Question

Which of the following option is always correct for $x\in (0,\frac{\pi }{2})?$

• A. $\frac{\sin \left(\frac{1}{10}\right)}{\left(\frac{1}{10}\right)}<\frac{\sin \left(\frac{1}{9}\right)}{\left(\frac{1}{9}\right)}$
• B. $\frac{\sin \left(\frac{1}{10}\right)}{\left(\frac{1}{9}\right)}>\frac{\sin \left(\frac{1}{9}\right)}{\left(\frac{1}{10}\right)}$
• C. $\frac{\sin \left(\frac{1}{10}\right)}{\left(\frac{1}{10}\right)}>\frac{\sin \left(\frac{1}{9}\right)}{\left(\frac{1}{9}\right)}$
• D. $\frac{\sin \left(\frac{1}{10}\right)}{\left(\frac{1}{9}\right)}=\frac{\sin \left(\frac{1}{9}\right)}{\left(\frac{1}{10}\right)}$

Answer 1

The correct option is C $\frac{\sin \left(\frac{1}{10}\right)}{\left(\frac{1}{10}\right)}>\frac{\sin \left(\frac{1}{9}\right)}{\left(\frac{1}{9}\right)}$

Let $f\left(x\right)=\frac{\sin x}{x}$

$\Rightarrow f^{\prime }\left(x\right)=\frac{x\cos x-\sin x}{x^2}$

$f^{\prime }\left(x\right)=\left(\frac{\cos x}{x^2}\right)(x-\tan x)$

$\Rightarrow f^{\prime }\left(x\right)<0\forall x\in (0,\frac{\pi }{2})$

$(\because x<\tan x\text{in}(0,\frac{\pi }{2}))$.

Hence $f\left(x\right)$ is strictly decreasing

We know, $\left(\frac{1}{10}\right)<\left(\frac{1}{9}\right)$

$\Rightarrow f\left(\frac{1}{10}\right)>f\left(\frac{1}{9}\right)$
$\Rightarrow \frac{\sin \left(\frac{1}{10}\right)}{\left(\frac{1}{10}\right)}>\frac{\sin \left(\frac{1}{9}\right)}{\left(\frac{1}{9}\right)}$

Let $f\left(x\right)=x\sin x$
$\Rightarrow f^{\prime }\left(x\right)=x\cos x+\sin x=\cos x(x+\tan x)$
$\because \cos x,\tan x>0$ in $x\in (0,\frac{\pi }{2})$
$\Rightarrow f^{\prime }\left(x\right)>0,x\in (0,\frac{\pi }{2})$
$\Rightarrow f\left(x\right)$ strictly increases
$\Rightarrow$ for $\frac{\pi }{2}>x>0,f\left(x\right)>f\left(0\right)$
$\Rightarrow \frac{1}{9}\sin \left(\frac{1}{9}\right)>\frac{1}{10}\sin \left(\frac{1}{10}\right)$
$\Rightarrow \frac{\sin \left(\frac{1}{10}\right)}{\left(\frac{1}{9}\right)}<\frac{\sin \left(\frac{1}{9}\right)}{\left(\frac{1}{10}\right)}$