OCR C2 2006 June — Question 9 11 marks

Exam BoardOCR
ModuleC2 (Core Mathematics 2)
Year2006
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicNumerical integration
TypeTrapezium rule with stated number of strips
DifficultyModerate -0.8 This is a straightforward multi-part C2 question testing routine skills: sketching an exponential curve (basic), applying the trapezium rule formula with given values (mechanical calculation), and manipulating logarithms to prove a given result. All parts are standard textbook exercises requiring recall and direct application rather than problem-solving or insight.
Spec1.06a Exponential function: a^x and e^x graphs and properties1.06d Natural logarithm: ln(x) function and properties1.06f Laws of logarithms: addition, subtraction, power rules1.09f Trapezium rule: numerical integration
9
  1. Sketch the curve \(y = \left( \frac { 1 } { 2 } \right) ^ { x }\), and state the coordinates of any point where the curve crosses an axis.
  2. Use the trapezium rule, with 4 strips of width 0.5 , to estimate the area of the region bounded by the curve \(y = \left( \frac { 1 } { 2 } \right) ^ { x }\), the axes, and the line \(x = 2\).
  3. The point \(P\) on the curve \(y = \left( \frac { 1 } { 2 } \right) ^ { x }\) has \(y\)-coordinate equal to \(\frac { 1 } { 6 }\). Prove that the \(x\)-coordinate of \(P\) may be written as $$1 + \frac { \log _ { 10 } 3 } { \log _ { 10 } 2 }$$
AnswerMarks Guidance
(i) [Graph of exponential curve, in at least first quadrant]M1, A1, B1 Attempt sketch of any exponential graph, in at least first quadrant. Correct graph – must be in both quadrants. For identification of (0, 1)
(ii) \(A = \frac{1}{2} \times 0.5 \times \{1 + 2(0.5^1 + 0.5 + 0.5^3) + 0.5^4\}\)
AnswerMarks Guidance
\(= 1.09\)B1, M1, A1, A1 State, or imply, at least three correct \(y\)-values. For correct use of trapezium rule, inc correct \(h\). For correct unsimplified expression. For the correct value 1.09, or better
(iii) \((\frac{1}{2})^x = \frac{1}{6} \Rightarrow x\log_{10}\frac{1}{2} = \log_{10}\frac{1}{6}\)
\(x = \frac{\log_{10}\frac{1}{6}}{\log_{10}\frac{1}{2}} = \frac{\log_{10}6}{\log_{10}2}\)
Hence \(= \frac{\log_{10}2 + \log_{10}3}{\log_{10}2}\)
AnswerMarks Guidance
\(= 1 + \frac{\log_{10}3}{\log_{10}2}\)M1, A1, M1, A1 For equation \((\frac{1}{2})^x = \frac{1}{6}\) and attempt logs. Obtain \(x\log(\frac{1}{2}) = \log(\frac{1}{6})\), or equivalent. For use of \(\log 6 = \log 2 + \log 3\). For showing the given answer correctly
OR
\((\frac{1}{2})^x = \frac{1}{6} \Rightarrow 2^x = 6\)
\(\Rightarrow x\log_{10}2 = \log_{10}6\)
\(x = \frac{\log_{10}6}{\log_{10}2}\)
\(= \frac{\log_{10}2 + \log_{10}3}{\log_{10}2}\)
AnswerMarks Guidance
\(= 1 + \frac{\log_{10}3}{\log_{10}2}\)M1, A1, M1, A1 For equation \(2^x = 6\) and attempt at logs. Obtain \(x\log 2 = \log 6\), or equivalent. For use of \(\log 6 = \log 2 + \log 3\). For showing the given answer correctly
OR
\((\frac{1}{2})^x = \frac{1}{6}\)
\((x - 1)\log_{10}2 = \log_{10}3\)
AnswerMarks Guidance
Hence \(x = 1 + \frac{\log_{10}3}{\log_{10}2}\)M1, A1, M1, A1 Attempt to rearrange equation to \(2^{x+1} = 3\). Obtain \(2^{x-1} = 3\). For attempt at logs. For showing the given answer correctly
OR
\(x = \frac{\log_{10}2 + \log_{10}3}{\log_{10}2}\)
\(= \frac{\log_{10}6}{\log_{10}2}\)
\(x\log_{10}2 = \log_{10}6\)
\(\log_{10}2^x = \log_{10}6\)
AnswerMarks Guidance
\(2^x = 6\)M1, A1, M1, A1 Use \(\log 2 + \log 3 = \log 6\). Obtain \(x\log 2 = \log 6\). Attempt to remove logarithms. Show \((\frac{1}{2})^x = \frac{1}{6}\) correctly
Total: 11 marks
**(i)** [Graph of exponential curve, in at least first quadrant] | M1, A1, B1 | Attempt sketch of any exponential graph, in at least first quadrant. Correct graph – must be in both quadrants. For identification of (0, 1) | 3 marks

**(ii)** $A = \frac{1}{2} \times 0.5 \times \{1 + 2(0.5^1 + 0.5 + 0.5^3) + 0.5^4\}$

$= 1.09$ | B1, M1, A1, A1 | State, or imply, at least three correct $y$-values. For correct use of trapezium rule, inc correct $h$. For correct unsimplified expression. For the correct value 1.09, or better | 4 marks

**(iii)** $(\frac{1}{2})^x = \frac{1}{6} \Rightarrow x\log_{10}\frac{1}{2} = \log_{10}\frac{1}{6}$

$x = \frac{\log_{10}\frac{1}{6}}{\log_{10}\frac{1}{2}} = \frac{\log_{10}6}{\log_{10}2}$

Hence $= \frac{\log_{10}2 + \log_{10}3}{\log_{10}2}$

$= 1 + \frac{\log_{10}3}{\log_{10}2}$ | M1, A1, M1, A1 | For equation $(\frac{1}{2})^x = \frac{1}{6}$ and attempt logs. Obtain $x\log(\frac{1}{2}) = \log(\frac{1}{6})$, or equivalent. For use of $\log 6 = \log 2 + \log 3$. For showing the given answer correctly | 4 marks

**OR**

$(\frac{1}{2})^x = \frac{1}{6} \Rightarrow 2^x = 6$

$\Rightarrow x\log_{10}2 = \log_{10}6$

$x = \frac{\log_{10}6}{\log_{10}2}$

$= \frac{\log_{10}2 + \log_{10}3}{\log_{10}2}$

$= 1 + \frac{\log_{10}3}{\log_{10}2}$ | M1, A1, M1, A1 | For equation $2^x = 6$ and attempt at logs. Obtain $x\log 2 = \log 6$, or equivalent. For use of $\log 6 = \log 2 + \log 3$. For showing the given answer correctly | 

**OR**

$(\frac{1}{2})^x = \frac{1}{6}$

$(x - 1)\log_{10}2 = \log_{10}3$

Hence $x = 1 + \frac{\log_{10}3}{\log_{10}2}$ | M1, A1, M1, A1 | Attempt to rearrange equation to $2^{x+1} = 3$. Obtain $2^{x-1} = 3$. For attempt at logs. For showing the given answer correctly |

**OR**

$x = \frac{\log_{10}2 + \log_{10}3}{\log_{10}2}$

$= \frac{\log_{10}6}{\log_{10}2}$

$x\log_{10}2 = \log_{10}6$

$\log_{10}2^x = \log_{10}6$

$2^x = 6$ | M1, A1, M1, A1 | Use $\log 2 + \log 3 = \log 6$. Obtain $x\log 2 = \log 6$. Attempt to remove logarithms. Show $(\frac{1}{2})^x = \frac{1}{6}$ correctly | 

**Total: 11 marks**
9 (i) Sketch the curve $y = \left( \frac { 1 } { 2 } \right) ^ { x }$, and state the coordinates of any point where the curve crosses an axis.\\
(ii) Use the trapezium rule, with 4 strips of width 0.5 , to estimate the area of the region bounded by the curve $y = \left( \frac { 1 } { 2 } \right) ^ { x }$, the axes, and the line $x = 2$.\\
(iii) The point $P$ on the curve $y = \left( \frac { 1 } { 2 } \right) ^ { x }$ has $y$-coordinate equal to $\frac { 1 } { 6 }$. Prove that the $x$-coordinate of $P$ may be written as

$$1 + \frac { \log _ { 10 } 3 } { \log _ { 10 } 2 }$$

\hfill \mbox{\textit{OCR C2 2006 Q9 [11]}}
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