Answer :
Certainly, let's solve these two parts step by step.
### Part (i): Solve the inequality [tex]\(|x-3| > |x-14|\)[/tex]
To investigate the inequality [tex]\(|x-3| > |x-14|\)[/tex], we need to consider various cases based on the properties of absolute values.
#### Case 1: [tex]\(x \geq 14\)[/tex]
- For [tex]\(x \geq 14\)[/tex], both [tex]\(|x-3|\)[/tex] and [tex]\(|x-14|\)[/tex] can be simplified directly:
[tex]\[ |x-3| = x - 3 \quad \text{and} \quad |x-14| = x - 14 \][/tex]
The inequality [tex]\(x - 3 > x - 14\)[/tex] simplifies to:
[tex]\[ -3 > -14 \][/tex]
This is always true. Hence, for [tex]\(x \geq 14\)[/tex], the inequality [tex]\( |x-3| > |x-14| \)[/tex] holds.
#### Case 2: [tex]\(3 \leq x < 14\)[/tex]
- For [tex]\(3 \leq x < 14\)[/tex], we analyze the given absolute values as follows:
[tex]\[ |x-3| = x - 3 \quad \text{and} \quad |x-14| = 14 - x \][/tex]
The inequality [tex]\(x - 3 > 14 - x\)[/tex] can be simplified:
[tex]\[ x - 3 > 14 - x \][/tex]
To solve this inequality, combine like terms:
[tex]\[ 2x > 17 \][/tex]
[tex]\[ x > 8.5 \][/tex]
Therefore, for [tex]\(x \in [3,14)\)[/tex], the inequality holds if [tex]\(x > 8.5\)[/tex].
#### Case 3: [tex]\(x < 3\)[/tex]
- For [tex]\(x < 3\)[/tex], the absolute values become:
[tex]\[ |x-3| = 3 - x \quad \text{and} \quad |x-14| = 14 - x \][/tex]
The inequality [tex]\(3 - x > 14 - x\)[/tex] simplifies to:
[tex]\[ 3 > 14 \][/tex]
This is never true, so there are no solutions for [tex]\(x < 3\)[/tex].
Combining the results of all cases, the solution to [tex]\(|x-3| > |x-14|\)[/tex] is:
[tex]\[ x > 14 \quad \text{or} \quad 8.5 < x < 14 \][/tex]
So the final solution for part (i) is:
[tex]\[ x > 8.5 \][/tex]
### Part (ii): Solve the inequality [tex]\(\frac{x^2-4}{x} \leq 0\)[/tex]
To solve [tex]\(\frac{x^2-4}{x} \leq 0\)[/tex], we first find the roots and analyze the sign changes. Notice:
[tex]\[ \frac{x^2-4}{x} = \frac{(x-2)(x+2)}{x} \][/tex]
Set the numerator equal to zero to find the critical points:
[tex]\[ x^2 - 4 = 0 \implies x = 2 \text{ or } x = -2 \][/tex]
Also include the denominator:
[tex]\[ x \neq 0 \][/tex]
The critical points divide the real line into intervals. We test these intervals:
- [tex]\( (-\infty, -2) \)[/tex]
- [tex]\( (-2, 0) \)[/tex]
- [tex]\( (0, 2) \)[/tex]
- [tex]\( (2, \infty) \)[/tex]
In each interval, choose a test point and determine the sign of [tex]\(\frac{(x-2)(x+2)}{x}\)[/tex]:
- For [tex]\(x \in (-\infty, -2)\)[/tex], choose [tex]\( x = -3 \)[/tex]:
[tex]\[ \frac{(-3-2)(-3+2)}{-3} = \frac{(-5)(-1)}{-3} = \frac{5}{-3} < 0 \][/tex]
- For [tex]\(x \in (-2, 0)\)[/tex], choose [tex]\( x = -1 \)[/tex]:
[tex]\[ \frac{(-1-2)(-1+2)}{-1} = \frac{(-3)(1)}{-1} = 3 > 0 \][/tex]
- For [tex]\(x \in (0, 2)\)[/tex], choose [tex]\( x = 1 \)[/tex]:
[tex]\[ \frac{(1-2)(1+2)}{1} = \frac{(-1)(3)}{1} = -3 < 0 \][/tex]
- For [tex]\(x \in (2, \infty)\)[/tex], choose [tex]\( x = 3 \)[/tex]:
[tex]\[ \frac{(3-2)(3+2)}{3} = \frac{(1)(5)}{3} > 0 \][/tex]
From our sign analysis, the function is non-positive on the intervals [tex]\((-\infty, -2)\)[/tex] and [tex]\((0, 2)\)[/tex].
Including the critical points where the expression equals zero:
[tex]\[ x = -2 \quad \text{and} \quad x = 2 \][/tex]
Combining these:
[tex]\[ x \in [-2, 0) \cup (0, 2] \][/tex]
So the final solution for part (ii) is:
[tex]\[ x \in [-2, 0) \cup (0, 2] \][/tex]
These are the values of [tex]\(x\)[/tex] that satisfy the given inequality [tex]\(\frac{x^2-4}{x} \leq 0\)[/tex].
### Part (i): Solve the inequality [tex]\(|x-3| > |x-14|\)[/tex]
To investigate the inequality [tex]\(|x-3| > |x-14|\)[/tex], we need to consider various cases based on the properties of absolute values.
#### Case 1: [tex]\(x \geq 14\)[/tex]
- For [tex]\(x \geq 14\)[/tex], both [tex]\(|x-3|\)[/tex] and [tex]\(|x-14|\)[/tex] can be simplified directly:
[tex]\[ |x-3| = x - 3 \quad \text{and} \quad |x-14| = x - 14 \][/tex]
The inequality [tex]\(x - 3 > x - 14\)[/tex] simplifies to:
[tex]\[ -3 > -14 \][/tex]
This is always true. Hence, for [tex]\(x \geq 14\)[/tex], the inequality [tex]\( |x-3| > |x-14| \)[/tex] holds.
#### Case 2: [tex]\(3 \leq x < 14\)[/tex]
- For [tex]\(3 \leq x < 14\)[/tex], we analyze the given absolute values as follows:
[tex]\[ |x-3| = x - 3 \quad \text{and} \quad |x-14| = 14 - x \][/tex]
The inequality [tex]\(x - 3 > 14 - x\)[/tex] can be simplified:
[tex]\[ x - 3 > 14 - x \][/tex]
To solve this inequality, combine like terms:
[tex]\[ 2x > 17 \][/tex]
[tex]\[ x > 8.5 \][/tex]
Therefore, for [tex]\(x \in [3,14)\)[/tex], the inequality holds if [tex]\(x > 8.5\)[/tex].
#### Case 3: [tex]\(x < 3\)[/tex]
- For [tex]\(x < 3\)[/tex], the absolute values become:
[tex]\[ |x-3| = 3 - x \quad \text{and} \quad |x-14| = 14 - x \][/tex]
The inequality [tex]\(3 - x > 14 - x\)[/tex] simplifies to:
[tex]\[ 3 > 14 \][/tex]
This is never true, so there are no solutions for [tex]\(x < 3\)[/tex].
Combining the results of all cases, the solution to [tex]\(|x-3| > |x-14|\)[/tex] is:
[tex]\[ x > 14 \quad \text{or} \quad 8.5 < x < 14 \][/tex]
So the final solution for part (i) is:
[tex]\[ x > 8.5 \][/tex]
### Part (ii): Solve the inequality [tex]\(\frac{x^2-4}{x} \leq 0\)[/tex]
To solve [tex]\(\frac{x^2-4}{x} \leq 0\)[/tex], we first find the roots and analyze the sign changes. Notice:
[tex]\[ \frac{x^2-4}{x} = \frac{(x-2)(x+2)}{x} \][/tex]
Set the numerator equal to zero to find the critical points:
[tex]\[ x^2 - 4 = 0 \implies x = 2 \text{ or } x = -2 \][/tex]
Also include the denominator:
[tex]\[ x \neq 0 \][/tex]
The critical points divide the real line into intervals. We test these intervals:
- [tex]\( (-\infty, -2) \)[/tex]
- [tex]\( (-2, 0) \)[/tex]
- [tex]\( (0, 2) \)[/tex]
- [tex]\( (2, \infty) \)[/tex]
In each interval, choose a test point and determine the sign of [tex]\(\frac{(x-2)(x+2)}{x}\)[/tex]:
- For [tex]\(x \in (-\infty, -2)\)[/tex], choose [tex]\( x = -3 \)[/tex]:
[tex]\[ \frac{(-3-2)(-3+2)}{-3} = \frac{(-5)(-1)}{-3} = \frac{5}{-3} < 0 \][/tex]
- For [tex]\(x \in (-2, 0)\)[/tex], choose [tex]\( x = -1 \)[/tex]:
[tex]\[ \frac{(-1-2)(-1+2)}{-1} = \frac{(-3)(1)}{-1} = 3 > 0 \][/tex]
- For [tex]\(x \in (0, 2)\)[/tex], choose [tex]\( x = 1 \)[/tex]:
[tex]\[ \frac{(1-2)(1+2)}{1} = \frac{(-1)(3)}{1} = -3 < 0 \][/tex]
- For [tex]\(x \in (2, \infty)\)[/tex], choose [tex]\( x = 3 \)[/tex]:
[tex]\[ \frac{(3-2)(3+2)}{3} = \frac{(1)(5)}{3} > 0 \][/tex]
From our sign analysis, the function is non-positive on the intervals [tex]\((-\infty, -2)\)[/tex] and [tex]\((0, 2)\)[/tex].
Including the critical points where the expression equals zero:
[tex]\[ x = -2 \quad \text{and} \quad x = 2 \][/tex]
Combining these:
[tex]\[ x \in [-2, 0) \cup (0, 2] \][/tex]
So the final solution for part (ii) is:
[tex]\[ x \in [-2, 0) \cup (0, 2] \][/tex]
These are the values of [tex]\(x\)[/tex] that satisfy the given inequality [tex]\(\frac{x^2-4}{x} \leq 0\)[/tex].
