Question

$Iff\left(x\right)={\int }_{-1}^x|t|dt,x\ge -1,then$ [MNR 1994]

• A.
  
• B.
• C.
• D.

Answer 1

The correct option is A

Let us divide interval into two sub - intervals
$I_1,-1\le x<0$ so that x is -ve and $I_2,x\ge 0$ so that x is +ve.
$For{I}_1,f\left(x\right)={\int }_{-1}^x(-t)dt=-\frac{1}{2}(x^2-1)\cdots \left(i\right)$
$For{I}_2,f\left(x\right)={\int }_{-1}^0(-t)dt+{\int }_0^{x}tdt$
$=-\frac{1}{2}[t^2{]}_{-1}^0+\frac{1}{2}[t^2{]}_0^x=\frac{1}{2}(1+x^2)\cdots \left(ii\right)$
Hence the function can be defined as the following
$f\left(x\right)=\{\begin{array}{c}-\frac{1}{2}(x^2-1),If-1\le x<0\\ \frac{1}{2}(x^2+1),ifx\ge 0\end{array}$
For f, L = R = V =$\frac{1}{2}$ at x =0, so f' is also continuous at x = 0. Thus both f and f' are continuous at x = 0 and hence both are continuous for x > -1 i.e., x + 1 > 0