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TMUA 2018 · Paper 1 · Question 5 of 20

TMUA 2018 Paper 1 Question 5

Algebra and functions — Remainder theorem · optimising over an assignment. Try it first; the answer and a full worked solution are below.

TMUA 2018 · Paper 1Algebra and functionsRemainder theorem · optimising over an assignment5 options
The function f is defined by f(x) = x3 + ax2 + bx + c.

a, b and c take the values 1, 2 and 3 with no two of them being equal and not necessarily in this order.

The remainder when f(x) is divided by (x + 2) is R.

The remainder when f(x) is divided by (x + 3) is S.

What is the largest possible value of R − S?

  1. A−26
  2. B5
  3. C7
  4. D17
  5. E29
Show the answer and worked solution
answer · D
  1. A−26
  2. B5
  3. C7
  4. D17
  5. E29
By the remainder theorem R = f(−2) = −8 + 4a − 2b + c and S = f(−3) = −27 + 9a − 3b + c. Subtracting, the c cancels and R − S = 19 − 5a + b. To make this as large as possible take a smallest and b largest, so a = 1 and b = 3 (leaving c = 2), giving R − S = 19 − 5 + 3 = 17.