| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2013 |
| Session | June |
| 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 Part (i) is routine algebraic manipulation of trigonometric expressions requiring standard double angle formulas. Part (ii) applies the result directly with straightforward integration of tan(2x) using reverse chain rule and evaluation at given limits. This is a standard C4 question testing identity verification followed by direct application—slightly easier than average due to the guided structure and routine techniques involved. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.08c Integrate e^(kx), 1/x, sin(kx), cos(kx) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{(1+\tan x)-(1-\tan x)}{(1-\tan x)(1+\tan x)}\) | M1 | Combine using common denominator (or write as 2 separate fractions). Accept \(1-\tan x \cdot 1+\tan x\) in denominator |
| \(= \frac{2\tan x}{1-\tan^2 x} = \tan 2x\) Answer Given | A1 | \(\frac{2\tan x}{1-\tan^2 x}\) essential stage. A0 for omission of necessary brackets |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int \tan 2x\,dx = \lambda\ln(\sec 2x)\) or \(\mu\ln(\cos 2x)\) \([=F(x)]\) | M1 | |
| \(\lambda = \frac{1}{2}\) or \(\mu = -\frac{1}{2}\) | A1 | |
| Their \(F\!\left[\frac{\pi}{6}\right] -\) their \(F\!\left[\frac{\pi}{12}\right]\) | M1 | Dependent on attempt at integration; not for \(\tan\!\left(\frac{\pi}{3}\right) - \tan\!\left(\frac{\pi}{6}\right)\) |
| \(\frac{1}{2}\ln 2 - \frac{1}{2}\ln\frac{2}{\sqrt{3}}\) oe | A1 | Any correct but probably unsimplified numerical version |
| \(\frac{1}{2}\ln\sqrt{3}\) or \(\frac{1}{4}\ln 3\) or \(\ln 3^{1/4}\) or \(\frac{1}{2}\ln\frac{6}{2\sqrt{3}}\) oe ISW | +A1 | Any correct version in the form \(a\ln b\) |
# Question 5(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{(1+\tan x)-(1-\tan x)}{(1-\tan x)(1+\tan x)}$ | M1 | Combine using common denominator (or write as 2 separate fractions). Accept $1-\tan x \cdot 1+\tan x$ in denominator |
| $= \frac{2\tan x}{1-\tan^2 x} = \tan 2x$ Answer Given | A1 | $\frac{2\tan x}{1-\tan^2 x}$ essential stage. A0 for omission of necessary brackets |
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# Question 5(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int \tan 2x\,dx = \lambda\ln(\sec 2x)$ or $\mu\ln(\cos 2x)$ $[=F(x)]$ | M1 | |
| $\lambda = \frac{1}{2}$ or $\mu = -\frac{1}{2}$ | A1 | |
| Their $F\!\left[\frac{\pi}{6}\right] -$ their $F\!\left[\frac{\pi}{12}\right]$ | M1 | Dependent on attempt at integration; not for $\tan\!\left(\frac{\pi}{3}\right) - \tan\!\left(\frac{\pi}{6}\right)$ |
| $\frac{1}{2}\ln 2 - \frac{1}{2}\ln\frac{2}{\sqrt{3}}$ oe | A1 | Any correct but probably unsimplified numerical version |
| $\frac{1}{2}\ln\sqrt{3}$ or $\frac{1}{4}\ln 3$ or $\ln 3^{1/4}$ or $\frac{1}{2}\ln\frac{6}{2\sqrt{3}}$ oe ISW | +A1 | Any correct version in the form $a\ln b$ |5 (i) Show that $\frac { 1 } { 1 - \tan x } - \frac { 1 } { 1 + \tan x } \equiv \tan 2 x$.\\
(ii) Hence evaluate $\int _ { \frac { 1 } { 12 } \pi } ^ { \frac { 1 } { 6 } \pi } \left( \frac { 1 } { 1 - \tan x } - \frac { 1 } { 1 + \tan x } \right) \mathrm { d } x$, giving your answer in the form $a \ln b$.
\hfill \mbox{\textit{OCR C4 2013 Q5 [7]}}