Mathematics 2014 Jamb Examination

To differentiate \(y = \cos 3x\), we must use the chain rule. The chain rule states that if \(y = f(g(x))\), then \(\dfrac{dy}{dx} = f'(g(x)) \times g'(x)\).
In this case: Let \(u = 3x\), so \(y = \cos u\). Differentiating \(u\) with respect to \(x\):\[ \frac{du}{dx} = 3 \] Differentiating \(y\) with respect to \(u\):\[ \frac{dy}{du} = -\sin u \] Now, multiply them together:\[ \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \]\[ \frac{dy}{dx} = (-\sin u) \times 3 \]\[ \frac{dy}{dx} = -3 \sin 3x \]Note that the derivative of cosine is negative sine, and the coefficient of the inner function (3) becomes a multiplier.

To find the minimum value of a quadratic function \(y = ax^2 + bx + c\), we can find the x-coordinate of the vertex using the formula \(x = \dfrac{-b}{2a}\).
From the equation \(y = x^2 - 2x - 3\), we have \(a = 1\) and \(b = -2\):\[ x = \frac{-(-2)}{2(1)} = \frac{2}{2} = 1 \] Now, substitute \(x = 1\) into the original equation to find the minimum value of \(y\):\[ y = (1)^2 - 2(1) - 3 \]\[ y = 1 - 2 - 3 \]\[ y = -4 \] Alternatively, using differentiation, we set the first derivative to zero:\[ \frac{dy}{dx} = 2x - 2 = 0 \Rightarrow x = 1 \] Substituting back gives the same minimum value of -4.

To integrate the trigonometric function \(\sin 2x\), we use the general integration rule for sine functions: \(\displaystyle \int \sin(ax) \, dx = -\dfrac{1}{a} \cos(ax) + k\).
In this problem, the coefficient \(a\) of \(x\) is 2. Therefore:\[ \int \sin 2x \, dx = -\frac{1}{2} \cos 2x + k \] This can be verified by differentiating the result. The derivative of \(-\dfrac{1}{2} \cos 2x\) is \(-\dfrac{1}{2} \times (-2 \sin 2x)\), which simplifies back to \(\sin 2x\). The constant \(k\) represents the constant of integration for an indefinite integral.

We use the power rule for integration combined with the chain rule (u-substitution). Let \(u = 2x + 3\), then \(du = 2dx\), which means \(dx = \dfrac{du}{2}\).
Substituting into the integral:\[ \int u^{\frac{1}{2}} \frac{du}{2} = \frac{1}{2} \int u^{\frac{1}{2}} \, du \] Applying the power rule \(\displaystyle \int u^n \, du = \frac{u^{n+1}}{n+1}\):\[ = \frac{1}{2} \left( \frac{u^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} \right) + k \]\[ = \frac{1}{2} \left( \frac{u^{\frac{3}{2}}}{\frac{3}{2}} \right) + k \]\[ = \frac{1}{2} \times \frac{2}{3} u^{\frac{3}{2}} + k = \frac{1}{3} u^{\frac{3}{2}} + k \] Substituting \(u = 2x + 3\) back in:\[ \frac{1}{3}(2x + 3)^{\frac{3}{2}} + k \]

The mean is calculated by summing all the values in the data set and dividing by the total number of values. There are 5 values in this set.
Sum of the values:\[ \Sigma = (2 - t) + (4 + t) + (3 - 2t) + (2 + t) + (t - 1) \] Let's group the constants and the 't' terms:
Constants: \(2 + 4 + 3 + 2 - 1 = 10\)
Variables: \(-t + t - 2t + t + t = 0t = 0\)
Total Sum: \(10 + 0 = 10\)
Number of terms (n) = 5.
Mean (\(\bar{x}\)):\[ \bar{x} = \frac{\Sigma}{n} = \frac{10}{5} = 2 \] The terms involving \(t\) cancel each other out completely, leaving a constant mean of 2.

The mode of a frequency distribution is the value that occurs most frequently. To identify the mode, we look at the frequency row of the table and find the highest number.
From the table provided:
- Value 0 has a frequency of 1
- Value 1 has a frequency of 2
- Value 2 has a frequency of 2
- Value 3 has a frequency of 1
- Value 4 has a frequency of 9
The highest frequency is 9, which corresponds to the value 4. Therefore, 4 is the modal value of this distribution, as it appears significantly more times than any other number in the set.

To find the median, we must first arrange the data set in ascending order. The data set contains 15 numbers.
Ordered set: 1, 2, 3, 3, 4, 5, 5, 5, 5, 6, 7, 8, 9, 9, 10.
Since the number of observations \(n = 15\) is odd, the median is the middle value at the position:\[ \text{Position} = \dfrac{n + 1}{2} = \dfrac{15 + 1}{2} = 8^{th} \text{ term} \]Counting to the 8th term in our ordered list:
1st: 1, 2nd: 2, 3rd: 3, 4th: 3, 5th: 4, 6th: 5, 7th: 5, 8th: 5.
The value at the 8th position is 5. Therefore, the median of this distribution is 5.

To find the standard deviation, we follow these steps:
Step 1: Find the mean (\(\bar{x}\)):\[ \bar{x} = \dfrac{5 + 4 + 3 + 2 + 1}{5} = \dfrac{15}{5} = 3 \]Step 2: Find the squared deviations from the mean \((x - \bar{x})^2\):\[ (5-3)^2 = 4 \]\[ (4-3)^2 = 1 \]\[ (3-3)^2 = 0 \]\[ (2-3)^2 = 1 \]\[ (1-3)^2 = 4 \]Step 3: Find the Variance (\(\sigma^2\)):\[ \sigma^2 = \dfrac{\sum (x - \bar{x})^2}{n} = \dfrac{4 + 1 + 0 + 1 + 4}{5} = \dfrac{10}{5} = 2 \]Step 4: Find the Standard Deviation (\(\sigma\)) by taking the square root of the variance:\[ \sigma = \sqrt{2} \]

This problem involves a combination because the order in which the girls are selected for the team does not matter. The formula for combinations of \(n\) items taken \(r\) at a time is given by:\[ ^nC_r = \dfrac{n!}{r!(n - r)!} \]In this case, \(n = 7\) and \(r = 3\). Substituting these values into the formula:\[ ^7C_3 = \dfrac{7!}{3!(7 - 3)!} = \dfrac{7!}{3!4!} \]This represents the total number of unique groups of three that can be formed from the original seven individuals.

Probability is defined as the number of successful outcomes divided by the total number of outcomes. The total number of throws is given as 100.
The phrase 'at least a 4' means the outcome can be 4, 5, or 6. We sum the frequencies of these outcomes:\[ \text{Successes} = \text{Freq}(4) + \text{Freq}(5) + \text{Freq}(6) \]\[ \text{Successes} = 16 + 10 + 14 = 40 \]Now, calculate the probability (P):\[ P = \dfrac{40}{100} \]Simplifying the fraction by dividing both the numerator and the denominator by 20:\[ P = \dfrac{2}{5} \]