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

Option: 1

\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

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


Answers (1)

best_answer

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

For continuity at  \mathrm{x=0, f(g(0))=f(1)=1-1=0}
\mathrm{\left.f\left(0^{+}\right)\right)=f(1)=1-1=0}
\mathrm{\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}

\mathrm{Thus\, f\left ( x \right )\, is\, contineous\, at\, x= 0}
For differentiability at \mathrm{ x= 0}.

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

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

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Gaurav

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