Question

For $0<x_1<x_2<1,$ which of the following is correct

• A. $\frac{\sin x_2}{\sin x_1}\ge \frac{\ln x_2}{\ln x_1}$
• B. $\frac{\sin x_2}{\sin x_1}>\frac{\ln x_2}{\ln x_1}$
• C. $\frac{\sin x_2}{\sin x_1}<\frac{\ln x_2}{\ln x_1}$
• D. $\frac{\sin x_2}{\sin x_1}\le \frac{\ln x_2}{\ln x_1}$

Answer 1

The correct option is B $\frac{\sin x_2}{\sin x_1}>\frac{\ln x_2}{\ln x_1}$

Let $f\left(x\right)=\frac{\ln x}{\sin x}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{\sin x-x\ln x\cdot \cos x}{x{\sin }^{2}x}>0$ for $0<x<1(\because \ln x<0,\sin x,\cos x>0)$
So, $f\left(x\right)$ is increasing.
$\Rightarrow$ for $x_1<x_2:$
$\Rightarrow \frac{\ln x_1}{\sin x_1}<\frac{\ln x_2}{\sin x_2}$
$\Rightarrow \frac{\sin x_2}{\sin x_1}>\frac{\ln x_2}{\ln x_1}(\because \ln x<0,0<x<1)$