A) point of minima
B) point of maxima
C) point of discontinuity
D) None of the above
Correct Answer: A
Solution :
Given, \[f(x)=\left\{ \begin{align} & \,\,\,\,\,\,\,\,\,{{x}^{2}},\,x\le 0 \\ & 2\sin x,\,x>0 \\ \end{align} \right.\] \[f(x)=\left\{ \begin{align} & \,\,\,\,\,\,\,\,\,2x,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\le 0 \\ & \text{non-differentiable,}\,\,x=0\,x=0\,\,\text{is} \\ & \,\,\,\,\,\,2\cos x,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x>0 \\ \end{align} \right.\] So, \[x=0\]is a critical point \[f({{0}^{-}})>0\]as well as \[f({{0}^{+}})>0\]and \[f(0)=0\] Hence, it is a point of minima.
You need to login to perform this action.
You will be redirected in
3 sec
