FP1 June 2009 Q4

EdexcelOld spec10 marksDifferentiation Numerical Methods

4. Given that \(\alpha\) is the only real root of the equation \[x^3 - x^2 - 6 = 0\]

(a) show that \(2.2 \lt \alpha \lt 2.3\) (2)

(b) Taking 2.2 as a first approximation to \(\alpha\), apply the Newton-Raphson procedure once to \(\mathrm{f}(x) = x^3 - x^2 - 6\) to obtain a second approximation to \(\alpha\), giving your answer to 3 decimal places. (5)

(c) Use linear interpolation once on the interval \([2.2,\ 2.3]\) to find another approximation to \(\alpha\), giving your answer to 3 decimal places. (3)

Mark scheme (a)

Scheme	Marks
\(\mathrm{f}(2.2) = 2.2^3 - 2.2^2 - 6 \qquad (= -0.192)\) \(\mathrm{f}(2.3) = 2.3^3 - 2.3^2 - 6 \qquad (= 0.877)\)	M1
Change of sign \(\Rightarrow\) Root need numerical values correct (to 1 s.f.).	A1
	(2)

Notes

(a) M1 for attempt at f(2.2) and f(2.3)

A1 need indication that there is a change of sign – (could be −0.19<0, 0.88>0) and need conclusion. (These marks may be awarded in other parts of the question if not done in part (a))

Mark scheme (b)

Scheme	Marks
\(\mathrm{f}'(x) = 3x^2 - 2x\)	B1
\(\mathrm{f}'(2.2) = 10.12\)	B1
\(x_1 = x_0 - \dfrac{\mathrm{f}(x_0)}{\mathrm{f}'(x_0)} = 2.2 - \dfrac{-0.192}{10.12}\)	M1 A1ft
\(= 2.219\)	A1cao
	(5)

Notes

(b) B1 for seeing correct derivative (but may be implied by later correct work)

B1 for seeing 10.12 or this may be implied by later work

M1 Attempt Newton-Raphson with their values

A1ft may be implied by the following answer (but does not require an evaluation)

Final A1 must 2.219 exactly as shown. So answer of 2.21897 would get 4/5

If done twice ignore second attempt

Mark scheme (c)

Scheme	Marks
\(\dfrac{\alpha - 2.2}{\pm'0.192'} = \dfrac{2.3 - \alpha}{\pm'0.877'}\) (or equivalent such as \(\dfrac{k}{\pm'0.192'} = \dfrac{0.1 - k}{\pm'0.877'}\).)	M1
\(\alpha(0.877 + 0.192) = 2.3 \times 0.192 + 2.2 \times 0.877\) or \(k(0.877 + 0.192) = 0.1 \times 0.192\), where \(\alpha = 2.2 + k\)	A1
so \(\alpha \approx 2.218\) (2.21796…) (Allow awrt)	A1
	(3)
	[10]

Alternative

Scheme	Marks
Uses equation of line joining (2.2, −0.192) to (2.3, 0.877) and substitutes \(y = 0\)	M1
\(y + 0.192 = \dfrac{0.192 + 0.877}{0.1}(x - 2.2)\) and \(y = 0\), so \(\alpha \approx 2.218\) or awrt as before (NB Gradient = 10.69)	A1, A1

Notes

N.B. If you see \(0.192 - \alpha\) or \(0.877 - \alpha\) in the fraction then this is M0

A1 correct linear expression and definition of variable if not \(\alpha\) (may be implied by final correct answer- does not need 3 dp accuracy)

A1 for awrt 2.218

If done twice ignore second attempt