IIT JEE 1968 Mathematics Question Paper with Answer and Solution

(B) Given that $x, y, z$ are in $G.P.$,we have $y^2 = xz$.
Let $a^x = b^y = c^z = m$.
Taking logarithm on both sides,we get $x \log a = y \log b = z \log c = \log m$.
This implies $x = \frac{\log m}{\log a} = \log_a m$,$y = \log_b m$,and $z = \log_c m$.
Since $x, y, z$ are in $G.P.$,we have $\frac{y}{x} = \frac{z}{y}$.
Substituting the values,we get $\frac{\log_b m}{\log_a m} = \frac{\log_c m}{\log_b m}$.
Using the change of base formula $\frac{\log_b m}{\log_a m} = \log_b a$,we get $\log_b a = \log_c b$.

(A) Let $\alpha$ be a root of the first equation,then $\frac{1}{\alpha}$ is a root of the second equation.
For the first equation: $a\alpha^2 + b\alpha + c = 0$.
For the second equation: $a'(\frac{1}{\alpha})^2 + b'(\frac{1}{\alpha}) + c' = 0$,which simplifies to $c'\alpha^2 + b'\alpha + a' = 0$.
Using the cross-multiplication method for the system of equations $a\alpha^2 + b\alpha + c = 0$ and $c'\alpha^2 + b'\alpha + a' = 0$:
$\frac{\alpha^2}{ba' - b'c} = \frac{\alpha}{cc' - aa'} = \frac{1}{ab' - bc'}$.
From the second and third terms: $\alpha = \frac{cc' - aa'}{ab' - bc'}$.
From the first and third terms: $\alpha^2 = \frac{ba' - b'c}{ab' - bc'}$.
Equating $\alpha^2 = (\alpha)^2$,we get: $\frac{ba' - b'c}{ab' - bc'} = \left(\frac{cc' - aa'}{ab' - bc'}\right)^2$.
$(cc' - aa')^2 = (ba' - b'c)(ab' - bc')$.

(A) Let the roots of the quadratic equation be $\alpha$ and $\beta$.
The $A.M.$ (Arithmetic Mean) of the roots is given by $A = \frac{\alpha + \beta}{2}$,which implies $\alpha + \beta = 2A$.
The $G.M.$ (Geometric Mean) of the roots is given by $G = \sqrt{\alpha \beta}$,which implies $\alpha \beta = G^2$.
$A$ quadratic equation with roots $\alpha$ and $\beta$ is given by $t^2 - (\text{sum of roots})t + (\text{product of roots}) = 0$.
Substituting the values,we get $t^2 - (2A)t + G^2 = 0$.

(B) The total number of ways to select $5$ members from $10$ people ($6$ men + $4$ women) is given by $^{10}C_5 = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252$.
The number of ways to form a committee with no women (i.e.,all $5$ members are men) is given by $^6C_5 = 6$.
The number of ways to have at least one woman is the total number of ways minus the number of ways with no women:
$252 - 6 = 246$.

(B) The equation of the normal to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ at point $(a \sec \theta, b \tan \theta)$ is given by $ax \cos \theta + by \cot \theta = a^2 + b^2$.
For point $P$,the normal is $ax \cos \theta + by \cot \theta = a^2 + b^2$.
For point $Q$,the normal is $ax \cos \phi + by \cot \phi = a^2 + b^2$.
Given $\phi = \frac{\pi}{2} - \theta$,we have $\cos \phi = \sin \theta$ and $\cot \phi = \tan \theta$.
So,the second normal is $ax \sin \theta + by \tan \theta = a^2 + b^2$.
Subtracting the two equations: $ax(\cos \theta - \sin \theta) + by(\cot \theta - \tan \theta) = 0$.
$ax(\cos \theta - \sin \theta) + by\left(\frac{\cos^2 \theta - \sin^2 \theta}{\sin \theta \cos \theta}\right) = 0$.
$ax(\cos \theta - \sin \theta) + by\frac{(\cos \theta - \sin \theta)(\cos \theta + \sin \theta)}{\sin \theta \cos \theta} = 0$.
Since $\cos \theta \neq \sin \theta$,we divide by $(\cos \theta - \sin \theta)$ to get $ax + by\frac{\cos \theta + \sin \theta}{\sin \theta \cos \theta} = 0$,which implies $x = -y\frac{\cos \theta + \sin \theta}{a \sin \theta \cos \theta} \cdot \frac{b}{a}$ (simplified).
Solving the system for $k$ yields $k = -\left(\frac{a^2+b^2}{b}\right)$.