June 2022 Paper 1 Q4
4 Using an appropriate expansion show that, for sufficiently small values of \(x\),
\(\dfrac{1-x}{(2+x)^2} \approx \frac{1}{4} - \frac{1}{2}x + \frac{7}{16}x^2\). [4]
| Scheme | Marks | AO |
|---|---|---|
| AG \(\dfrac{1}{(2+x)^2} = \dfrac{1}{4\left(1+\frac{x}{2}\right)^2} = \left[\dfrac{1}{4}\left(1+\dfrac{x}{2}\right)^{-2}\right]\) | B1 | 2.1 |
| \(= \dfrac{1}{4}\left(1 + (-2)\left(\dfrac{x}{2}\right) + \dfrac{(-2)(-3)}{2!}\left(\dfrac{x}{2}\right)^2 + \ldots\right)\) | M1 | 2.1 |
| \(\dfrac{1-x}{(2+x)^2} \approx \dfrac{(1-x)}{4}\left(1 - x + \dfrac{3}{4}x^2\right)\) | M1 | 2.1 |
| \(\approx \frac{1}{4} - \frac{1}{2}x + \frac{7}{16}x^2\) | A1 | 2.1 |
| [4] |
Notes
B1: Dealing correctly with the 2. Need not use negative powers for this mark
M1: Allow for expanding \((1+kx)^{-2}\) even where the B mark is not awarded
M1: Attempt to multiply their expansion by the numerator
A1: Convincing argument
Note there is mathematically nothing wrong with the direct expansion
\((2+x)^{-2} = 2^{-2} - 2\times 2^{-3}x + 3\times 2^{-4}x^2\)
Award B1M1 if seen
Alternative method
| Scheme | Marks |
|---|---|
| \(\dfrac{1-x}{(2+x)^2} = \dfrac{3}{(2+x)^2} - \dfrac{1}{2+x}\) | M1 |
| \(\dfrac{3}{(2+x)^2} = \dfrac{3}{4}\left(1 + (-2)\left(\dfrac{x}{2}\right) + \dfrac{(-2)(-3)}{2!}\left(\dfrac{x}{2}\right)^2 + \ldots\right)\) | B1 |
| \(-\dfrac{1}{2+x} = -\dfrac{1}{2}\left(1 - \dfrac{x}{2} + \dfrac{x^2}{4} \ldots\right)\) | M1 |
| \(\approx \frac{1}{4} - \frac{1}{2}x + \frac{7}{16}x^2\) | A1 |
M1: Using partial fractions – allow an arithmetic slip
B1: Dealing correctly with the 2. Need not use negative powers for this mark
M1: Allow for expanding both \((1+kx)^{-2}\) and \((1+kx)^{-1}\) even where the B mark is not awarded
A1: Adding terms to complete a convincing argument