Question

Let $f:[a,b]\to [1,\infty )$ be a continuous function and let $g:R\to R$ be defined as
$g\left(x\right)=\{\begin{array}{c}0\text{if}x<a,\\ \underset{a}{\overset{x}{\int }}f\left(t\right)dt\text{if}a\le x\le b,\\ \underset{a}{\overset{b}{\int }}f\left(t\right)dt\text{if}x>b.\end{array}$
Then

• A. $g\left(x\right)$ is continuous but not differentiable at $a$
• B. $g\left(x\right)$ is differentiable on $R$
• C. $g\left(x\right)$ is continuous but not differentiable at $b$
• D. $g\left(x\right)$ is continuous and differentiable at either $a$ or $b$ but not both

Answer 1

Answer: (A), (C)

$g\left(x\right)$ is continuous but not differentiable at $b$

$g\left(x\right)=\{\begin{array}{c}0\text{if}x<a,\\ \underset{a}{\overset{x}{\int }}f\left(t\right)dt\text{if}a\le x\le b,\\ \underset{a}{\overset{b}{\int }}f\left(t\right)dt\text{if}x>b.\end{array}$

$\text{LHL}=\underset{x\to a^{-}}{lim}g\left(x\right)=0$

$\text{RHL}=\underset{x\to a^{+}}{lim}g\left(x\right)=\underset{x\to a^{+}}{lim}\underset{a}{\overset{x}{\int }}f\left(t\right)dt$
$=\underset{a}{\overset{a}{\int }}f\left(t\right)dt=0$

$g\left(a\right)=\underset{a}{\overset{a}{\int }}f\left(t\right)dt=0$
Hence, $g\left(x\right)$ is continuous at $x=a.$
Similarly, $g\left(x\right)$ is continuous at $x=b.$

Differentiating $g\left(x\right)$ w.r.t. $x$,
$g^{\prime }\left(x\right)=\{\begin{array}{c}0\text{if}x<a,\\ f\left(x\right)\text{if}a\le x\le b,\\ 0\text{if}x>b\end{array}$
$g^{\prime }\left(a^{-}\right)=0g^{\prime }\left(a^{+}\right)=f\left(a\right)g^{\prime }\left(a^{-}\right)\ne g^{\prime }\left(a^{+}\right)$

$g^{\prime }\left(b^{-}\right)=f\left(b\right)g^{\prime }\left(b^{+}\right)=0g^{\prime }\left(b^{-}\right)\ne g^{\prime }\left(b^{+}\right)$

Hence, $g\left(x\right)$ is non differentiable at $a$ and $b.$