Q. Let $f(x) = \int_{1}^{x}{\sqrt{2 - t^{2}}\ dt.}$ Then, real roots of the equation $x^{2} - f^{'}(x) = 0$ are
A) $\pm 1$
B) $\pm \frac{1}{\sqrt{2}}$
C) $\pm \frac{1}{2}$
D) 0 and 1
Explanation
(a)
Given, $f(x) = \int_{1}^{x}\sqrt{2 - t^{2}}dt$
$\Rightarrow f^{'}(x) = \sqrt{2 - x^{2}}$
Now, $x^{2} - f^{'}(x) = 0$
$\Rightarrow x^{2} - \sqrt{2 - x^{2}} = 0$
$\Rightarrow x^{4} + x^{2} - 2 = 0$
$\Rightarrow \left( x^{2} + 2 \right)\left( x^{2} - 1 \right) = 0$
$\Rightarrow x^{2} - 1 = 0\ \Rightarrow x = \pm 1$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.