Question

If $f:R\to S$ defined by $f\left(x\right)=\sin x-\sqrt{3}\cos x+1$ is on to ,then the interval of $S$ is-

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

Answer 1

Answer: (A)

[-1,3]

Given f: R$\to$ s difined by

$f\left(x\right)=sinx-\sqrt{3}cosx+1$

$=\frac{2}{3(sinx-\sqrt{3cosx})}+1$

$=2\frac{1}{2}sinx-\sqrt{\frac{3}{2}}cosx+1$

$=2\{cos{60}^{\circ }sinx-sin{60}^{\circ }cosx\}+1$

[$\because cos{60}^{\circ }=\frac{1}{2}sin{60}^{\circ }=\frac{\sqrt{3}}{2}]$

$2\left\{sin\right(60-x^{\circ }\left)\right\}+1$

$=2sin({60}^{\circ }-x)+1$

we know that -1$\le sin\theta \le$1 for all volume of$\theta$

$\therefore -\le sin({60}^{\circ }-x)\le 1$

$\Rightarrow -2\le sin({60}^{\circ }-x)\le 2$

$\Rightarrow -2+1\le 2sin({60}^{\circ }-x)+1\le 2+1$

$\Rightarrow -1\le f\left(x\right)\le 3$

$\therefore f\left(x\right)ϵ[-1,3]$

$\therefore$interval of S is [-1,3]