If $f(x)=\cos (\log x)$, then $f(x) \cdot f(y)-\frac{1}{2}\left(f\left(\frac{x}{y}\right)+f(x y)\right)$ has…

If $f(x)=\cos (\log x)$, then $f(x) \cdot f(y)-\frac{1}{2}\left(f\left(\frac{x}{y}\right)+f(x y)\right)$ has the value
  1. $-2$
  2. $-1$
  3. $0$
  4. $\frac{1}{2}$

Solution

$\begin{aligned} & f(x)=\cos (\log x) \\ & \text { Now, } f(x) \cdot f(y)-\frac{1}{2}\left(f\left(\frac{x}{y}\right)+f(x y)\right) \\ & =\cos (\log x) \cdot \cos (\log y)-\frac{1}{2}\left(\cos \log \left(\frac{x}{y}\right)+\cos \log (x y)\right) \\ & =\cos (\log x) \cdot \cos (\log y)-\frac{1}{2}\{\cos (\log x-\log y)+\cos (\log x+\log y)\} \\ & =\cos (\log x) \cdot \cos (\log y)-\frac{1}{2} \times 2 \cos (\log x) \cdot \cos (\log y) \\ & =0\end{aligned}$

Asked in: MHT CET 2022 (05 Aug Shift 1)

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