If $f: R → S$, defined by $f(x) = \sin x-\sqrt{3} \cos x + 1$, is onto, then the interval of S, is
Answer & explanation
Correct answer: option 4
We have,
$-\sqrt{1+(3)^2}≤\sin x-\sqrt{3} \cos x≤\sqrt{1+(3)^2}$ for all $x ∈ R$
$⇒-2≤\sin x-\sqrt{3} \cos x≤2$ for all $x ∈ R$
$⇒-1≤ \sin x-\sqrt{3} \cos x + 1 ≤3$ for all $x ∈ R$
$⇒-1≤f(x)≤\sqrt{3}$ for all $x ∈ R$
If $f: R → S$ is onto, then S = Range (f) = [-1, 3]