Higher November 2022 Paper 5 Q21
21 The graph of \(y = \dfrac{1}{x - 2}\) is drawn on the grid for \(-2 \leqslant x \leqslant 6\).
(a) There are no values of \(x\) for which \(\dfrac{1}{x - 2} = k\).
Find the value of \(k\). [1]
(b)
(i) Use the graph to find approximate solutions to the equation \(\dfrac{1}{x - 2} = 3x - 1\).
Give your answers to 1 decimal place.
Show your working on the graph. [4]
Give your answers to 1 decimal place.
Show your working on the graph. [4]
(ii) Show algebraically that \(\dfrac{1}{x - 2} = 3x - 1\) has the same solutions as \(3x^2 - 7x + 1 = 0\). [4]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| [k = ] 0 | 1 | ||
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| (b)(i) | |||
| \(y = 3x - 1\) ruled | M2 | M1 for correct freehand or short line or for \(y = 3x - k\) ruled or \(y = ax - 1\) ruled but not \(y = -1\) | For M2 must cross curve twice Accuracy ± 1mm at (0, -1) and (1, 2) |
| 0.1 to 0.3 and 2.1 to 2.3 | A2 | A1 for each After A0, SC1 for both values correct | Only award if M2 scored previously 0.15287… , 2.1804…. |
| (b)(ii) | |||
| \(1 = (3x - 1)(x - 2)\) | M1 | Allow recovery from missing brackets for M1 or ‘ = 1’ | |
| \(3x^2 - x - 6x + 2\) | B2 | For correctly expansion of brackets B1 for 3 terms correct in expansion | For B2 accept \(3x^2 - 7x + 2\) For B1 \(-7x\) counts as two terms |
| \(3x^2 - 7x + 1 = 0\) | A1 | Dep on M1B2 with no errors or omissions | |