C3 June 2014 (R) Q4

Notes

B1 \(\dfrac{\mathrm{d}x}{\mathrm{d}y}=4\sec^2 2y\tan 2y\) or equivalent such as \(\dfrac{\mathrm{d}x}{\mathrm{d}y}=4\dfrac{\sin 2y\cos 2y}{\cos^4 2y}\)
Accept \(\dfrac{\mathrm{d}x}{\mathrm{d}y}=2\sec 2y\tan 2y\times\sec 2y+2\sec 2y\tan 2y\times\sec 2y\), \(1=4\sec^2 2y\tan 2y\dfrac{\mathrm{d}y}{\mathrm{d}x}\)

M1 Uses \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{1}{\frac{\mathrm{d}x}{\mathrm{d}y}}\) to get an expression for \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) in terms of \(y\).
It may be scored following the award of the next M1 if \(\dfrac{\mathrm{d}x}{\mathrm{d}y}\) has been written in terms of \(x\).
Follow through on their expression but condone errors on the coefficient.
For example \(\dfrac{\mathrm{d}x}{\mathrm{d}y}=2\sec^2 2y\tan 2y\Rightarrow\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{1}{2\sec^2 2y\tan 2y}\) is OK as is \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{2}{\sec^2 2y\tan 2y}\)
Do not accept \(y\)'s going to \(x\)'s. So for example \(\dfrac{\mathrm{d}x}{\mathrm{d}y}=2\sec^2 2y\tan 2y\Rightarrow\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{1}{2\sec^2 2x\tan 2x}\) is M0

M1 Uses \(\tan^2 2y=\sec^2 2y-1\) and \(x=\sec^2 2y\) to get their \(\dfrac{\mathrm{d}x}{\mathrm{d}y}\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) in terms of just \(x\)
\(\dfrac{\mathrm{d}x}{\mathrm{d}y}=2\sec^2 2y\tan 2y\Rightarrow\dfrac{\mathrm{d}x}{\mathrm{d}y}=2x\sqrt{(\sec^2 2y-1)}=2x\sqrt{x-1}\) is incorrect but scores M1
\(\dfrac{\mathrm{d}x}{\mathrm{d}y}=2\sec 2y\tan 2y\Rightarrow\dfrac{\mathrm{d}x}{\mathrm{d}y}=2\sec 2y\sqrt{(\sec^2 2y-1)}=2\sqrt{x}\sqrt{x-1}\) is incorrect but scores M1
The stating and use \(1+\tan^2x=\sec^2x\) is unlikely to score this mark.
Accept \(1+\tan^2 2y=\sec^2 2y\Rightarrow 1+\tan^2 2y=x\Rightarrow\tan 2y=\sqrt{x-1}\). So \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{1}{4\sec^2 2y\tan 2y}=\dfrac{1}{4x\sqrt{x-1}}\)
Condone examples where the candidate adapts something to get the given answer
Eg. \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{1}{4\sec^2 2y\tan^2 2y}=\dfrac{1}{4\sec^2 2y\left(\sec^2 2y-1\right)}=\dfrac{1}{4x\sqrt{(x-1)}}\)

A1* Completely correct solution. This is a ‘show that’ question and it is a requirement that all elements are seen.

Notes

M1 Uses the product rule to differentiate \((x^2+x^3)\ln 2x\). If the rule is stated it must be correct. It may be implied by their \(u=..,u'=..,v=..,v'=..\) followed by \(vu'+uv'\). If the rule is neither stated nor implied only accept expressions of the form \(\ln 2x\times(ax+bx^2)+(x^2+x^3)\times\dfrac{C}{x}\)
It is acceptable to multiply out the expression to get \(x^2\ln 2x+x^3\ln 2x\) but the product rule must be applied to both terms

A1 One term correct (unsimplified). Either \((x^2+x^3)\times\dfrac{2}{2x}\) or \((2x+3x^2)\ln 2x\)
If they have multiplied out before differentiating the equivalent would be two of the four terms correct.

A1 A completely correct (unsimplified) expression \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=(x^2+x^3)\times\dfrac{2}{2x}+(2x+3x^2)\ln 2x\)

dM1 Fully substitutes \(x=\dfrac{\mathrm{e}}{2}\) (dependent on previous M mark) into their expression for \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=\ldots\) Implied by awrt 11.5

A1 \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=3\left(\tfrac{\mathrm{e}}{2}\right)+\mathrm{e}^2\) Accept equivalent simplified forms such as \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=1.5\mathrm{e}+\mathrm{e}^2,\ \dfrac{\mathrm{d}y}{\mathrm{d}x}=\mathrm{e}(1.5+\mathrm{e}),\ \dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{\mathrm{e}(2\mathrm{e}+3)}{2}\)

Notes

M1 Uses quotient rule with \(u=3\cos x,\ v=(x+1)^{\frac{1}{3}},\ u'=\pm A\sin x\) and \(v'=B(x+1)^{-\frac{2}{3}}\).
If the rule is quoted it must be correct. It may be implied by their \(u=3\cos x,\ v=(x+1)^{\frac{1}{3}},\ u'=\pm A\sin x\) and \(v'=B(x+1)^{-\frac{2}{3}}\) followed by \(\dfrac{vu'-uv'}{v^2}\)
Additionally this could be scored by using the product rule with \(u=3\cos x,\ v=(x+1)^{-\frac{1}{3}}\ u'=\pm A\sin x\) and \(v'=B(x+1)^{-\frac{4}{3}}\). If the rule is quoted it must be correct. It may be implied by their \(u=3\cos x,\ v=(x+1)^{-\frac{1}{3}}\) \(u'=\pm A\sin x\) and \(v'=B(x+1)^{-\frac{4}{3}}\) followed by \(vu'+uv'\)
If it is not quoted nor implied only accept either of the two expressions
1) Using quotient form \(\dfrac{(x+1)^{\frac{1}{3}}\times\pm A\sin x-3\cos x\times B(x+1)^{-\frac{2}{3}}}{\left((x+1)^{\frac{1}{3}}\right)^2}\) or \(\dfrac{(x+1)^{\frac{1}{3}}\times\pm A\sin x-3\cos x\times B(x+1)^{-\frac{2}{3}}}{(x+1)^{\frac{1}{9}}}\)
2) Using product form \((x+1)^{-\frac{1}{3}}\times\pm A\sin x+3\cos x\times B(x+1)^{-\frac{4}{3}}\)

A1 A correct gradient. Accept \(\mathrm{f}^{\prime}(x)=\dfrac{(x+1)^{\frac{1}{3}}(-3\sin x)-3\cos x\left(\frac{1}{3}(x+1)^{-\frac{2}{3}}\right)}{\left((x+1)^{\frac{1}{3}}\right)^2}\)
or \(\mathrm{f}^{\prime}(x)=(x+1)^{-\frac{1}{3}}\times -3\sin x+3\cos x\times -\dfrac{1}{3}(x+1)^{-\frac{4}{3}}\)

A1 \(\mathrm{f}^{\prime}(x)=\dfrac{-3(x+1)(\sin x)-\cos x}{(x+1)^{\frac{4}{3}}}\) oe. or a statement that \(\mathrm{g}(x)=-3(x+1)(\sin x)-\cos x\) oe.