15:I[8216,["972","static/chunks/972-b28da0de92681928.js","47","static/chunks/47-a7f06e9eda1ac756.js","115","static/chunks/app/%5Bexam%5D/questions/%5Byear%5D/%5Bsubject%5D/%5Bchapter%5D/%5BquestionId%5D/page-7b6f0a12b9cc0206.js"],"BackButton"] 16:I[1553,["972","static/chunks/972-b28da0de92681928.js","47","static/chunks/47-a7f06e9eda1ac756.js","115","static/chunks/app/%5Bexam%5D/questions/%5Byear%5D/%5Bsubject%5D/%5Bchapter%5D/%5BquestionId%5D/page-7b6f0a12b9cc0206.js"],"QuestionViewer"] 17:T785,

Given,

$$f(x) = {{x - 1} \over {x + 1}}$$

Also given,

$${f^{n + 1}}(x) = f({f^n}(x))$$ ..... (1)

$$\therefore$$ For $$n = 1$$

$${f^{1 + 1}}(x) = f({f^1}(x))$$

$$ \Rightarrow {f^2}(x) = f(f(x))$$

$$ = f\left( {{{x - 1} \over {x + 1}}} \right)$$

$$ = {{{{x - 1} \over {x + 1}} - 1} \over {{{x - 1} \over {x + 1}} + 1}}$$

$$ = {{{{x - 1 - x - 1} \over {x + 1}}} \over {{{x - 1 + x + 1} \over {x + 1}}}}$$

$$ = {{ - 2} \over {2x}}$$

$$ = - {1 \over x}$$

From equation (1), when n = 2

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

$$ \Rightarrow {f^3}(x) = f({f^2}(x))$$

$$ = f\left( { - {1 \over x}} \right)$$

$$ = {{ - {1 \over x} - 1} \over { - {1 \over x} + 1}}$$

$$ = {{{{ - 1 - x} \over x}} \over {{{ - 1 + x} \over x}}}$$

$$ = {{ - 1 - x} \over { - 1 + x}} = {{ - (x + 1)} \over {x - 1}}$$

Similarly,

$${f^4}(x) = f({f^3}(x))$$

$$ = f\left( {{{ - x + 1} \over {x - 1}}} \right)$$

$$ = {{{{ - (x + 1)} \over {x - 1}} - 1} \over {{{ - (x + 1)} \over {x - 1}} + 1}}$$

$$ = {{{{ - x - 1 - x + 1} \over {x - 1}}} \over {{{ - x - 1 + x - 1} \over {x - 1}}}}$$

$$ = {{ - 2x} \over { - 2}} = x$$

$$\therefore$$ $${f^5}(x) = f({f^4}(x))$$

$$ = f(x)$$

$$ = {{x - 1} \over {x + 1}}$$

$${f^6}(x) = f({f^5}(x))$$

$$ = f\left( {{{x - 1} \over {x + 1}}} \right)$$

$$ = - {1 \over x}$$ (Already calculated earlier)

$${f^7}(x) = f({f^6}(x))$$

$$ =

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