A) [0, 1]
B) \[[-1,1]\]
C) [0, 3]
D) \[[-1,3]\]
Correct Answer: D
Solution :
Since, \[-2\le \sin x-\sqrt{3}\cos x\le 2\] \[\Rightarrow \] \[-1\le \sin x-\sqrt{3}\cos x+1\le 3\] Here, range of\[f(x)\]is\[[-1,\text{ }3]\]. For\[f\]to be onto, \[f=[-1,3]\]You need to login to perform this action.
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