The value of ordinate of the graph of y = 2 + cos x lies in the interval
[1, 3]
Understanding the Range of the Ordinate for y = 2 + cos x
The question asks for the interval in which the value of the ordinate (the y-value) of the graph of the function \(y = 2 + \cos x\) lies. To determine this, we need to understand the range of the basic cosine function and how it is affected by the addition of a constant.
Range of the Cosine Function
For any real number \(x\), the value of \(\cos x\) is always between -1 and 1, inclusive. Mathematically, this is represented as:
\(-1 \le \cos x \le 1\)
This is the fundamental range for the basic cosine function \(f(x) = \cos x\).
Finding the Range of y = 2 + cos x
Our given function is \(y = 2 + \cos x\). To find the range of \(y\), we can use the known range of \(\cos x\). Since \(-1 \le \cos x \le 1\), we can add 2 to all parts of this inequality:
\(2 + (-1) \le 2 + \cos x \le 2 + 1\)
Simplifying the inequality, we get:
\(1 \le 2 + \cos x \le 3\)
Since \(y = 2 + \cos x\), the inequality becomes:
\(1 \le y \le 3\)
This inequality tells us that the value of \(y\), the ordinate of the graph, lies between 1 and 3, including 1 and 3. In interval notation, this is represented as \([1, 3]\).
Therefore, the value of the ordinate of the graph of \(y = 2 + \cos x\) lies in the interval \([1, 3]\).
Comparing with Options
Let's compare our derived interval with the given options:
- [0, 1] - Incorrect. The minimum value is 1.
- [0, 3] - Incorrect. The minimum value is 1, not 0.
- [-1, 1] - Incorrect. This is the range of \(\cos x\), not \(2 + \cos x\).
- [1, 3] - Correct. This matches our calculated interval.
The range of the function \(y = 2 + \cos x\) is indeed \([1, 3]\).
| Function | Range (Interval) |
|---|---|
| \(f(x) = \sin x\) | [-1, 1] |
| \(f(x) = \cos x\) | [-1, 1] |
| \(f(x) = A \cos x\) | [-|A|, |A|] |
| \(f(x) = A \cos x + B\) | [-|A| + B, |A| + B] |
| \(y = 2 + \cos x\) (Here A=1, B=2) | [-1 + 2, 1 + 2] = [1, 3] |
Revision Table: Key Trigonometric Ranges
| Trigonometric Function | Domain (Real Numbers) | Range |
|---|---|---|
| \(\sin x\) | \((-\infty, \infty)\) | [-1, 1] |
| \(\cos x\) | \((-\infty, \infty)\) | [-1, 1] |
| \(\tan x\) | \(x \ne n\pi + \frac{\pi}{2}\), where \(n\) is an integer | \((-\infty, \infty)\) |
| \(\csc x\) | \(x \ne n\pi\), where \(n\) is an integer | \((-\infty, -1] \cup [1, \infty)\) |
| \(\sec x\) | \(x \ne n\pi + \frac{\pi}{2}\), where \(n\) is an integer | \((-\infty, -1] \cup [1, \infty)\) |
| \(\cot x\) | \(x \ne n\pi\), where \(n\) is an integer | \((-\infty, \infty)\) |
Additional Information: Transformations of Trigonometric Graphs
Understanding how transformations affect the range of trigonometric functions is crucial. For a function like \(y = A \cos(Bx + C) + D\):
- The value of \(A\) is the amplitude. It vertically stretches or compresses the graph and affects the range. The range of \(A \cos(Bx + C)\) is \([-|A|, |A|]\).
- The value of \(B\) affects the period of the function, but not the range.
- The value of \(C\) causes a horizontal shift (phase shift), but not the range.
- The value of \(D\) causes a vertical shift. This shift directly affects the range. If the range of the original function is \([m, M]\), the range of the shifted function \(f(x) + D\) is \([m+D, M+D]\).
In our specific problem, \(y = 2 + \cos x\), we can write it as \(y = 1 \cdot \cos(1 \cdot x + 0) + 2\). Here, \(A=1\) and \(D=2\). The range of \(\cos x\) is \([-1, 1]\). Applying the vertical shift \(+2\), the new range is \([-1+2, 1+2]\), which is \([1, 3]\). This confirms our calculation.
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What is the value of tan 18°?
What is the period of the function f(x) = ln(2 + sin 2x)?
If tan (45° + θ) = 1 + sin 2θ, where \(-\frac{\pi}{4}< θ <\frac{\pi}{4},\) then what is the value of cos 2θ?
