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Let f: \mathbb{R}-0,1 \rightarrow \mathbb{R} be a function such that f(x)+f\left(\frac{1}{1-x}\right)=1+x .Then f(2) is equal to:

Option: 1

\frac{9}{2}


Option: 2

\frac{7}{4}


Option: 3

\frac{9}{4}


Option: 4

\frac{7}{3}


Answers (1)

best_answer

For \: x=2, \Rightarrow f(2)+f(-1)=3\\ _________(I)

For \: x=\frac{1}{2} \Rightarrow f\left(\frac{1}{2}\right)+f(-1)=\frac{3}{2} _____________(II)

For \: x=-1 \Rightarrow f(-1)+f\left(\frac{1}{2}\right)=0 _____________(III)

\text { (2) }-(3) \Rightarrow f(2)-f(-1)=\frac{3}{2} ____________(IV)

(1)+(4) \Rightarrow 2 f(2)=\frac{9}{2} \Rightarrow f(2)=\frac{9}{4}

 

 

Posted by

Gautam harsolia

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