WAEC - Further Mathematics (2020 - No. 27)
A function f defined by f : x -> x\(^2\) + px + q is such that f(3) = 6 and f(3) = 0. Find the value of q.
- 9
- 6
15
21
Explanation
f(x) = x\(^3\) + px + q
f(3) 3\(^2\) + 3pf + q = 6
f\(^{1}\)(x) = 2x + p
f\(^1\)(3) = 2(3) + p = 0
p = -6
from 9 + 3p + q = 6
9 + 3(-6) + q = 6
9 - 18 + q = 6
-9 + q = 6
Q = 6 + 9
= 15
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