Understanding Onto Functions and Function Range

The question asks for the set S, which is the codomain of the function f(x) = 4 sin x – 3 cos x + 1. We are given that the function f is an onto function (also known as a surjective function). For an onto function, the codomain (the set S) is equal to the range of the function.

Therefore, to find S, we need to determine the range of the function f(x) = 4 sin x – 3 cos x + 1.

Finding the Range of \(a \sin x + b \cos x + c\)

Functions of the form \(a \sin x + b \cos x\) have a specific range. We can rewrite \(a \sin x + b \cos x\) in the form \(R \sin(x + \alpha)\) or \(R \cos(x - \beta)\), where \(R = \sqrt{a^2 + b^2}\). The range of \(R \sin(x + \alpha)\) or \(R \cos(x - \beta)\) is \([-R, R]\), because the range of \(\sin \theta\) and \(\cos \theta\) is \([-1, 1]\).

So, the range of \(a \sin x + b \cos x\) is \([-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]\).

If we add a constant \(c\) to this expression, the range shifts by \(c\). The range of \(a \sin x + b \cos x + c\) is \([-\sqrt{a^2 + b^2} + c, \sqrt{a^2 + b^2} + c]\).

Calculating the Range for f(x) = 4 sin x – 3 cos x + 1

In our function f(x) = 4 sin x – 3 cos x + 1, we have:

• a = 4
• b = -3
• c = 1

First, let's calculate the value of \(\sqrt{a^2 + b^2}\):

\( \sqrt{a^2 + b^2} = \sqrt{4^2 + (-3)^2} \)

\( = \sqrt{16 + 9} \)

\( = \sqrt{25} \)

\( = 5 \)

So, the range of \(4 \sin x - 3 \cos x\) is \([-5, 5]\).

Now, we add the constant term, c = 1, to this range:

Minimum value = \(-5 + 1 = -4\)

Maximum value = \(5 + 1 = 6\)

The range of f(x) = 4 sin x – 3 cos x + 1 is \([-4, 6]\).

Determining the Codomain S

Since the function f is onto, its codomain S is equal to its range.

Therefore, S = \([-4, 6]\).

Comparing with the Options

Let's compare our calculated range with the given options:

The calculated range \([-4, 6]\) matches Option 4.

Conclusion

Given that the function f(x) = 4 sin x – 3 cos x + 1 is onto, its codomain S must be equal to its range. By calculating the range of this trigonometric function, we found it to be \([-4, 6]\). Therefore, S = \([-4, 6]\).

Revision Table: Key Concepts for Function Range

Additional Information: Transforming \(a \sin x + b \cos x\)

The transformation of \(a \sin x + b \cos x\) into the form \(R \sin(x + \alpha)\) is a useful technique. Let \(a \sin x + b \cos x = R \sin(x + \alpha)\). Using the sine addition formula, \(\sin(x + \alpha) = \sin x \cos \alpha + \cos x \sin \alpha\), we get:

\( a \sin x + b \cos x = R (\sin x \cos \alpha + \cos x \sin \alpha) \)

\( a \sin x + b \cos x = (R \cos \alpha) \sin x + (R \sin \alpha) \cos x \)

Comparing the coefficients of \(\sin x\) and \(\cos x\):

• \(a = R \cos \alpha\)
• \(b = R \sin \alpha\)

Squaring and adding these equations gives \(a^2 + b^2 = R^2 \cos^2 \alpha + R^2 \sin^2 \alpha = R^2 (\cos^2 \alpha + \sin^2 \alpha) = R^2\). So, \(R = \sqrt{a^2 + b^2}\).

Dividing the second equation by the first gives \(\frac{b}{a} = \tan \alpha\). This transformation shows why the amplitude (and thus the range) depends on \(\sqrt{a^2 + b^2}\).