| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2012 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Determine if function is increasing/decreasing |
| Difficulty | Moderate -0.3 This is a straightforward differentiation and inequality problem requiring standard techniques: differentiate a polynomial, solve a quadratic inequality, find stationary points, and use the second derivative test. While it involves multiple parts and some algebraic manipulation, all steps are routine A-level procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02g Inequalities: linear and quadratic in single variable1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(3x^2 - 4x + 1 (<) 5\) | M1 | Attempt differentiate & put 5 on RHS |
| \((3x+2)(x-2) < 0\) | M1 | Attempt to factorise or solve |
| \(-\frac{2}{3} < x < 2\) or \(\left[-\frac{2}{3}, 2\right]\) or \(\left(-\frac{2}{3}, 2\right)\). Allow \(<\) | A2 [4] | SC Allow A1 for \(-\frac{2}{3}\) and 2 seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(3x^2 - 4x + 1 = 0 \Rightarrow (3x-1)(x-1) = 0\) | M1 | Derivative \(= 0\) & any attempt to solve |
| \(x = \frac{1}{3}\) or \(1\) | A1 | Both |
| \(y = \frac{4}{27}\) or \(0\) | A1 | Both |
| \(f''(x) = 6x - 4 \to f''\!\left(\frac{1}{3}\right) = -2 \; (<0)\); | ||
| \(f''(1) = 2 \; (>0)\) | M1 | Or other valid method |
| max at \(\left(\frac{1}{3}, \frac{4}{27}\right)\); min at \((1, 0)\) cao | A1 [5] | Allow just \(x\) values or just \(y\) values given for identification |
## Question 10:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $3x^2 - 4x + 1 (<) 5$ | M1 | Attempt differentiate & put 5 on RHS |
| $(3x+2)(x-2) < 0$ | M1 | Attempt to factorise or solve |
| $-\frac{2}{3} < x < 2$ or $\left[-\frac{2}{3}, 2\right]$ or $\left(-\frac{2}{3}, 2\right)$. Allow $<$ | A2 [4] | SC Allow A1 for $-\frac{2}{3}$ and 2 seen |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $3x^2 - 4x + 1 = 0 \Rightarrow (3x-1)(x-1) = 0$ | M1 | Derivative $= 0$ & **any** attempt to solve |
| $x = \frac{1}{3}$ or $1$ | A1 | Both |
| $y = \frac{4}{27}$ or $0$ | A1 | Both |
| $f''(x) = 6x - 4 \to f''\!\left(\frac{1}{3}\right) = -2 \; (<0)$; | | |
| $f''(1) = 2 \; (>0)$ | M1 | Or other valid method |
| max at $\left(\frac{1}{3}, \frac{4}{27}\right)$; min at $(1, 0)$ cao | A1 [5] | Allow just $x$ values or just $y$ values given for identification |10 It is given that a curve has equation $y = \mathrm { f } ( x )$, where $\mathrm { f } ( x ) = x ^ { 3 } - 2 x ^ { 2 } + x$.\\
(i) Find the set of values of $x$ for which the gradient of the curve is less than 5 .\\
(ii) Find the values of $\mathrm { f } ( x )$ at the two stationary points on the curve and determine the nature of each stationary point.
\hfill \mbox{\textit{CAIE P1 2012 Q10 [9]}}