Mathematics 2011 Jamb Examination

We solve this equation by calculating the determinant of the \(2 \times 2\) matrices on both sides. The determinant is found by subtracting the product of the off-diagonal elements from the product of the main diagonal elements.\[(2 \times 3x) - (3 \times 5) = (4 \times 2x) - (1 \times 3)\]\[6x - 15 = 8x - 3\]Next, we group terms with \(x\) on one side and the constant values on the other:\[-15 + 3 = 8x - 6x\]\[-12 = 2x\]Dividing both sides by 2 to solve for \(x\):\[x = -6\]The value of the unknown variable \(x\) is -6.

To evaluate a \(3 \times 3\) determinant, we expand along the first row using the cofactors of each element:\[\Delta = 4 \begin{vmatrix} 3 & 1 \\ -1 & 3 \end{vmatrix} - 2 \begin{vmatrix} 2 & 1 \\ -1 & 3 \end{vmatrix} + (-1) \begin{vmatrix} 2 & 3 \\ -1 & -1 \end{vmatrix}\]Now we calculate each of the \(2 \times 2\) determinants:\[\begin{vmatrix} 3 & 1 \\ -1 & 3 \end{vmatrix} = (3 \times 3) - (-1 \times 1) = 9 + 1 = 10\]\[\begin{vmatrix} 2 & 1 \\ -1 & 3 \end{vmatrix} = (2 \times 3) - (-1 \times 1) = 6 + 1 = 7\]\[\begin{vmatrix} 2 & 3 \\ -1 & -1 \end{vmatrix} = (2 \times -1) - (-1 \times 3) = -2 + 3 = 1\]Substituting these back into the expansion formula:\[\Delta = 4(10) - 2(7) - 1(1)\]\[\Delta = 40 - 14 - 1 = 25\]The determinant of the given matrix is 25.

The inverse of a \(2 \times 2\) matrix \(A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\) is given by the formula:\[A^{-1} = \frac{1}{\det(A)} \text{adj}(A) = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}\]First, we calculate the determinant of matrix \(N\):\[\det(N) = (2 \times 4) - (3 \times 1) = 8 - 3 = 5\]Next, we find the adjugate of matrix \(N\) by swapping the elements on the main diagonal and changing the signs of the elements on the off-diagonal:\[\text{adj}(N) = \begin{pmatrix} 4 & -1 \\ -3 & 2 \end{pmatrix}\]Combining these, the inverse of matrix \(N\) is:\[N^{-1} = \frac{1}{5} \begin{pmatrix} 4 & -1 \\ -3 & 2 \end{pmatrix}\]

In any polygon with \(n\) sides, the sum of the interior angles is given by the formula:\[\text{Sum} = (n - 2) \times 180^\circ\]For a 12-sided regular polygon (a dodecagon), we substitute \(n = 12\):\[\text{Sum} = (12 - 2) \times 180^\circ = 10 \times 180^\circ = 1800^\circ\]Since the polygon is regular, all its interior angles are equal. To find the measurement of a single interior angle, we divide the total sum by the number of sides:\[\text{Interior angle} = \frac{1800^\circ}{12} = 150^\circ\]Therefore, each interior angle of the regular 12-sided polygon is 150 degrees.

We can determine the length of the chord by considering the right-angled triangle formed by the radius of the circle, the perpendicular distance from the centre to the chord, and half the chord's length.

Let \(r = 7\) cm be the radius (hypotenuse), \(d = 5\) cm be the distance from the centre, and \(x\) be half the length of the chord. Using Pythagoras' theorem:\[x^2 + d^2 = r^2\]\[x^2 + 5^2 = 7^2\]\[x^2 + 25 = 49\]\[x^2 = 49 - 25 = 24\]\[x = \sqrt{24} = \sqrt{4 \times 6} = 2\sqrt{6} \text{ cm}\]Because the perpendicular from the centre bisects the chord, the total length is twice \(x\):\[\text{Chord length} = 2 \times 2\sqrt{6} = 4\sqrt{6} \text{ cm}\]

To find the volume of water the tank can hold, we first calculate the total volume of the rectangular tank and then subtract the volume of the metal cube occupying space within it.
The volume of the rectangular tank is:\[V_{tank} = \text{length} \times \text{width} \times \text{height} = 3 \times 4 \times 5 = 60 \text{ cm}^3\]The volume of the solid metal cube is:\[V_{cube} = \text{side}^3 = 3^3 = 27 \text{ cm}^3\]The volume of water is the difference between these two:\[V_{water} = 60 - 27 = 33 \text{ cm}^3\]

In geometry, the perpendicular bisector of a line segment \(XY\) is a line that divides \(XY\) into two equal halves at a 90-degree angle.
By definition, any point lying on this bisector is at the same distance from point \(X\) as it is from point \(Y\). Therefore, it is the locus of points that are equidistant from the two fixed endpoints.

The midpoint \(M\) of two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by the formula:\[M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)\]Given the midpoint is \((5, 8)\), we set up two equations:
For the x-coordinate:\[\frac{x + 8}{2} = 5 \implies x + 8 = 10 \implies x = 2\]For the y-coordinate:\[\frac{y + 6}{2} = 8 \implies y + 6 = 16 \implies y = 10\]Thus, the coordinates of point \(P\) are \((2, 10)\).

First, we find the gradient of the given line by rewriting it in the form \(y = mx + c\):\[2y = 5x + 4 \implies y = \frac{5}{2}x + 2\]The gradient \(m_1\) is \(\frac{5}{2}\). For a line to be perpendicular, its gradient \(m_2\) must satisfy \(m_1 \times m_2 = -1\), so \(m_2 = -\frac{2}{5}\).
Using the point-slope form with the point \((4, 2)\):\[y - 2 = -\frac{2}{5}(x - 4)\]\[5(y - 2) = -2(x - 4)\]\[5y - 10 = -2x + 8\]\[5y + 2x - 18 = 0\]

Given \(\tan \theta = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{3}{4}\), we can identify the sides of a right-angled triangle as opposite = 3 and adjacent = 4.
We find the hypotenuse using Pythagoras' theorem:\[\text{hypotenuse} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5\]Now, we determine the required trigonometric ratios:\[\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{4}{5}, \quad \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{3}{5}\]The value of \(\cos \theta - \sin \theta = \frac{4}{5} - \frac{3}{5} = \frac{1}{5}\).