Functions Question 569
Question: If $ f(x)=\frac{{{\cos }^{2}}x+{{\sin }^{4}}x}{{{\sin }^{2}}x+{{\cos }^{4}}x} $ for $ x\in R $ , then $ f(2002)= $
[EAMCET 2002]
Options:
A) 1
B) 2
C) 3
D) 4
Show Answer
Answer:
Correct Answer: A
Solution:
$ f(x)=\frac{{{\cos }^{2}}x+{{\sin }^{4}}x}{{{\sin }^{2}}x+{{\cos }^{4}}x} $
Þ $ f(x)=\frac{{{\cos }^{2}}x+{{\sin }^{2}}x(1-{{\cos }^{2}}x)}{{{\sin }^{2}}x+{{\cos }^{2}}x(1-{{\sin }^{2}}x)} $
Þ $ f(x)=\frac{{{\sin }^{2}}x+{{\cos }^{2}}x-{{\sin }^{2}}x{{\cos }^{2}}x}{{{\sin }^{2}}x+{{\cos }^{2}}x-{{\sin }^{2}}x{{\cos }^{2}}x} $
Þ $ f(x)=1 $
Þ $ f(2002)=1 $ .
BETA
