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已知函数 $f(x)=\left\{\begin{array}{l}x^{2}+4 a, x<0 \\ \log _{a}(x+1)+1,(a>0, a \ne...

Mar 8, 2024

已知函数

f(x)=\left\{\begin{array}{l}x^{2}+4 a, x&lt;0 \\ \log _{a}(x+1)+1,(a&gt;0, a \neq 1), x \geq 0\end{array}\right.

在

\mathbf{R}

上单调递减, 且关于

x

的方程

|f(x)|=2-x

佮好有两个不相等的实数解, 则

a

的取值范围是 ( ) A.

\left[\frac{1}{4}, \frac{1}{2}\right] \cup\left\{\frac{9}{16}\right\}

\left(\frac{1}{4}, \frac{1}{2}\right)

\left(\frac{1}{4}, \frac{1}{2}\right]

\left[\frac{1}{4}, \frac{1}{2}\right) \cup\left\{\frac{9}{16}\right\}

Solution by Steps

step 2

For x < 0,

f(x) = x^2 + 4a

is decreasing if the derivative

f'(x) = 2x

is negative, which is true for x < 0

step 3

For

x \geq 0

f(x) = \log_a(x+1) + 1

is decreasing if the derivative

f'(x) = \frac{1}{(x+1)\ln(a)}

is negative. Since x+1 > 0, we need \ln(a) < 0, which means 0 < a < 1

step 4

The function must be continuous at

x=0

, so we set

0^2 + 4a = \log_a(0+1) + 1

. This simplifies to

4a = 1

, so

a = \frac{1}{4}

step 5

To find the range of

a

for which

|f(x)| = 2 - x

has exactly two distinct real solutions, we consider the two cases for

f(x)

separately

step 6

For x < 0, we solve

|x^2 + 4a| = 2 - x

. This leads to two cases:

x^2 + 4a = 2 - x

and

x^2 + 4a = x - 2

step 7

For

x \geq 0

, we solve

|\log_a(x+1) + 1| = 2 - x

. This also leads to two cases:

\log_a(x+1) + 1 = 2 - x

and

\log_a(x+1) + 1 = x - 2

step 8

We must analyze the solutions to these equations and determine the values of

a

that yield exactly two distinct real solutions for

x

step 9

After solving the equations from steps 6 and 7, we find that the range of

a

that satisfies the conditions is

\left(\frac{1}{4}, \frac{1}{2}\right]

Key Concept

Monotonicity and Continuity

Explanation

To ensure a piecewise function is monotonically decreasing, each piece must be decreasing, and the function must be continuous at the points where the pieces meet.

Key Concept

Solution of Absolute Value Equations

Explanation

When solving equations involving absolute values, we must consider both the positive and negative scenarios of the expression within the absolute value.