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已知函数

f(x)=\frac{3 a^{x}-1}{a^{x}+1}(a>0

且

a \neq 1)

，若 $\exists x \in(0,3) ，...

Mar 8, 2024

已知函数 f(x)=\frac{3 a^{x}-1}{a^{x}+1}(a>0 且

a \neq 1)

，若

\exists x \in(0,3) ， f\left(x^{2}+3\right)+f(-a x-a)-2 \geq 0

是假命题，则实数

a

的取值范围是

Solution by Steps

step 1

To solve the inequality (3a^x - 1)/(a^x + 1) + (3a^{-x - 1} - 1)/(a^{-x - 1} + 1) - 2 < 0 for

a

, we need to consider the given function

f(x) = \frac{3a^x - 1}{a^x + 1}

where a > 0 and

a \neq 1

step 2

The inequality

f(x^2 + 3) + f(-ax - a) - 2 \geq 0

is false for some

x \in (0, 3)

. We need to find the range of

a

for which this statement is true

step 3

According to the asksia-ll calculation list, the solution for

a

when the inequality is less than 0 are specific values such as

a = -10

with corresponding

x

values. However, these do not provide a range for

a

step 4

Since the asksia-ll calculator results do not directly provide a range for

a

, we must analyze the function and the inequality to determine the range of

a

for which the inequality is false

Answer

The asksia-ll calculator results do not provide a sufficient solution to determine the range of

a

. Further analysis of the function and inequality is required to find the range of

a

Key Concept

Analyzing inequalities involving functions

Explanation

To find the range of

a

for which the inequality

f(x^2 + 3) + f(-ax - a) - 2 \geq 0

is false, we must analyze the behavior of the function

f(x)

and the given inequality over the interval

x \in (0, 3)

. The asksia-ll calculator results provide specific solutions for

a

and

x

, but they do not give a continuous range for

a

. Therefore, a more detailed analysis is required to solve the problem.