IIT JEE 1971 Mathematics Question Paper with Answer and Solution

(A) We know that $^n{P_r} = \frac{n!}{(n-r)!}$.
Consider the expression in option $A$:
$^{n - 1}{P_r} + r \cdot ^{n - 1}{P_{r - 1}} = \frac{(n - 1)!}{(n - 1 - r)!} + r \cdot \frac{(n - 1)!}{(n - 1 - (r - 1))!}$
$= \frac{(n - 1)!}{(n - r - 1)!} + r \cdot \frac{(n - 1)!}{(n - r)!}$
$= \frac{(n - 1)!}{(n - r - 1)!} \left( 1 + \frac{r}{n - r} \right)$
$= \frac{(n - 1)!}{(n - r - 1)!} \left( \frac{n - r + r}{n - r} \right)$
$= \frac{(n - 1)!}{(n - r - 1)!} \cdot \frac{n}{n - r} = \frac{n \cdot (n - 1)!}{(n - r) \cdot (n - r - 1)!} = \frac{n!}{(n - r)!} = ^n{P_r}$.

(A) Given the expansion $(1 + x)^n = \sum_{r=0}^{n} C_r x^r$.
We need to find the sum $S = \sum_{r=0}^{n} (r + 1) C_r$.
This can be written as $S = \sum_{r=0}^{n} r C_r + \sum_{r=0}^{n} C_r$.
We know that $\sum_{r=0}^{n} C_r = 2^n$ and $\sum_{r=0}^{n} r C_r = n 2^{n-1}$.
Substituting these values,$S = n 2^{n-1} + 2^n$.
$S = n 2^{n-1} + 2 \cdot 2^{n-1} = (n + 2) 2^{n-1}$.
Alternatively,for $n=1$,the expression is $C_0 + 2C_1 = 1 + 2(1) = 3$. Checking the options for $n=1$: $(1+2)2^{1-1} = 3(1) = 3$. Thus,option $(A)$ is correct.

(A) Given the equations $\sin \theta = \sin \alpha$ and $\cos \theta = \cos \alpha$.
From $\sin \theta = \sin \alpha$,the general solution is $\theta = n\pi + (-1)^n \alpha$,where $n \in \mathbb{Z}$.
From $\cos \theta = \cos \alpha$,the general solution is $\theta = 2m\pi \pm \alpha$,where $m \in \mathbb{Z}$.
For $\theta$ to satisfy both equations simultaneously,the angle must be coterminal with $\alpha$.
This implies $\theta = 2n\pi + \alpha$ for any integer $n$.

(A) The slopes of $AB$ and $BC$ are $m_1 = -4$ and $m_2 = \frac{3}{4}$ respectively. Let $\alpha$ be the angle between $AB$ and $BC$. Then,$\tan \alpha = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{-4 - \frac{3}{4}}{1 + (-4)(\frac{3}{4})} \right| = \left| \frac{-\frac{19}{4}}{1 - 3} \right| = \left| \frac{-\frac{19}{4}}{-2} \right| = \frac{19}{8}$.
Since $AB = AC$,the triangle $ABC$ is isosceles with $\angle ABC = \angle ACB = \alpha$. Thus,the line $AC$ also makes an angle $\alpha$ with the line $BC$. Let $m$ be the slope of line $AC$. The equation of line $AC$ passing through $A(2, -7)$ is $y + 7 = m(x - 2)$.
The angle between $AC$ and $BC$ is $\alpha$,so $\tan \alpha = \left| \frac{m - \frac{3}{4}}{1 + m(\frac{3}{4})} \right| = \frac{19}{8}$.
$\frac{4m - 3}{4 + 3m} = \pm \frac{19}{8}$.
Case $1$: $8(4m - 3) = 19(4 + 3m)$ $\Rightarrow 32m - 24 = 76 + 57m$ $\Rightarrow -25m = 100$ $\Rightarrow m = -4$ (This is the slope of line $AB$).
Case $2$: $8(4m - 3) = -19(4 + 3m)$ $\Rightarrow 32m - 24 = -76 - 57m$ $\Rightarrow 89m = -52$ $\Rightarrow m = -\frac{52}{89}$.
Using $m = -\frac{52}{89}$ in $y + 7 = m(x - 2)$,we get $89(y + 7) = -52(x - 2)$ $\Rightarrow 89y + 623 = -52x + 104$ $\Rightarrow 52x + 89y + 519 = 0$.

(D) The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is given by $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$.
Substituting the given points $(-4, 3)$ and $(12, -1)$:
$(x - (-4))(x - 12) + (y - 3)(y - (-1)) = 0$
$(x + 4)(x - 12) + (y - 3)(y + 1) = 0$
Expanding the terms:
$(x^2 - 12x + 4x - 48) + (y^2 + y - 3y - 3) = 0$
$x^2 + y^2 - 8x - 2y - 51 = 0$.

(C) Using $L-Hospital's$ rule,we differentiate the numerator and denominator with respect to $x$:
$\mathop {\lim }\limits_{x \to 0} \frac{{\tan 2x - x}}{{3x - \sin x}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{d}{{dx}}(\tan 2x - x)}}{{\frac{d}{{dx}}(3x - \sin x)}}$
$= \mathop {\lim }\limits_{x \to 0} \frac{{2\sec^2 2x - 1}}{{3 - \cos x}}$
Substituting $x = 0$:
$= \frac{{2\sec^2(0) - 1}}{{3 - \cos(0)}} = \frac{{2(1)^2 - 1}}{{3 - 1}} = \frac{{2 - 1}}{2} = \frac{1}{2}$.

(B) The given equation is $(2 + k)x + (1 + k)y = 5 + 7k$.
Rearranging the terms to separate $k$:
$2x + kx + y + ky = 5 + 7k$
$(2x + y - 5) + k(x + y - 7) = 0$.
This represents a family of lines passing through the intersection of the lines $2x + y - 5 = 0$ and $x + y - 7 = 0$.
Solving these two equations:
$2x + y = 5$
$x + y = 7$
Subtracting the second from the first: $(2x - x) + (y - y) = 5 - 7$,which gives $x = -2$.
Substituting $x = -2$ into $x + y = 7$: $-2 + y = 7$,so $y = 9$.
Thus,all lines pass through the point $(-2, 9)$.

(D) To solve the integral $I = \int \sqrt{\frac{1-x}{1+x}} \, dx$,we rationalize the numerator inside the square root:
$I = \int \sqrt{\frac{(1-x)(1-x)}{(1+x)(1-x)}} \, dx = \int \sqrt{\frac{(1-x)^2}{1-x^2}} \, dx$
$I = \int \frac{1-x}{\sqrt{1-x^2}} \, dx = \int \frac{1}{\sqrt{1-x^2}} \, dx - \int \frac{x}{\sqrt{1-x^2}} \, dx$
For the first part,$\int \frac{1}{\sqrt{1-x^2}} \, dx = \sin^{-1} x$.
For the second part,let $u = 1-x^2$,then $du = -2x \, dx$,so $x \, dx = -\frac{1}{2} du$.
$-\int \frac{x}{\sqrt{1-x^2}} \, dx = -\int \frac{-1/2 \, du}{\sqrt{u}} = \frac{1}{2} \int u^{-1/2} \, du = \frac{1}{2} \cdot \frac{u^{1/2}}{1/2} = \sqrt{u} = \sqrt{1-x^2}$.
Combining these,we get $I = \sin^{-1} x + \sqrt{1-x^2} + c$.

(C) To evaluate the integral $I = \int (\log x)^2 \, dx$,we use the method of substitution.
Let $\log x = t$,which implies $x = e^t$. Therefore,$dx = e^t \, dt$.
Substituting these into the integral,we get:
$I = \int t^2 e^t \, dt$.
Using integration by parts,$\int u \, dv = uv - \int v \, du$,where $u = t^2$ and $dv = e^t \, dt$:
$du = 2t \, dt$ and $v = e^t$.
$I = t^2 e^t - \int 2t e^t \, dt = t^2 e^t - 2 \int t e^t \, dt$.
Applying integration by parts again for $\int t e^t \, dt$:
$I = t^2 e^t - 2(t e^t - \int e^t \, dt) = t^2 e^t - 2t e^t + 2e^t + c$.
Substituting $t = \log x$ and $e^t = x$ back into the expression:
$I = x(\log x)^2 - 2x \log x + 2x + c$.

(B) Let $E$ be the event that the ball is white.
Let $B_1$ be the event of choosing bag $x$ and $B_2$ be the event of choosing bag $y$.
Since a bag is picked at random,$P(B_1) = P(B_2) = \frac{1}{2}$.
Probability of picking a white ball from bag $x$ is $P(E|B_1) = \frac{3}{3+2} = \frac{3}{5}$.
Probability of picking a white ball from bag $y$ is $P(E|B_2) = \frac{2}{2+4} = \frac{2}{6} = \frac{1}{3}$.
Using the law of total probability,$P(E) = P(B_1)P(E|B_1) + P(B_2)P(E|B_2)$.
$P(E) = \frac{1}{2} \times \frac{3}{5} + \frac{1}{2} \times \frac{1}{3} = \frac{3}{10} + \frac{1}{6} = \frac{9+5}{30} = \frac{14}{30} = \frac{7}{15}$.