Question

The number of integral values of x satisfying the inequation $\frac{x}{x+2} \leq \frac{1}{|x|}$ is

Answer 1

3

Answer 2

The inequation to solve is $\frac{x}{x+2} \leq \frac{1}{|x|}$.

1. Domain Restrictions: The denominators cannot be zero. $x+2 \neq 0 \implies x \neq -2$. $|x| \neq 0 \implies x \neq 0$. So, $x \in \mathbb{R} \setminus \{-2, 0\}$.

2. Case Analysis based on $|x|$:

Case 1: $x > 0$ In this case, $|x| = x$. The inequation becomes: $\frac{x}{x+2} \leq \frac{1}{x}$ Since $x > 0$, $x+2$ is also positive. We can multiply both sides by $x(x+2)$ (which is positive) without changing the inequality sign: $x^2 \leq x+2$ $x^2 - x - 2 \leq 0$ Factor the quadratic: $(x-2)(x+1) \leq 0$ The roots are $x=2$ and $x=-1$. Since the parabola $y = x^2-x-2$ opens upwards, it is less than or equal to zero between its roots: $-1 \leq x \leq 2$ Now, combine this with the condition for this case, $x > 0$: $x \in (0, 2]$ The integral values of $x$ in this range are $1, 2$.

Case 2: $x < 0$ In this case, $|x| = -x$. The inequation becomes: $\frac{x}{x+2} \leq \frac{1}{-x}$ $\frac{x}{x+2} \leq -\frac{1}{x}$ Bring all terms to one side to solve the rational inequality: $\frac{x}{x+2} + \frac{1}{x} \leq 0$ Find a common denominator: $\frac{x \cdot x + 1 \cdot (x+2)}{x(x+2)} \leq 0$ $\frac{x^2 + x + 2}{x(x+2)} \leq 0$

Now, analyze the numerator $x^2 + x + 2$. The discriminant is $\Delta = b^2 - 4ac = 1^2 - 4(1)(2) = 1 - 8 = -7$. Since the discriminant is negative ($\Delta < 0$) and the leading coefficient is positive ($1 > 0$), the quadratic expression $x^2 + x + 2$ is always positive for all real values of $x$.

For the fraction $\frac{x^2 + x + 2}{x(x+2)}$ to be less than or equal to zero, since the numerator ($x^2+x+2$) is always positive, the denominator $x(x+2)$ must be negative. So, we need to solve $x(x+2) < 0$. The roots of $x(x+2) = 0$ are $x=0$ and $x=-2$. Since $x(x+2)$ is an upward-opening parabola, it is negative between its roots: $-2 < x < 0$ Now, combine this with the condition for this case, $x < 0$: $x \in (-2, 0)$ The integral values of $x$ in this range are $-1$.

3. Combine Solutions: The integral values from Case 1 are $\{1, 2\}$. The integral values from Case 2 are $\{-1\}$. The set of all integral values satisfying the inequation is $\{-1, 1, 2\}$.

The number of integral values is 3.