MCQ
જો $f(x) = \frac{{x - 3}}{{x + 1}}$, તો $f[f\{ f(x)\} ] = . . ..$
- ✓$x$
- B$-x$
- C$\frac{x}{2}$
- D$ - \frac{1}{x}$
$ = \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$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.