If I is minimum then the ordered pair

Question:

Let $\mathrm{I}=\int_{\mathrm{a}}^{\mathrm{b}}\left(x^{4}-2 x^{2}\right) \mathrm{d} x$. If I is minimum then the ordered pair $(a, b)$ is:

  1. (1) $(0, \sqrt{2})$

  2. (2) $(-\sqrt{2}, 0)$

  3. (3) $(\sqrt{2},-\sqrt{2})$

  4. (4) $(-\sqrt{2}, \sqrt{2})$


Correct Option: , 4

Solution:

$I=\int_{a}^{b}\left(x^{4}-2 x^{2}\right) d x$

$\Rightarrow \frac{d I}{d x}=x^{4}-2 x^{2}=0$ (for minimum)

$\Rightarrow x=0, \pm \sqrt{2}$

Also  $I=\left[\frac{x^{5}}{5}-\frac{2 x^{3}}{3}\right]_{a}^{b}$

For $\mathrm{a}=0, \mathrm{~b}=\sqrt{2}$

$I=\frac{8 \sqrt{2}}{15}$

For $a=-\sqrt{2}, b=0$

$I=\frac{-8 \sqrt{2}}{15}$

For $\mathrm{a}=-\sqrt{2}, \mathrm{~b}=\sqrt{2}$

$I=\frac{16 \sqrt{2}}{15}$

For $a=-\sqrt{2}, b=\sqrt{2}$

$I=\frac{-16 \sqrt{2}}{15} .$

$\therefore \quad I$ is minimum when $(a, b)=(-\sqrt{2}, \sqrt{2})$

 

 

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