If `f(x)+2f(1/x)=3x , x!=0,` and `S={x in R :f(x)=f(-x)}` ; then S: (1) is an empty set. (2) contains exactly one element. (3) contains exactly two elements. (4) contains more than two elements

Options

A. contains exactly one element
B. contains exactly two elements
C. contains more than two elements
D. is an empty set

Explanation

To solve the given problem step by step, we start with the equation provided:

Given:
\[ f(x) + 2f\left(\frac{1}{x}\right) = 3x \quad (1) \]
for \( x \neq 0 \).

### Step 1: Substitute \( x \) with \( \frac{1}{x} \)
We replace \( x \) with \( \frac{1}{x} \) in equation (1):
\[ f\left(\frac{1}{x}\right) + 2f(x) = 3\left(\frac{1}{x}\right) \quad (2) \]

### Step 2: Rewrite the equations
Now we have two equations:
1. \( f(x) + 2f\left(\frac{1}{x}\right) = 3x \) (Equation 1)
2. \( f\left(\frac{1}{x}\right) + 2f(x) = \frac{3}{x} \) (Equation 2)

### Step 3: Multiply Equation (2) by 2
Multiply the second equation by 2:
\[ 2f\left(\frac{1}{x}\right) + 4f(x) = \frac{6}{x} \quad (3) \]

### Step 4: Rearranging the equations
Now we have:
1. \( f(x) + 2f\left(\frac{1}{x}\right) = 3x \) (1)
2. \( 2f\left(\frac{1}{x}\right) + 4f(x) = \frac{6}{x} \) (3)

### Step 5: Subtract Equation (1) from Equation (3)
Now subtract Equation (1) from Equation (3):
\[ (2f\left(\frac{1}{x}\right) + 4f(x)) - (f(x) + 2f\left(\frac{1}{x}\right)) = \frac{6}{x} - 3x \]

This simplifies to:
\[ 3f(x) = \frac{6}{x} - 3x \]

### Step 6: Solve for \( f(x) \)
Now, divide both sides by 3:
\[ f(x) = \frac{2}{x} - x \]

### Step 7: Use the property \( f(x) = f(-x) \)
Since it is given that \( f(x) = f(-x) \), we can substitute \(-x\) into the function:
\[ f(-x) = \frac{2}{-x} - (-x) = -\frac{2}{x} + x \]

### Step 8: Set the two expressions equal
Now set \( f(x) \) equal to \( f(-x) \):
\[ \frac{2}{x} - x = -\frac{2}{x} + x \]

### Step 9: Solve for \( x \)
Combine like terms:
\[ \frac{2}{x} + \frac{2}{x} = x + x \]
\[ \frac{4}{x} = 2x \]

Cross-multiply:
\[ 4 = 2x^2 \]
\[ x^2 = 2 \]
\[ x = \pm \sqrt{2} \]

### Step 10: Determine the set \( S \)
The solutions \( x = \sqrt{2} \) and \( x = -\sqrt{2} \) indicate that the set \( S \) contains exactly two elements.

### Conclusion
Thus, the answer is that \( S \) contains exactly two elements.