Question

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

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

Answer 1

The correct option is D

$\left[-1,3\right]$

Explanation for the correct option:

Find the interval of $S$:

Given, $f\left(x\right)=\sin x-\sqrt{3}\cos x+1$

As we know,

The range of the function $f\left(x\right)=a\cos x+b\sin x+c$ is:

$c–\sqrt{(a^2+b^2)}\le f\left(x\right)\le c+\sqrt{(a^2+b^2)}$

Here, $a=-\sqrt{3},b=1,c=1$

Now,

$\begin{array}{rcl}c–\sqrt{(a^2+b^2)}& =& 1–\sqrt{(3+1)}\\ & =& 1–2\\ & =& -1\end{array}$

$\begin{array}{rcl}c+\sqrt{(a^2+b^2)}& =& 1+\sqrt{(3+1)}\\ & =& 1+2\\ & =& 3\end{array}$

Given that $f:R\to S$ is an onto function.

$\therefore$the interval of $S$ will be equal to the range of $f\left(x\right)$.

Thus, the interval of $S$ is $[-1,3]$

Hence, Option ‘D’ is Correct.