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Date November 2018 Marks available 6 Reference code 18N.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 vectors a ( 3 2 p ) and b = ( p + 1 8 ) .

Find the possible values of p for which a and b are parallel.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1 (eliminating k)

recognizing parallel vectors are multiples of each other      (M1)

eg   a = kb,   ( 3 2 p ) k ( p + 1 8 ) ,   p + 1 3 = 8 2 p ,  3k = p + 1 and 2kp = 8

correct working (must be quadratic)       (A1)

eg   2p2 + 2p = 24,  p2 + p – 12,   3 = p 2 + p 4

valid attempt to solve their quadratic equation       (M1)

eg   factorizing, formula, completing the square

evidence of correct working      (A1)

eg   (p + 4)(p – 3),  x = 2 ± 4 4 ( 2 ) ( 24 ) 4

p = –4,  p = 3     A1A1 N4

 

METHOD 2 (solving for k)

recognizing parallel vectors are multiples of each other      (M1)

eg   a = kb,   ( 3 2 p ) = k ( p + 1 8 ) ,  3k = p + 1 and 2kp = 8

correct working (must be quadratic)       (A1)

eg   3k2 – k = 4,  3k2 – k – 4,  4k2 = 3 – k

one correct value for k      (A1)

eg   k = –1, k =  4 3 ,  k =  3 4

substituting their value(s) of k      (M1)

eg    ( 3 2 p ) = 3 4 ( p + 1 8 ) ,   3 ( 4 3 ) = p + 1 and  2 ( 4 3 ) p = 8 ,   ( 1 ) ( 3 2 p ) = ( p + 1 8 )

p = –4,  p = 3     A1A1 N4

 

METHOD 3 (working with angles and cosine formula)

recognizing angle between parallel vectors is 0 and/or 180°      M1

eg   cos θ = ±1,   a b = | a | | b |

correct substitution of scalar product and magnitudes into equation      (A1)

eg  3 ( p + 1 ) + 2 p ( 8 ) 3 2 + ( 2 p ) 2 ( p + 1 ) 2 + 8 2 = ± 1 ,   19 p + 3 = 4 p 2 + 9 p 2 + 2 p + 65

correct working (must include both ± )      (A1)

eg  3 ( p + 1 ) + 2 p ( 8 ) = ± 3 2 + ( 2 p ) 2 ( p + 1 ) 2 + 8 2 ,   19 p + 3 = ± 4 p 2 + 9 p 2 + 2 p + 65

correct quartic equation      (A1)

eg    361 p 2 + 114 p + 9 = 4 p 4 + 8 p 3 + 269 p 2 + 18 p + 585 ,   4 p 4 + 8 p 3 92 p 2 96 p + 576 = 0 ,   p 4 + 2 p 3 23 p 2 24 p + 144 = 0 ,    ( p + 4 ) 2 ( p 3 ) 2 = 0

p = –4,  p = 3     A2 N4

 

[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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Topic 2—Functions » SL 2.10—Solving equations graphically and analytically
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