The maximum and minimum values of `6 sin x cos x +4cos2x` are respectively

Options

A. `5 and -5`
B. `2sqrt13 and -2 sqrt13`
C. `10 and -10`
D. none of these

Explanation

To find the maximum and minimum values of the function \( f(x) = 6 \sin x \cos x + 4 \cos 2x \), we can follow these steps:

### Step 1: Rewrite the function
We start by rewriting \( 6 \sin x \cos x \) using the double angle identity for sine:
\[ 6 \sin x \cos x = 3 \cdot 2 \sin x \cos x = 3 \sin 2x \]
Thus, the function becomes:
\[ f(x) = 3 \sin 2x + 4 \cos 2x \]

### Step 2: Identify coefficients
Now we identify the coefficients \( a \) and \( b \):
- \( a = 3 \) (coefficient of \( \sin 2x \))
- \( b = 4 \) (coefficient of \( \cos 2x \))

### Step 3: Use the maximum and minimum value formula
The maximum and minimum values of a function of the form \( a \sin \theta + b \cos \theta \) can be found using the formulas:
- Maximum value = \( \sqrt{a^2 + b^2} \)
- Minimum value = \( -\sqrt{a^2 + b^2} \)

### Step 4: Calculate \( a^2 + b^2 \)
Now we calculate \( a^2 + b^2 \):
\[ a^2 + b^2 = 3^2 + 4^2 = 9 + 16 = 25 \]

### Step 5: Find the maximum and minimum values
Now we can find the maximum and minimum values:
- Maximum value = \( \sqrt{25} = 5 \)
- Minimum value = \( -\sqrt{25} = -5 \)

### Conclusion
Thus, the maximum and minimum values of the function \( f(x) = 6 \sin x \cos x + 4 \cos 2x \) are:
- Maximum value = \( 5 \)
- Minimum value = \( -5 \)

### Final Answer
The maximum and minimum values are \( 5 \) and \( -5 \), respectively.

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