Date | May 2019 | Marks available | 5 | Reference code | 19M.1.SL.TZ1.S_8 |
Level | Standard Level | Paper | Paper 1 | Time zone | Time zone 1 |
Command term | Find | Question number | S_8 | Adapted from | N/A |
Question
Let f(x)=9−x2, x∈R.
The following diagram shows part of the graph of f.
Rectangle PQRS is drawn with P and Q on the x-axis and R and S on the graph of f.
Let OP = b.
Consider another function g(x)=(x−3)2+k, x∈R.
Find the x-intercepts of the graph of f.
Show that the area of PQRS is 18b−2b3.
Hence find the value of b such that the area of PQRS is a maximum.
Show that when the graphs of f and g intersect, 2x2−6x+k=0.
Given that the graphs of f and g intersect only once, find the value of k.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
valid approach (M1)
eg f(x)=0, 9−x2=0 , one correct solution
x=−3, 3 (accept (3, 0), (−3, 0)) A1 N2
[2 marks]
valid approach (M1)
eg height = f(b), base = 2(OP) or 2b, 2b(9−x2), 2b×f(b)
correct working that clearly leads to given answer A1
eg 2b(9−b2)
Note: Do not accept sloppy notation eg 2b×9−b2.
area = 18b−2b3 AG N0
[2 marks]
setting derivative = 0 (seen anywhere) (M1)
eg A′=0, [18b−2b3]′=0
correct derivative (must be in terms of b only) (seen anywhere) A2
eg 18−6b2, 2b(−2b)+(9−b2)×2
correct working (A1)
eg 6b2=18, b=±√3
b=√3 A1 N3
[5 marks]
valid approach (M1)
eg f=g, 9−x2=(x−3)2+k
correct working (A1)
eg 9−x2=x2−6x+9+k, 9−x2−x2+6x−9−k=0
2x2−6x+k=0 AG N0
[2 marks]
METHOD 1 (discriminant)
recognizing to use discriminant (seen anywhere) (M1)
eg Δ, b2−4ac
discriminant = 0 (seen anywhere) M1
correct substitution into discriminant (do not accept only in quadratic formula) (A1)
eg (−6)2−4(2)(k), (6)2−4(2)(k)
correct working (A1)
eg 36−8k=0, 8k=36
k=368(=92,4.5) A1 N2
METHOD 2 (completing the square)
valid approach to complete the square (M1)
eg 2(x2−3x+94)=−k+184, x2−3x+94−94+k2=0
correct working (A1)
eg 2(x−32)2=−k+184, (x−32)2−94+k2=0
recognizing condition for one solution M1
eg (x−32)2=0, −94+k2=0
correct working (A1)
eg −k=184, k2=94
k=184(=92,4.5) A1 N2
METHOD 3 (using vertex)
valid approach to find vertex (seen anywhere) M1
eg (2x2−6x+k)′=0, −b2a
correct working (A1)
eg (2x2−6x+k)′=4x−6, −(−6)2(2)
x=64(=32) (A1)
correct substitution (A1)
eg 2(32)2−6(32)+k=0
k=184(=92,4.5) A1 N2
[5 marks]