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

$\begin{array}{1 1}(a)\;\text{f(x) is continuous but not differentiable at x=0}\\(b)\;\text{f(x)is differentiable at x=0}\\(c)\;\text{f(x)is not differentiable at x=0}\\(d)\;\text{none of these}\end{array}$

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We have $f(x)=x(\sqrt x-\sqrt{x+1}$
Let us check differentiability of $f(x)$ at $x=0$
$Lf'(0)=\lim\limits_{h\to 0}\large\frac{(0-h)[\sqrt{0-h}-\sqrt{0-h+1}]}{-h}$
$\Rightarrow \lim\limits_{h\to 0}\big[\sqrt{-h}-\sqrt{-h+1}\big]$
$\Rightarrow 0-\sqrt 1=-1$
$Rf'(0)=\lim\limits_{h\to 0}\large\frac{(0+h)[\sqrt{0+h}-\sqrt{0+h+1}-0]}{h}$
$\Rightarrow \lim\limits_{h\to 0}\sqrt{h}-\sqrt{h+1}=-1$
Since $Lf'(0)=Rf'(0)$
$\therefore f$ is differentiable at $x=0$
Hence (b) is the correct answer.
answered Jan 3, 2014 by sreemathi.v

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