Q. Let $f\left( x + \frac{1}{x} \right) = x^{2} + \frac{1}{x^{2}},\ x \neq 0$, then $f(x)$ is equal to
A) $x^{2}$
B) $x^{2} - 1$
C) $x^{2} - 2$
D) $x^{2} + 1$
Explanation
(c)
$f\left( x + \frac{1}{x} \right) = x^{2} + \frac{1}{x^{2}} = \left( x + \frac{1}{x} \right)^{2} - 2$
$\therefore\ \ \ \ \ f(x) = x^{2} - 2$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.