Q. The real number $x$ when added to its inverse gives the minimum value of the sum at $x$ which is equal to
A) $- 2$
B) 2
C) 1
D) $- 1$
Explanation
(c)
Let $A = x + \frac{1}{x}\ \Rightarrow \frac{dA}{dx} = 1 - \frac{1}{x^{2}}$
Put $\frac{dA}{dx} = 0$ for maxima or minima
$1 - \frac{1}{x^{2}} = 0\ \ \Rightarrow \ \ x = - 1,\ 1$
Again on differentiating w.r.t. $x$, we get
$\frac{d^{2}A}{dx^{2}} = \frac{2}{x^{3}} \Rightarrow \ \ \left( \frac{d^{2}A}{dx^{2}} \right)_{x = 1} = 2 > 0$
$\therefore A$ is minimum at $x = 1$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.