Question

The number of real solutions 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

$\tan ^ { - 1 } \sqrt { x ( x + 1 ) } + \sin ^ { - 1 } \sqrt { x ^ { 2 } + x + 1 } = \frac { \pi } { 2 }$ $\tan^{- 1}\sqrt{x(x + 1)}$is defined, when $x(x + 1) \geq 0$ ........(i)

$\sin^{- 1}\sqrt{x^{2} + x + 1}$ is defined, when $0 \leq x(x + 1) + 1 \leq 1$ or

$0 \leq x(x + 1) \leq 0$ ........(ii)

From (i) and (ii), $x(x + 1) = 0$ or $x = 0$ and –1.

Hence, number of solutions is 2