Q. Let $f(x) = \frac{\propto x^{2}}{x + 1},\ x \neq - 1$. The value of $\propto$ for which $f(a) = a,\ (a \neq 0)$ is
A) $1 - \frac{1}{a}$
B) $\frac{1}{a}$
C) $1 + \frac{1}{a}$
D) $\frac{1}{a} - 1$
Explanation
(c)
$\ \ \ \ \ \ \ \ \ \ \ f(a) = a$
$\Rightarrow \ \ \ \ \frac{\alpha a^{2}}{\alpha + 1} = a$
$\Rightarrow \ \ \ \ \alpha a^{2} = a^{2} + a$
$\Rightarrow \ \ \ \ \ \ \alpha = 1 + \frac{1}{a}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because a \neq 0)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.