OCR C3 2012 January — Question 1 3 marks

Exam BoardOCR
ModuleC3 (Core Mathematics 3)
Year2012
SessionJanuary
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStandard Integrals and Reverse Chain Rule
TypeDefinite integral with logarithmic form
DifficultyModerate -0.8 This is a straightforward application of the standard integral ∫(1/x)dx = ln|x|. Students need only recognize the form, integrate to get 2ln|x|, then evaluate at the limits: 2ln(√6) - 2ln(√2) = ln6 - ln2 = ln3. It's simpler than average C3 questions as it requires just one standard technique with no algebraic manipulation or problem-solving insight.
Spec1.06f Laws of logarithms: addition, subtraction, power rules1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)1.08d Evaluate definite integrals: between limits

1 Show that \(\int _ { \sqrt { 2 } } ^ { \sqrt { 6 } } \frac { 2 } { x } \mathrm {~d} x = \ln 3\).

Question 1:
AnswerMarks Guidance
AnswerMarks Guidance
State \(2\ln x\)B1 May be implied by immediate use of limits
Use both relevant logarithm properties correctlyM1 Either or both may be implied, e.g. \(2\ln\sqrt{6} = \ln 6\) or \(\ln 6 - \ln 2 = \ln 3\)
Obtain \(\ln 3\)A1 AG; with at least one property shown explicitly
# Question 1:

| Answer | Marks | Guidance |
|--------|-------|----------|
| State $2\ln x$ | B1 | May be implied by immediate use of limits |
| Use both relevant logarithm properties correctly | M1 | Either or both may be implied, e.g. $2\ln\sqrt{6} = \ln 6$ or $\ln 6 - \ln 2 = \ln 3$ |
| Obtain $\ln 3$ | A1 | AG; with at least one property shown explicitly |

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1 Show that $\int _ { \sqrt { 2 } } ^ { \sqrt { 6 } } \frac { 2 } { x } \mathrm {~d} x = \ln 3$.

\hfill \mbox{\textit{OCR C3 2012 Q1 [3]}}