Theory of Equations 5 Question 15
15. 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) $\log _e 2$
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Solution:
- Let $f(x)=k e^{x}-x$
$$ \begin{aligned} & & f^{\prime}(x) & =k e^{x}-1=0 \\ \Rightarrow \quad & & x & =-\ln k \end{aligned} $$
$$ \begin{aligned} f^{\prime \prime}(x) & =k e^{x} \\ \therefore \quad\left[f^{\prime \prime}(x)\right] _{x=-\ln k} & =1>0 \end{aligned} $$
Hence, $\quad f(-\ln k)=1+\ln k$
For one root of given equation
$1+\ln k=0$
$\Rightarrow \quad k=\frac{1}{e}$
