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

Answer: (A)

$\sqrt{2}$

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 $\sqrt{2}$.