C3 June 2009 Q7
7. The function f is defined by
\[\mathrm{f}(x) = 1 - \frac{2}{(x + 4)} + \frac{x - 8}{(x - 2)(x + 4)}, \quad x \in \mathbb{R},\ x \ne -4,\ x \ne 2\](a) Show that \(\mathrm{f}(x) = \dfrac{x - 3}{x - 2}\) (5)
The function g is defined by
\[\mathrm{g}(x) = \frac{\mathrm{e}^x - 3}{\mathrm{e}^x - 2}, \quad x \in \mathbb{R},\ x \ne \ln 2\](b) Differentiate \(\mathrm{g}(x)\) to show that \(\mathrm{g}'(x) = \dfrac{\mathrm{e}^x}{(\mathrm{e}^x - 2)^2}\) (3)
(c) Find the exact values of \(x\) for which \(\mathrm{g}'(x) = 1\) (4)
| Scheme | Marks |
|---|---|
| \(\mathrm{f}(x) = 1 - \dfrac{2}{(x + 4)} + \dfrac{x - 8}{(x - 2)(x + 4)}\) \(x \in \mathbb{R},\ x \ne -4,\ x \ne 2.\) | |
| \(\mathrm{f}(x) = \dfrac{(x - 2)(x + 4) - 2(x - 2) + x - 8}{(x - 2)(x + 4)}\) | M1 A1 |
| \(= \dfrac{x^2 + 2x - 8 - 2x + 4 + x - 8}{(x - 2)(x + 4)}\) | |
| \(= \dfrac{x^2 + x - 12}{[(x + 4)(x - 2)]}\) | A1 |
| \(= \dfrac{(x + 4)(x - 3)}{[(x + 4)(x - 2)]}\) | dM1 |
| \(= \dfrac{(x - 3)}{(x - 2)}\) | A1 cso AG |
| (5) |
Notes
M1: An attempt to combine to one fraction
A1: Correct result of combining all three fractions
A1: Simplifies to give the correct numerator. Ignore omission of denominator
dM1: An attempt to factorise the numerator.
A1 cso AG: Correct result
| Scheme | Marks |
|---|---|
| \(\mathrm{g}(x) = \dfrac{\mathrm{e}^x - 3}{\mathrm{e}^x - 2} \quad x \in \mathbb{R},\ x \ne \ln 2.\) | |
| Apply quotient rule: \(\left\{\begin{aligned} u &= \mathrm{e}^x - 3 & v &= \mathrm{e}^x - 2 \\ \frac{\mathrm{d}u}{\mathrm{d}x} &= \mathrm{e}^x & \frac{\mathrm{d}v}{\mathrm{d}x} &= \mathrm{e}^x \end{aligned}\right\}\) | |
| \(\mathrm{g}'(x) = \dfrac{\mathrm{e}^x(\mathrm{e}^x - 2) - \mathrm{e}^x(\mathrm{e}^x - 3)}{(\mathrm{e}^x - 2)^2}\) | M1 A1 |
| \(= \dfrac{\mathrm{e}^{2x} - 2\mathrm{e}^x - \mathrm{e}^{2x} + 3\mathrm{e}^x}{(\mathrm{e}^x - 2)^2}\) | |
| \(= \dfrac{\mathrm{e}^x}{(\mathrm{e}^x - 2)^2}\) | A1 AG cso |
| (3) |
Notes
M1: Applying \(\dfrac{vu' - uv'}{v^2}\)
A1: Correct differentiation
A1 AG cso: Correct result
| Scheme | Marks |
|---|---|
| \(\mathrm{g}'(x) = 1 \Rightarrow \dfrac{\mathrm{e}^x}{(\mathrm{e}^x - 2)^2} = 1\) | |
| \(\mathrm{e}^x = (\mathrm{e}^x - 2)^2\) \(\mathrm{e}^x = \mathrm{e}^{2x} - 2\mathrm{e}^x - 2\mathrm{e}^x + 4\) | M1 |
| \(\underline{\mathrm{e}^{2x} - 5\mathrm{e}^x + 4} = 0\) | A1 |
| \((\mathrm{e}^x - 4)(\mathrm{e}^x - 1) = 0\) | M1 |
| \(\mathrm{e}^x = 4\) or \(\mathrm{e}^x = 1\) | |
| \(x = \ln 4\) or \(x = 0\) | A1 |
| (4) | |
| (12 marks) |
Notes
M1: Puts their differentiated numerator equal to their denominator.
A1: \(\underline{\mathrm{e}^{2x} - 5\mathrm{e}^x + 4}\)
M1: Attempt to factorise or solve quadratic in \(\mathrm{e}^x\)
A1: both \(x = 0,\ \ln 4\)