Question

The absolute minimum value of $f\left(x\right)=\frac{x-2}{\sqrt{x^2+1}}$ is

• A. $\sqrt{5}$
• B. 1
• C. $-\sqrt{5}$
• D. $\frac{1}{2}$

Answer 1

Answer: (C)

$-\sqrt{5}$

Given $f\left(x\right)=\frac{x-2}{\sqrt{x^2+1}}$
$\therefore$ $f^{\prime }\left(x\right)=(\sqrt{x^1+1}\left(1\right)-\frac{(x-2)\left(2x\right)}{2\sqrt{x^2+1}})\frac{1}{x^2+1}=\frac{1+2x}{{(1+x^2)}^{\frac{3}{2}}}$
$\therefore$ $f\left(x\right)$ is an increasing function for $xϵ(-\frac{1}{2},\infty )$ & decreasing function for $xϵ(-\infty ,-\frac{1}{2})$
$\therefore$ at $x=\frac{-1}{2}$, $f\left(x\right)$ attains absolute minimum value
$\therefore$ $f(-\frac{1}{2})=\frac{-\frac{1}{2}-2}{\sqrt{\frac{5}{4}}}=-\sqrt{5}$
Ans: C