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  • The set of points of differentiability of the function <img alt="\mathrm{f(x)=\left(\begin{array}{r} \frac{\sqrt{x+1}-1}{\sqrt{x}}, \text { for } x \neq 0 \\ 0, \text { for } x=0 \end{array}

The set of points of differentiability of the function
\mathrm{f(x)=\left(\begin{array}{r} \frac{\sqrt{x+1}-1}{\sqrt{x}}, \text { for } x \neq 0 \\ 0, \text { for } x=0 \end{array}\right.}     is
 

Option: 1

R


Option: 2

[0, \infty)


Option: 3

(0, \infty)


Option: 4

\mathrm{R-|0|}


Answers (1)

best_answer

The given function is differentiable at all points in its domain except possibly at x=0.

Now, \mathrm{(RHD ~at ~x=0)=\lim _{h \rightarrow,} \frac{f(0+h)-f(0)}{h}=\lim _{h \rightarrow 0} \frac{\sqrt{h+1}-1}{h^{3 / 2}}}
\mathrm{ =\lim _{h \rightarrow 0 h h^{3 / 2}(\sqrt{h+1}+1)}=\lim _{h \rightarrow 0} \frac{1}{\sqrt{h}(\sqrt{h+1}+1)} \rightarrow \infty } 
So, the function is not differentiable at x=0. Hence, the requred set is  \mathrm{ (0, \infty) } 

Posted by

Divya Prakash Singh

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