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
The correct answer is Option (4) → [-1, 3]
We have,
$-\sqrt{1+(\sqrt{3})^2}≤\sin x-\sqrt{3}\cos x≤\sqrt{1+(\sqrt{3})^2}$ 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 $x∈R$
⇒ Range (f) = [-1, 3]
If $f: R→S$ is onto, then S = Range (f) = [-1, 3]