Q66 / 120 · VITEEE 2011 · Mathematics

If $\mathrm{f}(x)=\left\{\begin{array}{ll}x^{2}, & x \leq 0 \\ 2 \sin x, x>0 & \text {, then } x=0 \text { is }\end{array}\right.$

  • Correct answer
    A
    point of minima
  • B
    point of maxima
  • C
    point of discontinuity
  • D
    None of the above

Solution

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.

Hence, the correct answer is option: 'A'.

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