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JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Differentiation

If

y (x) = cot^{- 1} (\frac{1 + s i n x + 1 - s i n x}{1 + s i n x - 1 - s i n x}), x \in (\frac{π}{2}, π)

, then

\frac{d y}{d x}

at

x = \frac{5 π}{6}

is:

(A)-1
(B)0
(C) $- \frac{1}{2}$ ✓ Correct
(D) $\frac{1}{2}$

Answer: (C)

Solution

For

x \in (\frac{π}{2}, π)

write

1 \pm sin x = (cos \frac{x}{2} \pm sin \frac{x}{2})^{2}

. On this interval

sin \frac{x}{2} > cos \frac{x}{2} > 0

, so

1 + sin x = cos \frac{x}{2} + sin \frac{x}{2}

and

1 - sin x = sin \frac{x}{2} - cos \frac{x}{2}

. The argument becomes

\frac{2 sin \frac{x}{2}}{2 cos \frac{x}{2}} = tan \frac{x}{2} = cot (\frac{π}{2} - \frac{x}{2}),

so

y = cot^{- 1} (cot (\frac{π}{2} - \frac{x}{2})) = \frac{π}{2} - \frac{x}{2}

. Therefore

\frac{d y}{d x} = - \frac{1}{2}

, and at

x = \frac{5 π}{6}

its value is

- \frac{1}{2}

.

JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Single correct option

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