Question

Let the function be defined as follows: $f\left(x\right)=x^3+x^2-10x,-1\le x<0$
$\cos x,0\le x<\frac{\pi }{2}$
$1+\sin x,\frac{\pi }{2}\le x\le \pi .$ Then $f\left(x\right)$ has

• A. a local minimum at $x=\pi /2$
• B. a local maximum at $x=\pi /2$
• C. a local minimum at $x=-1$
• D. a local maximum at $x=\pi$

Answer 1

The correct option is B a local maximum at $x=\pi /2$

The function $f^{\prime }\left(x\right)$ is given by
$f^{\prime }\left(x\right)=\{\begin{array}{cc}3x^2+2x-10& \text{-}1\le x<0\\ -\sin x& \text{0}\le x<\pi /2\\ \cos x& \text{pi}/2\le x\le \pi \end{array}$
The function $f\left(x\right)$ is not differentiable at $x=0$, $x=\pi /2$
as $f^{\prime }(0-)=-10$, $f^{\prime }(0+)=0$; $f^{\prime }(\pi /2-)=-1$, $f^{\prime }(\pi /2+)=0$.
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, $f\left(x\right)$ does not have any extremum 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, $f\left(x\right)$ does not have any extremum at $x=\pi /2$