Let f(x) =cosx sin2x. Then , min `(f(x):-pilexlepi)` is

Options

A. `-9//7`
B. `9//7`
C. `-1//9`
D. `-2//9`

Answer

To find the minimum of the function \( f(x) = \cos x \sin 2x \) in the interval \( -\pi \leq x \leq \pi \), we can follow these steps:

### Step 1: Rewrite the Function
We start with the function:
\[ f(x) = \cos x \sin 2x \]
Using the double angle formula for sine, we can rewrite \( \sin 2x \) as \( 2 \sin x \cos x \):
\[ f(x) = \cos x (2 \sin x \cos x) = 2 \cos^2 x \sin x \]

### Step 2: Substitute \( t = \sin x \)
Let \( t = \sin x \). Since \( x \) ranges from \( -\pi \) to \( \pi \), \( t \) will range from \( -1 \) to \( 1 \). We can express \( \cos^2 x \) in terms of \( t \):
\[ \cos^2 x = 1 - \sin^2 x = 1 - t^2 \]
Thus, we can rewrite \( f(x) \) as:
\[ f(t) = 2 (1 - t^2) t = 2t - 2t^3 \]

### Step 3: Differentiate the Function
To find the critical points, we differentiate \( f(t) \):
\[ f'(t) = 2 - 6t^2 \]
Setting the derivative equal to zero to find critical points:
\[ 2 - 6t^2 = 0 \implies 6t^2 = 2 \implies t^2 = \frac{1}{3} \implies t = \pm \frac{1}{\sqrt{3}} \]

### Step 4: Determine the Nature of Critical Points
Next, we find the second derivative to determine whether these points are minima or maxima:
\[ f''(t) = -12t \]
Evaluating the second derivative at the critical points:
1. For \( t = \frac{1}{\sqrt{3}} \):
\[ f''\left(\frac{1}{\sqrt{3}}\right) = -12 \cdot \frac{1}{\sqrt{3}} < 0 \quad (\text{local maximum}) \]
2. For \( t = -\frac{1}{\sqrt{3}} \):
\[ f''\left(-\frac{1}{\sqrt{3}}\right) = -12 \cdot \left(-\frac{1}{\sqrt{3}}\right) > 0 \quad (\text{local minimum}) \]

### Step 5: Calculate the Minimum Value
Now we calculate the value of \( f(t) \) at \( t = -\frac{1}{\sqrt{3}} \):
\[ f\left(-\frac{1}{\sqrt{3}}\right) = 2\left(-\frac{1}{\sqrt{3}}\right) - 2\left(-\frac{1}{\sqrt{3}}\right)^3 \]
Calculating each term:
\[ = -\frac{2}{\sqrt{3}} + 2\left(-\frac{1}{3\sqrt{3}}\right) = -\frac{2}{\sqrt{3}} + \frac{2}{3\sqrt{3}} = -\frac{6}{3\sqrt{3}} + \frac{2}{3\sqrt{3}} = -\frac{4}{3\sqrt{3}} \]

### Step 6: Final Result
Thus, the minimum value of \( f(x) \) in the interval \( -\pi \leq x \leq \pi \) is:
\[ \text{Minimum } f(x) = -\frac{4}{3\sqrt{3}} \]