| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2003 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Indefinite & Definite Integrals |
| Type | Trapezium rule estimation |
| Difficulty | Moderate -0.3 This question involves routine differentiation using quotient/chain rule, standard trapezium rule application with only 2 intervals (minimal computation), and interpreting concavity to determine over/under-estimation. All parts are textbook-standard techniques with no problem-solving insight required, making it slightly easier than average. |
| Spec | 1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.07l Derivative of ln(x): and related functions1.08d Evaluate definite integrals: between limits1.09f Trapezium rule: numerical integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempt to apply the chain or quotient rule | M1 | |
| Obtain derivative of the form \(\dfrac{k\sec^2 x}{(1+\tan x)^2}\) or equivalent | A1 | |
| Obtain correct derivative \(-\dfrac{\sec^2 x}{(1+\tan x)^2}\) or equivalent | A1 | |
| Explain why derivative, and hence gradient of the curve, is always negative | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| State or imply correct ordinates: \(1, 0.7071..., 0.5\) | B1 | |
| Use correct formula, or equivalent, with \(h = \frac{1}{8}\pi\) and three ordinates | M1 | |
| Obtain answer \(0.57\) \((0.57220...) \pm 0.01\) (accept \(0.18\pi\)) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Justify the statement that the rule gives an over-estimate | B1 |
# Question 6(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempt to apply the chain or quotient rule | M1 | |
| Obtain derivative of the form $\dfrac{k\sec^2 x}{(1+\tan x)^2}$ or equivalent | A1 | |
| Obtain correct derivative $-\dfrac{\sec^2 x}{(1+\tan x)^2}$ or equivalent | A1 | |
| Explain why derivative, and hence gradient of the curve, is always negative | A1 | |
**Total: [4]**
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# Question 6(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| State or imply correct ordinates: $1, 0.7071..., 0.5$ | B1 | |
| Use correct formula, or equivalent, with $h = \frac{1}{8}\pi$ and three ordinates | M1 | |
| Obtain answer $0.57$ $(0.57220...) \pm 0.01$ (accept $0.18\pi$) | A1 | |
**Total: [3]**
---
# Question 6(iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Justify the statement that the rule gives an over-estimate | B1 | |
**Total: [1]**
---6 The equation of a curve is $y = \frac { 1 } { 1 + \tan x }$.\\
(i) Show, by differentiation, that the gradient of the curve is always negative.\\
(ii) Use the trapezium rule with 2 intervals to estimate the value of
$$\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \frac { 1 } { 1 + \tan x } \mathrm {~d} x$$
giving your answer correct to 2 significant figures.\\
(iii)\\
\includegraphics[max width=\textwidth, alt={}, center]{a31a4b4e-83a6-47d9-9679-3471b3da1b6e-3_556_802_1384_708}
The diagram shows a sketch of the curve for $0 \leqslant x \leqslant \frac { 1 } { 4 } \pi$. State, with a reason, whether the trapezium rule gives an under-estimate or an over-estimate of the true value of the integral in part (ii).
\hfill \mbox{\textit{CAIE P2 2003 Q6 [8]}}