| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2021 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Combined region areas |
| Difficulty | Standard +0.8 This question requires finding a normal equation (involving derivative of a function with negative fractional power), then calculating area between a curve and normal line. The derivative of (x-2)^(-1/3) requires chain rule and fractional indices, the normal requires negative reciprocal gradient, and the area calculation involves integrating the difference of two functions with one containing a fractional power. Multiple techniques combined with non-trivial algebra makes this moderately challenging but still within standard A-level scope. |
| Spec | 1.07m Tangents and normals: gradient and equations1.08d Evaluate definite integrals: between limits1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{dy}{dx} = \frac{1}{2} + \frac{1}{3(x-2)^{\frac{4}{3}}}\) | B1 | OE. Allow unsimplified. |
| Attempt at evaluating their \(\frac{dy}{dx}\) at \(x=3\): \(\left[\frac{1}{2} + \frac{1}{3(3-2)^{\frac{4}{3}}} = \frac{5}{6}\right]\) | *M1 | Substituting \(x=3\) into their differentiated expression – defined by one of 3 original terms with correct power of \(x\). |
| Gradient of normal \(= \frac{-1}{\text{their } \frac{dy}{dx}} \left[= -\frac{6}{5}\right]\) | *DM1 | Negative reciprocal of their evaluated \(\frac{dy}{dx}\). |
| Equation of normal: \(y - \frac{6}{5} = (\text{their normal gradient})(x-3)\); \(\left[y = -\frac{6}{5}x + 4.8 \Rightarrow 5y = -6x + 24\right]\) | DM1 | Using their normal gradient and \(A\) in equation of straight line. Dependent on *M1 and *DM1. |
| [When \(y=0\),] \(x=4\) | A1 | or \((4, 0)\) |
| Total: 5 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Area under curve \(= \int\left(\frac{1}{2}x + \frac{7}{10} - \frac{1}{(x-2)^{\frac{1}{3}}}\right)[dx]\) | M1 | For intention to integrate the curve (no need for limits). Condone inclusion of \(\pi\) for this mark. |
| \(\frac{1}{4}x^2 + \frac{7}{10}x - \frac{3(x-2)^{\frac{2}{3}}}{2}\) | A1 | For correct integral. Allow unsimplified. Condone inclusion of \(\pi\) for this mark. |
| \(\left(\frac{9}{4} + 2.1 - \frac{3}{2}\right) - \left(\frac{6.25}{4} + 1.75 - \frac{3 \times 0.5^{\frac{2}{3}}}{2}\right)\) | M1 | Clear substitution of 3 and 2.5 into their integrated expression (with at least one correct term) and subtracting. |
| \(0.48[24]\) | A1 | If M1A1M0 scored then SC B1 can be awarded for correct answer. |
| [Area of triangle \(=\)] \(0.6\) | B1 | OE |
| [Total area \(=\)] \(1.08\) | A1 | Dependent on the first M1 and WWW. |
| Total: 6 marks |
## Question 11(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{dy}{dx} = \frac{1}{2} + \frac{1}{3(x-2)^{\frac{4}{3}}}$ | B1 | OE. Allow unsimplified. |
| Attempt at evaluating their $\frac{dy}{dx}$ at $x=3$: $\left[\frac{1}{2} + \frac{1}{3(3-2)^{\frac{4}{3}}} = \frac{5}{6}\right]$ | *M1 | Substituting $x=3$ into their differentiated expression – defined by one of 3 original terms with correct power of $x$. |
| Gradient of normal $= \frac{-1}{\text{their } \frac{dy}{dx}} \left[= -\frac{6}{5}\right]$ | *DM1 | Negative reciprocal of their evaluated $\frac{dy}{dx}$. |
| Equation of normal: $y - \frac{6}{5} = (\text{their normal gradient})(x-3)$; $\left[y = -\frac{6}{5}x + 4.8 \Rightarrow 5y = -6x + 24\right]$ | DM1 | Using their normal gradient and $A$ in equation of straight line. Dependent on *M1 and *DM1. |
| [When $y=0$,] $x=4$ | A1 | or $(4, 0)$ |
| **Total: 5 marks** | | |
---
## Question 11(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Area under curve $= \int\left(\frac{1}{2}x + \frac{7}{10} - \frac{1}{(x-2)^{\frac{1}{3}}}\right)[dx]$ | M1 | For intention to integrate the curve (no need for limits). Condone inclusion of $\pi$ for this mark. |
| $\frac{1}{4}x^2 + \frac{7}{10}x - \frac{3(x-2)^{\frac{2}{3}}}{2}$ | A1 | For correct integral. Allow unsimplified. Condone inclusion of $\pi$ for this mark. |
| $\left(\frac{9}{4} + 2.1 - \frac{3}{2}\right) - \left(\frac{6.25}{4} + 1.75 - \frac{3 \times 0.5^{\frac{2}{3}}}{2}\right)$ | M1 | Clear substitution of 3 and 2.5 into their integrated expression (with at least one correct term) and subtracting. |
| $0.48[24]$ | A1 | If M1A1M0 scored then SC B1 can be awarded for correct answer. |
| [Area of triangle $=$] $0.6$ | B1 | OE |
| [Total area $=$] $1.08$ | A1 | Dependent on the first M1 and WWW. |
| **Total: 6 marks** | | |
---11\\
\includegraphics[max width=\textwidth, alt={}, center]{10b2ec29-adca-4313-ae24-bab8b2d9f8a4-16_505_1166_258_486}
The diagram shows the line $x = \frac { 5 } { 2 }$, part of the curve $y = \frac { 1 } { 2 } x + \frac { 7 } { 10 } - \frac { 1 } { ( x - 2 ) ^ { \frac { 1 } { 3 } } }$ and the normal to the curve at the point $A \left( 3 , \frac { 6 } { 5 } \right)$.
\begin{enumerate}[label=(\alph*)]
\item Find the $x$-coordinate of the point where the normal to the curve meets the $x$-axis.
\item Find the area of the shaded region, giving your answer correct to 2 decimal places.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2021 Q11 [11]}}