Mathematics 2009 Jamb Examination

The distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:\[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]Substituting the given points:\[3 = \sqrt{(4 - 1)^2 + (2 - r)^2}\]\[3 = \sqrt{9 + (2 - r)^2}\]Squaring both sides:\[9 = 9 + (2 - r)^2\]\[(2 - r)^2 = 0\]\[2 - r = 0\]\[r = 2\]This value makes the distance between the two points exactly 3 units.

Differentiation is carried out using the chain rule.
The derivative of \(\cos u\) is \(-\sin u\), multiplied by the derivative of \(u\).
Given:\[y = 3\cos 4x\]\[\dfrac{dy}{dx} = 3 \times (-\sin 4x) \times 4\]\[\dfrac{dy}{dx} = -12\sin 4x\]This follows directly from standard trigonometric differentiation rules.

Using standard trigonometric values:\[\sin 45^\circ = \dfrac{\sqrt{2}}{2}\]\[\cos 30^\circ = \dfrac{\sqrt{3}}{2}\]Subtracting:\[\sin 45^\circ - \cos 30^\circ = \dfrac{\sqrt{2}}{2} - \dfrac{\sqrt{3}}{2}\]\[= \dfrac{\sqrt{2} - \sqrt{3}}{2}\]This is obtained by expressing both terms with a common denominator.

The angle of depression is equal to the angle of elevation.
From the right-angled triangle formed:\[\tan 60^\circ = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{300}{\text{width}}\]Since \(\tan 60^\circ = \sqrt{3}\):\[\sqrt{3} = \dfrac{300}{\text{width}}\]\[\text{width} = \dfrac{300}{\sqrt{3}}\]\[\text{width} = 100\sqrt{3}\,\text{m}\]This result follows from basic trigonometric ratios.

First expand the expression for \(s\):\[s = (2 + 3t)(5t - 4)\]\[s = 15t^2 - 2t - 8\]Differentiate with respect to \(t\):\[\dfrac{ds}{dt} = 30t - 2\]Substitute \(t = \dfrac{4}{5}\):\[\dfrac{ds}{dt} = 30 \times \dfrac{4}{5} - 2\]\[= 24 - 2\]\[= 22\,\text{units per sec}\]This gives the rate of change of \(s\) with respect to time at the given instant.

To find the minimum distance, differentiate \(s\) with respect to \(t\):\[\dfrac{ds}{dt} = 3t^2 - 2t - 1\]Set the derivative equal to zero:\[3t^2 - 2t - 1 = 0\]Solving:\[t = \dfrac{2 \pm \sqrt{4 + 12}}{6} = \dfrac{2 \pm 4}{6}\]This gives \(t = 1\) or \(t = -\dfrac{1}{3}\).
Using the second derivative \(\dfrac{d^2s}{dt^2} = 6t - 2\), at \(t = 1\) this value is positive, so the distance is minimum.
Substitute \(t = 1\) into \(s\):\[s = 1 - 1 - 1 + 5 = 4\,\text{cm}\]This is the least possible distance from the fixed point.

From standard trigonometric integration rules:\[\dfrac{d}{d\theta}(\tan \theta) = \sec^2 \theta\]This means that the integral of \(\sec^2 \theta\) is simply \(\tan \theta\).
Therefore:\[\int \sec^2 \theta\, d\theta = \tan \theta + k\]where \(k\) is the constant of integration.

The total number of students is 260.
Adding all frequencies:\[20 + 2x + 60 + 40 + x + 50 = 260\]\[170 + 3x = 260\]\[3x = 90 \Rightarrow x = 30\]Students absent for at least four days are those absent for 4, 5, and 6 days:\[40 + 30 + 50 = 120\]This gives the required number of students.

The total angle in a full circle is 360°.
From the pie chart, the sector representing apples covers
360° - 100° - 100° - 115° = 45°.
The fraction representing apples is therefore:\[\dfrac{45}{360}\]The number of apples is:\[\dfrac{45}{360} \times 400 = 50\]This corresponds to the number of apples shown by the chart.

The probability of failing one examination is:\[1 - \dfrac{2}{3} = \dfrac{1}{3}\]If the examinations are independent, the probability of failing all three is:\[\left(\dfrac{1}{3}\right)^3 = \dfrac{1}{27}\]This represents the chance that the student does not pass any of the examinations.