FP1 June 2011 Q4
4. \[\mathrm{f}(x) = x^2 + \frac{5}{2x} - 3x - 1, \quad x \neq 0\]
(a) Use differentiation to find \(\mathrm{f}'(x)\). (2)
The root \(\alpha\) of the equation \(\mathrm{f}(x) = 0\) lies in the interval \([0.7,\ 0.9]\).
(b) Taking 0.8 as a first approximation to \(\alpha\), apply the Newton-Raphson process once to \(\mathrm{f}(x)\) to obtain a second approximation to \(\alpha\). Give your answer to 3 decimal places. (4)
| Scheme | Marks |
|---|---|
| \(\mathrm{f}(x) = x^2 + \dfrac{5}{2x} - 3x - 1,\quad x \neq 0\) | |
| \(\mathrm{f}(x) = x^2 + \dfrac{5}{2}x^{-1} - 3x - 1\) | |
| \(\mathrm{f}'(x) = 2x - \dfrac{5}{2}x^{-2} - 3\ \{+\,0\}\) At least two of the four terms differentiated correctly. Correct differentiation. (Allow any correct unsimplified form) | M1 A1 |
| \(\left\{\mathrm{f}'(x) = 2x - \dfrac{5}{2x^2} - 3\right\}\) | |
| (2) |
| Scheme | Marks |
|---|---|
| \(\mathrm{f}(0.8) = 0.8^2 + \dfrac{5}{2(0.8)} - 3(0.8) - 1\ (= 0.365)\ \left(= \dfrac{73}{200}\right)\) A correct numerical expression for \(\mathrm{f}(0.8)\) | B1 |
| \(\mathrm{f}'(0.8) = -5.30625\ \left(= \dfrac{-849}{160}\right)\) Attempt to insert \(x = 0.8\) into their \(\mathrm{f}'(x)\). Does not require an evaluation. (If \(\mathrm{f}'(0.8)\) is incorrect for their derivative and there is no working score M0) | M1 |
| \(\alpha_2 = 0.8 - \left(\dfrac{\text{"}0.365\text{"}}{\text{"}{-5.30625}\text{"}}\right)\) Correct application of Newton-Raphson using their values. Does not require an evaluation. | M1 |
| \(= 0.868786808\ldots\) | |
| \(= 0.869\ (3\text{dp})\) 0.869 | A1 cao |
| (4) | |
| (6 marks) |
Notes
A correct answer only with no working scores no marks. N-R must be seen.
Ignore any further applications of N-R
A derivative of \(2x - 5(2x)^{-2} - 3\) is quite common and leads to \(\mathrm{f}'(0.8) = -3.353125\) and a final answer of 0.909. This would normally score M1A0B1M1M1A0 (4/6)
Similarly for a derivative of \(2x - 10x^{-2} - 3\) where the corresponding values are \(\mathrm{f}'(0.8) = -17.025\) and answer 0.821
(corrected from the printed mark scheme: the value printed as “0868786808...” is 0.868786808...)