Date | May 2019 | Marks available | 2 | Reference code | 19M.1.AHL.TZ1.H_10 |
Level | Additional Higher Level | Paper | Paper 1 | Time zone | Time zone 1 |
Command term | Find | Question number | H_10 | Adapted from | N/A |
Question
The function p(x) is defined by p(x)=x3−3x2+8x−24 where x∈R.
Find the remainder when p(x) is divided by (x−2).
Find the remainder when p(x) is divided by (x−3).
Prove that p(x) has only one real zero.
Write down the transformation that will transform the graph of y=p(x) onto the graph of y=8x3−12x2+16x−24.
The random variable X follows a Poisson distribution with a mean of λ and 6P(X=3)=3P(X=2)−2P(X=1)+3P(X=0).
Find the value of λ.
Markscheme
p(2)=8−12+16−24 (M1)
Note: Award M1 for a valid attempt at remainder theorem or polynomial division.
= −12 A1
remainder = −12
[2 marks]
p(3)=27−27+24−24 = 0 A1
remainder = 0
[1 mark]
x=3 (is a zero) A1
Note: Can be seen anywhere.
EITHER
factorise to get (x−3)(x2+8) (M1)A1
x2+8≠0 (for x∈R) (or equivalent statement) R1
Note: Award R1 if correct two complex roots are given.
OR
p′(x)=3x2−6x+8 A1
attempting to show p′(x)≠0 M1
eg discriminant = 36 – 96 < 0, completing the square
no turning points R1
THEN
only one real zero (as the curve is continuous) AG
[4 marks]
new graph is y=p(2x) (M1)
stretch parallel to the x-axis (with x=0 invariant), scale factor 0.5 A1
Note: Accept “horizontal” instead of “parallel to the x-axis”.
[2 marks]
6λ3e−λ6=3λ2e−λ2−2λe−λ+3e−λ M1A1
Note: Allow factorials in the denominator for A1.
2λ3−3λ2+4λ−6=0 A1
Note: Accept any correct cubic equation without factorials and e−λ.
EITHER
4(2λ3−3λ2+4λ−6)=8λ3−12λ2+16λ−24=0 (M1)
2λ=3 (A1)
OR
(2λ−3)(λ2+2)=0 (M1)(A1)
THEN
λ = 1.5 A1
[6 marks]