Functions Question 633
Question: The range of $ f(x)=\sec ( \frac{\pi }{4}{{\cos }^{2}}x ),,\ -\infty <x<\infty $ is
Odisha JEE 2002
Options:
A) $ [1,\ \sqrt{2}] $
B) $ [1,\ \infty ) $
C) $ [-\sqrt{2},\ -1]\cup [1,\ \sqrt{2}] $
D) $ (-\infty ,\ -1]\cup [1,\ \infty ) $
Show Answer
Answer:
Correct Answer: A
Solution:
$ f(x)=\sec ( \frac{\pi }{4},{{\cos }^{2}}x ) $ We know that, $ 0\le {{\cos }^{2}}x\le 1 $ at $ \cos x=0,, $ $ f(x)=1 $ and at $ \cos x=1, $ $ f(x)=\sec ( \frac{\pi }{4} ) $; \ $ 1\le x\le \sqrt{2} $ Þ $ x\in [1,\sqrt{2}] $ .
BETA
