Mathematics · 2020
JEE Main · 6 September 2020, Shift 2 · Q51
For a suitably chosen real constant a, let a function, f: R -{- a } → R be defined by f(x)=(a -x)/(a +x). Further suppose that for any real number x…
For a suitably chosen real constant a, let a function, $\displaystyle f: \mathbf{R}-\{-\mathbf{a}\} \rightarrow \mathbf{R}$ be defined by $\displaystyle f(x)=\frac{\mathrm{a}-x}{\mathrm{a}+x}$. Further suppose that for any real number $\displaystyle x \neq-a$ and $\displaystyle f(x) \neq-a$, $\displaystyle (f \circ f)(x)=x$. Then $\displaystyle f\left(-\frac{1}{2}\right)$ is equal to :
Official answer
(1)
$\displaystyle 3$
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JEE Main 2020 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.