When has weighted sum of two specific unimodal functions two peaks? Consider the functions
$f(x) = x e^{-x}$ and $g(x) = xe^{-kx}$ for $x \geq 0$, where $e$ is the Euler constant and $k > 1$. It is easy to check that $f(x)$ is unimodal and has its global maximum at $x_f^* = 1$, while $g(x)$ is also unimodal and has its global maximum at $x_g^* = \dfrac{1}{k} \in (0,1)$. Notice further that both functions are strictly concave up to $\tilde{x}_f = 2$ (for $f(x)$) and $\tilde{x}_g = \dfrac{2}{k}$ (for $g(x)$), after which they are strictly convex.
What I am interested in is the following: Take the weighted sum of the functions
$$h_{\lambda}(x) := (1-\lambda)f(x) + \lambda g(x),$$
where $0 < \lambda < 1$. For which values of $k$ can we guarantee that there exists a range of weights $\lambda$ such that $h_{\lambda}(x)$ has two peaks?
What we can easily deduce is that for every $\lambda$, $h_\lambda(x)$ is strictly increasing on $[0, 1/k]$ and strictly decreasing on $[1, \infty)$. Moreover, due to the concavity property of the individual functions, $h_{\lambda}(x)$ is clearly strictly concave on $[0, 2/k]$. Taken together, this implies that for $k \leq 2$, $h_{\lambda}(x)$ is unimodal and has a single peak somewhere in $(1/k, 1)$.
What about the case where $k > 2$? Using Mathematica, it appears that two peaks can only emerge for $k \gtrapprox 6$. But can we somehow show this analytically? Or at least prove that for $k$ sufficiently large, the function has two peaks when $\lambda$ is set appropriately? Many thanks in advance!

$\def\e{\mathrm{e}}$Note that between any two adjacent local maximum points, there exists exactly one local minimum point, and $f(x) = (1 - λ)x\e^{-x} + λx\e^{-kx}$ is increasing on $\left[ 0, \dfrac{1}{k} \right]$ and decreasing on $[1, +\infty]$, thus$$
f \text{ has two peaks} \Longleftrightarrow f' \text{ has three zeros}.
$$
Also, the zeros of $f'$ lie on $\left( \dfrac{1}{k}, 1 \right)$, and $f'(x) = (1 - λ)(1 - x) \e^{-x} + λ(1 - kx) \e^{-kx}$, thus\begin{align*}
\mathrel{\phantom{\Longleftrightarrow}}{} f' \text{ has three zeros} &\Longleftrightarrow \frac{kx - 1}{1 - x} \e^{-(k - 1)x} = \frac{1 - λ}{λ} \text{ has three roots}\\
&\Longleftrightarrow \ln(kx - 1) - \ln(1 - x) - (k - 1)x = \ln\frac{1 - λ}{λ} \text{ has three roots}.
\end{align*}
Define $g(x) = \ln(kx - 1) - \ln(1 - x) - (k - 1)x$. Because the range of $\dfrac{1 - λ}{λ}$ is $(0, +\infty)$ given that $λ \in (0, 1)$, then the range of $\ln\dfrac{1 - λ}{λ}$ is $(-\infty, +\infty)$. Therefore,\begin{align*}
&\mathrel{\phantom{\Longleftrightarrow}}{} \exists λ \in (0, 1): g(x) = \ln\frac{1 - λ}{λ} \text{ has three roots}\\
&\Longleftrightarrow \exists c \in \mathbb{R}: g(x) = c \text{ has three roots}\\
&\Longleftrightarrow g' \text{ has two zeros}.
\end{align*}
Now, because\begin{align*}
g'(x) &= \frac{k}{kx - 1} + \frac{1}{1 - x} - (k - 1)\\
&= (k - 1) \frac{kx^2 - (k + 1)x + 2}{(kx - 1)(1 - x)}, \quad x \in \left( \frac{1}{k}, 1\right)
\end{align*}
thus\begin{align*}
&\mathrel{\phantom{\Longleftrightarrow}}{} g' \text{ has two zeros} \Longleftrightarrow kx^2 - (k + 1)x + 2 = 0 \text{ has two roots on } \left( \frac{1}{k}, 1 \right).
\end{align*}
Define $h(x) = kx^2 - (k + 1)x + 2$. Note that the symmetry axis of the graph of $y = h(x)$ is $x = \dfrac{k + 1}{2k}$ and $\dfrac{1}{k} < \dfrac{k + 1}{2k} < 1$, also $h\left( \dfrac{1}{k} \right) = h(1) = 1 > 0$, thus\begin{align*}
h(x) = 0 \text{ has two roots on } \left( \frac{1}{k}, 1 \right) &\Longleftrightarrow Δ = (k + 1)^2 - 8k > 0\\
&\stackrel{k > 1}{\Longleftrightarrow} k > 3 + 2\sqrt{2} ≈ 5.828427.
\end{align*}
