2. The first term of a GP is 2. The common ratio of the GP is also 2. The product of the last two terms of the GP is 512. Determine the number of terms in the GP

\begin{align*} ar^{n-1}ar^{n-2} =512\\ a^{2}r^{(n-1)+(n-2)}=512\\ 2^{2}\times2^{(2n-3)}=512\\ 2^{(2n-3)}=128\\ 2^{(2n-3)}=2^{7}\\ 2n-3=7\\ 2n=7+3\\ 2n=10\\ n=5\\ \end{align*}

3. The expression \(ax^{2}-30+9\) is a perfect square , where \(a\) is a constant. Find the value of \(a\)

\begin{align*} ax^{2}-30+9 \\ 4ac =b^{2}\\ 4\times a\times9=(-30)^{2}\\ 36a=900\\ a=\frac{900}{36}\\ a=25 \end{align*}

4. Make \(x\) the subject of the formula

\begin{align*} y=\frac{bx}{\sqrt{cx^{2}-a}} \end{align*} \begin{align*} y=\frac{bx}{\sqrt{cx^{2}-a}}\\ y({\sqrt{cx^{2}-a}})=bx\\ y^{2}(cx^{2}-a)=b^{2}x^{2}\\ cx^{2}y^{2}-ay^{2}=b^{2}x^{2}\\ cx^{2}y^{2}-b^{2}x^{2}=ay^{2}\\ x^{2}(cy^{2}- b^{2})=ay^{2}\\ x^{2}=\frac{ay^{2}}{(cy^{2}- b^{2})}\\ x= \sqrt{\frac{ay^{2}}{(cy^{2}- b^{2})}} \end{align*}

6. Four quantities P,Q,R and S are such that P varies directly as the square root of Q and inversely as the square of the difference of R and S. Quantity Q is increased by 44% while quantity Q nd R are each decreased by 10%

\begin{align*} P\propto\sqrt{Q}\\ P\propto\frac{1}{(R-S)^{2}}\\ P_{0}=K\frac{\sqrt{Q}}{(R-S)^{2}}\\ P_{1}=K\frac{\sqrt{Q}}{(0.9R-0.9S)^{2}}\\ P_{1}=K\frac{\sqrt{Q}}{[0.9(R-S)]^{2}}\\ P_{1}=K\frac{\sqrt{1.44Q}}{0.9^{2}(R-S)^{2}}\\ P_{1}-P_{0} =K \frac{1.2\sqrt{Q}}{0.81(R-S)^{2}}-K\frac{\sqrt{Q}}{(R-S)^{2}}\\ \% \triangle P=\frac{P_{1}-P_{0}}{P_{0}}\times 100=0.481\times100\\ =48.1 \% \end{align*}

11. A point Q is 2000nm to the west of a point P (\(40^\text{o}N ,155^\text{o}W\)). Find the longitude of Q to the nearest degree. 3mks

\begin{align*} Let \theta = longitude\\ difference\\ between P \\and \text{} Q\\ \theta \times 60\times \cos \alpha =2000nm\\ \theta \times 60 \cos 40^\text{o} =2000nm\\ \theta = 43.51^\text{o}\\ \text{ Point P has gone into the the eastern side of the globe by}\\ 198.51-180\\ =15.49\\ \text{Longitude of point P} =180-(15.49)\\ = 161.49^\text{o}E \end{align*}

10. The equation of a trigonometric wave is \(y=4sin(ax-70)^\text{o}\). The wave has a period of \(180^\text{o}\)

(a) Determine the value of \(a\)

\begin{align*} \frac{360}{a}=180\\ a=2\\ \end{align*}

(a) Deduce the phase angle of the wave

\begin{align*} \text{Phase angle }=+70\\ \end{align*}

14. The position vectors of point P,Q and R are \(OP=6i-2j+3k\), \(OQ = 12i-5j+6k\) and \(OR = 8i-3j+4k\). Show that \(P\), \(Q\) and \(R\) are collinear.

\begin{align*} \text{Think of a straight line with three points on it}\\ PQ=\left(\begin{array}{c}12\\ -5\\6 \end{array}\right)-\left(\begin{array}{c}6\\ -2\\ 3\end{array}\right)=\left(\begin{array}{c}6\\ -3\\3\end{array}\right)\\ PR=\left(\begin{array}{c}8\\ -3\\4 \end{array}\right)-\left(\begin{array}{c}6\\ -2\\3\end{array}\right)=\left(\begin{array}{c}2\\ -1\\1 \end{array}\right)\\ \text{If PQ and PR are parallel, then PQ=kPR}\\ \left(\begin{array}{c}6\\ -3\\3\end{array}\right)=k\left(\begin{array}{c}2\\ -1\\1\end{array}\right)\\ k=3 \therefore PQ=3PR\\ \text{P is common point.Points P,Q and R are collinear} \end{align*}

In a transformation, an object of area \(x cm^2 \) is mapped onto an image whose area is \(13xcm^2\).Given that the matrix of transformation is

\begin{align*} \left(\begin{array}{c}x& 7\\ x-1&3x\end{array}\right). \text{Find the possible values of \(x\). \(3mks\)} \\ Determinant = \text{Area Scale Factor}\\ 3x^2-7(x-1)=\frac{13x}{x}\\ 3x^2-7x+7=13\\ 3x^2-7x+7-13=0\\ 3x^2-7x-6=0\\ 3x^2-9x+2x-6=0\\ 3x(x-3)+2(x-3)=0\\ (3x+2)(x-3)=0\\ x=3 \\or\\ 3x=-2\\ x=\frac{-2}{3} \end{align*}