| Exam Board | OCR |
|---|---|
| Module | H240/01 (Pure Mathematics) |
| Year | 2018 |
| Session | March |
| Marks | 9 |
| Topic | Reciprocal Trig & Identities |
| Type | Differentiation of reciprocal functions |
| Difficulty | Standard +0.8 Part (i) is a standard proof of a derivative formula requiring chain rule application. Part (ii) requires expanding the squared expression, recognizing that sec²(2x) integrates to ½tan(2x), and carefully handling the algebra with exact values at π/12 and π/6. The 'detailed reasoning' requirement and exact value calculation with non-standard limits elevate this above routine integration questions, but it remains accessible with solid technique. |
| Spec | 1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sec x = (\cos x)^{-1}\); \(\frac{dy}{dx} = -(\cos x)^{-2}(-\sin x)\); \(\frac{dy}{dx} = \frac{1 \times \sin x}{\cos x \times \cos x} = \sec x \tan x\) A.G. | B1, M1, A1 | Correct definition for sec x soi; Attempt chain rule or quotient rule; Show given answer; At least one step needed |
| Answer | Marks | Guidance |
|---|---|---|
| DR: \(\sec^2 2x + 2\sec 2x \tan 2x + \tan^2 2x\); \(2\sec^2 2x + 2\sec 2x \tan 2x - 1\); \(\int f(x)dx = \tan 2x + \sec 2x - x\); \((\sqrt{3} + 2 - \frac{1}{3}\pi) - (\frac{1}{3}\sqrt{3} + \frac{2}{3}\sqrt{3} - \frac{1}{12}\pi)\); \(2 - \frac{1}{12}\pi\) | B1, M1, M1, A1, M1, A1 | Correct expansion of bracket; Use \(\tan^2 2x = \sec^2 2x - 1\); Attempt integration; Correct integral; Attempt use of limits; Obtain \(2 - \frac{1}{12}\pi\); One trig term correct; Correct order and subtraction |
**(i)**
| $\sec x = (\cos x)^{-1}$; $\frac{dy}{dx} = -(\cos x)^{-2}(-\sin x)$; $\frac{dy}{dx} = \frac{1 \times \sin x}{\cos x \times \cos x} = \sec x \tan x$ A.G. | B1, M1, A1 | Correct definition for sec x soi; Attempt chain rule or quotient rule; Show given answer; At least one step needed |
**(ii)**
| DR: $\sec^2 2x + 2\sec 2x \tan 2x + \tan^2 2x$; $2\sec^2 2x + 2\sec 2x \tan 2x - 1$; $\int f(x)dx = \tan 2x + \sec 2x - x$; $(\sqrt{3} + 2 - \frac{1}{3}\pi) - (\frac{1}{3}\sqrt{3} + \frac{2}{3}\sqrt{3} - \frac{1}{12}\pi)$; $2 - \frac{1}{12}\pi$ | B1, M1, M1, A1, M1, A1 | Correct expansion of bracket; Use $\tan^2 2x = \sec^2 2x - 1$; Attempt integration; Correct integral; Attempt use of limits; Obtain $2 - \frac{1}{12}\pi$; One trig term correct; Correct order and subtraction |8 (i) Given that $y = \sec x$, show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \sec x \tan x$.\\
(ii) In this question you must show detailed reasoning.
Find the exact value of $\int _ { \frac { 1 } { 12 } \pi } ^ { \frac { 1 } { 6 } \pi } ( \sec 2 x + \tan 2 x ) ^ { 2 } \mathrm {~d} x$.
\hfill \mbox{\textit{OCR H240/01 2018 Q8 [9]}}