Question

If \[\cos^{- 1} x > \sin^{- 1} x\], then

विकल्प

• \[\frac{1}{\sqrt{2}} < x \leq 1\]

•  \[0 \leq x < \frac{1}{\sqrt{2}}\]

•  \[- 1 \leq x < \frac{1}{\sqrt{2}}\]

•  x > 0

Answer 1

\[\cos^{- 1} x > \sin^{- 1} x\]
\[ \Rightarrow \cos^{- 1} x > \frac{\pi}{2} - \cos^{- 1} x\]
\[ \Rightarrow 2 \cos^{- 1} x > \frac{\pi}{2}\]
\[ \Rightarrow \cos^{- 1} x > \frac{\pi}{4}\]
\[ \Rightarrow x > \cos\frac{\pi}{4}\]
\[ \Rightarrow x > \frac{1}{\sqrt{2}}\]

We know that the maximum value of cosine fuction is 1.

\[\therefore \frac{1}{\sqrt{2}} < x \leq 1\]

Hence, the correct answer is option(a).