A-LevelMathematicsNumerical methods9709/21 · Oct/Nov 2022 · Q79 Marks

The diagram shows the curve with equation y = (2lnx)/(3x+1). The curve crosses the x-axis at the point A and has a maximum point B. The shaded region is bounded by the curve and the lines x = 3 and y = 0. [Figure 7.1]

(a)Find the gradient of the curve at A.[3]
(b)Show by calculation that the x-coordinate of B lies between 3.0 and 3.1.[3]
(c)Use the trapezium rule with two intervals to find an approximation to the area of the shaded region. Give your answer correct to 2 decimal places.[3]

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Mark scheme

Marking points from the official mark scheme (Oct/Nov 2022). The mark or mark type is shown in brackets.

[M1] Differentiate using quotient rule (or product rule)
[A1] Obtain ( (3x+1)(2/x) - 6lnx ) / (3x+1)² (OE)
[A1] Substitute x=1 to obtain 2 (OE)
[M1] Equate numerator of first derivative to zero (May be implied.)
[M1] Consider sign of (2(3x+1)/x - 6lnx) for 3.0 and 3.1 (OE)
[A1] Obtain 0.074... and -0.14... or equivalents and justify conclusion (AG – necessary detail needed. 0.00075 and – 0.001275.)
[B1] Use y-values [0], 2ln2/7 or 0.1980 and 2ln3/10 or 0.2197
[M1] Use correct formula, or equivalent, with h=1
[A1] Obtain 0.31

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