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1.Relation and Function
normal
જો $f(x) = \frac{{\alpha x}}{{x + 1}},x \ne - 1$, તો $\alpha $ ની . . . . કિમત માટે $f(f(x)) = x$ મળે.
A
$\sqrt 2 $
B
$ - \sqrt 2 $
C
$1$
D
$-1$
Solution
(d) $f(x) = \frac{{\alpha x}}{{x + 1}}$;
$f(f(x)) = f\left( {\frac{{\alpha x}}{{x + 1}}} \right) = \frac{{\alpha \left( {\frac{{\alpha x}}{{x + 1}}} \right)}}{{\frac{{\alpha x}}{{x + 1}} + 1}}$
But $f(f(x)) = x$,
$\therefore$ $\frac{{{\alpha ^2}x}}{{\alpha x + x + 1}} = x$
Put $\alpha = – 1$, $\frac{{{{( – 1)}^2}x}}{{( – 1)x + x + 1}} = \frac{x}{{ – x + x + 1}} = x$;
$\therefore$ $\alpha = – 1$.
Standard 12
Mathematics
