Question

If $f\left(x\right)=\frac{si{n}^{4}x+co{s}^{2}x}{si{n}^{2}x+co{s}^{4}x}$ for $xϵR,$ then $f\left(2002\right)$.

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is A

1

Given:
$f\left(x\right)=\frac{si{n}^{4}x+co{s}^{2}x}{si{n}^{2}x+co{s}^{4}x}$
On dividing the numerator and denominator by $co{s}^{4}x$, we get
$f\left(x\right)=\frac{ta{n}^{4}x+se{c}^{2}x}{1+ta{n}^{2}xse{c}^{2}x}=\frac{1+ta{n}^{4}x+ta{n}^{2}x}{1+ta{n}^{2}x(1+ta{n}^{2}x)}=\frac{1+ta{n}^{4}x+ta{n}^{2}x}{1+ta{n}^{4}x+ta{n}^{2}x}=1\text{For every}xϵR$
For x = 2002, we have
f(2002) = 1