# Find the real roots of the equation \[{{\text{x}}^{\text{2}}}{\text{ - f '(x) = 0}}\] if ${\text{f(x) = }}\int\limits_{\text{1}}^{\text{x}} {\sqrt {{\text{2 - }}{{\text{t}}^{\text{2}}}} {\text{dt}}} $

A. $ \pm 1$

B. $ \pm \dfrac{1}{2}$

C. $ \pm \dfrac{1}{2}$

D. 0 and 1

Last updated date: 27th Mar 2023

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Answer

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Hint: In order to solve this problem use the concept that differentiation is the inverse of integration and vice-versa. Using this concept you can get the roots of the equation given.

Complete step-by-step answer:

The given equations are ${\text{f(x) = }}\int\limits_{\text{1}}^{\text{x}} {\sqrt {{\text{2 - }}{{\text{t}}^{\text{2}}}} {\text{dt}}} $ and \[{{\text{x}}^{\text{2}}}{\text{ - f '(x) = 0}}\]

On solving ${\text{f(x) = }}\int\limits_{\text{1}}^{\text{x}} {\sqrt {{\text{2 - }}{{\text{t}}^{\text{2}}}} {\text{dt}}} $ we get f(x) in terms of x as in limit there is x so, t will be replaced by x.

So, f(x) can be written as ${\text{f(x) = }}\int\limits_{\text{1}}^{\text{x}} {\sqrt {{\text{2 - }}{{\text{x}}^{\text{2}}}} {\text{dx}}} $

And we know integration of f ‘(x) is f(x) similarly differentiation of f(x) is f ‘(x).

If ${\text{f(x) = }}\int\limits_{\text{1}}^{\text{x}} {\sqrt {{\text{2 - }}{{\text{x}}^{\text{2}}}} {\text{dx}}} $

So, ${\text{f '(x) = }}\sqrt {{\text{2 - }}{{\text{x}}^{\text{2}}}} $

The other equation is \[{{\text{x}}^{\text{2}}}{\text{ - f '(x) = 0}}\].

On putting the value of ${\text{f '(x) = }}\sqrt {{\text{2 - }}{{\text{x}}^{\text{2}}}} $ in the above equation we get the equation as:

\[{{\text{x}}^{\text{2}}} - \sqrt {2 - {x^2}} = 0\]

\[{{\text{x}}^{\text{2}}}{\text{ = }}\sqrt {{\text{2 - }}{{\text{x}}^{\text{2}}}} \]

On squaring both sides we get,

${{\text{x}}^{\text{4}}}{\text{ + }}{{\text{x}}^{\text{2}}}{\text{ - 2 = 0}}$

${{\text{x}}^{\text{2}}}{\text{ = 1, - 2}}$

So, the real values of ${\text{x} = \pm 1}$

So, the correct option for this question is A.

Note: Whenever you face such types of problems you have to use the concept that integration of f ‘(x) is f(x) similarly differentiation of f(x) is f ‘(x). Here in this question we have then found the roots of the equation obtained then eliminated the imaginary roots as only real roots have been asked in the question. Proceeding like this will take you to the right answer.

Complete step-by-step answer:

The given equations are ${\text{f(x) = }}\int\limits_{\text{1}}^{\text{x}} {\sqrt {{\text{2 - }}{{\text{t}}^{\text{2}}}} {\text{dt}}} $ and \[{{\text{x}}^{\text{2}}}{\text{ - f '(x) = 0}}\]

On solving ${\text{f(x) = }}\int\limits_{\text{1}}^{\text{x}} {\sqrt {{\text{2 - }}{{\text{t}}^{\text{2}}}} {\text{dt}}} $ we get f(x) in terms of x as in limit there is x so, t will be replaced by x.

So, f(x) can be written as ${\text{f(x) = }}\int\limits_{\text{1}}^{\text{x}} {\sqrt {{\text{2 - }}{{\text{x}}^{\text{2}}}} {\text{dx}}} $

And we know integration of f ‘(x) is f(x) similarly differentiation of f(x) is f ‘(x).

If ${\text{f(x) = }}\int\limits_{\text{1}}^{\text{x}} {\sqrt {{\text{2 - }}{{\text{x}}^{\text{2}}}} {\text{dx}}} $

So, ${\text{f '(x) = }}\sqrt {{\text{2 - }}{{\text{x}}^{\text{2}}}} $

The other equation is \[{{\text{x}}^{\text{2}}}{\text{ - f '(x) = 0}}\].

On putting the value of ${\text{f '(x) = }}\sqrt {{\text{2 - }}{{\text{x}}^{\text{2}}}} $ in the above equation we get the equation as:

\[{{\text{x}}^{\text{2}}} - \sqrt {2 - {x^2}} = 0\]

\[{{\text{x}}^{\text{2}}}{\text{ = }}\sqrt {{\text{2 - }}{{\text{x}}^{\text{2}}}} \]

On squaring both sides we get,

${{\text{x}}^{\text{4}}}{\text{ + }}{{\text{x}}^{\text{2}}}{\text{ - 2 = 0}}$

${{\text{x}}^{\text{2}}}{\text{ = 1, - 2}}$

So, the real values of ${\text{x} = \pm 1}$

So, the correct option for this question is A.

Note: Whenever you face such types of problems you have to use the concept that integration of f ‘(x) is f(x) similarly differentiation of f(x) is f ‘(x). Here in this question we have then found the roots of the equation obtained then eliminated the imaginary roots as only real roots have been asked in the question. Proceeding like this will take you to the right answer.

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