| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2016 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Prove identity then solve equation |
| Difficulty | Standard +0.3 This is a multi-part question combining modulus functions and trigonometric identities. Part (i) of the trig section requires proving a standard identity using basic manipulation (sin 2θ = 2sin θ cos θ, tan θ = sin θ/cos θ). The subsequent parts are routine applications of this result with straightforward algebraic manipulation. While it requires multiple techniques, all are standard C3 material with no novel insight needed, making it slightly easier than average. |
| Spec | 1.02l Modulus function: notation, relations, equations and inequalities1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence1.05h Reciprocal trig functions: sec, cosec, cot definitions and graphs1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05l Double angle formulae: and compound angle formulae |
| Answer | Marks | Guidance |
|---|---|---|
| State range of \(f\) is \(f(x) \ge 3a\) or \(y \ge 3a\) | B1 | Allow \(f \ge 3a\) or equiv expression in words but \(3a\) to be included |
| State range of \(g\) is all real numbers or equiv such as \(y \in \mathbb{R}\) (real numbers) | B1 | |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| State function is not 1 – 1 or different x-values give same y-value or equiv | B1 | no credit for 'no inverse due to modulus' nor for 'cannot be reflected across \(y = x\)' |
| Obtain form \(k(y + 4a)\) or \(k(y + 4a)\) | M1 | for non-zero constant \(k\) |
| Obtain \(\frac{1}{5}(x + 4a)\) or \(\frac{1}{5}x + \frac{4}{5}a\) | A1 | Must finally be in terms of \(x\) |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Either Attempt composition of functions the right way round | M1 | Earned for 5(what they think \(f(x)\) is) \(- 4a\) |
| Obtain \(5 | 2x + a | + 11a = 31a\) or equiv |
| Or Apply their \(g^{-1}\) to \(31a\) | M1 | |
| Obtain \( | 2x + a | = ka\) |
| Either Solve \(2x + a = 4a\) and obtain \(\frac{5}{2}a\) | B1 FT | Following their \( |
| Solve linear equation in which signs of (their) \(2x\) and (their) \(4a\) are different | M1 | |
| Obtain \(-\frac{5}{2}a\) | A1 | And no others; obtaining \(-\frac{5}{2}a\) and then concluding \(\frac{5}{2}a\) is A0 |
| Or Square both sides and obtain \(4x^2 + 4ax - 15a^2 = 0\) | B1 FT | Following their \( |
| Solve 3-term quadratic equation to obtain two values | M1 | Allow M1 if factorisation wrong but expansion gives correct first and third terms; allow M1 if incorrect use of formula involves only one error |
| Obtain \(-\frac{5}{2}a\), \(\frac{3}{2}a\) | A1 | And no others; continuing from two correct answers to conclude \(\frac{5}{2}a\), \(\frac{3}{2}a\) is A0 |
| [5] |
## Part i
State range of $f$ is $f(x) \ge 3a$ or $y \ge 3a$ | B1 | Allow $f \ge 3a$ or equiv expression in words but $3a$ to be included
State range of $g$ is all real numbers or equiv such as $y \in \mathbb{R}$ (real numbers) | B1 |
| [2] |
## Part ii
State function is not 1 – 1 or different x-values give same y-value or equiv | B1 | no credit for 'no inverse due to modulus' nor for 'cannot be reflected across $y = x$'
Obtain form $k(y + 4a)$ or $k(y + 4a)$ | M1 | for non-zero constant $k$
Obtain $\frac{1}{5}(x + 4a)$ or $\frac{1}{5}x + \frac{4}{5}a$ | A1 | Must finally be in terms of $x$
| [3] |
## Part iii
Either Attempt composition of functions the right way round | M1 | Earned for 5(what they think $f(x)$ is) $- 4a$
Obtain $5|2x + a| + 11a = 31a$ or equiv | A1 |
Or Apply their $g^{-1}$ to $31a$ | M1 |
Obtain $|2x + a| = ka$ | A1 |
Either Solve $2x + a = 4a$ and obtain $\frac{5}{2}a$ | B1 FT | Following their $|2x + a| = ka$
Solve linear equation in which signs of (their) $2x$ and (their) $4a$ are different | M1 |
Obtain $-\frac{5}{2}a$ | A1 | And no others; obtaining $-\frac{5}{2}a$ and then concluding $\frac{5}{2}a$ is A0
Or Square both sides and obtain $4x^2 + 4ax - 15a^2 = 0$ | B1 FT | Following their $|2x + a| = ka$
Solve 3-term quadratic equation to obtain two values | M1 | Allow M1 if factorisation wrong but expansion gives correct first and third terms; allow M1 if incorrect use of formula involves only one error
Obtain $-\frac{5}{2}a$, $\frac{3}{2}a$ | A1 | And no others; continuing from two correct answers to conclude $\frac{5}{2}a$, $\frac{3}{2}a$ is A0
| [5] |8 The functions f and g are defined for all real values of $x$ by
$$\mathrm { f } ( x ) = | 2 x + a | + 3 a \quad \text { and } \quad \mathrm { g } ( x ) = 5 x - 4 a$$
where $a$ is a positive constant.\\
(i) State the range of f and the range of g .\\
(ii) State why f has no inverse, and find an expression for $\mathrm { g } ^ { - 1 } ( x )$.\\
(iii) Solve for $x$ the equation $\operatorname { gf } ( x ) = 31 a$.\\
(i) Show that $\sin 2 \theta ( \tan \theta + \cot \theta ) \equiv 2$.\\
(ii) Hence
\begin{enumerate}[label=(\alph*)]
\item find the exact value of $\tan \frac { 1 } { 12 } \pi + \tan \frac { 1 } { 8 } \pi + \cot \frac { 1 } { 12 } \pi + \cot \frac { 1 } { 8 } \pi$,
\item solve the equation $\sin 4 \theta ( \tan \theta + \cot \theta ) = 1$ for $0 < \theta < \frac { 1 } { 2 } \pi$,
\item express $( 1 - \cos 2 \theta ) ^ { 2 } \left( \tan \frac { 1 } { 2 } \theta + \cot \frac { 1 } { 2 } \theta \right) ^ { 3 }$ in terms of $\sin \theta$.
\end{enumerate}
\hfill \mbox{\textit{OCR C3 2016 Q8 [10]}}