If $\sin ^{-1} x \leq \cos ^{-1} x$, then $x \in \ldots \ldots$

$\left[\frac{1}{\sqrt{2}}, 1\right]$

JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

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Application +4 | -1

\sin ^{-1} x \leq \cos ^{-1} x

, then

x \in \ldots \ldots

View answer and explanation

Correct option: B

\left[-1, \frac{1}{\sqrt{2}}\right]

Option B (

\left[-1, \frac{{1}}{{\sqrt{{2}}}}\right]

) is correct because solving the inequality

\sin^{-1}x \le \cos^{-1}x

leads to this interval.

Practice All Chapter MCQs →

\cos ^{-1}(\cos 12)-\sin ^{-1}(\sin 12)

\sin \left(2 \tan ^{-1}(1 / 3)\right)+

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\cos ^ {-1}\left\{\frac{1}{\sqrt{2}}\left(\cos \frac{9 \pi}{10}-\sin \frac{9 \pi}{10}\right)\right\}=

\sin \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=1

x

{x^2} + {y^2} + {z^2} = {r^2},

{\tan ^{ - 1}}\left( {\frac{{xy}}{{zr}}} \right) + {\tan ^{ - 1}}\left( {\frac{{yz}}{{xr}}} \right) + {\tan ^{ - 1}}\left( {\frac{{xz}}{{yr}}} \right)\;

2{\tan ^{ - 1}}\left( {\frac{1}{3}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{7}} \right)