Question

If $f\left(x\right)=\frac{co{s}^{2}x+si{n}^{4}x}{co{s}^{4}x+si{n}^{2}x}$ for all real x, then f(2016) =

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is A

1

$f\left(x\right)=\frac{co{s}^{4}x+(1-co{s}^{2}x{)}^2}{co{s}^{4}x+(1-co{s}^{2}x)}=\frac{co{s}^{4}x-co{s}^{2}x+1}{co{s}^{4}x-co{s}^{2}x+1}=1$

$\therefore f\left(2016\right)=1$