Functions Question 20
Question 20 - 01 February - Shift 2
Let $f: R-{0,1} \to R$ be a function such that $f(x)+f(\frac{1}{1-x})=1+x$. Then $f(2)$ is equal to :
(1) $\frac{9}{2}$
(2) $\frac{9}{4}$
(3) $\frac{7}{4}$
(4) $\frac{7}{3}$
Show Answer
Answer: (2)
Solution:
Formula: Operations on functions
$f(x)+f(\frac{1}{1-x})=1+x$
$x=2 \Rightarrow f(2)+f(-1)=3 \ldots \ldots (1)$
$x=-1 \Rightarrow f(-1)+f(\frac{1}{2})=0 \ldots \ldots (2)$
$x=\frac{1}{2} \Rightarrow f(\frac{1}{2})+f(2)=\frac{3}{2} \ldots \ldots (3)$
$(1)+(3)-(2) \Rightarrow 2 f(2)=\frac{9}{2}$
$\therefore f(2)=\frac{9}{4}$
