SATHEE: Definite Integration Question 11

Definite Integration Question 11

Question 11

$\int_{0}^{x} f (t) d t = x + \int_{x}^{1} t f (t) d t$ , then the value of $f (1)$ is (a) $\frac{1}{2}$ (b) 0 (c) 1 (d) $- \frac{1}{2}$

(1998, 2M)

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Answer:

Correct Answer: 11. (a)

Solution:

Given, $\int_{0}^{x} f (t) d t = x + \int_{x}^{1} t f (t) d t$

On differentiating both sides w.r.t. $x$ , we get

$\begin{aligned} f (x) 1 & = 1 - x f (x) \cdot 1 \Rightarrow (1 + x) f (x) = 1 \Rightarrow & f (x) = \frac{1}{1 + x} \Rightarrow f (1) = \frac{1}{1 + 1} = \frac{1}{2} \end{aligned}$

Definite Integration Question 11

Question 11

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Definite Integration Question 11

Question 11

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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