Question已知函数 $f(x)$ 满足 $f(x+1)=\left\{\begin{array}{ll}a x+a, & x \leq-1 \\ \ln (x+1), & x>-1\end{array}\right.$ ，函数 $g(x)=f(x)-f(-x)$ 有5个零点，求 $a$ 的取值范围。选项为： A. $\left(-\frac{1}{e}, 0\right)$ B. $\left(0, \frac{1}{e}\right)$ C. $\left(-\frac{1}{e}, \frac{1}{e}\right)$ D. $\left(\frac{1}{e},+\infty\right)$

Studdy Solution

STEP 1

Assumptions1. The function $f(x)$ is defined as $f(x+1)=\left\{\begin{array}{ll}a x+a, & x \leq-1 \\ \ln (x+1), & x>-1\end{array}\right.$ . The function $g(x)$ is defined as $g(x)=f(x)-f(-x)$ 3. The function $g(x)$ has5 roots4. We are asked to find the range of the real number $a$

STEP 2

First, we need to find the expressions of $f(x)$ and $f(-x)$ according to the definition of $f(x+1)$ .
For $x \leq -1$ , we have $f(x) = a(x-1) + a = ax$ and $f(-x) = a(-x-1) + a = -ax -2a$ .
For $x > -1$ , we have $f(x) = \ln(x+1)$ and $f(-x) = \ln(-x+1)$ .

STEP 3

Substitute $f(x)$ and $f(-x)$ into $g(x)$ , we get two expressions of $g(x)$ .
For $x \leq -1$ , $g(x) = f(x) - f(-x) = ax - (-ax -2a) =2ax +2a$ .
For $x > -1$ , $g(x) = f(x) - f(-x) = \ln(x+1) - \ln(-x+1)$ .

STEP 4

implify the expression of $g(x)$ for $x > -1$ using the property of logarithm.
$g(x) = \ln\left(\frac{x+1}{-x+1}\right)$

STEP 5

We know that $g(x)$ has5 roots, which means the function $g(x)$ intersects the x-axis at5 points.For $x \leq -1$ , $g(x) =2ax +2a =0$ has one root at $x = -1$ .
For $x > -1$ , $g(x) = \ln\left(\frac{x+1}{-x+1}\right) =0$ has roots when $\frac{x+1}{-x+1} =1$ , which gives us4 roots at $x =0, -\frac{1}{e}, \frac{1}{e},1$ .

STEP 6

To satisfy the condition that $g(x)$ has5 roots, the value of $a$ should make $g(x)$ continuous at $x = -1$ .
So we need to solve the equation $2a(-1) +2a = \ln\left(\frac{-1+1}{-(-1)+1}\right)$ .

STEP 7

olve the equation for $a$ .
$-2a = \ln(0)$

STEP 8

The natural logarithm of0 is undefined, which means there is no real number $a$ that can satisfy the equation.
Therefore, the range of the real number $a$ is an empty set.
The correct answer is none of the options A, B, C, and D.

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Studdy Solution

STEP 1

Assumptions1. The function $f(x)$ is defined as $f(x+1)=\left\{\begin{array}{ll}a x+a, & x \leq-1 \\ \ln (x+1), & x>-1\end{array}\right.$ . The function $g(x)$ is defined as $g(x)=f(x)-f(-x)$ 3. The function $g(x)$ has5 roots4. We are asked to find the range of the real number $a$

STEP 2

First, we need to find the expressions of $f(x)$ and $f(-x)$ according to the definition of $f(x+1)$ .
For $x \leq -1$ , we have $f(x) = a(x-1) + a = ax$ and $f(-x) = a(-x-1) + a = -ax -2a$ .
For $x > -1$ , we have $f(x) = \ln(x+1)$ and $f(-x) = \ln(-x+1)$ .

STEP 3

Substitute $f(x)$ and $f(-x)$ into $g(x)$ , we get two expressions of $g(x)$ .
For $x \leq -1$ , $g(x) = f(x) - f(-x) = ax - (-ax -2a) =2ax +2a$ .
For $x > -1$ , $g(x) = f(x) - f(-x) = \ln(x+1) - \ln(-x+1)$ .

STEP 4

implify the expression of $g(x)$ for $x > -1$ using the property of logarithm.
$g(x) = \ln\left(\frac{x+1}{-x+1}\right)$

STEP 5

STEP 6

To satisfy the condition that $g(x)$ has5 roots, the value of $a$ should make $g(x)$ continuous at $x = -1$ .
So we need to solve the equation $2a(-1) +2a = \ln\left(\frac{-1+1}{-(-1)+1}\right)$ .

STEP 7

olve the equation for $a$ .
$-2a = \ln(0)$

STEP 8

The natural logarithm of0 is undefined, which means there is no real number $a$ that can satisfy the equation.
Therefore, the range of the real number $a$ is an empty set.
The correct answer is none of the options A, B, C, and D.

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Studdy Solution

STEP 1

STEP 2

STEP 3

STEP 4

implify the expression of g(x)g(x)g(x) for x>−1x > -1x>−1 using the property of logarithm.g(x)=ln⁡(x+1−x+1)g(x) = \ln\left(\frac{x+1}{-x+1}\right)g(x)=ln(−x+1x+1​)

STEP 5

STEP 6

To satisfy the condition that g(x)g(x)g(x) has5 roots, the value of aaa should make g(x)g(x)g(x) continuous at x=−1x = -1x=−1.So we need to solve the equation 2a(−1)+2a=ln⁡(−1+1−(−1)+1)2a(-1) +2a = \ln\left(\frac{-1+1}{-(-1)+1}\right)2a(−1)+2a=ln(−(−1)+1−1+1​).

STEP 7

olve the equation for aaa.−2a=ln⁡(0)-2a = \ln(0)−2a=ln(0)

STEP 8

The natural logarithm of0 is undefined, which means there is no real number aaa that can satisfy the equation.Therefore, the range of the real number aaa is an empty set.The correct answer is none of the options A, B, C, and D.

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implify the expression of $g(x)$ for $x > -1$ using the property of logarithm.
$g(x) = \ln\left(\frac{x+1}{-x+1}\right)$

To satisfy the condition that $g(x)$ has5 roots, the value of $a$ should make $g(x)$ continuous at $x = -1$ .
So we need to solve the equation $2a(-1) +2a = \ln\left(\frac{-1+1}{-(-1)+1}\right)$ .

olve the equation for $a$ .
$-2a = \ln(0)$

The natural logarithm of0 is undefined, which means there is no real number $a$ that can satisfy the equation.
Therefore, the range of the real number $a$ is an empty set.
The correct answer is none of the options A, B, C, and D.