Question

Consider a real-valued continuous function $f\left(x\right)$ defined on the interval $[a,b].$ Which of the following statements does NOT hold good?

• A. If $f\left(x\right)\ge 0$ on $[a,b],$ then $\underset{a}{\overset{b}{\int }}f\left(x\right)dx\le \underset{a}{\overset{b}{\int }}{\left(f\right(x\left)\right)}^{2}dx$
• B. If $f\left(x\right)$ attains a minimum at $x=c$ where $a<c<b,$ then $f^{\prime }\left(c\right)=0$
• C. If $f\left(x\right)$ is increasing on $[a,b],$ then $f\left(x\right)\ge 0$ on $(a,b)$
• D. If $f\left(x\right)$ is increasing on $[a,b],$ then $\left(f\right(x){)}^2$ is increasing on $[a,b]$

Answer 1

Answer: (A), (B), (C), (D)

If $f\left(x\right)$ is increasing on $[a,b],$ then $\left(f\right(x){)}^2$ is increasing on $[a,b]$

If $0<f\left(x\right)<1,$
$\Rightarrow f^2\left(x\right)<f\left(x\right)$
$\Rightarrow \underset{a}{\overset{b}{\int }}f\left(x\right)dx>\underset{a}{\overset{b}{\int }}{f}^2\left(x\right)dx$

If $f\left(x\right)<0,$ $\frac{d}{dx}\left(f^2\right(x\left)\right)=2f\left(x\right)f^{\prime }\left(x\right)<0$ when $f^{\prime }\left(x\right)>0$ and so $f^2\left(x\right)$ is decreasing while $f\left(x\right)$ is increasing.

A function can be negative and increasing.
e.g. $f\left(x\right)=-\frac{1}{x},x\in [2,4]$

A function may not be differentiable at $x=c$ for which it attains its minimum. It may have a sharp edge at $x=c.$
e.g. $f\left(x\right)=|x-2|,x\in [1,3]$
Here, $f\left(x\right)$ takes minimum at $x=2\in (1,3)$ but $f^{\prime }\left(c\right)$ is not defined.