| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2013 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Find gradient at a point - direct evaluation |
| Difficulty | Moderate -0.8 This is a straightforward application of chain rule differentiation and integration by inspection/substitution. Part (i) requires rewriting as a power function and applying standard rules, while part (ii) is a routine reverse chain rule integration. Both parts are mechanical with no problem-solving required, making this easier than average for A-level. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.08b Integrate x^n: where n != -1 and sums1.08d Evaluate definite integrals: between limits |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\frac{dy}{dx} = 2 \times -\frac{1}{2} \times (5x - 6)^{-\frac{3}{2}} \times 5 \to -\frac{5}{(5x-6)^{\frac{3}{2}}}\) | B1 B1, B1 [3] | B1 without '×5'. B1 For '×5'. Use of 'uv' or 'u'v' ok. |
| (ii) integral \(= \frac{2\sqrt{5x - 6}}{\frac{5}{2}} \div 5\) Uses 2 to 3 \(\to\) 2.4 – 1.6 = 0.8 | B1 B1, M1 A1 [4] | B1 without '÷5'. B1 for '÷ 5'. Use of limits in an integral. |
$y = \frac{2}{\sqrt{5x - 6}}$
(i) $\frac{dy}{dx} = 2 \times -\frac{1}{2} \times (5x - 6)^{-\frac{3}{2}} \times 5 \to -\frac{5}{(5x-6)^{\frac{3}{2}}}$ | B1 B1, B1 [3] | B1 without '×5'. B1 For '×5'. Use of 'uv' or 'u'v' ok.
(ii) integral $= \frac{2\sqrt{5x - 6}}{\frac{5}{2}} \div 5$ Uses 2 to 3 $\to$ 2.4 – 1.6 = 0.8 | B1 B1, M1 A1 [4] | B1 without '÷5'. B1 for '÷ 5'. Use of limits in an integral.3 The equation of a curve is $y = \frac { 2 } { \sqrt { } ( 5 x - 6 ) }$.\\
(i) Find the gradient of the curve at the point where $x = 2$.\\
(ii) Find $\int \frac { 2 } { \sqrt { } ( 5 x - 6 ) } \mathrm { d } x$ and hence evaluate $\int _ { 2 } ^ { 3 } \frac { 2 } { \sqrt { } ( 5 x - 6 ) } \mathrm { d } x$.
\hfill \mbox{\textit{CAIE P1 2013 Q3 [7]}}