Let the range of the function \[ f(x) = \frac{1}{2 + \sin 3x + \cos 3x}, \, x \in \mathbb{R} \, \text{be } [a, b]. \] If \( \alpha \) and \( \beta \) are respectively the arithmetic mean (A.M.) and the geometric mean (G.M.) of \( a \) and \( b \), then \( \frac{\alpha}{\beta} \) is equal to:
- \( \sqrt{2} \)
- 2
- \( \sqrt{\pi} \)
- \( \pi \)
The Correct Option is A
Approach Solution - 1
To find the range of the function \(f(x) = \frac{1}{2 + \sin 3x + \cos 3x}\), where \(x \in \mathbb{R}\), we start by exploring the expression \(2 + \sin 3x + \cos 3x\).
We know:
- The range of \(\sin \theta \, \text{and} \, \cos \theta\) are both \([-1, 1]\).
Thus, the expression \(2 + \sin 3x + \cos 3x\) can be written as:
\(2 + r \sqrt{2} \sin(3x + \phi)\), where \(r = \sqrt{(\sqrt{2})^2 + (\sqrt{2})^2} = \sqrt{2}\)
This simplifies to:
\(2 + \sqrt{2} (\sin 3x + \cos 3x) = 2 + \sqrt{2} \sqrt{1} \sin(3x + \phi)\)
The maximum value of \(\sin(3x + \phi)\) is 1, and the minimum value is -1.
So, the minimum and maximum values of \(2 + \sin 3x + \cos 3x\) are:
- Minimum: \(2 - \sqrt{2}\)
- Maximum: \(2 + \sqrt{2}\)
Thus, the range of \(f(x)\) is the reciprocal of these maximum and minimum values:
- Maximum of \(f(x)\) is \(\frac{1}{2 - \sqrt{2}}\)
- Minimum of \(f(x)\) is \(\frac{1}{2 + \sqrt{2}}\)
Therefore, the range of \(f(x)\) is \(\left[\frac{1}{2 + \sqrt{2}}, \frac{1}{2 - \sqrt{2}}\right]\).
Now, let's calculate the arithmetic mean (\(\alpha\)) and geometric mean (\(\beta\)) of \(a\) and \(b\).
- \(\alpha = \frac{1}{2 + \sqrt{2}} \times \frac{1}{2 - \sqrt{2}} = \frac{\frac{1}{2 + \sqrt{2}} + \frac{1}{2 - \sqrt{2}}}{2} = \frac{(2 - \sqrt{2}) + (2 + \sqrt{2})}{2 \cdot ((2 + \sqrt{2}) (2 - \sqrt{2}))}\)
- \(\alpha = \frac{4}{2(2^2 - (\sqrt{2})^2)} = \frac{4}{2 \cdot 2} = 1\)
- \(\beta = \sqrt{\frac{1}{2 + \sqrt{2}} \cdot \frac{1}{2 - \sqrt{2}}}\)
- \(\beta = \sqrt{\frac{1}{(2 + \sqrt{2})(2 - \sqrt{2})}} = \sqrt{\frac{1}{4 - 2}}\)
- \(\beta = \sqrt{\frac{1}{2}}\)
We now calculate \(\frac{\alpha}{\beta}\):
\(\frac{\alpha}{\beta} = \frac{1}{\sqrt{\frac{1}{2}}} = \sqrt{2}\)
Therefore, \(\frac{\alpha}{\beta} = \sqrt{2}\), which is the correct option.
Approach Solution -2
We are given the function:
\[ f(x) = \frac{1}{2 + \sin 3x + \cos 3x} \]
The function involves a trigonometric expression inside the denominator. First, we analyze the range of the expression \(2 + \sin 3x + \cos 3x\).
Step 1: Analyze the term \(\sin 3x + \cos 3x\).
We know that:
\[ \sin 3x + \cos 3x = \sqrt{2} \sin \left(3x + \frac{\pi}{4}\right) \]
Thus, the maximum value of \(\sin 3x + \cos 3x\) is \(\sqrt{2}\), and the minimum value is \(-\sqrt{2}\).
Step 2: Find the range of \(2 + \sin 3x + \cos 3x\).
Adding 2 to the above expression, we get:
\[ 2 + \sin 3x + \cos 3x = 2 + \sqrt{2} \sin \left(3x + \frac{\pi}{4}\right) \]
The minimum value of \(2 + \sin 3x + \cos 3x\) occurs when \(\sin 3x + \cos 3x = -\sqrt{2}\), giving:
\[ 2 - \sqrt{2} \]
The maximum value occurs when \(\sin 3x + \cos 3x = \sqrt{2}\), giving:
\[ 2 + \sqrt{2} \]
Thus, the range of \(f(x)\) is:
\[ \frac{1}{2 + \sqrt{2}} \text{ to } \frac{1}{2 - \sqrt{2}} \]
Step 3: Find the A.M. and G.M.
Let \( a = 2 + \sqrt{2} \) and \( b = 2 - \sqrt{2} \). The A.M. (Arithmetic Mean) and G.M. (Geometric Mean) of \(a\) and \(b\) are given by:
\[ \alpha = \frac{a + b}{2} \quad \text{and} \quad \beta = \sqrt{a \cdot b} \]
Calculating \(a + b\):
\[ a + b = (2 + \sqrt{2}) + (2 - \sqrt{2}) = 4 \]
Thus,
\[ \alpha = \frac{4}{2} = 2 \]
Now, calculate \(a \cdot b\):
\[ a \cdot b = (2 + \sqrt{2})(2 - \sqrt{2}) = 4 - 2 = 2 \]
Thus,
\[ \beta = \sqrt{2} \]
Step 4: Calculate \(\frac{\alpha}{\beta}\)
Finally, we calculate:
\[ \frac{\alpha}{\beta} = \frac{2}{\sqrt{2}} = \sqrt{2} \]
Thus, the value of \(\frac{\alpha}{\beta}\) is \(\sqrt{2}\).
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