If `2f(x^2)+3f(1/x^2)=x^2-1`, then `f(x^2)` is

Options

A. `(1-x^(4))/(5x^(2))`
B. `(1- x^(2))/(5x)`
C. `(5x^(2))/(1-x^(4))`
D. `(3- x^(2) - 2x^(4))/(5x^(2))`

Answer

To solve the equation \( 2f(x^2) + 3f\left(\frac{1}{x^2}\right) = x^2 - 1 \) for \( f(x^2) \), we will follow these steps:

### Step 1: Write the given equation
We start with the equation:
\[ 2f(x^2) + 3f\left(\frac{1}{x^2}\right) = x^2 - 1 \tag{1} \]

### Step 2: Substitute \( x^2 \) with \( \frac{1}{x^2} \)
Next, we replace \( x^2 \) with \( \frac{1}{x^2} \) in the original equation:
\[ 2f\left(\frac{1}{x^2}\right) + 3f(x^2) = \frac{1}{x^2} - 1 \tag{2} \]

### Step 3: Multiply the equations
Now, we will multiply equation (1) by 2 and equation (2) by 3:
- From equation (1):
\[ 4f(x^2) + 6f\left(\frac{1}{x^2}\right) = 2(x^2 - 1) \implies 4f(x^2) + 6f\left(\frac{1}{x^2}\right) = 2x^2 - 2 \tag{3} \]
- From equation (2):
\[ 6f\left(\frac{1}{x^2}\right) + 9f(x^2) = 3\left(\frac{1}{x^2} - 1\right) \implies 6f\left(\frac{1}{x^2}\right) + 9f(x^2) = \frac{3}{x^2} - 3 \tag{4} \]

### Step 4: Subtract the equations
Now, we subtract equation (3) from equation (4):
\[ (6f\left(\frac{1}{x^2}\right) + 9f(x^2)) - (4f(x^2) + 6f\left(\frac{1}{x^2}\right)) = \left(\frac{3}{x^2} - 3\right) - (2x^2 - 2) \]
This simplifies to:
\[ (9f(x^2) - 4f(x^2)) = \left(\frac{3}{x^2} - 3 - 2x^2 + 2\right) \]
\[ 5f(x^2) = \frac{3}{x^2} - 2x^2 - 1 \]

### Step 5: Solve for \( f(x^2) \)
Now, we can isolate \( f(x^2) \):
\[ f(x^2) = \frac{1}{5}\left(\frac{3}{x^2} - 2x^2 - 1\right) \]

### Final Result
Thus, the function \( f(x^2) \) is given by:
\[ f(x^2) = \frac{3}{5x^2} - \frac{2}{5}x^2 - \frac{1}{5} \]