WAEC Further Mathematics Past Questions & Answers - Page 86

Given $y^{2} + xy - x = 0$

Using the method of implicit differentiation, we have

$2y\frac{\mathrm d y}{\mathrm d x} + x\frac{\mathrm d y}{\mathrm d x} + y - 1 = 0$

$\frac{\mathrm d y}{\mathrm d x}(2y + x) = 1 - y$

$\frac{\mathrm d y}{\mathrm d x} = \frac{1 - y}{x + 2y}$

$\lim \limits_{x \to 3} \frac{x + 3}{x^{2} - x - 12} = \frac{3 + 3}{3^{2} - 3 - 12}$

= $\frac{6}{-6} = -1$

$f(x) = (x^{2} + 3)^{2}$

Using the chain rule, $\frac{\mathrm d y}{\mathrm d x} = \frac{\mathrm d y}{\mathrm d u} \times \frac{\mathrm d u}{\mathrm d x}$

Let $u = x^{2} + 3$ so that $y = u^{2}$

$\frac{\mathrm d y}{\mathrm d u} = 2u$

$\frac{\mathrm d u}{\mathrm d x} = 2x$

$\therefore \frac{\mathrm d y}{\mathrm d x} = 2u(2x) = 4xu$

But $u = x^{2} + 3$,

$\frac{\mathrm d y}{\mathrm d x} = 4x(x^{2} + 3)$

At $x = \frac{1}{2}, \frac{\mathrm d y}{\mathrm d x} = 4(\frac{1}{2})((\frac{1}{2})^{2} + 3)$

= $2 \times \frac{13}{4} = \frac{13}{2} = 6.5$

Let the roots of the equation be $\alpha$ and $\beta$.

$\alpha + \beta = \frac{-b}{a} = \frac{4}{7}$

$\alpha \beta = \frac{c}{a} = \frac{5}{7}$

$\implies a = 7, b = -4, c = 5$

Equation: $ax^{2} + bx + c = 0$

= $7x^{2} - 4x + 5 = 0$

$f(x) = p + qx$

$f(1) = p + q(1) \implies p + q = 7 .... (1)$

$f(5) = p + 5q = 19 .....(2)$

Solving for p and q using simultaneous equation, p = 4, q = 3

$f(3) = 4 + 3(3) = 13$