The positive value of $k$ for which $k e^{x} - x = 0$ has only one real solution is

\frac{1}{e}

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{log}_{e} 2

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Solution

The correct option is B $\frac{1}{e}$
$k e^{x} - x = 0$ will have only one solution when $g (x) = x$ is tangent to $f (x) = k e^{x}$
let $P (a, b)$ be a point on $f (x)$ and it also satisfy $g (x)$
Therefore, $k e^{a} = a$ ...(1)
slope of $g (x) =$ slope of $f (x)$ at point P
$\Rightarrow 1 = \frac{d y}{d x}$ at point P $= k e^{a}$
From (1), we get $a = 1$
Therefore, $k = 1 / e$
Ans: A

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