$\mathrm{cot}\, x < \sqrt3$

Project ID: 
3000000095
Question: 

Task: Solve the inequality: $$\cot x < \sqrt3\ \mbox{ for }\ x\in\mathbb{R}.$$ Marie solved the task in the following steps:

(1) She identified the points where the cotangent function is undefined: $$x=k\cdot\pi,\ \mbox{ where }\ k\in\mathbb{Z}$$

(2) She set up and solved the equation $\cot x=\sqrt3$: $$x=\frac{\pi}{6}+k\cdot\pi,\ \mbox{ where }\ k\in\mathbb{Z}$$ (3) She stated that the cotangent function is decreasing on every open interval bounded by two consecutive points where it is undefined, i.e., on the intervals: $$\left(0+k\cdot\pi;\pi+k\cdot\pi\right),\ \mbox{ where }\ k\in\mathbb{Z}$$ (4) Marie further claimed that the previous two steps imply that for each $k\in\mathbb{Z}$: $$\cot x<\sqrt3\Leftrightarrow 0+k\cdot\pi< x < \frac{\pi}{6}+k\cdot\pi$$ (5) Finally, she wrote the solution of the inequality obtained in the previous step in the form: $$K=\bigcup_{k\in\mathbb{Z}}\left(0+k\cdot\pi;\frac{\pi}{6}+k\cdot\pi\right)$$ The solution is incorrect. In which step did Marie make a mistake?

Answer 1: 

The mistake is in step (1). Marie incorrectly identified the points where the cotangent function is undefined.

Answer 2: 

The mistake is in step (2). Marie incorrectly solved the equation $\cot x=\sqrt3$.

Answer 3: 

The mistake is in step (3). The cotangent function is decreasing over its domain.

Answer 4: 

The mistake is in step (4). Marie incorrectly applied the property of the cotangent function described in step (3) to solve the inequality $\cot x<\sqrt3$.

Fixed Answer: 
All Fixed
Correct Answer: 
Answer 4
Hint: 

The cotangent function is decreasing on every open interval bounded by two consecutive points where the function is undefined, i.e., on the intervals $$\left(0+k\cdot\pi; \pi+k\cdot\pi\right),\mbox{ where } k\in\mathbb{Z}.$$ Therefore, in each of these intervals (i.e. for each $k\in\mathbb{Z}$), it holds (see the figure): $$\cot x< \sqrt3\Leftrightarrow\frac{\pi}{6}+k\cdot\pi< x< \pi+k\cdot\pi$$

We can write the solution of the inequality $\cot x< \sqrt3$ as: $$K=\bigcup_{k \in \mathbb{Z}}\left(\frac{\pi}{6}+k\cdot\pi;\pi+k\cdot\pi\right)$$