JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Inverse Trigonometric Functions

Given that the inverse trigonometric functions take principal values only. Then, the number of real values of

x

which satisfy

sin^{- 1} (\frac{3 x}{5}) + sin^{- 1} (\frac{4 x}{5}) = sin^{- 1} x

is equal to :

Answer: (D)

Solution

The three inverse sines need

\frac{4 x}{5} \leq 1

and

∣ x ∣ \leq 1

, so

∣ x ∣ \leq 1

. Writing the equation as

sin^{- 1} \frac{3 x}{5} = sin^{- 1} x - sin^{- 1} \frac{4 x}{5}

and taking the sine of both sides,

\frac{3 x}{5} = x 1 - \frac{16 x ^{2}}{25} - \frac{4 x}{5} 1 - x^{2}

For

x \neq = 0

this becomes

25 - 16 x^{2} = 3 + 4 1 - x^{2}

, and squaring gives

25 - 16 x^{2} = 9 + 16 (1 - x^{2}) + 24 1 - x^{2} = 25 - 16 x^{2} + 24 1 - x^{2}

1 - x^{2} = 0

and

x = \pm 1

. Both work, since

sin^{- 1} \frac{3}{5} + sin^{- 1} \frac{4}{5} = \frac{π}{2} = sin^{- 1} 1

and the same holds with all signs reversed, and

x = 0

satisfies the equation as well. The equation has three real solutions.

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Single correct option