Functions Question 115
Question: If $ f:R\to S $ defined by $ f(x)=\sin x-\sqrt{3}\cos x+1 $ is onto, then the interval of S is
[AIEEE 2004; IIT Screening 2004]
Options:
A) [?1, 3]
B) [1, 1]
C) [0, 1]
D) [0, ?1]
Show Answer
Answer:
Correct Answer: A
Solution:
$ -\sqrt{1+{{(-\sqrt{3})}^{2}}}\le (\sin x-\sqrt{3}\cos x)\le \sqrt{1+{{(-\sqrt{3})}^{2}}} $ $ -2\le (\sin x-\sqrt{3}\cos x)\le 2 $ $ -2+1\le (\sin x-\sqrt{3}\cos x+1)\le 2+1 $ $ -1\le (\sin x-\sqrt{3}\cos x+1)\le 3 $ i.e., range $ =[-1,3] $ \ For f to be onto $ S=[-1,3] $ .
BETA
