The correct answer is option (B):
statement (2) alone is sufficient to answer the question
The question asks if $^4 \sqrt{x} > ^3 \sqrt{x}$ given that x is positive. This inequality can be rewritten as $x^{1/4} > x^{1/3}$.
To determine when this inequality holds, we can consider the relationship between the exponents. Since x is positive, we can analyze the behavior of the function $f(x) = x^a$.
If x > 1, then $x^{1/4} > x^{1/3}$ is false because the exponent 1/3 is greater than 1/4, and for x > 1, a larger exponent leads to a larger value. So, if x > 1, $x^{1/3} > x^{1/4}$.
If 0 < x < 1, then $x^{1/4} > x^{1/3}$ is true because for 0 < x < 1, a smaller exponent leads to a larger value. So, if 0 < x < 1, $x^{1/4} > x^{1/3}$.
If x = 1, then $x^{1/4} = 1$ and $x^{1/3} = 1$, so $x^{1/4} = x^{1/3}$, and the inequality $x^{1/4} > x^{1/3}$ is false.
Therefore, the inequality $^4 \sqrt{x} > ^3 \sqrt{x}$ is true if and only if 0 < x < 1.
Now let's analyze the statements.
Statement 1: 0.5 < x < 2
This statement tells us that x is in the range (0.5, 2).
If x is in the range (0.5, 1), then the inequality $^4 \sqrt{x} > ^3 \sqrt{x}$ is true.
If x is in the range [1, 2), then the inequality is false.
Since Statement 1 allows for values where the inequality is true and values where it is false, it is not sufficient.
Statement 2: $12x^2 - 7x + 1 = 0$
Solve using factoring:
We look for two numbers that multiply to 12 and add to -7. They are -3 and -4.
Rewrite the equation:
$12x^2 - 4x - 3x + 1 = 0$
$4x(3x - 1) - 1(3x - 1) = 0$
$(4x - 1)(3x - 1) = 0$
So the solutions are:
$4x - 1 = 0 \Rightarrow x = 1/4$
$3x - 1 = 0 \Rightarrow x = 1/3$
Both values 1/4 and 1/3 satisfy 0 < x < 1.
We know from the initial inequality analysis that for any x in this interval, $^4 \sqrt{x} > ^3 \sqrt{x}$ is true.
Therefore, Statement 2 alone guarantees a definite “YES” to the question. It is sufficient.
The final answer is: $\boxed{\text{statement (2) alone is sufficient to answer the question}}$.