IIT JEE 1976 Mathematics Question Paper with Answer and Solution

(C) Let $z = \frac{3 + 2i\sin \theta}{1 - 2i\sin \theta}$.
To simplify,multiply the numerator and denominator by the conjugate of the denominator $(1 + 2i\sin \theta)$:
$z = \frac{(3 + 2i\sin \theta)(1 + 2i\sin \theta)}{(1 - 2i\sin \theta)(1 + 2i\sin \theta)}$
$z = \frac{3 + 6i\sin \theta + 2i\sin \theta + 4i^2\sin^2 \theta}{1 + 4\sin^2 \theta}$
Since $i^2 = -1$,we have:
$z = \frac{3 - 4\sin^2 \theta + 8i\sin \theta}{1 + 4\sin^2 \theta} = \left( \frac{3 - 4\sin^2 \theta}{1 + 4\sin^2 \theta} \right) + i\left( \frac{8\sin \theta}{1 + 4\sin^2 \theta} \right)$
For $z$ to be real,the imaginary part must be zero:
$\text{Im}(z) = \frac{8\sin \theta}{1 + 4\sin^2 \theta} = 0$
This implies $\sin \theta = 0$.
The general solution for $\sin \theta = 0$ is $\theta = n\pi$,where $n$ is an integer.

(C) complex number is purely imaginary if its real part is $0$.
Multiply the numerator and denominator by the conjugate of the denominator:
$\frac{3 + 2i\sin \theta}{1 - 2i\sin \theta} \times \frac{1 + 2i\sin \theta}{1 + 2i\sin \theta} = \frac{3 + 6i\sin \theta + 2i\sin \theta + 4i^2\sin^2 \theta}{1^2 + (2\sin \theta)^2} = \frac{(3 - 4\sin^2 \theta) + i(8\sin \theta)}{1 + 4\sin^2 \theta}$.
For the expression to be purely imaginary,the real part must be $0$:
$\frac{3 - 4\sin^2 \theta}{1 + 4\sin^2 \theta} = 0$
$3 - 4\sin^2 \theta = 0$
$\sin^2 \theta = \frac{3}{4}$
$\sin \theta = \pm \frac{\sqrt{3}}{2} = \sin \left( \pm \frac{\pi}{3} \right)$
Therefore,$\theta = n\pi \pm \frac{\pi}{3}$.

(C) Let the common root be $y$.
Then,$y^2 + py + q = 0$ and $y^2 + \alpha y + \beta = 0$.
Subtracting the two equations:
$(y^2 + py + q) - (y^2 + \alpha y + \beta) = 0$
$y(p - \alpha) + (q - \beta) = 0$
$y(p - \alpha) = \beta - q$
$y = \frac{\beta - q}{p - \alpha} = \frac{q - \beta}{\alpha - p}$.
Alternatively,using the method of cross-multiplication for the system:
$\frac{y^2}{p\beta - q\alpha} = \frac{y}{q - \beta} = \frac{1}{\alpha - p}$.
From the second and third terms,$y = \frac{q - \beta}{\alpha - p}$.
From the first and second terms,$y = \frac{p\beta - q\alpha}{q - \beta}$.
Thus,the common root is $\frac{q - \beta}{\alpha - p}$ or $\frac{p\beta - \alpha q}{q - \beta}$.

(B) The numbers are $4$-digit numbers formed using the set $\{0, 1, 2, 3, 4\}$.
Since the number must be greater than $1000$ and less than or equal to $4000$,the first digit can be $1, 2, 3,$ or $4$.
Case $1$: First digit is $1, 2,$ or $3$.
For each of these $3$ choices,the remaining $3$ positions can each be filled in $5$ ways (digits $0, 1, 2, 3, 4$).
Total numbers for these cases $= 3 \times 5 \times 5 \times 5 = 375$.
However,we must exclude the case where the number is $1000$ (since it must be greater than $1000$).
So,$375 - 1 = 374$ numbers.
Case $2$: First digit is $4$.
The only number $\le 4000$ starting with $4$ is $4000$ itself.
Adding this to our count: $374 + 1 = 375$.
Thus,the total count is $375$.

(A) Let $H$ be the height of the cloud above the lake level and $h = 2500 \, m$ be the height of the observation point above the lake.
Let $x$ be the horizontal distance from the observation point to the cloud.
From the angle of elevation,we have $\tan(15^\circ) = \frac{H - h}{x}$,so $x = \frac{H - h}{\tan(15^\circ)} = (H - h) \cot(15^\circ)$.
From the angle of depression of the reflection,the depth of the reflection below the lake is $H$. The total vertical distance from the observation point to the reflection is $H + h$.
Thus,$\tan(45^\circ) = \frac{H + h}{x}$,so $x = \frac{H + h}{\tan(45^\circ)} = (H + h) \cot(45^\circ)$.
Equating the two expressions for $x$: $(H - h) \cot(15^\circ) = (H + h) \cot(45^\circ)$.
Since $\cot(45^\circ) = 1$ and $\cot(15^\circ) = 2 + \sqrt{3}$,we have $(H - 2500)(2 + \sqrt{3}) = H + 2500$.
$H(2 + \sqrt{3}) - 2500(2 + \sqrt{3}) = H + 2500$.
$H(1 + \sqrt{3}) = 2500(3 + \sqrt{3}) = 2500 \sqrt{3}(\sqrt{3} + 1)$.
$H = 2500 \sqrt{3} \, m$.

(A) Let the lines be $L_1: 4x - 7y + 10 = 0$,$L_2: x + y - 5 = 0$,and $L_3: 7x + 4y - 15 = 0$.
First,check the slopes of the lines:
Slope of $L_1$ $(m_1)$ = $4/7$.
Slope of $L_3$ $(m_3)$ = $-7/4$.
Since $m_1 \times m_3 = (4/7) \times (-7/4) = -1$,the lines $L_1$ and $L_3$ are perpendicular to each other.
Therefore,the triangle is a right-angled triangle with the right angle at the intersection of $L_1$ and $L_3$.
The orthocentre of a right-angled triangle is the vertex where the right angle is formed.
Solving $4x - 7y = -10$ and $7x + 4y = 15$:
Multiply the first by $4$ and the second by $7$: $16x - 28y = -40$ and $49x + 28y = 105$.
Adding these gives $65x = 65$,so $x = 1$.
Substituting $x = 1$ into $4(1) - 7y = -10$ gives $-7y = -14$,so $y = 2$.
The orthocentre is $(1, 2)$.

(A) The slope of line $BC$ passing through $(-3, -1)$ and $(-1, -3)$ is $m = \frac{-3 - (-1)}{-1 - (-3)} = \frac{-2}{2} = -1$.
Since the required line is parallel to $BC$,its equation is of the form $x + y + \lambda = 0$.
The distance of this line from the origin $(0, 0)$ is given by $\frac{|\lambda|}{\sqrt{1^2 + 1^2}} = \frac{1}{2}$.
This implies $\frac{|\lambda|}{\sqrt{2}} = \frac{1}{2}$,so $|\lambda| = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.
Thus,$\lambda = \pm \frac{1}{\sqrt{2}}$.
For the line to cut $OB$ and $OC$,it must lie between the origin and the line $BC$. The line $BC$ is $x + y + 4 = 0$. Since the origin $(0,0)$ gives $0+0+4 > 0$,we choose $\lambda$ such that the line lies between them,which gives $\lambda = \frac{1}{\sqrt{2}}$.
The equation is $x + y + \frac{1}{\sqrt{2}} = 0$,which simplifies to $2x + 2y + \sqrt{2} = 0$.

(A) Let the point be $(h, k)$. Since it lies on the line $x + y = 4$,we have $h + k = 4$ or $k = 4 - h$ $(i)$.
The perpendicular distance from $(h, k)$ to the line $4x + 3y - 10 = 0$ is given by $d = \frac{|4h + 3k - 10|}{\sqrt{4^2 + 3^2}} = 1$.
Thus,$|4h + 3k - 10| = 5$,which implies $4h + 3k - 10 = 5$ or $4h + 3k - 10 = -5$.
Case $1$: $4h + 3k = 15$. Substituting $k = 4 - h$,we get $4h + 3(4 - h) = 15$ $\Rightarrow 4h + 12 - 3h = 15$ $\Rightarrow h = 3$. Then $k = 4 - 3 = 1$. So,the point is $(3, 1)$.
Case $2$: $4h + 3k = 5$. Substituting $k = 4 - h$,we get $4h + 3(4 - h) = 5$ $\Rightarrow 4h + 12 - 3h = 5$ $\Rightarrow h = -7$. Then $k = 4 - (-7) = 11$. So,the point is $(-7, 11)$.
Therefore,the required points are $(3, 1)$ and $(-7, 11)$.

(C) The line $2x + 3y = 12$ meets the $y$-axis at $B$ where $x=0$,so $B = (0, 4)$.
The slope of the line $2x + 3y = 12$ is $m_1 = -\frac{2}{3}$.
The slope of the line perpendicular to $AB$ is $m_2 = -\frac{1}{m_1} = \frac{3}{2}$.
The equation of the line passing through $(5, 5)$ with slope $\frac{3}{2}$ is $y - 5 = \frac{3}{2}(x - 5)$,which simplifies to $3x - 2y = 5$.
This line meets the $x$-axis at $C$ where $y=0$,so $3x = 5 \implies C = (\frac{5}{3}, 0)$.
To find $E$,we solve the system of equations:
$2x + 3y = 12$ $(i)$
$3x - 2y = 5$ (ii)
Multiplying $(i)$ by $2$ and (ii) by $3$: $4x + 6y = 24$ and $9x - 6y = 15$.
Adding these gives $13x = 39 \implies x = 3$. Substituting $x=3$ into $(i)$,$6 + 3y = 12 \implies 3y = 6 \implies y = 2$. So $E = (3, 2)$.
The area of quadrilateral $OCEB$ can be calculated by splitting it into $\Delta OCE$ and $\Delta OEB$.
Area of $\Delta OCE = \frac{1}{2} |x_O(y_C - y_E) + x_C(y_E - y_O) + x_E(y_O - y_C)| = \frac{1}{2} |0 + \frac{5}{3}(2 - 0) + 3(0 - 0)| = \frac{5}{3}$.
Area of $\Delta OEB = \frac{1}{2} |x_O(y_E - y_B) + x_E(y_B - y_O) + x_B(y_O - y_E)| = \frac{1}{2} |0 + 3(4 - 0) + 0| = 6$.
Total area $= \frac{5}{3} + 6 = \frac{23}{3} \text{ sq. units}$.

(C) Given limit: $\mathop {\lim }\limits_{x \to 1} \frac{x - 1}{2x^2 - 7x + 5}$.
Substituting $x = 1$,we get the indeterminate form $\frac{0}{0}$.
Factorizing the denominator: $2x^2 - 7x + 5 = (x - 1)(2x - 5)$.
Thus,$\mathop {\lim }\limits_{x \to 1} \frac{x - 1}{(x - 1)(2x - 5)} = \mathop {\lim }\limits_{x \to 1} \frac{1}{2x - 5}$.
Substituting $x = 1$: $\frac{1}{2(1) - 5} = \frac{1}{-3} = -\frac{1}{3}$.
Alternatively,using $L$-Hospital's rule: $\mathop {\lim }\limits_{x \to 1} \frac{\frac{d}{dx}(x - 1)}{\frac{d}{dx}(2x^2 - 7x + 5)} = \mathop {\lim }\limits_{x \to 1} \frac{1}{4x - 7} = \frac{1}{4(1) - 7} = -\frac{1}{3}$.

(A) We need to evaluate $\mathop {\lim }\limits_{\theta \to \pi /2} (\sec \theta - \tan \theta )$.
Substituting $\sec \theta = \frac{1}{\cos \theta}$ and $\tan \theta = \frac{\sin \theta}{\cos \theta}$,we get:
$\mathop {\lim }\limits_{\theta \to \pi /2} \left( \frac{1 - \sin \theta}{\cos \theta} \right)$.
As $\theta \to \pi /2$,this is a $\frac{0}{0}$ indeterminate form.
Using the identity $1 - \sin \theta = \left( \cos \frac{\theta}{2} - \sin \frac{\theta}{2} \right)^2$ and $\cos \theta = \cos^2 \frac{\theta}{2} - \sin^2 \frac{\theta}{2} = (\cos \frac{\theta}{2} - \sin \frac{\theta}{2})(\cos \frac{\theta}{2} + \sin \frac{\theta}{2})$:
$\mathop {\lim }\limits_{\theta \to \pi /2} \frac{(\cos \frac{\theta}{2} - \sin \frac{\theta}{2})^2}{(\cos \frac{\theta}{2} - \sin \frac{\theta}{2})(\cos \frac{\theta}{2} + \sin \frac{\theta}{2})} = \mathop {\lim }\limits_{\theta \to \pi /2} \frac{\cos \frac{\theta}{2} - \sin \frac{\theta}{2}}{\cos \frac{\theta}{2} + \sin \frac{\theta}{2}}$.
Substituting $\theta = \pi / 2$,we get $\frac{\cos(\pi/4) - \sin(\pi/4)}{\cos(\pi/4) + \sin(\pi/4)} = \frac{1/\sqrt{2} - 1/\sqrt{2}}{1/\sqrt{2} + 1/\sqrt{2}} = \frac{0}{\sqrt{2}} = 0$.

(A) Given the equation $ax^2 + bx + c = 0$. Let the roots be $\alpha$ and $\beta$. Then $\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$.
The sum of the reciprocals of their squares is $\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{\alpha^2\beta^2} = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{(\alpha\beta)^2}$.
Substituting the values: $\frac{(-\frac{b}{a})^2 - 2(\frac{c}{a})}{(\frac{c}{a})^2} = \frac{\frac{b^2}{a^2} - \frac{2c}{a}}{\frac{c^2}{a^2}} = \frac{b^2 - 2ac}{c^2}$.
According to the given condition,$\alpha + \beta = \frac{1}{\alpha^2} + \frac{1}{\beta^2}$,so $-\frac{b}{a} = \frac{b^2 - 2ac}{c^2}$.
Cross-multiplying gives $-bc^2 = a(b^2 - 2ac) = ab^2 - 2a^2c$.
Rearranging the terms,we get $2a^2c = ab^2 + bc^2$.
Dividing by $a^2c^2$ is not necessary here; we observe that $2(ca^2) = bc^2 + ab^2$. This is the condition for $bc^2, ca^2, ab^2$ to be in $A.P.$

(C) To find the total numbers greater than $3000$ using digits ${0, 1, 2, 3, 4, 5}$,we consider $4, 5,$ and $6$ digit numbers.
$1$. $4$-digit numbers greater than $3000$:
The first digit can be $3, 4,$ or $5$ ($3$ choices).
The remaining $3$ positions can be filled by the remaining $5$ digits in $^5P_3 = 5 \times 4 \times 3 = 60$ ways.
Total $4$-digit numbers $= 3 \times 60 = 180$.
$2$. $5$-digit numbers:
The first digit cannot be $0$ ($5$ choices: $1, 2, 3, 4, 5$).
The remaining $4$ positions can be filled by the remaining $5$ digits in $^5P_4 = 5 \times 4 \times 3 \times 2 = 120$ ways.
Total $5$-digit numbers $= 5 \times 120 = 600$.
$3$. $6$-digit numbers:
The first digit cannot be $0$ ($5$ choices: $1, 2, 3, 4, 5$).
The remaining $5$ positions can be filled by the remaining $5$ digits in $5! = 120$ ways.
Total $6$-digit numbers $= 5 \times 120 = 600$.
Total numbers $= 180 + 600 + 600 = 1380$.

(D) Let $E_1$ be the event that the ball is drawn from bag $A$,$E_2$ be the event that it is drawn from bag $B$,and $E$ be the event that the drawn ball is red.
We need to find $P(E_2|E)$.
Since both bags are equally likely to be selected,we have $P(E_1) = P(E_2) = \frac{1}{2}$.
Also,the probability of drawing a red ball from bag $A$ is $P(E|E_1) = \frac{3}{5}$,and from bag $B$ is $P(E|E_2) = \frac{5}{9}$.
Using Bayes' theorem:
$P(E_2|E) = \frac{P(E_2) \cdot P(E|E_2)}{P(E_1) \cdot P(E|E_1) + P(E_2) \cdot P(E|E_2)}$
$P(E_2|E) = \frac{\frac{1}{2} \cdot \frac{5}{9}}{\frac{1}{2} \cdot \frac{3}{5} + \frac{1}{2} \cdot \frac{5}{9}}$
$P(E_2|E) = \frac{\frac{5}{18}}{\frac{3}{10} + \frac{5}{18}} = \frac{\frac{5}{18}}{\frac{27 + 25}{90}} = \frac{5}{18} \cdot \frac{90}{52} = \frac{5 \cdot 5}{52} = \frac{25}{52}$.

(D) Let $I = \int \sqrt{2 + \sin 3x} \cdot \cos 3x \, dx$.
Substitute $t = 2 + \sin 3x$.
Then,differentiating with respect to $x$,we get $dt = 3 \cos 3x \, dx$,which implies $\cos 3x \, dx = \frac{1}{3} dt$.
Substituting these into the integral:
$I = \int \sqrt{t} \cdot \frac{1}{3} dt = \frac{1}{3} \int t^{1/2} dt$.
Using the power rule $\int t^n dt = \frac{t^{n+1}}{n+1} + c$:
$I = \frac{1}{3} \cdot \frac{t^{3/2}}{3/2} + c = \frac{1}{3} \cdot \frac{2}{3} t^{3/2} + c = \frac{2}{9} t^{3/2} + c$.
Substituting back $t = 2 + \sin 3x$:
$I = \frac{2}{9}(2 + \sin 3x)^{3/2} + c$.