Question

3. A function f is defined by f(x) = 72/(px+r), where p and r are constants. Given that f(6)=12 and f(7)=9, find: (a) the value of p and r; (b) f⁻¹(5/3).

Accepted Answer

(a) f(6)=72/(6p+r)=12 → 6p+r=6. f(7)=72/(7p+r)=9 → 7p+r=8. Subtracting: p=2, so r=6-12=-6. Thus f(x) = 72/(2x-6).
(b) Let y=72/(2x-6). Solve for x: y(2x-6)=72 → x = (36+3y)/y, so f⁻¹(y) = (36+3y)/y. f⁻¹(5/3) = (36+5)/(5/3) = 41/(5/3) = 123/5.

WAEC Further Mathematics 2025 Theory — Question 3

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