A) f(x) is continuous but not differentiable at \[x=0\]
B) f(x) is not differentiable at \[x=0\]
C) f(x) is differentiable at \[x=0\]
D) None of these
Correct Answer: C
Solution :
Since, given function \[\operatorname{f}(x) =\,\,x\left[ \sqrt{x}-\sqrt{x+1} \right]\] is algebraic and every algebraic function is continuous. \[\therefore \,\, f(x) is continuous at x = 0.\] Now \[L.H.D.=Lf'(0)=\underset{h\to {{0}^{-}}}{\mathop{lim}}\,\frac{f(0-h)-f(0)}{0-h-0}\] \[=\,\,\underset{h\to 0}{\mathop{\lim }}\,\frac{-h\left[ \sqrt{-h}-\sqrt{-h+1} \right]}{-h}=-1\] \[R.H.D.=Rf'(0)=\underset{h\to {{0}^{+}}}{\mathop{lim}}\,\frac{f(0+h)-f(0)}{0+h-0}\] \[=\,\,\underset{h\to \,0}{\mathop{\lim }}\,\frac{h\left[ \sqrt{h}-\sqrt{h+1} \right]}{h}=-1\] \[\because \,\,\,Lf'(0)=Rf'(0)\] \[\therefore \,\,\, the function is differentiable at x = 0\]
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