If $f\left( x \right) = \mathop \smallint \nolimits_{{x^2}}^{{x^2} + 1} {e^{ - {t^2}}}dt,$ then $f\left( x \right)$ increases on

JEE MATHEMATICS MCQ: INCREASING & DECREASING

1 Medium

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Analyzing +4 | -1

f\left( x \right) = \int \nolimits_{{x^2}}^{{x^2} + 1} {e^{ - {t^2}}}dt,

then

f\left( x \right)

increases on

View answer and explanation

Correct option: C

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

Option C is correct because the function

f(x)

increases where its derivative

f'(x)

is positive, which occurs for

x \in (-\infty, 0)

Practice All Chapter MCQs →

f\left( x \right) = \frac{{K\sin x + 2\cos x}}{{\sin x + \cos x}}

x

f\left( x \right) = {x^2} + ax + 1

a

f(x)=

\sin x-\cos x-a x+b

f(x)=x e^{x(1-x)}, x \in R

f

g

fog

f\left( x \right) = {x^3} + b{x^2} + cx + d

0 < {b^2} < c

\left( { - \infty ,\;\infty } \right)