Question

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

• A. $(0,1]$
• B. $[1,\infty )$
• C. $\left(0,\infty \right)$
• D. None of these

Answer 1

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

$f\left(x\right)={\int }_1^x\frac{e^t}{t}dt\Rightarrow f\left(1\right)=0$ and
$f^{\prime }\left(x\right)=\frac{e^x}{x}$

$\mathrm{Let}g\left(x\right)=f\left(x\right)-\ln \left(x\right),x\in R^{+}$
$\Rightarrow g^{\prime }\left(x\right)=f^{\prime }\left(x\right)-\frac{1}{x}=\frac{e^x-1}{x}>0\forall x\in R^{+}$
$\Rightarrow g\left(x\right)$ is increasing for $x\in R^{+}$

$g\left(1\right)=f\left(1\right)-\ln 1=0-0=0$
$\Rightarrow g\left(x\right)>0\forall x>1$ and $g\left(x\right)\le 0\forall x\in (0,1]$

$\Rightarrow \ln x\ge f\left(x\right)\forall x\in (0,1]$