IIT JEE 1978 Mathematics Question Paper with Answer and Solution

(D) Let $z = (1 - \cos \theta ) + i(2\sin \theta )$. We want the real part of $z^{-1} = \frac{1}{z}$.
Using trigonometric identities,$1 - \cos \theta = 2\sin^2(\frac{\theta}{2})$ and $2\sin \theta = 4\sin(\frac{\theta}{2})\cos(\frac{\theta}{2})$.
So,$z = 2\sin(\frac{\theta}{2}) [\sin(\frac{\theta}{2}) + 2i\cos(\frac{\theta}{2})]$.
Then,$z^{-1} = \frac{1}{2\sin(\frac{\theta}{2})} \cdot \frac{1}{\sin(\frac{\theta}{2}) + 2i\cos(\frac{\theta}{2})}$.
Multiply numerator and denominator by the conjugate $\sin(\frac{\theta}{2}) - 2i\cos(\frac{\theta}{2})$:
$z^{-1} = \frac{1}{2\sin(\frac{\theta}{2})} \cdot \frac{\sin(\frac{\theta}{2}) - 2i\cos(\frac{\theta}{2})}{\sin^2(\frac{\theta}{2}) + 4\cos^2(\frac{\theta}{2})}$.
The real part is $\frac{\sin(\frac{\theta}{2})}{2\sin(\frac{\theta}{2}) [\sin^2(\frac{\theta}{2}) + 4\cos^2(\frac{\theta}{2})]} = \frac{1}{2[\sin^2(\frac{\theta}{2}) + 4\cos^2(\frac{\theta}{2})]}$.
Using $\sin^2(\frac{\theta}{2}) = \frac{1 - \cos \theta}{2}$ and $\cos^2(\frac{\theta}{2}) = \frac{1 + \cos \theta}{2}$:
Real part $= \frac{1}{2[\frac{1 - \cos \theta}{2} + 4(\frac{1 + \cos \theta}{2})]} = \frac{1}{1 - \cos \theta + 4 + 4\cos \theta} = \frac{1}{5 + 3\cos \theta }$.

(B) Given that $\alpha$ and $\beta$ are complex cube roots of unity,we have $\alpha = \omega$ and $\beta = \omega^2$ (or vice versa),where $1 + \omega + \omega^2 = 0$ and $\omega^3 = 1$.
We need to calculate $xyz = (a + b)(a\omega + b\omega^2)(a\omega^2 + b\omega)$.
First,multiply $y$ and $z$:
$yz = (a\omega + b\omega^2)(a\omega^2 + b\omega) = a^2\omega^3 + ab\omega^2 + ab\omega^4 + b^2\omega^3$.
Since $\omega^3 = 1$ and $\omega^4 = \omega$,this simplifies to:
$yz = a^2(1) + ab(\omega^2 + \omega) + b^2(1) = a^2 + ab(-1) + b^2 = a^2 - ab + b^2$.
Now,multiply by $x$:
$xyz = (a + b)(a^2 - ab + b^2) = a^3 + b^3$.

(A) Case $1$: If $9, x, y, z, a$ are in $A.P.$,then the sum of the terms is $9 + x + y + z + a = \frac{5}{2}(9 + a)$.
Given $x + y + z = 15$,we have $9 + 15 + a = \frac{5}{2}(9 + a)$.
$24 + a = \frac{45 + 5a}{2}$.
$48 + 2a = 45 + 5a$.
$3a = 3 \Rightarrow a = 1$.
Case $2$: If $9, x, y, z, a$ are in $H.P.$,then $\frac{1}{9}, \frac{1}{x}, \frac{1}{y}, \frac{1}{z}, \frac{1}{a}$ are in $A.P.$
The sum of these terms is $\frac{1}{9} + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} + \frac{1}{a} = \frac{5}{2}\left(\frac{1}{9} + \frac{1}{a}\right)$.
Given $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{5}{3}$,we have $\frac{1}{9} + \frac{5}{3} + \frac{1}{a} = \frac{5}{2}\left(\frac{1}{9} + \frac{1}{a}\right)$.
$\frac{16}{9} + \frac{1}{a} = \frac{5}{18} + \frac{5}{2a}$.
$\frac{1}{a} - \frac{5}{2a} = \frac{5}{18} - \frac{16}{9}$.
$-\frac{3}{2a} = \frac{5 - 32}{18} = -\frac{27}{18} = -\frac{3}{2}$.
$-\frac{3}{2a} = -\frac{3}{2} \Rightarrow a = 1$.

(A) Given that $\alpha, \beta$ are roots of $x^2 + px + 1 = 0$,we have $\alpha + \beta = -p$ and $\alpha \beta = 1$.
Given that $\gamma, \delta$ are roots of $x^2 + qx + 1 = 0$,we have $\gamma + \delta = -q$ and $\gamma \delta = 1$.
Consider the expression $(\alpha - \gamma)(\beta - \gamma)(\alpha + \delta)(\beta + \delta)$:
$= (\alpha \beta - \gamma(\alpha + \beta) + \gamma^2)(\alpha \beta + \delta(\alpha + \beta) + \delta^2)$
$= (1 + p\gamma + \gamma^2)(1 - p\delta + \delta^2)$.
Since $\gamma$ is a root of $x^2 + qx + 1 = 0$,$\gamma^2 + q\gamma + 1 = 0$,so $1 + \gamma^2 = -q\gamma$.
Similarly,$\delta^2 + 1 = -q\delta$.
Substituting these into the expression:
$= (-q\gamma + p\gamma)(-q\delta - p\delta)$
$= \gamma(p - q) \times \delta(-p - q)$
$= -\gamma \delta (p - q)(p + q)$
$= -1(p^2 - q^2) = q^2 - p^2$.

(B) We have $\tan \alpha = \frac{m}{m + 1}$ and $\tan \beta = \frac{1}{2m + 1}$.
Using the formula $\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$:
$\tan(\alpha + \beta) = \frac{\frac{m}{m + 1} + \frac{1}{2m + 1}}{1 - \left(\frac{m}{m + 1}\right) \left(\frac{1}{2m + 1}\right)}$
$= \frac{\frac{m(2m + 1) + (m + 1)}{(m + 1)(2m + 1)}}{\frac{(m + 1)(2m + 1) - m}{(m + 1)(2m + 1)}}$
$= \frac{2m^2 + m + m + 1}{2m^2 + m + 2m + 1 - m}$
$= \frac{2m^2 + 2m + 1}{2m^2 + 2m + 1} = 1$
Since $\tan(\alpha + \beta) = 1$,we have $\alpha + \beta = \frac{\pi}{4}$.

(A) Given limit: $L = \mathop {\lim }\limits_{x \to a} \frac{{\sqrt {a + 2x} - \sqrt {3x} }}{{\sqrt {3a + x} - 2\sqrt x }}$
Rationalizing the numerator and denominator:
$L = \mathop {\lim }\limits_{x \to a} \frac{(\sqrt {a + 2x} - \sqrt {3x})(\sqrt {a + 2x} + \sqrt {3x})}{(\sqrt {3a + x} - 2\sqrt x)(\sqrt {3a + x} + 2\sqrt x)} \times \frac{\sqrt {3a + x} + 2\sqrt x}{\sqrt {a + 2x} + \sqrt {3x}}$
$L = \mathop {\lim }\limits_{x \to a} \frac{(a + 2x - 3x)}{(3a + x - 4x)} \times \frac{\sqrt {3a + x} + 2\sqrt x}{\sqrt {a + 2x} + \sqrt {3x}}$
$L = \mathop {\lim }\limits_{x \to a} \frac{a - x}{3a - 3x} \times \frac{\sqrt {3a + x} + 2\sqrt x}{\sqrt {a + 2x} + \sqrt {3x}}$
$L = \mathop {\lim }\limits_{x \to a} \frac{a - x}{3(a - x)} \times \frac{\sqrt {3a + x} + 2\sqrt x}{\sqrt {a + 2x} + \sqrt {3x}}$
$L = \frac{1}{3} \times \frac{\sqrt {3a + a} + 2\sqrt a}{\sqrt {a + 2a} + \sqrt {3a}} = \frac{1}{3} \times \frac{2\sqrt {4a} + 2\sqrt a}{2\sqrt {3a}} = \frac{1}{3} \times \frac{2(2\sqrt a) + 2\sqrt a}{2\sqrt {3a}} = \frac{1}{3} \times \frac{6\sqrt a}{2\sqrt {3a}} = \frac{1}{3} \times \frac{3}{\sqrt 3} = \frac{1}{\sqrt 3} = \frac{\sqrt 3}{3}$.
Wait,re-evaluating the limit: $\frac{1}{3} \times \frac{2\sqrt{4a} + 2\sqrt{a}}{2\sqrt{3a}} = \frac{1}{3} \times \frac{4\sqrt{a} + 2\sqrt{a}}{2\sqrt{3a}} = \frac{1}{3} \times \frac{6\sqrt{a}}{2\sqrt{3a}} = \frac{1}{3} \times \frac{3}{\sqrt{3}} = \frac{1}{\sqrt{3}}$.
Correction: The provided option $B$ is $\frac{2}{3\sqrt{3}}$. Let's re-check the simplification: $\frac{1}{3} \times \frac{2\sqrt{4a} + 2\sqrt{a}}{2\sqrt{3a}} = \frac{1}{3} \times \frac{4\sqrt{a} + 2\sqrt{a}}{2\sqrt{3a}} = \frac{6\sqrt{a}}{6\sqrt{3a}} = \frac{1}{\sqrt{3}}$.
Actually,$\frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$. The correct value is $\frac{1}{\sqrt{3}}$.

(C) Let $L = \mathop {\lim }\limits_{x \to 1} (1 - x)\tan \left( {\frac{{\pi x}}{2}} \right)$.
Substitute $y = 1 - x$,so as $x \to 1$,$y \to 0$ and $x = 1 - y$.
Then $L = \mathop {\lim }\limits_{y \to 0} y \tan \left( {\frac{{\pi (1 - y)}}{2}} \right) = \mathop {\lim }\limits_{y \to 0} y \tan \left( {\frac{\pi }{2} - \frac{{\pi y}}{2}} \right)$.
Using the identity $\tan \left( {\frac{\pi }{2} - \theta } \right) = \cot \theta$,we get:
$L = \mathop {\lim }\limits_{y \to 0} y \cot \left( {\frac{{\pi y}}{2}} \right) = \mathop {\lim }\limits_{y \to 0} \frac{y}{\tan \left( {\frac{{\pi y}}{2}} \right)}$.
Multiply and divide by $\frac{\pi }{2}$:
$L = \mathop {\lim }\limits_{y \to 0} \frac{1}{\frac{\pi }{2}} \cdot \frac{{\frac{{\pi y}}{2}}}{{\tan \left( {\frac{{\pi y}}{2}} \right)}} = \frac{2}{\pi } \cdot 1 = \frac{2}{\pi }$.

(C) The figure consists of $8$ squares arranged in three rows: top row ($2$ squares),middle row ($4$ squares),and bottom row ($2$ squares).
We need to place $6$ '$X$'s in these $8$ squares.
The total number of ways to place $6$ '$X$'s in $8$ squares is $^8C_6 = \frac{8 \times 7}{2 \times 1} = 28$.
We must ensure each row contains at least one '$X$'.
Let $R_1, R_2, R_3$ be the top,middle,and bottom rows respectively.
Total squares = $2 + 4 + 2 = 8$.
If the top row $(R_1)$ has no '$X$',then all $6$ '$X$'s must be placed in the remaining $8 - 2 = 6$ squares. This can be done in $^6C_6 = 1$ way.
If the bottom row $(R_3)$ has no '$X$',then all $6$ '$X$'s must be placed in the remaining $8 - 2 = 6$ squares. This can be done in $^6C_6 = 1$ way.
Note that it is impossible for both the top and bottom rows to have no '$X$' simultaneously,as that would require placing $6$ '$X$'s in only $4$ squares,which is impossible.
Thus,the number of invalid ways is $1 + 1 = 2$.
The required number of ways is $28 - 2 = 26$.

(A) Given one side is $4x + 7y + 5 = 0.$ The vertices are $A(-3, 1)$ and $B(1, 1).$
Since $A$ and $B$ do not satisfy the given line equation,the side containing $A$ and $B$ is parallel to the given line.
Let the side containing $A$ and $B$ be $4x + 7y + k = 0.$
Since it passes through $(-3, 1),$ $4(-3) + 7(1) + k = 0 \implies -12 + 7 + k = 0 \implies k = 5.$
So,the side is $4x + 7y - 5 = 0.$
Wait,the side containing $A$ and $B$ is $4x + 7y - 5 = 0.$
The other side parallel to this is $4x + 7y + k' = 0.$
Passing through $(1, 1),$ $4(1) + 7(1) + k' = 0 \implies k' = -11.$
So,the opposite side is $4x + 7y - 11 = 0.$
The perpendicular sides have slope $7/4.$
Side through $(-3, 1): y - 1 = \frac{7}{4}(x + 3) \implies 4y - 4 = 7x + 21 \implies 7x - 4y + 25 = 0.$
Side through $(1, 1): y - 1 = \frac{7}{4}(x - 1) \implies 4y - 4 = 7x - 7 \implies 7x - 4y - 3 = 0.$
Thus,the equations are $4x + 7y - 11 = 0, 7x - 4y + 25 = 0$ and $7x - 4y - 3 = 0.$
Comparing with options,option $(A)$ matches the form $7x - 4y + 25 = 0, 4x + 7y - 11 = 0$ and $7x - 4y - 3 = 0.$

(C) The given system of homogeneous linear equations is:
$x - cy - bz = 0$
$-cx + y - az = 0$
$-bx - ay + z = 0$
For this system to have a non-trivial solution (where $x, y, z$ are not all zero),the determinant of the coefficient matrix must be zero:
$\Delta = \begin{vmatrix} 1 & -c & -b \\ -c & 1 & -a \\ -b & -a & 1 \end{vmatrix} = 0$
Expanding the determinant along the first row:
$1(1 - a^2) - (-c)(-c - ab) + (-b)(ac + b) = 0$
$1 - a^2 - c^2 - abc - abc - b^2 = 0$
$1 - a^2 - b^2 - c^2 - 2abc = 0$
Rearranging the terms,we get:
$a^2 + b^2 + c^2 + 2abc = 1$

(B) Let $t = x^2$.
Then,differentiating both sides with respect to $x$,we get $dt = 2x \, dx$,which implies $x \, dx = \frac{1}{2} \, dt$.
Substituting these into the integral:
$\int \frac{x}{1 + x^4} \, dx = \int \frac{1}{1 + (x^2)^2} \cdot (x \, dx) = \int \frac{1}{1 + t^2} \cdot \frac{1}{2} \, dt$.
$= \frac{1}{2} \int \frac{1}{1 + t^2} \, dt$.
Using the standard integral formula $\int \frac{1}{1 + t^2} \, dt = \tan^{-1}(t) + c$,we get:
$= \frac{1}{2} \tan^{-1}(t) + c$.
Substituting $t = x^2$ back,the final result is $\frac{1}{2} \tan^{-1}(x^2) + c$.

(B) Let $I = \int e^x \sin x \, dx$.
Using integration by parts,$\int u v \, dx = u \int v \, dx - \int (u' \int v \, dx) \, dx$.
Taking $u = \sin x$ and $v = e^x$,we get:
$I = \sin x \cdot e^x - \int \cos x \cdot e^x \, dx$.
Applying integration by parts again to $\int e^x \cos x \, dx$ with $u = \cos x$ and $v = e^x$:
$\int e^x \cos x \, dx = \cos x \cdot e^x - \int (-\sin x) \cdot e^x \, dx = e^x \cos x + \int e^x \sin x \, dx$.
Substituting this back into the equation for $I$:
$I = e^x \sin x - (e^x \cos x + I) + c$.
$I = e^x \sin x - e^x \cos x - I + c$.
$2I = e^x (\sin x - \cos x) + c$.
$I = \frac{1}{2} e^x (\sin x - \cos x) + c$.

(D) Let the third vertex be $(p, q)$. Since it lies on the line $y = x + 3$,we have $q = p + 3$ $(i)$.
Given the area of the triangle is $5$,the area formula is:
$\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 5$
$\frac{1}{2} |2(-2 - q) + 3(q - 1) + p(1 - (-2))| = 5$
$|2(-2 - q) + 3(q - 1) + 3p| = 10$
$|-4 - 2q + 3q - 3 + 3p| = 10$
$|q + 3p - 7| = 10$
Substituting $q = p + 3$ into the equation:
$|(p + 3) + 3p - 7| = 10$
$|4p - 4| = 10$
Case $1$: $4p - 4 = 10$ $\Rightarrow 4p = 14$ $\Rightarrow p = \frac{7}{2}$. Then $q = \frac{7}{2} + 3 = \frac{13}{2}$.
Case $2$: $4p - 4 = -10$ $\Rightarrow 4p = -6$ $\Rightarrow p = -\frac{3}{2}$. Then $q = -\frac{3}{2} + 3 = \frac{3}{2}$.
The possible vertices are $\left( \frac{7}{2}, \frac{13}{2} \right)$ or $\left( -\frac{3}{2}, \frac{3}{2} \right)$.
Comparing with the given options,the correct option is $D$.