# Nakafa Framework: LLM

URL: https://nakafa.com/en/exercises/high-school/snbt/quantitative-knowledge/try-out/set-9/9

Exercises: Try Out - Set 9: Real exam simulation to sharpen your skills and build confidence. - Problem 9

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## Exercise 9

### Question

export const metadata = {
  title: "Problem 9",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/26/2025",
};

For the function <InlineMath math="f(x) = \frac{\sqrt{5 - x}}{\sqrt{x^2 + x - 6}}" /> to be defined in its domain, the domain of function <InlineMath math="f" /> is...


### Choices

- [ ] $$Df = \{x | x \leq 5\}$$
- [ ] $$Df = \{x | 2 < x \leq 5\}$$
- [ ] $$Df = \{x | x < -3 \text{ or } 2 < x < 5\}$$
- [x] $$Df = \{x | x < -3 \text{ or } 2 < x \leq 5\}$$
- [ ] $$Df = \{x | x < -3 \text{ or } 2 \leq x \leq 5\}$$

### Answer & Explanation

export const metadata = {
  title: "Solution for Problem 9",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/26/2025",
};

To find the domain of the function, we need to ensure that all components of the function are defined (values inside square roots must be non-negative, and the denominator cannot be zero).

The function <InlineMath math="f(x) = \frac{\sqrt{5 - x}}{\sqrt{x^2 + x - 6}}" /> has two main conditions:

#### Condition 1: Numerator

The value inside the square root in the numerator must be non-negative:

<MathContainer>
  <BlockMath math="5 - x \geq 0" />
  <BlockMath math="-x \geq -5" />
  <BlockMath math="x \leq 5" />
</MathContainer>

#### Condition 2: Denominator

The value inside the square root in the denominator must be positive (since the square root in the denominator cannot be zero):

<MathContainer>
  <BlockMath math="x^2 + x - 6 > 0" />
  <BlockMath math="(x + 3)(x - 2) > 0" />
</MathContainer>

The zeros of this inequality are <InlineMath math="x = -3" /> and <InlineMath math="x = 2" />.

Using test points, we find the solution region for the denominator: <InlineMath math="x < -3" /> or <InlineMath math="x > 2" />.

#### Domain Intersection

The domain of function <InlineMath math="f" /> is the intersection of the two conditions above:
1.  <InlineMath math="x \leq 5" />
2.  <InlineMath math="x < -3" /> or <InlineMath math="x > 2" />

The intersection is:
-   For <InlineMath math="x < -3" />, it satisfies <InlineMath math="x \leq 5" />.
-   For <InlineMath math="x > 2" />, it is bounded by <InlineMath math="x \leq 5" />, becoming <InlineMath math="2 < x \leq 5" />.

Therefore, the domain of function <InlineMath math="f" /> is:

<MathContainer>
  <BlockMath math="Df = \{x | x < -3 \text{ or } 2 < x \leq 5\}" />
</MathContainer>


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