Logg på
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 $$x$$
$$\left( {0,\infty } \right)$$
$$\left( { - \infty ,0} \right)$$
Kommentarer (0)
Logg inn for å kommentere
Annonse
Tillat at javascript laster denne siden ordentlig
