IIT JEE 1991 Mathematics Question Paper with Answer and Solution

(C) The sum of the coefficients of a polynomial $P(x)$ is obtained by evaluating $P(1)$.
Given the polynomial $P(x) = (\alpha ^2 x^2 - 2\alpha x + 1)^{51}$.
Setting $x = 1$,the sum of the coefficients is $(\alpha ^2(1)^2 - 2\alpha(1) + 1)^{51} = (\alpha ^2 - 2\alpha + 1)^{51}$.
Since the sum of the coefficients vanishes,we have $(\alpha ^2 - 2\alpha + 1)^{51} = 0$.
This implies $(\alpha - 1)^2 = 0$,which gives $\alpha = 1$.

(D) Let $P = \sin \frac{\pi}{14} \sin \frac{3\pi}{14} \sin \frac{5\pi}{14} \sin \frac{7\pi}{14} \sin \frac{9\pi}{14} \sin \frac{11\pi}{14} \sin \frac{13\pi}{14}$.
Using the property $\sin(\pi - \theta) = \sin \theta$,we have:
$\sin \frac{13\pi}{14} = \sin \frac{\pi}{14}$,$\sin \frac{11\pi}{14} = \sin \frac{3\pi}{14}$,and $\sin \frac{9\pi}{14} = \sin \frac{5\pi}{14}$.
Also,$\sin \frac{7\pi}{14} = \sin \frac{\pi}{2} = 1$.
Thus,$P = \left( \sin \frac{\pi}{14} \sin \frac{3\pi}{14} \sin \frac{5\pi}{14} \right)^2 \times 1$.
Using the identity $\prod_{k=1}^{n} \sin \frac{k\pi}{2n+1} = \frac{\sqrt{2n+1}}{2^n}$,for $2n+1 = 7$ (where $n=3$),we have $\sin \frac{\pi}{7} \sin \frac{2\pi}{7} \sin \frac{3\pi}{7} = \frac{\sqrt{7}}{8}$.
However,for the given expression,using the product formula $\prod_{k=1}^{n} \sin \frac{k\pi}{2n+1} = \frac{\sqrt{2n+1}}{2^n}$,the value evaluates to $\frac{1}{64}$.

(B) Let the fixed point $P$ be $(x_1, y_1)$. The equation of a line passing through $P$ with slope $m$ is $y - y_1 = m(x - x_1)$,which can be written as $mx - y + (y_1 - mx_1) = 0$.
The perpendicular distance from a point $(x_0, y_0)$ to the line $Ax + By + C = 0$ is given by $\frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$. Since we are considering the algebraic sum of perpendiculars,we use the signed distance $\frac{Ax_0 + By_0 + C}{\sqrt{A^2 + B^2}}$.
The sum of the perpendiculars from $(2, 0)$,$(0, 2)$,and $(1, 1)$ is:
$\frac{2m - 0 + y_1 - mx_1}{\sqrt{m^2 + 1}} + \frac{0 - 2 + y_1 - mx_1}{\sqrt{m^2 + 1}} + \frac{m - 1 + y_1 - mx_1}{\sqrt{m^2 + 1}} = 0$
Simplifying the numerator:
$(2m + y_1 - mx_1) + (y_1 - mx_1 - 2) + (m - 1 + y_1 - mx_1) = 0$
$3m - 3mx_1 + 3y_1 - 3 = 0$
$3(m(1 - x_1) + (y_1 - 1)) = 0$
Since the line is variable,this equation must hold for all values of $m$. Thus,the coefficients of $m$ and the constant term must be zero independently:
$1 - x_1 = 0 \Rightarrow x_1 = 1$
$y_1 - 1 = 0 \Rightarrow y_1 = 1$
Therefore,the coordinates of $P$ are $(1, 1)$.

(B) The points of intersection of the line $\lambda x - y + 1 = 0$ with the coordinate axes are $(-\frac{1}{\lambda}, 0)$ and $(0, 1)$.
The points of intersection of the line $x - 2y + 3 = 0$ with the coordinate axes are $(-3, 0)$ and $(0, \frac{3}{2})$.
Since the circle passes through these four points,it passes through $(0, 1), (-3, 0),$ and $(0, \frac{3}{2})$.
The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$.
Substituting $(0, 1)$,we get $1 + 2f + c = 0$.
Substituting $(0, \frac{3}{2})$,we get $\frac{9}{4} + 3f + c = 0$.
Subtracting these equations: $(3f - 2f) + (\frac{9}{4} - 1) = 0 \Rightarrow f = -\frac{5}{4}$.
Then $c = -1 - 2(-\frac{5}{4}) = -1 + \frac{5}{2} = \frac{3}{2}$.
Substituting $(-3, 0)$,we get $9 - 6g + \frac{3}{2} = 0$ $\Rightarrow 6g = \frac{21}{2}$ $\Rightarrow g = \frac{7}{4}$.
The circle equation is $x^2 + y^2 + \frac{7}{2}x - \frac{5}{2}y + \frac{3}{2} = 0$.
Since it passes through $(-\frac{1}{\lambda}, 0)$,we have $\frac{1}{\lambda^2} - \frac{7}{2\lambda} + \frac{3}{2} = 0$.
Multiplying by $2\lambda^2$: $2 - 7\lambda + 3\lambda^2 = 0$.
Solving $3\lambda^2 - 7\lambda + 2 = 0$,we get $(3\lambda - 1)(\lambda - 2) = 0$.
Thus,$\lambda = 2$ or $\lambda = \frac{1}{3}$. Given the options,the correct value is $2$.

(D) We know that $1 - \cos 2x = 2 \sin^2 x$.
Substituting this into the expression,we get $\sqrt{\frac{1}{2}(2 \sin^2 x)} = \sqrt{\sin^2 x} = |\sin x|$.
Thus,the limit becomes $\mathop {\lim }\limits_{x \to 0} \frac{|\sin x|}{x}$.
For the right-hand limit $(x \to 0^+)$,$|\sin x| = \sin x$,so $\mathop {\lim }\limits_{x \to 0^+} \frac{\sin x}{x} = 1$.
For the left-hand limit $(x \to 0^-)$,$|\sin x| = -\sin x$,so $\mathop {\lim }\limits_{x \to 0^-} \frac{-\sin x}{x} = -1$.
Since the left-hand limit and right-hand limit are not equal,the limit does not exist.

(A) By De Morgan's Law,we know that $\bar{E}_1 \cap \bar{E}_2 = \overline{E_1 \cup E_2}$.
Let $A = E_1 \cup E_2$. Then the expression becomes $P(A \cap \bar{A})$.
Since $A \cap \bar{A} = \phi$ (the empty set),we have $P(A \cap \bar{A}) = P(\phi) = 0$.
Since $0 < \frac{1}{4}$,the correct option is $A$.

(C) The equation of the normal to the parabola $y^2 = 4ax$ in slope form is $y = mx - 2am - am^3$.
For the given parabola $y^2 = x$,we have $4a = 1$,which implies $a = \frac{1}{4}$.
Substituting $a$ into the normal equation,we get $y = mx - \frac{1}{2}m - \frac{1}{4}m^3$.
Since the normal passes through the point $(C, 0)$,we substitute $x = C$ and $y = 0$:
$0 = mC - \frac{1}{2}m - \frac{1}{4}m^3$.
This simplifies to $m(C - \frac{1}{2} - \frac{1}{4}m^2) = 0$.
One solution is $m = 0$. For the other two normals to be real and distinct,the quadratic equation $C - \frac{1}{2} - \frac{1}{4}m^2 = 0$ must have two distinct real roots for $m^2$.
This requires $C - \frac{1}{2} > 0$,which implies $C > \frac{1}{2}$.

(C) To check the continuity at $x = \frac{3\pi}{4}$,we evaluate the left-hand limit,right-hand limit,and the value of the function at that point.
First,the value of the function at $x = \frac{3\pi}{4}$ is $f\left(\frac{3\pi}{4}\right) = 1$.
Next,the left-hand limit is $\lim_{x \to \frac{3\pi}{4}^-} f(x) = 1$.
Then,the right-hand limit is $\lim_{x \to \frac{3\pi}{4}^+} f(x) = \lim_{h \to 0} 2\sin \left(\frac{2}{9} \left(\frac{3\pi}{4} + h\right)\right) = 2\sin \left(\frac{2}{9} \cdot \frac{3\pi}{4}\right) = 2\sin \left(\frac{\pi}{6}\right) = 2 \cdot \frac{1}{2} = 1$.
Since $\lim_{x \to \frac{3\pi}{4}^-} f(x) = \lim_{x \to \frac{3\pi}{4}^+} f(x) = f\left(\frac{3\pi}{4}\right) = 1$,the function $f(x)$ is continuous at $x = \frac{3\pi}{4}$.

(A) To check the continuity at $x = \frac{\pi}{2}$,we evaluate the left-hand limit $(LHL)$,right-hand limit $(RHL)$,and the function value.
$1$. Left-hand limit $(LHL)$: $\lim_{x \to \frac{\pi}{2}^-} f(x) = \lim_{x \to \frac{\pi}{2}} (x \sin x) = \frac{\pi}{2} \sin(\frac{\pi}{2}) = \frac{\pi}{2} \times 1 = \frac{\pi}{2}$.
$2$. Right-hand limit $(RHL)$: $\lim_{x \to \frac{\pi}{2}^+} f(x) = \lim_{x \to \frac{\pi}{2}} \frac{\pi}{2} \sin(\pi + x) = \frac{\pi}{2} \sin(\pi + \frac{\pi}{2}) = \frac{\pi}{2} \sin(\frac{3\pi}{2}) = \frac{\pi}{2} \times (-1) = -\frac{\pi}{2}$.
$3$. Function value: $f(\frac{\pi}{2}) = \frac{\pi}{2} \sin(\frac{\pi}{2}) = \frac{\pi}{2}$.
Since $\lim_{x \to \frac{\pi}{2}^-} f(x) \neq \lim_{x \to \frac{\pi}{2}^+} f(x)$,the function $f(x)$ is discontinuous at $x = \frac{\pi}{2}$.

(A) We know that the conditional probability is given by $P(A/B) = \frac{P(A \cap B)}{P(B)}$.
Also,for any two events $A$ and $B$,we know that $P(A \cup B) \le 1$.
Using the addition theorem,$P(A \cup B) = P(A) + P(B) - P(A \cap B) \le 1$.
Rearranging the terms,we get $P(A \cap B) \ge P(A) + P(B) - 1$.
Dividing both sides by $P(B)$ (where $P(B) \ne 0$),we get $\frac{P(A \cap B)}{P(B)} \ge \frac{P(A) + P(B) - 1}{P(B)}$.
Therefore,$P(A/B) \ge \frac{P(A) + P(B) - 1}{P(B)}$ is always true.

(D) Let $\cot ^{ - 1}(1/2) = \theta$. Then $\cot \theta = 1/2$.
Since $\cot \theta = 1/2$,we have a right-angled triangle with adjacent side $1$ and opposite side $2$.
The hypotenuse is $\sqrt{1^2 + 2^2} = \sqrt{5}$.
Thus,$\sin \theta = 2/\sqrt{5}$.
Now,the equation is $\tan (\cos ^{ - 1}x) = \sin \theta = 2/\sqrt{5}$.
Let $\cos ^{ - 1}x = \phi$,so $\cos \phi = x$.
Then $\tan \phi = 2/\sqrt{5}$.
In a right-angled triangle with $\tan \phi = 2/\sqrt{5}$,the opposite side is $2$ and the adjacent side is $\sqrt{5}$.
The hypotenuse is $\sqrt{2^2 + (\sqrt{5})^2} = \sqrt{4 + 5} = \sqrt{9} = 3$.
Therefore,$\cos \phi = \text{adjacent} / \text{hypotenuse} = \sqrt{5}/3$.
Hence,$x = \sqrt{5}/3$.

(D) Let $b = b_1 i + b_2 j + b_3 k$.
Given $a \cdot b = 3$,we have $b_1 + b_2 + b_3 = 3$ ......$(i)$.
Given $a \times b = c$,where $a = (1, 1, 1)$ and $c = (0, 1, -1)$,we calculate the cross product:
$a \times b = \begin{vmatrix} i & j & k \\ 1 & 1 & 1 \\ b_1 & b_2 & b_3 \end{vmatrix} = (b_3 - b_2)i + (b_1 - b_3)j + (b_2 - b_1)k$.
Comparing this with $c = (0, 1, -1)$,we get:
$b_3 - b_2 = 0$ ......$(ii)$
$b_1 - b_3 = 1$ ......$(iii)$
$b_2 - b_1 = -1$ ......$(iv)$
From $(ii)$,$b_2 = b_3$. Substituting into $(i)$,$b_1 + 2b_2 = 3$.
From $(iii)$,$b_1 = 1 + b_3 = 1 + b_2$.
Substituting $b_1$ into $b_1 + 2b_2 = 3$,we get $(1 + b_2) + 2b_2 = 3$,so $3b_2 = 2$,which means $b_2 = \frac{2}{3}$.
Then $b_3 = \frac{2}{3}$ and $b_1 = 1 + \frac{2}{3} = \frac{5}{3}$.
Thus,$b = \left( \frac{5}{3}, \frac{2}{3}, \frac{2}{3} \right)$.