Q. The number of real solution of $\tan^{- 1}{\sqrt{x(x + 1)} + \sin^{- 1}\ \sqrt{x^{2} + x + 1} = \frac{\pi}{2}}$is
A) 0
B) 1
C) 2
D) $\infty$
Explanation
(c)
Clearly, $x(x + 1) \geq 0\ and\ x^{2} + x + 1 \leq 1$
$\Rightarrow x(x + 1) = 0$
$\Rightarrow x = 0, - 1$
When $x = 0,$
LHS=$\tan^{- 1}{0 + \sin^{- 1}{1 = \frac{\pi}{2}}}$
When $x = - 1,$
LHS=$\tan^{- 1}{0 + \sin^{- 1}\sqrt{1 - 1 + 1}}$
$= 0 + \sin^{- 1}{(1) = \frac{\pi}{2}}$
Thus, the number of solution is 2
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.