If f : R → S defined by $f(x)=\sin x-\sqrt{3}\cos x+1$ is onto, then what is the interval of S?
Answer & explanation
Correct answer: option 3
$-\sqrt{1+(-\sqrt{3})^2}≤(\sin x-\sqrt{3}\cos x)≤\sqrt{1+(\sqrt{-3})^2}$
$-2≤(\sin x-\sqrt{3}\cos x)≤2$
$-2+1≤(\sin x-\sqrt{3}\cos x+1)≤2+1$
$-1≤(\sin x-\sqrt{3}\cos x+1)≤3$ i.e.,
Range = [-1, 3]
For f to be onto S = [-1, 3]