SATHEE: Definite Integration Question 9

Definite Integration Question 9

Question 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) $\pm \frac{1}{\sqrt{2}}$ (c) $\pm \frac{1}{2}$ (d) 0 and 1

Show Answer

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

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

Definite Integration Question 9

Question 9

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Definite Integration Question 9

Question 9

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu