| Exam Board | OCR MEI |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2021 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas Between Curves |
| Type | Curve-Line-Axis Bounded Region |
| Difficulty | Standard +0.3 This is a straightforward area calculation requiring students to find intersection points, set up integrals for two regions (curve-to-axis and line-to-curve), and evaluate them. While it involves multiple steps and 'show detailed reasoning,' the techniques are standard A-level integration with no conceptual surprises—slightly easier than average due to the routine nature of the setup and calculation. |
| Spec | 1.08e Area between curve and x-axis: using definite integrals1.08f Area between two curves: using integration |
| Answer | Marks | Guidance |
|---|---|---|
| [Graph: correct straight line through \((0, -1)\) and \((1, 4)\)] | B1 (1.1) | Correct line through \((0,-1)\) and \((1,4)\); If x-intercept marked and \((1,4)\) slightly out may award B1 BOD |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(3 - \sqrt{x} = 5x - 1\) or \(y = 5(3-y)^2 - 1\); \(4 - 5x = \sqrt{x} \Rightarrow (4-5x)^2 = x\) | M1 | 3.1a |
| \(25x^2 - 41x + 16 = 0\) or \(5y^2 - 31y + 44 = 0\) | M1 | 1.1 — Getting into suitable form for solution ie \(= 0\). Or as quadratic in \(\sqrt{x}\): \(5x + \sqrt{x} - 4 = 0\) |
| \((x-1)(25x-16) = 0\) or \((5y-11)(y-4) = 0\) | M1 | 1.1 — Attempt to solve quadratic by formula or factorising (oe, via \(\sqrt{x}\) quadratic). \((5\sqrt{x}-4)(\sqrt{x}+1)=0\) |
| \(x = \frac{16}{25} = 0.64\) or \(y = 2.2\) | A1 | 2.2a — Correct root chosen. Or replace M1A1 with SC1 if \(x=0.64\) or \(y=2.2\) seen with no method for solving. \(\sqrt{x} = \frac{4}{5}\) so \(x = 0.64\) |
| \(y = 5 \times 0.64 - 1 = 2.2\) or \(x = (3-2.2)^2 = 0.64\) | M1 | 1.1 — FT their positive root |
| \(\frac{1}{2}(0.64 - 0.2)(2.2) \left[= 0.484 = \frac{121}{250}\right]\) | M1 | 3.1a — Attempt to find area of triangle. \(\frac{1}{2} \times 0.64 \times 2.2\) is M0 |
| Alternative method for area of triangle | ||
| \(\int_{0.2}^{0.64}(5x-1)\,dx\) | M1 M1 | Correct integral; Correct limits (FT their positive root) |
| \(\int_{0.64}^{4}(3 - \sqrt{x})\,dx\) | M1* | 2.1 — Allow any limits \(0 \leq x \leq 4\). Allow if clearly embedded eg \(\int_{0.64}^{4}(5x - 4 + \sqrt{x})\,dx\) |
| \(\left[3x - \frac{2}{3}x^{\frac{3}{2}}\right]_{0.64}^{4}\) | M1** | 1.1 — Integration of M1* integral \(\sqrt{x}\) term correct (ignore limits). Dep on M1* only |
| \(\left(12 - \frac{16}{3}\right) - \left(1.92 - \frac{128}{375}\right) = \frac{636}{125} = 5.088\) | M1 | 1.1 — Evaluation of M1* integral substitution seen. Dep on M1** |
| Total area \(= 0.484 + 5.088 = 5.572 = \frac{1393}{250} = 5\frac{143}{250}\) | A1 | 1.1 — All correct, other partitions possible. Dep on all 9 previous marks |
| [10] |
## Question 9:
**Part (a):**
[Graph: correct straight line through $(0, -1)$ and $(1, 4)$] | B1 (1.1) | Correct line through $(0,-1)$ and $(1,4)$; If x-intercept marked and $(1,4)$ slightly out may award B1 BOD
[1 mark]
## Question 9(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $3 - \sqrt{x} = 5x - 1$ or $y = 5(3-y)^2 - 1$; $4 - 5x = \sqrt{x} \Rightarrow (4-5x)^2 = x$ | M1 | 3.1a |
| $25x^2 - 41x + 16 = 0$ or $5y^2 - 31y + 44 = 0$ | M1 | 1.1 — Getting into suitable form for solution ie $= 0$. Or as quadratic in $\sqrt{x}$: $5x + \sqrt{x} - 4 = 0$ |
| $(x-1)(25x-16) = 0$ or $(5y-11)(y-4) = 0$ | M1 | 1.1 — Attempt to solve quadratic by formula or factorising (oe, via $\sqrt{x}$ quadratic). $(5\sqrt{x}-4)(\sqrt{x}+1)=0$ |
| $x = \frac{16}{25} = 0.64$ or $y = 2.2$ | A1 | 2.2a — Correct root chosen. Or replace M1A1 with SC1 if $x=0.64$ or $y=2.2$ seen with no method for solving. $\sqrt{x} = \frac{4}{5}$ so $x = 0.64$ |
| $y = 5 \times 0.64 - 1 = 2.2$ or $x = (3-2.2)^2 = 0.64$ | M1 | 1.1 — FT their positive root |
| $\frac{1}{2}(0.64 - 0.2)(2.2) \left[= 0.484 = \frac{121}{250}\right]$ | M1 | 3.1a — Attempt to find area of triangle. $\frac{1}{2} \times 0.64 \times 2.2$ is M0 |
| **Alternative method for area of triangle** | | |
| $\int_{0.2}^{0.64}(5x-1)\,dx$ | M1 M1 | Correct integral; Correct limits (FT their positive root) |
| $\int_{0.64}^{4}(3 - \sqrt{x})\,dx$ | M1* | 2.1 — Allow any limits $0 \leq x \leq 4$. Allow if clearly embedded eg $\int_{0.64}^{4}(5x - 4 + \sqrt{x})\,dx$ |
| $\left[3x - \frac{2}{3}x^{\frac{3}{2}}\right]_{0.64}^{4}$ | M1** | 1.1 — Integration of M1* integral $\sqrt{x}$ term correct (ignore limits). Dep on M1* only |
| $\left(12 - \frac{16}{3}\right) - \left(1.92 - \frac{128}{375}\right) = \frac{636}{125} = 5.088$ | M1 | 1.1 — Evaluation of M1* integral substitution seen. Dep on M1** |
| Total area $= 0.484 + 5.088 = 5.572 = \frac{1393}{250} = 5\frac{143}{250}$ | A1 | 1.1 — All correct, other partitions possible. Dep on all 9 previous marks |
| | **[10]** | |
---9 The diagram shows the curve $\mathrm { y } = 3 - \sqrt { \mathrm { x } }$.\\
\includegraphics[max width=\textwidth, alt={}, center]{a0d9573f-8273-4562-a2d3-07f15d9da1af-6_810_1008_1155_283}
\begin{enumerate}[label=(\alph*)]
\item Draw the line $\mathrm { y } = 5 \mathrm { x } - 1$ on the copy of the diagram in the Printed Answer Booklet.
\item In this question you must show detailed reasoning.
Determine the exact area of the region bounded by the curve $y = 3 - \sqrt { x }$, the lines $y = 5 x - 1$ and $x = 4$ and the $x$-axis.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 3 2021 Q9 [11]}}