SATHEE: Definite Integration Question 33

Definite Integration Question 33

Question 33

Let $f$ be a positive function.

If $I_{1} = \int_{1 - k}^{k} x f [x (1 - x)] d x$ and

$I_{2} = \int_{1 - k}^{k} f [x (1 - x)] d x, where 2 k - 1 > 0$

Then, $\frac{I_{1}}{I_{2}}$ is

(1997C, 2M) (a) 2 (b) $k$ (c) $1 / 2$ (d) 1

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Solution:

Given, $I_{1} = \int_{1 - k}^{k} x f [x (1 - x)] d x$

$\begin{aligned} \Rightarrow I_{1} & = \int_{1 - k}^{k} (1 - x) f [(1 - x) x] d x & = \int_{1 - k}^{k} f [(1 - x)] d x] - \int_{1 - k}^{k} x f (1 - x)] d x \Rightarrow I_{1} & = I_{2} - I_{1} \Rightarrow \frac{I_{1}}{I_{2}} = \frac{1}{2} \end{aligned}$

Definite Integration Question 33

Question 33

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Definite Integration Question 33

Question 33

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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