Question
If $$f:R \to S,$$ defined by $$f\left( x \right) = \sin \,x - \sqrt 3 \,\cos \,x + 1,$$ is onto, then the interval of $$S$$ is :
A.
$$\left[ { - 1,\,3} \right]$$
B.
$$\left[ { - 1,\,1} \right]$$
C.
$$\left[ {0,\,1} \right]$$
D.
$$\left[ { 0,\,3} \right]$$
Answer :
$$\left[ { - 1,\,3} \right]$$
Solution :
$$\eqalign{
& f\left( x \right){\text{ is onto}}\,\,\,\therefore \,S = {\text{range of }}f\left( x \right) \cr
& {\text{Now }}f\left( x \right) = \sin \,x - \sqrt 3 \,\cos \,x + 1 = 2\,\sin \,\left( {x - \frac{\pi }{3}} \right) + 1 \cr
& \because \, - 1 \leqslant \sin \left( {x - \frac{\pi }{3}} \right) \leqslant 1 \cr
& - 1 \leqslant 2\,\sin \left( {x - \frac{\pi }{3}} \right) + 1 \leqslant 3 \cr
& \therefore \,f\left( x \right)\, \in \left[ { - 1,\,3} \right] = S\, \cr} $$