| Exam Board | AQA |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2020 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard Integrals and Reverse Chain Rule |
| Type | Definite integral with logarithmic form |
| Difficulty | Easy -1.2 This question tests understanding of the constant of integration rather than integration technique itself. Parts (a)(i) and (a)(ii) require only recognizing the missing constant of integration, while part (b) requires knowing that ln(Ax) = ln(A) + ln(x), where ln(A) can be absorbed into the arbitrary constant. This is conceptual recall with minimal calculation, significantly easier than a typical integration problem. |
| Spec | 1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Tom's solution has no constant of integration | E1 (AO 2.4) | Accept: "Tom forgot the \(+c\)", "There is no constant on the RHS" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Although there is a constant, it is in the wrong place | E1 (AO 2.4) | Accept: constant is in wrong place; \(k\) should not be there or \(k=1\); differentiating does not give \(\frac{1}{x}\); constant has been multiplied instead of added; should be \(\ln kx\) not \(k\ln x\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\ln Ax = \ln A + \ln x\) | M1 (AO 1.1a) | Rewrites \(\ln Ax\) as \(\ln A + \ln x\); condone use of any letter for \(A\); condone use of log without specified base |
| This is equivalent as \(c = \ln A\) | R1 (AO 2.2a) | Must deduce explicitly that \(c = \ln A\), clearly demonstrating equivalence |
## Question 6(a)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Tom's solution has no constant of integration | E1 (AO 2.4) | Accept: "Tom forgot the $+c$", "There is no constant on the RHS" |
---
## Question 6(a)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Although there is a constant, it is in the wrong place | E1 (AO 2.4) | Accept: constant is in wrong place; $k$ should not be there or $k=1$; differentiating does not give $\frac{1}{x}$; constant has been multiplied instead of added; should be $\ln kx$ not $k\ln x$ |
---
## Question 6(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\ln Ax = \ln A + \ln x$ | M1 (AO 1.1a) | Rewrites $\ln Ax$ as $\ln A + \ln x$; condone use of any letter for $A$; condone use of log without specified base |
| This is equivalent as $c = \ln A$ | R1 (AO 2.2a) | Must deduce explicitly that $c = \ln A$, clearly demonstrating equivalence |
---6 Four students, Tom, Josh, Floella and Georgia are attempting to complete the indefinite integral
$$\int \frac { 1 } { x } \mathrm {~d} x \quad \text { for } x > 0$$
Each of the students' solutions is shown below:
$$\begin{array} { l l }
\text { Tom } & \int \frac { 1 } { x } \mathrm {~d} x = \ln x \\
\text { Josh } & \int \frac { 1 } { x } \mathrm {~d} x = k \ln x \\
\text { Floella } & \int \frac { 1 } { x } \mathrm {~d} x = \ln A x \\
\text { Georgia } & \int \frac { 1 } { x } \mathrm {~d} x = \ln x + c
\end{array}$$
6
\begin{enumerate}[label=(\alph*)]
\item (i) Explain what is wrong with Tom's answer.
6 (a) (ii) Explain what is wrong with Josh's answer.\\
6
\item Explain why Floella and Georgia's answers are equivalent.
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 1 2020 Q6 [4]}}