Let fx-1x 2-t2 dt- Then the real roots of the equation x2-f-x-0 are

We have, f(x)=∫1x nbsp;2 t2 nbsp;dt⇒f'(x)=2 x2Now nbsp;the nbsp;given nbsp;equation nbsp;x2 f'(x)=0 nbsp;becomesx2 2 x2=0⇒x2=2 x2⇒x4+x2 2=0⇒x4+2x2 ...

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Standard XII

Mathematics

Differentiation under Integral Sign

Let fx=∫1x 2-...

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

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Solution

The correct option is B $- 1$
We have,
$f (x) = \int_{1}^{x} \sqrt{2 - t^{2}} dt \Rightarrow f' (x) = \sqrt{2 - x^{2}} Now the given equation x^{2} - f' (x) = 0 becomes x^{2} - \sqrt{2 - x^{2}} = 0 \Rightarrow x^{2} = \sqrt{2 - x^{2}} \Rightarrow x^{4} + x^{2} - 2 = 0 \Rightarrow x^{4} + 2 x^{2} - x^{2} - 2 = 0 \Rightarrow x^{2} (x^{2} + 2) - 1 (x^{2} + 2) = 0 \Rightarrow (x^{2} + 2) (x^{2} - 1) = 0 \Rightarrow x = \pm 1$

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Let f(x)=∫x1 √2−t2 dt. Then the real roots of the equation x2−f'(x)=0 are

The correct option is B −1We have, f(x)=∫x1 √2−t2 dt⇒f'(x)=√2−x2Now the given equation x2−f'(x)=0 becomesx2−√2−x2=0⇒x2=√2−x2⇒x4+x2−2=0⇒x4+2x2−x2−2=0⇒x2(x2+2)−1(x2+2)=0⇒(x2+2)(x2−1)=0⇒x=±1

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