Domain of One Variable Function – A function with sin inside ln – Exercise 2451

Exercise

Determine the domain of the function:

$f(x)=\ln (\sin \frac{\pi}{x})$

Final Answer

Show final answer

Solution

Let’s find the domain of the function:

$f(x)=\ln (\sin \frac{\pi}{x})$

Because there is a denominator, the denominator must be different from zero:

$x\neq 0$

Also, there is a ln function, so we need the expressions inside the ln to be positive:

$\sin \frac{\pi}{x}> 0$

Let’s look at inequality. Sin function is positive in the first half of its cycle. Therefore, in the first cycle we receive:

$0<\frac{\pi}{x}<\pi$

But it has endless cycles. We express this with parameter k (integer). Adding the cycle of sin function in both sides results in

$2\pi k < \frac{\pi}{x}< \pi+2\pi k$

$2\pi k< \frac{\pi}{x}< \pi (1+2k)$

$2 k< \frac{1}{x}< 1+2k$

Now one see that to isolate x one have to break up into cases:

where k = 0 we get

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

From this inequality we get that for x> 0 (positive) we get

$x>0\Longrightarrow 0<1<x\Longrightarrow x>1$

But for x <0 (negative) we get

$x<0\Longrightarrow 0>1>x$

Meaning, there is no solution.

Remember that x is different from zero, otherwise we get a denominator function equal to zero, which is not defined.

Let’s move to find thedomain for k that is different from zero. Recall the inequalities we received:

$2 k< \frac{1}{x}< 1+2k$

Again, to divide by x or k one have to determine whether they are positive or negative. Therefore, when k> 0 (positive) then for x <0 (negative) we get

$\frac{1}{2 k}< x<\frac{1}{ 1+2k}$

That is, there is no solution. But for x> 0 (positive) we get

$\frac{1}{1+2 k}< x<\frac{1}{ 2k}$

And when k <0 then for x> 0 we get

$\frac{1}{2 k}> x>\frac{1}{1+ 2k}$

That is, there is no solution. But for x <0 we get

$\frac{1}{1+2 k}>x>\frac{1}{ 2k}$

And the final answer is the union of the 3 results we got: for k = 0, k> 0 and k <0. Therefore, we get

$\frac{1}{1+2 k}< x<\frac{1}{ 2k}, k>0$

and

$\frac{1}{1+2 k}> x>\frac{1}{ 2k}, k<0$

and

$x>1, k=0$

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