20N.2.AHL.TZ0.H_3

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

the principal axis is $\frac{5+\left(-1\right)}{2}\left(=2\right)$

so $p=2$       A1

the amplitude is $\frac{5-\left(-1\right)}{2}\left(=3\right)$

so $q=3$       A1

EITHER

one period is $2\left(-\frac{3\mathrm{\pi }}{4}-\left(-\frac{9\mathrm{\pi }}{4}\right)\right)$       (M1)

$=3\mathrm{\pi }$

$\Rightarrow \frac{2\mathrm{\pi }}{r}=3\mathrm{\pi }$

OR

Substituting a point eg $-1=2+\sin \left(-\frac{3\mathrm{\pi }}{4}r\right)$

$\sin \left(-\frac{3\mathrm{\pi }}{4}r\right)=-1\Rightarrow -\frac{3\mathrm{\pi }}{4}r=\dots -\frac{5\mathrm{\pi }}{2}, -\frac{\mathrm{\pi }}{2}, \frac{3\mathrm{\pi }}{2},\dots$

Choice of correct solution $-\frac{3\mathrm{\pi }}{4}r=-\frac{\mathrm{\pi }}{2}$       (M1)

THEN

$\Rightarrow r=\frac{2}{3}$       A1

$\left(\Rightarrow y=2+3 \sin \left(\frac{2x}{3}\right)\right)$

Note: $q$ and $r$ can be both given as negatives for full marks

[4 marks]

roots are $x=-1.09459\dots , x=-3.617797\dots$       (A1)

${\int }_{-3.617797\dots }^{-1.09459\dots }\left(2+3 \sin \left(\frac{2x}{3}\right)\right)\text{d}x$       (M1)

$=-1.66\left(=-1.66179\dots \right)$       (A1)

so area $=1.66 \left({\text{units}}^2\right)$       A1

[4 marks]

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