| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/1 (Pre-U Mathematics Paper 1) |
| Year | 2014 |
| Session | June |
| Marks | 10 |
| Topic | Addition & Double Angle Formulae |
| Type | Derive triple angle then solve equation |
| Difficulty | Standard +0.8 This question requires applying the double angle formula twice to derive a quadruple angle formula (moderately challenging algebraic manipulation), then using the derived result to find an unknown constant by substituting a specific value and rearranging. While it involves multiple steps and careful algebra, the techniques are standard for Further Maths students and the path is clearly signposted. It's above average difficulty but not exceptionally hard. |
| Spec | 1.05l Double angle formulae: and compound angle formulae |
**(i)** Attempt to use $\tan 4x = \tan 2(2x)$ — M1
Obtain $\tan 4x = \frac{2\tan 2x}{1-\tan^2(2x)}$ or $\frac{\tan 2x + \tan 2x}{1 - \tan^2(2x)}$ — A1
Attempt to substitute for $\tan 2x$ — M1*
Obtain $\dfrac{\dfrac{2(2\tan x)}{1-\tan^2 x}}{1 - \left(\dfrac{2\tan x}{1-\tan^2 x}\right)^2}$ — dep A1
Correctly form a single term in the denominator by removing the 1 — dep M1
Obtain $\dfrac{4\tan x\left(1-\tan^2 x\right)}{1 - 6\tan^2 x + \tan^4 x}$ AG — A1 **[6]**
Stages in the argument must be consistent and clearly presented for A1.
**(ii)** State or imply that the root gives $\tan 4x = 1$ — B1*
Attempt to write the equation in the same structure as the identity — M1 dep
$\dfrac{4\tan x(1-\tan^2 x)}{1-6\tan^2 x + \tan^4 x}$
Obtain $p = 4$ — A1 dep
Convincing and clear argument with all stages shown including the statement that $\tan\!\left(\dfrac{\pi}{4}\right) = 1$ — B1 dep **[4]**
Evaluation using decimals 0/412 (i) Use the identity $\tan 2 x \equiv \frac { 2 \tan x } { 1 - \tan ^ { 2 } x }$ to show that $\tan 4 x \equiv \frac { 4 \left( 1 - \tan ^ { 2 } x \right) \tan x } { 1 - 6 \tan ^ { 2 } x + \tan ^ { 4 } x }$.\\
(ii) Hence, given that $x = \frac { 1 } { 16 } \pi$ is a root of the equation $\tan ^ { 4 } x + p \tan ^ { 3 } x - 6 \tan ^ { 2 } x - p \tan x + 1 = 0$ where $p$ is a positive constant, find the value of $p$.
\hfill \mbox{\textit{Pre-U Pre-U 9794/1 2014 Q12 [10]}}