IIT JEE 1967 Mathematics Question Paper with Answer and Solution

(B) The given equation is $\frac{1}{x + p} + \frac{1}{x + q} = \frac{1}{r}$.
Multiplying by $r(x + p)(x + q)$,we get $r(x + q) + r(x + p) = (x + p)(x + q)$.
$rx + rq + rx + rp = x^2 + px + qx + pq$.
$x^2 + x(p + q - 2r) + (pq - rp - rq) = 0$.
Let the roots be $\alpha$ and $-\alpha$. Since the sum of the roots is zero,the coefficient of $x$ must be zero:
$p + q - 2r = 0 \implies r = \frac{p + q}{2}$.
The product of the roots is $\alpha(-\alpha) = -\alpha^2 = pq - r(p + q)$.
Substituting $r = \frac{p + q}{2}$ into the product expression:
$-\alpha^2 = pq - \frac{p + q}{2}(p + q) = pq - \frac{(p + q)^2}{2}$.
$-\alpha^2 = \frac{2pq - (p^2 + q^2 + 2pq)}{2} = -\frac{p^2 + q^2}{2}$.

(A) We know the property of combinations: $^{n}C_{x} = ^{n}C_{y}$ implies either $x = y$ or $x + y = n$.
Given $^{15}C_{3r} = ^{15}C_{r+3}$.
Case $1$: $3r = r + 3$ $\Rightarrow 2r = 3$ $\Rightarrow r = 1.5$ (Not an integer,so reject).
Case $2$: $3r + (r + 3) = 15$ $\Rightarrow 4r + 3 = 15$ $\Rightarrow 4r = 12$ $\Rightarrow r = 3$.
Thus,the value of $r$ is $3$.

(A) The general term in the expansion of $(ax - \frac{1}{bx^2})^{11}$ is given by $T_{r+1} = ^{11}C_r (ax)^{11-r} (-\frac{1}{bx^2})^r$.
$T_{r+1} = ^{11}C_r a^{11-r} x^{11-r} (-1)^r b^{-r} x^{-2r}$
$T_{r+1} = ^{11}C_r a^{11-r} (-1)^r b^{-r} x^{11-3r}$
To find the coefficient of $x^{-7}$,we set the exponent of $x$ equal to $-7$:
$11 - 3r = -7$
$3r = 18 \Rightarrow r = 6$
Substituting $r = 6$ into the expression for the coefficient:
Coefficient $= ^{11}C_6 a^{11-6} (-1)^6 b^{-6}$
$= ^{11}C_6 a^5 \frac{1}{b^6}$
Since $^{11}C_6 = ^{11}C_5 = \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} = 462$
Therefore,the coefficient is $\frac{462a^5}{b^6}$.

(B) We are given $\tan A = -\frac{1}{2}$ and $\tan B = -\frac{1}{3}.$
Using the formula for the tangent of the sum of two angles:
$\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$
Substituting the values:
$\tan(A + B) = \frac{-\frac{1}{2} - \frac{1}{3}}{1 - (-\frac{1}{2})(-\frac{1}{3})}$
$\tan(A + B) = \frac{-\frac{5}{6}}{1 - \frac{1}{6}} = \frac{-\frac{5}{6}}{\frac{5}{6}} = -1$
Since $\tan(A + B) = -1,$ and knowing that $\tan(\frac{3\pi}{4}) = -1,$
$A + B = \frac{3\pi}{4}.$

(C) Let the height of the tower be $h$ and the distance from the tower to the first point be $x$.
From the right-angled triangle,we have:
$\tan(60^\circ) = \frac{h}{OA}$ $\Rightarrow \sqrt{3} = \frac{h}{OA}$ $\Rightarrow OA = \frac{h}{\sqrt{3}}$.
$\tan(30^\circ) = \frac{h}{OB}$ $\Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{20 + OA}$ $\Rightarrow 20 + OA = h\sqrt{3}$.
Substituting $OA = \frac{h}{\sqrt{3}}$ into the equation:
$20 + \frac{h}{\sqrt{3}} = h\sqrt{3}$
$20 = h\sqrt{3} - \frac{h}{\sqrt{3}}$
$20 = h \left( \frac{3 - 1}{\sqrt{3}} \right) = h \left( \frac{2}{\sqrt{3}} \right)$
$h = \frac{20 \times \sqrt{3}}{2} = 10\sqrt{3} \, m$.

(B) Let the vertices be $A(0, 0)$,$B(2, -1)$,and $C(1, 3)$.
$1$. Find the equation of altitude from $A$ to $BC$:
Slope of $BC = \frac{3 - (-1)}{1 - 2} = \frac{4}{-1} = -4$.
The slope of the altitude $AD$ is the negative reciprocal of the slope of $BC$,which is $\frac{1}{4}$.
The equation of $AD$ passing through $A(0, 0)$ is $y - 0 = \frac{1}{4}(x - 0)$,which simplifies to $x - 4y = 0$ ..... $(i)$.
$2$. Find the equation of altitude from $B$ to $AC$:
Slope of $AC = \frac{3 - 0}{1 - 0} = 3$.
The slope of the altitude $BE$ is the negative reciprocal of the slope of $AC$,which is $-\frac{1}{3}$.
The equation of $BE$ passing through $B(2, -1)$ is $y - (-1) = -\frac{1}{3}(x - 2)$,which simplifies to $3(y + 1) = -(x - 2)$ $\Rightarrow 3y + 3 = -x + 2$ $\Rightarrow x + 3y + 1 = 0$ ..... $(ii)$.
$3$. Solve the system of equations $(i)$ and $(ii)$:
From $(i)$,$x = 4y$. Substitute this into $(ii)$:
$4y + 3y + 1 = 0$ $\Rightarrow 7y = -1$ $\Rightarrow y = -\frac{1}{7}$.
Then $x = 4(-\frac{1}{7}) = -\frac{4}{7}$.
Thus,the orthocentre is $\left( -\frac{4}{7}, -\frac{1}{7} \right)$.

(C) The coordinates of point $A$ using the section formula are $\left( \frac{3k - 5}{k + 1}, \frac{5k + 1}{k + 1} \right)$.
Given $B = (1, 5)$ and $C = (7, -2)$,the area of $\triangle ABC$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 2$.
Substituting the coordinates: $\frac{1}{2} \left| \frac{3k - 5}{k + 1}(5 - (-2)) + 1(-2 - \frac{5k + 1}{k + 1}) + 7(\frac{5k + 1}{k + 1} - 5) \right| = 2$.
$\frac{1}{2} \left| \frac{7(3k - 5) - (2k + 2 + 5k + 1) + 7(5k + 1 - 5k - 5)}{k + 1} \right| = 2$.
$\frac{1}{2} \left| \frac{21k - 35 - 7k - 3 - 28}{k + 1} \right| = 2$.
$|14k - 66| = 4|k + 1|$.
Case $1$: $14k - 66 = 4k + 4$ $\Rightarrow 10k = 70$ $\Rightarrow k = 7$.
Case $2$: $14k - 66 = -4k - 4$ $\Rightarrow 18k = 62$ $\Rightarrow k = 31/9$.