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已知 $f(x)=\left\{\begin{array}{l}|x+2|, x \leq 0 \\ \left|\log _{3} x\right|, x>0...
Mar 8, 2024
已知 f(x)={∣x+2∣,x≤0∣log⁡3x∣,xgt;0f(x)=\left\{\begin{array}{l}|x+2|, x \leq 0 \\ \left|\log _{3} x\right|, x&gt;0\end{array}\right. ，若方程 f(x)−a=0f(x)-a=0 有四个根 x1,x2,x3,x4x_{1}, x_{2}, x_{3}, x_{4} ，且 x_{1}<x_{2}<x_{3}<x_{4} ，则 x3−x1+x4−x2x_{3}-x_{1}+x_{4}-x_{2} 的取值范围为
Solution by Steps
step 1
Solve the equation ∣x+2∣−a=0 |x+2| - a = 0 for x x when x≤0 x \leq 0
step 2
The solutions are x=−2+a x = -2 + a and x=−2−a x = -2 - a , but since x≤0 x \leq 0 , we only consider x=−2−a x = -2 - a with a > 0
step 3
Solve the equation ∣log⁡3(x)∣−a=0 |\log_3(x)| - a = 0 for x x when x > 0
step 4
The solutions are x=3a x = 3^a and x=3−a x = 3^{-a} , and since x > 0 , both solutions are valid for a≥0 a \geq 0
step 5
Combine the solutions from steps 2 and 4 to find the four roots x1,x2,x3,x4 x_1, x_2, x_3, x_4 where x_1 < x_2 < x_3 < x_4
step 6
The roots are x1=−2−a x_1 = -2 - a , x2=3−a x_2 = 3^{-a} , x3=−2+a x_3 = -2 + a , and x4=3a x_4 = 3^a
step 7
Calculate x3−x1+x4−x2 x_3 - x_1 + x_4 - x_2
step 8
Substitute the values of x1,x2,x3,x4 x_1, x_2, x_3, x_4 into the expression to get (−2+a)−(−2−a)+(3a)−(3−a) (-2 + a) - (-2 - a) + (3^a) - (3^{-a})
step 9
Simplify the expression to get 2a+3a−3−a 2a + 3^a - 3^{-a}
step 10
The range of x3−x1+x4−x2 x_3 - x_1 + x_4 - x_2 is the set of all possible values of 2a+3a−3−a 2a + 3^a - 3^{-a} as a a varies over [0,∞) [0, \infty)
Answer
The range of x3−x1+x4−x2 x_3 - x_1 + x_4 - x_2 is [2,∞) [2, \infty) .
Key Concept
Absolute Value Equations and Inequalities
Explanation
The key concept involves solving absolute value equations separately for their positive and negative cases and considering the domain restrictions for each case. The range of the expression is determined by evaluating the simplified expression over the allowed values of a a .
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