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.

  • A - 9
  • B - 6
  • C 15
  • D 21
waec 2020furthermathematics

The correct answer is C. 15

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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