^{13}$.
Putting $x = 1$:
$C_2 + 2C_3 + 3C_4 + .... + 14C_{15} = \frac{1 \cdot 15(2)^{14} - (2^{15} - 1)}{1^2}$.
$= 15 \cdot 2^{14} - 2^{15} + 1$.
$= 15 \cdot 2^{14} - 2 \cdot 2^{14} + 1$.
$= (15 - 2) \cdot 2^{14} + 1 = 13 \cdot 2^{14} + 1$.

(A) We know that for any real $\theta$,$\sec^2 \theta \ge 1$.
Given the equation $\sec^2 \theta = \frac{4xy}{(x + y)^2}$,it must satisfy $\frac{4xy}{(x + y)^2} \ge 1$.
Since $(x + y)^2 > 0$ (assuming $x, y \neq 0$),we have $4xy \ge (x + y)^2$.
$4xy \ge x^2 + 2xy + y^2$.
$0 \ge x^2 - 2xy + y^2$.
$0 \ge (x - y)^2$.
Since the square of a real number cannot be negative,$(x - y)^2$ must be $0$.
Therefore,$x - y = 0$,which implies $x = y$.

(B) Given expression: $E = \frac{\sin(B + A) + \cos(B - A)}{\sin(B - A) + \cos(B + A)}$
Using the identities $\cos \theta = \sin(90^\circ - \theta)$:
$E = \frac{\sin(B + A) + \sin(90^\circ - (B - A))}{\sin(B - A) + \sin(90^\circ - (B + A))}$
$E = \frac{\sin(B + A) + \sin(90^\circ - B + A)}{\sin(B - A) + \sin(90^\circ - B - A)}$
Using $\sin C + \sin D = 2 \sin(\frac{C+D}{2}) \cos(\frac{C-D}{2})$:
Numerator: $\sin(B+A) + \sin(90^\circ - B + A) = 2 \sin(\frac{B+A+90^\circ-B+A}{2}) \cos(\frac{B+A-90^\circ+B-A}{2}) = 2 \sin(45^\circ + A) \cos(B - 45^\circ)$
Denominator: $\sin(B-A) + \sin(90^\circ - B - A) = 2 \sin(\frac{B-A+90^\circ-B-A}{2}) \cos(\frac{B-A-90^\circ+B+A}{2}) = 2 \sin(45^\circ - A) \cos(B - 45^\circ)$
$E = \frac{2 \sin(45^\circ + A) \cos(B - 45^\circ)}{2 \sin(45^\circ - A) \cos(B - 45^\circ)} = \frac{\sin(45^\circ + A)}{\sin(45^\circ - A)}$
Using $\sin(x \pm y) = \sin x \cos y \pm \cos x \sin y$:
$E = \frac{\sin 45^\circ \cos A + \cos 45^\circ \sin A}{\sin 45^\circ \cos A - \cos 45^\circ \sin A} = \frac{\frac{1}{\sqrt{2}} \cos A + \frac{1}{\sqrt{2}} \sin A}{\frac{1}{\sqrt{2}} \cos A - \frac{1}{\sqrt{2}} \sin A} = \frac{\cos A + \sin A}{\cos A - \sin A}$.

(A) Given,$m \tan (\theta - 30^\circ) = n \tan (\theta + 120^\circ)$.
$\frac{m}{n} = \frac{\tan (\theta + 120^\circ)}{\tan (\theta - 30^\circ)}$.
Applying componendo and dividendo:
$\frac{m + n}{m - n} = \frac{\tan (\theta + 120^\circ) + \tan (\theta - 30^\circ)}{\tan (\theta + 120^\circ) - \tan (\theta - 30^\circ)}$.
Using $\tan A + \tan B = \frac{\sin(A+B)}{\cos A \cos B}$ and $\tan A - \tan B = \frac{\sin(A-B)}{\cos A \cos B}$:
$\frac{m + n}{m - n} = \frac{\sin((\theta + 120^\circ) + (\theta - 30^\circ))}{\sin((\theta + 120^\circ) - (\theta - 30^\circ))} = \frac{\sin(2\theta + 90^\circ)}{\sin(150^\circ)}$.
Since $\sin(2\theta + 90^\circ) = \cos 2\theta$ and $\sin(150^\circ) = \sin(180^\circ - 30^\circ) = \sin 30^\circ = \frac{1}{2}$:
$\frac{m + n}{m - n} = \frac{\cos 2\theta}{1/2} = 2 \cos 2\theta$.

(A) We know that $\cot(A) = \frac{1 + \cos(2A)}{\sin(2A)}$.
Let $A = 7\frac{1}{2}^{\circ}$. Then $2A = 15^{\circ}$.
$\cot(7\frac{1}{2}^{\circ}) = \frac{1 + \cos(15^{\circ})}{\sin(15^{\circ})}$.
We know $\cos(15^{\circ}) = \cos(45^{\circ} - 30^{\circ}) = \cos(45^{\circ})\cos(30^{\circ}) + \sin(45^{\circ})\sin(30^{\circ}) = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3} + 1}{2\sqrt{2}}$.
And $\sin(15^{\circ}) = \sin(45^{\circ} - 30^{\circ}) = \sin(45^{\circ})\cos(30^{\circ}) - \cos(45^{\circ})\sin(30^{\circ}) = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3} - 1}{2\sqrt{2}}$.
Substituting these values:
$\cot(7\frac{1}{2}^{\circ}) = \frac{1 + \frac{\sqrt{3} + 1}{2\sqrt{2}}}{\frac{\sqrt{3} - 1}{2\sqrt{2}}} = \frac{2\sqrt{2} + \sqrt{3} + 1}{\sqrt{3} - 1}$.
Rationalizing the denominator:
$= \frac{(2\sqrt{2} + \sqrt{3} + 1)(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{2\sqrt{6} + 2\sqrt{2} + 3 + \sqrt{3} + \sqrt{3} + 1}{3 - 1} = \frac{2\sqrt{6} + 2\sqrt{2} + 2\sqrt{3} + 4}{2} = \sqrt{6} + \sqrt{2} + \sqrt{3} + 2$.
Since $\sqrt{4} = 2$,the expression is $\sqrt{6} + \sqrt{2} + \sqrt{3} + \sqrt{4}$.

(D) Let the required line passing through $(1, 2)$ be inclined at an angle $\theta$ to the $x$-axis. The equation of the line is $\frac{x-1}{\cos \theta} = \frac{y-2}{\sin \theta} = r$,where $r$ is the distance from $(1, 2)$.
Any point on this line is $(1 + r \cos \theta, 2 + r \sin \theta)$.
Given $r = \frac{\sqrt{6}}{3}$,the point is $(1 + \frac{\sqrt{6}}{3} \cos \theta, 2 + \frac{\sqrt{6}}{3} \sin \theta)$.
Since this point lies on $x + y = 4$,we have $(1 + \frac{\sqrt{6}}{3} \cos \theta) + (2 + \frac{\sqrt{6}}{3} \sin \theta) = 4$.
$\frac{\sqrt{6}}{3} (\cos \theta + \sin \theta) = 1 \implies \sin \theta + \cos \theta = \frac{3}{\sqrt{6}} = \frac{\sqrt{6}}{2}$.
Dividing by $\sqrt{2}$,we get $\frac{1}{\sqrt{2}} \sin \theta + \frac{1}{\sqrt{2}} \cos \theta = \frac{\sqrt{6}}{2\sqrt{2}} = \frac{\sqrt{3}}{2}$.
$\sin(\theta + 45^\circ) = \sin 60^\circ$ or $\sin 120^\circ$.
$\theta + 45^\circ = 60^\circ \implies \theta = 15^\circ$ or $\theta + 45^\circ = 120^\circ \implies \theta = 75^\circ$.
Since $75^\circ$ is one of the options,the correct answer is $75^\circ$.