Practicing Success
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 |
$\pm 1$ $\pm \frac{1}{\sqrt{2}}$ $\pm \frac{1}{2}$ 0 and 1 |
$\pm 1$ |
We have, $f(x)=\int\limits_1^x \sqrt{2-t^2} d t \Rightarrow f'(x)=\sqrt{2-x^2}$ ∴ $x^2-f'(x)=0$ $\Rightarrow x^2-\sqrt{2-x^2}=0$ $\Rightarrow x^4=2-x^2$ $\Rightarrow x^4+x^2-2=0$ $\Rightarrow \left(x^2+2\right)\left(x^2-1\right)=0 \Rightarrow x= \pm 1 \quad\left[∵ x^2+2 \neq 0\right]$ |