Key Concepts and Formulas

• A quadratic expression $Ax^2 + Bx + C$ is positive for all real $x$ if and only if $A > 0$ and the discriminant $D = B^2 - 4AC < 0$.
• The quadratic formula for finding the roots of $ax^2 + bx + c = 0$ is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
• For a quadratic inequality $ax^2 + bx + c < 0$ with $a > 0$, the solution is the interval between the roots of $ax^2 + bx + c = 0$.

Step-by-Step Solution

Step 1: Identify Coefficients

We are given the quadratic expression $(1 + 2m)x^2 - 2(1 + 3m)x + 4(1 + m)$. We identify the coefficients $A$, $B$, and $C$ in terms of $m$:

• $A = 1 + 2m$
• $B = -2(1 + 3m)$
• $C = 4(1 + m)$ This step is crucial because the conditions for the quadratic expression to be always positive depend directly on these coefficients.

Step 2: Apply the Condition A > 0

For the expression to be always positive, the coefficient of $x^2$ must be positive: $1 + 2m > 0$ Subtracting 1 from both sides: $2m > -1$ Dividing by 2: $m > -\frac{1}{2}$ This gives us the first condition on $m$.

Step 3: Calculate the Discriminant

The discriminant $D$ is given by $D = B^2 - 4AC$. Substituting the values of $A$, $B$, and $C$, we get: $D = (-2(1 + 3m))^2 - 4(1 + 2m)(4(1 + m))$ $D = 4(1 + 3m)^2 - 16(1 + 2m)(1 + m)$ We need to expand and simplify this expression.

Step 4: Expand and Simplify the Discriminant

Expanding the squared term and the product term: $(1 + 3m)^2 = 1 + 6m + 9m^2$ $(1 + 2m)(1 + m) = 1 + m + 2m + 2m^2 = 1 + 3m + 2m^2$ Substituting these expansions back into the expression for $D$: $D = 4(1 + 6m + 9m^2) - 16(1 + 3m + 2m^2)$ $D = 4 + 24m + 36m^2 - 16 - 48m - 32m^2$ $D = 4m^2 - 24m - 12$ This simplified expression will be used in the next step.

Step 5: Apply the Condition D < 0

We require the discriminant to be negative: $4m^2 - 24m - 12 < 0$ Dividing the inequality by 4: $m^2 - 6m - 3 < 0$ Now, we need to solve this quadratic inequality.

Step 6: Solve the Quadratic Inequality

First, find the roots of the corresponding quadratic equation $m^2 - 6m - 3 = 0$ using the quadratic formula: $m = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-3)}}{2(1)}$ $m = \frac{6 \pm \sqrt{36 + 12}}{2}$ $m = \frac{6 \pm \sqrt{48}}{2}$ $m = \frac{6 \pm 4\sqrt{3}}{2}$ $m = 3 \pm 2\sqrt{3}$ So the roots are $m_1 = 3 - 2\sqrt{3}$ and $m_2 = 3 + 2\sqrt{3}$. Since the coefficient of $m^2$ is positive, the inequality $m^2 - 6m - 3 < 0$ is satisfied for values of $m$ between the roots: $3 - 2\sqrt{3} < m < 3 + 2\sqrt{3}$

Step 7: Find the Intersection of the Two Conditions

We need to find the values of $m$ that satisfy both $m > -\frac{1}{2}$ and $3 - 2\sqrt{3} < m < 3 + 2\sqrt{3}$. Since $2\sqrt{3} = \sqrt{12} \approx 3.464$, we have $3 - 2\sqrt{3} \approx 3 - 3.464 = -0.464$ and $3 + 2\sqrt{3} \approx 3 + 3.464 = 6.464$. Thus, the interval is approximately $-0.464 < m < 6.464$.

The first condition is $m > -0.5$. The intersection of the two conditions is $-0.464 < m < 6.464$, since $-0.464 > -0.5$. Thus, we have $3 - 2\sqrt{3} < m < 3 + 2\sqrt{3}$.

Step 8: Find the Integral Values of m

We need to find the integers $m$ such that $3 - 2\sqrt{3} < m < 3 + 2\sqrt{3}$. Since $-0.464 < m < 6.464$, the integral values of $m$ are $0, 1, 2, 3, 4, 5, 6$.

Step 9: Count the Integral Values

There are 7 integers in the list $0, 1, 2, 3, 4, 5, 6$.

Common Mistakes & Tips

• Remember to check both conditions: $A > 0$ and $D < 0$. Failing to check $A > 0$ is a common mistake.
• Be careful with algebraic manipulations, especially when expanding and simplifying the discriminant. Double-check your work to avoid errors.
• When solving quadratic inequalities, remember that the solution depends on the sign of the leading coefficient.

Summary

We found the conditions for the quadratic expression to be always positive: $A > 0$ and $D < 0$. These conditions led to the inequalities $m > -\frac{1}{2}$ and $3 - 2\sqrt{3} < m < 3 + 2\sqrt{3}$. The intersection of these intervals gives $3 - 2\sqrt{3} < m < 3 + 2\sqrt{3}$, which contains the integral values $0, 1, 2, 3, 4, 5, 6$. Therefore, there are 7 integral values of $m$.

Final Answer

The final answer is \boxed{7}, which corresponds to option (A).