A) \[\frac{1}{1-x}\]
B) \[\frac{1}{x-1}\]
C) \[\frac{1}{1+x}\]
D) \[\frac{1}{x}\]
Correct Answer: C
Solution :
[c] : We have, \[\int\limits_{0}^{x}{f}(t)dt=x+\int\limits_{x}^{1}{t}f(t)dt\] \[\Rightarrow \]\[\frac{d}{dx}\int\limits_{0}^{x}{f}(t)dt=\frac{d}{dx}(x)+\frac{d}{dx}\int\limits_{x}^{1}{t}\,f(t)dt\] \[\Rightarrow \]\[f(x)\frac{d}{dx}(x)-f(0).\frac{d}{dx}(0)\] \[=\frac{d}{dx}(x)+1\times f(1)\frac{d}{dx}(1)-xf(x)\frac{d}{dx}(x)\] \[\Rightarrow \]\[f(x)-0=1+0-x\,f(x)\Rightarrow f(x)=1-xf(x)\] \[\Rightarrow \]\[f(x)+x\,f(x)=1\Rightarrow f(x)(1+x)=1\] \[\Rightarrow \]\[f(x)=\frac{1}{1+x}\]
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