Date | November 2020 | Marks available | 6 | Reference code | 20N.1.SL.TZ0.S_5 |
Level | Standard Level | Paper | Paper 1 (without calculator) | Time zone | Time zone 0 |
Command term | Find | Question number | S_5 | Adapted from | N/A |
Question
Let f(x)=-x2+4x+5 and g(x)=-f(x)+k.
Find the values of k so that g(x)=0 has no real roots.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1 – (discriminant)
correct expression for g (A1)
eg -(-x2+4x+5)+k , x2-4x-5+k=0
evidence of discriminant (M1)
eg b2-4ac, Δ
correct substitution into discriminant of g (A1)
eg (-4)2-4(1)(-5+k) , 16-4(k-5)
recognizing discriminant is negative (M1)
eg Δ<0 , (-4)2-4(1)(-5+k)<0 , 16<4(k-5) , 16-4(-1)(5)<0
correct working (must be correct inequality) (A1)
eg -4k<-36 , k-5>4 , 16+20-4k<0
k>9 A1 N3
METHOD 2 – (transformation of vertex of f)
valid approach for finding f(x) vertex (M1)
eg -b2a=2 , f'
correct vertex of (A1)
eg
correct vertex of (A1)
eg
correct vertex of (A1)
eg
recognizing when vertex is above -axis (M1)
eg , sketch
A1 N3
METHOD 3 – (transformation of )
recognizing vertical reflection of (M1)
eg , sketch
correct expression for (A1)
eg
valid approach for finding vertex of (M1)
eg
correct coordinate of vertex of (A1)
eg
recognizing when vertex is above -axis (M1)
eg , sketch
A1 N3
[6 marks]