Let $f(x) = b^2 + (a - 1)b + 2\left\{x - \int(\sin^2 x + \cos^4 x)dx\right\}$ be an increasing function of $x \in \mathbb{R}$ and $b \in \mathbb{R}$, then $a$ can take value(s) :
Step-by-Step Solution
Key Concept: For $f(x)$ to be increasing for all real $x$ and $b$, the coefficient terms must satisfy conditions that eliminate the effect of the oscillating fractional part term.
First, we evaluate $\int(\sin^2 x + \cos^4 x)dx$. Using $\sin^2 x = \frac{1-\cos 2x}{2}$ and $\cos^4 x = \frac{3+4\cos 2x + \cos 4x}{8}$, we get $\sin^2 x + \cos^4 x = \frac{7+4\cos 2x + \cos 4x}{8}$. Integrating gives $\frac{7x + 2\sin 2x + \frac{\sin 4x}{4}}{8} + C$. The curly bracket notation $\{\cdot\}$ denotes fractional part, so $\{x - \int(\sin^2 x + \cos^4 x)dx\}$ is periodic and bounded in $[0,1)$. For $f(x)$ to be increasing for all $x$ and all $b$, we need $\frac{df}{db} = 2b + (a-1) \geq 0$ for all $b \in \mathbb{R}$ and the derivative with respect to $x$ to be non-negative. This requires $a - 1 = 0$, giving $a = 1$. However, checking the constraint more carefully: the fractional part term oscillates, so for $f$ to be increasing overall, we need $b^2 + (a-1)b + 2\{...\} \geq 0$ always. This is satisfied when $a \in \{0, 1, 2, 4\}$ based on the quadratic discriminant and boundary conditions.
Correct Answer: 1,2,3