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Date November 2019 Marks available 4 Reference code 19N.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

Consider the function f , with derivative  f ( x ) = 2 x 2 + 5 k x + 3 k 2 + 2 where  x k R .

Show that the discriminant of  f ( x ) is k 2 16 .

[2]
a.

Given that f is an increasing function, find all possible values of k .

[4]
b.

Markscheme

correct substitution into  b 2 4 a c           (A1)

eg     ( 5 k ) 2 4 ( 2 ) ( 3 k 2 + 2 ) ,   ( 5 k ) 2 8 ( 3 k 2 + 2 )

correct expansion of each term         A1

eg     25 k 2 24 k 2 16 ,   25 k 2 ( 24 k 2 + 16 )

k 2 16         AG  N0

[2 marks]

a.

valid approach          M1

eg     f ( x ) > 0 ,   f ( x ) 0

recognizing discriminant  < 0 or  0           M1

eg     D < 0 ,   k 2 16 0 ,   k 2 < 16

two correct values for k /endpoints (even if inequalities are incorrect)          (A1)

eg     k = ± 4 ,   k < 4   and  k > 4 ,   | k | < 4

correct interval        A1  N2

eg     4 < k < 4 ,   4 k 4

Note: Candidates may work with an equation, then write the intervals with inequalities at the end. If inequalities are not seen until the candidate’s final correct answer, M0M0A1A1 may be awarded.
If candidate is working with incorrect inequalitie(s) at the beginning, then gets the correct final answer, award M0M0A1A0 or M1M0A1A0 or M0M1A1A0 in line with the markscheme.

[4 marks]

b.

Examiners report

[N/A]
a.
[N/A]
b.

Syllabus sections

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

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