Let `f(x)=[b^(2)+(a-1)b+2]x-int(sin^(2)x+cos^(4)x)dx` be an increasing function of `x""inRandbinR`, then a can take value(s)

Options

A. 0
B. 1
C. 2
D. 4

Answer

To solve the problem, we need to analyze the function \( f(x) = [b^2 + (a-1)b + 2]x - \int (\sin^2 x + \cos^4 x) \, dx \) and determine the values of \( a \) for which \( f(x) \) is an increasing function.

### Step 1: Differentiate \( f(x) \)
To determine if \( f(x) \) is increasing, we need to find its derivative \( f'(x) \):
\[ f'(x) = b^2 + (a-1)b + 2 - (\sin^2 x + \cos^4 x) \]
The term \( \int (\sin^2 x + \cos^4 x) \, dx \) differentiates to \( \sin^2 x + \cos^4 x \).

### Step 2: Set the condition for increasing function
Since \( f(x) \) is an increasing function, we require:
\[ f'(x) > 0 \]
This implies:
\[ b^2 + (a-1)b + 2 - (\sin^2 x + \cos^4 x) > 0 \]

### Step 3: Analyze \( \sin^2 x + \cos^4 x \)
Next, we need to find the range of \( \sin^2 x + \cos^4 x \). We can express \( \sin^2 x \) in terms of \( \cos^2 x \):
\[ \sin^2 x = 1 - \cos^2 x \]
Thus,
\[ \sin^2 x + \cos^4 x = 1 - \cos^2 x + \cos^4 x \]
Let \( y = \cos^2 x \), then:
\[ \sin^2 x + \cos^4 x = 1 - y + y^2 \]
This is a quadratic function in \( y \).

### Step 4: Find the minimum value of \( 1 - y + y^2 \)
The vertex of the quadratic \( y^2 - y + 1 \) occurs at:
\[ y = -\frac{b}{2a} = \frac{1}{2} \]
Substituting \( y = \frac{1}{2} \):
\[ 1 - \frac{1}{2} + \left(\frac{1}{2}\right)^2 = 1 - \frac{1}{2} + \frac{1}{4} = \frac{3}{4} \]
Thus, the minimum value of \( \sin^2 x + \cos^4 x \) is \( \frac{3}{4} \).

### Step 5: Substitute the minimum value into the inequality
Now substituting the minimum value back into our inequality:
\[ b^2 + (a-1)b + 2 - \frac{3}{4} > 0 \]
This simplifies to:
\[ b^2 + (a-1)b + \frac{5}{4} > 0 \]

### Step 6: Analyze the quadratic inequality
For the quadratic \( b^2 + (a-1)b + \frac{5}{4} \) to be always positive, the discriminant must be less than zero:
\[ D = (a-1)^2 - 4 \cdot 1 \cdot \frac{5}{4} < 0 \]
This simplifies to:
\[ (a-1)^2 - 5 < 0 \]
\[ (a-1)^2 < 5 \]

### Step 7: Solve the inequality
Taking square roots:
\[ -\sqrt{5} < a - 1 < \sqrt{5} \]
Thus,
\[ 1 - \sqrt{5} < a < 1 + \sqrt{5} \]

### Step 8: Approximate the values
Calculating \( 1 - \sqrt{5} \) and \( 1 + \sqrt{5} \):
- \( \sqrt{5} \approx 2.236 \)
- Therefore, \( 1 - \sqrt{5} \approx -1.236 \) and \( 1 + \sqrt{5} \approx 3.236 \).

### Conclusion
The values of \( a \) that make \( f(x) \) an increasing function are:
\[ a \in (1 - \sqrt{5}, 1 + \sqrt{5}) \approx (-1.236, 3.236) \]