If ` f(x) = cos ( log x) , then f(x^2) f(y^2) -1/2 [f((x^2)/(y^2))+f(x^2 y^2)]` has the value

Options

A. `-2`
B. `-1`
C. `1/2`
D. None of these

Answer

To solve the problem, we need to evaluate the expression:

\[ f(x^2) f(y^2) - \frac{1}{2} \left[ f\left(\frac{x^2}{y^2}\right) + f(x^2 y^2) \right] \]

Given that \( f(x) = \cos(\log x) \), we can find \( f(x^2) \), \( f(y^2) \), \( f\left(\frac{x^2}{y^2}\right) \), and \( f(x^2 y^2) \).

### Step 1: Calculate \( f(x^2) \) and \( f(y^2) \)

Using the definition of \( f(x) \):

\[ f(x^2) = \cos(\log(x^2)) = \cos(2 \log x) \]

\[ f(y^2) = \cos(\log(y^2)) = \cos(2 \log y) \]

### Step 2: Calculate \( f(x^2) f(y^2) \)

Now, we can find \( f(x^2) f(y^2) \):

\[ f(x^2) f(y^2) = \cos(2 \log x) \cos(2 \log y) \]

### Step 3: Calculate \( f\left(\frac{x^2}{y^2}\right) \)

Next, we calculate \( f\left(\frac{x^2}{y^2}\right) \):

\[ f\left(\frac{x^2}{y^2}\right) = \cos\left(\log\left(\frac{x^2}{y^2}\right)\right) = \cos(\log(x^2) - \log(y^2)) = \cos(2 \log x - 2 \log y) = \cos(2(\log x - \log y)) \]

### Step 4: Calculate \( f(x^2 y^2) \)

Now, we calculate \( f(x^2 y^2) \):

\[ f(x^2 y^2) = \cos(\log(x^2 y^2)) = \cos(\log(x^2) + \log(y^2)) = \cos(2 \log x + 2 \log y) = \cos(2(\log x + \log y)) \]

### Step 5: Substitute into the expression

Now we substitute these values into the original expression:

\[ f(x^2) f(y^2) - \frac{1}{2} \left[ f\left(\frac{x^2}{y^2}\right) + f(x^2 y^2) \right] \]

This becomes:

\[ \cos(2 \log x) \cos(2 \log y) - \frac{1}{2} \left[ \cos(2(\log x - \log y)) + \cos(2(\log x + \log y)) \right] \]

### Step 6: Use the cosine addition formula

Using the cosine addition formula:

\[ \cos A \cos B = \frac{1}{2} \left[ \cos(A + B) + \cos(A - B) \right] \]

Let \( A = 2 \log x \) and \( B = 2 \log y \):

\[ \cos(2 \log x) \cos(2 \log y) = \frac{1}{2} \left[ \cos(2 \log x + 2 \log y) + \cos(2 \log x - 2 \log y) \right] \]

### Step 7: Substitute back

Now substituting back into our expression:

\[ \frac{1}{2} \left[ \cos(2(\log x + \log y)) + \cos(2(\log x - \log y)) \right] - \frac{1}{2} \left[ \cos(2(\log x - \log y)) + \cos(2(\log x + \log y)) \right] \]

### Step 8: Simplify

Notice that the two terms cancel out:

\[ \frac{1}{2} \left[ \cos(2(\log x + \log y)) + \cos(2(\log x - \log y)) \right] - \frac{1}{2} \left[ \cos(2(\log x - \log y)) + \cos(2(\log x + \log y)) \right] = 0 \]

### Final Answer

Thus, the value of the expression is:

\[ \boxed{0} \]