2. Determine for which values of the parameter $x$ the expression $\log _{\frac{1}{2}}\left(x^{2}-2 x-1\right)-2$ is defined, and solve the equation $\log _{\frac{1}{2}}\left(x^{2}-2 x-1\right)-2=0$.

17th Mathematics Knowledge Competition for Students of Secondary Technical and Vocational Schools, Selection Competition, March 16, 2017

Problems for 4th Year

Time for solving: 45 minutes.
2. The expression is defined for $x^{2}-2 x-1>0$. Solving the quadratic inequality, we get the solution $D_{f}=(-\infty, 1-\sqrt{2}) \cup(1+\sqrt{2}, \infty)$. The equation $\log _{\frac{1}{2}}\left(x^{2}-2 x-1\right)-2=0$ is transformed into the form $\log _{\frac{1}{2}}\left(x^{2}-2 x-1\right)=2$. We solve the equation using the definition of the logarithm. We get the quadratic equation $\frac{1}{4}=x^{2}-2 x-1$. The solutions to the quadratic equation are $x_{1}=\frac{5}{2}, x_{2}=-\frac{1}{2}$.

![](https://cdn.mathpix.com/cropped/2024_06_07_115a7be78276e1d4adcag-16.jpg?height=74&width=1756&top_left_y=1979&top_left_x=161)

![](https://cdn.mathpix.com/cropped/2024_06_07_115a7be78276e1d4adcag-16.jpg?height=60&width=1739&top_left_y=2032&top_left_x=176)

![](https://cdn.mathpix.com/cropped/2024_06_07_115a7be78276e1d4adcag-16.jpg?height=63&width=1739&top_left_y=2079&top_left_x=176)

Writing the domain $D_{f}=(-\infty, 1-\sqrt{2}) \cup(1+\sqrt{2}, \infty) \ldots \ldots \ldots \ldots \ldots \ldots . . .2$ points

![](https://cdn.mathpix.com/cropped/2024_06_07_115a7be78276e1d4adcag-16.jpg?height=80&width=1745&top_left_y=2179&top_left_x=170)

Solving the equation . ...................................................................................................

![](https://cdn.mathpix.com/cropped/2024_06_07_115a7be78276e1d4adcag-16.jpg?height=71&width=1742&top_left_y=2289&top_left_x=174)

## 17th Mathematics Knowledge Competition for Students of Secondary Technical and Vocational Schools

Preliminary Competition, March 16, 2017

## Solutions and Scoring Guide

(16. MARCH 2017, 00:04)

A competitor who arrives at the solution by any correct method (even if the scoring guide does not provide for it) receives all possible points.

A correct method is considered any procedure that:

- sensibly takes into account the wording of the problem,
- leads to the solution of the problem,
- is mathematically correct and complete.

If an intermediate or final result can be recognized, guessed, read from a diagram, or calculated mentally, the competitor generally receives all the points provided. If, however, the solution is guessed (it is not possible to arrive at it without calculation), and there is no written verification or proof, it is scored with 0 points.

A competitor who has only partially solved the problem, with correct procedures but no visible path to the final solution, cannot receive more than half of the possible points.

The mark ${ }^{\prime * \prime}$ next to the points means that the point or points can be awarded for a correct procedure, even if the calculation is incorrect.

## Fourth Year