MCQ
If $f(x) = \frac{{x - 3}}{{x + 1}}$, then $f[f\{ f(x)\} ]$ equals
  • $x$
  • B
    $-x$
  • C
    $\frac{x}{2}$
  • D
    $ - \frac{1}{x}$

Answer

Correct option: A.
$x$
a
(a) $f\,[f(x)] = \frac{{f(x) - 3}}{{f(x) + 1}}$

$ = \frac{{\left( {\frac{{x - 3}}{{x + 1}}} \right) - 3}}{{\left( {\frac{{x - 3}}{{x + 1}}} \right) + 1}} = \frac{{x - 3 - 3x - 3}}{{x - 3 + x + 1}} = \frac{{3 + x}}{{1 - x}}$

Now $f\,[f(f(x))] = f\,\left( {\frac{{3 + x}}{{1 - x}}} \right)$

$ = \frac{{\left( {\frac{{x - 3}}{{x + 1}}} \right) - 3}}{{\left( {\frac{{x - 3}}{{x + 1}}} \right) + 1}} = \frac{{x - 3 - 3x - 3}}{{x - 3 + x + 1}} = x$

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