Standard +0.3 This is a straightforward integration problem requiring expansion of the product, term-by-term integration, and using an initial condition to find the constant. While it involves some algebraic manipulation (expanding $(x+2)(2x-1)^2$), it's a standard AS-level calculus exercise with no conceptual challenges beyond routine technique, making it slightly easier than average.
Question 10:
10 | dy
Expands with at least one
dx
term correct | 1.1a | M1 | dy
= (x + 2)(4x2 – 4x +1)
dx
dy
= 4x3 + 4x2 – 7x + 2
dx
4 7
y = x4 + x3 – x2 + 2x + c
3 2
900 = 1296 + 288 – 126 + 12 + c
c = –570
4 7
y = x4 + x3 – x2 + 2x – 570
3 2
Obtains 4x3 + 4x2 – 7x + 2 | 1.1b | A1
Integrates their cubic expansion
with at least one term correct | 3.1a | M1
Integrates their expansion
correctly to obtain an expression
for y
dy
FT their cubic expansion of
dx
Condone missing + c | 1.1b | A1F
Substitutes x = 6 and y = 900
into their quartic equation and
finds a value for c | 1.1a | M1
Obtains
4 7
y = x4 + x3 – x2 + 2x – 570
3 2 | 2.2a | A1
Question 10 Total | 6
Q | Marking instructions | AO | Marks | Typical solution
It is given that
$$\frac{\mathrm{d}y}{\mathrm{d}x} = (x + 2)(2x - 1)^2$$
and when $x = 6$, $y = 900$
Find $y$ in terms of $x$
[6 marks]
\hfill \mbox{\textit{AQA AS Paper 1 2024 Q10 [6]}}