| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2011 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Differentiation of reciprocal functions |
| Difficulty | Standard +0.3 Part (i) is a standard bookwork proof requiring chain rule on sec x = 1/cos x. Part (ii) requires recognizing the double angle identity cos 2x = 2cos²x - 1 to simplify the denominator, then substituting u = cos x, making it a routine integration by substitution. This is slightly above average difficulty due to the identity manipulation and substitution steps, but follows standard C4 techniques without requiring novel insight. |
| Spec | 1.05h Reciprocal trig functions: sec, cosec, cot definitions and graphs1.07h Differentiation from first principles: for sin(x) and cos(x)1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| State/imply \(\frac{d}{dx}(\sec x) = \frac{d}{dx}(\frac{1}{\cos x})\) or \(\frac{d}{dx}(\cos x)^{-1}\) | B1 | Not just \(\sec x = \frac{1}{\cos x}\) |
| Attempt quotient rule or chain rule to power -1 | M1 | Allow \(\frac{u\,dv - v\,du}{v^2}\) & wrong trig signs |
| Obtain \(\frac{\sin x}{\cos^2 x}\) or \(--(\sin x)(\cos x)^{-2}\) | A1 | No inaccuracy allowed here |
| Simplify with suff evid to AG, e.g. \(\frac{1}{\cos x} \cdot \frac{\sin x}{\cos x}\) | A1 4 | Or vice versa. Not just = \(\sec x \tan x\) |
| Answer | Marks | Guidance |
|---|---|---|
| Use \(\cos 2x = +/-1 + /-2\cos^2 x\) or \(+/-1 +/- 2\sin^2 x\) | M1 | or \(\pm(\cos^2 x - \sin^2 x)\) |
| Correct denominator = \(\sqrt{2\cos^2 x}\) | A1 | \(\sqrt{2 - 2\sin^2 x}\) needs simplifying |
| Evidence that \(\frac{\tan x}{\cos x} = \sec x \tan x\) or \(\int \frac{\tan x}{\cos x}dx = \sec x\) | B1 | irrespective of any cosnt multiples |
| \(\frac{1}{\sqrt{2}}\sec x\) \((+ c)\) | A1 4 | Condone \(\theta\) for \(x\) except final line |
**(i)**
State/imply $\frac{d}{dx}(\sec x) = \frac{d}{dx}(\frac{1}{\cos x})$ or $\frac{d}{dx}(\cos x)^{-1}$ | B1 | Not just $\sec x = \frac{1}{\cos x}$
Attempt quotient rule or chain rule to power -1 | M1 | Allow $\frac{u\,dv - v\,du}{v^2}$ & wrong trig signs
Obtain $\frac{\sin x}{\cos^2 x}$ or $--(\sin x)(\cos x)^{-2}$ | A1 | No inaccuracy allowed here
Simplify with suff evid to AG, e.g. $\frac{1}{\cos x} \cdot \frac{\sin x}{\cos x}$ | A1 4 | Or vice versa. Not just = $\sec x \tan x$
**(ii)**
Use $\cos 2x = +/-1 + /-2\cos^2 x$ or $+/-1 +/- 2\sin^2 x$ | M1 | or $\pm(\cos^2 x - \sin^2 x)$
Correct denominator = $\sqrt{2\cos^2 x}$ | A1 | $\sqrt{2 - 2\sin^2 x}$ needs simplifying
Evidence that $\frac{\tan x}{\cos x} = \sec x \tan x$ or $\int \frac{\tan x}{\cos x}dx = \sec x$ | B1 | irrespective of any cosnt multiples
$\frac{1}{\sqrt{2}}\sec x$ $(+ c)$ | A1 4 | Condone $\theta$ for $x$ except final line
---3 (i) Show that the derivative of $\sec x$ can be written as $\sec x \tan x$.\\
(ii) Find $\int \frac { \tan x } { \sqrt { 1 + \cos 2 x } } \mathrm {~d} x$.
\hfill \mbox{\textit{OCR C4 2011 Q3 [8]}}