JEE Advanced · 2020 · 27 Sep 2020 · Paper 1 · Definite Integrals

Which of the following inequalities is/are TRUE?

Answer: (A), (B), (D)

Solution

Using Taylor series:

cos x = 1 - \frac{x ^{2}}{2} + \frac{x ^{4}}{4 !} - \dots \geq 1 - \frac{x ^{2}}{2}

for

x \in [0, 1]

.

(A)

\int_{0}^{1} x cos x d x \geq \int_{0}^{1} x (1 - \frac{x ^{2}}{2}) d x = \frac{1}{2} - \frac{1}{8} = \frac{3}{8}

. TRUE.

Using

sin x = x - \frac{x ^{3}}{6} + \dots \geq x - \frac{x ^{3}}{6}

for

x \in [0, 1]

:

(B)

\int_{0}^{1} x sin x d x \geq \int_{0}^{1} x (x - \frac{x ^{3}}{6}) d x = \frac{1}{3} - \frac{1}{30} = \frac{9}{30} = \frac{3}{10}

. TRUE.

(D)

\int_{0}^{1} x^{2} sin x d x \geq \int_{0}^{1} x^{2} (x - \frac{x ^{3}}{6}) d x = \frac{1}{4} - \frac{1}{36} = \frac{9 - 1}{36} = \frac{8}{36} = \frac{2}{9}

. TRUE.

(C) Since

cos x < 1

for

x \in (0, 1]

, we have

\int_{0}^{1} x^{2} cos x d x < \int_{0}^{1} x^{2} d x = \frac{1}{3} < \frac{1}{2}

. FALSE.

JEE Advanced · 2020 · 27 Sep 2020 · Paper 1 · One or more correct options