Question

Let $f\left(x\right)=\{\begin{array}{cc}{x}^3+x^2-10x& \text{-}1\le x<0\\ \sin x& \text{0}\le x<\pi /2\\ 1+\cos x& \text{pi}/2\le x\le \pi \end{array}$
then $f\left(x\right)$ has

• A. local minima at $x=\pi /2$
• B. local minima at $x=0$
• C. absolute maxima at $x=0$
• D. absolute maxima at $x=\pi /2$

Answer 1

Answer: (A), (C)

The correct options are
A local minima at $x=\pi /2$
C absolute maxima at $x=0$

The function $f^{\prime }\left(x\right)$ is given by
$f^{\prime }\left(x\right)=\{\begin{array}{cc}3x^2+2x-10& \text{-}1\le x<0\\ \cos x& \text{0}\le x<\pi /2\\ -\sin x& \text{pi}/2\le x\le \pi \end{array}$
The function $f\left(x\right)$ is not differentiable at $x=0$, $x=\pi /2$
as $f^{\prime }(0-)=-10$, $f^{\prime }(0+)=1$; $f^{\prime }(\pi /2-)=0$, $f^{\prime }(\pi /2+)=-1$.
The critical points of $f$ are given by $f^{\prime }\left(x\right)=0$ or $x=0$, $\pi /2$.
Since $f^{\prime }\left(x\right)<0$ for $-1\le x\le 0$ and $f^{\prime }\left(x\right)>0$ for $0\le x<\pi /2$
Therefore, $f\left(x\right)$ has local maximum at $x=0$
Also $f^{\prime }\left(x\right)>0$ for $0\le x<\pi /2$ and $f^{\prime }\left(x\right)<0$ for $\pi /2\le x\le \pi$
Therefore, $f\left(x\right)$ has local minimum at $x=\pi /2$
Ans: A,C