Higher June 2024 Paper 1R Q14
14
| Scheme | Marks |
|---|---|
| \((3x + 1)(2 - x) = 6x - 3x^2 + 2 - x\; (= -3x^2 + 5x + 2)\) or \((2 - x)(4 + x) = 8 + 2x - 4x - x^2\; (= -x^2 - 2x + 8)\) or \((3x + 1)(4 + x) = 12x + 3x^2 + 4 + x\; (= 3x^2 + 13x + 4)\) | M1 |
| \((-3x^2 + 5x + 2)(4 + x) = -12x^2 - 3x^3 + 20x + 5x^2 + 8 + 2x\) or \((-x^2 - 2x + 8)(3x + 1) = -3x^3 - x^2 - 6x^2 - 2x + 24x + 8\) or \((3x^2 + 13x + 4)(2 - x) = 6x^2 - 3x^3 + 26x - 13x^2 + 8 - 4x\) | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: \(-3x^3 - 7x^2 + 22x + 8\) | A1 |
| (3) |
Notes
M1: for a correct method to expand two brackets with at least 3 terms correct out of 4 terms (or 2 terms correct out of 3 terms). Do not award this mark for eg \(6x - 3x^2 + 2 - x + 8 + 2x - 4x - x^2\) or \(6x - 3x^2 + 2 - x + 4 + x\)
M1: ft dep on M1 and a quadratic for a correct method to multiply by the 3rd bracket – allow one further error
A1: oe but must be simplified
eg \(22x - 3x^3 - 7x^2 + 8\)
if no working shown then award B2 for 3 out of a maximum of 4 terms correct
14(a) ALT
| Scheme | Marks |
|---|---|
| \(24x + 6x^2 - 12x^2 - 3x^3 + 8 + 2x - 4x - x^2\) oe | M2 |
| \(-3x^3 - 7x^2 + 22x + 8\) | A1 |
| (3) |
Notes
M2: for a complete expansion with 8 terms present of which at least 4 are correct
(M1 for at least 4 correct terms from any number of terms)
A1: oe but must be simplified
eg \(22x - 3x^3 - 7x^2 + 8\)
if no working shown then award B2 for 3 out of a maximum of 4 terms correct
| Scheme | Marks |
|---|---|
\(\left(\dfrac{1}{a^6b^4}\right)^{-\frac{1}{2}}\) or \(\left(a^{-6}b^{-4}\right)^{-\frac{1}{2}}\) or \(\left(\dfrac{a^{1.5}b^{0.5}}{a^{4.5}b^{2.5}}\right)^{-1}\) or \(\left(\dfrac{a^9b^5}{a^3b}\right)^{\frac{1}{2}}\) or \(\left(\dfrac{a^{-3}b^{-1}}{a^{-9}b^{-5}}\right)^{\frac{1}{2}}\) oe | M1 |
\(\left(\dfrac{1}{a^3b^2}\right)^{-1}\) or \(\left(a^{-3}b^{-2}\right)^{-1}\) or \(\dfrac{a^{4.5}b^{2.5}}{a^{1.5}b^{0.5}}\) or \(\left(a^6b^4\right)^{\frac{1}{2}}\) or \(\left(\dfrac{a^{-1.5}b^{-0.5}}{a^{-4.5}b^{-2.5}}\right)\) oe | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: \(a^3b^2\) | A1 |
| (3) | |
| (6 marks) |
Notes
M1: for simplifying the \(a\) and the \(b\) term in the fraction
or for applying the power ½ to at least 3 out of 4 of \(a^3\), \(b\), \(a^9\), \(b^5\)
or for applying the negative power to at least 3 out of 4 of \(a^3\), \(b\), \(a^9\), \(b^5\)
M1: for two of
simplifying the \(a\) and the \(b\) term in the fraction
or for applying the power ½ to at least 3 out of 4 of \(a^3\), \(b\), \(a^9\), \(b^5\)
or for applying the negative power to at least 3 out of 4 of \(a^3\), \(b\), \(a^9\), \(b^5\)
A1: accept \(\dfrac{1}{a^{-3}b^{-2}}\)