| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Verify stationary point location |
| Difficulty | Standard +0.3 This is a structured multi-part question testing standard C3 techniques: product rule differentiation, second derivative test, algebraic substitution in integration, and integration by parts. Each part follows directly from the previous with clear guidance. While it requires multiple techniques, all are routine applications without requiring novel insight or complex problem-solving. |
| Spec | 1.07d Second derivatives: d^2y/dx^2 notation1.07q Product and quotient rules: differentiation1.08h Integration by substitution1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = 2x\ln(1+x)\) | M1 | Product rule; \(d/dx(\ln(1+x)) = 1/(1+x)\) soi |
| \(\frac{dy}{dx} = \frac{2x}{1+x} + 2\ln(1+x)\) | B1, A1 | |
| When \(x=0\), \(dy/dx = 0 + 2\ln 1 = 0 \Rightarrow\) origin is a stationary point | E1 [4] | www (i.e. from correct derivative) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{d^2y}{dx^2} = \frac{(1+x).2 - 2x.1}{(1+x)^2} + \frac{2}{1+x} = \frac{2}{(1+x)^2} + \frac{2}{1+x}\) | M1, A1ft | Quotient or product rule on their \(\frac{2x}{1+x}\); correctly applied to their \(\frac{2x}{1+x}\). o.e. e.g. \(\frac{4+2x}{(1+x)^2}\) cao |
| When \(x=0\), \(d^2y/dx^2 = 2+2 = 4 > 0 \Rightarrow (0,0)\) is a min point | A1, M1, E1 [5] | Substituting \(x=0\) into their \(d^2y/dx^2\); www – dep previous A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Let \(u = 1+x \Rightarrow du = dx\) | ||
| \(\int \frac{x^2}{1+x}\,dx = \int \frac{(u-1)^2}{u}\,du = \int \frac{u^2-2u+1}{u}\,du = \int\left(u - 2 + \frac{1}{u}\right)du\) | M1 | \(\frac{(u-1)^2}{u}\) |
| E1 | www (but condone \(du\) omitted except in final answer) | |
| \(\int_0^1 \frac{x^2}{1+x}\,dx = \int_1^2 \left(u-2+\frac{1}{u}\right)du\) | B1 | Changing limits (or substituting back for \(x\) and using 0 and 1) |
| \(= \left[\frac{1}{2}u^2 - 2u + \ln u\right]_1^2\) | B1 | \(\left[\frac{1}{2}u^2 - 2u + \ln u\right]\) |
| \(= 2 - 4 + \ln 2 - (\frac{1}{2} - 2 + \ln 1) = \ln 2 - \frac{1}{2}\) | M1, A1 [6] | Substituting limits (consistent with \(u\) or \(x\)); cao |
| Answer | Marks | Guidance |
|---|---|---|
| \(A = \int_0^1 2x\ln(1+x)\,dx\) | ||
| Parts: \(u = \ln(1+x)\), \(du/dx = 1/(1+x)\); \(dv/dx = 2x \Rightarrow v = x^2\) | M1 | soi |
| \(= \left[x^2\ln(1+x)\right]_0^1 - \int_0^1 \frac{x^2}{1+x}\,dx\) | A1 | |
| M1 | Substituting their \(\ln 2 - \frac{1}{2}\) for \(\int_0^1 \frac{x^2}{1+x}\,dx\) | |
| \(= \ln 2 - \ln 2 + \frac{1}{2} = \frac{1}{2}\) | A1 [4] | cao |
# Question 3:
## Part (i)
| $y = 2x\ln(1+x)$ | M1 | Product rule; $d/dx(\ln(1+x)) = 1/(1+x)$ soi |
|---|---|---|
| $\frac{dy}{dx} = \frac{2x}{1+x} + 2\ln(1+x)$ | B1, A1 | |
| When $x=0$, $dy/dx = 0 + 2\ln 1 = 0 \Rightarrow$ origin is a stationary point | E1 [4] | www (i.e. from correct derivative) |
## Part (ii)
| $\frac{d^2y}{dx^2} = \frac{(1+x).2 - 2x.1}{(1+x)^2} + \frac{2}{1+x} = \frac{2}{(1+x)^2} + \frac{2}{1+x}$ | M1, A1ft | Quotient or product rule on their $\frac{2x}{1+x}$; correctly applied to their $\frac{2x}{1+x}$. o.e. e.g. $\frac{4+2x}{(1+x)^2}$ cao |
|---|---|---|
| When $x=0$, $d^2y/dx^2 = 2+2 = 4 > 0 \Rightarrow (0,0)$ is a min point | A1, M1, E1 [5] | Substituting $x=0$ into their $d^2y/dx^2$; www – dep previous A1 |
## Part (iii)
| Let $u = 1+x \Rightarrow du = dx$ | | |
|---|---|---|
| $\int \frac{x^2}{1+x}\,dx = \int \frac{(u-1)^2}{u}\,du = \int \frac{u^2-2u+1}{u}\,du = \int\left(u - 2 + \frac{1}{u}\right)du$ | M1 | $\frac{(u-1)^2}{u}$ |
| | E1 | www (but condone $du$ omitted except in final answer) |
| $\int_0^1 \frac{x^2}{1+x}\,dx = \int_1^2 \left(u-2+\frac{1}{u}\right)du$ | B1 | Changing limits (or substituting back for $x$ and using 0 and 1) |
| $= \left[\frac{1}{2}u^2 - 2u + \ln u\right]_1^2$ | B1 | $\left[\frac{1}{2}u^2 - 2u + \ln u\right]$ |
| $= 2 - 4 + \ln 2 - (\frac{1}{2} - 2 + \ln 1) = \ln 2 - \frac{1}{2}$ | M1, A1 [6] | Substituting limits (consistent with $u$ or $x$); cao |
## Part (iv)
| $A = \int_0^1 2x\ln(1+x)\,dx$ | | |
|---|---|---|
| Parts: $u = \ln(1+x)$, $du/dx = 1/(1+x)$; $dv/dx = 2x \Rightarrow v = x^2$ | M1 | soi |
| $= \left[x^2\ln(1+x)\right]_0^1 - \int_0^1 \frac{x^2}{1+x}\,dx$ | A1 | |
| | M1 | Substituting their $\ln 2 - \frac{1}{2}$ for $\int_0^1 \frac{x^2}{1+x}\,dx$ |
| $= \ln 2 - \ln 2 + \frac{1}{2} = \frac{1}{2}$ | A1 [4] | cao |3 A curve is defined by the equation $y = 2 x \ln ( 1 + x )$.\\
(i) Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ and hence verify that the origin is a stationary point of the curve.\\
(ii) Find $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$, and use this to verify that the origin is a minimum point.\\
(iii) Using the substitution $u = 1 + x$, show that $\int \frac { x ^ { 2 } } { 1 + x } \mathrm {~d} x = \int \left( u - 2 + \frac { 1 } { u } \right) \mathrm { d } u$. Hence evaluate $\int _ { 0 } ^ { 1 } \frac { x ^ { 2 } } { 1 + x } \mathrm {~d} x$, giving your answer in an exact form.\\
(iv) Using integration by parts and your answer to part (iii), evaluate $\int _ { 0 } ^ { 1 } 2 x \ln ( 1 + x ) \mathrm { d } x$.
\hfill \mbox{\textit{OCR MEI C3 Q3 [19]}}