# Nakafa Framework: LLM

URL: https://nakafa.com/en/exercises/high-school/snbt/quantitative-knowledge/try-out/set-8/10

Exercises: Try Out - Set 8: Real exam simulation to sharpen your skills and build confidence. - Problem 10

---

## Exercise 10

### Question

export const metadata = {
    title: "Problem 10",
    authors: [{ name: "Nabil Akbarazzima Fatih" }],
    date: "12/25/2025",
};

For <InlineMath math="(2m + 1)x^2 - 4(m + 1)x + m > -4" /> to hold for every real <InlineMath math="x" />, the values of <InlineMath math="m" /> must satisfy...


### Choices

- [ ] $$m < 0 \lor m > \frac{1}{2}$$
- [ ] $$-\frac{1}{2} < m < \frac{1}{2}$$
- [x] $$0 < m < \frac{1}{2}$$
- [ ] $$0 \leq m < \frac{1}{2}$$
- [ ] $$m < -\frac{1}{2} \lor m > 0$$

### Answer & Explanation

export const metadata = {
  title: "Solution for Problem 10",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/25/2025",
};

Given the inequality:

<BlockMath math="(2m + 1)x^2 - 4(m + 1)x + m > -4" />

Rearrange it into the standard quadratic form by moving <InlineMath math="-4" /> to the left side:

<BlockMath math="(2m + 1)x^2 - 4(m + 1)x + (m + 4) > 0" />

For the quadratic function to be positive for every real <InlineMath math="x" /> (positive definite), its graph must lie entirely above the <InlineMath math="x" />-axis. The conditions are:

1. The coefficient of <InlineMath math="x^2" /> must be positive (<InlineMath math="a > 0" />).
2. The discriminant must be negative (<InlineMath math="D < 0" />).

**Condition 1: <InlineMath math="a > 0" />**

<BlockMath math="\begin{aligned} 2m + 1 &> 0 \\ 2m &> -1 \\ m &> -\frac{1}{2} \quad \dots \text{(1)} \end{aligned}" />

**Condition 2: <InlineMath math="D < 0" />**

Recall that <InlineMath math="D = b^2 - 4ac" />. With <InlineMath math="a = 2m + 1" />, <InlineMath math="b = -4(m + 1)" />, and <InlineMath math="c = m + 4" />:

<BlockMath math="\begin{aligned} (-4(m + 1))^2 - 4(2m + 1)(m + 4) &< 0 \\ 16(m^2 + 2m + 1) - 4(2m^2 + 9m + 4) &< 0 \end{aligned}" />

Divide both sides by <InlineMath math="4" /> to simplify:

<BlockMath math="\begin{aligned} 4(m^2 + 2m + 1) - (2m^2 + 9m + 4) &< 0 \\ 4m^2 + 8m + 4 - 2m^2 - 9m - 4 &< 0 \\ 2m^2 - m &< 0 \\ m(2m - 1) &< 0 \end{aligned}" />

The roots are <InlineMath math="m = 0" /> and <InlineMath math="m = \frac{1}{2}" />. Since the inequality is "&lt;", the solution lies between the roots:

<BlockMath math="0 < m < \frac{1}{2} \quad \dots \text{(2)}" />

**Conclusion**

We find the intersection of conditions <InlineMath math="(1)" /> and <InlineMath math="(2)" />:
- Condition <InlineMath math="(1)" />: <InlineMath math="m > -\frac{1}{2}" />
- Condition <InlineMath math="(2)" />: <InlineMath math="0 < m < \frac{1}{2}" />

The intersection of both regions is:

<BlockMath math="0 < m < \frac{1}{2}" />


---