Question

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:

• A. \( \sqrt{2} \)
• B. 2
• C. \( \sqrt{\pi} \)
• D. \( \pi \)

Answer 1

The Correct Option is A

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.

Answer 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 \(\sqrt{2}\).