Trigonometric Identities Question 374
Question: If $ f(x)={{\cos }^{2}}x+{{\sec }^{2}}x, $ then
[MNR 1986]
Options:
A) $ f(x)<1 $
B) $ f(x)=1 $
C) $ 1<f(x)<2 $
D) $ f(x)\ge 2 $
Show Answer
Answer:
Correct Answer: D
Solution:
Since $ {{( x-\frac{1}{x} )}^{2}}\ge 0,\text{}x\in R, $ we have $ x^{2}+\frac{1}{x^{2}}\ge 2 $ and Hence, $ f(x)={{\cos }^{2}}x+\frac{1}{{{\cos }^{2}}x}\ge 2 $ .
BETA
