IIT JEE 1977 Mathematics Question Paper with Answer and Solution

(B) The imaginary cube roots of unity are $\omega$ and $\omega^2$.
Let $\alpha = \omega$ and $\beta = \omega^2$.
We know that $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.
Substituting these values into the expression $\alpha^4 + \beta^4 + \frac{1}{\alpha\beta}$:
$= \omega^4 + (\omega^2)^4 + \frac{1}{\omega \cdot \omega^2}$
$= \omega^4 + \omega^8 + \frac{1}{\omega^3}$
$= \omega + \omega^2 + \frac{1}{1}$
$= \omega + \omega^2 + 1$
$= 0$.

(A) Given that $a, b, c$ are in $A.P.$
$2b = a + c$ ......$(i)$
Given that $a^2, b^2, c^2$ are in $H.P.$
$b^2 = \frac{2a^2c^2}{a^2 + c^2}$
$b^2(a^2 + c^2) = 2a^2c^2$
$b^2((a+c)^2 - 2ac) = 2a^2c^2$
Substituting $a+c = 2b$ from $(i)$:
$b^2(4b^2 - 2ac) = 2a^2c^2$
$4b^4 - 2acb^2 - 2a^2c^2 = 0$
$2b^4 - acb^2 - a^2c^2 = 0$
$(2b^2 + ac)(b^2 - ac) = 0$
Since $a, b, c$ are real,$2b^2 + ac = 0$ implies $a, b, c$ are $0$ or complex,so we consider $b^2 = ac$.
If $b^2 = ac$,then $a, b, c$ are in $G.P.$
Since $a, b, c$ are in both $A.P.$ and $G.P.$,we must have $a = b = c$.

(B) Let the incorrect equation be $x^2 + 17x + q = 0$.
Given the roots are $-2$ and $-15$,the product of the roots is $(-2) \times (-15) = 30$.
Since the constant term $q$ was not changed,$q = 30$.
The original equation is $x^2 + 13x + 30 = 0$.
Factoring the equation: $x^2 + 10x + 3x + 30 = 0$.
$x(x + 10) + 3(x + 10) = 0$.
$(x + 3)(x + 10) = 0$.
Thus,the roots are $x = -3$ and $x = -10$.

(B) Total number of persons $= 20 + 1 = 21$.
Let the host be $H$ and the two particular persons be $P_1$ and $P_2$.
Since $P_1$ and $P_2$ must sit on either side of the host,we treat the group $(P_1, H, P_2)$ as a single unit.
This leaves $21 - 3 = 18$ other persons plus the $1$ unit,making a total of $19$ entities to be arranged in a circle.
The number of ways to arrange $n$ items in a circle is $(n-1)!$,so $19$ entities can be arranged in $(19-1)! = 18!$ ways.
Within the unit $(P_1, H, P_2)$,the two persons $P_1$ and $P_2$ can be arranged in $2! = 2$ ways (i.e.,$P_1HP_2$ or $P_2HP_1$).
Therefore,the total number of ways is $2 \times 18!$.

(C) The word $BHARAT$ contains $6$ letters,where $A$ repeats $2$ times.
Total number of arrangements $ = \frac{6!}{2!} = \frac{720}{2} = 360$.
Now,consider the case where $B$ and $H$ always come together. Treat $(BH)$ as a single unit. The letters are $(BH), A, R, A, T$. There are $5$ units,where $A$ repeats $2$ times.
The number of arrangements where $B$ and $H$ are together $ = \frac{5!}{2!} \times 2! = 120 \times 1 = 120$.
Therefore,the number of words where $B$ and $H$ never come together $ = 360 - 120 = 240$.

(A) Let $P(n) = 3^{2n+2} - 8n - 9$.
For $n=1$,$P(1) = 3^4 - 8(1) - 9 = 81 - 17 = 64$,which is divisible by $16$.
For $n=2$,$P(2) = 3^6 - 8(2) - 9 = 729 - 16 - 9 = 704$.
$704 = 16 \times 44$,so it is divisible by $16$.
Using the Binomial Theorem: $3^{2n+2} = 9 \times 3^{2n} = 9 \times (1+8)^n$.
$9 \times (1+8)^n = 9 \times [1 + n(8) + \frac{n(n-1)}{2}(8^2) + \dots]$.
$9 \times [1 + 8n + 32n(n-1) + \dots] = 9 + 72n + 288n(n-1) + \dots$.
Thus,$P(n) = 9 + 72n + 288n(n-1) + \dots - 8n - 9 = 64n + 288n(n-1) + \dots$.
Since every term is divisible by $64$,the expression is divisible by $64$,and consequently by $16$.

(C) Given expression: $E = 2 \sin^2 \beta + 4 \cos(\alpha + \beta) \sin \alpha \sin \beta + \cos 2(\alpha + \beta)$
Using the identity $2 \sin^2 \beta = 1 - \cos 2\beta$ and $\cos 2(\alpha + \beta) = 2 \cos^2(\alpha + \beta) - 1$:
$E = (1 - \cos 2\beta) + 4 \cos(\alpha + \beta) \sin \alpha \sin \beta + (2 \cos^2(\alpha + \beta) - 1)$
$E = 2 \cos^2(\alpha + \beta) - \cos 2\beta + 4 \cos(\alpha + \beta) \sin \alpha \sin \beta$
Using $2 \sin \alpha \sin \beta = \cos(\alpha - \beta) - \cos(\alpha + \beta)$:
$E = 2 \cos^2(\alpha + \beta) - \cos 2\beta + 2 \cos(\alpha + \beta) [\cos(\alpha - \beta) - \cos(\alpha + \beta)]$
$E = 2 \cos^2(\alpha + \beta) - \cos 2\beta + 2 \cos(\alpha + \beta) \cos(\alpha - \beta) - 2 \cos^2(\alpha + \beta)$
$E = 2 \cos(\alpha + \beta) \cos(\alpha - \beta) - \cos 2\beta$
Using $2 \cos A \cos B = \cos(A + B) + \cos(A - B)$:
$E = (\cos 2\alpha + \cos 2\beta) - \cos 2\beta$
$E = \cos 2\alpha$

(C) To find the vertices of the triangle,we solve the equations in pairs:
$1$. Intersection of $7x - 2y + 10 = 0$ and $y + 2 = 0$:
Substituting $y = -2$ into the first equation: $7x - 2(-2) + 10 = 0 \implies 7x + 4 + 10 = 0 \implies 7x = -14 \implies x = -2$. Vertex: $(-2, -2)$.
$2$. Intersection of $7x + 2y - 10 = 0$ and $y + 2 = 0$:
Substituting $y = -2$ into the second equation: $7x + 2(-2) - 10 = 0 \implies 7x - 4 - 10 = 0 \implies 7x = 14 \implies x = 2$. Vertex: $(2, -2)$.
$3$. Intersection of $7x - 2y + 10 = 0$ and $7x + 2y - 10 = 0$:
Adding the two equations: $14x = 0 \implies x = 0$. Substituting $x = 0$ into $7x - 2y + 10 = 0$: $-2y = -10 \implies y = 5$. Vertex: $(0, 5)$.
The area of the triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
Area $= \frac{1}{2} |(-2)(-2 - 5) + 2(5 - (-2)) + 0(-2 - (-2))|$
Area $= \frac{1}{2} |(-2)(-7) + 2(7) + 0| = \frac{1}{2} |14 + 14| = \frac{28}{2} = 14 \, sq. \, units$.

(A) Let the two perpendicular lines be the coordinate axes. If $a$ and $b$ are the intercepts made by the moving line on the coordinate axes,then the equation of the line is $\frac{x}{a} + \frac{y}{b} = 1$ ..... $(i)$.
According to the problem,the sum of the reciprocals of the intercepts is constant,say $\frac{1}{a} + \frac{1}{b} = \frac{1}{k}$,where $k$ is a constant.
Multiplying by $k$,we get $\frac{k}{a} + \frac{k}{b} = 1$ ..... $(ii)$.
Comparing equation $(i)$ and equation $(ii)$,we can see that the line always passes through the fixed point $(k, k)$.

(A) We need to evaluate the limit: $\mathop {\lim }\limits_{x \to 1} \frac{(2x - 3)(\sqrt{x} - 1)}{2x^2 + x - 3}$.
First,factor the denominator: $2x^2 + x - 3 = (2x + 3)(x - 1)$.
Substitute this into the limit: $\mathop {\lim }\limits_{x \to 1} \frac{(2x - 3)(\sqrt{x} - 1)}{(2x + 3)(x - 1)}$.
Since $(x - 1) = (\sqrt{x} - 1)(\sqrt{x} + 1)$,the expression becomes: $\mathop {\lim }\limits_{x \to 1} \frac{(2x - 3)(\sqrt{x} - 1)}{(2x + 3)(\sqrt{x} - 1)(\sqrt{x} + 1)}$.
Cancel the common factor $(\sqrt{x} - 1)$: $\mathop {\lim }\limits_{x \to 1} \frac{2x - 3}{(2x + 3)(\sqrt{x} + 1)}$.
Now,substitute $x = 1$: $\frac{2(1) - 3}{(2(1) + 3)(\sqrt{1} + 1)} = \frac{-1}{(5)(2)} = -\frac{1}{10}$.

(A) Let $L = \mathop {\lim }\limits_{\alpha \to \pi /4} \frac{{\sin \alpha - \cos \alpha }}{{\alpha - \pi /4}}$.
We can rewrite the numerator as:
$\sin \alpha - \cos \alpha = \sqrt{2} \left( \sin \alpha \cdot \frac{1}{\sqrt{2}} - \cos \alpha \cdot \frac{1}{\sqrt{2}} \right) = \sqrt{2} \left( \sin \alpha \cos \frac{\pi}{4} - \cos \alpha \sin \frac{\pi}{4} \right) = \sqrt{2} \sin \left( \alpha - \frac{\pi}{4} \right)$.
Substituting this into the limit:
$L = \mathop {\lim }\limits_{\alpha \to \pi /4} \frac{\sqrt{2} \sin \left( \alpha - \frac{\pi}{4} \right)}{\alpha - \frac{\pi}{4}}$.
Using the standard limit $\mathop {\lim }\limits_{x \to 0} \frac{\sin x}{x} = 1$,where $x = \alpha - \frac{\pi}{4}$:
$L = \sqrt{2} \times 1 = \sqrt{2}$.
Alternatively,using $L$-Hospital's rule:
$L = \mathop {\lim }\limits_{\alpha \to \pi /4} \frac{\frac{d}{d\alpha}(\sin \alpha - \cos \alpha)}{\frac{d}{d\alpha}(\alpha - \pi/4)} = \mathop {\lim }\limits_{\alpha \to \pi /4} \frac{\cos \alpha + \sin \alpha}{1} = \cos \frac{\pi}{4} + \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$.

(A) Let $I = \int \frac{1 + x^2}{\sqrt{1 - x^2}} dx$.
Substitute $x = \sin \theta$,so $dx = \cos \theta d\theta$.
The integral becomes $I = \int \frac{1 + \sin^2 \theta}{\cos \theta} \cos \theta d\theta = \int (1 + \sin^2 \theta) d\theta$.
Using the identity $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$,we get $I = \int (1 + \frac{1 - \cos 2\theta}{2}) d\theta = \int (\frac{3}{2} - \frac{1}{2} \cos 2\theta) d\theta$.
Integrating,we get $I = \frac{3}{2} \theta - \frac{1}{4} \sin 2\theta + c = \frac{3}{2} \theta - \frac{1}{2} \sin \theta \cos \theta + c$.
Since $\theta = \sin^{-1} x$ and $\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - x^2}$,we have $I = \frac{3}{2} \sin^{-1} x - \frac{1}{2} x \sqrt{1 - x^2} + c$.

(A) Let $\sqrt{x} = t$. Then $x = t^2$,which implies $dx = 2t \, dt$.
Substituting these into the integral:
$\int \cos \sqrt{x} \, dx = \int \cos(t) \cdot 2t \, dt = 2 \int t \cos(t) \, dt$.
Using integration by parts,where $u = t$ and $dv = \cos(t) \, dt$:
$\int u \, dv = uv - \int v \, du = t \sin(t) - \int \sin(t) \, dt = t \sin(t) + \cos(t)$.
Multiplying by the constant $2$:
$2(t \sin(t) + \cos(t)) + c$.
Substituting $t = \sqrt{x}$ back:
$2[\sqrt{x} \sin \sqrt{x} + \cos \sqrt{x}] + c$.

(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:
$I = x(\log x)^2 - 2x \log x + 2x + c$.

(A) Let $I = \int \tan^3 2x \sec 2x \, dx$.
We can rewrite $\tan^3 2x$ as $\tan^2 2x \cdot \tan 2x$.
Since $\tan^2 2x = \sec^2 2x - 1$,the integral becomes:
$I = \int (\sec^2 2x - 1) \sec 2x \tan 2x \, dx$.
Let $u = \sec 2x$. Then $du = 2 \sec 2x \tan 2x \, dx$,which implies $\sec 2x \tan 2x \, dx = \frac{1}{2} du$.
Substituting these into the integral:
$I = \int (u^2 - 1) \cdot \frac{1}{2} du$
$I = \frac{1}{2} \int (u^2 - 1) \, du$
$I = \frac{1}{2} \left( \frac{u^3}{3} - u \right) + c$
$I = \frac{u^3}{6} - \frac{u}{2} + c$
Substituting $u = \sec 2x$ back into the equation:
$I = \frac{1}{6} \sec^3 2x - \frac{1}{2} \sec 2x + c$.