C3 January 2012 Q6

EdexcelOld spec12 marksDifferentiation Numerical Methods

6. \[\mathrm{f}(x) = x^2 - 3x + 2\cos\left(\tfrac{1}{2}x\right), \quad 0 \leqslant x \leqslant \pi\]

(a) Show that the equation \(\mathrm{f}(x) = 0\) has a solution in the interval \(0.8 \lt x \lt 0.9\) (2)

The curve with equation \(y = \mathrm{f}(x)\) has a minimum point \(P\).

(b) Show that the \(x\)-coordinate of \(P\) is the solution of the equation \[x = \frac{3 + \sin\left(\frac{1}{2}x\right)}{2}\] (4)

(c) Using the iteration formula \[x_{n+1} = \frac{3 + \sin\left(\frac{1}{2}x_n\right)}{2}, \quad x_0 = 2\] find the values of \(x_1\), \(x_2\) and \(x_3\), giving your answers to 3 decimal places. (3)

(d) By choosing a suitable interval, show that the \(x\)-coordinate of \(P\) is 1.9078 correct to 4 decimal places. (3)

Mark scheme (a)

Scheme	Marks
f(0.8) = 0.082, f(0.9)= -0.089	M1
Change of sign \(\Rightarrow\) root (0.8,0.9)	A1
	(2)

Notes

M1 Calculates both f(0.8) and f(0.9). Evidence of this mark could be, either, seeing both ‘x’ substitutions written out in the expression, or, one value correct to 1 sig fig, or the appearance of incorrect values of f(0.8)=awrt 0.2 or f(0.9)=awrt 0.1 from use of degrees

A1 This requires both values to be correct as well as a reason and a conclusion.
Accept f(0.8)= 0.08 truncated or rounded (2dp) or 0.1 rounded (1dp) and f(0.9)=-0.08 truncated or rounded as -0.09 (2dp) or -0.1(1dp)
Acceptable reasons are change of sign, <0 >0, +ve –ve, f(0.8)f(0.9)<0. Acceptable conclusion is hence root or □

Mark scheme (b)

Scheme	Marks
\(f'(x) = 2x - 3 - \sin\left(\dfrac{1}{2}x\right)\)	M1 A1
Sets \(f'(x) = 0 \Rightarrow \quad x = \dfrac{3 + \sin\left(\frac{1}{2}x\right)}{2}\)	M1A1*
	(4)

Notes

M1 Attempts to differentiate f(x). Seeing any of 2x,3 or \(\pm A\sin(\tfrac{1}{2}x)\) is sufficient evidence.

A1 f’(x) correct. Accept \(\dfrac{dy}{dx} = 2x - 3 - \sin\left(\frac{1}{2}x\right)\)

M1 Sets their f’(x)=0 and proceeds to x=….. You must be sure that they are setting what they think is f’(x)=0.
Accept \(2x = 3 + \sin\left(\frac{1}{2}x\right)\) going to x=..only if f’(x) =0 is stated first

A1 * \(x = \dfrac{3 + \sin\left(\frac{1}{2}x\right)}{2}\). This is a given answer so don’t accept just the sight of this answer. It is cso

Mark scheme (c)

Scheme	Marks
Sub \(x_0 = 2\) into \(x_{n+1} = \dfrac{3 + \sin\left(\frac{1}{2}x_n\right)}{2}\)	M1
\(x_1 = \text{awrt } 1.921\), \(x_2 = \text{awrt } 1.91(0)\) and \(x_3 = \text{awrt } 1.908\)	A1,A1
	(3)

Notes

M1 Substitutes \(x_0 = 2\) into \(x_{n+1} = \dfrac{3 + \sin\left(\frac{1}{2}x_n\right)}{2}\). Evidence of this mark could be awrt 1.9 or 1.5 (from degrees)

A1 \(x_1 = \text{awrt } 1.921\)

A1 \(x_2 = \text{awrt } 1.91(0)\) and \(x_3 = \text{awrt } 1.908\)

Mark scheme (d)

Scheme	Marks
[1.90775,1.90785]	M1
f’(1.90775)=-0.00016.. AND f’(1.90785)= 0.0000076..	M1
Change of sign \(\Rightarrow\) x=1.9078	A1
	(3)
	(12 marks)

Notes

Continued iteration is not acceptable for this part. Question states ‘By choosing a suitable interval…’

M1 Chooses the interval [1.90775,1.90785] or tighter containing the root= 1.907845522

M1 Calculates f’(1.90775) and f’(1.90785) or tighter with at least one correct, rounded or truncated
f’(1.90775)=-0.0001 truncated or awrt -0.0002 rounded
f’(1.90785)= 0.000007 truncated or awrt 0.000008 rounded
Accept versions of g(x)-x where \(g(x) = \dfrac{3 + \sin\left(\frac{1}{2}x\right)}{2}\).
When x= 1.90775, \(g(x) - x = 8 \times 10^{-5}\) rounded and truncated
When x= 1.90785, \(g(x) - x = -3 \times 10^{-6}\) truncated or \(= -4 \times 10^{-6}\) rounded

A1 Both values correct, rounded or truncated, a valid reason (see part a) and a minimal conclusion (see part a). Saying hence root is acceptable. There is no need to refer to the ‘turning point’.