| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2013 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard Integrals and Reverse Chain Rule |
| Type | Definite integral with trigonometric functions |
| Difficulty | Standard +0.3 This is a straightforward two-part question requiring standard differentiation of logarithms and trigonometric functions in part (i), followed by direct application of that result to evaluate a definite integral in part (ii). The question explicitly guides students through the process, making it slightly easier than average but still requiring competent execution of multiple techniques. |
| Spec | 1.07l Derivative of ln(x): and related functions1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Method I: \(\frac{\cos x}{1+\sin x} - \frac{-\sin x}{\cos x}\) or \(\frac{\cos x}{1+\sin x} + \frac{\sin x}{\cos x}\) | B2 | Each half (including 'middle' sign) scores B1 |
| \(\frac{+/-\cos^2 x +/- \sin x(1+\sin x)}{(1+\sin x)\cos x}\) | M1 | Combine, provided derivative was of form \(\frac{f'(x)}{f(x)}\); allow only variations in num signs |
| \(\frac{1+\sin x}{\cos x(1+\sin x)} = \frac{1}{\cos x}\) www AG | A1 | \(\cos^2 x + \sin^2 x = 1\) in intermediate step required |
| Method II: Change to \(\ln\left(\frac{1+\sin x}{\cos x}\right)\) | B1 | |
| Change to \(\ln(\sec x + \tan x)\) | B1 | Not \(\ln\left(\frac{1}{\cos x} + \tan x\right)\) |
| Diff as attempt at \(\frac{d}{dx}(\sec x + \tan x)\) over \((\sec x + \tan x)\) | M1 | |
| Reduce to \(\sec x = \frac{1}{\cos x}\) | A1 | |
| Method III: Change to \(\ln\left(\frac{1+\sin x}{\cos x}\right)\) | B1 | |
| Diff as attempt at quotient differentiation of \(\frac{1+\sin x}{\cos x}\) | M1 | |
| Fully correct differentiation | A1 | |
| Correct reduction to \(\frac{1}{\cos x}\) | A1 | |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Indef. integral \(= \ln(1+\sin x) - \ln(\cos x)\) [Method I] | B1 | Or \(\ln(\sec x + \tan x)\) [Method II] |
| Substitute limits and use log manipulation | M1 | Use of \(\ln A - \ln B = \ln\frac{A}{B}\) anywhere in question |
| Answer \(= \ln(2+\sqrt{3})\) | B1 | Accept \(\ln 3.73\) or \(\ln\frac{2+\sqrt{3}}{1}\); but not \(\ln\frac{1+\frac{\sqrt{3}}{2}}{\frac{1}{2}}\); answer has not been given |
| [3] |
# Question 7(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| **Method I:** $\frac{\cos x}{1+\sin x} - \frac{-\sin x}{\cos x}$ or $\frac{\cos x}{1+\sin x} + \frac{\sin x}{\cos x}$ | B2 | Each half (including 'middle' sign) scores B1 |
| $\frac{+/-\cos^2 x +/- \sin x(1+\sin x)}{(1+\sin x)\cos x}$ | M1 | Combine, provided derivative was of form $\frac{f'(x)}{f(x)}$; allow only variations in num signs |
| $\frac{1+\sin x}{\cos x(1+\sin x)} = \frac{1}{\cos x}$ www **AG** | A1 | $\cos^2 x + \sin^2 x = 1$ in intermediate step required |
| **Method II:** Change to $\ln\left(\frac{1+\sin x}{\cos x}\right)$ | B1 | |
| Change to $\ln(\sec x + \tan x)$ | B1 | Not $\ln\left(\frac{1}{\cos x} + \tan x\right)$ |
| Diff as attempt at $\frac{d}{dx}(\sec x + \tan x)$ over $(\sec x + \tan x)$ | M1 | |
| Reduce to $\sec x = \frac{1}{\cos x}$ | A1 | |
| **Method III:** Change to $\ln\left(\frac{1+\sin x}{\cos x}\right)$ | B1 | |
| Diff as attempt at quotient differentiation of $\frac{1+\sin x}{\cos x}$ | M1 | |
| Fully correct differentiation | A1 | |
| Correct reduction to $\frac{1}{\cos x}$ | A1 | |
| **[4]** | | |
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# Question 7(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Indef. integral $= \ln(1+\sin x) - \ln(\cos x)$ [Method I] | B1 | Or $\ln(\sec x + \tan x)$ [Method II] |
| Substitute limits and use log manipulation | M1 | Use of $\ln A - \ln B = \ln\frac{A}{B}$ anywhere in question |
| Answer $= \ln(2+\sqrt{3})$ | B1 | Accept $\ln 3.73$ or $\ln\frac{2+\sqrt{3}}{1}$; but not $\ln\frac{1+\frac{\sqrt{3}}{2}}{\frac{1}{2}}$; answer has **not** been given |
| **[3]** | | |7 (i) Given that $y = \ln ( 1 + \sin x ) - \ln ( \cos x )$, show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { \cos x }$.\\
(ii) Using this result, evaluate $\int _ { 0 } ^ { \frac { 1 } { 3 } \pi } \sec x \mathrm {~d} x$, giving your answer as a single logarithm.
\hfill \mbox{\textit{OCR C4 2013 Q7 [7]}}