Q. The value of $'a^{'}$for which $ax^{2} + \sin^{- 1}{\left( x^{2} - 2x + 2 \right) + \cos^{- 1}{\left( x^{2} - 2x + 2 \right) = 0}}$ has a real solution, is
A) $- \frac{2}{\pi}$
B) $\frac{2}{\pi}$
C) $- \frac{\pi}{2}$
D) $\frac{\pi}{2}$
Explanation
(c)
Here, $x^{2} - 2x + 2 = (x - 1)^{2} + 1 \geq 1$
But $- 1 \leq (x^{2} - 2x + 2) \leq 1$
Which is possible only when
$x^{2} - 2x + 2 = 1$
$\Rightarrow x = 1$
Then, $a(1)^{2} + \sin^{- 1}{(1) + \cos^{- 1}{(1) = 0}}$
$\Rightarrow a + \frac{\pi}{2} + 0 = 0$
$\Rightarrow a = - \frac{\pi}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.