OCR MEI C3 2016 June — Question 2 5 marks

Exam BoardOCR MEI
ModuleC3 (Core Mathematics 3)
Year2016
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComposite & Inverse Functions
TypeSolve equation involving composites
DifficultyStandard +0.8 This question requires composing functions, understanding that fg(x) = f(g(x)) = ln(2 + e^x), then solving the equation ln(2 + e^x) = 2x. This requires recognizing to exponentiate both sides to get 2 + e^x = e^{2x}, then rearranging to a quadratic in e^x. While the individual techniques are standard C3 content, the multi-step algebraic manipulation and substitution strategy elevate this above a routine exercise, making it moderately challenging for this level.
Spec1.02v Inverse and composite functions: graphs and conditions for existence1.06d Natural logarithm: ln(x) function and properties
The functions \(f(x)\) and \(g(x)\) are defined by \(f(x) = \ln x\) and \(g(x) = 2 + e^x\), for \(x > 0\). Find the exact value of \(x\), given that \(fg(x) = 2x\). [5]
AnswerMarks Guidance
Answer: \(fg(x) = \ln(2 + e^x)\)Marks: M1 Guidance: Condone missing brackets
Answer: \(\Rightarrow \ln(2 + e^x) = 2x\)Marks: A1 Guidance: (none)
Answer: \(\Rightarrow 2 + e^x = e^{2x}\)Marks: M1 Guidance: Rearranging into a quadratic in \(e^x\); may be implied from both correct roots
Answer: \(\Rightarrow e^{2x} - e^x - 2 = [0]\)Marks: M1 Guidance: (none)
Answer: \(\Rightarrow (e^x - 2)(e^x + 1) = 0\), \(e^x = 2, -1\)Marks: A1 Guidance: \(-1\) root may be inferred from factorising
Answer: \(\Rightarrow e^x = 2, x = \ln 2\)Marks: A1 Guidance: \(x = \ln(-1)\) is A0
Total Marks: [5]
**Answer:** $fg(x) = \ln(2 + e^x)$ | **Marks:** M1 | **Guidance:** Condone missing brackets

**Answer:** $\Rightarrow \ln(2 + e^x) = 2x$ | **Marks:** A1 | **Guidance:** (none)

**Answer:** $\Rightarrow 2 + e^x = e^{2x}$ | **Marks:** M1 | **Guidance:** Rearranging into a quadratic in $e^x$; may be implied from both correct roots

**Answer:** $\Rightarrow e^{2x} - e^x - 2 = [0]$ | **Marks:** M1 | **Guidance:** (none)

**Answer:** $\Rightarrow (e^x - 2)(e^x + 1) = 0$, $e^x = 2, -1$ | **Marks:** A1 | **Guidance:** $-1$ root may be inferred from factorising

**Answer:** $\Rightarrow e^x = 2, x = \ln 2$ | **Marks:** A1 | **Guidance:** $x = \ln(-1)$ is A0

**Total Marks:** [5]

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The functions $f(x)$ and $g(x)$ are defined by $f(x) = \ln x$ and $g(x) = 2 + e^x$, for $x > 0$.

Find the exact value of $x$, given that $fg(x) = 2x$. [5]

\hfill \mbox{\textit{OCR MEI C3 2016 Q2 [5]}}
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