Question

If $\mathrm{f}:\mathrm{R}\to \mathrm{S}$ defined by $\mathrm{f}\left(\mathrm{x}\right)=\mathrm{sinx}-\sqrt{3}\mathrm{cosx}+1$ is onto then the interval of $S$ is

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

Answer 1

The correct option is D

$[-1,3]$

Step 1: Given information

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

The given function is onto function.

Step 2: Determination of the above equation

$\therefore (\sin x-\sqrt{3}\cos x)+1$

Multiply and divide by $2$

$\begin{array}{l}\frac{2}{2}(\sin x-\sqrt{3}\cos x)+1\\ =2\left(\frac{1}{2}\sin x-\frac{\sqrt{3}}{2}\cos x\right)+1\\ =2\left(sin\left(\frac{\pi }{6}\right)\sin x-\cos \left(\frac{\pi }{6}\right)\cos x\right)+1;\left[\because \frac{1}{2}=\sin \frac{\pi }{6}\text{and}\frac{\sqrt{3}}{2}=\cos \left(\frac{\pi }{6}\right)\right]\\ =-2(\cos \left(\frac{\pi }{6}\right)\cos x-sin\left(\frac{\pi }{6}\right)\sin x)+1;\left[\because \cos (A+B)=\cos A\cos B-\sin A\sin B\right]\\ =-2\cos \left(\frac{\pi }{6}+x\right)+1\end{array}$

The expression $-2\cos \left(\frac{\pi }{6}+x\right)$ is very between $-2\mathrm{to}2$

The net value $-2+1$ represents the minimum value and $2+1$represents the maximum value

Therefore, the Range lies between $[-1,3]$

An interval of $S=$the Range of $f\left(x\right)=\left[-1,3\right]$

Hence, the correct option is an option (D).