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Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

MathematicsInverse Trigonometric FunctionsIdentities, eqations and inequations involving ITFMedium2 minQB

MathematicsMediumsingle choice

Passage / Comprehension

Use principal branches of $tan^{- 1}, sin^{- 1}$ and $cos^{- 1}$ . For $x > 1$ , $tan^{- 1} (2 x / (1 - x^{2})) = 2 tan^{- 1} x - π$ ; for $∣ x ∣ \leq 1, sin^{- 1} (2 x / (1 + x^{2})) = 2 tan^{- 1} x$ . Also $sin^{- 1} u + cos^{- 1} u = π /2$ on $[- 1, 1]$ .

Solve $cos^{- 1} (6 x / (1 + 9 x^{2})) = - π /2 + 2 tan^{- 1} (3 x)$ .

Options:

Answer:

Solution:

Let $z = 3 x$ . The right side is nonnegative exactly when $z \geq 1$ . For $z \geq 1$ , $sin^{- 1} (2 z / (1 + z^{2})) = π - 2 tan^{- 1} z$ , so $cos^{- 1} (2 z / (1 + z^{2})) = - π /2 + 2 tan^{- 1} z$ . Equality also holds at $z = 1$ , when both sides are 0 . Exact solution: $x \in [1/3, \infty)$ , which is option A.

Stream:JEESubject:MathematicsTopic:Inverse Trigonometric FunctionsSubtopic:Identities, eqations and inequations involving ITF

⏱ 2mℹ️ Source: QB

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