Question

If $f: R \rightarrow S$ defined by $f(x)=\sin x-\sqrt{3} \,\cos\,x+\,$ 1, is onto, function then S = ?

• A. [0, 3]
• B. [-1, 1]
• C. [0, 1]
• D. [-1, 3]

Answer 1

Answer: (D)

[-1, 3]

Answer 2

Since $- 2 \leq \,\sin \,x -\sqrt{3} \,\cos\, x\, \leq\, 2$ $\therefore -1 < \sin x - \sqrt{3} \cos\, x \,+ 1 \leq 3$ $\therefore R_f = [-1, 3]$ $\therefore$ [-1, 3] holds . $\begin{Bmatrix} \because \sin \, x - \sqrt{3} \,\cos \, x \\\[0.3em] = 2\left[\frac{1}{2}\sin \, x= \frac{\sqrt 3}{2}\cos \, x\right] \\\[0.3em] = 2\left[\sin\left(x- \frac{\pi}{6} \right)\right] \end{Bmatrix}$