JEE PYQ: Functions Question 19
Question 19 - 2020 (06 Sep Shift 2)
For a suitably chosen real constant $a$, let a function, $f : \mathbb{R} - {-a} \to \mathbb{R}$ 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:
(1) $\frac{1}{3}$
(2) $-\frac{1}{3}$
(3) $-3$
(4) $3$
Type: MCQ
Show Answer
Answer: (4) $3$
Solution
$a = 1$, $f(x) = \frac{1-x}{1+x}$, $f\left(-\frac{1}{2}\right) = 3$
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