The function f is defined by f(x) = \frac{e^{3x}}{4x^2 + k} where k is a positive constant - Edexcel - A-Level Maths Pure - Question 14 - 2022 - Paper 2

For the curve y = f(x) to have at least one stationary point, the derivative must equal zero at some point:

$12x^2 - 8x + 3k = 0$

This is a quadratic equation in x. A quadratic equation has at least one solution if the discriminant is non-negative:

$b^2 - 4ac \geq 0$

In this case, ( a = 12 ), ( b = -8 ), and ( c = 3k ):

$(-8)^2 - 4(12)(3k) \geq 0 \implies 64 - 144k \geq 0 \implies k \leq \frac{64}{144} = \frac{4}{9}$

Since k is a positive constant, the possible values of k must also satisfy:

$0 < k \leq \frac{4}{9}.$