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 the above
Correct Answer: C
Solution :
Key Idea:\[f(x)\]is differentiable function at\[x=a\]if\[\underset{x\to a}{\mathop{\lim }}\,\frac{F(x)-f(a)}{x-a}\]exist. Given, \[f(x)=x[\sqrt{x}-\sqrt{x+1}]\] Since, given function is algebraic and every algebraic function is continuous. \[\therefore \]\[f(x)\]is continuous at\[x=0\]. Now we will check differentiability of\[f(x)\]at \[x=0\]. \[LF(0)=\underset{h\to {{0}^{-}}}{\mathop{\lim }}\,\frac{f(0-h)-f(0)}{0-h-0}\] \[=\underset{h\to {{0}^{-}}}{\mathop{\lim }}\,\frac{-h[\sqrt{-h}-\sqrt{-h+1}]}{-h}=-1\] \[RF(0)=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\frac{f(0+h)-f(0)}{0+h-0}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{h(\sqrt{h}-\sqrt{h+1})}{h}=-1\] \[\therefore \]\[LF(0)=RF(0)\] \[\therefore \]The function is differentiable at\[x=0\]. Note: Every differentiable functions are continuous but every continuous functions are not differentiable.
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