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 find \( f_{2008}(x) + f_{2009}(x) \) where the function \( f(x) \) is defined as:

\[ f(x) = \frac{3x + 5}{2x - 3} \]

We also have the definitions:

- \( f_1(x) = f(x) \)
- \( f_n(x) = f(f_{n-1}(x)) \) for \( n \geq 2 \)

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

We start by calculating \( f(f(x)) \):

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

Substituting \( x \) in \( f(x) \):

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

Calculating the numerator:

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

Calculating the denominator:

\[ 2\left(\frac{3x + 5}{2x - 3}\right) - 3 = \frac{6x + 10 - 6x + 9}{2x - 3} = \frac{19}{2x - 3} \]

Thus, we have:

\[ f(f(x)) = \frac{19x}{19} = x \]

### Step 2: Identify the pattern

From the calculations above, we can observe:

- \( f_1(x) = f(x) \)
- \( f_2(x) = f(f(x)) = x \)

Continuing this pattern:

- \( f_3(x) = f(f_2(x)) = f(x) \)
- \( f_4(x) = f(f_3(x)) = f(f(x)) = x \)

This indicates that:

- \( f_{2n}(x) = x \) for even \( n \)
- \( f_{2n+1}(x) = f(x) \) for odd \( n \)

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

Since \( 2008 \) is even:

\[ f_{2008}(x) = x \]

And since \( 2009 \) is odd:

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

Now we can find:

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

### Step 4: Combine the expressions

To combine these two expressions, we need a common denominator:

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

This simplifies to:

\[ = \frac{2x^2 - 3x + 3x + 5}{2x - 3} = \frac{2x^2 + 5}{2x - 3} \]

### Final Answer

Thus, the final result is:

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