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1.Relation and Function
medium
If $f(x) = \frac{{\alpha \,x}}{{x + 1}},\;x \ne - 1$. Then, for what value of $\alpha $ is $f(f(x)) = x$
A
$\sqrt 2 $
B
$ - \sqrt 2 $
C
$1$
D
$-1$
(IIT-2001)
Solution
(d) $f(f(x)) = \frac{{\alpha \,f(x)}}{{f(x) + 1}} $
$= \frac{{\alpha \left( {\frac{{\alpha x}}{{x + 1}}} \right)}}{{\left( {\frac{{\alpha x}}{{x + 1}} + 1} \right)}} = \frac{{{\alpha ^2}.x}}{{\alpha x + x + 1}}$
$\therefore$ $x = \frac{{{\alpha ^2}.x}}{{(\alpha + 1)x + 1}}$ or $x((\alpha + 1)x + 1 – {\alpha ^2}) = 0$
==> $(\alpha + 1){x^2} + (1 – {\alpha ^2})x = 0$. This should hold for all $x.$
==> $\alpha + 1 = 0,\,\,1 – {\alpha ^2} = 0$,
$\because \alpha = – 1$.
Standard 12
Mathematics
