If a continuous $f$ defined on the real line $R$, assumes positive and negative values in $R$ then the equation $f(x)=0$ has a root in $R$. For example, if it is known that a continuous function $f$ on $R$ is positive at some point and its minimum values is negative then the equation $f(x)=0$ has a root in $R$.
Consider $f(x)=k e^x-x$ for all real $x$ where $k$ is a real constant.
$1.$ The line $\mathrm{y}=\mathrm{x}$ meets $\mathrm{y}=k e^{\mathrm{x}}$ for $\mathrm{k} \leq 0$ at
$(A)$ no point $(B)$ one point
$(C)$ two points $(D)$ more than two points
$2.$ The positive value of $\mathrm{k}$ for which $\mathrm{ke}^{\mathrm{x}}-\mathrm{x}=0$ has only one root is
$(A)$ $1 / \mathrm{e}$ $(B)$ $1$ $(C)$ e $(D)$ $\log _e 2$
$3.$ For $k>0$, the set of all values of $k$ for which $k e^x-x=0$ has two distinct roots is
$(A)$ $\left(0, \frac{1}{\mathrm{e}}\right)$ $(B)$ $\left(\frac{1}{\mathrm{e}}, 1\right)$ $(C)$ $\left(\frac{1}{e}, \infty\right)$ $(D)$ $(0,1)$
Give the answer question $1,2$ and $3.$