Question

Let $f\left(x\right)={\int }_0^x|x-1|dx,x\ge 0$.Then$f^{\prime }\left(x\right)$is

• A. continous at $x=1$
• B. continous at $x=2$
• C. differentiable at $x=1$
• D. differentiable at $x=2$

Answer 1

Answer: (A), (B), (D)

continous at $x=1$
differentiable at $x=2$
continous at $x=2$

$f\left(x\right)={\int }_0^x|x-1|dx$
Using leibnitz theorem
$f^{{}^{\prime }}\left(x\right)=1.|x-1|-0=|x-1|$
Therefore for $x=1$ $f^{{}^{\prime }}\left(x\right)$ is continuous, but not differentiable as moduls function is not differentiable at its critical point.
And for $x=2$, $f^{{}^{\prime }}\left(x\right)$ is continuous and differentiable