Question

The number of real solution of

tan^–1$\sqrt{x(x + 1)}$+ sin^–1$\sqrt{x^{2} + x + 1}$ = $\frac{\pi}{2}$ is–

• A. Zero
• B. One
• C. Two
• D. Infinite

Answer 1

Answer: (C)

Two

Answer 2

Clearly, x(x + 1) ³ 0 and x² + x + 1 £ 1. Together they imply x (x + 1) = 0.

\ x = 0, –1.

When x = 0,

L.H.S. = tan^–1 0 + sin^–1 1 = $\frac{\pi}{2}$.

When x = –1,

L.H.S. = tan^–1 0 + sin^–1 $\sqrt{1 - 1 + 1}$= 0 + sin^–1 1 =$\frac{\pi}{2}$,

Thus two solution.