JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

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Application +4 | -1

The equation

\sin ^{-1} x=2 \sin ^{-1} a

holds true for

View answer and explanation

Correct option: C

-\frac{1}{\sqrt{2}} \leq a \leq \frac{1}{\sqrt{2}}

Option C (

-\frac{{1}}{{\sqrt{{2}}}} \leq a \leq \frac{{1}}{{\sqrt{{2}}}}

) is correct because it defines the range of 'a' for which the equation

\sin^{-1} x = 2 \sin^{-1} a

holds true, considering the domain and range of inverse trigonometric functions.

More Questions from INVERSE TRIGONOMETRY

Practice All Chapter MCQs →

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

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

\cos \left(\tan ^{-1} 2 \sqrt{2}\right)

\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)