If f(x) = _ x^2 ^ x^2 + 1 e^ - t^2 dt, then f(x) increases in - Vidyadip

MCQ

If $f(x) = \int_{{x^2}}^{{x^2} + 1} {{e^{ - {t^2}}}} dt,$ then $f(x)$ increases in

A
$(2,\,\,2)$
B
No value of $x$
C
$(0,\,\,\infty )$
✓
$( - \infty ,\,\,0)$

✓

Answer

Correct option: D.

$( - \infty ,\,\,0)$

d
(d) $f'(x) = {e^{ - {{({x^2} + 1)}^2}}}.2x - {e^{ - {{({x^2})}^2}}}.2x $

$= 2x{e^{ - ({x^4} + 1 + 2{x^2})}}\left( {1 - {e^{2{x^2} + 1}}} \right)$

==> $f'(x) > 0,\forall x \in ( - \infty ,0).$

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