Date | May 2017 | Marks available | 2 | Reference code | 17M.1.hl.TZ1.12 |
Level | HL only | Paper | 1 | Time zone | TZ1 |
Command term | Write down | Question number | 12 | Adapted from | N/A |
Question
Consider the polynomial P(z)=z5−10z2+15z−6, z∈C.
The polynomial can be written in the form P(z)=(z−1)3(z2+bz+c).
Consider the function q(x)=x5−10x2+15x−6, x∈R.
Write down the sum and the product of the roots of P(z)=0.
Show that (z−1) is a factor of P(z).
Find the value of b and the value of c.
Hence find the complex roots of P(z)=0.
Show that the graph of y=q(x) is concave up for x>1.
Sketch the graph of y=q(x) showing clearly any intercepts with the axes.
Markscheme
sum=0 A1
product=6 A1
[2 marks]
P(1)=1−10+15−6=0 M1A1
⇒(z−1) is a factor of P(z) AG
Note: Accept use of division to show remainder is zero.
[2 marks]
METHOD 1
(z−1)3(z2+bz+c)=z5−10z2+15z−6 (M1)
by inspection c=6 A1
(z3−3z2+3z−1)(z2+bz+6)=z5−10z2+15z−6 (M1)(A1)
b=3 A1
METHOD 2
α, β are two roots of the quadratic
b=−(α+β), c=αβ (A1)
from part (a) 1+1+1+α+β=0 (M1)
⇒b=3 A1
1×1×1×αβ=6 (M1)
⇒c=6 A1
Note: Award FT if b=−7 following through from their sum =10.
METHOD 3
(z5−10z2+15z−6)÷(z−1)=z4+z3+z2−9z+6 (M1)A1
Note: This may have been seen in part (b).
z4+z3+z2−9z+6÷(z−1)=z3+2z2+3z−6 (M1)
z3+2z2+3z−6÷(z−1)=z2+3z+6 A1A1
[5 marks]
z2+3z+6=0 M1
z=−3±√9−4∙62 M1
=−3±√−152
z=−32±i√152 A1
(or z=1)
Notes: Award the second M1 for an attempt to use the quadratic formula or to complete the square.
Do not award FT from (c).
[3 marks]
d2ydx2=20x3−20 M1A1
for x>1, 20x3−20>0⇒ concave up R1AG
[3 marks]
x-intercept at (1, 0) A1
y-intercept at (0, −6) A1
stationary point of inflexion at (1, 0) with correct curvature either side A1
[3 marks]