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Mathematics Question on composite of functions

Let
f(x)=\frac{x−1}{x+1},x∈R− \left\\{0,−1,1\right\\}
If ƒn+1(x) = ƒ(ƒn(x)) for all n∈N, then ƒ6(6) + ƒ7(7) is equal to :

A

76\frac{7}{6}

B

−32-\frac{3}{2}

C

712\frac{7}{12}

D

−1112-\frac{11}{12}

Answer

−32-\frac{3}{2}

Explanation

Solution

The correct answer is (B) : −32-\frac{3}{2}
f(x)=x−1x+1⇒f(f(x))=(x−1x+1)−1(x−1x+1)+1=−1xf(x) = \frac{x - 1}{x + 1} \Rightarrow f(f(x)) = \frac{\left(\frac{x - 1}{x + 1}\right) - 1}{\left(\frac{x - 1}{x + 1}\right) + 1} = -\frac{1}{x}
⇒f3(x)=−x+1x−1⇒f4(x)=−(x−1x+1+1)(x−1x+1−1)=x\Rightarrow f^3(x) = -\frac{x + 1}{x - 1} \Rightarrow f^4(x) = -\frac{\left(\frac{x - 1}{x + 1} + 1\right)}{\left(\frac{x - 1}{x + 1} - 1\right)} = x
So, ƒ6(6) + ƒ7(7) = ƒ2(6) + ƒ3(7)
=−16−7+17−1=−96=−32= -\frac{1}{6}-\frac{7+1}{7-1}= - \frac{9}{6}=-\frac{3}{2}