Question

$\underset{x\to \frac{\pi }{4}}{lim}\frac{\sin x}{{\cos }^{-1}\left[\frac{1}{4}(3\sin x-\sin 3x)\right]}$ where $\left[\right]$ denotes integer function, is-

• A. $\frac{\sqrt{2}}{\pi }$
• B. $1$
• C. $\frac{4}{\pi }$
• D. $doesnotexist$

Answer 1

The correct option is A $\frac{\sqrt{2}}{\pi }$

We have,

${lim}_{x\to \frac{\pi }{4}}\frac{\sin x}{{\cos }^{-1}\left[\frac{1}{4}\left(3\sin x-\sin 3x\right)\right]}$

${lim}_{x\to \frac{\pi }{4}}\frac{\sin x}{{\cos }^{-1}\left[\frac{1}{4}\left(3\sin x-3\sin x+4{\sin }^{3}x\right)\right]}$

${lim}_{x\to \frac{\pi }{4}}\frac{\sin x}{{\cos }^{-1}\left[{\sin }^{3}x\right]}$

$=\frac{\sin \frac{\pi }{4}}{{\cos }^{-1}\left[{\left(\sin \frac{\pi }{4}\right)}^3\right]}$

$=\frac{\frac{1}{\sqrt{2}}}{{\cos }^{-1}\left[{\left(\frac{1}{\sqrt{2}}\right)}^3\right]}$

Since, $[.]$ is greatest integer function.

Therefore,

$=\frac{\frac{1}{\sqrt{2}}}{{\cos }^{-1}0}$

$=\frac{\frac{1}{\sqrt{2}}}{\left(\frac{\pi }{2}\right)}=\frac{2}{\pi }\times \frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{\pi }$

Hence, this is the answer.