| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2019 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Finding stationary points after integration |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question requiring standard techniques: integration by substitution (or recognition of chain rule reverse), finding second derivative, and locating/classifying a stationary point. While it involves multiple steps, each component is routine for A-level—slightly easier than average due to the predictable structure and lack of conceptual challenges. |
| Spec | 1.07d Second derivatives: d^2y/dx^2 notation1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives1.08a Fundamental theorem of calculus: integration as reverse of differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y = [(5x-1)^{1/2} \div \frac{3}{2} \div 5]\ [-2x]\) | B1, B1 | |
| \(3 = \frac{27}{(3/2)\times 5} - 4 + c\) | M1 | Substitute \(x=2,\ y=3\) |
| \(c = 7 - \frac{18}{5} = \frac{17}{5} \rightarrow y = \frac{2(5x-1)^{\frac{3}{2}}}{15} - 2x + \frac{17}{5}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(d^2y/dx^2 = [\frac{1}{2}(5x-1)^{-1/2}]\ [\times 5]\) | B1, B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((5x-1)^{1/2} - 2 = 0 \rightarrow 5x - 1 = 4,\quad x = 1\) | M1A1 | Set \(\frac{dy}{dx} = 0\) and attempt solution (M1) |
| \(y = \frac{16}{25} - 2 + \frac{17}{5} = \frac{37}{15}\) | A1 | Or 2.47 or \(\left(1, \frac{37}{15}\right)\) |
| \(\frac{d^2y}{dx^2} = \frac{5}{2}\times\frac{1}{2} = \frac{5}{4}\ (> 0)\) hence minimum | A1 | OE |
## Question 9(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = [(5x-1)^{1/2} \div \frac{3}{2} \div 5]\ [-2x]$ | B1, B1 | |
| $3 = \frac{27}{(3/2)\times 5} - 4 + c$ | M1 | Substitute $x=2,\ y=3$ |
| $c = 7 - \frac{18}{5} = \frac{17}{5} \rightarrow y = \frac{2(5x-1)^{\frac{3}{2}}}{15} - 2x + \frac{17}{5}$ | A1 | |
## Question 9(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $d^2y/dx^2 = [\frac{1}{2}(5x-1)^{-1/2}]\ [\times 5]$ | B1, B1 | |
## Question 9(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(5x-1)^{1/2} - 2 = 0 \rightarrow 5x - 1 = 4,\quad x = 1$ | M1A1 | Set $\frac{dy}{dx} = 0$ and attempt solution (M1) |
| $y = \frac{16}{25} - 2 + \frac{17}{5} = \frac{37}{15}$ | A1 | Or 2.47 or $\left(1, \frac{37}{15}\right)$ |
| $\frac{d^2y}{dx^2} = \frac{5}{2}\times\frac{1}{2} = \frac{5}{4}\ (> 0)$ hence minimum | A1 | OE |9 A curve for which $\frac { \mathrm { d } y } { \mathrm {~d} x } = ( 5 x - 1 ) ^ { \frac { 1 } { 2 } } - 2$ passes through the point ( 2,3 ).\\
(i) Find the equation of the curve.\\
(ii) Find $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$.\\
(iii) Find the coordinates of the stationary point on the curve and, showing all necessary working, determine the nature of this stationary point.\\
\hfill \mbox{\textit{CAIE P1 2019 Q9 [10]}}