Higher November 2021 Paper 1 Q19
19 \(ABCED\) is a five-sided shape.
Diagram NOT accurately drawn
\(ABCD\) is a rectangle.
\(CED\) is an equilateral triangle.
\(AB = x\) cm\(\qquad BC = y\) cm
The perimeter of \(ABCED\) is 100 cm.
The area of \(ABCED\) is \(R\) cm2
(a) Show that \(R = \dfrac{x}{4}\left(200 - \left[6 - \sqrt{3}\right]x\right)\) (3)
(b)
(i) Find the value of \(x\) for which \(R\) has its maximum value.
Give your answer in the form \(\dfrac{p}{q - \sqrt{3}}\) where \(p\) and \(q\) are integers. (2)
Give your answer in the form \(\dfrac{p}{q - \sqrt{3}}\) where \(p\) and \(q\) are integers. (2)
(ii) Explain why the maximum value of \(R\) is given by this value of \(x\). (1)
| Scheme | Marks |
|---|---|
E.g. \(x + y + x + y + x = 100\) oe or \(3x + 2y = 100\) oe \(\left(y = \dfrac{100 - 3x}{2}\right)\) E.g. \(\dfrac{1}{2} \times x \times x \times \sin 60\) \(\left(= \dfrac{1}{2} \times x \times x \times \dfrac{\sqrt{3}}{2}\right)\) \(\left(= \dfrac{x^2\sqrt{3}}{4}\right)\) or E.g. \(x^2 = \left(\dfrac{x}{2}\right)^2 + h^2\) and \(= \dfrac{1}{2} \times x \times \dfrac{x\sqrt{3}}{2}\ \left(= \dfrac{x^2\sqrt{3}}{4}\right)\) | M1 |
| \(x\,\text{``}\left(\dfrac{100 - 3x}{2}\right)\text{''} + \text{``}\dfrac{x^2\sqrt{3}}{4}\text{''}\) oe | M1 |
E.g. \(x\left(\dfrac{200 - 6x}{4}\right) + \dfrac{x^2\sqrt{3}}{4}\) or \(\dfrac{x}{4}(200 - 6x + x\sqrt{3})\) or \(\dfrac{200x - 6x^2}{4} + \dfrac{x^2\sqrt{3}}{4}\) or \(\dfrac{x}{4}(200 - 6x^2 + x^2\sqrt{3})\) Answer: Shown | A1 |
| (3) |
Notes
M1: for a correct equation for the perimeter of the shape or for a correct expression for the area of triangle CED
M1: for the area of the shape in terms of \(x\) only
A1: for the area given in correct form with full working shown (at least one intermediate step before the answer)
| Scheme | Marks |
|---|---|
| \(\left(\dfrac{\mathrm{d}R}{\mathrm{d}x} =\right) 50 - \dfrac{3}{2} \times 2 \times x + 2 \times \dfrac{x\sqrt{3}}{4} = 0\) oe | M1 |
Correct answer scores full marks (unless from obvious incorrect working) Answer: \(\dfrac{100}{6 - \sqrt{3}}\) | A1 |
| (2) |
Notes
M1: for differentiation of correct expression with 2 out of 3 terms correct and equated to 0 (can be implied by subsequent working)
A1: for a correct expression
| Scheme | Marks |
|---|---|
| Correct reason | B1 |
| (1) | |
| (6 marks) |
Notes
B1: for correct reason
\(R\) is a quadratic with negative coefficient of \(x^2\)
E.g. the graph of \(R\) is \(\cap\) shaped or
(allow \(\dfrac{\mathrm{d}^2R}{\mathrm{d}x^2} \lt 0\) oe)