Q112 / 160 · AP EAMCET 2023 (15 May Shift 1) · Mathematics

If $I=\int_{-a}^a\left(x^4-2 x^2\right) d x$, then $I$ is minimum at $a=$

  • A
    2
  • B
    $-\sqrt{2}$
  • Correct answer
    C
    $\sqrt{2}$
  • D
    -2

Solution

$$ \begin{aligned} & \text { } I=\int_{-a}^a\left(x^4-2 x^2\right) d x \\ & \frac{d I}{d a}=\frac{d}{d a}\left[\int_{-a}^a\left(x^4-2 x^2\right) d x\right] \\ & =2 a^4-4 a^2 \\ & \text { and } \frac{d^2 I}{d a^2}=8 a^3-8 a \end{aligned} $$ For minima, $\frac{d I}{d \alpha}=0$ $$ \Rightarrow a=0, \pm \sqrt{2} $$ but when $c=\sqrt{2}$ then $\frac{d^2 I}{d a^2}=8 \sqrt{2}>0$ Hence $a=\sqrt{2}$ is minima

Hence, the correct answer is option: 'C'.

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