C3 June 2012 Q2

2. \[\mathrm{f}(x) = x^3 + 3x^2 + 4x - 12\]

(a) Show that the equation \(\mathrm{f}(x) = 0\) can be written as \[x = \sqrt{\left(\frac{4(3 - x)}{(3 + x)}\right)}, \qquad x \neq -3\] (3)

The equation \(x^3 + 3x^2 + 4x - 12 = 0\) has a single root which is between 1 and 2

(b) Use the iteration formula \[x_{n+1} = \sqrt{\left(\frac{4(3 - x_n)}{(3 + x_n)}\right)}, \ n \geqslant 0\] with \(x_0 = 1\) to find, to 2 decimal places, the value of \(x_1\), \(x_2\) and \(x_3\). (3)

The root of \(\mathrm{f}(x) = 0\) is \(\alpha\).

Mark scheme (a)

Scheme	Marks
\(x^3 + 3x^2 + 4x - 12 = 0 \Rightarrow x^3 + 3x^2 = 12 - 4x\)
\(\Rightarrow x^2(x + 3) = 12 - 4x\)	M1
\(\Rightarrow x^2 = \dfrac{12 - 4x}{(x + 3)} \Rightarrow x = \sqrt{\dfrac{4(3 - x)}{(x + 3)}}\)	dM1A1*
	(3)

Notes

M1 Moves from f(x)=0, which may be implied by subsequent working, to \(x^2(x \pm 3) = \pm 12 \pm 4x\) by separating terms and factorising in either order. No need to factorise rhs for this mark.

dM1 Divides by ‘(x+3)’ term to make \(x^2\) the subject, then takes square root. No need for rhs to be factorised at this stage

A1* CSO. This is a given solution. Do not allow sloppy algebra or notation with root on just numerator for instance. The 12-4x needs to have been factorised.

Alternative to (a) working backwards

Scheme	Marks
\(x = \sqrt{\dfrac{4(3 - x)}{(x + 3)}} \Rightarrow x^2 = \dfrac{4(3 - x)}{(x + 3)} \Rightarrow x^2(x + 3) = 4(3 - x)\)	M1
\(x^3 + 3x^2 = 12 - 4x \Rightarrow x^3 + 3x^2 + 4x - 12 = 0\)	dM1
States that this is f(x)=0	A1*
	(3)

Alternative starting with the given result and working backwards

M1 Square (both sides) and multiply by (x+3)

dM1 Expand brackets and collect terms on one side of the equation =0

A1 A statement to the effect that this is f(x)=0

Mark scheme (b)

Scheme	Marks
\(x_1 = 1.41,\quad \text{awrt } x_2 = 1.20 \quad x_3 = 1.31\)	M1A1,A1
	(3)

Notes

Note that this appears B1,B1,B1 on EPEN

M1 An attempt to substitute \(x_0 = 1\) into the iterative formula to calculate \(x_1\).
This can be awarded for the sight of \(\sqrt{\dfrac{4(3 - 1)}{(3 + 1)}}, \sqrt{\dfrac{8}{4}}, \sqrt{2}\) and even 1.4

A1 \(x_1 = 1.41\). The subscript is not important. Mark as the first value found, \(\sqrt{2}\) is A0

A1 \(x_2 = \text{awrt } 1.20 \quad x_3 = \text{awrt } 1.31\). Mark as the second and third values found. Condone 1.2 for \(x_2\)

Mark scheme (c)

Scheme	Marks
Choosing (1.2715,1.2725) or tighter containing root 1.271998323	M1
\(\mathrm{f}(1.2725) = (+)0.00827\ldots \quad \mathrm{f}(1.2715) = -0.00821\ldots\)	M1
Change of sign\(\Rightarrow \alpha = 1.272\)	A1
	(3)
	(9 marks)

Notes

Note that this appears M1A1A1 on EPEN

M1 Choosing the interval (1.2715,1.2725) or tighter containing the root 1.271998323.
Continued iteration is not allowed for this question and is M0

M1 Calculates f(1.2715) and f(1.2725), or the tighter interval with at least 1 correct to 1 sig fig rounded or truncated.
Accept f(1.2715) = -0.008 1sf rounded or truncated. Also accept f(1.2715) = -0.01 2dp
Accept f(1.2725) = (+)0.008 1sf rounded or truncated. Also accept f(1.2725) = (+)0.01 2dp

A1 Both values correct (see above),
A valid reason; Accept change of sign, or >0 <0, or f(1.2715) \(\times\)f(1.2725)<0
And a (minimal) conclusion; Accept hence root or \(\alpha\)=1.272 or QED or □

An acceptable answer to (c) with an example of a tighter interval

M1 Choosing the interval (1.2715, 1.2720). This contains the root 1.2719(98323)

M1 Calculates f(1.2715) and f(1.2720), with at least 1 correct to 1 sig fig rounded or truncated.
Accept f(1.2715) = -0.008 1sf rounded or truncated f(1.2715) = -0.01 2dp
Accept f(1.2720) = (+)0.00003 1sf rounded or f(1.2720) = (+)0.00002 truncated 1sf

A1 Both values correct (see above),
A valid reason; Accept change of sign, or >0 <0, or f(1.2715) \(\times\)f(1.2720)<0
And a (minimal) conclusion; Accept hence root or \(\alpha\)=1.272 or QED or □

\(x\)	f(\(x\))
1.2715	-0.00821362
1.2716	-0.00656564
1.2717	-0.00491752
1.2718	-0.00326927
1.2719	-0.00162088
1.2720	+0.00002765
1.2721	+0.00167631
1.2722	+0.00332511
1.2723	+0.00497405
1.2724	+0.00662312
1.2725	+0.00827233

An acceptable answer to (c) using g(x) where \(\mathrm{g}(x) = \sqrt{\dfrac{4(3 - x)}{(x + 3)}} - x\)

2^nd M1 Calculates g(1.2715) and g(1.2725), or the tighter interval with at least 1 correct to 1 sig fig rounded or truncated.
g(1.2715) = 0.0007559. Accept g(1.2715) = awrt (+)0.0008 1sf rounded or awrt 0.0007 truncated.
g(1.2725)=-0.00076105. Accept g(1.2725) = awrt -0.0008 1sf rounded or awrt -0.0007 truncated.