Which of the following inequalities is/are TRUE?

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$\int_{0}^{1} x \sin x d x \geq \frac{3}{10}$

$\int_{0}^{1} x^{2} \cos x d x \geq \frac{1}{2}$

$\int_{0}^{1} x^{2} \sin x d x \geq \frac{2}{9}$

$\int_{0}^{1} x \cos x d x \geq \frac{3}{8}$

answer is A, B, D.

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Detailed Solution

(1) We know that $\cos x = 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - ...$

So,option (1) is correct.
(2) $\sin x = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - ...$
$∴ \sin x \geq x - \frac{x^{3}}{6}$
$\Rightarrow \int_{0}^{1} x \sin x \geq \int_{0}^{1} x (x - \frac{x^{3}}{6}) d x$
$\Rightarrow \int_{0}^{1} x \sin x \geq \frac{1}{3} - \frac{1}{30} \Rightarrow \int_{0}^{1} x \sin x d x \geq \frac{9}{30} = \frac{3}{10}$
So, option (2) is correct.
(3) Now, $\cos x < 1$
$\Rightarrow x^{2} \cos x < x^{2}$
$\Rightarrow \int_{0}^{1} x^{2} \cos x d x < \frac{1}{3}$
So option (3) is incorrect.
(4) $\int_{0}^{1} x^{2} \sin x d x \geq \int_{0}^{1} x^{2} (x - \frac{x^{3}}{6}) d x$
$\int_{0}^{1} x^{2} \sin x d x \geq \frac{1}{4} - \frac{1}{36}$
$\int_{0}^{1} x^{2} \sin x d x \geq \frac{2}{9} .$
So, option (4) is correct.