Indefinite Integration 3 Question 11

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:

  1. 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

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



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