June 2022 Paper 2 Q5
5 The binomial expansion of \((2 + 5x)^4\) is given by
\[(2 + 5x)^4 = A + 160x + Bx^2 + 1000x^3 + 625x^4\](a) Find the value of \(A\) and the value of \(B\). [2 marks]
(b) Show that\[(2 + 5x)^4 - (2 - 5x)^4 = Cx + Dx^3\]where \(C\) and \(D\) are constants to be found. [2 marks]
(c) Hence, or otherwise, find\[\int \left((2 + 5x)^4 - (2 - 5x)^4\right)\mathrm{d}x\] [2 marks]
| Scheme | Marks | AO |
|---|---|---|
| Obtains 16 Not incorrectly labelled | B1 | 1.1b |
| Obtains 600 Not incorrectly labelled | B1 | 1.1b |
| (2) |
Typical solution
\[(2 + 5x)^4 = 16 + 160x + 600x^2 + 1000x^3 + 625x^4\]\[A = 16, \quad B = 600\]| Scheme | Marks | AO |
|---|---|---|
| Obtains the expansion of \((2 - 5x)^4 = A - 160x + Bx^2 - 1000x^3 + 625x^4\) Accept \(A\) and \(B\) unsubstituted or their \(A\) and \(B\) Or Uses a valid method and obtains one of \(C = 320\) or \(D = 2000\) | M1 | 1.1a |
| Completes reasoned argument to show \((2 + 5x)^4 - (2 - 5x)^4 = 320x + 2000x^3\) Accept \(A\) and \(B\) unsubstituted or their \(A\) and \(B\) Must finish with \(320x + 2000x^3\) don’t accept just \(C = 320\) and \(D = 2000\) | R1F | 2.1 |
| (2) |
Typical solution
\[(2 + 5x)^4 - (2 - 5x)^4\]\[= 16 + 160x + 600x^2 + 1000x^3 + 625x^4 - (16 - 160x + 600x^2 - 1000x^3 + 625x^4)\]\[= 320x + 2000x^3\]| Scheme | Marks | AO |
|---|---|---|
| Integrates one term correctly Accept \(C\) and \(D\) unsubstituted or their \(C\) and \(D\) Or Uses reverse of chain rule to obtain at least one term of the form \(P(2 \pm 5x)^5\), \(P = \pm\dfrac{1}{5}\) or \(\pm\dfrac{1}{25}\) | M1 | 1.1a |
| Obtains \(\dfrac{320}{2}x^2 + \dfrac{2000}{4}x^4 + c\) FT \(C\) and \(D\) unsubstituted or their \(C\) and \(D\) Or \(\dfrac{(2 + 5x)^5}{5 \times 5} + \dfrac{(2 - 5x)^5}{5 \times 5} + c\) Condone missing \(+c\) | A1F | 1.1b |
| (2) | ||
| (6 marks) |