If $f(x) = \frac{{\alpha \,x}}{{x + 1}},\;x \ne - 1$. Then, for what value of $\alpha $ is $f(f(x)) = x$
If $f(x) = \frac{{\alpha \,x}}{{x + 1}},\;x \ne - 1$. Then, for what value of $\alpha $ is $f(f(x)) = x$
- [IIT 2001]
- A
$\sqrt 2 $
- B
$ - \sqrt 2 $
- C
$1$
- D
$-1$
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Statement $1$ : If $A$ and $B$ be two sets having $p$ and $q$ elements respectively, where $q > p$. Then the total number of functions from set $A$ to set $B$ is $q^P$.
Statement $2$ : The total number of selections of $p$ different objects out of $q$ objects is ${}^q{C_p}$.
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Statement $2$ : The total number of selections of $p$ different objects out of $q$ objects is ${}^q{C_p}$.
- [AIEEE 2012]
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- [KVPY 2013]