WAEC Further Mathematics Past Questions & Answers - Page 132

Given \(M(4, -1)\) and \(N(x, y)\) with midpoint \(P(3, -4)\).

\(\implies \frac{4 + x}{2} = 3; \frac{-1 + y}{2} = -4\)

\(\therefore x = 2; y = -7\)

N = (2, -7)

To find the maximum value, we can use the second derivative test where, given \(f(x)\), the second derivative < 0, makes it a maximum value.

\(x(x + 1)^{2} = x(x^{2} + 2x + 1) = x^{3} + 2x^{2} + x\)

\(\frac{\mathrm d y}{\mathrm d x} = 3x^{2} + 4x + 1 = 0\)

Solving, we have \( x = \frac{-1}{3}\) or \(-1\).

\(\frac{\mathrm d^{2} y}{\mathrm d x^{2}} = 6x + 4\)

When \(x = \frac{-1}{3}, \frac{\mathrm d^{2} y}{\mathrm d x^{2}} = 2 > 0\)

When \(x = -1, \frac{\mathrm d^{2} y}{\mathrm d x^{2}} = -2 < 0\)

At maximum value of x being -1, \(y = -1(-1 + 1)^{2} = 0\)

We are given the equation \(x^{2} + y^{2} - 4x - 2y = 0\)

\(y = x^{2} + y^{2} - 4x - 2y \)

Using the method of implicit differentiation, 

\(\frac{\mathrm d y}{\mathrm d x} = 2x + 2y\frac{\mathrm d y}{\mathrm d x} - 4 - 2\frac{\mathrm d y}{\mathrm d x}\)

For the tangent, \(\frac{\mathrm d y}{\mathrm d x} = 0\),

\(\therefore 2x + 2y\frac{\mathrm d y}{\mathrm d x} - 4 - 2\frac{\mathrm d y}{\mathrm d x} = 0\)

\((2y - 2)\frac{\mathrm d y}{\mathrm d x} = 4 - 2x \implies \frac{\mathrm d y}{\mathrm d x} = \frac{4 - 2x}{2y - 2}\)

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

Equation: \(\frac{y - 3}{x - 1} = \frac{1}{2} \implies 2y - 6 = x - 1\)

= \(2y - x - 6 + 1 = 2y - x - 5 = 0\)

\(180° = \pi radian\)

\(\frac{13}{4}\pi = \frac{13}{4} \times 180° = 585°\)

\(\begin{pmatrix} 2 & x \\ 3 & 5 \end{pmatrix} = 13\)

\(\begin{vmatrix} 2 & x \\ 3 & 5 \end{vmatrix} = (2 \times 5) - (3 \times x) = 13\)

\(10 - 3x = 13 \implies -3x = 3; x = -1\)