JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

1 Easy

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Analyzing +4 | -1

Let

f\left( x \right) = {\left( {\sin \left( {{{\tan }^{ - 1}}x} \right) + \sin \left( {{{\cot }^{ - 1}}x} \right)} \right)^2} - 1,\,\left| x \right| > 1.\,\,\,\,\mathrm{If }\,\,\frac{{dy}}{{dx}} = \frac{1}{2}\frac{d}{{dx}}\left( {{{\sin }^{ - 1}}\left( {f\left( x \right)} \right)} \right)\,\,\,\,\mathrm{and}\,\,\,\mathrm{ }y\left( {\sqrt 3 } \right) = \frac{\pi }{6}

then

y\left( { - \sqrt 3 } \right)

is equal to

View answer and explanation

Correct option: B

- \frac{\pi }{6}

Option B (

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

) is correct because after simplifying

f(x)

and using the given initial condition, we can find

y(-\sqrt{3})

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