Question

Let a, b, c be the real numbers such that a + b + c = 2 and ab + bc + ca = 1, where a≤ b ≤ c, then

The complete set of values of 'a' is :

• A. $0, \frac{1}{3}$
• B. $0, \frac{1}{2}$
• C. [0, 1]
• D. $-\frac{1}{3}, \frac{2}{3}$

Answer 1

The mathematically derived set of values for 'a' is [0, 1/3]. None of the provided options match this result. Option C is [0, 1]. Since [0, 1/3] ⊂ [0, 1], and intervals are expected for a "complete set of values", option C might be the intended answer, possibly due to a typo in the question or options. However, based on the rigorous derivation, values of 'a' in (1/3, 1] are not possible. Given the necessity to choose one option, and assuming a potential error in the options, option C is the most plausible candidate if the intended range was close to [0, 1] or [0, 1/3].

Answer 2

Let a, b, c be real numbers such that a + b + c = 2 and ab + bc + ca = 1, with the constraint a ≤ b ≤ c. Consider the quadratic equation whose roots are b and c. From the given equations, we have b + c = 2 - a and bc = 1 - a(b + c) = 1 - a(2 - a) = 1 - 2a + a² = (1 - a)². The quadratic equation is $t^2 - (2 - a)t + (1 - a)^2 = 0$. For real roots b and c, the discriminant must be non-negative: $\Delta = (-(2 - a))^2 - 4(1 - a)^2 = (2 - a)^2 - 4(1 - a)^2 = (4 - 4a + a^2) - 4(1 - 2a + a^2) = 4 - 4a + a^2 - 4 + 8a - 4a^2 = 4a - 3a^2$. $\Delta \ge 0 \implies 4a - 3a^2 \ge 0 \implies a(4 - 3a) \ge 0$. This inequality holds for $0 \le a \le 4/3$.

The roots are $b = \frac{(2 - a) - \sqrt{4a - 3a^2}}{2}$ and $c = \frac{(2 - a) + \sqrt{4a - 3a^2}}{2}$. The condition a ≤ b ≤ c must be satisfied. The condition b ≤ c is satisfied if $\sqrt{4a - 3a^2} \ge 0$, which means $0 \le a \le 4/3$.

The condition a ≤ b is: $a \le \frac{(2 - a) - \sqrt{4a - 3a^2}}{2}$ $2a \le 2 - a - \sqrt{4a - 3a^2}$ $3a - 2 \le - \sqrt{4a - 3a^2}$ $\sqrt{4a - 3a^2} \le 2 - 3a$

For this inequality to hold, we need $2 - 3a \ge 0$ (since the left side is non-negative) and $4a - 3a^2 \ge 0$ (for the square root to be real). $2 - 3a \ge 0 \implies 3a \le 2 \implies a \le 2/3$. $4a - 3a^2 \ge 0 \implies 0 \le a \le 4/3$. The intersection of these two conditions is $0 \le a \le 2/3$.

If $0 \le a \le 2/3$, we can square both sides of $\sqrt{4a - 3a^2} \le 2 - 3a$: $(\sqrt{4a - 3a^2})^2 \le (2 - 3a)^2$ $4a - 3a^2 \le 4 - 12a + 9a^2$ $0 \le 12a^2 - 16a + 4$ Divide by 4: $0 \le 3a^2 - 4a + 1$ Factor the quadratic: $3a^2 - 4a + 1 = (3a - 1)(a - 1)$. The inequality $(3a - 1)(a - 1) \ge 0$ holds when $a \le 1/3$ or $a \ge 1$.

We need to satisfy all conditions for a ≤ b: $0 \le a \le 2/3$ AND ($a \le 1/3$ or $a \ge 1$). The intersection of $[0, 2/3]$ and $(-\infty, 1/3] \cup [1, \infty)$ is $[0, 1/3]$. So the condition a ≤ b holds for $a \in [0, 1/3]$.

For a ≤ b ≤ c to hold, we need a ≤ b AND b ≤ c, and b, c must be real. b, c are real when $a \in [0, 4/3]$. a ≤ b holds when $a \in [0, 1/3]$. b ≤ c holds when $a \in [0, 4/3]$.

The complete set of values of 'a' is the intersection of these conditions: $[0, 4/3] \cap [0, 1/3] \cap [0, 4/3] = [0, 1/3]$.

The mathematically derived set of values for 'a' is $[0, 1/3]$. Comparing this with the given options, none of the options exactly match this interval. Option C is $[0, 1]$. Since $[0, 1/3] \subset [0, 1]$, and intervals are expected for a "complete set of values", option C might be the intended answer, possibly due to a typo in the question or options. However, based on the rigorous derivation, values of 'a' in $(1/3, 1]$ are not possible.

Given the necessity to choose one option, and assuming a potential error in the options, option C is the most plausible candidate if the intended range was close to [0, 1] or [0, 1/3].