Let $f(x) = {{x - 1} \over {x + 1}},\,x \in R - \{ 0, - 1,1\}$. If ${f^{n + 1}}(x) = f({f^n}(x))$ for all n $\in$ N, then ${f^6}(6) + {f^7}(7)$ is equal to :

<p>Given,</p> <p>$f(x) = {{x - 1} \over {x + 1}}$</p> <p>Also given,</p> <p>${f^{n + 1}}(x) = f({f^n}(x))$ ..... (1)</p> <p>$\therefore$ For $n = 1$</p> <p>${f^{1 + 1}}(x) = f({f^1}(x))$</p> <p>$\Rightarrow {f^2}(x) = f(f(x))$</p> <p>$= f\left( {{{x - 1} \over {x + 1}}} \right)$</p> <p>$= {{{{x - 1} \over {x + 1}} - 1} \over {{{x - 1} \over {x + 1}} + 1}}$</p> <p>$= {{{{x - 1 - x - 1} \over {x + 1}}} \over {{{x - 1 + x + 1} \over {x + 1}}}}$</p> <p>$= {{ - 2} \over {2x}}$</p> <p>$= - {1 \over x}$</p> <p>From equation (1), when n = 2</p> <p>${f^{2 + 1}}(x) = f({f^2}(x))$</p> <p>$\Rightarrow {f^3}(x) = f({f^2}(x))$</p> <p>$= f\left( { - {1 \over x}} \right)$</p> <p>$= {{ - {1 \over x} - 1} \over { - {1 \over x} + 1}}$</p> <p>$= {{{{ - 1 - x} \over x}} \over {{{ - 1 + x} \over x}}}$</p> <p>$= {{ - 1 - x} \over { - 1 + x}} = {{ - (x + 1)} \over {x - 1}}$</p> <p>Similarly,</p> <p>${f^4}(x) = f({f^3}(x))$</p> <p>$= f\left( {{{ - x + 1} \over {x - 1}}} \right)$</p> <p>$= {{{{ - (x + 1)} \over {x - 1}} - 1} \over {{{ - (x + 1)} \over {x - 1}} + 1}}$</p> <p>$$ = {{{{ - x - 1 - x + 1} \over {x - 1}}} \over {{{ - x - 1 + x - 1} \over {x - 1}}}}$$</p> <p>$= {{ - 2x} \over { - 2}} = x$</p> <p>$\therefore$ ${f^5}(x) = f({f^4}(x))$</p> <p>$= f(x)$</p> <p>$= {{x - 1} \over {x + 1}}$</p> <p>${f^6}(x) = f({f^5}(x))$</p> <p>$= f\left( {{{x - 1} \over {x + 1}}} \right)$</p> <p>$= - {1 \over x}$ (Already calculated earlier)</p> <p>${f^7}(x) = f({f^6}(x))$</p> <p>$= f\left( { - {1 \over x}} \right)$</p> <p>$= {{ - {1 \over x} - 1} \over { - {1 \over x} + 1}}$</p> <p>$= {{ - (x + 1)} \over {x - 1}}$</p> <p>$\therefore$ ${f^6}(6) = - {1 \over 6}$</p> <p>and ${f^7}(7) = {{ - (7 + 1)} \over {7 - 1}} = - {8 \over 6}$</p> <p>So, ${f^6}(6) + {f^7}(7)$</p> <p>$= - {1 \over 6} - {8 \over 6}$</p> <p>$= - {3 \over 2}$</p>