For a suitably chosen real constant $a$, let a function, $f: \mathbb{R}-\{-a\} \rightarrow \mathbb{R}$ be defined by $f(x)=\frac{a-x}{a+x}$. Further suppose that for any real number $x \neq -a$ and $f(x) \neq -a$, $(f \circ f)(x)=x$. Then $f\left(-\frac{1}{2}\right)$ is equal to:

JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Sets Relations and Functions

For a suitably chosen real constant

a

, let a function,

f : R - {- a} \to R

be defined by

f (x) = \frac{a - x}{a + x}

. Further suppose that for any real number

x \neq = - a

and

f (x) \neq = - a

(f \circ f) (x) = x

. Then

f (- \frac{1}{2})

is equal to:

Answer: (D)

The requirement

(f \circ f) (x) = x

pins down the constant. Computing

f (f (x)) = \frac{a - \frac{a - x}{a + x}}{a + \frac{a - x}{a + x}}

and demanding it equal

x

forces

a = 1

, so

f (x) = \frac{1 - x}{1 + x}

. Then

f (- \frac{1}{2}) = \frac{1 + \frac{1}{2}}{1 - \frac{1}{2}} = \frac{3/2}{1/2} = 3

JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Single correct option