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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 fx      (A1)

eg    2, 9

correct vertex of -fx      (A1)

eg    2, -9

correct vertex of gx      (A1)

eg    2-9+0k , 2, -9+k

recognizing when vertex is above x-axis      (M1)

eg    -9+k>0, sketch

k>9        A1 N3

 

METHOD 3 – (transformation of f)

recognizing vertical reflection of fx      (M1)

eg    -fx , x2-4x-5 , sketch

correct expression for gx      (A1)

eg    x2-4x-5+k

valid approach for finding vertex of gx      (M1)

eg    -b2a=2 , g'x=0

correct y coordinate of vertex of gx      (A1)

eg    y=-9+k , 2, -9+k

recognizing when vertex is above x-axis      (M1)

eg    -9+k>0 , sketch

k>9        A1 N3

 

[6 marks]

Examiners report

[N/A]

Syllabus sections

Topic 2—Functions » SL 2.7—Solutions of quadratic equations and inequalities, discriminant and nature of roots
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