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JEE MAIN 2020
06-09-2020 S2
Question
For a suitably chosen real constant a , let a function, $\mathrm{f}: \mathrm{R}-\{-\mathrm{a}\} \rightarrow \mathrm{R}$ be defined by $f(x)=\frac{a-x}{a+x}$. Further suppose that for any real number $\mathrm{x} \neq-\mathrm{a}$ and $\mathrm{f}(\mathrm{x}) \neq-\mathrm{a}$, $(\mathrm{fof})(\mathrm{x})=\mathrm{x}$. Then $f\left(-\frac{1}{2}\right)$ is equal to:
Select the correct option:
A
-3
B
$\frac{1}{3}$
C
$-\frac{1}{3}$
D
3
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
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Question Tags
JEE Main
Mathematics
Medium
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