OCR MEI C2 2013 January — Question 5 3 marks

Exam BoardOCR MEI
ModuleC2 (Core Mathematics 2)
Year2013
SessionJanuary
Marks3
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 (rise over run) followed by a conceptual question about approximating derivatives. The calculation is simple arithmetic with no algebraic manipulation, and recognizing that a closer x-value gives a better approximation requires only basic understanding of differentiation concepts rather than actual calculus.
Spec1.07a Derivative as gradient: of tangent to curve

A and B are points on the curve \(y = 4\sqrt{x}\). Point A has coordinates \((9, 12)\) and point B has \(x\)-coordinate \(9.5\). Find the gradient of the chord AB. The gradient of AB is an approximation to the gradient of the curve at A. State the \(x\)-coordinate of a point C on the curve such that the gradient of AC is a closer approximation. [3]

Answer: \(\text{gradient} = \frac{4\sqrt{9.5} - 12}{9.5 - 9}\); \(0.6577\) to \(0.66\); \(9 < xc < 9.5\)
AnswerMarks Guidance
Marks: M1A1 B1
Guidance: \(4\sqrt{38} - 244\sqrt{38} - 24\); or \(0.657656...isw\); allow \(8.53 \le xc < 9\)
**Answer:** $\text{gradient} = \frac{4\sqrt{9.5} - 12}{9.5 - 9}$; $0.6577$ to $0.66$; $9 < xc < 9.5$

**Marks:** M1 | A1 | B1 | [3]

**Guidance:** $4\sqrt{38} - 244\sqrt{38} - 24$; or $0.657656...isw$; allow $8.53 \le xc < 9$

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A and B are points on the curve $y = 4\sqrt{x}$. Point A has coordinates $(9, 12)$ and point B has $x$-coordinate $9.5$. Find the gradient of the chord AB.

The gradient of AB is an approximation to the gradient of the curve at A. State the $x$-coordinate of a point C on the curve such that the gradient of AC is a closer approximation. [3]

\hfill \mbox{\textit{OCR MEI C2 2013 Q5 [3]}}