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}' \left(0\right) =\displaystyle \lim_{h \to 0} \frac{\left(0-h\right)\left[\sqrt{0-h} - \sqrt{0-h+1}\right] - 0}{-h} $
$=\displaystyle \lim_{h \to 0} \left[\sqrt{-h} - \sqrt{-h+1}\right]$
$ = 0- \sqrt{1} = - 1$
$ Rf' \left(0\right) =\displaystyle \lim_{h \to 0} \frac{\left(0+h\right)\left[\sqrt{0+h} - \sqrt{0+h +1}\right] - 0}{h}$
$ =\displaystyle \lim_{h \to 0} \sqrt{h} - \sqrt{h+1} = - 1$
Since $ Lf'\left(0\right) = Rf'\left(0\right)$
$ \therefore f $ is differentiable at $x = 0$.