Question

Let $f\left(x\right)=\{\begin{array}{c}{\int }_0^x|1-t|dt,x>1\\ x-\frac{1}{2},x\le 1\end{array}$ Then.

• A. $f\left(x\right)$ is continuous at $x=1$
• B. $f\left(x\right)$ is not continuous at $x=1$
• C. $f\left(x\right)$ is differentiable at $x=1$
• D. $f\left(x\right)$ is not differentiable at $x=1$

Answer 1

Answer: (A), (D)

$f\left(x\right)$ is not differentiable at $x=1$
$f\left(x\right)$ is continuous at $x=1$

$f\left(x\right)=\{{\int }_0^x|1-t|dtx>1,(x-\frac{1}{2}),x\le 1\}$

$RHL$

$\underset{x\to 1^{+}}{lim}f\left(x\right)=\underset{x\to 1^{+}}{lim}[{\int }_0^1|1-t|dt-{\int }_1^x|1-t|dt]=\underset{x\to 1^{+}}{lim}-{\left|\frac{{(1-t)}^2}{2}\right|}_0^1+{\left|\frac{{(1-t)}^2}{2}\right|}_1^x$

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

$LHL$

$=\underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1^{-}}{lim}(x-\frac{1}{2})=\underset{h\to 0}{lim}1-h-\frac{1}{2}=\frac{1}{2}$

At $x=1f\left(1\right)=1-\frac{1}{2}=\frac{1}{2}$

Therefore $x$ is continuous at $x=1$

$LHD$

$\underset{x\to 1^{-}}{lim}\frac{f\left(x\right)-f\left(1\right)}{(x-1)}=\underset{h\to 0}{lim}\frac{f(1-h)-f\left(1\right)}{(1-h-1)}\underset{h\to 0}{lim}\frac{1-h-\frac{1}{2}-\frac{1}{2}}{-h}=1$

$RHD$

$\underset{x\to 1^{+}}{lim}\frac{f\left(x\right)-f\left(1\right)}{(x-1)}=\underset{h\to 0}{lim}\frac{f(1+h)-f\left(1\right)}{(1+h-1)}\underset{h\to 0}{lim}\frac{\frac{1}{2}+\left(\frac{{(1-1-h)}^2}{2}\right)-\frac{1}{2}}{h}=\frac{h^2}{2h}=h=0$

Therefore $f$ is not differentiable at $x=1$