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Date November 2013 Marks available 3 Reference code 13N.2.hl.TZ0.7
Level HL only Paper 2 Time zone TZ0
Command term Find Question number 7 Adapted from N/A

Question

The vectors a and b are such that  a =(3cosθ+6)i +7 j and b =(cosθ2)i +(1+sinθ)j.

Given that a and b are perpendicular,

show that 3sin2θ7sinθ+2=0;

[3]
a.

find the smallest possible positive value of θ.

[3]
b.

Markscheme

attempting to form (3cosθ+6)(cosθ2)+7(1+sinθ)=0     M1

3cos2θ12+7sinθ+7=0     A1

3(1sin2θ)+7sinθ5=0     M1

3sin2θ7sinθ+2=0     AG

[3 marks]

a.

attempting to solve algebraically (including substitution) or graphically for sinθ     (M1)

sinθ=13     (A1)

θ=0.340 (=19.5)     A1

[3 marks]

b.

Examiners report

Part (a) was very well done. Most candidates were able to use the scalar product and cos2θ=1sin2θ to show the required result.

a.

Part (b) was reasonably well done. A few candidates confused ‘smallest possible positive value’ with a minimum function value. Some candidates gave θ=0.34 as their final answer.

b.

Syllabus sections

Topic 2 - Core: Functions and equations » 2.6 » Solving quadratic equations using the quadratic formula.

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