A) \[[-1,3]\]
B) \[[-1,1]\]
C) \[[0,1]\]
D) \[[0,3]\]
Correct Answer: A
Solution :
[d] \[f(x)\] is onto \[\therefore S=range\,of\,f(x)\] Now \[f(x)=\] \[\sin x-\sqrt{3}\cos x+1=2\sin \left( x-\frac{\pi }{3} \right)+1\] \[\because -1\le \sin \left( x-\frac{\pi }{3} \right)\le 1\] \[-1\le 2\sin \left( x-\frac{\pi }{3} \right)+1\le 3\] \[\therefore f(x)\in [-1,3]=S\]
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