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Problem 109

AMC 10/12, early questions

Algebra Difficulty 3.3 Find the answer

The function $f(x)= \begin{cases} ax^{2}-6x+a^{2}+1 & (x < 1)\\ x^{5-2a} & (x\geqslant 1) \end{cases}$ is a monotonically decreasing function on $\mathbb{R}$ , then the range of the real number $a$ is \_\_\_\_\_\_.

Your answer

Official solution

Since the function $f(x)= \begin{cases} ax^{2}-6x+a^{2}+1 & (x 0 \\ \frac {3}{a}\geqslant 1 \\ 5-2a < 0 \\ 2a^{2}-5\geqslant 1 \end{cases}$ , solving these we get $\frac {5}{2} < a\leqslant 3$ ,

Therefore, the answer is: $\boxed{\left( \frac {5}{2},3\right]}$ .

By utilizing the properties of monotonicity of functions, quadratic functions, and power functions, we find the range of the real number $a$ .

This problem mainly examines the properties of monotonicity of functions, quadratic functions, and power functions, and is considered a basic question.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.