Question

Let f(x) be a increasing function defined on (0, ∞), $f(2a^2 + a + 1) > f(3a^2 - 4a + 1)$, then the range of a is

• A. $a \in (-\infty, \frac{1}{3}) \cup (1, \infty)$
• B. $a \in (0, 5)$
• C. $a \in (\frac{1}{3}, \infty)$
• D. $a \in (0, \frac{1}{3}) \cup (1, 5)$

Answer 1

Answer: (D)

$a \in (0, \frac{1}{3}) \cup (1, 5)$

Answer 2

Since f(x) is increasing, $2a^2 + a + 1 > 3a^2 - 4a + 1$, which simplifies to $a^2 - 5a < 0$, or $a(a-5) < 0$. This inequality holds for $0 < a < 5$. Also, the arguments of f must be positive:

1. $2a^2 + a + 1 > 0$. This is always true as the discriminant is negative and the leading coefficient is positive.
2. $3a^2 - 4a + 1 > 0$. The roots of $3a^2 - 4a + 1 = 0$ are $a = 1/3$ and $a = 1$. Since the parabola opens upwards, this inequality holds for $a < 1/3$ or $a > 1$. Combining $0 < a < 5$ with ($a < 1/3$ or $a > 1$) gives the intersection $(0, 1/3) \cup (1, 5)$.