Question

If $f\left(x\right)=\frac{{\cos }^{2}x+{\sin }^{4}x}{{\sin }^{2}x+{\cos }^{4}x}\forall x\in R$, then $f\left(2020\right)$ is equal to

• A. $2020$
• B. $0$
• C. $1$
• D. $\frac{1}{2020}$

Answer 1

The correct option is C $1$

$f\left(x\right)=\frac{{\cos }^{2}x+{\sin }^{4}x}{{\sin }^{2}x+{\cos }^{4}x}=\frac{{\cos }^{2}x+(1-{\cos }^{2}x{)}^2}{{\sin }^{2}x+{\cos }^{4}x}$
$=\frac{{\cos }^{2}x+1+{\cos }^{4}x-2{\cos }^{2}x}{{\sin }^{2}x+{\cos }^{4}x}=\frac{1+{\cos }^{4}x-{\cos }^{2}x}{{\sin }^{2}x+{\cos }^{4}x}$
$=\frac{{\sin }^{2}x+{\cos }^{4}x}{{\sin }^{2}x+{\cos }^{4}x}=1$
Clearly $f\left(x\right)$ is always $1$
$\therefore f\left(2020\right)=1$