Question

If \(\mathrm{f}(x)=\cos (\log x)\) then \(\mathrm{f}\left(x^2\right) \cdot \mathrm{f}\left(y^2\right)-\frac{1}{2}\left[\mathrm{f}\left(\frac{x^2}{y^2}\right)+\mathrm{f}\left(x^2 y^2\right)\right]\) has the value

• A. -2
• B. -1
• C. \(\frac{1}{2}\)
• D. 0

Hint:

Put \(a=\log x^2\) and \(b=\log y^2\), then use the product to sum identity.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
\(f(x)=\cos(\log x)\), so \(f(x^2)=\cos(\log x^2)\).

Step 2: Substitute
Let \(a=\log x^2\) and \(b=\log y^2\). Then \(f(x^2/y^2)=\cos(a-b)\) and \(f(x^2y^2)=\cos(a+b)\).

Step 3: Simplify
Expression \(=\cos a\cos b-\dfrac{1}{2}[\cos(a-b)+\cos(a+b)]\)

Step 4: Calculation
\(\cos(a-b)+\cos(a+b)=2\cos a\cos b\), so the expression is \(\cos a\cos b-\cos a\cos b=0\).

Final Answer:
The expression equals 0. \[ \boxed{\text{(D) }0} \]