C3 June 2015 Q6

EdexcelOld spec8 marksLogs & Exponentials Numerical Methods

Figure 1: the curve y = 2^(x+1) - 3 and the line y = 17 - x intersecting at the point A — Figure 1

Figure 1 is a sketch showing part of the curve with equation \(y = 2^{x+1} - 3\) and part of the line with equation \(y = 17 - x\).

The curve and the line intersect at the point \(A\).

(a) Show that the \(x\) coordinate of \(A\) satisfies the equation\[x = \frac{\ln(20 - x)}{\ln 2} - 1\] (3)

(b) Use the iterative formula\[x_{n+1} = \frac{\ln(20 - x_n)}{\ln 2} - 1, \quad x_0 = 3\]to calculate the values of \(x_1\), \(x_2\) and \(x_3\), giving your answers to 3 decimal places. (3)

(c) Use your answer to part (b) to deduce the coordinates of the point \(A\), giving your answers to one decimal place. (2)

Mark scheme (a)

Scheme	Marks
\(2^{x+1} - 3 = 17 - x \Rightarrow 2^{x+1} = 20 - x\)	M1
\((x + 1)\ln 2 = \ln(20 - x) \Rightarrow x = \ldots\)	dM1
\(x = \dfrac{\ln(20 - x)}{\ln 2} - 1\)	A1*
	(3)

6.(a) Alt

Scheme	Marks
\(2^{x+1} - 3 = 17 - x \Rightarrow 2^x = \dfrac{20 - x}{2}\)	M1
\(x\ln 2 = \ln\dfrac{20 - x}{2} \Rightarrow x = \ldots\)	dM1
\(x = \dfrac{\ln(20 - x)}{\ln 2} - 1\)	A1*
	(3)

6.(a) backwards

Scheme	Marks
\(x = \dfrac{\ln(20 - x)}{\ln 2} - 1 \Rightarrow (x + 1)\ln 2 = \ln(20 - x)\)	M1
\(\Rightarrow 2^{x+1} = 20 - x\)	dM1
Hence \(y = 2^{x+1} - 3\) meets \(y = 17 - x\)	A1*
	(3)

Notes

M1: Setting equations in \(x\) equal to each other and proceeding to make \(2^{x+1}\) the subject

dM1: Take ln’s or logs of both sides, use the power law and proceed to \(x = ..\)

A1*: This is a given answer and all aspects must be correct including ln or \(\log_{\mathrm{e}}\) rather than \(\log_{10}\)
Bracketing on both \((x + 1)\) and \(\ln(20 - x)\) must be correct.
Eg \(x + 1\ln 2 = \ln(20 - x) \Rightarrow x = \dfrac{\ln(20 - x)}{\ln 2} - 1\) is A0*

Special case: Students who start from the point \(2^{x+1} = 20 - x\) can score M1 dM1A0*

Mark scheme (b)

Scheme	Marks
Sub \(x_0 = 3\) into \(x_{n+1} = \dfrac{\ln(20 - x_n)}{\ln 2} - 1, \Rightarrow x_1 = 3.087\) (awrt)	M1A1
\(x_2 = 3.080, x_3 = 3.081\) (awrt)	A1
	(3)

Notes

M1: Sub \(x_0 = 3\) into \(x_{n+1} = \dfrac{\ln(20 - x_n)}{\ln 2} - 1\) to find \(x_1 = ..\)
Accept as evidence \(x_1 = \dfrac{\ln(20 - 3)}{\ln 2} - 1\), awrt \(x_1 = 3.1\)
Allow \(x_0 = 3\) into the miscopied iterative equation \(x_1 = \dfrac{\ln(20 - 3)}{\ln 2}\) to find \(x_1 = ..\)
Note that the answer to this, 4.087, on its own without sight of \(\dfrac{\ln(20 - 3)}{\ln 2}\) is M0

A1: awrt 3 dp \(x_1 = 3.087\)

A1: awrt \(x_2 = 3.080, x_3 = 3.081\). Tolerate 3.08 for 3.080
Note that the subscripts are not important, just mark in the order seen

Mark scheme (c)

Scheme	Marks
\(A = (3.1, 13.9)\) cao	M1,A1
	(2)
	(8 marks)

Notes

Note that this appears as B1B1 on e pen. It is marked M1A1

M1: For sight of 3.1
Alternatively it can be scored for substituting their value of \(x\) or a rounded value of \(x\) from (b) into either \(2^{x+1} - 3\) or \(17 - x\) to find the \(y\) coordinate.

A1: \((3.1, 13.9)\)