Question

Q.1) $\left|1+\frac{3}{x}\right|>2$

$\Rightarrow \left(1+\frac{3}{x}\right)^2>4$

$\Rightarrow 1+\frac{9}{x^2}+\frac{6}{x}>4$

$\Rightarrow \frac{x^2+9+6x}{x^2}>4$

$\Rightarrow x^2+9+6x>4x^2$

$\Rightarrow 4x^2-x^2-9-6x<0$

$\Rightarrow 3x^2-6x-9<0$

$\Rightarrow 3x^2-9x+3x-9<0$

$\Rightarrow 3x(x-3)+3(x-3)<0$

$\Rightarrow (3x+3)(x-3)<0$

$\Rightarrow (x+1)(x-3)<0$

Answer 1

$x\in(-1,3)$

Answer 2

The problem is to solve the inequality $\left|1+\frac{3}{x}\right|>2$.

Method 1: Using the definition of absolute value.

$|A| > b$ is equivalent to $A > b$ or $A < -b$. In this case, $A = 1 + \frac{3}{x}$ and $b = 2$. So, we have two cases:

Case 1: $1 + \frac{3}{x} > 2$

$\frac{3}{x} > 2 - 1$

$\frac{3}{x} > 1$

$\frac{3}{x} - 1 > 0$

$\frac{3 - x}{x} > 0$

To solve this inequality, we find the critical points where the numerator or denominator is zero. These are $x=3$ and $x=0$. We analyze the sign of the expression $\frac{3-x}{x}$ in the intervals $(-\infty, 0)$, $(0, 3)$, and $(3, \infty)$.

• For $x \in (-\infty, 0)$, let $x=-1$. $\frac{3 - (-1)}{-1} = \frac{4}{-1} = -4 < 0$.

• For $x \in (0, 3)$, let $x=1$. $\frac{3 - 1}{1} = \frac{2}{1} = 2 > 0$.

• For $x \in (3, \infty)$, let $x=4$. $\frac{3 - 4}{4} = \frac{-1}{4} < 0$.

The inequality $\frac{3 - x}{x} > 0$ is satisfied for $x \in (0, 3)$.

Case 2: $1 + \frac{3}{x} < -2$

$\frac{3}{x} < -2 - 1$

$\frac{3}{x} < -3$

$\frac{3}{x} + 3 < 0$

$\frac{3 + 3x}{x} < 0$

$\frac{3(1 + x)}{x} < 0$

$\frac{1 + x}{x} < 0$

The critical points are where the numerator or denominator is zero. These are $x=-1$ and $x=0$. We analyze the sign of the expression $\frac{1+x}{x}$ in the intervals $(-\infty, -1)$, $(-1, 0)$, and $(0, \infty)$.

• For $x \in (-\infty, -1)$, let $x=-2$. $\frac{1 + (-2)}{-2} = \frac{-1}{-2} = \frac{1}{2} > 0$.

• For $x \in (-1, 0)$, let $x=-0.5$. $\frac{1 + (-0.5)}{-0.5} = \frac{0.5}{-0.5} = -1 < 0$.

• For $x \in (0, \infty)$, let $x=1$. $\frac{1 + 1}{1} = \frac{2}{1} = 2 > 0$.

The inequality $\frac{1 + x}{x} < 0$ is satisfied for $x \in (-1, 0)$.

The solution to the original inequality is the union of the solutions from Case 1 and Case 2. Solution set = $(0, 3) \cup (-1, 0)$. This can also be written as $x \in (-1, 3)$ and $x \neq 0$.

Method 2: Squaring both sides.

$\left|1+\frac{3}{x}\right|>2$ Square both sides: $\left(1+\frac{3}{x}\right)^2 > 4$.

$1 + 2 \cdot 1 \cdot \frac{3}{x} + \left(\frac{3}{x}\right)^2 > 4$

$1 + \frac{6}{x} + \frac{9}{x^2} > 4$

Combine terms on the left side: $\frac{x^2}{x^2} + \frac{6x}{x^2} + \frac{9}{x^2} > 4$

$\frac{x^2 + 6x + 9}{x^2} > 4$

Subtract 4 from both sides: $\frac{x^2 + 6x + 9}{x^2} - 4 > 0$

$\frac{x^2 + 6x + 9 - 4x^2}{x^2} > 0$

$\frac{-3x^2 + 6x + 9}{x^2} > 0$

Factor the numerator: $\frac{-3(x^2 - 2x - 3)}{x^2} > 0$

$\frac{-3(x+1)(x-3)}{x^2} > 0$

Divide by -3 and reverse the inequality sign: $\frac{(x+1)(x-3)}{x^2} < 0$

For this inequality to hold, the numerator $(x+1)(x-3)$ and the denominator $x^2$ must have opposite signs. Since $x^2 \ge 0$, and $x \neq 0$ (from the original inequality), we have $x^2 > 0$. Therefore, the numerator must be negative: $(x+1)(x-3) < 0$

The critical points are $x=-1$ and $x=3$. The quadratic expression $(x+1)(x-3)$ is a parabola opening upwards, so it is negative between the roots. The solution to $(x+1)(x-3) < 0$ is $x \in (-1, 3)$. However, we must also satisfy the condition that the denominator $x^2$ is non-zero, which means $x \neq 0$. So, the solution is $x \in (-1, 3)$ and $x \neq 0$. This gives the solution set $(-1, 0) \cup (0, 3)$.

The provided solution incorrectly concludes that the solution is $x \in (-1, 3)$ after obtaining $(x+1)(x-3) < 0$. It failed to consider the restriction $x \neq 0$ imposed by the original inequality.

The final answer is $\boxed{(-1, 0) \cup (0, 3)}$.

Explanation of the solution: The inequality is $\left|1+\frac{3}{x}\right|>2$. This is equivalent to $1+\frac{3}{x} > 2$ or $1+\frac{3}{x} < -2$.

Case 1: $1+\frac{3}{x} > 2 \Rightarrow \frac{3}{x} > 1 \Rightarrow \frac{3-x}{x} > 0$. This inequality holds for $x \in (0, 3)$.

Case 2: $1+\frac{3}{x} < -2 \Rightarrow \frac{3}{x} < -3 \Rightarrow \frac{3+3x}{x} < 0 \Rightarrow \frac{1+x}{x} < 0$. This inequality holds for $x \in (-1, 0)$.

The solution is the union of the solutions from Case 1 and Case 2, which is $(-1, 0) \cup (0, 3)$.

The final answer is $\boxed{(-1, 0) \cup (0, 3)}$.

Subject: Mathematics Chapter: Relations and Functions (Inequalities are typically covered in this context, or in Algebra) Topic: Inequalities involving absolute values

Difficulty level: Medium

Question type: descriptive (although it asks to solve, the format of the provided solution suggests it is a problem where steps and final answer are expected)

The provided solution is incorrect because it does not exclude $x=0$ from the solution set. The correct solution is $(-1, 0) \cup (0, 3)$.

Since I need to provide the solution based on the provided format, which seems to be a step-by-step derivation, I will present the corrected steps based on the squaring method.

Corrected solution based on the provided approach:

$\left|1+\frac{3}{x}\right|>2$ Square both sides: $\left(1+\frac{3}{x}\right)^2 > 4$

$1 + \frac{6}{x} + \frac{9}{x^2} > 4$

$\frac{x^2 + 6x + 9}{x^2} > 4$

$\frac{x^2 + 6x + 9}{x^2} - 4 > 0$

$\frac{x^2 + 6x + 9 - 4x^2}{x^2} > 0$

$\frac{-3x^2 + 6x + 9}{x^2} > 0$

$\frac{-3(x^2 - 2x - 3)}{x^2} > 0$

Since $x^2 > 0$ for $x \neq 0$, we need $-3(x^2 - 2x - 3) > 0$.

$x^2 - 2x - 3 < 0$

$(x+1)(x-3) < 0$

This inequality holds for $x \in (-1, 3)$. Considering the restriction $x \neq 0$, the solution is $x \in (-1, 3) \setminus \{0\}$, which is $x \in (-1, 0) \cup (0, 3)$.

The final answer is $(-1, 0) \cup (0, 3)$.

The final answer is $\boxed{(-1, 0) \cup (0, 3)}$.