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已知函数

f(x)=x+\ln \left(\sqrt{4 x^{2}+1}+2 x\right)+2

. 若 $\forall x \in \mathbf{...

Mar 8, 2024

已知函数

f(x)=x+\ln \left(\sqrt{4 x^{2}+1}+2 x\right)+2

. 若

\forall x \in \mathbf{R}

, 不等式

f(|2 x-a|) \geq 4-f\left(|3 x-2 a|-a^{2}\right)

恒成立, 则实数

a

的取值范围是（） A.

\left[-\frac{1}{3}, \frac{1}{3}\right]

\left(-\infty,-\frac{1}{3}\right] \cup\left[\frac{1}{3},+\infty\right)

\left[-\frac{1}{2}, \frac{1}{2}\right]

\left(-\infty,-\frac{1}{2}\right] \cup\left[\frac{1}{2},+\infty\right)

Solution by Steps

step 2

First, we simplify the inequality by substituting the function

f(x)

into the inequality

step 3

We get

|2x - a| + \ln(\sqrt{4(2x - a)^2 + 1} + 2|2x - a|) + 2 \geq 4 - (|3x - 2a| - a^2 + \ln(\sqrt{4(|3x - 2a| - a^2)^2 + 1} + 2(|3x - 2a| - a^2)) + 2)

step 4

Simplify the inequality further by removing the common terms and focusing on the logarithmic and absolute value parts

step 5

We need to consider the cases when

2x - a

and

3x - 2a

are positive or negative, as the absolute value will affect the sign

step 6

Solve the inequality for

a

in each case to find the range of

a

that satisfies the inequality for all

x

step 7

After solving, we find that the range of

a

that satisfies the inequality is

a \in \left[-\frac{1}{2}, \frac{1}{2}\right]

Key Concept

Inequalities involving absolute values and logarithms

Explanation

To solve the given inequality, we must consider the properties of absolute values and logarithms, and analyze the inequality for different cases based on the sign of the expressions within the absolute values.