IIT JEE 1964 Mathematics Question Paper with Answer and Solution

(C) Given expression: $1 + \cos 56^\circ + \cos 58^\circ - \cos 66^\circ $
Using the identity $1 + \cos 2\theta = 2 \cos^2 \theta$,we have $1 + \cos 56^\circ = 2 \cos^2 28^\circ$.
Now,the expression becomes $2 \cos^2 28^\circ + (\cos 58^\circ - \cos 66^\circ)$.
Using the identity $\cos C - \cos D = 2 \sin \frac{C+D}{2} \sin \frac{D-C}{2}$,we get $\cos 58^\circ - \cos 66^\circ = 2 \sin \frac{58^\circ + 66^\circ}{2} \sin \frac{66^\circ - 58^\circ}{2} = 2 \sin 62^\circ \sin 4^\circ$.
Since $\sin 62^\circ = \cos 28^\circ$,the expression is $2 \cos^2 28^\circ + 2 \cos 28^\circ \sin 4^\circ = 2 \cos 28^\circ (\cos 28^\circ + \sin 4^\circ)$.
Since $\sin 4^\circ = \cos 86^\circ$,we have $2 \cos 28^\circ (\cos 28^\circ + \cos 86^\circ)$.
Using $\cos C + \cos D = 2 \cos \frac{C+D}{2} \cos \frac{C-D}{2}$,we get $2 \cos 28^\circ [2 \cos \frac{28^\circ + 86^\circ}{2} \cos \frac{86^\circ - 28^\circ}{2}] = 2 \cos 28^\circ [2 \cos 57^\circ \cos 29^\circ]$.
Since $\cos 57^\circ = \sin 33^\circ$,the final result is $4 \cos 28^\circ \cos 29^\circ \sin 33^\circ$.

(B) Let the total height of the pole be $H$. The lower part is $\frac{H}{3}$ and the upper part is $\frac{2H}{3}$. Let the point be at a distance $d = 20 \, m$ from the base. Let $\alpha$ be the angle subtended by the lower part and $\beta$ be the angle subtended by the whole pole at the point. The angle subtended by the upper part is $\theta = \beta - \alpha$,where $\tan \theta = \frac{1}{2}$.
We have $\tan \alpha = \frac{H/3}{d} = \frac{H}{3d}$ and $\tan \beta = \frac{H}{d}$.
Using the formula $\tan(\beta - \alpha) = \frac{\tan \beta - \tan \alpha}{1 + \tan \beta \tan \alpha}$,we get:
$\frac{1}{2} = \frac{\frac{H}{d} - \frac{H}{3d}}{1 + (\frac{H}{d})(\frac{H}{3d})} = \frac{\frac{2H}{3d}}{1 + \frac{H^2}{3d^2}} = \frac{2Hd}{3d^2 + H^2}$.
Thus,$3d^2 + H^2 = 4Hd$,which simplifies to $H^2 - 4dH + 3d^2 = 0$.
Substituting $d = 20$,we get $H^2 - 4(20)H + 3(20^2) = 0$,so $H^2 - 80H + 1200 = 0$.
Factoring the quadratic equation: $(H - 20)(H - 60) = 0$.
Therefore,the possible heights of the pole are $H = 20 \, m$ and $H = 60 \, m$.

(B) The formula for the centroid $(G)$ of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by $G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)$.
Given vertices are $(2, 1)$,$(5, 2)$,and $(3, 4)$.
Calculating the $x$-coordinate: $x = \frac{2 + 5 + 3}{3} = \frac{10}{3}$.
Calculating the $y$-coordinate: $y = \frac{1 + 2 + 4}{3} = \frac{7}{3}$.
Thus,the centroid is $\left( \frac{10}{3}, \frac{7}{3} \right)$.

(A) Let the coordinates of the moving point $P$ be $(x, y)$.
The coordinates of the given points are $O(0, 0)$,$A(0, 4)$,and $B(6, 0)$.
The area of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ 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 of $\Delta POA = \frac{1}{2} |0(4 - y) + 0(y - 0) + x(0 - 4)| = \frac{1}{2} |-4x| = 2|x|$.
Area of $\Delta POB = \frac{1}{2} |0(0 - y) + 6(y - 0) + x(0 - 0)| = \frac{1}{2} |6y| = 3|y|$.
According to the problem,Area of $\Delta POA = 2 \times$ Area of $\Delta POB$.
Therefore,$2|x| = 2 \times 3|y|$.
$|x| = 3|y|$.
Squaring both sides,we get $x^2 = 9y^2$,which implies $x^2 - 9y^2 = 0$.
This can be factored as $(x - 3y)(x + 3y) = 0$.

(B) Let the circumcentre be $O(x, y)$ and the given vertices be $A(2, 1), B(5, 2),$ and $C(3, 4).$
Since the circumcentre is equidistant from all vertices,$OA^2 = OB^2 = OC^2.$
From $OA^2 = OB^2: (x - 2)^2 + (y - 1)^2 = (x - 5)^2 + (y - 2)^2.$
Expanding this: $x^2 - 4x + 4 + y^2 - 2y + 1 = x^2 - 10x + 25 + y^2 - 4y + 4.$
Simplifying: $6x + 2y = 24 \implies 3x + y = 12$......$(i)$
From $OA^2 = OC^2: (x - 2)^2 + (y - 1)^2 = (x - 3)^2 + (y - 4)^2.$
Expanding this: $x^2 - 4x + 4 + y^2 - 2y + 1 = x^2 - 6x + 9 + y^2 - 8y + 16.$
Simplifying: $2x + 6y = 20 \implies x + 3y = 10$......(ii)
Solving equations $(i)$ and (ii): Multiply $(i)$ by $3$: $9x + 3y = 36.$
Subtracting (ii) from this: $(9x - x) + (3y - 3y) = 36 - 10 \implies 8x = 26 \implies x = \frac{13}{4}.$
Substituting $x = \frac{13}{4}$ into $(i)$: $3(\frac{13}{4}) + y = 12 \implies y = 12 - \frac{39}{4} = \frac{48 - 39}{4} = \frac{9}{4}.$
Thus,the circumcentre is $\left( \frac{13}{4}, \frac{9}{4} \right).$