Question

Let $f\left(x\right)=\left\{\begin{array}{cc}\underset{0}{\overset{x}{\int }}(5+|1-t|)dt,& x>2\\ 5x+1,& x\le 2\end{array}\right\}$
Which of the following is true for $f\left(x\right)$?

• A. The right hand derivative of $f\left(x\right)$ at $x=2$ doesn’t exist
• B. $f\left(x\right)$ is continuous at $x=2$
• C. $f\left(x\right)$ is continuous but not differentiable at $x=2$
• D. $f\left(x\right)$ is everywhere differentiable

Answer 1

Answer: (B), (C)

$f\left(x\right)$ is continuous but not differentiable at $x=2$

$f\left(x\right)=\left\{\begin{array}{cc}\underset{0}{\overset{x}{\int }}(5+|1-t|)dt,& x>2\\ 5x+1,& x\le 2\end{array}\right\}$
For $x>2$
$f\left(x\right)=\underset{0}{\overset{x}{\int }}(5+|1-t|)dt=\underset{0}{\overset{1}{\int }}5+(1-t)dt+\underset{1}{\overset{x}{\int }}(5-(1-t\left)\right)dt$
$=\underset{0}{\overset{1}{\int }}(6-t)dt+\underset{1}{\overset{x}{\int }}(4+t)dt$
$=1+4x+\frac{x^2}{2}$
$\therefore f\left(x\right)=\left\{\begin{array}{cc}\frac{x^2}{2}+4x+1,& x>2\\ 5x+1,& x\le 2\end{array}\right\}$
$\underset{x\to 2^{-}}{lim}f\left(x\right)\underset{x\to 2^{-}}{lim}(5x+1)=11$,
$\underset{x\to 2^{+}}{lim}f\left(x\right)=\underset{x\to 2^{+}}{lim}(\frac{x^2}{2}+4x+1)=11$
$f\left(2\right)=11$
$\therefore f\left(x\right)$ is continuous at $x=2$
Hence $f\left(x\right)$ is everywhere continuous.
$f^{\prime }\left(x\right)=\left\{\begin{array}{cc}x+4,& x>2\\ 5,& x\le 2\end{array}\right\}$
At $x=2$
L.H.D. $=5$, R.H.D. $=6$
$\therefore f\left(x\right)$ is not differentiable at $x=2$