Let `f :R -{(3)/(2)}to R, f (x) = (3x+5)/(2x-3).Let f _(1) (x)=f (x), f_(n) (x)=f (f _(n-1) (x)))` for ` pige 2, n in N,` then `f _(2008) (x)+ f _(2009) (x)=`

Options

A. `(2x ^(2) +5)/(2x-3)`
B. `(x^(2) +5)/(2x-3)`
C. `(2x ^(2) -5)/(2x-3)`
D. `(x ^(2) -5)/(2x-3)`

Answer

To solve the problem, we need to evaluate the expression \( f_{2008}(x) + f_{2009}(x) \) where \( f(x) = \frac{3x + 5}{2x - 3} \).

### Step 1: Calculate \( f(f(x)) \)

We start by finding \( f(f(x)) \).

1. Substitute \( f(x) \) into itself:
\[ f(f(x)) = f\left(\frac{3x + 5}{2x - 3}\right) \]
This means replacing \( x \) in \( f(x) \) with \( \frac{3x + 5}{2x - 3} \).

2. Calculate \( f(f(x)) \):
\[ f\left(\frac{3x + 5}{2x - 3}\right) = \frac{3\left(\frac{3x + 5}{2x - 3}\right) + 5}{2\left(\frac{3x + 5}{2x - 3}\right) - 3} \]

3. Simplify the numerator:
\[ = \frac{\frac{9x + 15}{2x - 3} + 5}{2\left(\frac{3x + 5}{2x - 3}\right) - 3} \]
\[ = \frac{\frac{9x + 15 + 5(2x - 3)}{2x - 3}}{2\left(\frac{3x + 5}{2x - 3}\right) - 3} \]
\[ = \frac{\frac{9x + 15 + 10x - 15}{2x - 3}}{2\left(\frac{3x + 5}{2x - 3}\right) - 3} \]
\[ = \frac{\frac{19x}{2x - 3}}{2\left(\frac{3x + 5}{2x - 3}\right) - 3} \]

4. Simplify the denominator:
\[ = \frac{19x}{\frac{6x + 10 - 6x + 9}{2x - 3}} = \frac{19x(2x - 3)}{19} = x \]

Thus, we find that:
\[ f(f(x)) = x \]

### Step 2: Determine \( f_{n}(x) \)

From the result \( f(f(x)) = x \), we can conclude:
- \( f_{1}(x) = f(x) \)
- \( f_{2}(x) = f(f(x)) = x \)
- \( f_{3}(x) = f(f(f(x))) = f(x) \)
- \( f_{4}(x) = f(f(f(f(x)))) = x \)

This pattern shows that:
- If \( n \) is odd, \( f_{n}(x) = f(x) \)
- If \( n \) is even, \( f_{n}(x) = x \)

### Step 3: Calculate \( f_{2008}(x) + f_{2009}(x) \)

Now we can evaluate:
- \( f_{2008}(x) = x \) (since 2008 is even)
- \( f_{2009}(x) = f(x) \) (since 2009 is odd)

Thus:
\[ f_{2008}(x) + f_{2009}(x) = x + f(x) = x + \frac{3x + 5}{2x - 3} \]

### Step 4: Combine the expressions

Combine the two terms:
\[ f_{2008}(x) + f_{2009}(x) = x + \frac{3x + 5}{2x - 3} \]
To combine, we need a common denominator:
\[ = \frac{x(2x - 3)}{2x - 3} + \frac{3x + 5}{2x - 3} \]
\[ = \frac{2x^2 - 3x + 3x + 5}{2x - 3} \]
\[ = \frac{2x^2 + 5}{2x - 3} \]

### Final Answer
\[ f_{2008}(x) + f_{2009}(x) = \frac{2x^2 + 5}{2x - 3} \]