Mathematics 2004 Jamb Examination

We are given that the operation \(*\) on integers is defined as:\[p * q = \sqrt{pq}\]We are to evaluate:\[4 * (8 * 32)\]Compute the inner operation\[8 * 32 = \sqrt{8 \cdot 32} = \sqrt{256} = 16\]Compute the outer operation\[4 * 16 = \sqrt{4 \cdot 16} = \sqrt{64} = \boxed{8}\]

We are given the matrix:\[A = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}\]To find the inverse of a \(2 \times 2\) matrix:\[A^{-1} = \dfrac{1}{\det(A)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}\quad \text{for} \quad A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\]Identify the elements \[a = 2, \quad b = 1, \quad c = 1, \quad d = 1\]Compute the determinant \[\det(A) = ad - bc = (2)(1) - (1)(1) = 2 - 1 = 1\]Compute the inverse \[A^{-1} = \dfrac{1}{1} \begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix}= \begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix}\]\[\boxed{\begin{pmatrix}1 & -1 \\-1 & 2\end{pmatrix}}\]

We are given the matrix:\[P = \begin{bmatrix}1 & 0 & -1 \\3 & 4 & 5 \\-1 & 0 & 1\end{bmatrix}\]We are to find the determinant \(|P|\).Expand along the first row:\[|P| = 1 \cdot \begin{vmatrix}4 & 5 \\0 & 1\end{vmatrix}- 0 \cdot \begin{vmatrix}3 & 5 \\-1 & 1\end{vmatrix}+ (-1) \cdot \begin{vmatrix}3 & 4 \\-1 & 0\end{vmatrix}\]Compute the 2×2 determinants:\[\begin{vmatrix}4 & 5 \\0 & 1\end{vmatrix} = (4)(1) - (5)(0) = 4\]\[\begin{vmatrix}3 & 4 \\-1 & 0\end{vmatrix} = (3)(0) - (4)(-1) = 0 + 4 = 4\]Now substitute back:\[|P| = 1(4) - 0 + (-1)(4) = 4 - 4 = \boxed{0}\]

The sum of interior angles of an \( n \)-sided polygon is given by: \[ \text{Sum} = (n - 2) \times 180^\circ \] For a pentagon \( (n = 5) \), we have: \[ \text{Sum} = (5 - 2) \times 180 = 3 \times 180 = 540^\circ \] Given: \[ 6x + 6y = 540 \] Divide through by 6: \[ x + y = 90 \] Solve for \( y \): \[ y = 90 - x \] \[ \boxed{y = 90 - x} \]

We are told that \(PQRSTV\) is a regular polygon inscribed in a circle.Since there are 6 letters, it is a regular hexagon:\[\text{Number of sides} = 6\]Each side is given as:\[\text{Side length} = 7\, \text{cm}\]Compute the perimeter of the hexagon:\[\text{Perimeter} = 6 \times 7 = 42\, \text{cm}\]Since the regular hexagon is inscribed in a circle, its perimeter approximates the circumference of the circle.\[\text{Circumference of the circle} = \boxed{42\, \text{cm}}\]

If \( \angle PRQ = 35^\circ \), then: \[ \angle POS = 2 \times 35^\circ = 70^\circ \] Since \( \triangle POS \) is an isosceles triangle (radii \( OP = OS \)): \[ \angle PSO = \angle SPO = \frac{180^\circ - 70^\circ}{2} = \frac{110^\circ}{2} = \boxed{55^\circ} \]

Recall that:
Area of parallelogram A = base * height
PQTU: 32 = 4 * h
h = ³²/₄
h = 8
∴ = Area of Trapezium PQRU = (¹/₂)(4 + 14) *8
= ¹/₂ * (18 * 8)
= 72 cm²

We are given: \[l = 22\,\text{cm}, \quad \theta = 3x^\circ, \quad \text{Diameter} = 14\,\text{cm} \Rightarrow r = \dfrac{14}{2} = 7\,\text{cm}\]Using the arc length formula: \[l = \dfrac{\theta}{360} \times 2\pi r\]Substitute values: \[22 = \dfrac{3x}{360} \times 2\pi \times 7 = \dfrac{42\pi x}{360}\]Multiply both sides by 360: \[22 \times 360 = 42\pi x \Rightarrow 7920 = 42\pi x\]Solve for \(x\): \[x = \dfrac{7920}{42\pi} = \dfrac{7920}{131.9469} \approx 60\]Since \(\theta = 3x = 3 \times 20 = \boxed{60^\circ}\), \[\boxed{x = 20^\circ}\]

We are given a square \( PQRS \), and we are to determine the locus of a point inside the square that is equidistant from sides \( PQ \) and \( QR \). The point lies on the angle bisector of \( \angle PQR \), since it is equidistant from the two lines meeting at point \( Q \). In a square, the angle bisector of a right angle at vertex \( Q \) lies along the diagonal from \( Q \) to \( S \) (note: in standard square labeling \( PQRS \), vertex \( Q \) is opposite \( S \)). \[ \boxed{\text{The locus is the diagonal } QS} \]

The locus of a point that is a fixed distance (e.g. 5 cm) from a line LM is:

A set of two lines, each parallel to LM, 
One on each side of LM,
Each at a distance of 5 cm from LM.

A. pair of lines on opposite sides of LM and parallel to it, each distances 5cm form LM