IIT JEE 1962 Mathematics Question Paper with Answer and Solution

(C) The $k$-th term of the series is $T_k = 1 + x + x^2 + \dots + x^{k-1} = \frac{1 - x^k}{1 - x}$.
The sum of $n$ terms is $S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} \frac{1 - x^k}{1 - x}$.
$S_n = \frac{1}{1 - x} \sum_{k=1}^{n} (1 - x^k) = \frac{1}{1 - x} \left[ \sum_{k=1}^{n} 1 - \sum_{k=1}^{n} x^k \right]$.
$S_n = \frac{1}{1 - x} \left[ n - \frac{x(1 - x^n)}{1 - x} \right]$.
$S_n = \frac{n(1 - x) - x(1 - x^n)}{(1 - x)^2}$.

(B) The general term of the series is given by $k \frac{C_k}{C_{k-1}}$.
We know that $C_k = \binom{n}{k} = \frac{n!}{k!(n-k)!}$.
Thus,$\frac{C_k}{C_{k-1}} = \frac{n! / (k!(n-k)!)}{n! / ((k-1)!(n-k+1)!)} = \frac{(k-1)!(n-k+1)!}{k!(n-k)!} = \frac{n-k+1}{k}$.
Therefore,the $k$-th term is $k \times \frac{n-k+1}{k} = n-k+1$.
The sum is $\sum_{k=1}^{n} (n-k+1) = n + (n-1) + \dots + 1 = \frac{n(n+1)}{2}$.
For $n=15$,the sum is $\frac{15(15+1)}{2} = \frac{15 \times 16}{2} = 15 \times 8 = 120$.

(C) Let the diagonal be $BD$ along the line $8x - 15y = 0$. The slope of $BD$ is $m_1 = \frac{8}{15}$.
Let the vertex be $D(1, 2)$. The sides passing through $D$ are $AD$ and $CD$.
In a square,the diagonal makes an angle of $45^\circ$ with the sides.
Let the slope of the sides be $m$. Then,$\tan(45^\circ) = \left| \frac{m - \frac{8}{15}}{1 + m \cdot \frac{8}{15}} \right|$.
$1 = \left| \frac{15m - 8}{15 + 8m} \right|$.
This gives two cases:
Case $1$: $15m - 8 = 15 + 8m$ $\Rightarrow 7m = 23$ $\Rightarrow m = \frac{23}{7}$.
Case $2$: $15m - 8 = -(15 + 8m)$ $\Rightarrow 15m - 8 = -15 - 8m$ $\Rightarrow 23m = -7$ $\Rightarrow m = -\frac{7}{23}$.
The equations of the sides passing through $(1, 2)$ are:
$y - 2 = \frac{23}{7}(x - 1)$ $\Rightarrow 7y - 14 = 23x - 23$ $\Rightarrow 23x - 7y - 9 = 0$.
$y - 2 = -\frac{7}{23}(x - 1)$ $\Rightarrow 23y - 46 = -7x + 7$ $\Rightarrow 7x + 23y - 53 = 0$.

(D) Let the vertices be $A(2, 1)$,$B(5, 2)$,and $C(4, 4)$.
The equation of side $AB$ is $y - 1 = \frac{2 - 1}{5 - 2}(x - 2) \implies y - 1 = \frac{1}{3}(x - 2) \implies x - 3y + 1 = 0$.
The length of the perpendicular from $C(4, 4)$ to $AB$ is $D_C = \frac{|4 - 3(4) + 1|}{\sqrt{1^2 + (-3)^2}} = \frac{|4 - 12 + 1|}{\sqrt{10}} = \frac{7}{\sqrt{10}}$.
The equation of side $BC$ is $y - 2 = \frac{4 - 2}{4 - 5}(x - 5) \implies y - 2 = -2(x - 5) \implies 2x + y - 12 = 0$.
The length of the perpendicular from $A(2, 1)$ to $BC$ is $D_A = \frac{|2(2) + 1 - 12|}{\sqrt{2^2 + 1^2}} = \frac{|4 + 1 - 12|}{\sqrt{5}} = \frac{7}{\sqrt{5}}$.
The equation of side $AC$ is $y - 1 = \frac{4 - 1}{4 - 2}(x - 2) \implies y - 1 = \frac{3}{2}(x - 2) \implies 3x - 2y - 4 = 0$.
The length of the perpendicular from $B(5, 2)$ to $AC$ is $D_B = \frac{|3(5) - 2(2) - 4|}{\sqrt{3^2 + (-2)^2}} = \frac{|15 - 4 - 4|}{\sqrt{13}} = \frac{7}{\sqrt{13}}$.
Thus,the lengths are $\frac{7}{\sqrt{5}}, \frac{7}{\sqrt{13}}, \frac{7}{\sqrt{10}}$.