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WAEC Further Mathematics 2014 Paper

Question : 4

Total: 40

Given that

x \cdot y = \frac{x + y}{2}, x \circ y = \frac{x^{2}}{y}

(3 \cdot b) \circ 48 = \frac{1}{3}

, find b, where b > 0.

Solution:

Concept:

Substitute the first operation result into the second operation, then solve the resulting equation for

b

Explanation:

The first operation gives:

3 \cdot b = \frac{3+b}{2}

Now apply the second operation:

\left( \frac{3+b}{2} \right) \circ 48 = \frac{\left( \frac{3+b}{2} \right)^{2}}{48}

Since the given value is

\frac{1}{3}

\frac{\left( \frac{3+b}{2} \right)^{2}}{48} = \frac{1}{3}

Simplify the fraction:

\frac{(3+b)^2}{4 \times 48} = \frac{1}{3}

\frac{(3+b)^2}{192} = \frac{1}{3}

Cross-multiply:

(3+b)^2 = \frac{192}{3} = 64

Since

b > 0

, we take the positive square root:

3 + b = 8

b = 5

Answer:

b = 5

(Option C).

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