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If $\mathrm{f}(x)=\left\{\begin{array}{ll}x^{2}, & x \leq 0 \\ 2 \sin x, x>0 & \text {, then } x=0 \text { is }\end{array}\right.$
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The correct answer is:
point of minima
Given $f(x)=\left\{\begin{array}{rc}x^{2}, & x \leq 0 \\ 2 \sin x, & x>0,\end{array}\right.$
$$
f^{\prime}(x)=\left\{\begin{array}{cc}
2 x, & x < 0 \\
\text { nondifferentiable, } & x=0 \\
2 \cos x, & x>0
\end{array}\right.
$$
So, $x=0$ is a critical point $f\left(0^{-}\right)>0$ as well as $\mathrm{f}\left(0^{+}\right)>0$ and $\mathrm{f}(0)=0$
Hence, it is a point of minima.
$$
f^{\prime}(x)=\left\{\begin{array}{cc}
2 x, & x < 0 \\
\text { nondifferentiable, } & x=0 \\
2 \cos x, & x>0
\end{array}\right.
$$
So, $x=0$ is a critical point $f\left(0^{-}\right)>0$ as well as $\mathrm{f}\left(0^{+}\right)>0$ and $\mathrm{f}(0)=0$
Hence, it is a point of minima.
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