Mathematics 2011 Jamb Examination

Let X be the origin. Walking 100m West to Y and 100m North to Z forms a right-angled isosceles triangle.
At point Z, the direction back to Y is South (180°). To find the bearing of X, we need the clockwise angle from North at Z. Since \(ZY = YX = 100\)m, the internal angle \(\angle YZX\) is \(45^\circ\).
Looking from Z, X is to the East of the South line. The bearing is measured clockwise from North:\[\text{Bearing} = 180^\circ - 45^\circ = 135^\circ\]

To find the derivative, we first expand the product of the two linear factors:\[y = (2x + 1)(3x + 1) = 6x^2 + 2x + 3x + 1 = 6x^2 + 5x + 1\]Now, we apply the power rule for differentiation, \(\frac{d}{dx}(ax^n) = anx^{n-1}\):\[\frac{dy}{dx} = \frac{d}{dx}(6x^2) + \frac{d}{dx}(5x) + \frac{d}{dx}(1)\]\[\frac{dy}{dx} = 12x + 5\]

The mode for grouped data is found using the formula:\[\text{Mode} = L + \left( \frac{d_1}{d_1 + d_2} \right) w\]The modal class is 6–8 (highest frequency of 5).
\(L\) (lower boundary) = 5.5
\(d_1\) (frequency difference with previous class) = 5 - 2 = 3
\(d_2\) (frequency difference with next class) = 5 - 3 = 2
\(w\) (class width) = 3\[\text{Mode} = 5.5 + \left( \frac{3}{3 + 2} \right) 3 = 5.5 + \frac{9}{5} = 5.5 + 1.8 = 7.3\]The closest whole number option provided is 7.

To find the minimum point, we first find the stationary points by setting the first derivative to zero:\[y' = 3x^2 + 2x - 1 = 0\]Factorising the quadratic:\[(3x - 1)(x + 1) = 0 \implies x = \frac{1}{3} \text{ or } x = -1\]To determine which is the minimum, we use the second derivative test:\[y'' = 6x + 2\]At \(x = \frac{1}{3}\), \(y'' = 6(\frac{1}{3}) + 2 = 4 > 0\) (Minimum).
At \(x = -1\), \(y'' = 6(-1) + 2 = -4 < 0\) (Maximum).
Thus, the minimum occurs at \(x = \frac{1}{3}\).

We evaluate the definite integral by first finding the anti-derivative of the function:\[\int (3 - 2x) dx = 3x - x^2\]Now, we apply the limits from 0 to 1:\[[3x - x^2]_{0}^{1} = (3(1) - (1)^2) - (3(0) - (0)^2)\]\[= (3 - 1) - (0) = 2\]The value of the integral is 2.

To integrate the function \(\cos 4x\), we apply the general rule for integrating trigonometric functions of the form \(\cos(ax)\), which is:\[\int \cos(ax) \, dx = \frac{1}{a} \sin(ax) + C\]In this specific problem, the constant \(a\) is equal to 4. Applying the rule:\[\int \cos 4x \, dx = \frac{1}{4} \sin 4x + k\]Where \(k\) represents the constant of integration. Alternatively, using \(u\)-substitution where \(u = 4x\), then \(du = 4dx\) or \(dx = \frac{du}{4}\). Substituting these into the integral gives:\[\int \cos u \frac{du}{4} = \frac{1}{4} \sin u + k = \frac{1}{4} \sin 4x + k\]

Let the four consecutive integers be represented as \(n\), \(n+1\), \(n+2\), and \(n+3\). According to the question, their sum is 34:\[n + (n + 1) + (n + 2) + (n + 3) = 34\]Combining the like terms on the left-hand side of the equation:\[4n + 6 = 34\]To solve for \(n\), we first subtract 6 from both sides:\[4n = 34 - 6\]\[4n = 28\]Dividing both sides by 4 gives:\[n = 7\]The four integers are therefore 7, 8, 9, and 10. The least of these numbers is 7.

To find the median, we first determine the total frequency (\(N\)) by summing all the frequencies:\[N = 1 + 4 + 3 + 8 + 2 + 5 = 23\]The median position for an odd number of data points is calculated as:\[\text{Position} = \frac{N + 1}{2} = \frac{23 + 1}{2} = 12^{\text{th}} \text{ term}\]By calculating the cumulative frequencies (1, 5, 8, 16, 18, 23), we see that the 12th term falls within the frequency of the number 3. Thus, the median is 3.
The range is calculated as the difference between the highest and lowest values in the data set:\[\text{Range} = \text{Maximum} - \text{Minimum} = 5 - 0 = 5\]Therefore, the median and range are 3 and 5 respectively.

To find the standard deviation, we first calculate the midpoint (\(x\)) of each class interval:
For 3–5, \(x = 4\); for 6–8, \(x = 7\); for 9–11, \(x = 10\).
Next, we find the mean (\(\bar{x}\)):\[\bar{x} = \dfrac{\sum fx}{\sum f} = \dfrac{(2 \times 4) + (2 \times 7) + (2 \times 10)}{2 + 2 + 2} \]\[= \dfrac{8 + 14 + 20}{6} = \dfrac{42}{6} = 7\]Now, we calculate the variance (\(\sigma^2\)) using the formula \(\dfrac{\sum f(x - \bar{x})^2}{\sum f}\):\[\sigma^2 = \dfrac{2(4 - 7)^2 + 2(7 - 7)^2 + 2(10 - 7)^2}{6}\]\[\sigma^2 = \dfrac{2(-3)^2 + 2(0)^2 + 2(3)^2}{6} = \dfrac{2(9) + 0 + 2(9)}{6} = \dfrac{36}{6} = 6\]The standard deviation is the square root of the variance:\[\sigma = \sqrt{6}\]

The word ELATION consists of 7 distinct letters: E, L, A, T, I, O, and N. Since there are no repeated letters, the number of possible permutations is determined by the factorial of the total number of letters.
The formula for the number of ways to arrange \(n\) distinct objects is \(n!\). In this case, \(n = 7\), so the number of arrangements is:\[7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5,040\]The correct mathematical expression for this value is \(7!\).