At `x=(5pi)/6,\ \ f(x)=2sin3x+3cos3x` is (a) 0 (b) maximum (c) minimum (d) none of these

Options

A. maximum
B. minimum
C. zero
D. neither maximum nor minimum

Explanation

To solve the problem, we need to analyze the function \( f(x) = 2\sin(3x) + 3\cos(3x) \) at the point \( x = \frac{5\pi}{6} \) to determine if it is zero, a maximum, a minimum, or none of these.

### Step 1: Evaluate the function at \( x = \frac{5\pi}{6} \)

We start by substituting \( x = \frac{5\pi}{6} \) into the function:

\[ f\left(\frac{5\pi}{6}\right) = 2\sin\left(3 \cdot \frac{5\pi}{6}\right) + 3\cos\left(3 \cdot \frac{5\pi}{6}\right) \]

Calculating \( 3 \cdot \frac{5\pi}{6} \):

\[ 3 \cdot \frac{5\pi}{6} = \frac{15\pi}{6} = \frac{5\pi}{2} \]

Now we can evaluate the sine and cosine:

\[ f\left(\frac{5\pi}{6}\right) = 2\sin\left(\frac{5\pi}{2}\right) + 3\cos\left(\frac{5\pi}{2}\right) \]

Using the periodic properties of sine and cosine:

\[ \sin\left(\frac{5\pi}{2}\right) = \sin\left(\frac{5\pi}{2} - 2\pi\right) = \sin\left(\frac{\pi}{2}\right) = 1 \]

\[ \cos\left(\frac{5\pi}{2}\right) = \cos\left(\frac{5\pi}{2} - 2\pi\right) = \cos\left(\frac{\pi}{2}\right) = 0 \]

Substituting these values back into the function:

\[ f\left(\frac{5\pi}{6}\right) = 2 \cdot 1 + 3 \cdot 0 = 2 \]

### Step 2: Differentiate the function

Next, we differentiate \( f(x) \):

\[ f'(x) = \frac{d}{dx}(2\sin(3x) + 3\cos(3x)) \]

Using the chain rule:

\[ f'(x) = 2 \cdot 3\cos(3x) - 3 \cdot 3\sin(3x) = 6\cos(3x) - 9\sin(3x) \]

### Step 3: Find critical points

Set the derivative equal to zero to find critical points:

\[ 6\cos(3x) - 9\sin(3x) = 0 \]

Rearranging gives:

\[ 6\cos(3x) = 9\sin(3x) \]

Dividing both sides by \( 3\cos(3x) \) (assuming \( \cos(3x) \neq 0 \)):

\[ \tan(3x) = \frac{2}{3} \]

### Step 4: Solve for \( x \)

Taking the arctangent:

\[ 3x = \tan^{-1}\left(\frac{2}{3}\right) + n\pi \quad (n \in \mathbb{Z}) \]

Thus,

\[ x = \frac{1}{3}\tan^{-1}\left(\frac{2}{3}\right) + \frac{n\pi}{3} \]

### Step 5: Determine if \( x = \frac{5\pi}{6} \) is a critical point

To check if \( x = \frac{5\pi}{6} \) is a critical point, we note that:

\[ \frac{5\pi}{6} \neq \frac{1}{3}\tan^{-1}\left(\frac{2}{3}\right) + \frac{n\pi}{3} \]

This means that \( x = \frac{5\pi}{6} \) is not a critical point.

### Conclusion

Since \( f\left(\frac{5\pi}{6}\right) = 2 \) (which is not zero) and \( x = \frac{5\pi}{6} \) is not a critical point, we conclude that the function is neither zero, maximum, nor minimum at this point.

Thus, the answer is **(d) none of these**.

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