Moderate -0.3 This is a straightforward separable variables question requiring standard integration techniques (power rule and logarithm), followed by substituting a boundary condition and solving for q. The algebra is routine and the question follows a predictable template, making it slightly easier than average for C4 level.
7 The gradient of a curve at the point \(( x , y )\), where \(x > - 2\), is given by
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 3 y ^ { 2 } ( x + 2 ) }$$
The points \(( 1,2 )\) and \(( q , 1.5 )\) lie on the curve. Find the value of \(q\), giving your answer correct to 3 significant figures.
Attempt to separate variables in format \(\int py^2\,(dy)=\int\frac{q}{x+2}\,(dx)\)
M1
where constants \(p\) and/or \(q\) may be wrong
Either \(y^3\) & \(\ln(x+2)\) or \(\frac{1}{3}y^3\) & \(\frac{1}{3}\ln(x+2)\)
A1+A1
Accept \(\frac{1}{3}\ln(3x+6)\) for \(\frac{1}{3}\ln(x+2)\) & \(
If indefinite integrals are being used (most likely scenario):
Substitute \(x=1, y=2\) into an eqn containing '+const'
M1
Sub \(y=1.5\) and their value of 'const' & solve for \(x\) or \(q\)
M1
\(x\) or \(q=-1.97\) only
A2
[SC \(x\) or \(q=-1.970\) or \(-1.971\) or \(-1.9705\) or \(-1.9706\)
A1] 7
If definite integrals are used (less likely scenario):
Use \(\int_{1.5}^{2}\ldots dy=\int_{q}^{1}\ldots dx\) where 2 corresponds with 1
M2
& 1.5 corresp with \(q\) (at top/bottom or v.v.)
Then A2 or SC A1 as above
Use \(\int_{1.5}^{2}\ldots dy=\int_{1}^{q}\ldots dx\) where 2 corresponds with \(q\)
M1
& 1.5 corresp with 1 (at top/bottom or v.v.)
Then A1 for 1.97 only
# Question 7:
| Answer/Working | Marks | Guidance |
|---|---|---|
| Attempt to separate variables in format $\int py^2\,(dy)=\int\frac{q}{x+2}\,(dx)$ | M1 | where constants $p$ and/or $q$ may be wrong |
| Either $y^3$ & $\ln(x+2)$ or $\frac{1}{3}y^3$ & $\frac{1}{3}\ln(x+2)$ | A1+A1 | Accept $\frac{1}{3}\ln(3x+6)$ for $\frac{1}{3}\ln(x+2)$ & $|\,|$ for () |
| **If indefinite integrals are being used (most likely scenario):** | | |
| Substitute $x=1, y=2$ into an eqn containing '+const' | M1 | |
| Sub $y=1.5$ and their value of 'const' & solve for $x$ or $q$ | M1 | |
| $x$ or $q=-1.97$ only | A2 | |
| [SC $x$ or $q=-1.970$ or $-1.971$ or $-1.9705$ or $-1.9706$ | A1] **7** | |
| **If definite integrals are used (less likely scenario):** | | |
| Use $\int_{1.5}^{2}\ldots dy=\int_{q}^{1}\ldots dx$ where 2 corresponds with 1 | M2 | & 1.5 corresp with $q$ (at top/bottom or v.v.) |
| Then A2 or SC A1 as above | | |
| Use $\int_{1.5}^{2}\ldots dy=\int_{1}^{q}\ldots dx$ where 2 corresponds with $q$ | M1 | & 1.5 corresp with 1 (at top/bottom or v.v.) |
| Then A1 for 1.97 only | | |
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7 The gradient of a curve at the point $( x , y )$, where $x > - 2$, is given by
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 3 y ^ { 2 } ( x + 2 ) }$$
The points $( 1,2 )$ and $( q , 1.5 )$ lie on the curve. Find the value of $q$, giving your answer correct to 3 significant figures.
\hfill \mbox{\textit{OCR C4 2011 Q7 [7]}}