June 2022 Paper 1 Q11
11 The gradient function of a curve is given by \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{3x^2\ln x}{\mathrm{e}^{3y}}\).
The curve passes through the point \((\mathrm{e}, 1)\).
| Scheme | Marks | AO |
|---|---|---|
| \(\displaystyle\int \mathrm{e}^{3y}\,\mathrm{d}y = \int 3x^2\ln x\,\mathrm{d}x\) | M1 | 3.1a |
| \(\displaystyle\int \mathrm{e}^{3y}\,\mathrm{d}y = \tfrac{1}{3}\mathrm{e}^{3y}\) | B1 | 1.1 |
| \(\displaystyle\int 3x^2\ln x\,\mathrm{d}x = x^3\ln x - \int x^2\,\mathrm{d}x\) | M1 | 3.1a |
| \(= x^3\ln x - \tfrac{1}{3}x^3 + c\) | A1 | 1.1 |
| \(\tfrac{1}{3}\mathrm{e}^3 = \mathrm{e}^3\ln\mathrm{e} - \tfrac{1}{3}\mathrm{e}^3 + c\) so \(c = -\tfrac{1}{3}\mathrm{e}^3\) | M1 | 1.1a |
| \(\tfrac{1}{3}\mathrm{e}^{3y} = x^3\ln x - \tfrac{1}{3}x^3 - \tfrac{1}{3}\mathrm{e}^3\) \(\mathrm{e}^{3y} = 3x^3\ln x - x^3 - \mathrm{e}^3\) | A1 | 1.1 |
| [6] |
Notes
M1: Separate variables and attempt integration of at least one side
Allow \(k\mathrm{e}^{3y}\), with \(k \ne 1\), as ‘attempt’ at integration of LHS
‘Attempt’ at RHS may not be use of integration by parts
Allow BOD on missing integral sign / missing \(\mathrm{d}y\) / missing \(\mathrm{d}x\) as long as intention clear
B1: Correct LHS
B0 if still part of an expression also involving \(x\)
M1: Attempt integration by parts on RHS – must have correct parts
As far as attempt at \(x^3\ln x - \displaystyle\int x^2\,\mathrm{d}x\), possibly with \(\displaystyle\int \frac{1}{x}x^3\,\mathrm{d}x\) not yet simplified
A1: Correct RHS (condone no \(+ c\))
Condone no modulus sign on \(\ln x\)
M1: Attempt use of \((\mathrm{e}, 1)\) to find \(c\)
Used in an equation involving \(x\), \(y\) and \(c\), following some integration attempt of both sides
As far as finding \(c\), either exact or as a decimal
M1 can be implied by sight of \(-\frac{1}{3}\mathrm{e}^3\) or \(-6.695\ldots\) following a correct equation
A1: Obtain correct equation, in required form
Any equivalent form on the RHS, but must be \(\mathrm{e}^{3y} = \ldots\)
A0 if decimal approximation for \(\mathrm{e}^3\)
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{e}^{3y} = 3\mathrm{e}^6\ln\mathrm{e}^2 - \mathrm{e}^6 - \mathrm{e}^3\) \(= 6\mathrm{e}^6 - \mathrm{e}^6 - \mathrm{e}^3\) \(= 5\mathrm{e}^6 - \mathrm{e}^3\) | M1* | 2.1 |
| \(3y = \ln(\mathrm{e}^3(5\mathrm{e}^3 - 1))\) \(= 3 + \ln(5\mathrm{e}^3 - 1)\) | M1dep* | 2.1 |
| \(y = 1 + \tfrac{1}{3}\ln(5\mathrm{e}^3 - 1)\) | A1 | 2.1 |
| [3] |
Notes
M1*: Substitute \(x = \mathrm{e}^2\), into their integral involving \(\ln x\), and attempt to simplify
\(\ln x\) may be \(\ln x^p\) if any coefficient has been taken into the ln term
As far as correctly simplifying the ln term to remove ln
Must be working exactly, so M0 if decimals seen before ln dealt with
M1dep*: Introduce logs correctly, and attempt to rearrange to given form
Their equation must have two terms, or possibly more, with the terms having a common factor of \(\mathrm{e}^k\)
Attempt must go as far as splitting into the sum of two terms, with \(\ln\mathrm{e}^k\) simplified to \(k\)
A1: Obtain \(y = 1 + \frac{1}{3}\ln(5\mathrm{e}^3 - 1)\)
No need to state \(a\), \(b\) and \(c\) explicitly