Standard +0.3 This is a straightforward parametric-to-Cartesian conversion using the double angle formula (cos 2t = 2cos²t - 1), followed by basic curve sketching and range analysis. Part (a) is routine manipulation, part (b) requires recognizing domain restrictions from the parameter range, and part (c) involves finding intersection conditions—all standard A-level techniques with no novel insight required. Slightly easier than average due to the guided 'show that' format.
14. A curve \(C\) has parametric equations
$$x = 1 - \cos t , \quad y = 2 \cos 2 t , \quad 0 \leq t < \pi$$
a. Show that the cartesian equation of the curve can be written as \(y = k ( 1 - x ) ^ { 2 } - 2\) where \(k\) is an integer.
b. i. Sketch the curve C .
ii. Explain briefly why C does not include all points of \(y = k ( 1 - x ) ^ { 2 } - 2 , x \in \mathbb { R }\).
The line with equation \(y = k - x\), where \(k\) is a constant, intersects C at two distinct points.
(c) State the range of values of \(k\), writing your answer in set notation.
Uses discriminant condition; accept \(b^2 - 4ac \geq 0\) or \(b^2 - 4ac = 0\) leading to critical value of \(k\)
\(k = \left\{k: -\frac{17}{16} < k \leq 2\right\}\)
A1
Correct range in set notation
Total: 5 marks
Total for Question 14: 10 marks
## Question 14:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Using $\cos 2t = 2\cos^2 t - 1$: $\frac{y}{2} = 2(1-x)^2 - 1$ | M1 | Attempts to use $\cos 2t = 2\cos^2 t - 1 \Rightarrow \frac{y}{2} = 2(1-x)^2 - 1$ |
| $\Rightarrow y = 4(1-x)^2 - 2$ | A1 | Proceeds to correct answer, $k=4$ |
**Total: 2 marks**
---
### Part (b)(i) and (b)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Sketch: $\cup$ parabola in quadrants one and four with minimum in quadrant four | M1 | For sketching a $\cup$ parabola in quadrant one and four with a minimum in quadrant four |
| $\cup$ parabola with minimum in quadrant four, end points $(0, 2)$ and $(5, 2)$ | A1 | Correct end points required |
| C does not include all points because $-1 \leq \cos 2t \leq 1$ so $-2 \leq y \leq 2$, or $-1 \leq \sin t \leq 1$ so $0 \leq x \leq 2$ | B1 | Reference to limits on $\sin$ or $\cos$ with a link to a restriction on $x$ or $y$ |
**Total: 3 marks**
---
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Lower bound: $k = -\frac{17}{16}$ or upper bound $k=2$ via substituting $(0,2)$ into $y = k - x$ | B1 | Deduces either correct lower value $k = -\frac{17}{16}$ or deduces $k = 2$ by substituting $(0,2)$ into $y = k-x$ |
| Sets $k - x = 4(1-x)^2 - 2$, proceeds to 3-term quadratic: $4x^2 - 7x + 2 - k = 0$ | M1 | Attempt at upper value for $k$; finds where $y = k-x$ meets $y = 4(1-x)^2 - 2$ and proceeds to a 3-term quadratic |
| Correct quadratic: $4x^2 - 7x + 2 - k = 0$ | A1 | Correct 3-term quadratic |
| Applies discriminant condition $b^2 - 4ac > 0$ (or $\geq 0$): $49 - 4(4)(2-k) > 0 \Rightarrow k > -\frac{17}{16}$ | M1 | Uses discriminant condition; accept $b^2 - 4ac \geq 0$ or $b^2 - 4ac = 0$ leading to critical value of $k$ |
| $k = \left\{k: -\frac{17}{16} < k \leq 2\right\}$ | A1 | Correct range in set notation |
**Total: 5 marks**
**Total for Question 14: 10 marks**
14. A curve $C$ has parametric equations
$$x = 1 - \cos t , \quad y = 2 \cos 2 t , \quad 0 \leq t < \pi$$
a. Show that the cartesian equation of the curve can be written as $y = k ( 1 - x ) ^ { 2 } - 2$ where $k$ is an integer.\\
b. i. Sketch the curve C .\\
ii. Explain briefly why C does not include all points of $y = k ( 1 - x ) ^ { 2 } - 2 , x \in \mathbb { R }$.
The line with equation $y = k - x$, where $k$ is a constant, intersects C at two distinct points.\\
(c) State the range of values of $k$, writing your answer in set notation.
\hfill \mbox{\textit{Edexcel PMT Mocks Q14 [10]}}