Practicing Success
If $f: R→S$, defined by $f(x) = \sin x-\sqrt{3}\cos x + 1$, is onto, then the interval of S, is |
[0, 1] [-1, 1] [0, 3] [-1, 3] |
[-1, 3] |
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] |