Mathematics 2019 Jamb Examination

First, find the gradient (slope) \( \displaystyle m \) using the two points: \[ \displaystyle m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{22 - 12}{5 - 3} = \frac{10}{2} = 5 \] The equation of a straight line is \( \displaystyle y = mx + c \). Given the intercept \( \displaystyle c = 2 \): \[ \displaystyle y = 5x + 2 \]

Using the midpoint formula \( \displaystyle (x, y) = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \):
For x-coordinate: \( \displaystyle 2 = \frac{m + n}{2} \implies m + n = 4 \)
For y-coordinate: \( \displaystyle m = \frac{n - 4}{2} \implies 2m = n - 4 \implies 2m - n = -4 \)
Solving simultaneously:
\( \displaystyle n = 4 - m \)
\( \displaystyle 2m - (4 - m) = -4 \implies 3m - 4 = -4 \implies 3m = 0 \implies m = 0 \)
Substitute \( \displaystyle m = 0 \)
\( \displaystyle n = 4 - 0 = 4 \).

Evaluate the determinant by cross-multiplying: \[ \displaystyle (2 \times 9) - (-4 \times x) = 58 \] \[ \displaystyle 18 + 4x = 58 \] \[ \displaystyle 4x = 58 - 18 \] \[ \displaystyle 4x = 40 \implies x = 10 \]

Apply the power rule for differentiation \( \displaystyle \frac{d}{dx}(ax^n) = anx^{n-1} \): 1. \( \displaystyle \frac{d}{dx}(6x^3) = 18x^2 \) 2. \( \displaystyle \frac{d}{dx}(2x^{-2}) = -4x^{-3} \) 3. \( \displaystyle \frac{d}{dx}(-x^{-3}) = -(-3)x^{-4} = 3x^{-4} \) \[ \displaystyle \frac{dy}{dx} = 18x^2 - 4x^{-3} + 3x^{-4} \]

Use the rule \( \displaystyle \frac{d}{dx}[\log(u)] = \frac{1}{u} \cdot \frac{du}{dx} \): Let \( \displaystyle u = 4x^3 - 2x \). Then \( \displaystyle \frac{du}{dx} = 12x^2 - 2 \). \[ \displaystyle \frac{d}{dx} = \frac{1}{4x^3 - 2x} \cdot (12x^2 - 2) = \frac{12x^2 - 2}{4x^3 - 2x} \]

Substitute \( x = -2 \) into the function: \[ f(-2) = 3(-2)^3 + 4(-2)^2 + (-2) - 8 \] \[ f(-2) = 3(-8) + 4(4) - 2 - 8 \] \[ f(-2) = -24 + 16 - 2 - 8 = -18 \]

Multiply both sides by the LCM of 3 and 2, which is 6: \[ 2(4x - 6) \le 3(3 + 2x) \] \[ 8x - 12 \le 9 + 6x \] \[ 8x - 6x \le 9 + 12 \] \[ 2x \le 21 \implies x \le \frac{21}{2} \]

Split into two parts: 1) \( -7 \le 9 - 8x \implies 8x \le 16 \implies x \le 2 \) 2) \( 9 - 8x < 16 - x \implies -7 < 7x \implies -1 < x \) Combining both: \( -1 < x \le 2 \)

Calculate each term for \( n = 1, 2, 3, 4 \): - \( n=1: 2^{2(1)-1} = 2^1 = 2 \) - \( n=2: 2^{2(2)-1} = 2^3 = 8 \) - \( n=3: 2^{2(3)-1} = 2^5 = 32 \) - \( n=4: 2^{2(4)-1} = 2^7 = 128 \) Sum \( = 2 + 8 + 32 + 128 = 170 \)

Integrate the expression: \[ \left[ \frac{2x^3}{3} + \frac{x^2}{2} \right]_{-1}^{2} \] Substitute limits: \[ \left( \frac{2(8)}{3} + \frac{4}{2} \right) - \left( \frac{2(-1)}{3} + \frac{1}{2} \right) \] \[ \left( \frac{16}{3} + 2 \right) - \left( -\frac{2}{3} + \frac{1}{2} \right) = \frac{22}{3} - (-\frac{1}{6}) = \frac{44+1}{6} = \frac{45}{6} = 7\frac{1}{2} \]