4MA0 Higher January 2017 Paper 3 Q18
18 The curve with equation \(\;y = 10x^2 + 9x + 5\;\) has a minimum at point \(A\).
Find the coordinates of \(A\).
Show your working clearly.
(4)
| Scheme | Marks |
|---|---|
| \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x} =\right) 20x + 9\) | M1 |
| \(20x + 9 = 0\) | M1 |
| \(x = -0.45\) oe | A1 |
| (−0.45, 2.975) oe | A1ft |
| (4) | |
| (4 marks) |
Notes
M1: for differentiating \(10x^2\) or \(9x\) correctly
M1: equating their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) (of the form \(ax + b\)) to zero, dep on previous M1
A1: dep on at least M1
for \(x = -0.45\) oe
A1ft: dep on M2
accept fractions
\(\left(-\dfrac{9}{20}, 2\dfrac{39}{40}\right)\) or \(\left(-\dfrac{9}{20}, \dfrac{119}{40}\right)\)
or M2 for \(\dfrac{-9}{2 \times 10}\) (from \(\dfrac{-b}{2a}\))
Alternative scheme (completing the square)
| Scheme | Marks |
|---|---|
| \(\left(x + \dfrac{9}{20}\right)^2 + \ldots\) | M1 |
| \(\left(x + \dfrac{9}{20}\right)^2 - \left(\dfrac{9}{20}\right)^2 + \dfrac{5}{10} = 0\) | M1 |
| \(x = -0.45\) oe | A1 |
| (−0.45, 2.975) oe | A1ft |
Notes
A1: dep on at least M1
for \(x = -0.45\) oe
A1ft: dep on M2
accept fractions
\(\left(-\dfrac{9}{20}, 2\dfrac{39}{40}\right)\) or \(\left(-\dfrac{9}{20}, \dfrac{119}{40}\right)\)