при каких значениях параметра `a` уравнение `(2abs...

سؤال

при каких значениях параметра a уравнение (2absx+2)/(absx-1)=a не имеет корней.

الفكرة

Analyzing the equation for solution existence

We are solving for values of $a$ where the equation has no solutions. By analyzing the function $g(t) = \frac{2(t+1)}{t-1}$ , we determine its range for $t \geq 0, t \neq 1$ . $g(t)$ decreases to values greater than 2 from infinity, meaning values of $a$ greater than 2 result in no solutions.

Analyzing solution intervals for the equation

For $t \in [0,1)$ , the range of $g(t)$ is $(-\infty, -2]$ . As $t$ approaches 1 from the left, the function heads toward negative infinity. Combining this with the range $(2, \infty)$ for $t > 1$ , the solutions exist if $a \in (-\infty, -2] \cup (2, \infty)$ , with no solutions for $a \in (-2, 2]$ .

Determining solution conditions for $a$

If $a < -2$ , there is a solution. At $a = -2$ , $x = 0$ is a solution, so $a = -2$ works. However, for $a$ between -2 and 2 (excluding -2 but including 2), there are no solutions. Using the formula and the range method is more efficient for this analysis.

Thought for 19s

الإجابة

Положим $t=|x|$ , тогда $t\ge 0$ и $t\ne 1$ . Уравнение

\frac{2|x|+2}{|x|-1}=a \quad\Longleftrightarrow\quad \frac{2t+2}{t-1}=a.

Если $a\ne 2$ , решаем по $t$ :

2t+2 = a(t-1)=at-a \;\Longrightarrow\; (2-a)t=-(a+2) \;\Longrightarrow\; t=\frac{a+2}{a-2}.

Нужны условия существования $t$ из области: $t\ge 0$ (и автоматически $t\ne 1$ , т.к. $t=1$ тут невозможен).

\frac{a+2}{a-2}\ge 0 \;\Longleftrightarrow\; a\le -2 \;\text{или}\; a>2.

Отдельно $a=2$ : $\frac{2t+2}{t-1}=2 \Rightarrow 2t+2=2t-2$ , противоречие, решений нет.

Значит, корней нет при

\boxed{\, -2<a\le 2 \, }.