OCR MEI C2 2014 June — Question 3 2 marks

Exam BoardOCR MEI
ModuleC2 (Core Mathematics 2)
Year2014
SessionJune
Marks2
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTangents, normals and gradients
TypeFind derivative of simple polynomial (integer powers)
DifficultyEasy -1.2 This is a straightforward application of the chord gradient formula to estimate a derivative. Students simply calculate (2.4-3.6)/(2.2-2) = -6, requiring only basic arithmetic and understanding that chord gradient approximates curve gradient. No conceptual depth or problem-solving required.
Spec1.07a Derivative as gradient: of tangent to curve
The points P\((2, 3.6)\) and Q\((2.2, 2.4)\) lie on the curve \(y = f(x)\). Use P and Q to estimate the gradient of the curve at the point where \(x = 2\). [2]
AnswerMarks Guidance
\(\frac{2.4 - 3.6}{2.2 - 2}\) oeM1 M1 may be embedded eg in equation of straight line
\(-6\) caoA1 B2 if unsupported; ignore subsequent work irrelevant to finding the gradient
Total: [2]
Question 4(i)
AnswerMarks Guidance
\((6, -1.5)\) oeB2 B1 for each value; allow \(x = 6, y = -1.5\); SC0 for \((6, -3)\)
Total: [2]
Question 4(ii)
AnswerMarks Guidance
\((2, -3)\)B2 B1 for each value; allow \(x = 2, y = -3\); SC0 for \((6, -3)\)
Total: [2]
$\frac{2.4 - 3.6}{2.2 - 2}$ oe | M1 | M1 may be embedded eg in equation of straight line
$-6$ cao | A1 | B2 if unsupported; ignore subsequent work irrelevant to finding the gradient
**Total: [2]**

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# Question 4(i)

$(6, -1.5)$ oe | B2 | B1 for each value; allow $x = 6, y = -1.5$; SC0 for $(6, -3)$
**Total: [2]**

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# Question 4(ii)

$(2, -3)$ | B2 | B1 for each value; allow $x = 2, y = -3$; SC0 for $(6, -3)$
**Total: [2]**

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The points P$(2, 3.6)$ and Q$(2.2, 2.4)$ lie on the curve $y = f(x)$. Use P and Q to estimate the gradient of the curve at the point where $x = 2$. [2]

\hfill \mbox{\textit{OCR MEI C2 2014 Q3 [2]}}
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Q1 3 Q2 5 Q3 2 Q4 4 Q5 3 Q6 4 Q7 5 Q8 3 Q9 3 Q10 4 Q11 13 Q12 10 Q13 13