JEE Main 2020 (Online) 6th September Evening ShiftQuestion 1
For a suitably chosen real constant $a$, let a function, $f:\mathbf{R}-\{-a\}\to\mathbf{R}$ be defined by $f(x)=\frac{a-x}{a+x}$. Further suppose that for any real number $x\ne-a$ and $f(x)\ne-a$, $(f\circ f)(x)=x$. Then $f\left(-\frac12\right)$ is equal to :
Sets, Relations and Function
Solution
$f(f(x))=\frac{a-f}{a+f}=\frac{a^2+ax-a+x}{a^2+ax+a-x}=x$ for all $x$. Cross-multiplying: $a^2+ax-a+x=a^2x+ax^2+ax-x^2$; comparing $x^2$ coefficients: $0=a-1\Rightarrow a=1$ (other coefficients then agree). $f(x)=\frac{1-x}{1+x}$, $f(-\tfrac12)=\frac{3/2}{1/2}=3$. Option (A).