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