question_answer

Let \[I=\int\limits_{a}^{b}{\left( {{x}^{4}}-2{{x}^{2}} \right)dx}\]. If I is minimum then the ordered pair (a, b) is -   [JEE Main Online Paper (Held On 10-Jan-2019 Morning]

A) \[d\in R\]                                   

B) \[\,(0,\,\,\sqrt{2})\]

C) \[(-\,\sqrt{2},\,\,\sqrt{2})\]           

D)                  \[\,(-\sqrt{2},\,\,0)\]

Correct Answer: C

Solution :

\[I\,=\,\int\limits_{a}^{b}{({{x}^{4}}-2{{x}^{2}})\,dx}\] \[f\left( x \right)={{x}^{4}}-2{{x}^{2}}={{x}^{2}}\left( x-\sqrt{2} \right)\left( x+\sqrt{2} \right)\] \[-\sqrt{2}\,\le \,x\,\le \,\sqrt{2}\,\] integration will be greatest negative so for minimum value of I             \[\left. \begin{align}   & a=-\sqrt{2} \\  & b=\sqrt{2} \\ \end{align} \right)\]