OCR MEI C3 2010 June — Question 6 6 marks

Exam BoardOCR MEI
ModuleC3 (Core Mathematics 3)
Year2010
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComposite & Inverse Functions
TypeFind inverse function
DifficultyStandard +0.3 This is a straightforward inverse function question requiring standard algebraic manipulation (swap x and y, rearrange) and application of inverse trigonometric functions. The domain restriction is given, making it easier than questions where students must determine it themselves. Slightly above average difficulty due to the composite nature (linear transformation of sine) and the need to correctly identify the range of f as the domain of f^(-1), but still a routine C3 exercise.
Spec1.02v Inverse and composite functions: graphs and conditions for existence1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^2

6 The function \(\mathrm { f } ( x )\) is defined by $$f ( x ) = 1 + 2 \sin 3 x , \quad - \frac { \pi } { 6 } \leqslant x \leqslant \frac { \pi } { 6 }$$ You are given that this function has an inverse, \(\mathrm { f } ^ { - 1 } ( x )\).
Find \(\mathrm { f } ^ { - 1 } ( x )\) and its domain.

Question 6:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(f(x) = 1 + 2\sin 3x = y \Leftrightarrow x = 1 + 2\sin 3y\)M1 attempt to invert; at least one step attempted
\(\sin 3y = (x-1)/2\)A1
\(3y = \arcsin\left[\frac{x-1}{2}\right]\)A1
\(f^{-1}(x) = \frac{1}{3}\arcsin\left[\frac{x-1}{2}\right]\)A1 must be \(y = \ldots\) or \(f^{-1}(x) = \ldots\); (or any other variable provided same used on each side)
Range of f is \(-1\) to \(3\), so \(-1 \leq x \leq 3\)M1 or \(-1 \leq (x-1)/2 \leq 1\); condone \(<\)'s for M1
A1 [6]must be '\(x\)', not \(y\) or \(f(x)\); allow unsupported correct answers
## Question 6:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(x) = 1 + 2\sin 3x = y \Leftrightarrow x = 1 + 2\sin 3y$ | M1 | attempt to invert; at least one step attempted |
| $\sin 3y = (x-1)/2$ | A1 | |
| $3y = \arcsin\left[\frac{x-1}{2}\right]$ | A1 | |
| $f^{-1}(x) = \frac{1}{3}\arcsin\left[\frac{x-1}{2}\right]$ | A1 | must be $y = \ldots$ or $f^{-1}(x) = \ldots$; (or any other variable provided same used on each side) |
| Range of f is $-1$ to $3$, so $-1 \leq x \leq 3$ | M1 | or $-1 \leq (x-1)/2 \leq 1$; condone $<$'s for M1 |
| | A1 [6] | must be '$x$', not $y$ or $f(x)$; allow unsupported correct answers |

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6 The function $\mathrm { f } ( x )$ is defined by

$$f ( x ) = 1 + 2 \sin 3 x , \quad - \frac { \pi } { 6 } \leqslant x \leqslant \frac { \pi } { 6 }$$

You are given that this function has an inverse, $\mathrm { f } ^ { - 1 } ( x )$.\\
Find $\mathrm { f } ^ { - 1 } ( x )$ and its domain.

\hfill \mbox{\textit{OCR MEI C3 2010 Q6 [6]}}