Question

Let $f(x)=\frac{2025x+2026}{2027x-2025},x\in\mathbb{R},x\ne\frac{2025}{2027}$ be a function. Then $f^{1000}(100)$, where $f^{2}(x)=f(f(x))$ is equal to

• A. 10
• B. 20028
• C. 100
• D. 1000
• E. 202846

Hint:

Math Tip: For any rational function of the form $f(x) = \frac{ax+b}{cx-a}$, applying the function twice always results in $x$. That is, $f(f(x)) = x$. Recognizing this standard form saves you from doing heavy algebraic calculations!

Answer 1

The Correct Option is C

Concept:
Functions - Composition of Functions and Involutions.
Step 1: Identify the function form and objective.
We are given $f(x) = \frac{2025x+2026}{2027x-2025}$.
We need to find the value of $f^{1000}(100)$, which means applying the function $f$ 1000 times consecutively.
Step 2: Evaluate the first composition $f^2(x) = f(f(x))$.
Substitute $f(x)$ into itself: $$ f(f(x)) = \frac{2025\left(\frac{2025x+2026}{2027x-2025}\right) + 2026}{2027\left(\frac{2025x+2026}{2027x-2025}\right) - 2025} $$
Step 3: Simplify the numerator.
Multiply out the numerator terms: $$ \text{Numerator} = 2025(2025x + 2026) + 2026(2027x - 2025) $$ $$ = 2025^2 x + 2025(2026) + 2026(2027)x - 2026(2025) $$ Notice the constant terms cancel out: $$ = x(2025^2 + 2026 \times 2027) $$
Step 4: Simplify the denominator.
Multiply out the denominator terms: $$ \text{Denominator} = 2027(2025x + 2026) - 2025(2027x - 2025) $$ $$ = 2027(2025)x + 2027(2026) - 2025(2027)x + 2025^2 $$ Notice the $x$ terms cancel out: $$ = 2026 \times 2027 + 2025^2 $$
Step 5: Determine the simplified form of $f(f(x))$.
Divide the simplified numerator by the simplified denominator: $$ f(f(x)) = \frac{x(2025^2 + 2026 \times 2027)}{2026 \times 2027 + 2025^2} = x $$ Thus, $f^2(x) = x$. The function is an involution (its own inverse).
Step 6: Apply the pattern to find $f^{1000}(100)$.
Since $f^2(x) = x$, any even number of compositions will return $x$.
$$ f^{\text{even}}(x) = x $$ $$ f^{\text{odd}}(x) = f(x) $$ Because 1000 is an even number, $f^{1000}(x) = x$.
Therefore, $f^{1000}(100) = 100$.