| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2013 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Find stationary points and nature |
| Difficulty | Moderate -0.3 This is a straightforward multi-part calculus question requiring standard techniques: solving f'(x)=0 using a suggested substitution, finding the second derivative to classify stationary points, and integrating to find f(x) using a boundary condition. All steps are routine A-level procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx1.08b Integrate x^n: where n != -1 and sums |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(3u + \frac{3}{u} - 10 = 0\) | B1 | Or \(3x - 10\sqrt{x} + 3 = 0\) Or \((3\sqrt{x}-1)(\sqrt{x}-3)\) or apply formula etc. |
| \(3u^2 - 10u + 3 = 0 \Rightarrow (3u-1)(u-3) = 0\) | M1 | |
| \(\sqrt{x} = \frac{1}{3}\) or \(3\) | A1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| (ii) \(f''(x) = \frac{3}{2}x^{-\frac{1}{2}} - \frac{3}{2}x^{-\frac{3}{2}}\) | B1 | Allow anywhere |
| At \(x = \frac{1}{9}\) \(f''(x) = \frac{3}{2}(3) - \frac{3}{2}(27)(= -36) < 0 \rightarrow\) Max | M1 | Valid method. Allow inac subs, even 3, \(\frac{1}{3}\) |
| At \(x = 9\) \(f''(x) = \frac{3}{2} \cdot \frac{1}{3} - \frac{3}{2} \cdot \frac{1}{27}(= -\frac{4}{9}) > 0 \rightarrow\) Min | A1 | Fully correct. No working, no marks. |
| Answer | Marks | Guidance |
|---|---|---|
| (iii) \(f(x) = 2x^{\frac{3}{2}} + 6x^{\frac{1}{2}} - 10x\) (+ c) | B2 M1 A1 | B1 for 2/3 terms correct. Allow in (i) Sub \((-7, 16 + 12 - 40 + c\) \(c = 5\) |
(i) $3u + \frac{3}{u} - 10 = 0$ | B1 | Or $3x - 10\sqrt{x} + 3 = 0$ Or $(3\sqrt{x}-1)(\sqrt{x}-3)$ or apply formula etc.
$3u^2 - 10u + 3 = 0 \Rightarrow (3u-1)(u-3) = 0$ | M1 |
$\sqrt{x} = \frac{1}{3}$ or $3$ | A1 A1 |
[4]
(ii) $f''(x) = \frac{3}{2}x^{-\frac{1}{2}} - \frac{3}{2}x^{-\frac{3}{2}}$ | B1 | Allow anywhere
At $x = \frac{1}{9}$ $f''(x) = \frac{3}{2}(3) - \frac{3}{2}(27)(= -36) < 0 \rightarrow$ Max | M1 | Valid method. Allow inac subs, even 3, $\frac{1}{3}$
At $x = 9$ $f''(x) = \frac{3}{2} \cdot \frac{1}{3} - \frac{3}{2} \cdot \frac{1}{27}(= -\frac{4}{9}) > 0 \rightarrow$ Min | A1 | Fully correct. No working, no marks.
[3]
(iii) $f(x) = 2x^{\frac{3}{2}} + 6x^{\frac{1}{2}} - 10x$ (+ c) | B2 M1 A1 | B1 for 2/3 terms correct. Allow in (i) Sub $(-7, 16 + 12 - 40 + c$ $c = 5$
[4]9 A curve has equation $y = \mathrm { f } ( x )$ and is such that $\mathrm { f } ^ { \prime } ( x ) = 3 x ^ { \frac { 1 } { 2 } } + 3 x ^ { - \frac { 1 } { 2 } } - 10$.\\
(i) By using the substitution $u = x ^ { \frac { 1 } { 2 } }$, or otherwise, find the values of $x$ for which the curve $y = \mathrm { f } ( x )$ has stationary points.\\
(ii) Find $\mathrm { f } ^ { \prime \prime } ( x )$ and hence, or otherwise, determine the nature of each stationary point.\\
(iii) It is given that the curve $y = \mathrm { f } ( x )$ passes through the point $( 4 , - 7 )$. Find $\mathrm { f } ( x )$.
\hfill \mbox{\textit{CAIE P1 2013 Q9 [11]}}