Question

Let $f\left(x\right)=\cos x$ and $H\left(x\right)=\{\begin{array}{c}Min\left[f\right(t)/0\le t\le x]for0\le x\le \frac{\pi }{2}\\ \frac{\pi }{2}-xfor\frac{\pi }{2}<x\le 3\end{array}$, then-

• A. $H\left(x\right)$ is continuous and derivable in $[0,3]$
• B. $H\left(x\right)$ is continuous but not derivable at $x=\pi /2$
• C. $H\left(x\right)$ is neither continuous nor derivable at $x=\pi /2$
• D. Maximum value of $H\left(x\right)$ in $[0,3]$ is $1$

Answer 1

Answer: (A), (D)

$H\left(x\right)$ is continuous and derivable in $[0,3]$
Maximum value of $H\left(x\right)$ in $[0,3]$ is $1$

$H\left(x\right)=\{\begin{array}{c}\cos x;0\le x\le \frac{\pi }{2}\\ \frac{\pi }{2}-x;\frac{\pi }{2}<x\le 3\end{array}$
$H^{\prime }\left(\frac{{\pi }^{-}}{2}\right)=-\sin x=-1$
$H^{\prime }\left(\frac{{\pi }^{+}}{2}\right)=-1$
Hence $H\left(x\right)$ is continuous and derivable in $[0,3]$ and has maximum value $1$ in $[0,3]$