Mathematics 2012 Jamb Examination

To evaluate the definite integral of \(\sec^2 \theta\), we recall the fundamental trigonometric derivatives. The derivative of \(\tan \theta\) is \(\sec^2 \theta\), which means the antiderivative of \(\sec^2 \theta\) is \(\tan \theta\).
Applying the Fundamental Theorem of Calculus: \[\int_{0}^{\frac{\pi}{4}} \sec^2 \theta d\theta = [\tan \theta]_{0}^{\frac{\pi}{4}}\]Evaluating at the upper and lower limits: \[= \tan\left(\frac{\pi}{4}\right) - \tan(0)\]Since \(\frac{\pi}{4}\) radians is equivalent to 45°, we know that \(\tan(45°) = 1\) and \(\tan(0) = 0\): \[= 1 - 0 = 1\]The area under the curve of the squared secant function from 0 to \(\frac{\pi}{4}\) is exactly 1.

The mean is calculated as the sum of all values divided by the number of values. We can use this to find the total sum before and after the eighth number is added.
Step 1: Find the sum of the first seven numbers: \[\text{Sum}_7 = 7 \times 96 = 672\]Step 2: Find the sum of all eight numbers when the mean is 112: \[\text{Sum}_8 = 8 \times 112 = 896\]Step 3: The eighth number is the difference between the new total sum and the original sum: \[\text{Eighth number} = \text{Sum}_8 - \text{Sum}_7\]\[= 896 - 672 = 224\]By adding 224 to the set, the average increases significantly from 96 to 112.

The median is the middle value of a data set when it is arranged in ascending or descending order. First, let us count the total number of items (n). There are 16 numbers in the list.
Arranging the numbers in ascending order:
2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 7, 8, 9, 9
Since \(n = 16\) (an even number), the median is the average of the 8th and 9th terms: \[\text{8th term} = 4\]\[\text{9th term} = 4\]\[\text{Median} = \frac{4 + 4}{2} = 4\]The central value of this distribution is 4.

The range is a measure of statistical dispersion defined as the difference between the highest (maximum) value and the lowest (minimum) value in a data set.
Looking at the set {4, 9, 6, 3, 2, 8, 10, 11}: \[\text{Highest value} = 11\]\[\text{Lowest value} = 2\]The calculation is as follows: \[\text{Range} = \text{Maximum} - \text{Minimum}\]\[= 11 - 2 = 9\]The spread of the data from the smallest to the largest value is 9 units.

To find the standard deviation, we follow these steps:
1. Calculate the mean (\(\bar{x}\)): \[\bar{x} = \frac{2+3+8+10+12}{5} = \frac{35}{5} = 7\]2. Calculate the squares of the deviations from the mean \((x - \bar{x})^2\): \[(2-7)^2 = (-5)^2 = 25\]\[(3-7)^2 = (-4)^2 = 16\]\[(8-7)^2 = (1)^2 = 1\]\[(10-7)^2 = (3)^2 = 9\]\[(12-7)^2 = (5)^2 = 25\]3. Find the variance (\(\sigma^2\)), which is the average of these squared deviations: \[\sigma^2 = \frac{25 + 16 + 1 + 9 + 25}{5} = \frac{76}{5} = 15.2\]4. The standard deviation (\(\sigma\)) is the square root of the variance: \[\sigma = \sqrt{15.2} \approx 3.8987\]Rounding to one decimal place gives 3.9.

First, substitute \(n = 15\) into the combination expression: \[^{15+1}C_{15-2} = ^{16}C_{13}\]Using the property of combinations where \(^nC_r = ^nC_{n-r}\): \[^{16}C_{13} = ^{16}C_{16-13} = ^{16}C_3\]The formula for combinations is \(^nC_r = \dfrac{n!}{r!(n-r)!}\): \[^{16}C_3 = \dfrac{16 \times 15 \times 14}{3 \times 2 \times 1}\]\[= 16 \times \dfrac{15}{3} \times \dfrac{14}{2}\]\[= 16 \times 5 \times 7\]\[= 80 \times 7 = 560\]There are 560 ways to choose 13 items from a set of 16.

To find the number of arrangements (permutations) of a word with repeating letters, we use the formula: \[\frac{n!}{n_1! n_2!n_k!}\] where \(n\) is the total number of letters and \(n_1, n_2\), etc., are the frequencies of repeating letters.
In the word TOTALITY:
Total letters (n) = 8
T appears 3 times
O appears 1 time
A appears 1 time
L appears 1 time
I appears 1 time
Y appears 1 time
The calculation is: \[\frac{8!}{3!} = \frac{40320}{6} = 6720\]The letters of the word TOTALITY can be arranged in 6,720 distinct ways.

This is a binomial probability problem. Let \(p\) be the probability of passing and \(q\) be the probability of failing.
\(p = \frac{2}{3}\)
\(q = 1 - \frac{2}{3} = \frac{1}{3}\)
We want to find the probability of exactly 2 passes (k=2) in 3 attempts (n=3). The formula is: \[P(X=k) = ^nC_k \cdot p^k \cdot q^{n-k}\]\[P(X=2) = ^3C_2 \cdot \left(\frac{2}{3}\right)^2 \cdot \left(\frac{1}{3}\right)^{3-2}\]\[= 3 \cdot \left(\frac{4}{9}\right) \cdot \left(\frac{1}{3}\right)\]\[= 3 \cdot \frac{4}{27} = \frac{12}{27} = \frac{4}{9}\]Note: Checking original options, \(\frac{4}{9}\) is mathematically correct. Since the provided options list \(\frac{4}{27}\) (likely a distractor before multiplying by 3) and \(\frac{4}{9}\) isn't explicitly listed in the text provided but the logic holds, I have updated the correct answer mapping to the simplified fraction if available or the corrected calculation.

To find the probability that 'at least one' lives, it is easier to calculate the complement: the probability that 'none' of them live up to 80 years, and subtract that from 1.
Let \(P(M)\) be the probability the man lives = \(\frac{2}{3}\).
Let \(P(W)\) be the probability the wife lives = \(\frac{3}{5}\).
The probability the man does NOT live (\(P(M')\)) = \(1 - \frac{2}{3} = \frac{1}{3}\).
The probability the wife does NOT live (\(P(W')\)) = \(1 - \frac{3}{5} = \frac{2}{5}\).
Probability that both do NOT live: \[P(M' \cap W') = \frac{1}{3} \times \frac{2}{5} = \frac{2}{15}\]Probability that at least one lives: \[1 - P(M' \cap W') = 1 - \frac{2}{15} = \frac{13}{15}\]This covers the scenarios where only the man lives, only the wife lives, or both live.

To determine the inequality representing the shaded region, we first find the equation of the boundary line. The line passes through the y-axis at (0, 4) and the x-axis at (1, 0).
Step 1: Calculate the gradient (m) of the line: \[m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - 4}{1 - 0} = -4\]Step 2: Use the slope-intercept form \(y = mx + c\). Since the y-intercept is 4, the equation is: \[y = -4x + 4\]Step 3: Determine the inequality sign. The region is shaded below the line, and the line is solid. This indicates the "less than or equal to" relationship (\(\le\)). We can verify this by testing a point in the shaded region, such as the origin (0, 0): \[0 \le -4(0) + 4\]\[0 \le 4\]Since this statement is true, the inequality is indeed \(y \le -4x + 4\).