Let f(x)=axx+1,x≠−1 Then value of the constant a such that f(f(x))=x is

f(x)=axx+1⇒f(f(x))=af(x)f(x)+1=aaxx+1axx+1+1=a2x(a+1)x+1=x if a = -1

Let f(x)=axx+1,x≠−1 Then value of the constant a such that f(f(x))=x is

f(x)=axx+1⇒f(f(x))=af(x)f(x)+1=aaxx+1axx+1+1=a2x(a+1)x+1=x if a = -1

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Let $f (x) = \frac{a x}{x + 1}, x \neq - 1$ Then value of the constant a such that $f (f (x)) = x$ is

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$\sqrt{2}$

$- \sqrt{2}$

-1

answer is D.

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$f (x) = \frac{a x}{x + 1} \Rightarrow f (f (x)) = \frac{a f (x)}{f (x) + 1} = \frac{a (\frac{a x}{x + 1})}{\frac{a x}{x + 1} + 1}$

$= \frac{a^{2} x}{(a + 1) x + 1} = x$ if a = -1

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Let f(x)=axx+1,x≠−1 Then value of the constant a such that f(f(x))=x is

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Let $f (x) = \frac{a x}{x + 1}, x \neq - 1$ Then value of the constant a such that $f (f (x)) = x$ is