Higher June 2025 Paper 6 Q24
24
(a) Show that the equation \(x^4 + 3x^2 - 2x - 5 = 0\) has a solution between \(x = 1\) and \(x = 2\). [3]
(b) Find this solution correct to 1 decimal place.
You must show calculations to support your answer. [4]
You must show calculations to support your answer. [4]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(1^4 + 3(1)^2 - 2(1) - 5 = -3\) \(2^4 + 3(2)^2 - 2(2) - 5 = 19\) Sign change so solution between \(x\) = 1 and \(x\) = 2 | 3 | M2 for \(1^4 + 3(1)^2 - 2(1) - 5 = -3\) and \(2^4 + 3(2)^2 - 2(2) - 5 = 19\) or M1 for \(1^4 + 3(1)^2 - 2(1) - 5\) or \(2^4 + 3(2)^2 - 2(2) - 5\) soi by –3 or 19 | Accept a minimum of 1 + 3 – 2 – 5 = –3 and 16 + 12 – 4 – 5 = 19 Accept other values of x used between 1 and 2 (see table in part (b)). For full marks, we need to see correct substitutions and evaluations that produce a sign change and a correct statement. For M2, we can see correct substitutions and evaluations, or two correct answers producing a sign change with a correct statement. |
| Alternative Method 1 After \(x^4 + 3x^2 - 2x = 5\) seen M2 for \(1^4 + 3(1)^2 - 2(1) = 2\) and \(2^4 + 3(2)^2 - 2(2) = 24\) A1 for 24 > 5 and 2 < 5 so solution between \(x\) = 1 and \(x\) = 2 OR M1 for \(1^4 + 3(1)^2 - 2(1)\) or \(2^4 + 3(2)^2 - 2(2)\) soi by 2 or 24 | Examples just sufficient for third mark include:
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| Alternative Method 2 SC3 for using an iterative equation that converges to a value in the range 1.25 to 1.35 and concluding statement that 1 < 1.25 to 1.35 < 2 oe or SC2 for using an iterative equation that converges to a value in the range 1.25 to 1.35 | If within part (a) a candidate refers to their iterative equation work in part (b) then award marks for Alternative Method 2. | ||
| Answer | Marks | Part marks and guidance | |||||||||||||||||||||||||||||||||||||||||||||
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| Two correct evaluations for \(1.25 \leqslant x \leqslant 1.35\), one of which gives a positive answer and the other giving a negative answer | M3 | M2 for two correct evaluations for \(1 \lt x \lt 2\), one which gives a positive answer and the other giving a negative answer or M1 for one correct evaluation for \(1 \lt x \lt 2\) | Likely values: accept rot to 1+sf
If a candidate refers to relevant working in part (a) then award up to full marks for part (b). | ||||||||||||||||||||||||||||||||||||||||||||
| 1.3 | A1 | OR If 0 scored instead award SC1 for 1.3 with no worthwhile working Alternative Method by Iteration M1 rearranges to a correct iterative formula (converging or diverging, condone missing subscripts) M1 attempts first iteration (either a substitution seen or evaluated to at least 2dp rot) M1 continues further iteration(s) to reach \(x\) in the interval \(1.25 \leqslant x \lt 1.35\) A1 for 1.3 | |||||||||||||||||||||||||||||||||||||||||||||