IIT JEE 1963 Mathematics Question Paper with Answer and Solution

(B) Let $T_n$ be the $n^{th}$ term and $S_n$ be the sum up to $n$ terms.
$S_n = 1 + 3 + 7 + 15 + 31 + \dots + T_n$
We can write the $n^{th}$ term as $T_n = 2^n - 1$.
Thus,$S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} (2^k - 1)$.
$S_n = \sum_{k=1}^{n} 2^k - \sum_{k=1}^{n} 1$.
Using the sum formula for a geometric progression,$\sum_{k=1}^{n} 2^k = 2(2^n - 1) / (2 - 1) = 2^{n+1} - 2$.
Therefore,$S_n = (2^{n+1} - 2) - n = 2^{n+1} - n - 2$.

(D) Given $\sec x \cos 5x + 1 = 0$,we have $\cos 5x = -\cos x = \cos(\pi - x)$.
General solution for $\cos \theta = \cos \alpha$ is $\theta = 2n\pi \pm \alpha$.
Case $1$: $5x = 2n\pi + (\pi - x)$ $\Rightarrow 6x = (2n + 1)\pi$ $\Rightarrow x = \frac{(2n + 1)\pi}{6}$.
For $n=0, 1, 2, 3, 4, 5$,$x = \frac{\pi}{6}, \frac{3\pi}{6} (\frac{\pi}{2}), \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{9\pi}{6} (\frac{3\pi}{2}), \frac{11\pi}{6}$.
Case $2$: $5x = 2n\pi - (\pi - x)$ $\Rightarrow 4x = 2n\pi - \pi$ $\Rightarrow x = \frac{(2n - 1)\pi}{4}$.
For $n=1, 2, 3, 4$,$x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$.
Since the provided options do not match the complete solution set,the correct choice is $D$.

(B) Given the equation: $2\sin^2 \theta = 3\cos \theta$
Using the identity $\sin^2 \theta = 1 - \cos^2 \theta$,we get:
$2(1 - \cos^2 \theta) = 3\cos \theta$
$2 - 2\cos^2 \theta = 3\cos \theta$
$2\cos^2 \theta + 3\cos \theta - 2 = 0$
Let $x = \cos \theta$. Then $2x^2 + 3x - 2 = 0$.
Solving the quadratic equation using the quadratic formula:
$x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-2)}}{2(2)} = \frac{-3 \pm \sqrt{9 + 16}}{4} = \frac{-3 \pm 5}{4}$
This gives two values for $x$:
$x = \frac{-3 + 5}{4} = \frac{2}{4} = \frac{1}{2}$ or $x = \frac{-3 - 5}{4} = -2$
Since $-1 \le \cos \theta \le 1$,we discard $x = -2$.
Thus,$\cos \theta = \frac{1}{2}$.
In the interval $0 \le \theta \le 2\pi$,$\cos \theta = \frac{1}{2}$ at $\theta = \frac{\pi}{3}$ and $\theta = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$.

(C) Given: $\sec 4\theta - \sec 2\theta = 2$
$\Rightarrow \frac{1}{\cos 4\theta} - \frac{1}{\cos 2\theta} = 2$
$\Rightarrow \frac{\cos 2\theta - \cos 4\theta}{\cos 4\theta \cos 2\theta} = 2$
$\Rightarrow \cos 2\theta - \cos 4\theta = 2 \cos 4\theta \cos 2\theta$
Using the formula $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$:
$\Rightarrow \cos 2\theta - \cos 4\theta = \cos 6\theta + \cos 2\theta$
$\Rightarrow - \cos 4\theta = \cos 6\theta$
$\Rightarrow \cos 6\theta + \cos 4\theta = 0$
Using the formula $\cos C + \cos D = 2 \cos(\frac{C+D}{2}) \cos(\frac{C-D}{2})$:
$\Rightarrow 2 \cos 5\theta \cos \theta = 0$
Case $1$: $\cos 5\theta = 0$ $\Rightarrow 5\theta = (2n + 1)\frac{\pi}{2}$ $\Rightarrow \theta = (2n + 1)\frac{\pi}{10}$
Case $2$: $\cos \theta = 0 \Rightarrow \theta = (2n + 1)\frac{\pi}{2} = n\pi + \frac{\pi}{2}$
Thus,the general solution is $\theta = n\pi + \frac{\pi}{2}$ or $\theta = \frac{n\pi}{5} + \frac{\pi}{10}$.

(C) First,find the point of intersection of the lines $x - y + 1 = 0$ and $2x - 3y + 5 = 0$. Solving these,we get $x = 2$ and $y = 3$. The point of intersection is $(2, 3)$.
The equation of any line passing through $(2, 3)$ is given by $y - 3 = m(x - 2)$,which simplifies to $mx - y + (3 - 2m) = 0$.
The distance of this line from the point $(3, 2)$ is given as $\frac{7}{5}$. Using the perpendicular distance formula:
$\frac{|m(3) - 2 + 3 - 2m|}{\sqrt{m^2 + (-1)^2}} = \frac{7}{5}$
$\frac{|m + 1|}{\sqrt{m^2 + 1}} = \frac{7}{5}$
Squaring both sides: $25(m^2 + 2m + 1) = 49(m^2 + 1)$
$25m^2 + 50m + 25 = 49m^2 + 49$
$24m^2 - 50m + 24 = 0$
$12m^2 - 25m + 12 = 0$
$(3m - 4)(4m - 3) = 0$
Thus,$m = \frac{4}{3}$ or $m = \frac{3}{4}$.
For $m = \frac{4}{3}$,the line is $y - 3 = \frac{4}{3}(x - 2)$ $\Rightarrow 3y - 9 = 4x - 8$ $\Rightarrow 4x - 3y + 1 = 0$.
For $m = \frac{3}{4}$,the line is $y - 3 = \frac{3}{4}(x - 2)$ $\Rightarrow 4y - 12 = 3x - 6$ $\Rightarrow 3x - 4y + 6 = 0$.
Therefore,the equations are $3x - 4y + 6 = 0$ and $4x - 3y + 1 = 0$.

(D) The area of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
Substituting the given points $(a, b + c)$,$(b, c + a)$,and $(c, a + b)$:
Area $= \frac{1}{2} |a(c + a - (a + b)) + b(a + b - (b + c)) + c(b + c - (c + a))|$
Area $= \frac{1}{2} |a(c - b) + b(a - c) + c(b - a)|$
Area $= \frac{1}{2} |ac - ab + ba - bc + cb - ca|$
Area $= \frac{1}{2} |0| = 0$.
Alternatively,using the determinant method:
Area $= \frac{1}{2} \left| \begin{matrix} a & b+c & 1 \\ b & c+a & 1 \\ c & a+b & 1 \end{matrix} \right|$
Applying $C_2 \to C_1 + C_2$:
Area $= \frac{1}{2} \left| \begin{matrix} a & a+b+c & 1 \\ b & a+b+c & 1 \\ c & a+b+c & 1 \end{matrix} \right| = \frac{a+b+c}{2} \left| \begin{matrix} a & 1 & 1 \\ b & 1 & 1 \\ c & 1 & 1 \end{matrix} \right| = 0$ (since two columns are identical).