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  • Let <img alt="\mathrm{f(x)= \begin{cases}{[x]} & x \notin I \\ x-1 & x \in I\end{cases}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bf%28x%29%3D%20%5Cbegin%7Bcases

Let \mathrm{f(x)= \begin{cases}{[x]} & x \notin I \\ x-1 & x \in I\end{cases} , (where,[.] denotes the greatest integer function) and 

 \mathrm{g(x)= \begin{cases}\sin x+\cos x, & x<0 \\ 1, & x \geq 0\end{cases}}.Then for \mathrm{f(g(x)) \text { at } x=0}

Option: 1

\mathrm{\lim _{x \rightarrow 0} f(g(x))} exists but not continuous


Option: 2

continuous but not differentiable at x=0


Option: 3

Differentiable at x=0


Option: 4

\mathrm{\lim _{x \rightarrow 0} f(g(x))} does not exists


Answers (1)

best_answer

\mathrm{f(x)= \begin{cases}{[x]} & x \in I \\ x-1 & x \in I\end{cases}} (where,[.] denotes the greatest integer function) and \mathrm{g(x)=\left\{\begin{array}{cc} \sin x+\cos x, & x<0 \\ 1, & x \geq 0 \end{array}\right.},  Then for \mathrm{f(g(x))} at x=0

For continuity at \mathrm{x=0, f(g(0))=f(1)=1-1=0}

\mathrm{\begin{aligned} & f\left(g\left(0^{+}\right)\right)=f(1)=1-1=0 \\ & \left.f\left(g\left(0^{-}\right)\right)=f\left(\sin \left(0^{-}\right)+\cos \left(0^{-}\right)\right)=f\left(0^{-}\right)+\left(1^{-}\right)\right)=f\left(1^{-}\right)=\left[1^{-}\right]=0 \end{aligned}}

Thus \mathrm{f(x)} is continuous at x=0

For differentiability at x=0

\mathrm{ \begin{aligned} f^{\prime}\left(g\left(0^{+}\right)\right) & =\lim _{h \rightarrow 0} \frac{f(g(h))-f(g(0))}{h} \\ & =\lim _{h \rightarrow 0} \frac{0-0}{h}=0 \\ f^{\prime}\left(g\left(0^{-}\right)\right) & =\lim _{h \rightarrow 0} \frac{\left[1^{-}\right]-0}{-h}=0 \end{aligned}}

\mathrm{\therefore f(x) \: is \: \: differentiable \: \: at\: \: x=0.}

Posted by

Pankaj Sanodiya

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