1. Key Concepts and Formulas

• Condition for a Quadratic to be Always Positive: For a quadratic expression $Ax^2 + Bx + C$ to be strictly positive for all real values of $x$ (i.e., $Ax^2 + Bx + C > 0$ for all $x \in \mathbf{R}$), two conditions must be met:
  1. The leading coefficient must be positive: $A > 0$. This ensures the parabola opens upwards.
  2. The discriminant must be negative: $\Delta = B^2 - 4AC < 0$. This ensures the parabola does not intersect or touch the x-axis, thus remaining entirely above it.
• Probability Formula: For an event $E$ in a sample space of equally likely outcomes, the probability $P(E)$ is given by: $P(E) = \frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Possible Outcomes}}$
• Sample Space for Die Throws: When a fair six-faced die is thrown twice, the outcomes are ordered pairs $(\alpha, \beta)$ where $\alpha, \beta \in \{1, 2, 3, 4, 5, 6\}$. Each throw is independent.

2. Step-by-Step Solution

Step 1: Determine the algebraic condition for the quadratic to be always positive. We are given the quadratic expression $x^2 + \alpha x + \beta$. Comparing this with the standard form $Ax^2 + Bx + C$, we identify the coefficients:

• $A = 1$ (coefficient of $x^2$)
• $B = \alpha$ (coefficient of $x$)
• $C = \beta$ (constant term)

Now, we apply the two conditions for a quadratic to be always positive:

1. Leading Coefficient ($A > 0$): Here, $A = 1$. Since $1 > 0$, this condition is satisfied. This confirms that the parabola representing the quadratic opens upwards, which is necessary for it to be always positive.

2. Discriminant ($\Delta < 0$): The discriminant is calculated as $\Delta = B^2 - 4AC$. Substituting the coefficients from our quadratic: $\Delta = (\alpha)^2 - 4(1)(\beta) = \alpha^2 - 4\beta$ For the quadratic to be always positive for all $x \in \mathbf{R}$, the discriminant must be strictly negative: $\alpha^2 - 4\beta < 0$ Rearranging this inequality, we get the crucial condition: $\alpha^2 < 4\beta$ This is the algebraic condition that the values of $\alpha$ and $\beta$ must satisfy.

Step 2: Define the sample space and total number of outcomes. The numbers $\alpha$ and $\beta$ are the results of two independent throws of a fair six-faced die. Therefore:

• $\alpha$ can be any integer from the set $\{1, 2, 3, 4, 5, 6\}$.
• $\beta$ can be any integer from the set $\{1, 2, 3, 4, 5, 6\}$. Since there are 6 possible outcomes for the first throw and 6 possible outcomes for the second throw, the total number of distinct ordered pairs $(\alpha, \beta)$ in our sample space is: $\text{Total Number of Outcomes} = 6 \times 6 = 36$ Each of these 36 outcomes is equally likely.

Step 3: Identify the favorable outcomes. We need to find all pairs $(\alpha, \beta)$ from our sample space (where $\alpha, \beta \in \{1, 2, 3, 4, 5, 6\}$) that satisfy the condition $\alpha^2 < 4\beta$. We will do this systematically by considering each possible value of $\alpha$:

• Case 1: If $\alpha = 1$ The condition becomes $1^2 < 4\beta \implies 1 < 4\beta \implies \beta > \frac{1}{4}$. The possible integer values for $\beta$ from $\{1, 2, 3, 4, 5, 6\}$ that satisfy this are $\{1, 2, 3, 4, 5, 6\}$. Number of favorable outcomes for $\alpha=1$: 6 pairs.

• Case 2: If $\alpha = 2$ The condition becomes $2^2 < 4\beta \implies 4 < 4\beta \implies \beta > 1$. The possible integer values for $\beta$ from $\{1, 2, 3, 4, 5, 6\}$ that satisfy this are $\{2, 3, 4, 5, 6\}$. Number of favorable outcomes for $\alpha=2$: 5 pairs.

• Case 3: If $\alpha = 3$ The condition becomes $3^2 < 4\beta \implies 9 < 4\beta \implies \beta > \frac{9}{4} \implies \beta > 2.25$. The possible integer values for $\beta$ from $\{1, 2, 3, 4, 5, 6\}$ that satisfy this are $\{3, 4, 5, 6\}$. Number of favorable outcomes for $\alpha=3$: 4 pairs.

• Case 4: If $\alpha = 4$ The condition becomes $4^2 < 4\beta \implies 16 < 4\beta \implies \beta > 4$. The possible integer values for $\beta$ from $\{1, 2, 3, 4, 5, 6\}$ that satisfy this are $\{5, 6\}$. Number of favorable outcomes for $\alpha=4$: 2 pairs.

• Case 5: If $\alpha = 5$ The condition becomes $5^2 < 4\beta \implies 25 < 4\beta \implies \beta > \frac{25}{4} \implies \beta > 6.25$. There are no integer values for $\beta$ in the set $\{1, 2, 3, 4, 5, 6\}$ that satisfy this condition. Number of favorable outcomes for $\alpha=5$: 0 pairs.

• Case 6: If $\alpha = 6$ The condition becomes $6^2 < 4\beta \implies 36 < 4\beta \implies \beta > 9$. There are no integer values for $\beta$ in the set $\{1, 2, 3, 4, 5, 6\}$ that satisfy this condition. Number of favorable outcomes for $\alpha=6$: 0 pairs.

Summing up the number of favorable outcomes from all cases: $\text{Total Number of Favorable Outcomes} = 6 + 5 + 4 + 2 + 0 + 0 = 17$

Step 4: Calculate the probability. Using the probability formula, we divide the total number of favorable outcomes by the total number of possible outcomes: $P(\text{Quadratic always positive}) = \frac{\text{Total Number of Favorable Outcomes}}{\text{Total Number of Possible Outcomes}}$ $P(\text{Quadratic always positive}) = \frac{17}{36}$

3. Common Mistakes & Tips

• Common Mistake 1: Incorrect Discriminant Condition: It's crucial to remember that for a quadratic $Ax^2+Bx+C > 0$, the discriminant must be strictly negative ($\Delta < 0$). If the problem had asked for $Ax^2+Bx+C \ge 0$, then $\Delta \le 0$ would be the correct condition.
• Common Mistake 2: Errors in Counting Integer Solutions: When solving the inequality for $\beta$ (e.g., $\beta > k$), be careful with fractional values of $k$. For instance, $\beta > 2.25$ means the smallest integer value for $\beta$ is 3, not 2.
• Tip 1: Systematic Listing: For problems involving discrete outcomes like die rolls, always list the possibilities systematically (e.g., by iterating through one variable and then the other). This methodical approach minimizes errors and ensures no valid outcomes are missed.

4. Summary

This problem effectively tests the understanding of conditions for a quadratic expression to be always positive, combined with basic probability principles. The first step was to translate the condition "$x^2+\alpha x+\beta>0$ for all $x \in \mathbf{R}$" into an algebraic inequality involving $\alpha$ and $\beta$, which resulted in $\alpha^2 < 4\beta$. Following this, we systematically enumerated all possible pairs of die rolls $(\alpha, \beta)$ that satisfied this inequality. There were 17 such favorable outcomes. Dividing this by the total number of possible outcomes for two die rolls (36) gave the final probability.

The final answer is \boxed{\frac{17}{36}} which corresponds to option (A).