यदि f(x) = frac{{alpha x}}{{x + 1}},x ne - 1, | vedclass

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Question

(e) $f(x) = \frac{{\alpha x}}{{x + 1}}$;

$f(f(x)) = f\left( {\frac{{\alpha x}}{{x + 1}}} \right) = \frac{{\alpha \left( {\frac{{\alpha x}}{{x + 1}}

Vedclass · Accepted Answer

$-1$

Vedclass · Answer

(e) $f(x) = \frac{{\alpha x}}{{x + 1}}$;

$f(f(x)) = f\left( {\frac{{\alpha x}}{{x + 1}}} ight) = \frac{{\alpha \left( {\frac{{\alpha x}}{{x + 1}}} ight)}}{{\frac{{\alpha x}}{{x + 1}} + 1}}$

लेकिन $f(f(x)) = x$,

$	herefore$  $\frac{{{\alpha ^2}x}}{{\alpha x + x + 1}} = x$

$\alpha = - 1$ रखने पर

$\frac{{{{( - 1)}^2}x}}{{( - 1)x + x + 1}} = \frac{x}{{ - x + x + 1}} = x$;

$	herefore$ $\alpha = - 1$.

यदि $f(x) = \frac{{\alpha x}}{{x + 1}},x \ne - 1$, $f(f(x)) = x$, $\alpha $ का मान क्या है

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