Draft Version 1 Mark scheme Further Maths Core Pure (AS/Year 1) Unit Test 1: Complex numbers 1
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1 1 w z k k States or implies that 4 i TBC Uses the definition of argument to write 4 k π tan 1 k 4 Makes an attempt to solve for k, for example 4 + k = k is seen. M1.a Finds k = 6 (4 marks) Pearson Education Ltd 017. Copying permitted for purchasing institution only. This material is not copyright free. 1
2 a Finds r = 1, using r 6 6 r 144 r 1 M1.a TBC π 6 π arg z tan Finds. Likely states 6 and π π arg z π then deduces M1.a Writes π π z 1cos isin A1.a () b States z w 1 π π π π cos isin 4. Award one 1 4 seen and one method mark for method mark for π π π π or seen. M.a TBC States a fully correct answer: z π π cos isin w 6 6 () (6 marks) Pearson Education Ltd 017. Copying permitted for purchasing institution only. This material is not copyright free.
3 a Deduces that the midpoint of ( 8, 6) and (4, ) is (, ) M1.a TBC Calculates that the slope of the line joining ( 8, 6) and (4, ) is Deduces that the slope of the perpendicular bisector is M1.a Finds the correct equation of the locus (perpendicular bisector): y x 5 (4) b Figure Draws a straight line with a positive slope. Fully correct answer with (0, 5) 10, 0 and labelled. TBC c Demonstrates an understanding of the need to find the point of y x y x 5 intersection of and () M1.a TBC Solves to find 0 x 1 and 0 y 1 Finds the distance: d min 0 0 d 10 1 min A1.1 () (9 marks) a An alternative algebraic approach is acceptable. Pearson Education Ltd 017. Copying permitted for purchasing institution only. This material is not copyright free.
4 Pearson Education Ltd 017. Copying permitted for purchasing institution only. This material is not copyright free. 4
5 4a Figure Circle drawn with centre (6, 1). Circle should clearly cross the real axis and not touch the imaginary axis. TBC A1.a 4b Draws a line from the point (11, 10) that is tangential to the circle with centre (6, 1) and radius 5. States or implies that length of the opposite side is 5 (the radius of the circle). () M1.a TBC Calculates the length of the hypotenuse of this triangle is arcsin Deduces that 106 Figure 4 Clearly explains that the minimum angle is π 5 arcsin 106 with explanation referring to a diagram or providing a clear explanation. For example, as shown in the diagram opposite. A1.1 (5) (7 marks) Pearson Education Ltd 017. Copying permitted for purchasing institution only. This material is not copyright free. 5
6 5 Figure 5 Circle drawn with centre (1, ). TBC Circle should just touch the real axis and clearly cross the imaginary axis. Points (, ) and (, 4) indicated on the diagram. M1* 1.1b Line drawn at y = 1. A1.a Shades correct region. M1.1a Fully correct solution. 5 (6 marks) Award the method mark providing the line y = 1 is drawn correctly, even if the points (, ) and (, 4) are not indicated. Pearson Education Ltd 017. Copying permitted for purchasing institution only. This material is not copyright free. 6
7 6 Figure 6 Circle drawn with centre (, 5). Circle should just touch the imaginary axis and clearly not touch the real axis. Two half lines drawn on the diagram. Half lines start at ( 6, 5) and intersect the circle at the top and the bottom. TBC A1.a Shades correct region. M1.1a Fully correct solution. (6 marks) Pearson Education Ltd 017. Copying permitted for purchasing institution only. This material is not copyright free. 7
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