Question

Let $f\left(x\right)=\{\begin{array}{c}5x+1,x\le 2\\ \underset{0}{\overset{x}{\int }}(5+|1-t|)dt,x>2\end{array}$
then which of the following statement(s) is /are incorrect?

• A. $f\left(x\right)$ is continuous but not differentiable at $x=2$
• B. $f\left(x\right)$ is not continuous at $x=2$
• C. $f\left(x\right)$ is differentiable for all $x\in R$
• D. $f\left(x\right)$ is differentiable for $x>2$

Answer 1

Answer: (B), (C)

$f\left(x\right)$ is differentiable for all $x\in R$

For $x>2,$
$f\left(x\right)=\underset{0}{\overset{x}{\int }}(5+|1-t|)dt=\underset{0}{\overset{1}{\int }}(5+|1-t|)dt+\underset{1}{\overset{x}{\int }}(5+|1-t|)dt$
$=\underset{0}{\overset{1}{\int }}(6-t)dt+\underset{1}{\overset{x}{\int }}(4+t)dt$
$={[6t-\frac{t^2}{2}]}_0^1+{[4t+\frac{t^2}{2}]}_1^x$
$=6-\frac{1}{2}+4x+\frac{x^2}{2}-4-\frac{1}{2}$
$=\frac{x^2}{2}+4x+1$

$\therefore f\left(x\right)=\{\begin{array}{c}5x+1,x\le 2\\ \frac{x^2}{2}+4x+1,x>2\end{array}$

At $x=2,f\left(2\right)=11$
$\underset{x\to 2^{-}}{lim}f\left(x\right)=11$
$\underset{x\to 2^{+}}{lim}f\left(x\right)=2+8+1=11$
So $f\left(x\right)$ is continuous at $x=2.$

$f^{\prime }\left(x\right)=\{\begin{array}{c}5,x\le 2\\ x+4,x>2\end{array}$

At $x=2$
$\underset{x\to 2^{-}}{lim}{f}^{\prime }\left(x\right)=5$
$\underset{x\to 2^{+}}{lim}{f}^{\prime }\left(x\right)=2+4=6$
$\underset{x\to 2^{-}}{lim}{f}^{\prime }\left(x\right)\ne \underset{x\to 2^{+}}{lim}{f}^{\prime }\left(x\right)$
So $f\left(x\right)$ is not differentiable at $x=2$
Therefore $f\left(x\right)$ is differentiable for all $x\in R-\left\{2\right\}$