The number of real roots of the equation $\tan ^{-1} \sqrt{x(x+1)}+\sin ^{-1} \sqrt{x^{2}+x+1}=\frac{\pi}{4}$ is :

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Inverse Trigonometric Functions

The number of real roots of the equation

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

is :

Answer: (C)

The first term needs

x (x + 1) \geq 0

, that is

x \leq - 1

x \geq 0

, while the second needs

x^{2} + x + 1 \leq 1

, that is

x^{2} + x \leq 0

- 1 \leq x \leq 0

. The two requirements overlap only at

x = - 1

and

x = 0

, and at each of these

x (x + 1) = 0

and

x^{2} + x + 1 = 1

, so

tan^{- 1} 0 + sin^{- 1} 1 = 0 + \frac{π}{2} = \frac{π}{2}

which is larger than

\frac{π}{4}

. The equation has no real root.

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