June 2019 Paper 1 Q3
3.
\[y = \frac{5x^2 + 10x}{(x+1)^2} \qquad x \neq -1\]| Scheme | Marks | AO |
|---|---|---|
| Correct method used in attempting to differentiate \(y = \dfrac{5x^2 + 10x}{(x+1)^2}\) | M1 | 3.1a |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{(x+1)^2 \times (10x+10) - (5x^2+10x) \times 2(x+1)}{(x+1)^4}\) oe | A1 | 1.1b |
| Factorises/Cancels term in \((x+1)\) and attempts to simplify \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{(x+1) \times (10x+10) - (5x^2+10x) \times 2}{(x+1)^3} = \dfrac{A}{(x+1)^3}\) | M1 | 2.1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{10}{(x+1)^3}\) | A1 | 1.1b |
| (4) |
Notes
M1: Attempts to use a correct rule to differentiate Eg: Use of quotient (& chain) rules on \(y = \dfrac{5x^2 + 10x}{(x+1)^2}\)
Alternatively uses the product (and chain) rules on \(y = (5x^2 + 10x)(x+1)^{-2}\)
Condone slips but expect \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right) = \dfrac{(x+1)^2 \times (Ax+B) - (5x^2+10x) \times (Cx+D)}{(x+1)^4}\) \((A, B, C, D \gt 0)\) or
\(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right) = \dfrac{(x+1)^2 \times (Ax+B) - (5x^2+10x) \times (Cx+D)}{\left((x+1)^2\right)^2}\) \((A, B, C, D \gt 0)\) using the quotient rule
or \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right) = (x+1)^{-2} \times (Ax+B) + (5x^2+10x) \times C(x+1)^{-3}\) \((A, B, C \neq 0)\) using the product rule.
Condone missing brackets and slips for the M mark. For instance if they quote \(u = 5x^2 + 10\), \(v = (x+1)^2\) and don’t make the differentiation easier, they can be awarded this mark for applying the correct rule. Also allow where they quote the correct formula, give values of \(u\) and \(v\), but only have \(v\) rather than \(v^2\) the denominator.
A1: A correct (unsimplified) answer
Eg. \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right) = \dfrac{(x+1)^2 \times (10x+10) - (5x^2+10x) \times 2(x+1)}{(x+1)^4}\) or equivalent via the quotient rule.
OR \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right) = (x+1)^{-2} \times (10x+10) + (5x^2+10x) \times -2(x+1)^{-3}\) or equivalent via the product rule
M1: A valid attempt to proceed to the given form of the answer.
It is dependent upon having a quotient rule of \(\pm\dfrac{v\mathrm{d}u - u\mathrm{d}v}{v^2}\) and proceeding to \(\dfrac{A}{(x+1)^3}\)
It can also be scored on a quotient rule of \(\pm\dfrac{v\mathrm{d}u - u\mathrm{d}v}{v}\) and proceeding to \(\dfrac{A}{(x+1)}\)
You may see candidates expanding terms in the numerator. FYI \(10x^3 + 30x^2 + 30x + 10 - 10x^3 - 30x^2 - 20x\) but under this method they must reach the same expression as required by the main method.
Using the product rule expect to see a common denominator being used correctly before the above
A1: \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{10}{(x+1)^3}\) There is no requirement to see \(\dfrac{\mathrm{d}y}{\mathrm{d}x} =\) and they can recover from missing brackets/slips.
Alt via division
| Scheme | Marks | AO |
|---|---|---|
| Writes \(y = \dfrac{5x^2 + 10x}{(x+1)^2}\) in form \(y = A \pm \dfrac{B}{(x+1)^2}\) \(A, B \neq 0\) | M1 | 3.1a |
| Writes \(y = \dfrac{5x^2 + 10x}{(x+1)^2}\) in the form \(y = 5 - \dfrac{5}{(x+1)^2}\) | A1 | 1.1b |
| Uses the chain rule \(\Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{C}{(x+1)^3}\) (May be scored from \(A = 0\)) | M1 | 2.1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{10}{(x+1)^3}\) which cannot be awarded from incorrect value of \(A\) | A1 | 1.1b |
| (4) |
| Scheme | Marks | AO |
|---|---|---|
| For \(x \lt -1\) Follow through on their \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{A}{(x+1)^n},\ n = 1, 3\) | B1ft | 2.2a |
| (1) | ||
| (5 marks) |
Notes
B1ft: Score for deducing the correct answer of \(x \lt -1\) This can be scored independent of their answer to part (a). Alternatively score for a correct ft answer for their \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{A}{(x+1)^n}\) where \(A \lt 0\) and \(n = 1, 3\) award for \(x \gt -1\). So for example if \(A \gt 0\) and \(n = 1, 3 \Rightarrow x \lt -1\)
Alt via division
| Scheme | Marks | AO |
|---|---|---|
| For \(x \lt -1\) or correct follow through | B1ft | 2.2a |
| (1) |