June 2022 Paper 1 Q6
6
| Scheme | Marks | AO |
|---|---|---|
| \(3^5 + 5 \times 3^4 \times (2x) = 243 + 810x\) | B1 | 1.1 |
| \(10 \times 3^3 \times (2x)^2\) or \(10 \times 3^2 \times (2x)^3\) | M1 | 1.1a |
| \(+ 1080x^2\) | A1 | 1.1 |
| \(+ 720x^3\) | A1 | 1.1 |
| [4] |
Notes
B1: Obtain \(243 + 810x\)
Condone \(3^5 + 810x\)
Allow terms not written as a sum eg written separately, or linked with a comma
M1: Attempt at least one further term – product of correct binomial coeff, power of 3 and attempted power of \(2x\), with powers totalling 5
Binomial coeff must be numerical; \({}^5\mathrm{C}_2\) is not yet enough
Allow BOD if brackets missing when index is applied to \(2x\), even if never recovered eg \(540x^2\) or \(180x^3\)
A1: Obtain correct third term
Coefficient simplified
Terms separate, listed or summed
A1: Obtain correct fourth term
Coefficient simplified
Could be separate term, part of a list or part of a sum
If expanding brackets then mark as above, but all 5 sets of brackets must be considered (allow irrelevant terms to be discarded)
Alternative method: expanding \(\left[3\left(1 + \frac{2}{3}x\right)\right]^5\)
| Scheme | Marks |
|---|---|
| \(243 + 810x\) or \(243\left(1 + \frac{10}{3}x\right)\) | B1 |
| \(243\left(\frac{40}{9}x^2\right)\) or \(243\left(\frac{80}{27}x^3\right)\) | M1 |
| Either 3rd or 4th term correct | A1 |
| \(243 + 810x + 1080x^2 + 720x^3\) | A1 |
B1: First two terms correct
Allow with 243 still outside the bracket
M1: Attempt one further term
Condone just 3 not \(3^5\) being used, but must be the correct binomial coeff and an attempt at the correct power of \(\frac{2}{3}x\), but allow BOD if no brackets
A1: Either 3rd or 4th term correct
Allow with 243 still outside the bracket
A1: Fully correct expansion
With the 243 now multiplied into the expansion
| Scheme | Marks | AO |
|---|---|---|
| \(x = y + 2y^2\) | B1 | 3.1a |
| \(1080(y + 2y^2)^2 + 720(y + 2y^2)^3\) | M1 | 1.1a |
| \(4320y^3 + 720y^3\) | M1 | 1.1a |
| coeff of \(y^3\) is 5040 | A1 | 1.1 |
| [4] |
Notes
B1: Identify correct substitution
Could be stated, or implied by use in their binomial expansion
M1: Attempt to use binomial from (a) with their 2 term substitution
Must substitute into at least the \(x^2\) and \(x^3\) terms from their (a)
Allow M1 if using \(2y + 4y^2\) as their substitution
M1: Attempt expansion to obtain the two relevant terms in \(y^3\)
M0 if any other \(y^3\) terms
Expect 4(their 1080) and (their 720)
Allow M1 if using \(2y + 4y^2\) as their substitution - expect 16(their 1080) and 8(their 720)
A1: Allow \(5040y^3\)
Ignore any other non-cubic terms
Alternative method 1: attempting binomial expansion of \((3 + (2y + 4y^2))^5\) or \(((3 + 2y) + 4y^2)^5\)
| Scheme | Marks |
|---|---|
| eg \(((3 + 2y) + (4y^2))\) | B1 |
| eg \((3 + 2y)^5 + 5(3 + 2y)^4(4y^2)\) | M1 |
| eg \((\ldots + 720y^3 + \ldots) + 5(\ldots 4 \cdot 3^3 \cdot 2y \ldots)(4y^2)\) \(= 720y^3 + 4320y^3\) | M1 |
| coeff of \(y^3\) is 5040 | A1 |
B1: Group into two expressions, and attempt to use them
M1: Use their groups to obtain the appropriate two elements of their binomial expansion (ie those that would give \(y^3\) terms)
M1: Expand to attempt the two \(y^3\) terms, and no others
A1: Obtain 5040
Alternative method 2: attempting to expand all 5 brackets
| Scheme | Marks |
|---|---|
| eg \((3 + 2y + 4y^2)^5 =\) \((81 + 216y + 648y^2 + 960y^3 \ldots)(3 + 2y + 4y^2)\) | M1 |
| \((216y \times 4y^2) + (648y^2 \times 2y) + (960y^3 \times 3)\) | M1 |
| \(864y^3 + 1296y^3 + 2880y^3\) | A1 |
| coeff of \(y^3\) is 5040 | A1 |
M1: Attempt to use all 5 brackets
An attempt to use all 5 is sufficient
M1: Attempt all products that would give a \(y\)-cubed term
Condone additional terms, even those that would give another \(y^3\) term
Irrelevant terms (ie powers greater than 3) may never be seen
A1: Obtain correct terms or coefficients, with no more than one incorrect
They must have attempted all of the expected \(y^3\) terms, and no more, with no more than one coefficient error
If \((3 + 2y + 4y^2)^4 \times (3 + 2y + 4y^2)\) then expect \(2880 + 1296 + 864\),
If \((3 + 2y + 4y^2)^3 \times (3 + 2y + 4y^2)^2\) then expect \(1368 + 1728 + 1512 + 432\)
If they have not yet combined like terms then this A mark can only be implied by a later correct answer or relevant correct combination of terms
A1: Obtain 5040