Question

If $f(x)=x(\sqrt{x}-\sqrt{x+1})$ , then-

• A. f(x) is continuous but not differentiable at x = 0
• B. f(x) is differentiable at x = 0
• C. f(x) is not differentiable at x = 0
• D. none of these

Answer 1

Answer: (B)

We have $f(x)=x(\sqrt{x}-\sqrt{x+1})$
Let us check differentiability of f (x) at x = 0
${\text{Lf}}^{\prime }\left(0\right)=\underset{h\to 0}{\lim }\frac{\left(0-h\right)\left[\sqrt{0-h}-\sqrt{0-h+1}\right]-0}{-h}$
$=\underset{h\to 0}{\lim }\left[\sqrt{-h}-\sqrt{-h+1}\right]$
$=0-\sqrt{1}=-1$
$R{f}^{\prime }\left(0\right)=\underset{h\to 0}{\lim }\frac{\left(0+h\right)\left[\sqrt{0+h}-\sqrt{0+h+1}\right]-0}{h}$
$=\underset{h\to 0}{\lim }\sqrt{h}-\sqrt{h+1}=-1$
Since $L{f}^{\prime }\left(0\right)=R{f}^{\prime }\left(0\right)$
$\therefore f$ is differentiable at $x=0$.