Standard +0.8 This question requires integration of a product (needing integration by parts), finding the constant of integration using the minimum point condition (requiring differentiation to verify x=1 gives the minimum), and determining the range constant c. The 8-mark allocation and 'fully justify' requirement indicate students must show the minimum occurs at x=1 and verify all conditions, making this more demanding than a routine integration question but still within standard A-level techniques.
A function f has domain \(\mathbb{R}\) and range \(\{y \in \mathbb{R} : y \geq c\}\)
The graph of \(y = f(x)\) is shown.
\includegraphics{figure_2}
The gradient of the curve at the point \((x, y)\) is given by \(\frac{dy}{dx} = (x - 1)e^x\)
Find an expression for f(x).
Fully justify your answer.
[8 marks]
Question 7:
7 | Integrates using integration by
parts | AO3.1a | M1 | y (x1)exdx
dv
u x1 1
dx
dv
ex vex
dx
y (x1)ex exdx
y (x1)ex ex c
Range eat min y e
dy
Min point when 0x1
dx
So curve passes through (1,e)
e(11)e1e1c
c 2e
fx(x2)ex 2e
Applies integration by parts formula
correctly to either of (x1)ex or
xex | AO1.1a | M1
Obtains fully correct integral,
condone missing constant. | AO1.1b | A1
Explains clearly why the minimum
y value is e with reference to the
range of the function OE | AO2.4 | E1
dy
Uses 0to find x coordinate of
dx
minimum | AO1.1a | M1
Deduces that the curve passes
through the point (1,e) | AO2.2a | A1
Uses their minimum point to find
their c | AO1.1a | M1
States the correct equation in any
correct form
fx
Condone y instead of
CAO | AO1.1b | A1
Total | 8
Q | Marking Instructions | AO | Marks | Typical Solution
A function f has domain $\mathbb{R}$ and range $\{y \in \mathbb{R} : y \geq c\}$
The graph of $y = f(x)$ is shown.
\includegraphics{figure_2}
The gradient of the curve at the point $(x, y)$ is given by $\frac{dy}{dx} = (x - 1)e^x$
Find an expression for f(x).
Fully justify your answer.
[8 marks]
\hfill \mbox{\textit{AQA Paper 2 2018 Q7 [8]}}