Mathematics 2010 Jamb Examination

TUVW is a cyclic quad
3x + 20 + 88 = 180 (opp ∠ s of a cyclic quad are supplementary)
3x + 108 = 180
3x = 180 - 108
3x = 72
x = 72/3
x = 24^o

In a right-angled triangle, the two smaller sides are the base and the perpendicular height (the legs), while the longest side is the hypotenuse. The area of any triangle is calculated as half the product of the base and the vertical height. For a right-angled triangle, this is expressed as: \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \] Given the sides: \[ \text{base} = 4\text{ cm} \] \[ \text{height} = 5\text{ cm} \] Substituting these values into the formula: \[ \text{Area} = \frac{1}{2} \times 4 \times 5 \] \[ \text{Area} = 2 \times 5 \] \[ \text{Area} = 10\text{ cm}^2 \] This calculation confirms that the space enclosed by the triangle is exactly 10 square centimetres.

The area of a sector of a circle is a fraction of the total area of the circle, determined by the ratio of the central angle to the full 360 degrees of the circle. The formula for the area of a sector is: \[ \text{Area} = \frac{\theta}{360} \times \pi r^2 \] Given: \[ \theta = 50^\circ \] \[ r = 6\text{ cm} \] \[ \pi \approx \frac{22}{7} \] Plugging the values into the equation: \[ \text{Area} = \frac{50}{360} \times \frac{22}{7} \times 6^2 \] \[ \text{Area} = \frac{5}{36} \times \frac{22}{7} \times 36 \] \[ \text{Area} = 5 \times \frac{22}{7} \] \[ \text{Area} = \frac{110}{7} \] Thus, the area of the sector is 110/7 square centimetres.

The locus of points equidistant from two fixed points P and Q is the perpendicular bisector of the line segment PQ. First, we find the midpoint of PQ: \[ \text{Midpoint} = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{1 + 3}{2}, \frac{3 + 5}{2} \right) = (2, 4) \] Next, we find the gradient (slope) of line PQ: \[ m_1 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - 3}{3 - 1} = \frac{2}{2} = 1 \] The gradient of the perpendicular bisector (\( m_2 \)) must satisfy \( m_1 \times m_2 = -1 \): \[ 1 \times m_2 = -1 \Rightarrow m_2 = -1 \] Now, using the point-slope form with the midpoint (2, 4) and gradient -1: \[ y - y_1 = m(x - x_1) \] \[ y - 4 = -1(x - 2) \] \[ y - 4 = -x + 2 \] \[ y = -x + 6 \] The resulting linear equation defines all points that maintain an equal distance from both P and Q.

Area of a triangle = \(\frac{1}{2}\) ab Sinθ
12√3 = 12 x 8 x q sin 60
12√3 = 4 x q x 3√2
12√3 = 2q√3

q = \(\frac{12√3}{2√3}\)

q = 6cm.

To differentiate the function \( y = (2x + 1)^3 \), we must use the Chain Rule. The Chain Rule states that if we have a composite function \( y = f(g(x)) \), then the derivative is: \[ \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \] Let \( u = 2x + 1 \). Then \( y = u^3 \). First, find the derivative of the outer function with respect to \( u \): \[ \frac{dy}{du} = 3u^2 \] Next, find the derivative of the inner function \( u \) with respect to \( x \): \[ \frac{du}{dx} = 2 \] Now, multiply the two derivatives together: \[ \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \] \[ \frac{dy}{dx} = 3u^2 \times 2 \] \[ \frac{dy}{dx} = 6u^2 \] Substituting back \( u = 2x + 1 \): \[ \frac{dy}{dx} = 6(2x + 1)^2 \] This result matches the standard power rule application for composite linear functions.

To solve the integral \( \int(\sin x + 2)dx \), we can use the sum rule for integration, which allows us to integrate each term separately: \[ \int(\sin x + 2)dx = \int \sin x \, dx + \int 2 \, dx \] Integrating the first term: The derivative of \( \cos x \) is \( -\sin x \), so the integral of \( \sin x \) is \( -\cos x \). \[ \int \sin x \, dx = -\cos x \] Integrating the second term: The integral of a constant \( a \) with respect to \( x \) is \( ax \). \[ \int 2 \, dx = 2x \] Combining these results and adding the constant of integration \( K \): \[ \int(\sin x + 2)dx = -\cos x + 2x + K \] This provides the complete general antiderivative of the function.

In this context, the pass mark is defined as 5. This means that any student who obtained a mark less than 5 is considered to have failed the test. To find the total number of students who failed, we must sum the frequencies (number of students) for all marks below 5. Looking at the table, the marks below 5 are: 1. Mark 2: 3 students 2. Mark 3: 1 student 3. Mark 4: 5 students We add these frequencies together: \[ \text{Total failed} = 3 + 1 + 5 \] \[ \text{Total failed} = 9 \] Therefore, 9 students did not reach the minimum pass mark of 5.

To determine the total number of students who took the test, we need to calculate the sum of the frequencies provided in the table. Each frequency represents the number of students who achieved a specific mark. The frequencies are as follows: - For mark 1: 2 students - For mark 2: 2 students - For mark 3: 8 students - For mark 4: 4 students - For mark 5: 4 students Summing these values together: \[ \sum f = 2 + 2 + 8 + 4 + 4 \] \[ \sum f = 4 + 8 + 4 + 4 \] \[ \sum f = 12 + 8 \] \[ \sum f = 20 \] The total frequency corresponds to the total population of students in the dataset, which is 20.

The mean of a frequency distribution is calculated using the formula: \[ \bar{x} = \frac{\sum f x}{\sum f} \] Where \( x \) is the mark and \( f \) is the frequency of that mark. First, we calculate the sum of the products of marks and frequencies (\( \sum f x \)): \[ (1 \times 2) + (2 \times 2) + (3 \times 8) + (4 \times 4) + (5 \times 4) \] \[ = 2 + 4 + 24 + 16 + 20 \] \[ = 6 + 24 + 16 + 20 \] \[ = 30 + 16 + 20 \] \[ = 66 \] Next, we calculate the total number of students (\( \sum f \)): \[ 2 + 2 + 8 + 4 + 4 = 20 \] Finally, divide the total sum of marks by the number of students: \[ \text{Mean} = \frac{66}{20} \] \[ \text{Mean} = 3.3 \] This value represents the average performance of the students in the test.