Let $f(x)=\int\limits_1^x \sqrt{2-t^2} d t$. Then the real roots of the equation $x^2-f'(x)=0$ are
Answer & explanation
Correct answer: option 1
We have,
$f(x)=\int\limits_1^x \sqrt{2-t^2} d t \Rightarrow f'(x)=\sqrt{2-x^2}$
∴ $x^2-f'(x)=0$
$\Rightarrow x^2-\sqrt{2-x^2}=0$
$\Rightarrow x^4=2-x^2$
$\Rightarrow x^4+x^2-2=0$
$\Rightarrow \left(x^2+2\right)\left(x^2-1\right)=0 \Rightarrow x= \pm 1 \quad\left[∵ x^2+2 \neq 0\right]$