Higher June 2021 Paper 2 Q21
21 The point \(A\) is the only stationary point on the curve with equation \(y = kx^2 + \dfrac{16}{x}\) where \(k\) is a constant.
Given that the coordinates of \(A\) are \(\left(\dfrac{2}{3}, a\right)\)
find the value of \(a\).
Show your working clearly.
(5)
| Scheme | Marks |
|---|---|
| \(\left[\dfrac{\mathrm{d}y}{\mathrm{d}x} =\right] 2 \times kx - 16x^{-2}\) or \(2kx - \dfrac{16}{x^2}\) oe | M2 |
| \(\text{``}2kx - 16x^{-2}\text{''} = 0\) oe | M1 |
| eg \(\dfrac{8}{27}k = 8\) or \(\dfrac{4}{3}k = 36\) or \(k = 27\) oe | M1 |
| Working must be seen Answer: 36 | A1 |
| (5) | |
| (5 marks) |
Notes
M2: for both terms differentiated correctly
(M1) for one term differentiated correctly
M1: ft dep on M1
M1: (not ft) for substituting \(x = \dfrac{2}{3}\) into their correct equation for \(k\) and getting as far as one step from the value of \(k\) or the correct value of \(k\)
A1: dep on M4