Mathematics 2017 Jamb Examination

First, recognize that \( x^2 + 2xy + y^2 \) is a perfect square: \( (x + y)^2 \). Next, factor out 3 from the terms \( 3x + 3y \) to get \( 3(x + y) \). The expression becomes: \( (x + y)^2 + 3(x + y) - 18 \). Let \( u = x + y \). The quadratic is \( u^2 + 3u - 18 \). Factor the quadratic: \( (u + 6)(u - 3) \). Substituting back: \( (x + y + 6)(x + y - 3) \).

Factor out \( s \) from the right-hand side: \[ p = s \left( 1 + \frac{m^2}{nr} \right) \] Find a common denominator for the expression inside the bracket: \[ p = s \left( \frac{nr + m^2}{nr} \right) \] Multiply both sides by \( nr \): \[ p(nr) = s(nr + m^2) \] Isolate \( s \): \[ s = \frac{nrp}{nr + m^2} \]

Substitute \( x = 3 \) and \( y = -1 \) into the given binary operation formula: \[ 3 * -1 = 3(3) + 2(-1) - 1 \] \[ = 9 - 2 - 1 \] \[ = 6 \]

Use the gradient formula \( m = \dfrac{y_2 - y_1}{x_2 - x_1} \) with points \( (3, 2) \) and \( (1, 4) \): \[ m = \frac{4 - 2}{1 - 3} \] \[ m = \frac{2}{-2} = -1 \]

First, evaluate the cube root inside the bracket: \[ \sqrt[3]{64a^3} = \sqrt[3]{4^3 \cdot a^3} = 4a \] Now apply the negative exponent (which indicates the reciprocal): \[ (4a)^{-1} = \frac{1}{4a} \]

To solve this, rationalize the denominator by multiplying the numerator and denominator by the conjugate \( \sqrt{3} - 2\sqrt{2} \): \[ \frac{(2\sqrt{3} - \sqrt{2})(\sqrt{3} - 2\sqrt{2})}{(\sqrt{3} + 2\sqrt{2})(\sqrt{3} - 2\sqrt{2})} \] Numerator: \( 2(3) - 4\sqrt{6} - \sqrt{6} + 2(2) = 6 - 5\sqrt{6} + 4 = 10 - 5\sqrt{6} \) Denominator: \( 3 - 4(2) = 3 - 8 = -5 \) \[ \frac{10 - 5\sqrt{6}}{-5} = -2 + \sqrt{6} \] Comparing with \( m + n\sqrt{6} \), we get \( m = -2 \) and \( n = 1 \).

From the equation \( ax^2 + bx + c = 0 \):
Sum of roots \( \alpha + \beta = -b/a = -5/3 \)
Product of roots \( \alpha\beta = c/a = -2/3 \)
The expression \( \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} \) \[ \frac{-5/3}{-2/3} = \frac{-5}{3} \times \frac{3}{-2} = \frac{5}{2} \]

The range is the difference between the highest and lowest values in the set. Highest value = \( 0.47 \) Lowest value = \( -0.53 \) \[ \text{Range} = 0.47 - (-0.53) = 0.47 + 0.53 = 1.00 \]

Convert mixed fractions to improper fractions: \( 1\frac{2}{3} = \frac{5}{3} \). 1. \( \frac{1}{5} \times \frac{5}{3} = \frac{1}{3} \) 2. \( 5 + \frac{5}{3} = \frac{15+5}{3} = \frac{20}{3} \) Substitute back: \( 1 - \frac{1}{3} + \frac{20}{3} \) \[ = \frac{3 - 1 + 20}{3} = \frac{22}{3} \]

Expand the expression by multiplying each term: \[ (3 - 2x)(2x^2 - x + 1) \] \[ = 3(2x^2 - x + 1) - 2x(2x^2 - x + 1) \] \[ = 6x^2 - 3x + 3 - 4x^3 + 2x^2 - 2x \] Group like terms: \[ = -4x^3 + (6x^2 + 2x^2) + (-3x - 2x) + 3 \] \[ = -4x^3 + 8x^2 - 5x + 3 \]