The Minimum value of `27^cosx +81^sinx` is equal to

Options

A. `(2)/(3sqrt3)`
B. `(2)/(9sqrt3)`
C. `(4)/(3sqrt3)`
D. none of these

Explanation

To find the minimum value of the expression \(27^{\cos x} + 81^{\sin x}\), we can follow these steps:

### Step 1: Rewrite the expression
First, we can rewrite the bases in terms of powers of 3:
\[ 27^{\cos x} = (3^3)^{\cos x} = 3^{3 \cos x} \]
\[ 81^{\sin x} = (3^4)^{\sin x} = 3^{4 \sin x} \]
Thus, we can rewrite the expression as:
\[ 27^{\cos x} + 81^{\sin x} = 3^{3 \cos x} + 3^{4 \sin x} \]

### Step 2: Apply the AM-GM Inequality
Using the Arithmetic Mean-Geometric Mean (AM-GM) inequality, we know that:
\[ \frac{a + b}{2} \geq \sqrt{ab} \]
where \(a = 3^{3 \cos x}\) and \(b = 3^{4 \sin x}\). Therefore, we have:
\[ \frac{3^{3 \cos x} + 3^{4 \sin x}}{2} \geq \sqrt{3^{3 \cos x} \cdot 3^{4 \sin x}} \]
This simplifies to:
\[ \frac{3^{3 \cos x} + 3^{4 \sin x}}{2} \geq 3^{\frac{3 \cos x + 4 \sin x}{2}} \]
Thus, we can express the original inequality as:
\[ 3^{3 \cos x} + 3^{4 \sin x} \geq 2 \cdot 3^{\frac{3 \cos x + 4 \sin x}{2}} \]

### Step 3: Minimize the exponent
Next, we need to minimize the exponent \(\frac{3 \cos x + 4 \sin x}{2}\). We can use the known result that for any \(a\) and \(b\):
\[ a \sin x + b \cos x \text{ has a minimum value of } -\sqrt{a^2 + b^2} \]
In our case, \(a = 4\) and \(b = 3\):
\[ \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]
Thus, the minimum value of \(3 \cos x + 4 \sin x\) is \(-5\).

### Step 4: Substitute back into the inequality
Substituting this minimum value back into our inequality gives:
\[ 3^{3 \cos x + 4 \sin x} \geq 3^{-5} \]
So we have:
\[ 3^{3 \cos x + 4 \sin x} \geq \frac{1}{3^5} = \frac{1}{243} \]

### Step 5: Final expression
Thus, we can conclude:
\[ 27^{\cos x} + 81^{\sin x} \geq 2 \cdot 3^{-\frac{5}{2}} = 2 \cdot \frac{1}{\sqrt{3^5}} = \frac{2}{9\sqrt{3}} \]

### Conclusion
The minimum value of \(27^{\cos x} + 81^{\sin x}\) is:
\[ \frac{2}{9\sqrt{3}} \]