IIT JEE 1972 Mathematics Question Paper with Answer and Solution

(B) Given $x = -5 + 2\sqrt{-4} = -5 + 4i$.
Then $x + 5 = 4i$.
Squaring both sides,we get $(x + 5)^2 = (4i)^2$.
$x^2 + 10x + 25 = -16$,which implies $x^2 + 10x + 41 = 0$.
Now,we perform polynomial division of $x^4 + 9x^3 + 35x^2 - x + 4$ by $x^2 + 10x + 41$.
$x^4 + 9x^3 + 35x^2 - x + 4 = (x^2 + 10x + 41)(x^2 - x + 4) - 160$.
Since $x^2 + 10x + 41 = 0$,the expression becomes $0 \times (x^2 - x + 4) - 160 = -160$.

(B) Let $S = \sum_{k=0}^{n} (-1)^k {^nC_k} (a - k)$.
This can be written as $S = a \sum_{k=0}^{n} (-1)^k {^nC_k} - \sum_{k=0}^{n} (-1)^k k {^nC_k}$.
We know that $\sum_{k=0}^{n} (-1)^k {^nC_k} = (1 - 1)^n = 0$ for $n \ge 1$.
Also, $\sum_{k=0}^{n} (-1)^k k {^nC_k} = 0$ for $n > 1$ because $k {^nC_k} = n {^{n-1}C_{k-1}}$, and the sum becomes $n \sum_{k=1}^{n} (-1)^k {^{n-1}C_{k-1}} = n(1 - 1)^{n-1} = 0$.
Thus, $S = a(0) - 0 = 0$.

(C) Let the center of the circle be $(h, k)$.
Since the circle touches the $y$-axis at $(0,3)$,the $y$-coordinate of the center is $k = 3$ and the radius $r = |h|$.
The equation of the circle is $(x - h)^2 + (y - 3)^2 = h^2$.
This circle cuts an intercept of $8$ units on the $x$-axis. Setting $y = 0$ in the equation,we get $(x - h)^2 + (0 - 3)^2 = h^2$,which simplifies to $(x - h)^2 + 9 = h^2$,or $(x - h)^2 = h^2 - 9$.
Thus,$x = h \pm \sqrt{h^2 - 9}$.
The length of the intercept on the $x$-axis is the difference between these two $x$-values: $(h + \sqrt{h^2 - 9}) - (h - \sqrt{h^2 - 9}) = 2\sqrt{h^2 - 9}$.
Given the intercept is $8$,we have $2\sqrt{h^2 - 9} = 8$,so $\sqrt{h^2 - 9} = 4$.
Squaring both sides gives $h^2 - 9 = 16$,so $h^2 = 25$,which means $h = \pm 5$.
Since the radius $r = |h|$,we have $r = 5$.

(C) To check the continuity of $f(x)$ at $x = 1$,we evaluate the left-hand limit,right-hand limit,and the value of the function at $x = 1$.
Given $f(1) = 2$.
For $x \ne 1$,$f(x) = \frac{x^2 - 4x + 3}{x^2 - 1} = \frac{(x - 1)(x - 3)}{(x - 1)(x + 1)} = \frac{x - 3}{x + 1}$.
Now,calculate the limit as $x \to 1$:
$\lim_{x \to 1} f(x) = \lim_{x \to 1} \frac{x - 3}{x + 1} = \frac{1 - 3}{1 + 1} = \frac{-2}{2} = -1$.
Since $\lim_{x \to 1^+} f(x) = -1$ and $\lim_{x \to 1^-} f(x) = -1$,the limit exists and is equal to $-1$.
However,$f(1) = 2$.
Since $\lim_{x \to 1} f(x) \ne f(1)$,the function $f(x)$ is discontinuous at $x = 1$.

(B) We use the identity $\cos^2 x = \frac{1 + \cos 2x}{2}$.
$\int x \cos^2 x \, dx = \int x \left( \frac{1 + \cos 2x}{2} \right) \, dx = \frac{1}{2} \int x \, dx + \frac{1}{2} \int x \cos 2x \, dx$.
Integrating the first part: $\frac{1}{2} \int x \, dx = \frac{x^2}{4}$.
For the second part,use integration by parts: $\int u \, dv = uv - \int v \, du$,where $u = x$ and $dv = \cos 2x \, dx$. Then $du = dx$ and $v = \frac{\sin 2x}{2}$.
$\frac{1}{2} \int x \cos 2x \, dx = \frac{1}{2} \left( \frac{x \sin 2x}{2} - \int \frac{\sin 2x}{2} \, dx \right) = \frac{x \sin 2x}{4} - \frac{1}{4} \int \sin 2x \, dx$.
$= \frac{x \sin 2x}{4} - \frac{1}{4} \left( -\frac{\cos 2x}{2} \right) = \frac{x \sin 2x}{4} + \frac{\cos 2x}{8}$.
Combining both parts: $\frac{x^2}{4} + \frac{x \sin 2x}{4} + \frac{\cos 2x}{8} + c$.

(D) We know that $\sin^2 x = \frac{1 - \cos 2x}{2}$.
Substituting this into the integral:
$\int x \sin^2 x \, dx = \int x \left( \frac{1 - \cos 2x}{2} \right) dx$
$= \frac{1}{2} \int x \, dx - \frac{1}{2} \int x \cos 2x \, dx$
$= \frac{1}{2} \left( \frac{x^2}{2} \right) - \frac{1}{2} \left[ x \left( \frac{\sin 2x}{2} \right) - \int 1 \cdot \frac{\sin 2x}{2} \, dx \right]$
$= \frac{x^2}{4} - \frac{x}{4} \sin 2x + \frac{1}{2} \int \frac{\sin 2x}{2} \, dx$
$= \frac{x^2}{4} - \frac{x}{4} \sin 2x + \frac{1}{4} \left( -\frac{\cos 2x}{2} \right) + c$
$= \frac{x^2}{4} - \frac{x}{4} \sin 2x - \frac{1}{8} \cos 2x + c$.