Question

Let the function $f\left(x\right)$ be defined as follows:$f\left(x\right)=\{\begin{array}{c}{x}^3+x^2-10x-1\le x<0\\ \cos x0\le x<\frac{\pi }{2}\\ 1+\sin x\frac{\pi }{2}\le x\le \pi \end{array}$ , then which of the following statement(s) is/are correct

• A. Local maximum at $x=0$
• B. Local maximum at $x=\frac{\pi }{2}$
• C. Absolute maxima at $x=-1$
• D. Absolute minima at $x=\pi$

Answer 1

Answer: (B), (C)

Local maximum at $x=\frac{\pi }{2}$
Absolute maxima at $x=-1$

The function $f^{\prime }\left(x\right)$ is given by
$f^{\prime }\left(x\right)=\{\begin{array}{cc}3x^2+2x-10& \text{-}1\le x<0\\ -\sin x& \text{0}\le x<\pi /2\\ \cos x& \pi /2\le x\le \pi \end{array}$
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 $x=\pi /2$
Since, $f^{\prime }\left(x\right)<0$ for $x\in [-1.\pi ]$
Therefore, $f\left(x\right)$ have absolute maximum at $x=-1$ and absolute minimum at $x=\pi$
Ans: C,D