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

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