Higher November 2017 Paper 6 Q18
18 The diagram below shows a 1 cm coordinate grid.
(a) Find an inequality that defines region A and another inequality that defines region B.
Region A: ................................................
Region B: ................................................ [4]
(b) Shade the region on the grid given by the inequality \(y \geqslant 6\). [2]
(c) A fourth shaded region, given by the inequality\[y \geqslant kx + 2,\]
is added to the grid.
The unshaded region now has area 23 cm².
Find the value of \(k\). [5]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(y \leqslant 2\) | 1 | If both inequalities are wrong way round, condone once (max penalty 1 mark) | |
| and | |||
| \(y \geqslant -2x + 18\) oe | 3 | B1 for [‘gradient’=] −2 soi and M1 for suitable method to find equation of line eg. \(y - 8 = (\textit{their } {-2}) \times (x - 5)\) or \(y - 2 = (\textit{their } {-2}) \times (x - 8)\) | Or M1 for \(y = \textit{their } {-2}x + c\) with a point from the line substituted in to find \(c\) For M1 allow use of an inequality symbol in place of = |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(y = 6\) shown as a solid line and correct region shaded | 2 | B1 for line drawn at \(y = 6\) OR B1 for correct squares shaded but no line | Accept dashed line for B1 |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(\dfrac{8}{5}\) oe | 5 | M1 for \(\dfrac{1}{2} \times 4 \times (8 + 6)\) soi by 28 M1 for \(\dfrac{1}{2} \times 4h = \textit{their } 28 - 23\) oe A1 for [\(h\) =] 2.5 AND M1 for [\(k\) =] 4 ÷ their 2.5 oe | ‘h’ is ‘top of triangle’ |
| Alternative method M1 for \(\dfrac{1}{2} \times 4 \times (8 + \text{‘}t\text{’})\) M1 for their \(\dfrac{1}{2} \times 4 \times (8 + \text{‘}t\text{’}) = 23\) oe A1 for [\(t\) =] 3.5 AND M1 for [\(k\) =] 4 ÷ (6 – their 3.5) oe | ‘t’ is ‘top of trapezium’ Must be a trapezium | ||