Edexcel Paper 1 Specimen — Question 4 4 marks

Exam BoardEdexcel
ModulePaper 1 (Paper 1)
SessionSpecimen
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicLaws of Logarithms
TypeSimplify or prove logarithmic identity
DifficultyStandard +0.8 This question requires integration involving logarithms, applying logarithm laws to simplify the result, and solving an equation to find a constant. It combines multiple techniques (splitting the integrand, evaluating definite integrals, logarithm manipulation) in a non-routine way that goes beyond standard textbook exercises, though it's still within reach for well-prepared students.
Spec1.03g Parametric equations: of curves and conversion to cartesian

4. Given that \(a\) is a positive constant and $$\int _ { a } ^ { 2 a } \frac { t + 1 } { t } \mathrm {~d} t = \ln 7$$ show that \(a = \ln k\), where \(k\) is a constant to be found.

Question 4:
AnswerMarks Guidance
Answer/WorkingMark Guidance
Writes \(\int\frac{t+1}{t}\,dt = \int 1+\frac{1}{t}\,dt\) and attempts to integrateM1 Attempts to divide each term by \(t\) or multiply by \(t^{-1}\)
\(= t + \ln t \ (+c)\)M1 Integrates each term; knows \(\int\frac{1}{t}\,dt = \ln t\); \(+c\) not required
\((2a + \ln 2a) - (a + \ln a) = \ln 7\)M1 Substitutes both limits, subtracts, sets equal to \(\ln 7\)
\(a = \ln\frac{7}{2}\) with \(k = \frac{7}{2}\)A1 Proceeds to \(a=\ln\frac{7}{2}\); states \(k=\frac{7}{2}\) or exact equivalent such as \(3.5\)
## Question 4:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Writes $\int\frac{t+1}{t}\,dt = \int 1+\frac{1}{t}\,dt$ and attempts to integrate | M1 | Attempts to divide each term by $t$ or multiply by $t^{-1}$ |
| $= t + \ln t \ (+c)$ | M1 | Integrates each term; knows $\int\frac{1}{t}\,dt = \ln t$; $+c$ not required |
| $(2a + \ln 2a) - (a + \ln a) = \ln 7$ | M1 | Substitutes both limits, subtracts, sets equal to $\ln 7$ |
| $a = \ln\frac{7}{2}$ with $k = \frac{7}{2}$ | A1 | Proceeds to $a=\ln\frac{7}{2}$; states $k=\frac{7}{2}$ or exact equivalent such as $3.5$ |

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4. Given that $a$ is a positive constant and

$$\int _ { a } ^ { 2 a } \frac { t + 1 } { t } \mathrm {~d} t = \ln 7$$

show that $a = \ln k$, where $k$ is a constant to be found.\\

\hfill \mbox{\textit{Edexcel Paper 1  Q4 [4]}}