O-LevelMathematicsAlgebra and graphs4024/21 · Oct/Nov 2019 · Q810 Marks

Rectangle ABCD has area 80 cm². Triangle PCQ is removed from one corner of the rectangle. BQ = DP = 4 cm. AB = x cm. [Figure 8 Diagram of rectangle ABCD with triangle PCQ removed]

(a)Write down an expression, in terms of x, for CP.[1]
(b)Explain why CQ = `(80/x - 4)` cm.[1]
(c)Show that the shaded area, y cm², is given by `y = 32+2x+160/x`.[3]
(d)Complete the table for `y = 32+2x+160/x`. Values are given to 1 decimal place where appropriate. x | 4 | 6 | 8 | 10 | 12 | 14 | 16 | 18 | 20 ---|---|---|---|---|---|---|---|---|--- y | 80 | 70.7 | 68 | 68 | 69.3 | 71.4 | | 76.9 | 80[1]
(e)On the grid, draw the graph of `y = 32+2x+160/x` for 4 ≤ x ≤ 20. [Figure 8e Grid for graph][3]
(f)Use your graph to find the minimum possible shaded area.[1]

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Mark scheme

Marking points from the official mark scheme (Oct/Nov 2019). The mark or mark type is shown in brackets.

[1] x-4
[1] CB = area / length = 80/x and CQ = CB – 4 oe
[3] [=] 80 - 1/2 (x-4) (80/x - 4) M1 FT their expression from (a) 80 - 1/2 (80 - 4x - 320/x + 16) M1 FT their expression from (a) of the form ax + b Correct working leading to y = 32 + 2x + 160/x AG A1
[1] 74
[3] Correct smooth curve B2FT for 8 or 9 points correctly plotted or B1FT for 6 or 7 points correctly plotted
[1] 67.4 up to but not including 68

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