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已知函数 $f(x)=\left\{\begin{array}{l}x^{2}-4 x+2, x
Mar 8, 2024
已知函数 f(x)={x2−4x+2,xlt;a2x,x≥af(x)=\left\{\begin{array}{l}x^{2}-4 x+2, x&lt;a \\ 2^{x}, x \geq a\end{array}\right. ，其中 0<a<4. 若方程 f(f(x))=2f(f(x))=2 有且只有一个解，则 aa 的取值范围是
Solution by Steps
step 1
Consider the piecewise function f(x)f(x) and the condition f(f(x))=2f(f(x)) = 2
step 2
For x < a, f(x)=x2−4x+2f(x) = x^2 - 4x + 2. For x≥ax \geq a, f(x)=2xf(x) = 2^x
step 3
To find the value of aa such that f(f(x))=2f(f(x)) = 2 has exactly one solution, we need to analyze the behavior of f(x)f(x) and ensure it intersects the line y=2y = 2 at only one point
step 4
For x < a, set f(x)=2f(x) = 2 to get x2−4x+2=2x^2 - 4x + 2 = 2. Simplify to find the potential intersection points for x < a
step 5
Solving x2−4x=0x^2 - 4x = 0 gives us x(x−4)=0x(x - 4) = 0, so x=0x = 0 or x=4x = 4. Since 0 < a < 4, we must have a > 0 to avoid multiple solutions
step 6
For x≥ax \geq a, set f(x)=2f(x) = 2 to get 2x=22^x = 2. This implies x=1x = 1
step 7
Since x=1x = 1 is the only solution for x≥ax \geq a, we must ensure that a≤1a \leq 1 to avoid additional solutions for x < a
step 8
Combining the conditions from steps 5 and 7, we find that the range for aa is 0 < a \leq 1
Answer
0 < a \leq 1
Key Concept
Piecewise Function Intersection with a Constant
Explanation
To ensure the piecewise function f(x)f(x) intersects the line y=2y = 2 at only one point, we analyze the behavior of f(x)f(x) for x < a and x≥ax \geq a and find the range of aa that satisfies the condition.
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