COMPASSLearning

Talk to us Register Login

JEE Main · 2020 · 08 Jan 2020 · Shift 1 · Inverse Trigonometric Functions

Let

f (x) = (sin (tan^{- 1} x) + sin (cot^{- 1} x))^{2} - 1, ∣ x ∣ > 1

. If

\frac{dy}{dx} = \frac{1}{2} \frac{d}{dx} (sin^{- 1} (f (x)))

and

y (3) = \frac{π}{6}

, then

y (- 3)

is equal to

(A) $\frac{5 π}{6}$ ✓ Correct
(B) $\frac{π}{3}$
(C) $\frac{2 π}{3}$
(D) $- \frac{π}{6}$

Answer: (A)

Solution

Writing

θ = tan^{- 1} x

gives

sin θ = \frac{x}{1 + x ^{2}}

and

sin (cot^{- 1} x) = \frac{1}{1 + x ^{2}}

, so

f (x) = (\frac{x + 1}{1 + x ^{2}})^{2} - 1 = \frac{2 x}{1 + x ^{2}}

. With

\frac{d y}{d x} = \frac{1}{2} \cdot \frac{f ^{'} ( x )}{1 - f ^{2} ( x )}

, the simplification gives

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

, so

y = - tan^{- 1} x + c

. Using

y (3) = \frac{π}{6}

fixes

c = \frac{π}{2}

, whence

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

, and

y (- 3) = cot^{- 1} (- 3) = π - \frac{π}{6} = \frac{5 π}{6}

.

JEE Main · 2020 · 08 Jan 2020 · Shift 1 · Single correct option

Practise more: all JEE Main PYQs on Inverse Trigonometric Functions →

More JEE Main questions on Inverse Trigonometric Functions