JEE MAIN - Mathematics (2023 - 1st February Evening Shift - No. 15)
Let $$f:\mathbb{R}-{0,1}\to \mathbb{R}$$ be a function such that $$f(x)+f\left(\frac{1}{1-x}\right)=1+x$$. Then $$f(2)$$ is equal to
$$\frac{9}{4}$$
$$\frac{7}{4}$$
$$\frac{7}{3}$$
$$\frac{9}{2}$$
Explanation
$\begin{aligned} & \mathrm{f}(\mathrm{x})+\mathrm{f}\left(\frac{1}{1-\mathrm{x}}\right)=1+\mathrm{x} \\\\ & \mathrm{x}=2 \Rightarrow \mathrm{f}(2)+\mathrm{f}(-1)=3 ........(1) \\\\ & \mathrm{x}=-1 \Rightarrow \mathrm{f}(-1)+\mathrm{f}\left(\frac{1}{2}\right)=0 .........(2) \\\\ & \mathrm{x}=\frac{1}{2} \Rightarrow \mathrm{f}\left(\frac{1}{2}\right)+\mathrm{f}(2)=\frac{3}{2} ........(3) \\\\ & (1)+(3)-(2) \Rightarrow 2 \mathrm{f}(2)=\frac{9}{2} \\\\ & \therefore \mathrm{f}(2)=\frac{9}{4}\end{aligned}$
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