Functions Question 703
Question: If $ f:R\to S, $ defined by $ f(x)=sinx-\sqrt{3}\cos x+1, $ is onto, then the interval of S is
Options:
A) $ [-1,3] $
B) $ [-1,1] $
C) $ [0,1] $
D) $ [0,3] $
Show Answer
Answer:
Correct Answer: A
Solution:
[d] $ f(x) $ is onto
$ \therefore S=range,of,f(x) $ Now $ f(x)= $ $ \sin x-\sqrt{3}\cos x+1=2\sin ( x-\frac{\pi }{3} )+1 $ $ \because -1\le \sin ( x-\frac{\pi }{3} )\le 1 $ $ -1\le 2\sin ( x-\frac{\pi }{3} )+1\le 3 $
$ \therefore f(x)\in [-1,3]=S $
BETA
