| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2005 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Finding maximum/minimum on curve |
| Difficulty | Standard +0.3 This is a straightforward multi-part question requiring standard techniques: differentiation using quotient rule to find the maximum (part i), integration by substitution with u = x² + 1 (part ii), and solving a logarithmic equation numerically (part iii). All steps are routine A-level procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Use quotient or product rule | M1 | |
| Obtain derivative in any correct form | A1 | |
| Equate derivative to zero and solve for \(x\) or \(x^2\) | M1 | |
| Obtain \(x = 1\) correctly | A1 | 4 |
| [Differentiating \((x^2 + 1)y = x\) using the product rule can also earn the first M1A1.] | ||
| [SR: if the quotient rule is misused, with a 'reversed' numerator or \(\nabla\) instead of \(\nabla^2\) in the denominator, award M0A0 but allow the following M1A1.] | ||
| (ii) Obtain indefinite integral of the form \(k \ln(x^2 + 1)\), where \(k = \frac{1}{2}, 1\) or \(2\) | M1* | |
| Use limits \(x = 0\) and \(x = p\) correctly, or equivalent | M1(dep*) | |
| Obtain answer \(\frac{1}{2}\ln(p^2 + 1)\) | A1 | 3 |
| [Also accept \(-\ln \cos \theta\) or \(\ln \cos \theta\), where \(x = \tan \theta\), for the first M1*.] | ||
| (iii) Equate to 1 and convert equation to the form \(p^2 + 1 = \exp(1/k)\) | M1 | |
| Obtain answer \(p = 2.53\) | A1 | 2 |
**(i)** Use quotient or product rule | M1 |
Obtain derivative in any correct form | A1 |
Equate derivative to zero and solve for $x$ or $x^2$ | M1 |
Obtain $x = 1$ correctly | A1 | 4
[Differentiating $(x^2 + 1)y = x$ using the product rule can also earn the first M1A1.] | |
[SR: if the quotient rule is misused, with a 'reversed' numerator or $\nabla$ instead of $\nabla^2$ in the denominator, award M0A0 but allow the following M1A1.] | |
**(ii)** Obtain indefinite integral of the form $k \ln(x^2 + 1)$, where $k = \frac{1}{2}, 1$ or $2$ | M1* |
Use limits $x = 0$ and $x = p$ correctly, or equivalent | M1(dep*) |
Obtain answer $\frac{1}{2}\ln(p^2 + 1)$ | A1 | 3
[Also accept $-\ln \cos \theta$ or $\ln \cos \theta$, where $x = \tan \theta$, for the first M1*.] | |
**(iii)** Equate to 1 and convert equation to the form $p^2 + 1 = \exp(1/k)$ | M1 |
Obtain answer $p = 2.53$ | A1 | 29\\
\includegraphics[max width=\textwidth, alt={}, center]{208eab3e-a78c-43b4-918f-a9efc9b4f47a-4_429_748_264_699}
The diagram shows part of the curve $y = \frac { x } { x ^ { 2 } + 1 }$ and its maximum point $M$. The shaded region $R$ is bounded by the curve and by the lines $y = 0$ and $x = p$.\\
(i) Calculate the $x$-coordinate of $M$.\\
(ii) Find the area of $R$ in terms of $p$.\\
(iii) Hence calculate the value of $p$ for which the area of $R$ is 1 , giving your answer correct to 3 significant figures.
\hfill \mbox{\textit{CAIE P3 2005 Q9 [9]}}