NORTH SYDNEY BOYS. Mathematics. I Student Name: HIGH SCHOOL. Extension 1. o MrWeiss Year 11 Task 2

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1 NORTH SYNEY OYS HIGH SCHOOL Mathematics Etension 00 Year Task General Instruc"tions Working time - 50 minutes Write on the lined paper in the booklet provided Write using blue or black pen oard approved calculators may be used ll necessary working should be shown in every question Each new question is to be started on a new page. ttempt all questions Class Teacher: (Please tick or highlight) o Mr erry o Mr Ireland o Mr Fletcher o Mr Lam o Mr Rezcallah o Mr Trenwith o MrWeiss I Student Name: (To be used by the eam markers only.) Question 4 5 Total Total No 0/0 Mark

2 Marks (-4, 5) C has vertices (-4, 5), (5, 7) and C(, -). that the equation of the straight line through and C is +y-l=o the length of side C. divides side internally in the ratio :. the co-ordinates of. the perpendicular distance of from C. calculate the area of triangle flc line through parallel to C meets C at point E. C:alculate the area of triangle fl E. (9 marks) equation of the line which passes through the point of intersection 6y - 0 = 0 and + y - = 0, and through the point P(l, ). angle that the line + y + = 0 makes with the positive direction of the -ais. the region where y ~ - 4 and y > - are true simultaneously. 4 any boundary points clearly.

3 Question (7 marks) (a) Epress in simplest form as a single fraction, with no negative indices: (i) 4- - ( ") - I - - (b) Solve for : (i) 9 = 7-5 (ii) 4 = 4 ( c) Evaluate: (') I ogara (ii) IOg 6 - IOg 7 (d) Evaluate IOg4 7 to two decimal places. ( e) Solve for : (i) IOgl + IOgll X = IOgll 7 (ii) 0gs = logs (4 + 5) 5 Question 4 ( 0 marks) (a) Epress in terms of a and h. (no reasons needed) E c (b) , C e is a square. P is an equilateral triangle. Calculate the size of LX, giving geometrical reasons. (c) Find the size of each interior angle in a regular 0-sided polygon.

4 (d) Find the values of andy. Give the geometrical reason. (Not to scale) Question 5 (8 marks) (a) (i) State the geometrical reason why IE is similar to IC in the diagram below. (ii) Given that E = 5, calculate C. (Not to scale) (b) In the following diagram PQRS is a rhombus. (i) Prove LSPQ = LSP (ii) Prove LC = 90 0 L...---f--' ' ' C p EN OF EXMINTION

5 00 Preliminary Mathematics (Etension ) ssessment Task SOLUTIONS Suggested Solutions Question (a) ( marks) [] for correct gradient m =. [] for setting up equation. (d) ( marks) [] for correct substitution into dist. formula. ( 4, 5) C(, ) y + = 5+ 4 = 7 7 = y + = + y = + +y = 0 lternatively, substitute coordinates of points and C into equation. (b) ( mark) (e) ( marks) C : +y = 0 d = a +by +c a +b = (5)+(7) + = = [] for correct substitution. (5,7) d C = (5 ( )) +( 4 ) (c) ( marks) = 7 +7 = 49 = 7 [] for each value of and y correctly found. = bh = 7 = 7.5 ( 4, 5) (5, 7) (f) ( mark) = ( 4)+(5) + = 4+0 = (,) y = (5)+(7) + = 5+4 = E = C = 9 E = 9 47 = 47 8 = 49 6 = 8 6 Other methods eist, though this one shown here would be the easiest. NORTH SYNEY OYS HIGH SCHOOL LST UPTE JUNE 9, 00

6 00 Preliminary Mathematics (Etension ) ssessment Task SOLUTIONS Question (a) ( marks) [] for correctly placing equations into k method equation. [] for k =. [] for final factorised equation. 5 6y 0 = 0 +y = 0 5 6y 0+k(+y ) = 0 =,y= ( 4, ) y (, 4) k(+ ) = 0 k = = 5 6y 0++y 9 = 0 8+7y 49 = 0 7(4+y 7) = 0 4+y 7 = 0 (b) ( marks) [] for tanθ =. +y + = 0 y = y = Question (a) i. ( mark) 4 = 4 ii. ( marks) [] for correctly multiplying numerator & denominator by to obtain. = = ( )(+) ( ) = + m = = tanθ θ = 80 4 = 46. /46 9 (c) (4 marks) [] for dotted line y = [] for parabola y = 4. [] for boundary points [] for correct regions. { y = 4 y = 4 = + 4 = 0 (+4)( ) = 0 = 4, (b) i. ( marks) [] for =. 9 = 7 = = = ii. ( marks) [] for = 4/5. ( = ) 4 5 = () 4 5 = 4 = 6 LST UPTE JUNE 9, 00 NORTH SYNEY OYS HIGH SCHOOL

7 00 Preliminary Mathematics (Etension ) ssessment Task SOLUTIONS (c) i. ( mark) log a a = log a a = Question 4 (a) ( marks) ii. ( marks) ( ) 6 log 6 log 7 = log 7 = log 9 = a a b C E (d) ( marks) [] for correct change of base. a = +(80 b) = a+b 80 log 4 7 = log 07 log 0 4 =.8 ( d.p.) (b) ( marks) [] for correct reasoning. (e) i. ( marks) [] for correct use of addition rule for logarithms. 45 X P C log +log = log 7 log () = log = 7 = 7 ii. ( marks) [] for = 4+5. [] for solutions of. [] for justification of =. P = 60 ( in equilateral ) X = 0. = 45 (diagonals of a square bisect the ) y the sum of X, X = = 05 log 5 = log 5 (4+5) log 5 ( ) = log 5 (4+5) = = 0 ( 5)(+) = 0 = 5, ut as > 0 in the domain of the logarithmic function. Hence = 5. (c) ( marks) = (n ) 80 = 8 80 = 40 = 40 0 = 6 NORTH SYNEY OYS HIGH SCHOOL LST UPTE JUNE 9, 00

8 4 00 Preliminary Mathematics (Etension ) ssessment Task SOLUTIONS (d) ( marks) [] each for correct and y. [] for correct reasoning P Q 8 4 R y C y the intercepts of transversals over parallel lines PQ and C, = 0 8 = 0 = 5 y the same reason as above, y 4 = 0 Question 5 (a) y = = = i. ( mark) [] for final statement. Proof shown for completeness. Splitting the triangles into separate shapes: 6 4 E (5) C C = 4 6 = E = SS will not be accepted for this part LST UPTE JUNE 9, 00 E C 4 (b) E = C (common from diagram) Hence E C (two matching sides in same ratio plus included angle equal). ii. ( marks) E = 5 C E = C 5 = C = 0 i. ( marks) [] for correct proof shown. Let SPQ = α. Since PS = PS (base of isos ), and PS + PS = SPQ (eterior of ) PS = SPQ PS = SPQ = ii. ( marks) [] for each matching bullet point shown. S P Q C R y y y the same reasoning as the previouspart, if RQP = y then RCQ = y. Since SPQ + RQP = 80 (cointerior, SR PQ), then + y = (+y) = 90 y the sum of C, if C + C = 90, then C = 90 NORTH SYNEY OYS HIGH SCHOOL