Mathematics 2010 Jamb Examination

The gradient (m) of a straight line passing through two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by the formula: \[ m = \frac{y_2 - y_1}{x_2 - x_1} \] Using the points P(1, 1) and Q(2, 5): \[ x_1 = 1, y_1 = 1, x_2 = 2, y_2 = 5 \] Substituting these values into the formula: \[ m = \frac{5 - 1}{2 - 1} \] \[ m = \frac{4}{1} = 4 \] The gradient of the line is 4.

Cotangent is the ratio of the adjacent side to the opposite side in a right-angled triangle. Given: \[ \cot \theta = \frac{\text{Adjacent}}{\text{Opposite}} = \frac{8}{15} \] We can find the hypotenuse (h) using Pythagoras' theorem: \[ h^2 = 8^2 + 15^2 \] \[ h^2 = 64 + 225 = 289 \] \[ h = \sqrt{289} = 17 \] The sine of an angle is the ratio of the opposite side to the hypotenuse: \[ \sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{15}{17} \]

To differentiate \( y = x \sin x \), we must use the product rule because the expression is a product of two functions of x, namely \( u = x \) and \( v = \sin x \). The product rule is: \[ \frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx} \] Let: \[ u = x \implies \frac{du}{dx} = 1 \] \[ v = \sin x \implies \frac{dv}{dx} = \cos x \] Substituting these into the formula: \[ \frac{dy}{dx} = (x)(\cos x) + (\sin x)(1) \] \[ \frac{dy}{dx} = x \cos x + \sin x \] This is equivalent to \( \sin x + x \cos x \).

To evaluate the definite integral, we first find the antiderivative of the function: \[ \int (x^3 + x^2) dx = \frac{x^4}{4} + \frac{x^3}{3} \] Now we apply the limits from 0 to 2: \[ \left[ \frac{x^4}{4} + \frac{x^3}{3} \right]_0^2 = \left( \frac{2^4}{4} + \frac{2^3}{3} \right) - \left( \frac{0^4}{4} + \frac{0^3}{3} \right) \] \[ = \left( \frac{16}{4} + \frac{8}{3} \right) - 0 \] \[ = 4 + \frac{8}{3} = \frac{12 + 8}{3} = \frac{20}{3} \] Converting the improper fraction to a mixed number: \[ \frac{20}{3} = 6\frac{2}{3} \]

To form the committee, we need to select women and men independently. Since the order of selection does not matter, we use combinations: \[ ^nC_r = \frac{n!}{(n-r)!r!} \] Ways to choose 2 women from 5: \[ ^5C_2 = \frac{5 \times 4}{2 \times 1} = 10 \] Ways to choose 3 men from 6: \[ ^6C_3 = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \] To find the total number of ways to form the committee, multiply the two results: \[ \text{Total ways} = 10 \times 20 = 200 \]

When three coins are tossed, each coin has 2 possible outcomes (Head or Tail). The total number of possible outcomes in the sample space is: \[ 2 \times 2 \times 2 = 8 \] The sample space S is {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}.
The event of interest is all three coins showing heads: {HHH}.
There is only 1 such outcome. Therefore, the probability P(all heads) is: \[ P = \frac{\text{Number of successful outcomes}}{\text{Total number of outcomes}} = \frac{1}{8} \]

First, calculate the mean (\(\bar{x}\)) of the data set: \[ \bar{x} = \frac{2 + 3 + 5 + 6}{4} = \frac{16}{4} = 4 \] Next, find the squared deviations from the mean: \[ (2-4)^2 = (-2)^2 = 4 \] \[ (3-4)^2 = (-1)^2 = 1 \] \[ (5-4)^2 = (1)^2 = 1 \] \[ (6-4)^2 = (2)^2 = 4 \] Sum of squared deviations: \[ \sum (x - \bar{x})^2 = 4 + 1 + 1 + 4 = 10 \] The variance (\(\sigma^2\)) is the mean of these squared deviations: \[ \sigma^2 = \frac{10}{4} = \frac{5}{2} \] The standard deviation (\(\sigma\)) is the square root of the variance: \[ \sigma = \sqrt{\frac{5}{2}} \]

Since W is directly proportional to U, the relationship is: \[ W = kU \] where k is the constant of proportionality. First, find k: \[ 5 = k(3) \implies k = \frac{5}{3} \] Now, find U when \( W = \frac{2}{7} \): \[ \frac{2}{7} = \left(\frac{5}{3}\right)U \] To solve for U, multiply both sides by the reciprocal of k (which is \(\frac{3}{5}\)): \[ U = \frac{2}{7} \times \frac{3}{5} \] \[ U = \frac{6}{35} \]

We can factor the quadratic inequality: \[ x^2 - 1 > 0 \implies (x - 1)(x + 1) > 0 \] The critical points are \( x = 1 \) and \( x = -1 \). These points divide the number line into three intervals:
1) \( x < -1 \): Pick \( x = -2 \); \( (-2)^2 - 1 = 3 > 0 \) (True)
2) \( -1 < x < 1 \): Pick \( x = 0 \); \( (0)^2 - 1 = -1 < 0 \) (False)
3) \( x > 1 \): Pick \( x = 2 \); \( (2)^2 - 1 = 3 > 0 \) (True)
Therefore, the solution is the set of values where \( x < -1 \) or \( x > 1 \).