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

JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Definite Integrals

Let

I = \int_{a}^{b} (x^{4} - 2 x^{2}) dx

. If I is minimum then the ordered pair

(a, b)

is:

Answer: (D)

The integrand

x^{4} - 2 x^{2} = x^{2} (x^{2} - 2)

is negative exactly on

(- 2, 2)

and positive outside it. To make

\int_{a}^{b}

as negative as possible I should gather every bit of negative area and none of the positive, which means integrating right across the dip from

- 2

2

. Stopping short would leave negative area behind, and going further would start adding positive area back, so the minimising limits are

(a, b) = (- 2, 2)

JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Single correct option

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