Key Concepts and Formulas

1. Onto Function (Surjective Function): A function $f: A \to B$ is onto if for every element $y$ in the codomain $B$, there exists at least one element $x$ in the domain $A$ such that $f(x) = y$. This implies that the range of an onto function is equal to its codomain.
2. Range of $a \sin x + b \cos x$: An expression of the form $a \sin x + b \cos x$ can be rewritten as $R \sin(x + \alpha)$ or $R \cos(x - \alpha)$, where $R = \sqrt{a^2 + b^2}$. The range of $a \sin x + b \cos x$ is $[-\sqrt{a^2+b^2}, \sqrt{a^2+b^2}]$.
3. Range of $\sin \theta$: For any real number $\theta$, the value of $\sin \theta$ lies in the interval $[-1, 1]$.

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Step-by-Step Solution

1. Understanding the "Onto" Condition

• We are given a function $f: R \to S$ defined by $f\left( x \right) = \sin x - \sqrt 3 \cos x + 1$.
• We are told that this function is onto.
• Explanation: The definition of an onto function states that its codomain is exactly the same as its range. Therefore, to find the interval of $S$, we need to determine the range of the function $f(x)$.

2. Rewriting the Trigonometric Expression

• The function is $f\left( x \right) = \sin x - \sqrt 3 \cos x + 1$.

• Explanation: The term $\sin x - \sqrt 3 \cos x$ is a linear combination of $\sin x$ and $\cos x$. We can rewrite this in the form $R \sin(x - \alpha)$ or $R \cos(x + \alpha)$ to simplify finding its range. This involves identifying $a$ and $b$ from the expression $a \sin x + b \cos x$. Here, $a = 1$ and $b = -\sqrt{3}$.

• Step 2a: Calculate the amplitude ($R$). The amplitude $R$ is found using the formula $R = \sqrt{a^2 + b^2}$. $R = \sqrt{(1)^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2$

• Step 2b: Express the trigonometric part in the form $R \sin(x - \alpha)$. We factor out $R=2$: $\sin x - \sqrt 3 \cos x = 2 \left( \frac{1}{2} \sin x - \frac{\sqrt{3}}{2} \cos x \right)$ Explanation: We want to express $\frac{1}{2}$ and $\frac{\sqrt{3}}{2}$ as trigonometric values of a standard angle $\alpha$, so we can use the sine subtraction formula: $\sin(A - B) = \sin A \cos B - \cos A \sin B$. We look for an angle $\alpha$ such that $\cos \alpha = \frac{1}{2}$ and $\sin \alpha = \frac{\sqrt{3}}{2}$. This angle is $\alpha = \frac{\pi}{3}$.

• Step 2c: Apply the sine subtraction formula. Substituting the trigonometric values: $2 \left( \cos \frac{\pi}{3} \sin x - \sin \frac{\pi}{3} \cos x \right)$ Using the formula $\sin(x - \frac{\pi}{3}) = \sin x \cos \frac{\pi}{3} - \cos x \sin \frac{\pi}{3}$: $\sin x - \sqrt 3 \cos x = 2 \sin \left( x - \frac{\pi}{3} \right)$

• Step 2d: Substitute the simplified expression back into $f(x)$. Now, we replace the original trigonometric part with its simplified form in the function definition: $f\left( x \right) = 2 \sin \left( x - \frac{\pi}{3} \right) + 1$

3. Determining the Range of $f(x)$

• Explanation: We will now determine the range of $f(x)$ by applying transformations to the known range of the sine function.

• Step 3a: Establish the range of the sine term. Since the domain of $f(x)$ is $R$, the argument $x - \frac{\pi}{3}$ can take any real value. The range of the sine function for any real argument is $[-1, 1]$. $-1 \le \sin \left( x - \frac{\pi}{3} \right) \le 1$

• Step 3b: Scale the range by the amplitude. Multiply all parts of the inequality by the amplitude $R=2$: $2 \times (-1) \le 2 \sin \left( x - \frac{\pi}{3} \right) \le 2 \times 1$ $-2 \le 2 \sin \left( x - \frac{\pi}{3} \right) \le 2$ Explanation: Multiplying an inequality by a positive number preserves the direction of the inequality signs.

• Step 3c: Shift the range by the constant term. Add 1 to all parts of the inequality to account for the constant term in $f(x)$: $-2 + 1 \le 2 \sin \left( x - \frac{\pi}{3} \right) + 1 \le 2 + 1$ $-1 \le f\left( x \right) \le 3$ Explanation: Adding a constant to all parts of an inequality shifts the entire interval without changing the direction of the inequalities.

4. Concluding the Interval of S

• From the previous steps, we found that the range of $f(x)$ is $[-1, 3]$.
• Since the function $f: R \to S$ is given to be onto, its codomain $S$ must be equal to its range.
• Therefore, the interval of $S$ is $[-1, 3]$.

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Common Mistakes & Tips

• Definition of Onto: Always remember that for an onto function, the codomain is identical to the range.
• Trigonometric Transformations: Master the conversion of $a \sin x + b \cos x$ into $R \sin(x \pm \alpha)$ or $R \cos(x \pm \alpha)$. This is a crucial skill for finding ranges of trigonometric functions.
• Inequality Rules: Be meticulous when manipulating inequalities. Adding or subtracting a constant does not change the inequality direction, but multiplying or dividing by a negative number reverses it.
• Constant Term: Do not forget to include the constant term when calculating the final range of the function.

Summary

The problem asks for the codomain $S$ of an onto function $f(x) = \sin x - \sqrt{3} \cos x + 1$. Since the function is onto, its codomain must be equal to its range. We first rewrite the trigonometric part $\sin x - \sqrt{3} \cos x$ into the form $2 \sin(x - \frac{\pi}{3})$. By considering the range of the sine function, $[-1, 1]$, and applying the scaling factor of 2 and the addition of 1, we determine the range of $f(x)$ to be $[-1, 3]$. Therefore, the codomain $S$ is $[-1, 3]$.

The final answer is $\boxed{[-1, 3]}$.