| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2011 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Find gradient at a point - given gradient condition |
| Difficulty | Standard +0.8 This is a multi-part question involving integration by parts (twice) for part (i), differentiation with product rule for part (ii), and a more challenging algebraic problem in part (iii) requiring setting up and solving a transcendental equation where the tangent passes through the origin. Part (iii) requires insight to set up dy/dx at P equals y/x, then solve the resulting equation, making it above average difficulty overall. |
| Spec | 1.07m Tangents and normals: gradient and equations1.07q Product and quotient rules: differentiation1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Attempt integration by parts and reach \(\pm x^2 e^{-x} \pm \int 2xe^{-x}dx\) | M1* | |
| Obtain \(-x^2 e^{-x} + \int 2xe^{-x}dx\), or equivalent | A1 | |
| Integrate and obtain \(-x^2 e^{-x} - 2xe^{-x} - 2e^{-x}\), or equivalent | A1 | |
| Use limits \(x = 0\) and \(x = 3\), having integrated by parts twice | M1(dep*) | |
| Obtain the given answer correctly | A1 | [5] |
| (ii) Use correct product or quotient rule | M1 | |
| Obtain correct derivative in any form | A1 | |
| Equate derivative to zero and solve for non-zero \(x\) | M1 | |
| Obtain \(x = 2\) with no errors send | A1 | [4] |
| (iii) Carry out a complete method for finding the \(x\)-coordinate of \(P\) | M1 | |
| Obtain answer \(x = 1\) | A1 | [2] |
**(i)** Attempt integration by parts and reach $\pm x^2 e^{-x} \pm \int 2xe^{-x}dx$ | M1* |
Obtain $-x^2 e^{-x} + \int 2xe^{-x}dx$, or equivalent | A1 |
Integrate and obtain $-x^2 e^{-x} - 2xe^{-x} - 2e^{-x}$, or equivalent | A1 |
Use limits $x = 0$ and $x = 3$, having integrated by parts twice | M1(dep*) |
Obtain the given answer correctly | A1 | [5]
**(ii)** Use correct product or quotient rule | M1 |
Obtain correct derivative in any form | A1 |
Equate derivative to zero and solve for non-zero $x$ | M1 |
Obtain $x = 2$ with no errors send | A1 | [4]
**(iii)** Carry out a complete method for finding the $x$-coordinate of $P$ | M1 |
Obtain answer $x = 1$ | A1 | [2]10\\
\includegraphics[max width=\textwidth, alt={}, center]{76371b0f-0145-4cc4-a147-27bcd749816a-3_451_933_1777_605}
The diagram shows the curve $y = x ^ { 2 } \mathrm { e } ^ { - x }$.\\
(i) Show that the area of the shaded region bounded by the curve, the $x$-axis and the line $x = 3$ is equal to $2 - \frac { 17 } { \mathrm { e } ^ { 3 } }$.\\
(ii) Find the $x$-coordinate of the maximum point $M$ on the curve.\\
(iii) Find the $x$-coordinate of the point $P$ at which the tangent to the curve passes through the origin.
\hfill \mbox{\textit{CAIE P3 2011 Q10 [11]}}