WAEC Further Mathematics Past Questions & Answers - Page 135

\(\frac{3x - 1}{(x - 2)^{2}} = \frac{A}{(x - 2)} + \frac{Bx}{(x - 2)^{2}}\)

\(\frac{3x - 1}{(x - 2)^{2}} = \frac{A(x - 2) + Bx}{(x - 2)^{2}}\)

Comparing, we have

\(3x - 1 = Ax - 2A + Bx  \implies -2A = -1;  A + B = 3\)

\(\therefore A = \frac{1}{2}; B = \frac{5}{2}\)

= \(\frac{1}{2(x - 2)} + \frac{5x}{2(x - 2)^{2}}\)

Note: Given the sum of the roots and its product, we can get the equation using the formula:

\(x^{2} - (\alpha + \beta)x + (\alpha\beta) = 0\). This will be used later on in the course of our solution.

Given equation: \(2x^{2} - 5x + 6 = 0; a = 2, b = -5, c = 6\).

\(\alpha + \beta = \frac{-b}{a} = \frac{-(-5)}{2} = \frac{5}{2}\)

\(\alpha\beta = \frac{c}{a} = \frac{6}{2} = 3\)

Given the roots of the new equation as \((\alpha + 1)\) and \((\beta + 1)\), their sum and product will be

\((\alpha + 1) + (\beta + 1) = \alpha + \beta + 2 = \frac{5}{2} + 2 = \frac{9}{2} = \frac{-b}{a}\)

\((\alpha + 1)(\beta + 1) = \alpha\beta + \alpha + \beta + 1 = 3 + \frac{5}{2} + 1 = \frac{13}{2} = \frac{c}{a}\)

The new equation is given by: \(x^{2} - (\frac{-b}{a})x + (\frac{c}{a}) = 0\)

= \(x^{2} - (\frac{9}{2})x + \frac{13}{2} = 2x^{2} - 9x + 13 = 0\)

\(\log_{3}a - 2 = 3\log_{3}b\)

Using the laws of logarithm, we know that \( 2 = 2\log_{3}3 = \log_{3}3^{2}\)

\(\therefore \log_{3}a - \log_{3}3^{2} = \log_{3}b^{3}\)

= \(\log_{3}(\frac{a}{3^{2}}) = \log_{3}b^{3}   \implies  \frac{a}{9} = b^{3}\)

\(\implies a = 9b^{3}\)

All the statements are false except option C.

\(R \cap P = {3, 5, 7} and R \cap U = {2, 3, 5, 7, 11}\) 

\(\therefore (R \cap P) \subset (R \cap U)\)

\(f(x) = 3x^{3} - 2x^{2} + 7x + 5\). 

\(x - 1 = 0, x = 1\)

\(f(1) = 3(1)^{3} - 2(1)^{2} + 7(1) + 5 = 13\)