Question

Given the function $f\left(x\right)=e^x+\ln (x+1)-ax$, where $a\in R$. If there exists two distinct roots $x_1,x_2(x_1<x_2)$ of $f^{\prime }\left(x\right)=0$, then

• A. $f\left(x_2\right)-f\left(x_1\right)<0$
• B. $f\left(x_2\right)-f\left(x_1\right)>0$
• C. $a>2$
• D. $a<2$

Answer 1

Answer: (A), (C)

The correct options are
A $f\left(x_2\right)-f\left(x_1\right)<0$
C $a>2$

$f\left(x\right)=e^x+\ln (x+1)-ax$
Clearly, $x>-1$
$f^{\prime }\left(x\right)=e^x+\frac{1}{x+1}-a$
We can easily conclude that $f(-1^{+})\to -\infty$ and $f^{\prime }(-1^{+})\to +\infty$.
That means that $f^{\prime }$ has positive values for these regions.

$f^{\prime \prime }\left(x\right)=e^x-\frac{1}{(x+1{)}^2}$
Critical point of $f^{\prime }:$ $f^{\prime \prime }\left(x\right)=0\Rightarrow x=0$
$f^{\prime \prime \prime }\left(x\right)=e^x+\frac{2}{(x+1{)}^3}>0$ for all $x>-1$
$\Rightarrow f^{\prime \prime }$ is strictly increasing.
This means $f^{\prime \prime }\left(x\right)=0$ only when $x=0$.
So, $f^{\prime }$ reaches it's minimum at $x=0$.

If $f^{\prime }\left(0\right)>0$, then there are no $x_1,x_2$.
So, we have $f^{\prime }\left(0\right)<0$
$\Rightarrow e^0+\frac{1}{0+1}-a<0$
$\Rightarrow 1+1<a$
$\Rightarrow a>2$

Since $x_1<x_2$ and $\left\{f^{\prime }\right(0)<0\}$ while $f^{\prime }\left(0\right)>x$ for $x\to -1$ and $x\to +\infty$, then $x_1<0<x_2$ but also $f\left(x_1\right)>f\left(0\right)>f\left(x_2\right)$ which means $f\left(x_2\right)-f\left(x_1\right)<0.$
Thus, $2\ln a>2\ln 2>0>f\left(x_2\right)-f\left(x_1\right)$