Let f(x)=kxx+1,x≠−1 Then, the value of k for which ff(x)=x is.

ff(x)=k2x(x+1)kxx+1+1=k2x(k+1)x+1Since f f(x) must be = x⇒k+l=0,k=−1

Let f(x)=kxx+1,x≠−1 Then, the value of k for which ff(x)=x is.

ff(x)=k2x(x+1)kxx+1+1=k2x(k+1)x+1Since f f(x) must be = x⇒k+l=0,k=−1

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Let $f (x) = \frac{k x}{x + 1}, x \neq - 1$ Then, the value of k for which $f (f (x)) = x$ is.

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$\frac{1}{2}$

$- 1$

answer is C.

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$f f (x) = \frac{\frac{k^{2} x}{(x + 1)}}{\frac{k x}{x + 1} + 1} = \frac{k^{2} x}{(k + 1) x + 1}$

Since f f(x) must be = x

$\Rightarrow k + l = 0, k = - 1$

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Let f(x)=kxx+1,x≠−1 Then, the value of k for which ff(x)=x is.

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Let $f (x) = \frac{k x}{x + 1}, x \neq - 1$ Then, the value of k for which $f (f (x)) = x$ is.