Let f(x)=∫1x 2−t2dt, Then the real roots of the equation x2−f′(x)=0 are

A
$\pm 1$
B
$\pm \frac{1}{\sqrt{2}}$
C
$\pm \frac{1}{2}$
D
0 and 1

Solution:

We have,

$\begin{array}{l} f (x) = \int_{1}^{x} \sqrt{2 - t^{2}} d t \Rightarrow f^{'} (x) = \sqrt{2 - x^{2}} \\ ∴ x^{2} - f^{'} (x) = 0 \\ \Rightarrow x^{2} - \sqrt{2 - x^{2}} = 0 \\ \Rightarrow x^{4} = 2 - x^{2} \\ \Rightarrow x^{4} + x^{2} - 2 = 0 \\ \Rightarrow (x^{2} + 2) (x^{2} - 1) = 0 \Rightarrow x = \pm 1 [∵ x^{2} + 2 \neq 0] \end{array}$

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