Question

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

• A. local maxima at x = $\frac{\pi }{2}$
• B. local minima at x = $\frac{\pi }{2}$
• C. absolute maxima at x = -1
• D. absolute maxima at $x=\frac{\pi }{2}$

Answer 1

Answer: (A), (C)

The correct options are
A local maxima at x = $\frac{\pi }{2}$
C absolute maxima at x = -1

The function $f$ is not differentiable at x = 0,

$x=\frac{\pi }{2}as{f}^{\prime }\left(0^{-}\right)=-10,f^{\prime }\left(0^{+}\right)=1;$

$f^{\prime }\left({\frac{\pi }{2}}^{-}\right)=0,f^{\prime }\left({\frac{\pi }{2}}^{+}\right)=-1.\text{The function}$

$f^{\prime }\left(x\right)\text{is given by}$

$f^{\prime }\left(x\right)=\{\begin{array}{cc}3x^2+2x-10;& -1\le x<0\\ \cos x;& 0<x<\frac{\pi }{2}\\ -\sin x;& \frac{\pi }{2}<x\le \pi \end{array}$

The critical points of $f$ are given by

$f^{\prime }\left(x\right)=0$ or $x=0,\frac{\pi }{2},\pi$.

Since $f^{\prime }\left(x\right)>0$, for $0<x<\frac{\pi }{2}$ and $f^{\prime }\left(x\right)<0$, for

$\frac{\pi }{2}$ < x < $\pi$ so f has local maxima at
$x=\frac{\pi }{2}$.

Also $f^{\prime }\left(x\right)<0$ for $-1\le x<0$ and $f^{\prime }\left(x\right)>0$ for
$0<x<\frac{\pi }{2}$ so f has local minima at x = 0.

Since $f(-1)=-10,f\left(\frac{\pi }{2}\right)=1,f\left(0\right)=0$ and $f\left(\pi \right)=0$.

Thus f has absolute maximum at $x=-1$ and absolute minimum at $x=0$