| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2015 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Determine increasing/decreasing intervals |
| Difficulty | Moderate -0.3 This is a straightforward application of differentiation using the chain rule (or quotient rule) on simple rational functions, followed by sign analysis of the derivative. The stationary point calculation is routine. Slightly easier than average due to the simple algebraic forms and standard techniques, though it requires careful attention to the domain restrictions. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(-(x+1)^{-2} - 2(x+1)^{-3}\) | M1A1 A1 [3] | M1 for recognisable attempt at differentiation. Allow \(\frac{-x^2-4x-3}{(x+1)^4}\) from Q rule. (A2,1,0) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f'(x) < 0\) hence decreasing | B1 [1] | Dep. on *their* (i) \(< 0\) for \(x > 1\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{-1}{(x+1)^2} - \frac{2}{(x+1)^3} = 0\) or \(\frac{-x^2-4x-3}{(x+1)^4} = 0\) | M1* | Set \(\frac{dy}{dx}\) to 0 |
| \(\frac{-(x+1)-2}{(x+1)^3} = 0 \to -x-1-2=0\) or \(-x^2-4x-3=0\) | M1 Dep* | OR mult by \((x+1)^3\) or \((x+1)^5\) (i.e. \(\times\)mult); \(\times\) multn \(\to -(x+1)^3 - 2(x+1)^2 = 0\) |
| \(x = -3\), \(y = -1/4\) | A1A1 [4] | \((-3, -1/4)\) www scores 4/4 |
## Question 8(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $-(x+1)^{-2} - 2(x+1)^{-3}$ | M1A1 A1 [3] | M1 for recognisable attempt at differentiation. Allow $\frac{-x^2-4x-3}{(x+1)^4}$ from Q rule. (A2,1,0) |
## Question 8(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f'(x) < 0$ hence decreasing | B1 [1] | Dep. on *their* (i) $< 0$ for $x > 1$ |
## Question 8(iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{-1}{(x+1)^2} - \frac{2}{(x+1)^3} = 0$ or $\frac{-x^2-4x-3}{(x+1)^4} = 0$ | M1* | Set $\frac{dy}{dx}$ to 0 |
| $\frac{-(x+1)-2}{(x+1)^3} = 0 \to -x-1-2=0$ or $-x^2-4x-3=0$ | M1 Dep* | OR mult by $(x+1)^3$ or $(x+1)^5$ (i.e. $\times$mult); $\times$ multn $\to -(x+1)^3 - 2(x+1)^2 = 0$ |
| $x = -3$, $y = -1/4$ | A1A1 [4] | $(-3, -1/4)$ www scores 4/4 |
---8 The function f is defined by $\mathrm { f } ( x ) = \frac { 1 } { x + 1 } + \frac { 1 } { ( x + 1 ) ^ { 2 } }$ for $x > - 1$.\\
(i) Find $\mathrm { f } ^ { \prime } ( x )$.\\
(ii) State, with a reason, whether f is an increasing function, a decreasing function or neither.
The function g is defined by $\mathrm { g } ( x ) = \frac { 1 } { x + 1 } + \frac { 1 } { ( x + 1 ) ^ { 2 } }$ for $x < - 1$.\\
(iii) Find the coordinates of the stationary point on the curve $y = \mathrm { g } ( x )$.
\hfill \mbox{\textit{CAIE P1 2015 Q8 [8]}}