June 2025 Paper 1 Q10
10 The graph of \(y = \mathrm{e}^x\) can be transformed to the graph of \(y = \mathrm{e}^{2x-1}\) by a stretch parallel to the \(x\)-axis followed by a translation.
Alternatively the graph of \(y = \mathrm{e}^x\) can be transformed to the graph of \(y = \mathrm{e}^{2x-1}\) by a stretch parallel to the \(x\)-axis and a stretch parallel to the \(y\)-axis.
The point \(P\) lies on the curve \(y = \mathrm{e}^{2x-1}\) and has \(x\)-coordinate of \(\frac{1}{2}\).
| Scheme | Marks | AO |
|---|---|---|
| (i) \(\frac{1}{2}\) oe | B1 | 1.2 |
| [1] | ||
| (ii) parallel to the \(x\)-axis / in the \(x\)-direction | B1 | 2.5 |
| \(\frac{1}{2}\) oe | B1 | 1.2 |
| [2] |
Notes
(a)(i)
B1: State correct scale factor
(a)(ii)
B1: Identify direction – correct language needed. Allow ‘horizontal’
B0 for ‘in’, ‘on’ or ‘along’ the \(x\)-axis, or for right / left
Could be implied by \(\begin{pmatrix} k \\ 0 \end{pmatrix}\)
B1: Correct magnitude of translation. Not dep on previous B1, but must have indicated horizontal translation in some way, including informal language
Do not penalise mentioning the stretch from (a)(i) again in (a)(ii)
\(\begin{pmatrix} \frac{1}{2} \\ 0 \end{pmatrix}\) is B2
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{e}^{-1}\) | B1 | 3.1a |
| [1] |
Notes
B1: State correct scale factor. Any exact equiv
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2\mathrm{e}^{2x-1}\) | M1 | 1.1 |
| at \(x = \frac{1}{2}\), gradient \(= 2\) oe | A1 | 2.1 |
| \(y = \mathrm{e}^0 = 1\) | B1 | 1.1 |
| \(y - 1 = 2\left(x - \frac{1}{2}\right)\) \(y = 2x\) AG | A1 | 2.1 |
| [4] |
Notes
M1: State either correct derivative or \(\frac{1}{2}\mathrm{e}^{2x-1}\)
A1: Obtain correct gradient, following correct derivative. ‘Show that’ so correct derivative must be seen first
B1: State, or use, correct \(y\)-value
A1: Use correct numerical \(m\), \(x\) and \(y\) values to obtain given equation of tangent. Could also use \(y = mx + c\) to obtain \(c = 0\) and hence given equation
| Scheme | Marks | AO |
|---|---|---|
| \(\displaystyle\int \mathrm{e}^{2x-1}\,\mathrm{d}x = \frac{1}{2}\mathrm{e}^{2x-1}\) | B1 | 1.1 |
| \(\left[\frac{1}{2}\mathrm{e}^{2x-1}\right]_0^{\frac{1}{2}} = \frac{1}{2}\left(\mathrm{e}^0 - \mathrm{e}^{-1}\right)\) | M1* | 1.1a |
| \(\frac{1}{2}\left(1 - \mathrm{e}^{-1}\right) - \frac{1}{2} \times 1 \times \frac{1}{2}\) | M1dep* | 3.1a |
| \(\frac{1}{4} - \frac{1}{2\mathrm{e}}\) | A1 | 1.1 |
| [4] |
Notes
B1: State correct integral. Any equiv exact form. Could be in terms of \(u\) if using substitution
M1*: Attempt use of correct limits. Correct order and subtraction, working with exact values. Could be using \(u\) limits (must be \(-1\) and 0)
M1dep*: Attempt area of region ie their definite integral minus attempt at area of triangle. Could be implied by their definite integral minus 0.25 www. Must be using exact values
Area of a triangle must either be \(\frac{1}{2} \times \frac{1}{2} \times\) their \(y\) or from \(\displaystyle\int_0^{0.5} 2x\,\mathrm{d}x\)
A1: Correct area, in any exact form with like terms combined. Must have worked exactly throughout
Alternative method (integrating wrt \(y\))
| Scheme | Marks | AO |
|---|---|---|
| \(\displaystyle\int \left(\frac{1}{2}\ln y + \frac{1}{2}\right)\mathrm{d}y = \frac{1}{2}y\ln y\) | M1 | |
| \(\left[\frac{1}{2}y\ln y\right]_{\mathrm{e}^{-1}}^{1} = \left(\frac{1}{2}\ln 1 - \frac{1}{2}\mathrm{e}^{-1}\ln \mathrm{e}^{-1}\right)\) | M1* | |
| \(\frac{1}{2} \times 1 \times \frac{1}{2} - \frac{1}{2}\mathrm{e}^{-1}\) | M1dep* | |
| \(\frac{1}{4} - \frac{1}{2\mathrm{e}}\) | A1 | |
| [4] |
M1: Change subject and attempt full integration by parts(using correct parts)
M1*: Attempt use of correct limits. Correct order and subtraction, working with exact values. Must use correct \(y\) limits
M1dep*: Attempt area of region ie attempt at area of triangle minus their definite integral. Could be implied by 0.25 minus their definite integral www. Must be using exact values
Area of a triangle must either be \(\frac{1}{2} \times \frac{1}{2} \times\) their \(y\) or from \(\displaystyle\int_0^{1} \frac{1}{2}y\,\mathrm{d}y\)
A1: Correct area, in any exact form with like terms combined. Must have worked exactly throughout