1.Relation and Function
normal

If $f(x) = \frac{{\alpha x}}{{x + 1}},x \ne - 1$, for what value of $\alpha $ is $f(f(x)) = x$

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

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