Question

If a continuous function $f$ 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 real constant.
The positive value of $k$ for which $k{e}^x-x=0$ has only one root is
(1) $\frac{1}{e}$
(2) $1$
(3) $e$
(4) ${\log }_{e}2$

• A. $\frac{1}{e}$
• B. 1
• C. e
• D. $lo{g}_e$ 2

Answer 1

Answer: (A)

Let $f(x)=k{e}^x-x$
$f^{\prime }(x)=k{e}^x-1=0\Rightarrow x=-\ln k$
$f^{\prime \prime }(x)=k{e}^x$
$\therefore [f^{\prime \prime }(x){]}_x=-\ln k=1>0$
Hence, $f(-\ln k)=1+\ln k$
For one root of given equation
$1+\ln k=0\Rightarrow k<\frac{1}{e}$
Hence, $k\in (0,\frac{1}{e})$.