IIT JEE 1970 Mathematics Question Paper with Answer and Solution

(B) Given $x = a + b$,$y = a\omega + b\omega^2$,and $z = a\omega^2 + b\omega$.
We know that $1 + \omega + \omega^2 = 0$ and $\omega^3 = 1$.
Expanding the cubes:
$x^3 = (a + b)^3 = a^3 + b^3 + 3ab(a + b)$
$y^3 = (a\omega + b\omega^2)^3 = a^3\omega^3 + b^3\omega^6 + 3a^2b\omega^4 + 3ab^2\omega^5 = a^3 + b^3 + 3a^2b\omega + 3ab^2\omega^2$
$z^3 = (a\omega^2 + b\omega)^3 = a^3\omega^6 + b^3\omega^3 + 3a^2b\omega^5 + 3ab^2\omega^4 = a^3 + b^3 + 3a^2b\omega^2 + 3ab^2\omega$
Adding these:
$x^3 + y^3 + z^3 = 3(a^3 + b^3) + 3a^2b(1 + \omega + \omega^2) + 3ab^2(1 + \omega^2 + \omega)$
Since $1 + \omega + \omega^2 = 0$,the terms with $a^2b$ and $ab^2$ vanish.
Thus,$x^3 + y^3 + z^3 = 3(a^3 + b^3)$.

(D) Given: $m = \tan \theta + \sin \theta$ and $n = \tan \theta - \sin \theta$.
Calculate $m^2 - n^2$:
$m^2 - n^2 = (m + n)(m - n)$
$m + n = (\tan \theta + \sin \theta) + (\tan \theta - \sin \theta) = 2 \tan \theta$
$m - n = (\tan \theta + \sin \theta) - (\tan \theta - \sin \theta) = 2 \sin \theta$
$m^2 - n^2 = (2 \tan \theta)(2 \sin \theta) = 4 \tan \theta \sin \theta$ ... $(i)$
Calculate $4\sqrt{mn}$:
$mn = (\tan \theta + \sin \theta)(\tan \theta - \sin \theta) = \tan^2 \theta - \sin^2 \theta$
$mn = \frac{\sin^2 \theta}{\cos^2 \theta} - \sin^2 \theta = \sin^2 \theta \left( \frac{1}{\cos^2 \theta} - 1 \right) = \sin^2 \theta \left( \frac{1 - \cos^2 \theta}{\cos^2 \theta} \right) = \sin^2 \theta \frac{\sin^2 \theta}{\cos^2 \theta} = \sin^2 \theta \tan^2 \theta$
$\sqrt{mn} = \sqrt{\sin^2 \theta \tan^2 \theta} = \sin \theta \tan \theta$
$4\sqrt{mn} = 4 \sin \theta \tan \theta$ ... $(ii)$
From $(i)$ and $(ii)$,we get ${m^2} - {n^2} = 4\sqrt {mn}$.

(D) Given that $ABCD$ is a cyclic quadrilateral.
Since the sum of opposite angles in a cyclic quadrilateral is $180^\circ$,we have $A + C = 180^\circ$ and $B + D = 180^\circ$.
From $A + C = 180^\circ$,we get $A = 180^\circ - C$.
Therefore,$\cos A = \cos(180^\circ - C) = -\cos C$,which implies $\cos A + \cos C = 0$.
Similarly,from $B + D = 180^\circ$,we get $B = 180^\circ - D$.
Therefore,$\cos B = \cos(180^\circ - D) = -\cos D$,which implies $\cos B + \cos D = 0$.
Adding these two results,we get $\cos A + \cos B + \cos C + \cos D = 0 + 0 = 0$.

(D) The vertex $O$ is the intersection of $4x + 5y = 0$ and $7x + 2y = 0,$ which is the origin $(0, 0).$
Since the diagonal $11x + 7y = 9$ does not pass through the origin $(0, 0),$ it cannot be the diagonal $OB.$ Thus,$11x + 7y = 9$ is the equation of the diagonal $AC.$
Let the sides be $OA: 4x + 5y = 0$ and $OC: 7x + 2y = 0.$
Solving $AC$ with $OA$: $4x + 5y = 0$ and $11x + 7y = 9.$ Multiplying the first by $7$ and the second by $5$: $28x + 35y = 0$ and $55x + 35y = 45.$ Subtracting gives $27x = 45 \implies x = \frac{5}{3}.$ Then $y = -\frac{4}{3}.$ So $A = \left(\frac{5}{3}, -\frac{4}{3}\right).$
Solving $AC$ with $OC$: $7x + 2y = 0$ and $11x + 7y = 9.$ Multiplying the first by $7$ and the second by $2$: $49x + 14y = 0$ and $22x + 14y = 18.$ Subtracting gives $27x = -18 \implies x = -\frac{2}{3}.$ Then $y = \frac{7}{3}.$ So $C = \left(-\frac{2}{3}, \frac{7}{3}\right).$
The midpoint $M$ of diagonal $AC$ is $\left(\frac{5/3 - 2/3}{2}, \frac{-4/3 + 7/3}{2}\right) = \left(\frac{1/3}{2}, \frac{3/3}{2}\right) = \left(\frac{1}{6}, \frac{1}{2}\right).$
The other diagonal $OB$ passes through the origin $(0, 0)$ and the midpoint $M\left(\frac{1}{6}, \frac{1}{2}\right).$
The slope of $OB$ is $m = \frac{1/2 - 0}{1/6 - 0} = \frac{1/2}{1/6} = 3.$
The equation of $OB$ is $y = 3x,$ or $3x - y = 0.$ Since this is not among the options,the correct answer is $D$ (None of these).

(A) Let $P$ be the eye of the observer and $B$ be the centre of the spherical balloon. Let $Q$ be a point of tangency on the balloon from $P$. Then $\angle QPB = \frac{\alpha}{2}$.
In the right-angled triangle $\triangle PQB$,we have $\sin\left(\frac{\alpha}{2}\right) = \frac{BQ}{PB} = \frac{r}{l}$,where $l = PB$ is the distance from the observer to the centre of the balloon.
Thus,$l = r \csc\left(\frac{\alpha}{2}\right)$.
In the right-angled triangle formed by the height $H$ of the centre $B$ from the ground,we have $\sin \beta = \frac{H}{l}$.
Therefore,$H = l \sin \beta = r \csc\left(\frac{\alpha}{2}\right) \sin \beta$.