$f(x)=\int\limits_1^x \frac{e^t}{t} d t, x \in R^{+}$. Then complete set of values of x for which f(x) ≤ ln x is

(0, 1]

$f(x)=\int\limits_0^x \frac{e^t}{t} d t \Rightarrow f(1)=0$  and  $f'(x)=\frac{e^x}{x}$

Let $g(x)=f(x)-\ln (x) . x \in R^{+}$

$\Rightarrow g'(x)=f'(x)-\frac{1}{x}=\frac{e^x-1}{x}>0 ~\forall~ x \in R^{+}$

⇒ g(x) is increasing for $\forall~ x \in R^{+}$

$g(1)=f(1)-\ln 1=0-0=0$

$\Rightarrow g(x)>0 ~\forall~ x>1$  and  $g(x) \geq 0 ~\forall~ x \in(0,1]$

$\Rightarrow \ln x \geq f(x) \forall x \in(0,1]$