Mathematics 2009 Jamb Examination

Substitute \(m = 3\) and \(n = -5\) into the given operation.\[ m * n = n - (m + 2) \]\[ 3 * (-5) = -5 - (3 + 2) \]\[ = -5 - 5 = -10 \]The operation simply requires careful substitution and simplification.

The inverse of \(-5\) is a number \(x\) such that:\[ -5 \text{ plus } x = 0 \]Using the operation definition:\[ -5 \text{ plus } x = x + (-5)x = x(1 - 5) = -4x \]Set this equal to zero:\[ -4x = 0 \Rightarrow x = 0 \]Therefore, the inverse of \(-5\) under this operation is \(0\).

For a \(2 \times 2\) matrix \(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\), the determinant is given by:\[ |Q| = ad - bc \]Substitute the values:\[ |Q| = (9)(4) - (-2)(-7) \]\[ = 36 - 14 = 22 \]The determinant is positive because the first product is greater than the second.

First find the determinant of the matrix:\[ |P| = (x + 3)(x - 1) - (x + 2)(x + 1) \]Expand both terms:\[ = (x^2 + 2x - 3) - (x^2 + 3x + 2) \]\[ = -x - 5 \]Given that \(|P| = -10\):\[ -x - 5 = -10 \Rightarrow -x = -5 \Rightarrow x = 5 \]

The slopes of the lines are \(m_1 = 1\) and \(m_2 = \sqrt{3}\).
The formula for the angle \(\theta\) between two lines is:\[ \tan \theta = \left| \dfrac{m_2 - m_1}{1 + m_1 m_2} \right| \]Substitute the values:\[ \tan \theta = \left| \dfrac{\sqrt{3} - 1}{1 + \sqrt{3}} \right| \]This simplifies to \(\tan 15^\circ\).
Hence, the acute angle between the lines is \(15^\circ\).

In a right-angled isosceles triangle, the two equal sides are the perpendicular sides.
If each perpendicular side is \(a\), then the hypotenuse is \(a\sqrt{2}\).
Given hypotenuse \(= 2\):\[ a\sqrt{2} = 2 \Rightarrow a = \dfrac{2}{\sqrt{2}} = \sqrt{2} \]The area of a right-angled triangle is:\[ \dfrac{1}{2} \times \text{base} \times \text{height} \]\[ = \dfrac{1}{2} \times \sqrt{2} \times \sqrt{2} = \dfrac{1}{2} \times 2 = 1\,\text{cm}^2 \]

The perpendicular distance from the centre to the chord is \(5\,\text{cm}\).
The radius of the circle is \(13\,\text{cm}\).
Using the right-angled triangle formed by the radius, half of the chord, and the perpendicular distance:\[ \left(\dfrac{\text{chord}}{2}\right)^2 + 5^2 = 13^2 \]\[ \left(\dfrac{\text{chord}}{2}\right)^2 = 169 - 25 = 144 \]\[ \dfrac{\text{chord}}{2} = 12 \Rightarrow \text{chord} = 24\,\text{cm} \]

Recall the standard trigonometric values:\[ \sin 45^\circ = \dfrac{\sqrt{2}}{2}, \quad \cos 30^\circ = \dfrac{\sqrt{3}}{2} \]Substitute into the expression:\[ \sin 45^\circ - \cos 30^\circ = \dfrac{\sqrt{2}}{2} - \dfrac{\sqrt{3}}{2} \]\[ = \dfrac{\sqrt{2} - \sqrt{3}}{2} \]The value is negative because \(\sqrt{3} > \sqrt{2}\).

The angle of depression is equal to the angle of elevation, which is \(60^\circ\).
The height of the cliff is the opposite side, and the width of the river is the adjacent side.
Using the tangent ratio:\[ \tan 60^\circ = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{300}{\text{width}} \]Since \(\tan 60^\circ = \sqrt{3}\):\[ \sqrt{3} = \dfrac{300}{\text{width}} \]\[ \text{width} = \dfrac{300}{\sqrt{3}} = 100\sqrt{3}\,\text{m} \]

Differentiate step by step.
The derivative of \(\cos kx\) is \(-k\sin kx\).
Given \(y = 3\cos 4x\):\[ \dfrac{dy}{dx} = 3 \times (-4\sin 4x) \]\[ = -12\sin 4x \]The negative sign appears because the derivative of cosine is negative sine.