Mathematics 2016 Jamb Examination

First, simplify the integrand by dividing each term in the numerator by \(x\): \[ \int \dfrac{2x^3 + 2x}{x} dx = \int (2x^2 + 2) dx \] Now, integrate each term using the power rule \( \int x^n dx = \dfrac{x^{n+1}}{n+1} \): \[ \int 2x^2 dx + \int 2 dx = \dfrac{2x^3}{3} + 2x + k \]

The mean is calculated by summing the numbers and dividing by the count of numbers: \[ \text{Mean} = \dfrac{4 + y + 8 + 10}{4} = 7 \] Multiply both sides by 4: \[ 22 + y = 28 \] Subtract 22 from both sides: \[ y = 28 - 22 = 6 \]

Using the chain rule, where \( \dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx} \). Let \( u = \dfrac{1}{3}x + 6 \), then \( y = u^2 \): \[ \dfrac{dy}{du} = 2u \quad \text{and} \quad \dfrac{du}{dx} = \dfrac{1}{3} \] Multiply the derivatives: \[ \dfrac{dy}{dx} = 2\left(\dfrac{1}{3}x + 6\right) \cdot \dfrac{1}{3} = \dfrac{2}{3}\left(\dfrac{1}{3}x + 6\right) \]

Let \( x \) be the number of students who offer both subjects. Using the set formula: \[ \text{Total} = n(F) + n(A) - n(F \cap A) + n(F \cup A)' \] \[ 150 = 80 + 60 - x + 20 \] \[ 150 = 160 - x \] \[ x = 160 - 150 = 10 \]

In a Geometric Progression, \( T_n = ar^{n-1} \). Given \( T_2 = ar = \dfrac{8}{9} \) and \( T_6 = ar^5 = \dfrac{9}{2} \). Divide \( T_6 \) by \( T_2 \): \[ \dfrac{ar^5}{ar} = \dfrac{9/2}{8/9} \Rightarrow r^4 = \dfrac{9}{2} \cdot \dfrac{9}{8} = \dfrac{81}{16} \] Take the fourth root of both sides: \[ r = \sqrt[4]{\dfrac{81}{16}} = \dfrac{3}{2} \]

According to circle geometry theorems, the angle subtended by an arc at the center is twice the angle subtended at the circumference. Therefore: \[ \angle OTQ = \frac{1}{2} \times \angle OPQ = \frac{1}{2} \times 130^\circ = 65^\circ \]

For any polygon with \(n\) sides, the sum of its interior angles is given by the formula: \[ S = (n - 2) \times 180^\circ \] Given that the sum of the interior angles is: \[ S = 1080^\circ \] Substitute into the formula: \[ (n - 2) \times 180 = 1080 \] Divide both sides by \(180\): \[ n - 2 = \frac{1080}{180} \] \[ n - 2 = 6 \] Add 2 to both sides: \[ n = 8 \] A polygon with 8 sides is called an octagon. The formula works because any polygon can be divided into \((n-2)\) triangles from one vertex, and each triangle contributes \(180^\circ\) to the total interior angle sum. When the total is \(1080^\circ\), the polygon must therefore contain 6 triangles, implying 8 sides.

< MPT = 180^o - 118^o = 62^o
< PML = 62^o ( Alternative angle)
y + 2y + 10^o + 62^o = 180^o (Angles on a straight line)
3y = 180 - 72\[dfrac{3y}{3} = \dfrac{108}{3}\] y = 36^o
< PMT = 2y + 10 = 2(36) + 10 = 82^o

< ROP = 95^o (Exterior angle of cyclic quadrilateral)

Number of students that plays only two games = 4 + 8 + 3 = 15