Higher June 2019 Paper 2R Q18
18 The functions f and g are defined as
\[\text{f}(x) = \dfrac{x}{4x - 3} \quad \text{and} \quad \text{g}(x) = x - 5\](a) State which value of \(x\) must be excluded from any domain of the function f. (1)
(b) Find \(\text{fg}(x)\).
Simplify your answer. (2)
Simplify your answer. (2)
(c) Express the inverse function \(\text{f}^{-1}\) in the form \(\text{f}^{-1}(x) = \ldots\) (3)
Part of the curve with equation \(y = \text{h}(x)\) is shown on the grid.
(d) Find an estimate for the gradient of the curve at the point where \(x = -0.5\)
Show your working clearly. (3)
Show your working clearly. (3)
| Scheme | Marks |
|---|---|
| \(\dfrac{3}{4}\) oe | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| \(\dfrac{x - 5}{4(x - 5) - 3}\) | M1 |
| \(\dfrac{x - 5}{4x - 23}\) | A1 |
| (2) |
Notes
A1: cao
| Scheme | Marks |
|---|---|
\(y = \dfrac{x}{4x - 3}\) or \(x = \dfrac{y}{4y - 3}\) \(y(4x - 3) = x\) or \(x(4y - 3) = y\) | M1 |
| \(4xy - 3y = x\) or \(4xy - 3x = y\) \(4xy - x = 3y\) or \(4xy - y = 3x\) \(x(4y - 1) = 3y\) or \(y(4x - 1) = 3x\) | M1 |
| \(\dfrac{3x}{4x - 1}\) oe | A1 |
| (3) |
Notes
M1: Moving the denominator to the other side of the equation
M1: Factorising the variable on one side in a correct expression
| Scheme | Marks |
|---|---|
| Tangent drawn at \(x = -0.5\) | M1 |
| (G =) 18 ÷ 3 oe | M1 |
| Working required Answer: 5 → 7 | A1 |
| (3) | |
| (9 marks) |
Notes
M1: Drawing a tangent at \(x = -0.5\)
M1: Correct method to work out the gradient of the tangent at \(x = -0.5\) or \(x = +0.5\)
A1: Dep on 1st M1
SC B1 B1 for drawing a tangent at \(x = +0.5\) and gradient = −3 → −4