WAEC Further Mathematics Past Questions & Answers - Page 135

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.

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

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

$f(x) = 3x^{3} - 2x^{2} + 7x + 5$. 

$x - 1 = 0, x = 1$

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

$4x^{2} + 5kx + 10 = (2x + \sqrt{10})^{2}$

Expanding the right hand side equation, we have

$4x^{2} + 4x\sqrt{10} + 10$

Comparing with the left hand side, we have

$5k = 4\sqrt{10}  \implies k = \frac{4}{5}\sqrt{10}$