C3 January 2011 Q5

Figure 1 shows a sketch of part of the curve with equation \(y = \mathrm{f}(x)\), where

\[\mathrm{f}(x) = (8 - x)\ln x, \quad x \gt 0\]

The curve cuts the \(x\)-axis at the points \(A\) and \(B\) and has a maximum turning point at \(Q\), as shown in Figure 1.

(a) Write down the coordinates of \(A\) and the coordinates of \(B\). (2)

(b) Find \(\mathrm{f}'(x)\). (3)

(d) Show that the \(x\)-coordinate of \(Q\) is the solution of\[x = \frac{8}{1 + \ln x}\] (3)

To find an approximation for the \(x\)-coordinate of \(Q\), the iteration formula

\[x_{n+1} = \frac{8}{1 + \ln x_n}\]

is used.

(e) Taking \(x_0 = 3.55\), find the values of \(x_1\), \(x_2\) and \(x_3\). Give your answers to 3 decimal places. (3)

Mark scheme (a)

Scheme	Marks
Crosses \(x\)-axis \(\Rightarrow \mathrm{f}(x) = 0 \Rightarrow (8 - x)\ln x = 0\)
Either \((8 - x) = 0\) or \(\ln x = 0 \Rightarrow x = 8, 1\)	B1
Coordinates are \(A(1, 0)\) and \(B(8, 0)\).	B1
	(2)

B1: Either one of \(\{x\} = 1\) OR \(x = \{8\}\)

B1: Both \(A(1, \{0\})\) and \(B(8, \{0\})\)

Mark scheme (b)

Scheme	Marks
Apply product rule: \(\left\{\begin{aligned} u &= (8 - x) & v &= \ln x \\ \frac{\mathrm{d}u}{\mathrm{d}x} &= -1 & \frac{\mathrm{d}v}{\mathrm{d}x} &= \frac{1}{x} \end{aligned}\right\}\)	M1
\(\mathrm{f}'(x) = -\ln x + \dfrac{8 - x}{x}\)	A1 A1
	(3)

M1: \(vu' + uv'\)

A1: Any one term correct

A1: Both terms correct

Mark scheme (c)

Scheme	Marks
\(\mathrm{f}'(3.5) = 0.032951317\ldots\) \(\mathrm{f}'(3.6) = -0.058711623\ldots\)	M1
Sign change (and as \(\mathrm{f}'(x)\) is continuous) therefore the \(x\)-coordinate of \(Q\) lies between 3.5 and 3.6.	A1
	(2)

M1: Attempts to evaluate both \(\mathrm{f}'(3.5)\) and \(\mathrm{f}'(3.6)\)

A1: both values correct to at least 1 sf, sign change and conclusion

Mark scheme (d)

Scheme	Marks
At \(Q\), \(\mathrm{f}'(x) = 0 \Rightarrow -\ln x + \dfrac{8 - x}{x} = 0\)	M1
\(\Rightarrow -\ln x + \dfrac{8}{x} - 1 = 0\)	M1
\(\Rightarrow \dfrac{8}{x} = \ln x + 1 \Rightarrow 8 = x(\ln x + 1)\)
\(\Rightarrow x = \dfrac{8}{\ln x + 1}\) (as required)	A1
	(3)

M1: Setting \(\mathrm{f}'(x) = 0\).

M1: Splitting up the numerator and proceeding to \(x =\)

A1: For correct proof. No errors seen in working.

Mark scheme (e)

Scheme	Marks
Iterative formula: \(x_{n+1} = \dfrac{8}{\ln x_n + 1}\)
\(x_1 = \dfrac{8}{\ln(3.55) + 1}\)	M1
\(x_1 = 3.528974374\ldots\) \(x_2 = 3.538246011\ldots\) \(x_3 = 3.534144722\ldots\)	A1
\(x_1 = 3.529,\ x_2 = 3.538,\ x_3 = 3.534\), to 3 dp.	A1
	(3)
	(13 marks)

M1: An attempt to substitute \(x_0 = 3.55\) into the iterative formula. Can be implied by \(x_1 = 3.528(97)\ldots\)

A1: Both \(x_1 = \text{awrt } 3.529\) and \(x_2 = \text{awrt } 3.538\)

A1: \(x_1\), \(x_2\), \(x_3\) all stated correctly to 3 dp