SATHEE: Indefinite Integration 3 Question 9

Indefinite Integration 3 Question 9

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

(a) \pm 1

(b) $\pm \frac{1}{\sqrt{2}}$

(d) 0 and 1

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Answer:

Correct Answer: 9. (a)

Solution:

Given, $f (x) = \int_{1}^{x} \sqrt{2 - t^{2}} d t \Rightarrow f^{'} (x) = \sqrt{2 - x^{2}}$

Also, $x^{2} - f^{'} (x) = 0$

$\begin{aligned} ∴ & x^{2} = \sqrt{2 - x^{2}} \Rightarrow x^{4} = 2 - x^{2} \\ \Rightarrow & x^{4} + x^{2} - 2 & = 0 \Rightarrow x = \pm 1 \end{aligned}$

Indefinite Integration 3 Question 9

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

Answer:

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Indefinite Integration 3 Question 9

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

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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