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. The positive value of k for which $k{e}^x–x=0$ has only one root is

• A. $\frac{1}{e}$
• B. $1$
• C. $e$
• D. ${\log }_e\left(2\right)$

Answer 1

The correct option is A

$\frac{1}{e}$

Explanation for the correct option:

Step 1: First find the derivative of $\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}$:

We have been given that, a function $f\left(x\right)=k{e}^x–x$

We need to find the positive value of k which $k{e}^x–x=0$ has only one root.

Consider, $f\left(x\right)=k{e}^x–x$ then,

$f^{\text{'}}\left(x\right)=k\left(e^x\right)-1$

Step 2: Find the critical point by solving ${\mathit{f}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathbf{0}$:

Put $f^{\text{'}}\left(x\right)=0$

$$\begin{gathered} \Rightarrow k\left(e^x\right)-1=0 \\ \Rightarrow k{e}^x=1 \\ \Rightarrow e^x=\frac{1}{k} \\ \Rightarrow x=\ln \left(\frac{1}{k}\right) \\ \Rightarrow x=\ln \left(1\right)-\ln \left(k\right) \\ \Rightarrow x=0-\ln \left(k\right) \\ \Rightarrow x=-\ln \left(k\right) \end{gathered}$$

Now, find $f\text{'}\text{'}\left(x\right)$

$f\text{'}\text{'}\left(x\right)=k{e}^x$

For $x=-\ln k$ the value of $f\text{'}\text{'}\left(x\right)=k{e}^{-\ln k}$ is

$$\begin{gathered} \Rightarrow f\text{'}\text{'}\left(x\right)=k{e}^{-\ln k} \\ \Rightarrow f\text{'}\text{'}\left(x\right)=k{e}^{\ln \left(\frac{1}{k}\right)}\left(\because e^{lnx}=x\right) \\ \Rightarrow f\text{'}\text{'}\left(x\right)=k\left(\frac{1}{k}\right) \\ \Rightarrow f\text{'}\text{'}\left(x\right)=1>0 \end{gathered}$$

So, $f\left(x\right)$ has a minimum value at $x=-\ln \left(k\right)$

Step 3: Find the value of $\mathit{k}$:

Put $x=-\ln k$ in $k{e}^x–x=0$

$$\begin{gathered} \Rightarrow k\times e^{-\ln \left(k\right)}-(-\ln \left(k\right))=0 \\ \Rightarrow k{e}^{-\ln \left(k\right)}+\ln \left(k\right)=0 \\ \Rightarrow 1+\ln \left(k\right)=0 \\ \Rightarrow \mathit{k}\mathbf{=}\frac{\mathbf{1}}{\mathbf{e}} \end{gathered}$$

Therefore, option (A) is the correct answer.