June 2025 Paper 3 Q9
9 The function \(\mathrm{f}(x)\) is defined on all real numbers by \(\mathrm{f}(x) = x^3 + \mathrm{e}^{3x}\).
(a) Differentiate \(x^3 + \mathrm{e}^{3x}\). [2]
(b) Write down the coordinates of the point where the curve \(y = \mathrm{f}(x)\) crosses the \(y\)-axis. [1]
(c) Find the gradient of the curve of the inverse function, \(y = \mathrm{f}^{-1}(x)\), at the point where it crosses the \(x\)-axis. [2]
| Scheme | Marks | AO |
|---|---|---|
| \((\mathrm{f}^\prime(x) =)\ \ 3x^2 + 3\mathrm{e}^{3x}\) | B1 B1 | 1.1 1.1 |
| [2] |
Notes
B1: \(3x^2\) or \(k\mathrm{e}^{3x}\) where \(k = 1, 3\) or \(\frac{1}{3}\)
Implied by correct answer
B1: Fully correct derivative
| Scheme | Marks | AO |
|---|---|---|
| \((0, 1)\) | B1 | 1.1 |
| [1] |
Notes
B1: Or \(x = 0\), \(y = 1\).
1 alone is not sufficient.
Condone omission of brackets
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{\frac{\mathrm{d}x}{\mathrm{d}y}}\) so gradient is \(\dfrac{1}{\textit{their}\text{ gradient of } y = \mathrm{f}(x) \text{ (where it crosses } y\text{-axis)}}\) | M1 | 3.1a |
| \(\dfrac{1}{3}\) | A1 | 1.1 |
| [2] |
Notes
M1: Consideration of relationship between gradients of the two curves where they cross the relevant axes. May be shown in a diagram with sketch of any increasing function reflected in \(y = x\)
Implied by correct answer
FT their derivative
Alternative method for M1
| Scheme | Marks |
|---|---|
| \(x = y^3 + \mathrm{e}^{3y}\) \(1 = 3y^2\dfrac{\mathrm{d}y}{\mathrm{d}x} + 3\mathrm{e}^{3y}\dfrac{\mathrm{d}y}{\mathrm{d}x}\) \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{3y^2 + 3\mathrm{e}^{3y}}\) | M1 |
M1: Clear attempt to use implicit differentiation to find \(\frac{\mathrm{d}y}{\mathrm{d}x}\)
Implied by correct answer
A1: nfww