IIT JEE 1965 Mathematics Question Paper with Answer and Solution

(B) Given that $\omega$ is a cube root of unity,we know that $1 + \omega + \omega^2 = 0$ and $\omega^3 = 1$.
Using $1 + \omega^2 = -\omega$,the first term becomes $(1 - \omega + \omega^2)^5 = (-\omega - \omega)^5 = (-2\omega)^5 = -32\omega^5$.
Since $\omega^5 = \omega^3 \cdot \omega^2 = \omega^2$,this simplifies to $-32\omega^2$.
Using $1 + \omega = -\omega^2$,the second term becomes $(1 + \omega - \omega^2)^5 = (-\omega^2 - \omega^2)^5 = (-2\omega^2)^5 = -32\omega^{10}$.
Since $\omega^{10} = (\omega^3)^3 \cdot \omega = \omega$,this simplifies to $-32\omega$.
Adding the two terms: $-32\omega^2 - 32\omega = -32(\omega^2 + \omega)$.
Since $1 + \omega + \omega^2 = 0$,we have $\omega^2 + \omega = -1$.
Therefore,$-32(-1) = 32$.

(C) Given that $x, 1, z$ are in $A.P.$,the middle term is the arithmetic mean: $1 = \frac{x + z}{2}$,which implies $x + z = 2$......$(i)$
Given that $x, 2, z$ are in $G.P.$,the middle term is the geometric mean: $2^2 = xz$,which implies $xz = 4$......$(ii)$
For $x, 4, z$ to be in $H.P.$,the middle term must be the harmonic mean of $x$ and $z$,which is $\frac{2xz}{x + z}$.
Substituting the values from $(i)$ and $(ii)$ into the expression for the harmonic mean:
$\text{Harmonic Mean} = \frac{2(4)}{2} = \frac{8}{2} = 4$.
Since the middle term is $4$,$x, 4, z$ are in $H.P.$

(B) The general term $T_{r+1}$ in the expansion of $(a+b)^n$ is given by $^{n}C_{r} a^{n-r} b^{r}$.
Here,$a = (x/3)^{1/2}$,$b = 3/(2x^2)$,and $n = 10$.
$T_{r+1} = ^{10}C_{r} (x/3)^{(10-r)/2} (3/(2x^2))^r$
$T_{r+1} = ^{10}C_{r} (1/3)^{(10-r)/2} (3/2)^r x^{(10-r)/2 - 2r}$
For the term to be independent of $x$,the exponent of $x$ must be $0$:
$(10-r)/2 - 2r = 0$
$5 - r/2 - 2r = 0$
$5 = 5r/2 \Rightarrow r = 2$
Substituting $r = 2$ into the expression:
$T_{3} = ^{10}C_{2} (1/3)^{(10-2)/2} (3/2)^2$
$T_{3} = 45 \times (1/3)^4 \times (9/4)$
$T_{3} = 45 \times (1/81) \times (9/4) = 45/36 = 5/4$.

(D) We know that $\sin 108^\circ = \sin(180^\circ - 72^\circ) = \sin 72^\circ$ and $\sin 144^\circ = \sin(180^\circ - 36^\circ) = \sin 36^\circ$.
So,the expression becomes $\sin^2 36^\circ \sin^2 72^\circ$.
Using the identity $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$,we have:
$\sin^2 36^\circ \sin^2 72^\circ = \left(\frac{1 - \cos 72^\circ}{2}\right) \left(\frac{1 - \cos 144^\circ}{2}\right)$
$= \frac{1}{4} (1 - \cos 72^\circ)(1 - \cos 144^\circ)$
$= \frac{1}{4} (1 - \sin 18^\circ)(1 + \cos 36^\circ)$
Substituting the values $\sin 18^\circ = \frac{\sqrt{5}-1}{4}$ and $\cos 36^\circ = \frac{\sqrt{5}+1}{4}$:
$= \frac{1}{4} \left(1 - \frac{\sqrt{5}-1}{4}\right) \left(1 + \frac{\sqrt{5}+1}{4}\right)$
$= \frac{1}{4} \left(\frac{5-\sqrt{5}}{4}\right) \left(\frac{5+\sqrt{5}}{4}\right)$
$= \frac{1}{4} \left(\frac{25-5}{16}\right) = \frac{1}{4} \times \frac{20}{16} = \frac{5}{16}$.

$(B)$ Let the height of the aeroplane be $H = 1 \ km$. Let the distance covered by the aeroplane in $10$ seconds be $d$.
From the geometry of the problem,we have:
$d = H \cot(30^\circ) - H \cot(60^\circ)$
$d = 1 \times \sqrt{3} - 1 \times \frac{1}{\sqrt{3}} = \sqrt{3} - \frac{1}{\sqrt{3}} = \frac{3-1}{\sqrt{3}} = \frac{2}{\sqrt{3}} \ km$.
Time taken $t = 10 \ seconds = \frac{10}{3600} \ hours = \frac{1}{360} \ hours$.
Speed $= \frac{\text{Distance}}{\text{Time}} = \frac{2/\sqrt{3}}{1/360} = \frac{2}{\sqrt{3}} \times 360 = \frac{720}{\sqrt{3}} = 240\sqrt{3} \ km/h$.

(D) Let $H = 150 \, m$ be the height of the second chimney.
Let $d$ be the distance between the two chimneys.
From the geometry of the problem,the angle of depression to the foot of the first chimney is $\phi$,so $\tan \phi = \frac{H}{d}$.
Given $\tan \phi = \frac{5}{2}$,we have $\frac{150}{d} = \frac{5}{2}$,which gives $d = \frac{150 \times 2}{5} = 60 \, m$.
Let $h$ be the height of the first chimney. The angle of depression to the top of the first chimney is $\theta$,so $\tan \theta = \frac{H - h}{d}$.
Given $\tan \theta = \frac{4}{3}$,we have $\frac{150 - h}{60} = \frac{4}{3}$.
$150 - h = 60 \times \frac{4}{3} = 80$.
$h = 150 - 80 = 70 \, m$.
The vertical distance between the tops is $H - h = 150 - 70 = 80 \, m$.
The horizontal distance between the tops is $d = 60 \, m$.
The distance between the tops is $\sqrt{(H - h)^2 + d^2} = \sqrt{80^2 + 60^2} = \sqrt{6400 + 3600} = \sqrt{10000} = 100 \, m$.

(B) Let the line $x + y - 4 = 0$ divide the line segment joining $A(-1, 1)$ and $B(5, 7)$ in the ratio $k : 1$.
Using the section formula,the coordinates of the point of division are $\left( \frac{5k - 1}{k + 1}, \frac{7k + 1}{k + 1} \right)$.
Since this point lies on the line $x + y = 4$,we have:
$\frac{5k - 1}{k + 1} + \frac{7k + 1}{k + 1} = 4$
$5k - 1 + 7k + 1 = 4(k + 1)$
$12k = 4k + 4$
$8k = 4$
$k = \frac{4}{8} = \frac{1}{2}$.
Thus,the ratio is $1 : 2$.

(B) The slope of the line is $m = \frac{3}{4} = \tan \theta$. Thus,$\cos \theta = \frac{4}{5}$ and $\sin \theta = \frac{3}{5}$.
The coordinates of points at a distance $r = 5$ from $A(x_1, y_1) = (3, 2)$ are given by $(x_1 \pm r \cos \theta, y_1 \pm r \sin \theta)$.
For the positive direction: $(3 + 5 \times \frac{4}{5}, 2 + 5 \times \frac{3}{5}) = (3 + 4, 2 + 3) = (7, 5)$.
For the negative direction: $(3 - 5 \times \frac{4}{5}, 2 - 5 \times \frac{3}{5}) = (3 - 4, 2 - 3) = (-1, -1)$.
Thus,the required points are $(7, 5)$ and $(-1, -1)$.