Question

Which of the following is correct for $0<x_1<x_2<\frac{\pi }{2}$

• A. $x_1^2\sin x_1-x_2^2\sin x_2>x_2\cos x_2-x_1\cos x_1$
• B. $x_1^2\sin x_1-x_2^2\sin x_2<x_2\cos x_2-x_1\cos x_1$
• C. $x_1^2\sin x_1-x_2^2\sin x_2\le x_2\cos x_2-x_1\cos x_1$
• D. $x_1^2\sin x_1-x_2^2\sin x_2\ge x_2\cos x_2-x_1\cos x_1$

Answer 1

The correct option is B $x_1^2\sin x_1-x_2^2\sin x_2<x_2\cos x_2-x_1\cos x_1$

Let $f\left(x\right)=x^2\sin x+x\cos x$
$\Rightarrow f^{\prime }\left(x\right)=2x\sin x+x^2\cos x+\cos x-x\sin x$
$\Rightarrow f^{\prime }\left(x\right)=x\sin x+(x^2+1)\cos x>0$ $\Rightarrow f\left(x\right)$ strictly increases in $x\in (0,\frac{\pi }{2})$
for $\frac{\pi }{2}>x_2>x_1>0$
$\Rightarrow x_1^2\sin x_1+x_1\cos x_1<x_2\cos x_2+x_2^2\sin x_2$