| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2019 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Find minimum domain for inverse |
| Difficulty | Moderate -0.3 This is a straightforward question on inverse functions requiring knowledge that a function needs to be one-to-one for an inverse to exist. Parts (i)-(iii) involve routine identification of range, sketching a cosine transformation, and recognizing that restricting the domain to [0,π] makes it monotonic. Part (iv) requires standard inverse function technique (swap x and y, solve for y using arccos). Slightly easier than average due to the standard nature of all parts and clear signposting. |
| Spec | 1.02v Inverse and composite functions: graphs and conditions for existence1.05a Sine, cosine, tangent: definitions for all arguments1.05f Trigonometric function graphs: symmetries and periodicities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(-1 \leqslant f(x) \leqslant 5\) or \([-1, 5]\) (may use \(y\) or \(f\) instead of \(f(x)\)) | B1 B1 | \(-1 < f(x) \leqslant 5\) or \(-1 \leqslant x \leqslant 5\) or \((-1,5)\) or \([5,-1]\) B1 only |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Graph as shown: \(g(x) = 2 - 3\cos x\) for \(0 \leqslant x \leqslant p\) | *B1 | Start and end at \(-\)ve \(y\), symmetrical, centre \(+\)ve |
| DB1 | Shape all ok. Curves not lines. One cycle \([0, 2\pi]\). Flattens at each end | |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Greatest value of \(p = \pi\) | B1 | |
| Total: 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x = 2 - 3\cos x \rightarrow \cos x = \frac{1}{3}(2-x)\) | M1 | Attempt to make \(\cos x\) the subject. Use of \(\cos^{-1}\) |
| \(g^{-1}(x) = \cos^{-1}\dfrac{2-x}{3}\) (may use '\(y =\)') | A1 | Must be a function of \(x\) |
| Total: 2 |
## Question 9(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $-1 \leqslant f(x) \leqslant 5$ or $[-1, 5]$ (may use $y$ or $f$ instead of $f(x)$) | B1 B1 | $-1 < f(x) \leqslant 5$ or $-1 \leqslant x \leqslant 5$ or $(-1,5)$ or $[5,-1]$ B1 only |
| **Total: 2** | | |
## Question 9(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Graph as shown: $g(x) = 2 - 3\cos x$ for $0 \leqslant x \leqslant p$ | *B1 | Start and end at $-$ve $y$, symmetrical, centre $+$ve |
| | DB1 | Shape all ok. Curves not lines. One cycle $[0, 2\pi]$. Flattens at each end |
| **Total: 2** | | |
## Question 9(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Greatest value of $p = \pi$ | B1 | |
| **Total: 1** | | |
## Question 9(iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x = 2 - 3\cos x \rightarrow \cos x = \frac{1}{3}(2-x)$ | M1 | Attempt to make $\cos x$ the subject. Use of $\cos^{-1}$ |
| $g^{-1}(x) = \cos^{-1}\dfrac{2-x}{3}$ (may use '$y =$') | A1 | Must be a function of $x$ |
| **Total: 2** | | |9 The function f is defined by $\mathrm { f } ( x ) = 2 - 3 \cos x$ for $0 \leqslant x \leqslant 2 \pi$.\\
(i) State the range of f .\\
(ii) Sketch the graph of $y = \mathrm { f } ( x )$.
The function g is defined by $\mathrm { g } ( x ) = 2 - 3 \cos x$ for $0 \leqslant x \leqslant p$, where $p$ is a constant.\\
(iii) State the largest value of $p$ for which g has an inverse.\\
(iv) For this value of $p$, find an expression for $\mathrm { g } ^ { - 1 } ( x )$.\\
\hfill \mbox{\textit{CAIE P1 2019 Q9 [7]}}