| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2023 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Find stationary points and nature |
| Difficulty | Standard +0.3 This is a straightforward multi-part calculus question requiring chain rule differentiation, finding stationary points by setting dy/dx=0, verifying a normal line using perpendicular gradients, and computing a definite integral. All techniques are standard P1 procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{dy}{dx} = \{9\} + \left\{-\dfrac{3}{2}(2x+1)^{1/2} \times 2\right\}\) | B1, B1 | Including '\(+c\)' makes the second term B0 |
| \(9 - 3(2x+1)^{1/2} = 0\) leading to \(2x+1 = 9\) | M1 | Set differential to zero and solve by squaring SOI. Beware \(9^2 - 3^2(2x+1) = 0\) M0A0. \(2x+1 = \sqrt{3}\) or \(2x+1 = \pm 9\) get M0 |
| Max point \(= (4, 9)\) | A1 | WWW \(y=9\) must come from original equation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| When \(x = 1\frac{1}{2}\), shows substitution or \(\dfrac{dy}{dx} = 3\) | M1 | Substituting \(x = 1\frac{1}{2}\) into their \(\dfrac{dy}{dx}\) |
| Gradient of \(AB\) is \(\dfrac{5\frac{1}{2} - 3\frac{1}{2}}{1\frac{1}{2} - 7\frac{1}{2}} = \dfrac{-1}{3}\) | M1 | Substituting into a correct expression for \(m_{AB}\) |
| \(-\dfrac{1}{3} \times 3 = -1\). [Hence \(AB\) is the normal] | A1 | |
| Alternative: When \(x=1\frac{1}{2}\), \(\dfrac{dy}{dx} = 3\), [perpendicular gradient is \(-1/3\)] | M1 | |
| Perpendicular through A has equation \(y = \dfrac{-x}{3} + 6\) which contains B\((7.5, 3.5)\) leading to AB is a normal to the curve at A | M1, A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\left\{\dfrac{9x^2}{2}\right\} + \left\{\dfrac{-(2x+1)^{\frac{5}{2}}}{\frac{5}{2} \times 2}\right\}\) | B1 B1 | Integrating \(y\) with respect to \(x\) |
| \(\left\{\dfrac{9}{2}(7.5)^2 - \dfrac{1}{5}(2\times7.5+1)^{2.5}\right\} - \left\{\dfrac{9}{2}(1.5)^2 - \dfrac{1}{5}(2\times1.5+1)^{2.5}\right\}\) or \(\left(\dfrac{9}{2}\times\dfrac{225}{4} - \dfrac{1024}{5}\right) - \left(\dfrac{81}{8} - \dfrac{32}{5}\right)\) or \(\dfrac{1933}{40} - \dfrac{149}{40}\) or \(48.325 - 3.725\) | M1 | OE. Apply limits \(1\frac{1}{2}\) to \(7\frac{1}{2}\) to an integral. Working must be seen. Expect 44.6 |
| \(\dfrac{1}{2}\left(5\dfrac{1}{2} + 3\dfrac{1}{2}\right) \times 6\) or \(\displaystyle\int_{\frac{3}{2}}^{\frac{15}{2}}\left(\dfrac{-1}{3}x+6\right)dx = \left(\dfrac{-1}{6}\times\left(\dfrac{15}{2}\right)^2 + 6\times\dfrac{15}{2}\right) - \left(\dfrac{-1}{6}\times\left(\dfrac{3}{2}\right)^2 + 6\times\dfrac{3}{2}\right)\) or \(\dfrac{285}{8} - \dfrac{69}{8} = 27\) | B1 | SOI. Area of trapezium. May be seen combined with the area under the curve integral |
| [Shaded area \(= 44.6 - 27 =\)] 17.6 | A1 | SC B1 if no substitution of the limits seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(A = \displaystyle\int_{\frac{3}{2}}^{\frac{15}{2}}\left((9x-(2x+1)^{\frac{3}{2}}) - \left(\dfrac{-1}{3}x+6\right)\right)dx = \int_{\frac{3}{2}}^{\frac{15}{2}}\left(\dfrac{28}{3}x - (2x+1)^{\frac{3}{2}} - 6\right)dx\) | M1 | Finding the equation of AB and subtracting from the equation of the curve |
| \(= \left\{\dfrac{28}{3\times2}x^2 - 6x\right\} + \left\{\dfrac{-(2x+1)^{\frac{5}{2}}}{\frac{5}{2}\times2}\right\}\) | A1 A1 | |
| \(\dfrac{127}{10} - \dfrac{-49}{10}\) | M1 | Apply limits \(1\frac{1}{2}\) to \(7\frac{1}{2}\) to an integral. Working must be seen |
| 17.6 | A1 | SC B1 if no substitution of limits seen |
## Question 10(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{dy}{dx} = \{9\} + \left\{-\dfrac{3}{2}(2x+1)^{1/2} \times 2\right\}$ | B1, B1 | Including '$+c$' makes the second term B0 |
| $9 - 3(2x+1)^{1/2} = 0$ leading to $2x+1 = 9$ | M1 | Set differential to zero and solve by squaring SOI. Beware $9^2 - 3^2(2x+1) = 0$ M0A0. $2x+1 = \sqrt{3}$ or $2x+1 = \pm 9$ get M0 |
| Max point $= (4, 9)$ | A1 | WWW $y=9$ must come from original equation |
## Question 10(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| When $x = 1\frac{1}{2}$, shows substitution or $\dfrac{dy}{dx} = 3$ | M1 | Substituting $x = 1\frac{1}{2}$ into their $\dfrac{dy}{dx}$ |
| Gradient of $AB$ is $\dfrac{5\frac{1}{2} - 3\frac{1}{2}}{1\frac{1}{2} - 7\frac{1}{2}} = \dfrac{-1}{3}$ | M1 | Substituting into a correct expression for $m_{AB}$ |
| $-\dfrac{1}{3} \times 3 = -1$. [Hence $AB$ is the normal] | A1 | |
| **Alternative:** When $x=1\frac{1}{2}$, $\dfrac{dy}{dx} = 3$, [perpendicular gradient is $-1/3$] | M1 | |
| Perpendicular through A has equation $y = \dfrac{-x}{3} + 6$ which contains B$(7.5, 3.5)$ leading to AB is a normal to the curve at A | M1, A1 | |
## Question 10(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left\{\dfrac{9x^2}{2}\right\} + \left\{\dfrac{-(2x+1)^{\frac{5}{2}}}{\frac{5}{2} \times 2}\right\}$ | B1 B1 | Integrating $y$ with respect to $x$ |
| $\left\{\dfrac{9}{2}(7.5)^2 - \dfrac{1}{5}(2\times7.5+1)^{2.5}\right\} - \left\{\dfrac{9}{2}(1.5)^2 - \dfrac{1}{5}(2\times1.5+1)^{2.5}\right\}$ or $\left(\dfrac{9}{2}\times\dfrac{225}{4} - \dfrac{1024}{5}\right) - \left(\dfrac{81}{8} - \dfrac{32}{5}\right)$ or $\dfrac{1933}{40} - \dfrac{149}{40}$ or $48.325 - 3.725$ | M1 | OE. Apply limits $1\frac{1}{2}$ to $7\frac{1}{2}$ to an integral. Working must be seen. Expect 44.6 |
| $\dfrac{1}{2}\left(5\dfrac{1}{2} + 3\dfrac{1}{2}\right) \times 6$ or $\displaystyle\int_{\frac{3}{2}}^{\frac{15}{2}}\left(\dfrac{-1}{3}x+6\right)dx = \left(\dfrac{-1}{6}\times\left(\dfrac{15}{2}\right)^2 + 6\times\dfrac{15}{2}\right) - \left(\dfrac{-1}{6}\times\left(\dfrac{3}{2}\right)^2 + 6\times\dfrac{3}{2}\right)$ or $\dfrac{285}{8} - \dfrac{69}{8} = 27$ | B1 | SOI. Area of trapezium. May be seen combined with the area under the curve integral |
| [Shaded area $= 44.6 - 27 =$] 17.6 | A1 | **SC B1** if no substitution of the limits seen |
**Alternative for 10(c):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $A = \displaystyle\int_{\frac{3}{2}}^{\frac{15}{2}}\left((9x-(2x+1)^{\frac{3}{2}}) - \left(\dfrac{-1}{3}x+6\right)\right)dx = \int_{\frac{3}{2}}^{\frac{15}{2}}\left(\dfrac{28}{3}x - (2x+1)^{\frac{3}{2}} - 6\right)dx$ | M1 | Finding the equation of AB and subtracting from the equation of the curve |
| $= \left\{\dfrac{28}{3\times2}x^2 - 6x\right\} + \left\{\dfrac{-(2x+1)^{\frac{5}{2}}}{\frac{5}{2}\times2}\right\}$ | A1 A1 | |
| $\dfrac{127}{10} - \dfrac{-49}{10}$ | M1 | Apply limits $1\frac{1}{2}$ to $7\frac{1}{2}$ to an integral. Working must be seen |
| 17.6 | A1 | **SC B1** if no substitution of limits seen |10\\
\includegraphics[max width=\textwidth, alt={}, center]{51bd3ba6-e1d1-4c07-81cd-d99dd77f9306-14_832_830_276_653}
The diagram shows the points $A \left( 1 \frac { 1 } { 2 } , 5 \frac { 1 } { 2 } \right)$ and $B \left( 7 \frac { 1 } { 2 } , 3 \frac { 1 } { 2 } \right)$ lying on the curve with equation $y = 9 x - ( 2 x + 1 ) ^ { \frac { 3 } { 2 } }$.
\begin{enumerate}[label=(\alph*)]
\item Find the coordinates of the maximum point of the curve.
\item Verify that the line $A B$ is the normal to the curve at $A$.
\item Find the area of the shaded region.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2023 Q10 [12]}}