MCQ
If $f(x) = \int_{ - 1}^x {|t|\,dt,} $ $x \ge - 1,$ then
  • $f$ and $f'$ are continous for $x + 1 > 0$
  • B
    $f$ is continous but $f'$ is not continous for $x + 1 > 0$
  • C
    $f$ and $f'$ are not continous at $x = 0$
  • D
    $f$ is continous at $x = 0$ but $f'$ is not so

Answer

Correct option: A.
$f$ and $f'$ are continous for $x + 1 > 0$
a
(a) Let us divide the 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(x) = \int_{ - 1}^x {( - t)dt = - \frac{1}{2}({x^2} - 1)} $.....$(i)$

For ${I_2},f(x) = \int_{ - 1}^x {( - t)dt  + \int_{ 0 }^x {(  t)dt }}$

$ = - \frac{1}{2}[{t^2}]_{ - 1}^0 + \frac{1}{2}[{t^2}]_0^x = \frac{1}{2}(1 + {x^2})$.....$(ii)$

Hence the function can be defined as the following

$f(x) = \left\{ \begin{array}{l} - \frac{1}{2}({x^2} - 1),{\rm{If}} - 1 \le x < 0\\\frac{1}{2}({x^2} + 1),{\rm{if}}\,\,x \ge 0\end{array} \right.$ ,

$f'(x) = \left\{ \begin{array}{l} - x,\,\,{\rm{if}} - 1 < x < 0\\\,\,0,\,\,{\rm{if}}\,\,x = 0\\\,\,x,\,{\rm{if}}\,\,x > 0\end{array} \right.$

For $f,\,\,L = R = V = \frac{1}{2}$ at $x = 0$, so $f$ is continuous at $x = 0$.

For $f',\,\,L = R = V = 0$ 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 $x > - 1$

$i.e.,$ $x + 1 > 0$..

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