Question

If $f\left(x\right)=\frac{{\cos }^{2}x+{\sin }^{4}x}{{\sin }^{2}x+{\cos }^{4}x},x\in R$, then $f\left(2002\right)$ is equal to.

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A $1$

$f\left(x\right)=\frac{{\cos }^{2}x+{\sin }^{4}x}{{\sin }^{2}x+{\cos }^{4}x}$
$\Rightarrow f\left(x\right)=\frac{{\cos }^{2}x+{\sin }^{2}x(1-{\cos }^{2}x)}{{\sin }^{2}x+{\cos }^{2}x(1-{\sin }^{2}x]}$
$\Rightarrow f\left(x\right)=\frac{{\cos }^{2}x+{\sin }^{2}x-{\sin }^{2}x{\cos }^{2}x}{{\sin }^{2}x+{\cos }^{2}x-{\sin }^{2}x{\cos }^{2}x}$
$\Rightarrow f\left(x\right)=\frac{1-{\sin }^{2}x{\cos }^{2}x}{1-{\sin }^{2}x{\cos }^{2}x}$

$\Rightarrow f\left(x\right)=1\forall x\in R$

$\Rightarrow f\left(2002\right)=1$.