Question

Which of the following inequalities is/are TRUE?

• A. $\underset{0}{\overset{1}{\int }}x\cos xdx\ge \frac{3}{8}$
• B. $\underset{0}{\overset{1}{\int }}x\sin xdx\ge \frac{3}{10}$
• C. $\underset{0}{\overset{1}{\int }}{x}^2\cos xdx\ge \frac{1}{2}$
• D. $\underset{0}{\overset{1}{\int }}{x}^2\sin xdx\ge \frac{2}{9}$

Answer 1

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

We know that $\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\dots$
So, $\cos x>1-\frac{x^2}{2}$ for all $x\in (0,1)$
$\Rightarrow x\cos x>x-\frac{x^3}{2}$
$\Rightarrow \underset{0}{\overset{1}{\int }}x\cos xdx>\underset{0}{\overset{1}{\int }}\left(x-\frac{x^3}{2}\right)dx$
$\Rightarrow \underset{0}{\overset{1}{\int }}x\cos xdx>\frac{3}{8}($ Option $A$ is correct $)$ $\because x^2\cos x<x$ for all $x\in (0,1)$
$\Rightarrow \underset{0}{\overset{1}{\int }}{x}^2\cos xdx<\underset{0}{\overset{1}{\int }}xdx$
$\Rightarrow \underset{0}{\overset{1}{\int }}{x}^2\cos xdx<\frac{1}{2}($ Option $C$ is incorrect $)$
Again we know that, $\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\dots$
So, $\sin x>x-\frac{x^3}{6}$ for all $x\in (0,1)$
$\Rightarrow x\sin x>x^2-\frac{x^4}{6}$
$\underset{0}{\overset{1}{\int }}x\sin xdx>\underset{0}{\overset{1}{\int }}\left(x^2-\frac{x^4}{6}\right)dx$
$\Rightarrow \underset{0}{\overset{1}{\int }}x\sin dx>\frac{3}{10}$ (Option $B$ is correct) $\because x^2\sin x>x^3-\frac{x^5}{6}$
$\Rightarrow \underset{0}{\overset{1}{\int }}{x}^2\sin xdx>\frac{2}{9}$ (Option $D$ is correct)