Find \(s'(t)\) .

In this interval, there are only two values of t for which the object is not moving. One value is \(t = \frac{\pi }{6}\) .

Find the other value.

Show that \(s'(t) > 0\) between these two values of t .

Find the distance travelled between these two values of t .

\(s'(t) = 1 - 2\cos 2t\)    A1A2     N3

Note: Award A1 for 1, A2 for \(- 2\cos 2t\) .

[3 marks]

evidence of valid approach     (M1)

e.g. setting \(s'(t) = 0\)

correct working     A1

e.g. \(2\cos 2t = 1\) , \(\cos 2t = \frac{1}{2}\)

\(2t = \frac{\pi }{3}\) , \(\frac{{5\pi }}{3}\) , \(\ldots \)     (A1)

\(t = \frac{{5\pi }}{6}\)     A1     N3 

[4 marks]

evidence of valid approach     (M1)

e.g. choosing a value in the interval \(\frac{\pi }{6} < t < \frac{{5\pi }}{6}\)

correct substitution     A1

e.g. \(s'\left( {\frac{\pi }{2}} \right) = 1 - 2\cos \pi \)

\(s'\left( {\frac{\pi }{2}} \right) = 3\)     A1

\(s'(t) > 0\)     AG     N0

[3 marks]

evidence of approach using s or integral of \(s'\)     (M1)

e.g. \(\int {s'(t){\rm{d}}t} \) ; \(s\left( {\frac{{5\pi }}{6}} \right)\) , \(s\left( {\frac{\pi }{6}} \right)\) ; \(\left[ {t - \sin 2t} \right]_{\frac{\pi }{6}}^{\frac{{5\pi }}{6}}\)

substituting values and subtracting     (M1)

e.g. \(s\left( {\frac{{5\pi }}{6}} \right) - s\left( {\frac{\pi }{6}} \right)\) , \(\left( {\frac{\pi }{6} - \frac{{\sqrt 3 }}{2}} \right) - \left( {\frac{{5\pi }}{6} - \left( { - \frac{{\sqrt 3 }}{2}} \right)} \right)\)

correct substitution     A1

e.g. \(\frac{{5\pi }}{6} - \sin \frac{{5\pi }}{3} - \left[ {\frac{\pi }{6} - \sin \frac{\pi }{3}} \right]\) , \(\left( {\frac{{5\pi }}{6} - \left( { - \frac{{\sqrt 3 }}{2}} \right)} \right) - \left( {\frac{\pi }{6} - \frac{{\sqrt 3 }}{2}} \right)\)

distance is \(\frac{{2\pi }}{3} + \sqrt 3 \)     A1A1     N3

Note: Award A1 for \(\frac{{2\pi }}{3}\) , A1 for \(\sqrt 3 \) .

[5 marks]

The derivative in part (a) was reasonably well done, but errors here often caused trouble in later parts. Candidates occasionally attempted to use the double angle identity for \(\sin 2t\) before differentiating, but they rarely were successful in then applying the product rule.

In part (b), most candidates understood that they needed to set their derivative equal to zero, but fewer were able to take the next step to solve the resulting double angle equation. Again, some candidates over-complicated the equation by using the double angle identity. Few ended up with the correct answer \(\frac{{5\pi }}{6}\) .

In part (c), many candidates knew they needed to test a value between \(\pi /6\) and their value from part (b), but fewer were able to successfully complete that calculation. Some candidates simply tested their boundary values while others unsuccessfully attempted to make use of the second derivative.

Although many candidates did not attempt part (d), those who did often demonstrated a good understanding of how to use the displacement function s or the integral of their derivative from part (a). Candidates who had made an error in part (b) often could not finish, as \(\sin (2t)\) could not be evaluated at their value without a calculator. Of those who had successfully found the other boundary of \(5\pi /6\) , a common error was giving the incorrect sign of the value of \(\sin (5\pi /3)\) . Again, this part was a good discriminator between the grade 6 and 7 candidates.