| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2011 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Derive stationary point equation |
| Difficulty | Standard +0.8 This question requires quotient rule differentiation, algebraic manipulation to derive a fixed-point iteration formula from dy/dx = 0, and then applying the iteration. The algebraic rearrangement to isolate x in the required form is non-trivial and requires careful manipulation of a cubic equation. While the iteration itself is routine, the derivation and multi-step nature elevate this above average difficulty. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Attempt use of quotient rule | M1 | or equiv; allow numerator wrong way round but needs minus sign in numerator; for M1 condone 'minor' errors such as sign slips, absence of square in denominator, and absence of some brackets |
| Obtain \(\frac{3(x^3 - 4x^2 + 2) - (3x + 4)(3x^2 - 8x)}{(x^3 - 4x^2 + 2)^2}\) | A1 | or equiv; allow A1 if brackets absent from \(3x^2 - 8x\) term but not from both |
| Equate numerator to 0 and attempt simplification | M1 | at least as far as removing brackets, condoning sign or coeff slips; or equiv |
| Obtain \(-6x^3 + 32x + 6 = 0\) or equiv and hence \(x = \sqrt[3]{\frac{16}{3} x + 1}\) | A1 | 4 AG; necessary detail needed (i.e. at least one intermediate step) and following first derivative with correct numerator |
| (ii) Obtain correct first iterate having used initial value 2.4 | B1 | showing at least 3 dp (2.398 or 2.399 or greater accuracy 2.39861...) |
| Apply iterative process | M1 | to obtain at least 3 iterates in all; implied by plausible, converging sequence of values; having started with any initial non-negative value |
| Obtain at least 3 correct iterates from their starting point | A1 | allowing recovery after error |
| Obtain 2.398 | A1 | value required to exactly 3 dp; allow if apparently obtained by substitution of 2.4; answers only with no iterates shown gets 0/5 |
| Obtain \(-1.552\) | A1 | 5 value required to exactly 3 dp; allow if apparently obtained by substitution of 2.4; answers only with no iterates shown gets 0/5 |
**(i)** Attempt use of quotient rule | M1 | or equiv; allow numerator wrong way round but needs minus sign in numerator; for M1 condone 'minor' errors such as sign slips, absence of square in denominator, and absence of some brackets
Obtain $\frac{3(x^3 - 4x^2 + 2) - (3x + 4)(3x^2 - 8x)}{(x^3 - 4x^2 + 2)^2}$ | A1 | or equiv; allow A1 if brackets absent from $3x^2 - 8x$ term but not from both
Equate numerator to 0 and attempt simplification | M1 | at least as far as removing brackets, condoning sign or coeff slips; or equiv
Obtain $-6x^3 + 32x + 6 = 0$ or equiv and hence $x = \sqrt[3]{\frac{16}{3} x + 1}$ | A1 | 4 AG; necessary detail needed (i.e. at least one intermediate step) and following first derivative with correct numerator
**(ii)** Obtain correct first iterate having used initial value 2.4 | B1 | showing at least 3 dp (2.398 or 2.399 or greater accuracy 2.39861...)
Apply iterative process | M1 | to obtain at least 3 iterates in all; implied by plausible, converging sequence of values; having started with any initial non-negative value
Obtain at least 3 correct iterates from their starting point | A1 | allowing recovery after error
Obtain 2.398 | A1 | value required to exactly 3 dp; allow if apparently obtained by substitution of 2.4; answers only with no iterates shown gets 0/5
Obtain $-1.552$ | A1 | 5 value required to exactly 3 dp; allow if apparently obtained by substitution of 2.4; answers only with no iterates shown gets 0/5
[ 2.4 → 2.3986103 → 2.3981808 → 2.3980480 ]
---6 The curve with equation $y = \frac { 3 x + 4 } { x ^ { 3 } - 4 x ^ { 2 } + 2 }$ has a stationary point at $P$. It is given that $P$ is close to the point with coordinates $( 2.4 , - 1.6 )$.\\
(i) Find an expression for $\frac { \mathrm { d } y } { \mathrm {~d} x }$ and show that the $x$-coordinate of $P$ satisfies the equation
$$x = \sqrt [ 3 ] { \frac { 16 } { 3 } x + 1 }$$
(ii) By first using an iterative process based on the equation in part (i), find the coordinates of $P$, giving each coordinate correct to 3 decimal places.
\hfill \mbox{\textit{OCR C3 2011 Q6 [9]}}