Question

$\mathrm{If}f\left(x\right)=\cos \left(\log x\right),x>0;\mathrm{then}\mathrm{the}\mathrm{value}\mathrm{of}f\left(x\right)f\left(4\right)-\frac{1}{2}\left(f\left(\frac{x}{4}\right)+f\left(4x\right)\right\}=$

• A. 1
• B. -1
• C. 0.0
• D. $\pm 1$

Answer 1

The correct option is C 0.0


$\mathrm{Given}\mathrm{that},f\left(x\right)=\cos \left(\log x\right).\mathrm{Now},f\left(x\right)f\left(4\right)-\frac{1}{2}\left(f\left(\frac{x}{4}\right)+f\left(4x\right)\right\}=\cos \left(\log x\right)\cos \left(\log 4\right)-\frac{1}{2}\left(\cos \left(\log \frac{x}{4}\right)+\cos \left(\log 4x\right)\right\}=\cos \left(\log x\right)\cos \left(\log 4\right)-\frac{1}{2}\left(\cos \left(\log x-\log 4\right)+\cos \left(\log x+\log 4\right)\right\}=\cos \left(\log x\right)\cos \left(\log 4\right)-\frac{1}{2}\left(2\cos \left(\log x\right)\cos \left(\log 4\right)\right\}=0(\because \cos (A+B)+\cos (A-B)=2\cos A\cos B)$