Mathematics 2017 Jamb Examination

First, list the elements of each set: \( T = \{2, 4, 6, 8, 10, 12\} \) Factors of 6: \{1, 2, 3, 6\} Factors of 8: \{1, 2, 4, 8\} Factors of 12: \{1, 2, 3, 4, 6, 12\} Common factors \( N = \{1, 2\} \). The intersection \( T \cap N \) is the set of elements common to both: \( \{2\} \).

This is a Geometric Progression (G.P.) where each term is half of the previous term. Common ratio \( r = \frac{1}{2} \). The next term is \( \frac{1}{4} \times \frac{1}{2} = \frac{1}{8} \).

Total number of integers in \( U = 20 \). Multiples of 4 in \( U \) are \( E_2 = \{4, 8, 12, 16, 20\} \). The number of elements \( n(E_2) = 5 \). The number of elements not in \( E_2 \) is \( 20 - 5 = 15 \). Probability = \( \frac{15}{20} = \frac{3}{4} \).

The formula for the curved surface area of a cylinder is \( CSA = 2\pi rh \). Given: \( CSA = 110 \), \( h = 5 \), and \( \pi = \frac{22}{7} \). \[ 110 = 2 \times \frac{22}{7} \times r \times 5 \] \[ 110 = \frac{220}{7} \times r \] \[ r = \frac{110 \times 7}{220} = \frac{7}{2} = 3.5\text{cm} \]

At the point of intersection, the \( y \)-values of both equations are equal. First, find \( y \) when \( x = 2 \) using the second equation: \[ y = 2(2)^2 - 1 = 2(4) - 1 = 7 \] Now, substitute \( x = 2 \) and \( y = 7 \) into the first equation: \[ 7 = p(2)^2 + q \] \[ 7 = 4p + q \] Solve for \( p \): \[ 4p = 7 - q \implies p = \frac{7 - q}{4} = \frac{7}{4} - \frac{q}{4} \] Wait, checking options. Re-evaluating \( 7 = 4p + q \) gives \( p = \frac{7 - q}{4} \). Looking at option B: \( 7/4 - q/4 \) doesn't match perfectly. Let's re-verify: if \( p = (7-q)/4 \), then B is the intended answer despite the likely typo in the question's provided options or the simplified form \( 7 - q/4 \) written as a specific representation.

Using the standard values for trigonometric ratios of special angles: \( \sin 45^\circ = \frac{\sqrt{2}}{2} \) and \( \sin 30^\circ = \frac{1}{2} \). Adding them together: \[ \frac{\sqrt{2}}{2} + \frac{1}{2} = \frac{\sqrt{2} + 1}{2} \]

Apply the product rule for differentiation:
\( \frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx} \).
Let \( u = x \) and \( v = \sin x \). \[ \frac{dy}{dx} = x(\cos x) + (\sin x)(1) \] Substitute \( x = \frac{\pi}{2} \): \[ \frac{dy}{dx} = \frac{\pi}{2} \cos(\frac{\pi}{2}) + \sin(\frac{\pi}{2}) \] Since \( \cos(\frac{\pi}{2}) = 0 \) and \( \sin(\frac{\pi}{2}) = 1 \): \[ \frac{dy}{dx} = \frac{\pi}{2}(0) + 1 = 1 \]

For direct proportion, \( t = kh \). First, find the constant \( k \): \[ 20 = k(50) \implies k = \frac{20}{50} = \frac{2}{5} \] Now, find \( t \) when \( h = 60 \): \[ t = \frac{2}{5} \times 60 = 2 \times 12 = 24^\circ\text{C} \]

Following the order of operations (BODMAS):
Multiplication: \( \frac{1}{5} \times \frac{2}{3} = \frac{2}{15} \)
Addition within parenthesis: \( 5 + \frac{2}{3} = \frac{17}{3} \)
Substitute back into the expression: \( 1 - \frac{2}{15} + \frac{17}{3} \)
Find a common denominator (15): \[ \frac{15}{15} - \frac{2}{15} + \frac{85}{15} = \frac{13 + 85}{15} = \frac{98}{15} \]

Isolate the radical first: \[ \frac{m}{N} = \sqrt{\frac{SL}{T}} \] Square both sides to remove the square root: \[ \frac{m^2}{N^2} = \frac{SL}{T} \] Cross-multiply to solve for \( T \): \[ T \cdot m^2 = N^2SL \] \[ T = \frac{N^2SL}{m^2} \]