Considering only the principal values of inverse functions, the set $A=\left\{x \geq 0: \tan ^{-1}(2 x)+\tan ^{-1}(3 x)=\frac{\pi}{4}\right\}$

(B) contains more than two elements

JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Inverse Trigonometric Functions

Considering only the principal values of inverse functions, the set

A = {x \geq 0 : tan^{- 1} (2 x) + tan^{- 1} (3 x) = \frac{π}{4}}

Answer: (C)

Take the tangent of both sides of

tan^{- 1} (2 x) + tan^{- 1} (3 x) = \frac{π}{4}

. Using the addition formula,

\frac{2 x + 3 x}{1 - 6 x ^{2}} = 1

, so

5 x = 1 - 6 x^{2}

, i.e.

6 x^{2} + 5 x - 1 = 0

, which factors as

(6 x - 1) (x + 1) = 0

and gives

x = \frac{1}{6}

x = - 1

. The condition

x \geq 0

rules out

x = - 1

, and

x = \frac{1}{6}

keeps

1 - 6 x^{2} > 0

so the tangent step is valid. Only one value survives, so the set is a singleton.

JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Single correct option

If the domain of the function $f (x) = cos^{- 1} (\frac{2 x - 5}{11 - 3 x}) + sin^{- 1} (2 x^{2} - 3 x + 1)$ is the interval $[α, β]$ , then $α + 2 β$ is equal to :
JEE Main 2026View solution →
The number of solutions of $tan^{- 1} 4 x + tan^{- 1} 6 x = \frac{π}{6}$ , where $- \frac{1}{2 6} < x < \frac{1}{2 6}$ is equal to
JEE Main 2026View solution →
If the domain of the function $f (x) = sin^{- 1} (\frac{1}{x ^{2} - 2 x - 2})$ , is $(- \infty, α] \cup [β, γ] \cup [δ, \infty)$ , then $α + β + γ + δ$ is equal to
JEE Main 2026View solution →
If $k = tan (\frac{π}{4} + \frac{1}{2} cos^{- 1} (\frac{2}{3})) + tan (\frac{1}{2} sin^{- 1} (\frac{2}{3}))$ then the number of solutions of the equation …
JEE Main 2026View solution →
Considering the principal values of inverse trigonometric functions, the value of the expression $tan (2 sin^{- 1} (\frac{2}{13}) - 2 cos^{- 1} (\frac{3}{10}))$ is equal to :
JEE Main 2026View solution →
Let [ • ] denote the greatest integer function. If the domain of the function $f (x) = cos^{- 1} (\frac{4 x + 2 [ x ]}{3})$ is $[α, β]$ , then $12 (α + β)$ is equal to :
JEE Main 2026View solution →