Mathematics 2006 Jamb Examination

Area of 6 m square room = \(6^2 = 36\).
Cost per m² = \(540 \div 36 = 15\).
Area of 9 m square room = \(9^2 = 81\).
Cost = \(81 \times 15 = 1215\).

The formula for the sum of the first n positive integers is:\[ S = \dfrac{n(n+1)}{2} \].

Start with:\[ T = 2\pi \sqrt{\dfrac{l}{g}} \]Divide both sides by \(2\pi\):\[ \dfrac{T}{2\pi} = \sqrt{\dfrac{l}{g}} \]Square both sides:\[ \dfrac{T^2}{4\pi^2} = \dfrac{l}{g} \]So, \( g = \dfrac{4\pi^2 l}{T^2} \).

Factorize:
\( y = x^2 - x - 12 = (x-4)(x+3) \).
For \(y \geq 0\), parabola opens upwards, so solution is outside the roots:
\( x \leq -3 \) or \( x \geq 4 \).

The series is geometric with first term \(a = 3\), ratio \(r = -2\).
Sum of n terms:\[ S_n = a \dfrac{r^n - 1}{r - 1} = 3 \dfrac{(-2)^n - 1}{-3} \]\[ S_n = -(((-2)^n - 1)) = 1 - (-2)^n \]Given total = \(1 - 2^8 = -255\).
So, \(1 - (-2)^n = -255 \Rightarrow (-2)^n = 256 \).
\( (-2)^n = 2^8 \) ⇒ n = 8.

Multiply: \(\begin{pmatrix}2p & 8\\3 & -5q\end{pmatrix}\begin{pmatrix}1\\2\end{pmatrix}=\begin{pmatrix}2p+16\\3-10q\end{pmatrix}\). Equating: \(2p+16=24\), \(3-10q=-17\). Solve: \(p=4, q=2\).
So the correct option is C.

Compute 2x: \(\begin{pmatrix}2 & 0 & 2\\4 & -2 & 0\\-2 & 0 & 2\end{pmatrix}\). Subtract y: \(\begin{pmatrix}3 & -1 & 0\\4 & -1 & 1\\-4 & 1 & 1\end{pmatrix}\).
So the correct option is B.

Test x=1: it satisfies. Divide: \((x-1)(x^2-x-6)\). Factor quadratic: \((x-3)(x+2)\). Roots: 1, -2, 3.
So the correct option is A.

English = 50, Health = 38. Total = 88. Percentage = (88/160)*100 = 55%.
So the correct option is A.

Complement of E is all outside E in U. Diagram IV shows that. So the correct option is D.