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Let $f (x) = \int_{1}^{x} \sqrt{2 - t^{2}} d t$ Then the real roots of the equation $x^{2} - f^{'} (x) = 0$

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f′(x)=2−x2 Now, x2−f′(x)=0⇒x2=2−x2⇒x=±1

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

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Ready to Test Your Skills?

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Let $f (x) = \int_{1}^{x} \sqrt{2 - t^{2}} d t$ Then the real roots of the equation $x^{2} - f^{'} (x) = 0$