If `a^(2)gtb^(2),` then the minimum value of `f(x)=a^(2) cos^(2)x+b^(2) sin^(2) x` is

Options

A. `a^(2)-b^(2)`
B. `a^(2)+b^(2)`
C. `a^(2)`
D. `b^(2)`

Answer

To find the minimum value of the function \( f(x) = a^2 \cos^2 x + b^2 \sin^2 x \) given that \( a^2 > b^2 \), we can follow these steps:

### Step 1: Differentiate the function
We start by differentiating \( f(x) \) with respect to \( x \):
\[ f'(x) = \frac{d}{dx}(a^2 \cos^2 x + b^2 \sin^2 x) \]
Using the chain rule, we get:
\[ f'(x) = a^2 \cdot 2 \cos x (-\sin x) + b^2 \cdot 2 \sin x \cos x \]
This simplifies to:
\[ f'(x) = -2a^2 \sin x \cos x + 2b^2 \sin x \cos x \]
Factoring out \( 2 \sin x \cos x \):
\[ f'(x) = 2 \sin x \cos x (b^2 - a^2) \]

### Step 2: Set the derivative to zero
To find the critical points, we set \( f'(x) = 0 \):
\[ 2 \sin x \cos x (b^2 - a^2) = 0 \]
This gives us two cases:
1. \( \sin x = 0 \)
2. \( \cos x = 0 \)

### Step 3: Solve for critical points
From \( \sin x = 0 \), we have:
\[ x = n\pi \quad (n \in \mathbb{Z}) \]
From \( \cos x = 0 \), we have:
\[ x = \frac{\pi}{2} + n\pi \quad (n \in \mathbb{Z}) \]

### Step 4: Evaluate the function at critical points
Now we evaluate \( f(x) \) at the critical points:
1. For \( x = 0 \):
\[ f(0) = a^2 \cos^2(0) + b^2 \sin^2(0) = a^2 \cdot 1 + b^2 \cdot 0 = a^2 \]
2. For \( x = \frac{\pi}{2} \):
\[ f\left(\frac{\pi}{2}\right) = a^2 \cos^2\left(\frac{\pi}{2}\right) + b^2 \sin^2\left(\frac{\pi}{2}\right) = a^2 \cdot 0 + b^2 \cdot 1 = b^2 \]

### Step 5: Determine the minimum value
Since \( a^2 > b^2 \), we conclude that:
\[ f(0) = a^2 \quad \text{and} \quad f\left(\frac{\pi}{2}\right) = b^2 \]
Thus, the minimum value of \( f(x) \) occurs at \( x = \frac{\pi}{2} \) and is:
\[ \text{Minimum value} = b^2 \]

### Final Answer
The minimum value of \( f(x) = a^2 \cos^2 x + b^2 \sin^2 x \) is \( b^2 \).

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