Question

Let f(x) =$\int_1^x \sqrt{2-t^2}dt$ Then, the 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

Answer 1

Answer: (A)

$\pm$1

Answer 2

Given, f(x) =$\int_1^x \sqrt{2-t^2}dt \, \Rightarrow \, \, f'(x)=\sqrt{2-x^2}$
Also, $x^2-f'(x)=0$
$\therefore \, \, \, \, \, x62=\sqrt{2-x^2} \, \Rightarrow \, \, x^4=2-x^2$
$\Rightarrow \, \, \, \, x^4+x^2-2=0 \, \, \, \Rightarrow \, \, x=\pm 1$