Question

Let $f\left(x\right)=\{\begin{array}{c}\begin{array}{cc}{\int }_0^x(5+|1-t|)dt,& x>2\end{array}\\ \begin{array}{cc}5x+1,& x\le 2\end{array}\end{array}$, then at $x=2$

• A. $f\left(x\right)$ is continuous
• B. $f\left(x\right)$ is not continuous
• C. $f\left(x\right)$ is differentiable
• D. $f\left(x\right)$ is not differentiable

Answer 1

Answer: (A), (D)

$f\left(x\right)$ is continuous
$f\left(x\right)$ is not differentiable

We have, for $x>2$

$f\left(x\right)={\int }_0^5\{5+|1-t|\}dt$$={\int }_0^1\{5+1-t\}dt+{\int }_1^x\{5+t-1\}dt$ [Since $x>2$]

$={(6t-\frac{t^2}{2})}_0^1+{(4t+\frac{t^2}{2})}_1^x$$=6-\frac{1}{2}+4x+\frac{x^2}{2}-4-\frac{1}{2}$$=1+4x+\frac{x^2}{2}$

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

We have, $R{f}^{\prime }\left(2\right)=\underset{h\to 0}{lim}\frac{f(2+h)-h\left(2\right)}{h}$

$=\underset{h\to 0}{lim}\frac{1+4(2+h)+\frac{{(2+h)}^2}{2}-11}{h}$$=\underset{h\to 0}{lim}\frac{11+6h+\frac{h^2}{2}-11}{h}=6$

We have, $L{f}^{\prime }\left(2\right)=\underset{h\to 0}{lim}\frac{f(2-h)-f\left(2\right)}{-h}$

$=\underset{h\to 0}{lim}\frac{5(2-h)-1-11}{h}$$=\underset{h\to 0}{lim}\frac{11-5h-11}{h}=-5$

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

Since $R{f}^{\prime }\left(2\right)$ and $L{f}^{\prime }\left(2\right)$ are finite, therefore $f\left(x\right)$ is contnuous at $x=2$