Solution — 2025 Question 5 Exponential & Logarithmic Functions| Graphs| Quadratic Equations & Inequalities

(a)

$\begin{matrix} y & = k ∎ \end{matrix}$

[horizontal asymptote of $C$ ]

(b)

(c)

(i)

$\begin{matrix} k e^{2 k x} - e^{k x} + k & = 0 \\ k u^{2} - u + k & = 0 ∎ \end{matrix}$

[use substitution to rewrite exponential equation into quadratic]

(ii)

For the equation to have two distinct real roots,

$\begin{matrix} Discriminant > 0 \\ b^{2} - 4 a c > 0 \\ {(- 1)}^{2} - 4 k (k) > 0 \\ 1 - 4 k^{2} > 0 \\ 4 k^{2} - 1 < 0 \\ (2 k + 1) (2 k - 1) < 0 \end{matrix}$

[set up discriminant > 0 for two distinct real roots]

$\begin{matrix} - \frac{1}{2} < k < \frac{1}{2} \end{matrix}$

[solution to quadratic inequality in k]

Since $k$ is a positive constant,

$\begin{matrix} 0 < k < \frac{1}{2} ∎ \end{matrix}$

[final answer for k]