Question

$\frac{1-\sqrt{1-4x^2}}{x}<3$

Answer 1

$\left[-\frac{1}{2}, 0\right) \cup \left(0, \frac{1}{2}\right]$

Answer 2

To solve the inequality $\frac{1-\sqrt{1-4x^2}}{x}<3$, we first determine the domain of the expression.

1. Domain of the expression:
   For $\sqrt{1-4x^2}$ to be defined, we must have $1-4x^2 \ge 0$.
   $1 \ge 4x^2 \implies x^2 \le \frac{1}{4} \implies -\frac{1}{2} \le x \le \frac{1}{2}$.
   Also, the denominator $x$ cannot be zero, so $x \ne 0$.
   Combining these, the domain for $x$ is $x \in \left[-\frac{1}{2}, 0\right) \cup \left(0, \frac{1}{2}\right]$.

2. Solve the inequality by considering cases based on the sign of $x$:

   Case 1: $x > 0$
   In this case, $x \in \left(0, \frac{1}{2}\right]$. Multiply both sides of the inequality by $x$ (which is positive), so the inequality sign remains the same:
   $1-\sqrt{1-4x^2} < 3x$
   $1-3x < \sqrt{1-4x^2}$

   We need to consider two sub-cases for $1-3x$:

   • Subcase 1.1: $1-3x < 0$
     This implies $3x > 1$, so $x > \frac{1}{3}$.
     For these values, $x \in \left(\frac{1}{3}, \frac{1}{2}\right]$ (intersecting with the domain $x \in \left(0, \frac{1}{2}\right]$).
     If $1-3x$ is negative, then $1-3x < \sqrt{1-4x^2}$ is always true, because $\sqrt{1-4x^2}$ is always non-negative.
     So, $x \in \left(\frac{1}{3}, \frac{1}{2}\right]$ is part of the solution.

   • Subcase 1.2: $1-3x \ge 0$
     This implies $3x \le 1$, so $x \le \frac{1}{3}$.
     For these values, $x \in \left(0, \frac{1}{3}\right]$ (intersecting with the domain $x \in \left(0, \frac{1}{2}\right]$).
     Since both sides of $1-3x < \sqrt{1-4x^2}$ are non-negative, we can square both sides:
     $(1-3x)^2 < (\sqrt{1-4x^2})^2$
     $1 - 6x + 9x^2 < 1 - 4x^2$
     $13x^2 - 6x < 0$
     $x(13x - 6) < 0$
     This inequality holds when $0 < x < \frac{6}{13}$.
     Intersecting this with $x \in \left(0, \frac{1}{3}\right]$: Since $\frac{1}{3} = \frac{13}{39}$ and $\frac{6}{13} = \frac{18}{39}$, we have $\frac{1}{3} < \frac{6}{13}$.
     So, the intersection is $x \in \left(0, \frac{1}{3}\right]$.

   Combining the solutions from Subcase 1.1 and Subcase 1.2 for $x>0$:
   $\left(\frac{1}{3}, \frac{1}{2}\right] \cup \left(0, \frac{1}{3}\right] = \left(0, \frac{1}{2}\right]$.

   Case 2: $x < 0$
   In this case, $x \in \left[-\frac{1}{2}, 0\right)$. Multiply both sides of the inequality by $x$ (which is negative), so the inequality sign reverses:
   $1-\sqrt{1-4x^2} > 3x$
   $1-3x > \sqrt{1-4x^2}$
   Since $x < 0$, $1-3x$ will be positive (e.g., if $x=-0.1$, $1-3(-0.1) = 1.3$). The right side $\sqrt{1-4x^2}$ is non-negative. Since both sides are non-negative (LHS is positive, RHS is non-negative), we can square both sides:
   $(1-3x)^2 > (\sqrt{1-4x^2})^2$
   $1 - 6x + 9x^2 > 1 - 4x^2$
   $13x^2 - 6x > 0$
   $x(13x - 6) > 0$
   This inequality holds when $x < 0$ or $x > \frac{6}{13}$.
   Intersecting this with the domain for this case, $x \in \left[-\frac{1}{2}, 0\right)$:
   The part $x > \frac{6}{13}$ does not intersect with $x \in \left[-\frac{1}{2}, 0\right)$.
   The part $x < 0$ intersects with $x \in \left[-\frac{1}{2}, 0\right)$ to give $x \in \left[-\frac{1}{2}, 0\right)$.

3. Combine the solutions from Case 1 and Case 2:
   The solution for $x>0$ is $x \in \left(0, \frac{1}{2}\right]$.
   The solution for $x<0$ is $x \in \left[-\frac{1}{2}, 0\right)$.
   The overall solution is the union of these two sets:
   $x \in \left[-\frac{1}{2}, 0\right) \cup \left(0, \frac{1}{2}\right]$.

The solution covers all values within the domain of the expression, except for $x=0$.