Question

It is a continuous function $f$ defined on the real line $R$, assume positive and negative values in $R$ then the equation $f\left(x\right)=0$ has root in $R$. For example, if it is known that a continuous function $f$ on $R$ is positive at some point and its minimum value is negative then the equation $f\left(x\right)=0$ has a root in $R$. Consider $f\left(x\right)=k{e}^x–x$ for all real $x$ where $k$ is a real constant. For$k>0$, the set of all values of $k$ for which $k{e}^x-x=0$ has two distinct roots is

• A. $\left(0,\frac{1}{e}\right)$
• B. $\left(\frac{1}{e},1\right)$
• C. $\left(\frac{1}{e},\infty \right)$
• D. $(0,1)$

Answer 1

The correct option is A

$\left(0,\frac{1}{e}\right)$

Explanation for the correct answer:

Finding the values of $k$:

If $k>0$, then for $2$ distinct roots, local minima must be negative

$$\begin{gathered} f(-\ln k)<0 \\ \Rightarrow k{e}^{-\ln k}+\ln k \\ \Rightarrow 1+\ln k<0\mathbf{}\mathbf{[}\mathbf{\because }\mathit{f}\mathbf{(}\mathit{x}\mathbf{)}\mathbf{=}\mathit{k}{\mathit{e}}^{\mathbf{x}}\mathbf{-}\mathit{x}\mathbf{]} \\ \Rightarrow k<\frac{1}{e} \\ \Rightarrow 0<k<\frac{1}{e} \end{gathered}$$

The point is $\left(0,\frac{1}{e}\right)$.

Therefore, option (A) is the correct answer.