`If 0 le x le pi/2 then `

Options

A. `2 sin x + tan x lt 3x `
B. `2 sin x+tan x lt 2 x `
C. `2 sin x + tan x le 3x `
D. `2 sin x+ tan x le 3x `

Explanation

To solve the problem, we need to analyze the function \( f(x) = 2 \cos x + \sec^2 x - 3 \) on the interval \( [0, \frac{\pi}{2}] \). We will find the first and second derivatives to determine the behavior of the function.

### Step 1: Define the function
We start with the function:
\[ f(x) = 2 \cos x + \sec^2 x - 3 \]

### Step 2: Find the first derivative
Next, we differentiate \( f(x) \) to find \( f'(x) \):
\[ f'(x) = \frac{d}{dx}(2 \cos x) + \frac{d}{dx}(\sec^2 x) - \frac{d}{dx}(3) \]
Using the derivatives:
- The derivative of \( 2 \cos x \) is \( -2 \sin x \).
- The derivative of \( \sec^2 x \) is \( 2 \sec^2 x \tan x \).

Thus, we have:
\[ f'(x) = -2 \sin x + 2 \sec^2 x \tan x \]

### Step 3: Find the critical points
To find critical points, we set \( f'(x) = 0 \):
\[ -2 \sin x + 2 \sec^2 x \tan x = 0 \]
This simplifies to:
\[ \sec^2 x \tan x = \sin x \]
Rearranging gives:
\[ \tan x = \sin x \cos^2 x \]

### Step 4: Analyze the second derivative
Now, we find the second derivative \( f''(x) \) to check the concavity:
\[ f''(x) = -2 \cos x + 2 \sec^2 x \tan^2 x + 2 \sec^4 x \]
This expression will help us determine if \( f(x) \) is increasing or decreasing.

### Step 5: Determine the intervals of increase/decrease
We need to check the sign of \( f'(x) \) over the interval \( [0, \frac{\pi}{2}] \):
- At \( x = 0 \):
\[ f'(0) = -2 \sin(0) + 2 \sec^2(0) \tan(0) = 0 \]
- At \( x = \frac{\pi}{2} \):
\[ f'(\frac{\pi}{2}) \text{ is undefined since } \sec^2(\frac{\pi}{2}) \text{ is undefined.} \]

### Step 6: Evaluate the function at endpoints
Evaluate \( f(x) \) at the endpoints:
- At \( x = 0 \):
\[ f(0) = 2 \cos(0) + \sec^2(0) - 3 = 2 + 1 - 3 = 0 \]
- As \( x \) approaches \( \frac{\pi}{2} \):
\[ f(x) \to \infty \text{ since } \sec^2 x \to \infty. \]

### Conclusion
Since \( f(0) = 0 \) and \( f(x) \to \infty \) as \( x \) approaches \( \frac{\pi}{2} \), and \( f'(x) \) is positive in the interval, we conclude that \( f(x) \) is increasing on \( [0, \frac{\pi}{2}] \).

### Final Answer
Thus, \( f(x) > 0 \) for all \( x \in (0, \frac{\pi}{2}) \).