# Domain of One Variable Function – A function with square root and ln – Exercise 5736

Exercise

Determine the domain of the function:

$$y=\sqrt{x^2-1}+\ln\sqrt{1-4x^2}$$

There is no real solution

Solution

Let’s find the domain of the function:

$$y=\sqrt{x^2-1}+\ln\sqrt{1-4x^2}$$

Because there are square roots, the expressions inside the roots must be non-negative:

$$x^2-1\geq 0$$

$$1-4x^2\geq 0$$

Also, there is an ln function, so we need the expression inside the ln to be positive:

$$\sqrt{1-4x^2}>0$$

We got 3 inequalities. Let’s solve the first inequality:

$$x^2-1\geq 0$$

It is a square inequality. The roots of the quadratic equation:

$$x^2-1=0$$

are

$$x=\pm 1$$

Because we are looking for the section above the x-axis or on it and the parabola “smiles”, we get that the solution of the inequality is

$$-1\geq x \text{ or } x\geq 1$$

The other two inequalities are equivalent to the inequality:

$$1-4x^2>0$$

Again, it is a square inequality. The roots of the quadratic equation:

$$1-4x^2=0$$

are

$$x=\pm \frac{1}{2}$$

Because we are looking for the section above the x-axis and the parabola “cries”, we get that the solution of the inequality is

$$-\frac{1}{2}

We intersect both results, meaning

$$-1\geq x \text{ or } x\geq 1$$

and

$$-\frac{1}{2}

The intersection gives the empty group (there is no point sustaining the two inequalities).

Therefore, the final answer is that there is no real solution.

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