If f(x) = int_x^2^x^2+1 e^-t^2 dt, then f(x) increases in — JEE Main 2003 Mathematics PYQ

AIEEE 2003

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

Show answer and solution

Correct option: D

Solution

f'(x) = e^(-(x²+1)²)·2x - e^(-x⁴)·2x = 2x·e^(-x⁴)[e^(-(x²+1)²+x⁴) - 1].

Since (x²+1)² - x⁴ = 2x²+1 > 0 always, the bracket e^(-(2x²+1)) - 1 < 0 always.

So f'(x) > 0 requires 2x < 0, i.e., x < 0, giving the interval (-∞, 0).

[AIEEE 2003]

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