JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

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Application +4 | -1

\int \nolimits_{\sin x}^1 \,\,{t^2}\,f\left( t \right)\,\,dt = 1 - \sin \,\,x,\;\,\,\forall \;x \in \,\,\left[ {0,\,\,\pi /2} \right]

, then

f\,\,\left( {\frac{1}{{\sqrt 3 }}} \right)

View answer and explanation

Correct option: A

Option A (3) is correct because differentiating both sides using Leibniz's rule and then substituting

x

such that

\sin x = \frac{1}{\sqrt{3}}

gives the value of

f(\frac{1}{\sqrt{3}})

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