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

JEE Main · 2022 · 26 Jun 2022 · Shift 1 · Sets Relations and Functions

Let

f (x) = \frac{x - 1}{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:

Answer: (B)

Compute successive iterations of

f

f (x) = \frac{x - 1}{x + 1}

f^{2} (x) = f (f (x)) = - \frac{1}{x}

f^{3} (x) = \frac{x + 1}{1 - x}

f^{4} (x) = x

(the function returns to the identity after 4 iterations).

Since the period is 4:

f^{6} (x) = f^{2} (x) = - \frac{1}{x}

, so

f^{6} (6) = - \frac{1}{6}

f^{7} (x) = f^{3} (x) = \frac{x + 1}{1 - x}

, so

f^{7} (7) = \frac{8}{- 6} = - \frac{4}{3}

.

Therefore,

f^{6} (6) + f^{7} (7) = - \frac{1}{6} - \frac{4}{3} = - \frac{1}{6} - \frac{8}{6} = - \frac{9}{6} = - \frac{3}{2}

JEE Main · 2022 · 26 Jun 2022 · Shift 1 · Single correct option