Question

Let $f\left(x\right)={\int }_0^{1}t|t-x|dt,xϵR$
The function is discontinuous at

• A. $x=0$
• B. $x=1$
• C. $x=\frac{1}{2}$
• D. None of the above points

Answer 1

The correct option is D None of the above points

${\int }_0^{1}t|t-x|dx$
$x<0\Rightarrow f\left(x\right){\int }_0^{1}t(t-x)dx={(\frac{t^3}{3}-\frac{t^2}{2}x)}_0^1=\frac{1}{3}-\frac{x}{2}$
$x>0\Rightarrow f\left(x\right){\int }_0^{1}t(x-t)dt={(\frac{t^{2}x}{2}-\frac{t^2}{3})}_0^1=\frac{x}{2}-\frac{1}{3}$
$0\le x\le 1={\int }_0^x(x-t)dt+{\int }_0^{x}t(x-t)dt$
${(\frac{t^{2}x}{2}-\frac{t^3}{3})}_0^x+{(\frac{t^3}{3}-\frac{x{t}^2}{2})}_x^1$
$=(\frac{x^2}{2}-\frac{x^3}{3})+(\frac{1}{3}-\frac{x}{2})-(\frac{x^3}{3}-\frac{x^3}{2})$
$f\left(x\right)=\frac{x^3}{3}-\frac{x}{2}+\frac{1}{3}$
Minimum value is possible at $x=\frac{1}{\sqrt{2}}$
$f\left(6\right)=\frac{208}{3}$