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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
Let f(x) =$\int_1^x \sqrt{2-t^2}dt $ Then, the real roots of the
equation $x^2-f'(x)=0$ are
Updated On: Jun 14, 2022
- $\pm$1
- $\pm\frac{1}{\sqrt 2}$
- $\pm\frac{1}{ 2}$
- 0 and 1
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The Correct Option is A
Solution and Explanation
Given, f(x) =$\int_1^x \sqrt{2-t^2}dt \, \Rightarrow \, \, f'(x)=\sqrt{2-x^2}$
Also, $x^2-f'(x)=0$
$\therefore \, \, \, \, \, x62=\sqrt{2-x^2} \, \Rightarrow \, \, x^4=2-x^2$
$\Rightarrow \, \, \, \, x^4+x^2-2=0 \, \, \, \Rightarrow \, \, x=\pm 1$
Also, $x^2-f'(x)=0$
$\therefore \, \, \, \, \, x62=\sqrt{2-x^2} \, \Rightarrow \, \, x^4=2-x^2$
$\Rightarrow \, \, \, \, x^4+x^2-2=0 \, \, \, \Rightarrow \, \, x=\pm 1$
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A relation between involved variables, which satisfy the given differential equation is called its solution. The solution which contains as many arbitrary constants as the order of the differential equation is called the general solution and the solution free from arbitrary constants is called particular solution.
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