Mathematics 2012 Jamb Examination

To find the value of x, we equate the determinants of the two 2x2 matrices. The determinant of a 2x2 matrix is calculated by finding the difference between the product of the elements in the leading diagonal and the product of the elements in the secondary diagonal.
For the left-hand side:\[\begin{vmatrix} 5 & x \\ 3 & 2 \end{vmatrix} = (5 \times 2) - (3 \times x) = 10 - 3x\]For the right-hand side:\[\begin{vmatrix} 3 & 4 \\ 5 & 5 \end{vmatrix} = (3 \times 5) - (5 \times 4) = 15 - 20 = -5\]Equating both sides results in the linear equation:\[10 - 3x = -5\]Subtracting 10 from both sides gives:\[-3x = -5 - 10\]\[-3x = -15\]Dividing both sides by -3 yields:\[x = \frac{-15}{-3}\]\[x = 5\]

A unit matrix (or identity matrix) of order 3, denoted as \(I_3\), is defined as a square matrix where all the elements of the principal diagonal are ones and all other elements are zeros. It is represented as:\[I_3 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\]The determinant of any identity matrix, regardless of its dimension, is always equal to 1. This can be verified by expanding along the first row:\[|I_3| = 1\begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} - 0\begin{vmatrix} 0 & 0 \\ 0 & 1 \end{vmatrix} + 0\begin{vmatrix} 0 & 1 \\ 0 & 0 \end{vmatrix}\]\[|I_3| = 1(1 - 0) - 0 + 0 = 1\]

The problem describes a polygon with five distinct interior angles, which identifies it as a pentagon (n = 5). The sum of the interior angles of any polygon is calculated using the formula:\[\text{Sum} = (n - 2) \times 180^\circ\]Substituting \(n = 5\) into the formula gives:\[\text{Sum} = (5 - 2) \times 180^\circ = 3 \times 180^\circ = 540^\circ\]Summing the given angles in terms of x results in:\[x + 2x + 3x + 4x + 5x = 15x\]Setting the sum of the variables equal to the total interior sum:\[15x = 540^\circ\]\[x = \frac{540^\circ}{15}\]\[x = 36^\circ\]

The volume of a cuboid is determined by the product of its three dimensions: length (l), breadth (b), and height (h). The formula used is:\[V = l \times b \times h\]Substituting the provided measurements into the formula:\[V = 0.76 \times 2.6 \times 0.82\]Performing the multiplication step-by-step:\[0.76 \times 2.6 = 1.976\]\[1.976 \times 0.82 = 1.62032\]Approximating to two decimal places, the volume is 1.62 \(\text{cm}^3\).

The gradient (slope) of a line passing through two points \(P(x_1, y_1)\) and \(Q(x_2, y_2)\) is defined as the change in the y-coordinate divided by the change in the x-coordinate. The formula is:\[m = \frac{y_2 - y_1}{x_2 - x_1}\]Identifying the coordinates from the points P(5, -7) and Q(-2, -3):\[x_1 = 5, \; y_1 = -7\]\[x_2 = -2, \; y_2 = -3\]Substituting these values into the gradient formula:\[m = \frac{-3 - (-7)}{-2 - 5}\]\[m = \frac{-3 + 7}{-7}\]\[m = \frac{4}{-7} = -\frac{4}{7}\]

First, we determine the coordinates of the point of intersection by solving the simultaneous equations \(y = 2x + 4\) and \(y = 7 - x\). Since both equal y, we set them equal to each other:\[2x + 4 = 7 - x\]\[2x + x = 7 - 4\]\[3x = 3 \implies x = 1\]To find the corresponding y-value, substitute \(x = 1\) into \(y = 7 - x\):\[y = 7 - 1 = 6\]The intersection point is (1, 6). Next, we calculate the distance between the given point (4, 3) and the intersection point (1, 6) using the distance formula:\[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]\[d = \sqrt{(1 - 4)^2 + (6 - 3)^2}\]\[d = \sqrt{(-3)^2 + 3^2}\]\[d = \sqrt{9 + 9} = \sqrt{18}\] \(\sqrt{18}\) (simplifies to \(3\sqrt{2}\))

To find the equation of the line, we first calculate the gradient (m) using the two given points \(x_1, y_1 = -2, 1\) and \(x_2, y_2 = -0.5, 4\):\[m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{4 - 1}{-0.5 - (-2)} = \frac{3}{1.5} = 2\]With the gradient \(m = 2\), we use the point-slope equation formula \(y - y_1 = m(x - x_1)\) with the point (-2, 1):\[y - 1 = 2(x - (-2))\]\[y - 1 = 2(x + 2)\]\[y - 1 = 2x + 4\]Rearranging the equation to solve for y:\[y = 2x + 4 + 1\]\[y = 2x + 5\]

The angle 135° is located in the second quadrant of the Cartesian plane. In the second quadrant, the cosine of an angle is always negative. To find the exact value, we use the reference angle (the acute angle made with the x-axis):\[\text{Reference Angle} = 180^\circ - 135^\circ = 45^\circ\]The cosine of the angle is given by:\[\cos 135^\circ = -\cos 45^\circ\]Knowing that the trigonometric value for \(\cos 45^\circ\) is \(\frac{1}{\sqrt{2}}\), we rationalise the denominator:\[\frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}\]Applying the negative sign for the second quadrant, we obtain:\[\cos 135^\circ = -\frac{\sqrt{2}}{2}\]

To differentiate the function \(y = x^2 - \dfrac{1}{x}\), we first express the terms in power form to facilitate the use of the power rule:\[y = x^2 - x^{-1}\]The power rule for differentiation states that if \(f(x) = x^n\), then \(f'(x) = nx^{n-1}\). Applying this to each term individually:\[\frac{dy}{dx} = \frac{d}{dx}(x^2) - \frac{d}{dx}(x^{-1})\]\[\frac{dy}{dx} = 2x^{2-1} - (-1)x^{-1-1}\]\[\frac{dy}{dx} = 2x + x^{-2}\]Converting the negative exponent back into fractional form results in:\[\frac{dy}{dx} = 2x + \frac{1}{x^2}\]

To evaluate this definite integral, we find the antiderivative of the function \(x^2 - 4x\) and then apply the limits of integration from 1 to 2. The integration is as follows:\[\int (x^2 - 4x) dx = \frac{x^{2+1}}{3} - \frac{4x^{1+1}}{2} = \frac{x^3}{3} - 2x^2\]Now, substitute the upper and lower limits:\[\left[ \frac{x^3}{3} - 2x^2 \right]_1^2 = \left( \frac{2^3}{3} - 2(2^2) \right) - \left( \frac{1^3}{3} - 2(1^2) \right)\]Calculating the value at the upper limit (x=2):\[\frac{8}{3} - 8 = \frac{8 - 24}{3} = -\frac{16}{3}\]Calculating the value at the lower limit (x=1):\[\frac{1}{3} - 2 = \frac{1 - 6}{3} = -\frac{5}{3}\]Subtracting the lower limit result from the upper limit result:\[-\frac{16}{3} - \left( -\frac{5}{3} \right) = -\frac{16}{3} + \frac{5}{3} = -\frac{11}{3}\]