Question

The solution set of the inequality $\frac{|x+2|-|x|}{\sqrt{4-x^3}}\ge 0$ is

• A. [-1, $\sqrt[3]{4}$)
• B. [1, $\sqrt[3]{4}$)
• C. [-1, $\sqrt[3]{2}$)
• D. [0, $\sqrt[3]{4}$)

Answer 1

Answer: (A)

[-1, $\sqrt[3]{4}$)

Answer 2

The inequality is given by $\frac{|x+2|-|x|}{\sqrt{4-x^3}}\ge 0$.

For the expression to be defined, the term under the square root must be non-negative, and the denominator must be non-zero.

So, $4-x^3 > 0$, which implies $x^3 < 4$. Taking the cube root of both sides, we get $x < \sqrt[3]{4}$.

Thus, the domain of the inequality is $x \in (-\infty, \sqrt[3]{4})$.

For the inequality $\frac{|x+2|-|x|}{\sqrt{4-x^3}}\ge 0$ to hold, since the denominator $\sqrt{4-x^3}$ is strictly positive for $x < \sqrt[3]{4}$, the numerator must be non-negative.

So, we need to solve $|x+2|-|x| \ge 0$ under the condition $x < \sqrt[3]{4}$.

The inequality $|x+2| \ge |x|$ can be solved by squaring both sides (since both sides are non-negative):

$(x+2)^2 \ge x^2$

$x^2 + 4x + 4 \ge x^2$

$4x + 4 \ge 0$

$4x \ge -4$

$x \ge -1$.

So, we need to find the values of $x$ that satisfy both $x \ge -1$ and $x < \sqrt[3]{4}$.

The intersection of these two conditions is $[-1, \sqrt[3]{4})$.