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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 = $\left( {\begin{array}{*{20}{c}} 3 \\ {2p} \end{array}} \right)$ and b = $\left( {\begin{array}{*{20}{c}} {p + 1} \\ 8 \end{array}} \right)$ .

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, $\left( {\begin{array}{*{20}{c}} 3 \\ {2p} \end{array}} \right)$ = k $\left( {\begin{array}{*{20}{c}} {p + 1} \\ 8 \end{array}} \right)$ , $\frac{{p + 1}}{3} = \frac{8}{{2p}}$ , 3k = p + 1 and 2kp = 8

correct working (must be quadratic) (A1)

eg 2p² + 2p = 24, p² + p – 12, $3 = \frac{{{p^2} + p}}{4}$

valid attempt to solve their quadratic equation (M1)

eg factorizing, formula, completing the square

evidence of correct working (A1)

eg (p + 4)(p – 3), $x = \frac{{ - 2 \pm \sqrt {4 - 4\left( 2 \right)\left( { - 24} \right)} }}{4}$

p = –4, p = 3 A1A1 N4

METHOD 2 (solving for k)

recognizing parallel vectors are multiples of each other (M1)

eg a = kb, $\left( {\begin{array}{*{20}{c}} 3 \\ {2p} \end{array}} \right)$ = k $\left( {\begin{array}{*{20}{c}} {p + 1} \\ 8 \end{array}} \right)$ , 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 = $\frac{4}{3}$ , k = $\frac{3}{4}$

substituting their value(s) of k (M1)

eg $\left( {\begin{array}{*{20}{c}} 3 \\ {2p} \end{array}} \right) = \frac{3}{4}\left( {\begin{array}{*{20}{c}} {p + 1} \\ 8 \end{array}} \right)$ , $3\left( {\frac{4}{3}} \right) = p + 1$ and $2\left( {\frac{4}{3}} \right)p = 8$ , $\left( { - 1} \right)\left( {\begin{array}{*{20}{c}} 3 \\ {2p} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {p + 1} \\ 8 \end{array}} \right)$

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, $a \bullet b = \left| a \right|\left| b \right|$

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

eg $\frac{{3\left( {p + 1} \right) + 2p\left( 8 \right)}}{{\sqrt {{3^2} + {{\left( {2p} \right)}^2}} \sqrt {{{\left( {p + 1} \right)}^2} + {8^2}} }} = \pm 1$ , $19p + 3 = \sqrt {4{p^2} + 9} \sqrt {{p^2} + 2p + 65}$

correct working (must include both ± ) (A1)

eg $3\left( {p + 1} \right) + 2p\left( 8 \right) = \pm \sqrt {{3^2} + {{\left( {2p} \right)}^2}} \sqrt {{{\left( {p + 1} \right)}^2} + {8^2}}$ , $19p + 3 = \pm \sqrt {4{p^2} + 9} \sqrt {{p^2} + 2p + 65}$

correct quartic equation (A1)

eg $361\,{p^2} + 114p + 9 = 4{p^4} + 8{p^3} + 269{p^2} + 18p + 585$ , $4{p^4} + 8{p^3} - 92{p^2} - 96p + 576 = 0$ , ${p^4} + 2{p^3} - 23{p^2} - 24p + 144 = 0$ , ${\left( {p + 4} \right)^2}{\left( {p - 3} \right)^2} = 0$

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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