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Let $f(x) = \dfrac{x-1}{x+1}, \: f^{k+1}(x)=f\left(f^k(x)\right)$ for all $k=1, 2, 3, \dots , 99$.  Then $f^{100}(10)$ is

1. $1$
2. $10$
3. $100$
4. $101$
in Calculus
recategorized | 36 views
0

@ankitgupta.1729

+2

@`JEET

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

Now, $f^2(x) = f(f(x)) = \frac{f(x)-1}{f(x)+1} = \frac{\frac{x-1}{x+1}-1}{\frac{x-1}{x+1} + 1} = -1/x$

$f^3(x) =f(f^2(x) = f(-1/x) = \frac{-x-1}{x-1}$

$f^4(x) = f(f^3(x) = \frac{\frac{-x-1}{x-1}-1}{\frac{-x-1}{x-1} + 1} = x$

So, $f^4(x) = x$

It means $f^5(x) = f(x)$ and $f^6(x) = f^2(x)$ and $f^7(x) = f^3(x)$ and $f^8(x) = f^4(x) = x$

Now, observe the pattern, $f^4(x) = f^8(x) = f^{12}(x) = ......= f^{4k}(x) = x$ where $k\in I^+$

So, $f^{100}(x) = x$ which implies $f^{100}(10) = 10$

+1
Thanks.
+1
This was a beautiful question.

Looked so tough :D
0
Copy that in the answer part as you have fully answered it.

Answer: $\textbf {B}$

Solution:

Given: $\mathrm {f(x) = \frac{x-1}{x+1}}$

\begin {align} \Rightarrow \mathrm {f^2(x)} &= \mathrm{f(f(x))} \\&= \mathrm{\frac{f(x)-1}{f(x) + 1}} \\&= \mathrm{\dfrac{\frac{x-1}{x+1} + 1}{\frac{x-1}{x+1} + 1}} \\&= \mathrm {\frac{-1}{x}}\end {align}

\begin {align} \mathrm {f^3(x)} &= \mathrm {f(f^2(x))}\\&=\mathrm {f\bigg(\frac{-1}{x}\bigg)}\\&=\mathrm{\frac{-x-1}{x-1}} \end {align}

\begin{align} \mathrm {f^4(x)} &= \mathrm{f(f^3(x))} \\&= \mathrm{\dfrac{\frac{-x-1}{x-1}-1}{\frac{-x-1}{x-1}+1}} \\&=\mathrm x \end {align}

$\therefore \mathrm {f^4(x) = x}\tag{1}$
Similarly,
$\mathrm {f^5(x) = f(x)}$
$\mathrm {f^6(x) = f^2(x)}$
$\mathrm {f^7(x) = f^3(x)}$
$\mathrm {f^8(x) = f^4(x) =x}$ $\tag{[From (1)]}$
Now the pattern goes like:
\begin {align}\mathrm {f^4(x) = f^8(x) = f^{12}(x)= \cdots \cdots=f^{4k}(x) = x}~ \text{where\mathrm k\in\mathrm {I^+}} \end {align}
$\therefore \mathrm {f^{100}(x) = x} \\ \Rightarrow \mathrm {f^{100}(10) = 10}$

$\therefore \textbf B$ is the correct option.

by Boss (13.4k points)
edited by
+1
Here's a typing mistake. It will be $f^3(x)=f(f^2(x))$. Please correct that.
+1
Done!

Thanks