Mathematics 2006 Jamb Examination

The set of points equidistant from two parallel lines can be determined by the perpendicular distance from a point to each line.

Given the lines are of the form ax + by + k₁ = 0 and ax + by + k₂ = 0, the locus of points equidistant from these lines is another line parallel to them.

To find this line, which represents the equation of the locus, we need to find the average  k₁ and  k₂, which we shall be denoted by q

This line will have the same coefficients for x and y as the original lines.

Thus, the equation of the locus is: ax + by + q = 0

For a regular polygon, each interior angle is calculated as:$$\text{Interior angle} = \dfrac{(n - 2) \times 180^\circ}{n}$$Substituting $n = 6$:$$\dfrac{(6 - 2) \times 180}{6} = \dfrac{720}{6} = 120^\circ$$In $\triangle QSR$, which is isosceles, we have$$\angle Q = \angle S = 30^\circ$$Likewise, in $\triangle TSR$ (also isosceles),$$\angle T = \angle R = 30^\circ$$Now, considering the straight angle at $S$:$$\angle TSV + \angle VSR = 120^\circ$$$$\angle TSV + 30^\circ = 120^\circ$$$$\angle TSV = 90^\circ$$Next, in $\triangle VTS$, the sum of angles is:$$\angle VTS + \angle VST + \angle TVS = 180^\circ$$$$30^\circ + 90^\circ + \angle TVS = 180^\circ$$$$\angle TVS = 60^\circ$$

The locus equidistant from P and Q is the perpendicular bisector of PQ.
Thus, N is the midpoint of PQ, so PN = NQ.

\[ \text{In } \triangle PST, \quad \cos 60^\circ = \dfrac{ST}{8} \] \[ \Rightarrow ST = 8 \times \cos 60^\circ = 8 \times \dfrac{1}{2} = 4 \] \[ \text{Since } PQRT \text{ is a rectangle, } TR = 10 \] \[ \therefore \; SR = ST + TR = 4 + 10 = 14 \,\text{cm} \]

From the graph, the shaded region is bounded by y ≥ 0, y ≤ x, and y + x ≤ 4.
Thus, the solution set is y ≤ x, y + x ≤ 4 and y ≥ 0.

Given \(x ⊕ y = \dfrac{xy}{6}\). The identity is 6, so for any element \(a\):\[ a ⊕ e = \dfrac{ae}{6} = a \]This holds when \(e = 6\).
The inverse of 20 is \(b\) such that:\[ 20 ⊕ b = \dfrac{20b}{6} = 6 \]\[ 20b = 36 \Rightarrow b = \dfrac{36}{20} = \dfrac{9}{5} \].

Since \(p \propto \dfrac{1}{q^3}\) and \(q \propto r^2\), then \(q^3 \propto r^6\).
So, \(p \propto \dfrac{1}{r^6}\).

Compute:\[ (-5) ∗ 3 = \dfrac{(-5)^2 - 3^2}{2(-5)(3)} = \dfrac{25 - 9}{-30} = \dfrac{16}{-30} = \dfrac{-8}{15} \].

If divisor is \(2x + 1\), then root is \(x = -\tfrac{1}{2}\).
Substitute into \(kx^3 + x^2 - 5x - 2\):\[ k(-\tfrac{1}{2})^3 + (-\tfrac{1}{2})^2 - 5(-\tfrac{1}{2}) - 2 = 2 \]\[ -\tfrac{k}{8} + \tfrac{1}{4} + \tfrac{5}{2} - 2 = 2 \]\[ -\tfrac{k}{8} + \tfrac{3}{4} = 2 \]\[ -\tfrac{k}{8} = \tfrac{5}{4} \Rightarrow k = -10 \].

Simplify:\[ \dfrac{x+4}{3} - \dfrac{x-3}{2} < 4 \]LCM = 6:\[ \dfrac{2(x+4) - 3(x-3)}{6} < 4 \]\[ \dfrac{2x+8 - 3x + 9}{6} < 4 \]\[ \dfrac{-x+17}{6} < 4 \]\[ -x+17 < 24 \Rightarrow -x < 7 \Rightarrow x > -7 \].