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. $ log_e $ 2

Answer 1

Answer: (A)

Let $ f (x) = ke^x - x $
$f ' (x) = ke^x - 1 = 0 \Rightarrow x = - \ln k$
$f '' (x) = ke^x $
$\therefore [ f '' (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 \bigg( 0, \frac{1}{e}\bigg)$.