Higher January 2022 Paper 1R Q19
19 Given that \(\left(\sqrt[3]{\dfrac{1}{x}}\right)^4 = x^m\)
(a) find the value of \(m\) (1)
Given that \(a\), \(b\) and \(c\) are integers,
(b) express \(3x^2 + 12x + 19\) in the form \(a(x + b)^2 + c\) (2)
| Scheme | Marks |
|---|---|
| \(-\dfrac{4}{3}\) | B1 |
| (1) |
| Scheme | Marks |
|---|---|
\(3(x^2 + 4x) + 19\) and \(3[(x + 2)^2 - 2^2] + 19\) or \(3\left(x^2 + 4x + \dfrac{19}{3}\right)\) and \(3\left((x + 2)^2 - 2^2 + \dfrac{19}{3}\right)\) or \(a = 3\) and \(2ab = 12\) oe and \(b^2a + c = 19\) oe or \(a = 3\) and \(b = \dfrac{12}{2 \times 3}\) oe and \(c = -\dfrac{12^2}{4 \times 3} + 19\) oe | M1 |
| \(3(x + 2)^2 + 7\) | A1 |
| (2) | |
| (3 marks) |
Notes
M1: for correctly taking out a factor of 3 and correctly completing the square
or
for equating coefficients by expanding \(a(x + b)^2 + c = ax^2 + 2abx + b^2a + c\)
or
for equating coefficients by using \(ax^2 + bx + c = a\left(x + \dfrac{b}{2a}\right)^2 - \dfrac{b^2}{4a} + c\)
A1: accept \(a = 3\), \(b = 2\), \(c = 7\)