JEE Main · 2024 · 27 Jan 2024 · Shift 2 · Inverse Trigonometric Functions

Considering only the principal values of inverse trigonometric functions, the number of positive real values of

x

satisfying

tan^{- 1} (x) + tan^{- 1} (2 x) = \frac{π}{4}

is :

Answer: (B)

Solution

For

x > 0

tan^{- 1} x + tan^{- 1} (2 x) = \frac{π}{4}

tan^{- 1} (2 x) = \frac{π}{4} - tan^{- 1} x .

Taking tangent of both sides:

2 x = tan (\frac{π}{4} - tan^{- 1} x) = \frac{1 - x}{1 + x} .

2 x (1 + x) = 1 - x \Rightarrow 2 x^{2} + 3 x - 1 = 0 \Rightarrow x = \frac{- 3 \pm 17}{4} .

Since

17 > 3

, only

x = \frac{- 3 + 17}{4}

is positive; the other root is negative and rejected. For this value of

x

, both

tan^{- 1} x

and

tan^{- 1} 2 x

lie in

(0, \frac{π}{2})

with a sum less than

\frac{π}{2}

, so the tangent step is reversible and the root is genuine.

There is exactly one positive real value of

x

JEE Main · 2024 · 27 Jan 2024 · Shift 2 · Single correct option