June 2024 Paper 1 Q5
5 The line \(x + 13y = 108\) is the normal to the curve \(y = ax^2 + b\sqrt{x}\) at the point (4, 8).
Determine the values of the constants \(a\) and \(b\). [8]
| Scheme | Marks | AO |
|---|---|---|
| \(16a + 2b = 8\) | B1 | 3.1a |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2ax + \dfrac{b}{2\sqrt{x}}\) | M1 | 2.1 |
| Obtain correct derivative | A1 | 1.1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 8a + \tfrac{1}{4}b\) | M1 | 1.1 |
| gradient of tangent is 13 OR gradient of normal is \(-\dfrac{1}{8a + \frac{1}{4}b}\) or gradient of normal is \(-\dfrac{1}{2ax + \dfrac{b}{2\sqrt{x}}}\) OR \(\left(8a + \tfrac{1}{4}b\right) \times -\dfrac{1}{13} = -1\) | M1 | 1.1 |
| eg \(8a + \tfrac{1}{4}b = 13\) or \(-\dfrac{1}{8a + \frac{1}{4}b} = -\dfrac{1}{13}\) | M1 | 2.2a |
| \(8a + \tfrac{1}{4}b = 13\) oe | A1 | 1.1 |
| \(a = 2,\ b = -12\) | A1 | 1.1 |
| [8] |
Notes
B1: Substitute (4, 8) into the equation of the curve.
Seen anywhere in solution
Allow for unsimplified equation, even if error then occurs
M1: Attempt differentiation.
To obtain derivative of the form \(px + qx^{-0.5}\)
Can still be awarded if \(p\) and \(q\) now incorrect numerical values
M1: Use \(x = 4\) correctly in their derivative.
Must be an attempt at differentiation, but could still follow M0
Their derivative could now be part of an equation or an attempt at a perpendicular gradient
M1: Attempt to use the relationship between the gradients of perpendicular lines.
Attempt the gradient of the tangent, using attempt at gradient of given normal (condone \(-\tfrac{1}{13}x\) if recovered)
OR
Attempt gradient of the normal using their derivative (either algebraic or in terms of \(a\) and \(b\))
Condone slips with fractions within fractions as long as intent is clear
NB \(\left(8a + \tfrac{1}{4}b\right) \times -\dfrac{1}{13} = -1\) would imply next M1 as well
M1: Equate expressions / values for normals or for tangents, with \(x\) now substituted. Could be using gradients or equations.
Must be comparing like with like ie tangent with tangent, or normal with normal
If using equations then would need to equate expressions for either the gradients or the intercepts
A1: Obtain correct linear equation.
If using eg \(-\dfrac{1}{8a + \frac{1}{4}b} = -\dfrac{1}{13}\) then terms in the denominators must have been dealt with correctly
A1: Obtain \(a = 2,\ b = -12\)
BC so no method needed