FP1 June 2012 Q1
1. \[\mathrm{f}(x) = 2x^3 - 6x^2 - 7x - 4\]
(a) Show that \(\mathrm{f}(4) = 0\) (1)
(b) Use algebra to solve \(\mathrm{f}(x) = 0\) completely. (4)
| Scheme | Marks |
|---|---|
| \(\mathrm{f}(x) = 2x^3 - 6x^2 - 7x - 4\) | |
| \(\mathrm{f}(4) = \underline{128 - 96 - 28 - 4 = 0}\) \(\underline{128 - 96 - 28 - 4 = 0}\) | B1 |
| [1] |
Notes
Just \(2(4)^3 - 6(4)^2 - 7(4) - 4 = 0\) or \(2(64) - 6(16) - 7(4) - 4 = 0\) is B0But \(2(64) - 6(16) - 7(4) - 4 = 128 - 128 = 0\) or \(2(4)^3 - 6(4)^2 - 7(4) - 4 = 4 - 4 = 0\) is B1
There must be sufficient working to show that f(4) = 0
| Scheme | Marks |
|---|---|
| \(\mathrm{f}(4) = 0 \Rightarrow (x - 4)\) is a factor. | |
| \(\mathrm{f}(x) = (x - 4)(2x^2 + 2x + 1)\) M1: \((2x^2 + kx + 1)\) Uses inspection or long division or compares coefficients and \((x - 4)\) (not \((x + 4)\)) to obtain a quadratic factor of this form. A1: \((2x^2 + 2x + 1)\) cao | M1A1 |
| So, \(x = \dfrac{-2 \pm \sqrt{4 - 4(2)(1)}}{2(2)}\) \((2)\left(x^2 + x + \dfrac{1}{2}\right) = 0 \Rightarrow (2)\left(\left(x \pm \dfrac{1}{2}\right)^2 \pm k \pm \dfrac{1}{2}\right),\ k \neq 0 \Rightarrow x = \) Use of correct quadratic formula for their 3TQ or completes the square. | M1 |
| \(\Rightarrow x = \dfrac{-2 \pm \sqrt{-4}}{2(2)}\) | |
| \(\Rightarrow x = 4,\ \dfrac{-2 \pm 2\mathrm{i}}{4}\) All three roots stated somewhere in (b). Complex roots must be at least as given but apply isw if necessary. | A1 |
| [4] | |
| 5 marks |
Notes
Allow an attempt at factorisation provided the usual conditions are satisfied and proceeds as far as \(x = \ldots\)