| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2017 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard Integrals and Reverse Chain Rule |
| Type | Find stationary points from derivative |
| Difficulty | Moderate -0.8 This is a straightforward multi-part question requiring basic integration to find a curve equation (using a point to find the constant), completing the square (a routine algebraic manipulation), and solving a simple quadratic inequality. All techniques are standard P1 procedures with no problem-solving insight required, making it easier than average. |
| Spec | 1.02e Complete the square: quadratic polynomials and turning points1.02g Inequalities: linear and quadratic in single variable1.08a Fundamental theorem of calculus: integration as reverse of differentiation1.08b Integrate x^n: where n != -1 and sums |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y = 7x - \frac{x^3}{3} - \frac{6x^2}{2}\ (+c)\) | B1 | CAO |
| Uses \((3, -10) \rightarrow c = 5\) | M1 A1 | Uses the given point to find \(c\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(7 - x^2 - 6x = 16 - (x+3)^2\) | B1 B1 | B1 \(a = 16\), B1 \(b = 3\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(16 - (x+3)^2 > 0 \rightarrow (x+3)^2 < 16\), and solve | M1 | or factors \((x+7)(x-1)\) |
| End-points \(x = 1\) or \(-7\) | A1 | |
| \(\rightarrow -7 < x < 1\) | A1 | needs \(<\), not \(\leqslant\). (SR \(x < 1\) only, or \(x > -7\) only B1 i.e. 1/3) |
## Question 7:
**Part 7(i):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = 7x - \frac{x^3}{3} - \frac{6x^2}{2}\ (+c)$ | B1 | CAO |
| Uses $(3, -10) \rightarrow c = 5$ | M1 A1 | Uses the given point to find $c$ |
**Part 7(ii):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $7 - x^2 - 6x = 16 - (x+3)^2$ | B1 B1 | B1 $a = 16$, B1 $b = 3$ |
**Part 7(iii):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $16 - (x+3)^2 > 0 \rightarrow (x+3)^2 < 16$, and solve | M1 | or factors $(x+7)(x-1)$ |
| End-points $x = 1$ or $-7$ | A1 | |
| $\rightarrow -7 < x < 1$ | A1 | needs $<$, not $\leqslant$. (SR $x < 1$ only, or $x > -7$ only B1 i.e. 1/3) |
---7 A curve for which $\frac { \mathrm { d } y } { \mathrm {~d} x } = 7 - x ^ { 2 } - 6 x$ passes through the point $( 3 , - 10 )$.\\
(i) Find the equation of the curve.\\
(ii) Express $7 - x ^ { 2 } - 6 x$ in the form $a - ( x + b ) ^ { 2 }$, where $a$ and $b$ are constants.\\
(iii) Find the set of values of $x$ for which the gradient of the curve is positive.\\
\hfill \mbox{\textit{CAIE P1 2017 Q7 [8]}}