Mathematics 2009 Jamb Examination

First expand the expression for \( s \):\[(2 + 3t)(5t - 4) = 10t - 8 + 15t^2 - 12t = 15t^2 - 2t - 8\]Differentiate with respect to \( t \):\[\dfrac{ds}{dt} = 30t - 2\]Substitute \( t = \dfrac{4}{5} \):\[\dfrac{ds}{dt} = 30 \times \dfrac{4}{5} - 2 = 24 - 2 = 22\]This gives the rate of change of displacement with time at the given instant.

To find the minimum distance, differentiate \( s \) with respect to \( t \):\[\dfrac{ds}{dt} = 3t^2 - 2t - 1\]Set \( \dfrac{ds}{dt} = 0 \):\[3t^2 - 2t - 1 = 0\]Solving gives \( t = 1 \) or \( t = -\dfrac{1}{3} \). Time cannot be negative, so \( t = 1 \).
Substitute into \( s \):\[s = 1 - 1 - 1 + 5 = 4\]The minimum distance covered by the particle is 4.0 cm.

From basic trigonometric differentiation, the derivative of \( \tan \theta \) is \( \sec^2 \theta \).
Therefore, the integral of \( \sec^2 \theta \) is simply \( \tan \theta \) plus a constant of integration.

First find the value of \( x \) using the total number of students:\[20 + 2x + 60 + 40 + x + 50 = 260\]\[170 + 3x = 260 \Rightarrow 3x = 90 \Rightarrow x = 30\]Students absent for at least four days are those absent for 4, 5, and 6 days:\[40 + x + 50 = 40 + 30 + 50 = 120\]This represents all students who were absent for four days or more.

The probability of failing one examination is:\[1 - \dfrac{2}{3} = \dfrac{1}{3}\]If the three 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.

The perpendicular distance from the centre of a circle to a chord bisects the chord.
Let half of the chord be \(x\). Using Pythagoras' theorem in the right-angled triangle formed:\[x^2 + 5^2 = 13^2\]\[x^2 + 25 = 169\]\[x^2 = 144\]\[x = 12\]The full length of the chord is \(2x = 24\) cm.
This method applies because the radius drawn to the midpoint of a chord is perpendicular to the chord.

The formula for the surface area of a sphere is \(4\pi r^2\).
Given that the surface area is 154 cm²:\[4 \pi r^2 = 154\]Using \(\pi = \dfrac{22}{7}\):\[4 \times \dfrac{22}{7} \times r^2 = 154\]\[\dfrac{88}{7} r^2 = 154\]\[r^2 = \dfrac{154 \times 7}{88}\]\[r^2 = 12.25\]\[r = 3.5\,\text{cm}\]This value satisfies the given surface area.

The line \(x = 0\) is the y-axis, while \(y = 0\) is the x-axis.
A point that is equidistant from both axes must have equal x- and y-coordinates.
This condition gives \(x = y\), which can be written as:\[x - y = 0\]Since the motion is restricted to the first quadrant, both x and y are positive, which agrees with this equation.

The midpoint of any chord in a circle lies on the perpendicular drawn from the centre to that chord.
For chords of equal length, the distance of each midpoint from the centre is constant.
Using Pythagoras' theorem, if half the chord is 3 cm and the radius is 5 cm:\[r^2 = d^2 + 3^2\]\[25 = d^2 + 9\]\[d^2 = 16\]\[d = 4\,\text{cm}\]Hence, the locus of the midpoints is a circle of radius 4 cm with centre O.

The gradient of a line joining two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:\[m = \dfrac{y_2 - y_1}{x_2 - x_1}\]Substituting the given points:\[\dfrac{4 - p}{p - (-1)} = \dfrac{2}{3}\]\[\dfrac{4 - p}{p + 1} = \dfrac{2}{3}\]Cross-multiplying:\[3(4 - p) = 2(p + 1)\]\[12 - 3p = 2p + 2\]\[10 = 5p\]\[p = 2\]This value satisfies the required gradient.