June 2025 Paper 1 Q15
15 A curve has equation
\[y = x^2\]The point \(Q\) has coordinates (3, 2.5)
The point \(P\) on the curve which is closest to the point \(Q\) is shown on the diagram below.
(a) Show that the \(x\)-coordinate of \(P\) satisfies the equation\[2x^3 - 4x - 3 = 0\] [4 marks]
(b) The Newton–Raphson method is to be used to find an approximate solution to the equation\[2x^3 - 4x - 3 = 0\]Show that the Newton–Raphson method generates the iterative formula\[x_{n+1} = \frac{4x_n^3 + 3}{6x_n^2 - 4}\] [4 marks]
(c) Starting with \(x_0 = 3\), use the iterative formula given in part (b) to find the value of \(x_3\)
Give your answer to three decimal places.
[2 marks](d) Hence find the distance \(PQ\)
Give your answer to two decimal places.
[2 marks]| Scheme | Marks | AO |
|---|---|---|
| Obtains \(\pm\dfrac{1}{2x} = \dfrac{y - 2.5}{x - 3}\) or \(\pm 2x = \dfrac{y - 2.5}{x - 3}\) OE eg in the form of the equation of the straight line. Or Uses distance formula (may be in terms of \(x\) and \(y\)) \(\left(d^2 =\right)(x - 3)^2 + (y - 2.5)^2\) | M1 | 3.1a |
| Obtains \(-\dfrac{1}{2x} = \dfrac{y - 2.5}{x - 3}\) OE May have \(y\) replaced with \(x^2\) Or Obtains \((x - 3)^2 + \left(x^2 - 2.5\right)^2\) | A1 | 3.1a |
| Obtains \(\pm\dfrac{1}{2x} = \dfrac{x^2 - 2.5}{x - 3}\) OE Or \(\pm 2x = \dfrac{x^2 - 2.5}{x - 3}\) OE Or Differentiates their distance or distance2 expression \(\dfrac{\mathrm{d}\left(d^2\right)}{\mathrm{d}x} = 2x - 6 + 2\left(x^2 - 2.5\right) \times 2x = 4x^3 - 8x - 6\) | M1 | 3.1a |
| Completes reasoned argument to show \(2x^3 - 4x - 3 = 0\) Must see at least one line of correct intermediate working. | R1 | 2.1 |
| (4) |
Typical solution
\[\frac{\mathrm{d}y}{\mathrm{d}x} = 2x\]\[-\frac{1}{2x} = \frac{y - 2.5}{x - 3}\]\[-\frac{1}{2x} = \frac{x^2 - 2.5}{x - 3}\]\[-(x - 3) = 2x^3 - 5x\]\[2x^3 - 4x - 3 = 0\]| Scheme | Marks | AO |
|---|---|---|
| Obtains \(6x^2 - 4\) May be explicit or seen as denominator in their \(\dfrac{2x_n^3 - 4x_n - 3}{6x_n^2 - 4}\) | B1 | 1.1b |
| Forms \(x_n - \dfrac{2x_n^3 - 4x_n - 3}{\text{Their ``}6x_n^2 - 4\text{''}}\) Condone missing or inconsistent subscripts FT their \(\mathrm{f}^{\prime}(x) = ax^2 - 4\) if stated explicitly | M1 | 1.1a |
| Obtains two correct fractions eg \(\dfrac{x_n\left(6x_n^2 - 4\right)}{6x_n^2 - 4} - \dfrac{2x_n^3 - 4x_n - 3}{6x_n^2 - 4}\) Or obtains a single correct fraction with five terms in the numerator Condone missing or inconsistent subscripts | A1 | 3.1a |
| Completes reasoned argument with a single correct fraction before obtaining \(x_{n+1} = \dfrac{4x_n^3 + 3}{6x_n^2 - 4}\) Condone missing or inconsistent subscripts but \(x_{n+1} = \dfrac{4x_n^3 + 3}{6x_n^2 - 4}\) must be stated. AG | R1 | 2.1 |
| (4) |
Typical solution
\[\mathrm{f}(x) = 2x^3 - 4x - 3 = 0\]\[\mathrm{f}^{\prime}(x) = 6x^2 - 4\]\[x_{n+1} = x_n - \frac{2x_n^3 - 4x_n - 3}{6x_n^2 - 4}\]\[= \frac{x_n\left(6x_n^2 - 4\right)}{6x_n^2 - 4} - \frac{2x_n^3 - 4x_n - 3}{6x_n^2 - 4}\]\[= \frac{6x_n^3 - 4x_n - 2x_n^3 + 4x_n + 3}{6x_n^2 - 4}\]\[= \frac{4x_n^3 + 3}{6x_n^2 - 4}\]| Scheme | Marks | AO |
|---|---|---|
| Obtains 2.22 or AWRT 1.829 or AWRT 1.709 For 2.22 accept 2.220 or correct fraction \(\dfrac{111}{50}\) | M1 | 1.1a |
| Obtains AWRT 1.709 as their final answer. | A1 | 1.1b |
| (2) |
Typical solution
\[x_1 = 2.22\]\[x_2 = 1.828840\ldots\]\[x_3 = 1.709\]| Scheme | Marks | AO |
|---|---|---|
| Uses expression for distance with their \(x_3\) or a better approximation. | M1 | 3.1a |
| Deduces AWRT 1.36 | A1 | 1.1b |
| (2) | ||
| (12 marks) |