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

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

Now, for $x>1,{\int }_0^x|1-t|dt$

$={\int }_0^1(1-t)dt+{\int }_1^x(t-1)dt$

$={[t-\frac{t^2}{2}]}_0^1+{[\frac{t^2}{2}-t]}_1^x$

$=[1-\frac{1}{2}-0]+[\frac{x^2}{2}-x-\frac{1}{2}+1]$

$=\frac{1}{2}+\frac{x^2}{2}-x-\frac{1}{2}+1$

$\Rightarrow \frac{x^2}{2}-x+1$

$\therefore$ Equation (i) becomes,

$f\left(x\right)=\{\begin{array}{c}\frac{x^2}{2}-x+1;x>1\\ x-\frac{1}{2};x\le 1\end{array}$
Continuity at $x=1$,

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

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

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

$\therefore f\left(1\right)=\underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1^{+}}{lim}f\left(x\right)$

Hence, $f\left(x\right)$ is continuous at $x=1$.

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

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

$\therefore LHD\ne RHD$

Hence, $f\left(x\right)$ is not differentiable at $x=1$.