| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2011 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Indefinite & Definite Integrals |
| Type | Find curve from gradient |
| Difficulty | Moderate -0.8 This is a straightforward C2 integration question requiring basic power rule application (including fractional powers) and finding a constant using a given point. Both parts are routine textbook exercises with no problem-solving or novel insight required, making it easier than average. |
| Spec | 1.08b Integrate x^n: where n != -1 and sums |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\int (6x^2 - 1) dx = 4x^{\frac{1}{2}} - x + c\) | M1 | Obtain \(kx^{\frac{1}{2}}\). Any \(k\), as long as numerical. Allow both M1 and A1 for equiv eg \(x\sqrt{x}\) |
| Obtain \(4x^{\frac{1}{2}}\) | A1 | Allow for unsimplified coefficient as well (ie \(^6/_{1.5}\)). |
| Obtain \(-x\) (don't penalise lack of \(+c\)) | B1 | 3 |
| (ii) \(y = 4x^{\frac{1}{2}} - x + c\) with \(17 = 32 - 4 + c\) so \(c = -11\), hence \(y = 4x^{\frac{1}{2}} - x - 11\) | M1* | State or imply \(y =\) their integral from (i). Must have come from integration attempt ie increase in power by 1 for at least one term, but allow if -1 disappeared in part (i) at least one of the M1 and the B1 must have been awarded in part (i). Can still get this M1 if no \(+c\). The \(y\) does not have to be explicit – it could be implied by eg \(17 = f(4)\). |
| Answer | Marks | Guidance |
|---|---|---|
| Attempt to find \(c\) using (4, 17) | M1d* | M0 if no \(+c\). M0 if using \(x = 17, y = 4\). |
| Obtain \(y = 4x^{\frac{1}{2}} - x - 11\) | A1 | 3 |
**(i)** $\int (6x^2 - 1) dx = 4x^{\frac{1}{2}} - x + c$ | M1 | Obtain $kx^{\frac{1}{2}}$. Any $k$, as long as numerical. Allow both M1 and A1 for equiv eg $x\sqrt{x}$
**Obtain $4x^{\frac{1}{2}}$** | A1 | Allow for unsimplified coefficient as well (ie $^6/_{1.5}$).
**Obtain $-x$ (don't penalise lack of $+c$)** | B1 | 3 | Allow -1x. Maximum of 2 marks if $\int$ or $dx$ still present in final answer. Maximum of 2 marks if not given as one expression – eg the two terms are integrated separately and never combined.
**(ii)** $y = 4x^{\frac{1}{2}} - x + c$ with $17 = 32 - 4 + c$ so $c = -11$, hence $y = 4x^{\frac{1}{2}} - x - 11$ | M1* | State or imply $y =$ their integral from (i). Must have come from integration attempt ie increase in power by 1 for at least one term, but allow if -1 disappeared in part (i) at least one of the M1 and the B1 must have been awarded in part (i). Can still get this M1 if no $+c$. The $y$ does not have to be explicit – it could be implied by eg $17 = f(4)$.
M0 if they start with $y =$ their integral from (i), but then attempt to use $y - 17 = m(x - 4)$. This is a re-start and gains no credit.
**Attempt to find $c$ using (4, 17)** | M1d* | M0 if no $+c$. M0 if using $x = 17, y = 4$.
**Obtain $y = 4x^{\frac{1}{2}} - x - 11$** | A1 | 3 | Coefficients now need to be simplified, so -1x is A0. Allow A1 for equiv eg $x\sqrt{x}$. Must be an equation ie $y = ...$, so A0 for 'equation = ...' or 'f(x) = ...'.
---2 (i) Find $\int \left( 6 x ^ { \frac { 1 } { 2 } } - 1 \right) \mathrm { d } x$.\\
(ii) Hence find the equation of the curve for which $\frac { \mathrm { d } y } { \mathrm {~d} x } = 6 x ^ { \frac { 1 } { 2 } } - 1$ and which passes through the point $( 4,17 )$.
\hfill \mbox{\textit{OCR C2 2011 Q2 [6]}}