C3 June 2014 Q1

Notes

M1 Applies the quotient rule to \(\mathrm{f}(x)=\dfrac{4x+1}{x-2}\) with \(u=4x+1\) and \(v=x-2\). If the rule is quoted it must be correct. It may be implied by their \(u=4x+1,v=x-2,u'=..,v'=..\) followed by \(\dfrac{vu'-uv'}{v^2}\).
If neither quoted nor implied only accept expressions of the form \(\dfrac{(x-2)\times A-(4x+1)\times B}{(x-2)^2}\ A,B>0\) allowing for a sign slip inside the brackets.
Condone missing brackets for the method mark but not the final answer mark.
Alternatively they could apply the product rule with \(u=4x+1\) and \(v=(x-2)^{-1}\). If the rule is quoted it must be correct. It may be implied by their \(u=4x+1,v=(x-2)^{-1},u'=..,v'=..\) followed by \(vu'+uv'\).
If it is neither quoted nor implied only accept expressions of the form/ or equivalent to the form \((x-2)^{-1}\times C+(4x+1)\times D(x-2)^{-2}\)
A third alternative is to use the Chain rule. For this to score there must have been some attempt to divide first to achieve \(\mathrm{f}(x)=\dfrac{4x+1}{x-2}=..+\dfrac{..}{x-2}\) before applying the chain rule to get \(\mathrm{f}^{\prime}(x)=A(x-2)^{-2}\)

A1 A correct and unsimplified form of the answer.
Accept \(\dfrac{(x-2)\times 4-(4x+1)\times 1}{(x-2)^2}\) from the quotient rule
Accept \(\dfrac{4x-8-4x-1}{(x-2)^2}\) from the quotient rule even if the brackets were missing in line 1
Accept \((x-2)^{-1}\times 4+(4x+1)\times-1(x-2)^{-2}\) or equivalent from the product rule
Accept \(9\times-1(x-2)^{-2}\) from the chain rule

A1* Proceeds to achieve the given answer \(=\dfrac{-9}{(x-2)^2}\). Accept \(-9(x-2)^{-2}\)
All aspects must be correct including the bracketing.
If they differentiated using the product rule the intermediate lines must be seen.
Eg. \((x-2)^{-1}\times 4+(4x+1)\times-1(x-2)^{-2}=\dfrac{4}{(x-2)}-\dfrac{4x+1}{(x-2)^2}=\dfrac{4(x-2)-(4x+1)}{(x-2)^2}=\dfrac{-9}{(x-2)^2}\)

Alt 1.(a)

Notes

M1 Sets \(\dfrac{-9}{(x-2)^2}=-1\) and proceeds to \(x=\ldots.\)
The minimum expectation is that they multiply by \((x-2)^2\) and then either, divide by -1 before square rooting or multiply out before solving a 3TQ equation.
A correct answer of \(x=5\) would also score this mark following \(\dfrac{-9}{(x-2)^2}=-1\) as long as no incorrect work is seen.

A1 \(x=5\)

A1 \((5,7)\) or \(x=5,\ y=7\). Ignore any reference to \(x=-1\) (and \(y=1\)). Do not accept 21/3 for 7
If there is an extra solution, \(x>2\), then withhold this final mark.