Mathematics 2007 Jamb Examination

The two lines are \(y-5=0\) and \(y-3=0\), i.e. \(y=5\) and \(y=3\).
Any point equidistant from two parallel lines lies on the line midway between them (the midline).
The midline's y-coordinate is the average: \(y=\dfrac{5+3}{2}=4\).
Therefore the locus is \(y-4=0\).

Use the midpoint formula \(\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)\).
For the points \((1-k,-4)\) and \((2,k+1)\) the midpoint is \(\left(\dfrac{1-k+2}{2},\dfrac{-4+(k+1)}{2}\right)=\left(\dfrac{3-k}{2},\dfrac{k-3}{2}\right)\).
Set this equal to \((-k,k)\) and equate coordinates:
\(\dfrac{3-k}{2}=-k\) \(\Rightarrow 3-k=-2k\) \(\Rightarrow k=-3\).
Checking the second coordinate: \(\dfrac{k-3}{2}=k\) \(\Rightarrow k-3=2k\) \(\Rightarrow k=-3\), consistent.
Hence \(k=-3\).

For a regular polygon with \(n\) sides each exterior angle equals \(\dfrac{360^\circ}{n}\).
For an octagon \(n=8\), so each exterior angle is \(\dfrac{360^\circ}{8}=45^\circ\).
Therefore each exterior angle is \(45^\circ\).

Use the exact values \(\tan 60^\circ=\sqrt{3}\) and \(\tan 30^\circ=\dfrac{1}{\sqrt{3}}\).
Numerator: \(\sqrt{3}-\dfrac{1}{\sqrt{3}}=\dfrac{3-1}{\sqrt{3}}=\dfrac{2}{\sqrt{3}}\).
Denominator: \(\sqrt{3}+\dfrac{1}{\sqrt{3}}=\dfrac{3+1}{\sqrt{3}}=\dfrac{4}{\sqrt{3}}\).
Divide numerator by denominator: \(\dfrac{2/\sqrt{3}}{4/\sqrt{3}}=\dfrac{2}{4}=\dfrac{1}{2}\).
Therefore the value is \(\dfrac{1}{2}\).

If the side of the square is \(s\) then \(s^2=144\), so \(s=\sqrt{144}=12\).
The diagonal \(d\) of a square satisfies \(d=s\sqrt{2}\), hence \(d=12\sqrt{2}\).
Thus the diagonal length is \(12\sqrt{2}\) cm.

The general derivative is \(\dfrac{dy}{dx}=R(1+x)^{R-1}\).
None of the options matches this general form unless a specific value of \(R\) is intended. The set of options suggests \(R=2\). If \(R=2\), then \(y=(1+x)^2\) and \(\dfrac{dy}{dx}=2(1+x)=2+2x\), which is option B.
So under the (implicit) assumption \(R=2\) the derivative is \(2+2x\).

Musa's share : Lawal's share = \(600:800=3:4\).
Let their ages be \(3k\) and \(4k\). Then \(3k+4k=7k=28\) so \(k=4\).
Lawal's age is \(4k=4\times 4=16\) years.

The cumulative frequency 5\(\leq\) x \(\leq\) 12 = 10 + 12 + 10 + 8 + 6 + 8 + 10 + 8 = 72

$$\begin{pmatrix}3 & -2 \\ -7 & 5 \end{pmatrix} + 2 \begin{pmatrix}-2 & 4 \\ 3 & -1 \end{pmatrix}$$ $$\begin{pmatrix}-2 & 4 \\ 3 & -1 \end{pmatrix} = \begin{pmatrix}2 \cdot -2 & 2 \cdot 4 \\ 2 \cdot 3 & 2 \cdot -1\end{pmatrix} = \begin{pmatrix}-4 & 8 \\ 6 & -2\end{pmatrix}$$ $$\begin{pmatrix}3 & -2 \\ -7 & 5 \end{pmatrix} + \begin{pmatrix}-4 & 8 \\ 6 & -2\end{pmatrix} = \begin{pmatrix}3 + (-4) & -2 + 8 \\ -7 + 6 & 5 + (-2)\end{pmatrix}$$ $$= \begin{pmatrix}-1 & 6 \\ -1 & 3\end{pmatrix}$$  

Compute \(f(P)=3P-2I\).
First \(3P=3\begin{pmatrix}2 & 1 \\ -1 & 0 \end{pmatrix}=\begin{pmatrix}6 & 3 \\ -3 & 0 \end{pmatrix}\).
Subtract \(2I=\begin{pmatrix}2 & 0 \\ 0 & 2 \end{pmatrix}\) to get \(\begin{pmatrix}6-2 & 3-0 \\ -3-0 & 0-2 \end{pmatrix}=\begin{pmatrix}4 & 3 \\ -3 & -2 \end{pmatrix}\).
So \(f(P)=\begin{pmatrix}4 & 3 \\ -3 & -2 \end{pmatrix}\), which is option D.