Question

$f\left(x\right)=\underset{1}{\overset{x}{\int }}\frac{e^t}{t}dt,$ where $x\in R^{+},$ then the complete set of values of $x$ for which $f\left(x\right)<\ln x$ is

• A. $(0,1)$
• B. $(1,\infty )$
• C. $(0,\infty )$
• D. $x\in \varphi$

Answer 1

The correct option is A $(0,1)$

$f\left(x\right)=\underset{1}{\overset{x}{\int }}\frac{e^t}{t}dt\Rightarrow f\left(1\right)=0$ and $f^{\prime }\left(x\right)=\frac{e^x}{x}$
Let $g\left(x\right)=f\left(x\right)-\ln x,x\in R^{+}$
$\Rightarrow g^{\prime }\left(x\right)=f^{\prime }\left(x\right)-\frac{1}{x}=\frac{e^x-1}{x}>0\forall x\in R^{+}$
$\Rightarrow g\left(x\right)$ is increasing $\forall x\in R^{+},$
$g\left(1\right)=f\left(1\right)-\ln 1=0-0=0$
$\Rightarrow g\left(x\right)>0\forall x>1$ and $g\left(x\right)<0\forall x\in (0,1)$
$\Rightarrow f\left(x\right)<\ln x\forall x\in (0,1)$