WAEC Further Mathematics Past Questions & Answers - Page 99

\(s = 3i - j; t = 2i + 3j\)

\( t - 3s = (2i + 3j) - 3(3i - j) = 2i + 3j - 9i + 3j = -7i + 6j\)

\(t + 3s = (2i + 3j) + 3(3i - j) = 2i + 3j + 9i - 3j = 11i\)

\((t - 3s).(t + 3s) = (-7i + 6j).(11i) = -77\)

No explanation has been provided for this answer.

The equation of a circle is given as \((x - a)^{2} + (y - b)^{2} = r^{2}\).

Expanding, we have: \(x^{2} - 2ax + a^{2} + y^{2} - 2by + b^{2} = r^{2}\)

\(x^{2} + y^{2} - 2ax - 2by + a^{2} + b^{2} = r^{2}\)

Comparing with the equation, \(x^{2} + y^{2} - 8x + 9y = -15\), we have

\(2a = 8; 2b = -9; r^{2} - a^{2} - b^{2} = -15\)

\(a = 4; b = \frac{-9}{2}\)

\(\therefore  r^{2} = -15 + 4^{2} + (\frac{-9}{2})^{2}\)

= \(-15 + 16 + \frac{81}{4} = \frac{85}{4}\)

\(r = \sqrt{\frac{85}{4} = \frac{1}{2}\sqrt{85}\)

Since the bodies are in an opposite direction, one takes the negative velocity.

\(m_{1}v_{1} + m_{2}v_{2} = (m_{1} + m_{2})v\) (Momentum when the two bodies move in the same direction after collision)

\(3(-2) + 5(V) = (3 + 5)3.5\)

\(-6 + 5V = 28 \implies 5V = 34; V = 6.8 m/s\)

\(\frac{x^{2} + x + 4}{(1 - x)(x^{2} + 1)} = \frac{A}{1 -  x} + \frac{Bx + C}{x^{2} + 1}\)

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

\(\implies x^{2} + x + 4 = A(x^{2} + 1) + (Bx + C)(1 - x)\)

\(x^{2} + x + 4 = Ax^{2} + A + Bx - Bx^{2} - Cx + C\)

\(\implies (A - B)x^{2} = x^{2}; A - B = 1 ...... (i)\)

\((B - C)x = x; B - C = 1 ..... (ii)\)

\(A + C = 4 ...... (iii)\)

Solving the above simultaneous equations by any of the known methods, we get

\(A = 3, B = 2, C = 1\)

\(\therefore  \frac{x^{2} + x + 4}{(1 - x)(x^{2} + 1)} = \frac{3}{1 - x} + \frac{2x + 1}{x^{2} + 1}\)