JEE Advance - Mathematics (2003 - No. 7)

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

$$(-2, 2)$$

no value of $$x$$

$$\left( {0,\infty } \right)$$

$$\left( { - \infty ,0} \right)$$