JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

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Analyzing +4 | -1

Let

f(x)=\left(\sin \left(\tan ^{-1} x\right)+\sin \left(\cot ^{-1} x\right)\right)^2-1,|x| > 1

. If

\frac{d y}{d x}=\frac{1}{2} \frac{d}{d x}\left(\sin ^{-1}(f(x))\right)

and

y(\sqrt{3})=\frac{\pi}{6}

, then

y(-\sqrt{3})

is equal to:

View answer and explanation

Correct option: B

-\frac{\pi}{6}

Option B (-

\frac{{\pi}}{{6}}

) is correct because it's the value of y(-

\sqrt{{3}}

) obtained after simplification and substitution into the given equation.

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