Question

If a continuous function $f$ defined on the real line $R$, assumes positive and negative value 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 value 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.
The positive value of $k$ for which $k e^{x}-x=0$ has only one root is

• A. $1 / e$
• B. $1$
• C. $e$
• D. $\log _{e} 2$

Answer 1

Answer: (A)

Let $f(x)=k e^{x}-x$
$f^{\prime}(x)=k e^{x}-1$
Substituting $f^{\prime}(x)=0$
$ \Rightarrow x=-$ logk, we get
$f^{\prime \prime}(x)=k e^{x}$
$f^{\prime \prime}(-\log k)=1 > 0$
which implies that $f(x)$ has one minima at point
$x=-\log k$
Since the equation has only one root, we get
$f(-\log k)=0$
$\Rightarrow 1+\log k=0$
$ \Rightarrow k=\frac{1}{e}$