Higher June 2019 Paper 6 Q20
20
(a) Show that the equation \(x^4 - x^2 - 9 = 0\) has a solution between \(x = 1\) and \(x = 2\). [3]
(b) Find this solution correct to 1 decimal place.
Show your working. [4]
Show your working. [4]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(1^4 - 1^2 - 9 = -9\) \(2^4 - 2^2 - 9 = 3\) Sign change, solution between \(x = 1\) and \(x = 2\) | 3 | M2 for \(1^4 - 1^2 - 9 = -9\) and \(2^4 - 2^2 - 9 = 3\) or M1 for \(1^4 - 1^2 - 9\) or \(2^4 - 2^2 - 9\) soi by –9 or 3 Alternative method After \(x^4 - x^2 = 9\) seen M2 for \(2^4 - 2^2 = 12\) and \(1^4 - 1^2 = 0\) A1 for 12 > 9 and 0 < 9 so solution between \(x = 1\) and \(x = 2\) OR M1 for \(2^4 - 2^2\) or \(1^4 - 1^2\) soi by 12 or 0 Alternative method SC3 for using an iterative equation that converges to a value in the range 1.85 to 1.95 and concluding statement that 1 < 1.85 to 1.95 < 2 oe or SC2 for using an iterative equation that converges to a value in the range 1.85 to 1.95 Alternative method SC3 for using quadratic formula (see (b)) leading to a value in the range 1.88 to 1.89 and concluding statement that 1 < 1.88 to 1.89 < 2 oe or SC2 for using quadratic formula (see (b)) leading to a value in the range 1.88 to 1.89 | Accept other values of \(x\) used between 1 and 2 (see table in part (b)). For full marks, the two values need to produce a sign change. Examples just sufficient for third mark include: sign change –9 < 0 < 3 \(x = 1\) gives an answer < 0 and \(x = 2\) gives an > 0 Examples insufficient for third mark: so \(x\) lies between 1 and 2 If candidates refer to their working in part (b) within part (a), award marks for any of the final 2 alternative methods. |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| Two correct evaluations in the range 1.85 to 1.95, one which gives a positive value and the other giving a negative value | M3 | M2 for two correct evaluations between 1 and 2, one which gives a positive value and the other giving a negative value or M1 for one correct evaluation between 1 and 2 | Likely values: accept rot to 1 or more dp (see table below) If candidates refer to or use their working in part (a) within part (b), award up to full marks for part (b). |
| 1.9 | A1dep | Dependent on achieving at least M2 OR SC1 for 1.9 with no worthwhile working Alternative method by iteration M1 rearranges to a correct iterative formula (converging or diverging) M1 attempts first iteration (either substitution of \(1 \leqslant x \leqslant 2\) seen or found to at least 2dp rot) M1 continues further iteration(s) to reach \(x\) in the range 1.85 to 1.95 A1 for 1.9 Alternative method by quadratic formula M2 for [\(x^2 =\)] \(\dfrac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-9)}}{2(1)}\) soi by 3.54[1..] or M1 for this formula with at most two errors AND M1 for \(x = \sqrt{\textit{their } 3.54[1..]}\) soi by 1.88 to 1.89 A1 for 1.9 | Alternative iteration method notes condone missing subscripts e.g. M1 for \(x = \sqrt{\sqrt{9 + x^2}}\) and M1 for \(\sqrt{\sqrt{9 + 1^2}}\) or 1.77[8..] or 1.78 |
Table of likely values for Q20(b)
| \(x\) | \(x^4 - x^2 - 9\) |
|---|---|
| 1.1 | –8.7459 |
| 1.2 | –8.3664 |
| 1.25 | –8.12109… |
| 1.3 | –7.8339 |
| 1.4 | –7.1184 |
| 1.5* | –6.1875 |
| 1.6 | –5.0064 |
| 1.7 | –3.5379 |
| 1.75* | –2.68359… |
| 1.8 | –1.7424 |
| 1.85 | –0.70899… |
| 1.875* | –0.1560… |
| 1.9 | 0.4221 |
| 1.9375* | 1.3379… |
| 1.95 | 1.656506 |
| 2 | 3 |