If `f_(1)(x)=2x, f_(2)x(=3 sin x-x cos x,` for `x in (0, (pi)/(2))`

Options

A. `f_(1)(x) lt f_(2)(x)`
B. `f_(1)(x) gt f_(2)(x)`
C. `f_(1)(|x|)ltf_(2)(|x|)`
D. none of these

Answer

To solve the problem, we need to compare the two functions \( f_1(x) = 2x \) and \( f_2(x) = 3 \sin x - x \cos x \) for \( x \) in the interval \( (0, \frac{\pi}{2}) \).

### Step-by-Step Solution:

1. **Define the Difference of Functions**:
We define a new function \( f(x) \) as the difference between \( f_1(x) \) and \( f_2(x) \):
\[ f(x) = f_1(x) - f_2(x) = 2x - (3 \sin x - x \cos x) \]
Simplifying this gives:
\[ f(x) = 2x - 3 \sin x + x \cos x \]

2. **Differentiate \( f(x) \)**:
To analyze the behavior of \( f(x) \), we differentiate it:
\[ f'(x) = \frac{d}{dx}(2x) - \frac{d}{dx}(3 \sin x) + \frac{d}{dx}(x \cos x) \]
Using the product rule for \( x \cos x \):
\[ f'(x) = 2 - 3 \cos x + (\cos x - x \sin x) \]
Thus,
\[ f'(x) = 2 - 3 \cos x + \cos x - x \sin x = 2 - 2 \cos x - x \sin x \]

3. **Analyze the Sign of \( f'(x) \)**:
We need to determine whether \( f'(x) \) is positive or negative in the interval \( (0, \frac{\pi}{2}) \).

Since \( \cos x \) is positive and decreasing in this interval, and \( \sin x \) is also positive and increasing, we can analyze the behavior of \( f'(x) \):
- At \( x = 0 \):
\[ f'(0) = 2 - 2 \cdot 1 - 0 = 0 \]
- As \( x \) increases from \( 0 \) to \( \frac{\pi}{2} \), \( \cos x \) decreases and \( \sin x \) increases, which implies \( -2 \cos x \) becomes less negative and \( -x \sin x \) becomes more negative.

4. **Determine the Second Derivative**:
To confirm whether \( f'(x) \) is increasing or decreasing, we can find the second derivative \( f''(x) \):
\[ f''(x) = \frac{d}{dx}(f'(x)) = 0 + 2 \sin x - (\sin x + x \cos x) \]
Simplifying gives:
\[ f''(x) = \sin x - x \cos x \]

5. **Check the Sign of \( f''(x) \)**:
We need to check if \( f''(x) \) is positive or negative in the interval \( (0, \frac{\pi}{2}) \):
- At \( x = 0 \):
\[ f''(0) = \sin(0) - 0 \cdot \cos(0) = 0 \]
- For \( x > 0 \), \( \sin x \) is positive and \( x \cos x \) is also positive but grows slower than \( \sin x \). Hence, \( f''(x) \) is likely to be positive for small \( x \).

6. **Conclusion**:
Since \( f'(x) \) starts at 0 and is positive for \( x \) in \( (0, \frac{\pi}{2}) \), we conclude that \( f(x) \) is increasing. Thus, \( f(x) > 0 \) for \( x \in (0, \frac{\pi}{2}) \), which implies:
\[ f_1(x) > f_2(x) \]

### Final Result:
Thus, the relation between the two functions is:
\[ f_1(x) > f_2(x) \quad \text{for } x \in (0, \frac{\pi}{2}) \]