If `f(x)=cos(logx)`, then `f(x^(2))f(y^(2))-(1)/(2)[f(x^(2)y^(2))+f((x^(2))/(y^(2)))]=`

Options

A. `-2`
B. `-1`
C. `(1)/(2)`
D. 0

Answer

To solve the problem, we start with the function \( f(x) = \cos(\log x) \) and need to evaluate the expression:

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

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

Using the definition of the function:

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

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

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

Using the property of logarithms:

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

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

Using the property of logarithms again:

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

### Step 4: Substitute into the expression

Now we substitute back into the original expression:

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

The expression 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 5: Use the cosine addition formula

We can use the cosine addition formula:

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

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

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

### Step 6: Substitute back into the expression

Now we can substitute this back into our expression:

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

### Step 7: Simplify the expression

Notice that:

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

Thus, the two terms cancel out, leading to:

\[ 0 \]

### Final Answer

The final result is:

\[ \boxed{0} \]