Question

The function $f\left(x\right)=-1+\frac{x^2}{2}+\cos x$ has , where $x\in [-\frac{\pi }{2},\frac{\pi }{2}]$

• A. The minimum value $-2$
• B. No extremum at $x=0$
• C. The minimum value $0$
• D. None of these

Answer 1

The correct option is C The minimum value $0$

$f^{\prime }\left(x\right)=-\sin x+x=0\Rightarrow x=0$ in $[-\frac{\pi }{2},\frac{\pi }{2}]$
$f\left(0\right)=0$
$f(-\frac{\pi }{2})=\frac{{\pi }^2}{8}-1$ and $f\left(\frac{\pi }{2}\right)=\frac{{\pi }^2}{8}-1$
$\Rightarrow$at $x=0$, It will have a minimum.