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JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

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Application +4 | -1

Let

x \in(0,1)

. The set of all

x

such that

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

, is the interval:

View answer and explanation

Correct option: B

\left(\frac{1}{\sqrt{2}}, 1\right)

Option B (

\left(\frac{{1}}{{\sqrt{{2}}}}, 1\right)

) is correct because solving the inequality

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

leads to

x > \frac{{1}}{{\sqrt{{2}}}}

and

x < 1

.

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