Mathematics 2013 Jamb Examination

The locus of points at a constant distance from a straight line consists of two lines, one on either side of the original line, that never meet it. Since every point on these lines maintains the same perpendicular distance from PQ, they are by definition parallel to PQ. Therefore, the resulting shape is a pair of parallel lines.

The midpoint formula is given by: \(\displaystyle M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \).
Given \( M = (2, 3) \) and \( P = (-2, 1) \), let \( Q = (x, y) \).
For the x-coordinate: \(\displaystyle 2 = \frac{-2 + x}{2} \implies 4 = -2 + x \implies x = 6 \).
For the y-coordinate: \( \displaystyle3 = \frac{1 + y}{2} \implies 6 = 1 + y \implies y = 5 \).
Thus, the coordinates of Q are (6, 5).

Step 1: Find the midpoint of PQ: \(\displaystyle \left( \frac{2-5}{2}, \frac{-3+1}{2} \right) = (-1.5, -1) \).
Step 2: Find the gradient of PQ (\( m_1 \)): \(\displaystyle \frac{1 - (-3)}{-5 - 2} = \frac{4}{-7} \).
Step 3: The gradient of the perpendicular bisector (\( m_2 \)) is the negative reciprocal: \( m_2 = \frac{7}{4} \).
Step 4: Use the point-slope form \( y - y_1 = m(x - x_1) \):\[ y - (-1) = \frac{7}{4}(x - (-1.5)) \]\[ y + 1 = \frac{7}{4}(x + 1.5) \]\[ 4(y + 1) = 7(x + 1.5) \implies 4y + 4 = 7x + 10.5 \]Multiply by 2 to clear decimals: \[ 8y + 8 = 14x + 21 \implies 8y - 14x - 13 = 0 \].

We use the Cosine Rule: \( p^2 = q^2 + r^2 - 2qr \cos P \).
Substitute the given values:\[ p^2 = 8^2 + 6^2 - 2(8)(6)\left(\frac{1}{12}\right) \]\[ p^2 = 64 + 36 - 96\left(\frac{1}{12}\right) \]\[ p^2 = 100 - 8 = 92 \]There seems to be a discrepancy in the options provided in the source text. Based on the calculation \( p = \sqrt{92} \), option C (\( 9\sqrt{2} \)) is approximately 12.7, while \( \sqrt{92} \) is roughly 9.6. However, if \( \cos P = \frac{1}{12} \) was meant to be a different value or the calculation results in 100, the answer would change. Re-checking: \( 100 - 8 = 92 \). If the question intended \( p^2 = 100 \), then \( p = 10 \).

If \( \displaystyle\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{3}{4} \), we can find the hypotenuse using Pythagoras: \[ \text{Hypotenuse} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5 \]Now find sine and cosine:
\(\displaystyle \sin \theta = \frac{3}{5} \) and \(\displaystyle \cos \theta = \frac{4}{5} \).
Adding them together: \[ \frac{3}{5} + \frac{4}{5} = \frac{7}{5} = 1\frac{2}{5} \].

To integrate this expression, we first split the fraction into two separate terms:\[ \int \left( \frac{1}{x^3} + \frac{x}{x^3} \right) dx = \int \left( x^{-3} + x^{-2} \right) dx \]Next, we apply the power rule for integration, \(\int x^n dx = \frac{x^{n+1}}{n+1} + k\):\[ \frac{x^{-3+1}}{-3+1} + \frac{x^{-2+1}}{-2+1} + k \]\[ = \frac{x^{-2}}{-2} + \frac{x^{-1}}{-1} + k \]\[ = -\frac{1}{2x^2} - \frac{1}{x} + k \]This result accounts for both the inverse square and inverse linear components of the function.

To evaluate a definite integral, we first find the antiderivative of the function. The integral of \(\sin x\) is \(-\cos x\).

Now, we apply the Fundamental Theorem of Calculus by evaluating at the upper and lower limits:\[ [-\cos x]_{0}^{\pi/2} = (-\cos(\pi/2)) - (-\cos(0)) \]Knowing that \(\cos(\pi/2) = 0\) and \(\cos(0) = 1\), we substitute these values:\[ (0) - (-1) = 1 \]

To identify the correct Venn diagram for the expression \((P \cap Q) \cup R\), we can break the operation into two steps:
1. Identify \(P \cap Q\): This represents the intersection (overlap) between set P and set Q.
2. Identify the Union with R (\(\cup R\)): The union operation means we take everything identified in the first step and add everything that belongs to set R.
Looking at the diagrams:
* In Diagram A, the entire circle R is shaded, along with the specific region where P and Q overlap. This perfectly represents the union of the intersection \(P \cap Q\) and the set R.