Mathematics Grade 11 March Test 2018 Paper & Memo

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Downloaded from hlayiso.com Mathematics March 2018 Common Test INSTRUCTIONS AND INFORMATION Read the following instructions carefully before answering the questions: This question paper consists of 4 questions. 2. Answer ALL the questions. ue Number the answers correctly according to the numbering system used in this question paper. 4, Clearly show ALL calculations, diagrams, graphs, et cetera, which you have used in determining the answers. 6. You may use an approved scientific calculator (non-programmable and non-graphical) unless stated otherwise. 7. If necessary, round off answers to TWO decimal places, unless stated otherwise. 8. Diagrams are NOT necessarily drawn to scale. 9. Write neatly and legibly. Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics 3 March 2018 Common Test NSC QUESTION 1 I.1 Solve tory in cach of the following: Lt 3x? 4+10x-5=0 (correct to TWO decimal places) (4) Lio seta (4) 113) Savev4rel (5) 1.2 Solve forx and y if: V4+7=2y and x -yy43y° =15 (6) 130 Given: f(x) =99° - Sy 44 1.3.1 Solve forx if f(x) 50. (3) 1.3.2 Show that /(v) = -3 has no real roots. (4) [26] QUESTION 2 Simplify fully, WITHOUT using a calculator: yl gg 21 ee 8" isl /g§ (5) 2.2 Tf y =8/100.000, WITHOUT USING A CALCULATOR, determine the value of V16 x 4/625 x V0 in terms of ¥. (4) [9] Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics 4 March 2018 Common Test NSC QUESTION 3 3.1. Consider the following quadratic number pattern: 64; 42:24)... 3.1.1 Write down the next TWO terms of the number pattern. (2) 3.1.2. Determine an expression for the general term, T,, in the form T,=an +bn+c. (4) 3.1.3 Calculate the value of the 20" term of this number pattern. (2) 3.1.4 Determine the general term of the sequence of first differences of this number pattern. (2) 3.1.5 Between which two consecutive terms of the quadratic number pattern will the difference be equal to 174? (2) If py tl: 20 3 6p forma quadratic number pattern. Calculate the value of p. (4) [16] Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics March 2018 Common Test NSC QUESTION 4 4.1 A(psq), B(234), C(3;0) and D(—1; -2) are the vertices of parallelogram ABCD. O is the origin and OE is perpendicular to AB. \ BGA) , (0 CG0) x DE-L-2) 4.1.1 Calculate the length of DC. (leave your answer in surd form) (2) 4.1.2 Hence, write down the length of AB. (1) 4.1.3 Calculate the values of pand g. (2) 4.1.4 Determine the equation of OE. (4) 4.1.5 Calculate the coordinates of E. (4) Copyright Reserved Please turn over

Downloaded from hlayiso.com March 2018 Common Test 6 NSC Mathematics In the diagram P (3 ; 7), Q (-3 ; 4) and R (1: -6) are the vertices ofa triangle. PQ is produced to cut the x-axis at S. PR cuts the x-axis at T. QR cuts the y-axis 4.2 al A. y PGs 7) Q63 34) —_ “@ i a : yo } . S A\ |O {T x R(I 5-6) 4.2.1 Calculate 6 the angle of inclination of the line PS. (3) 4.2.2 Calculate the size of ROS. (4) (4) [24] N is the point (3. ;-11). Are the points N. Rand Q collinear? 4.2.3 Justify your answer by means of calculations. TOTAL MARKS: 75 Copyright Reserved

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Downloaded from hlayiso.com 3 GRADE II -omon Test March 2018 Mathematics QUESTION 2 4 GRADE 11 Marking Guide Common Test March 2018 4-DAY | v "51 Y writing as prime bases ¥ converting surd to exponent Y simplification using laws Y answer (5) 13.1 9x° -15x44>0 (x-4)(3e-) > 0 Y factors 4 1 vst oor x=t 3 3 £ = | 3 3 c= OF plea a y | 3 3 (3) 13.2 | 9 -15x+4=3 >_< Y standard form 9° -15x+7=0 _ ~b#Vb* -4ac = 2a A=b'-4ac Y substituting in A or formula < = _ 2 for roots of quadratic _ 4 ¥C15) -40)7) = (-15Y -4(9(7) equation 2(9) =-27 1stJ-27 is+v—27 = OR Vixe or A=-2 18 Because A< 0, the 18 se V=27 is not real, equation has no Y conclusion both roots are not real real roots. @ [26] Copyright Reserved Please turn over 22 | YiexVo2sxVi0 =V2 x¥5* x10 =(2)' x3) x(10)! =(10) x10)? = (10) 5 =10'x (10) =10y writing with bases of 2 and 5 ¥ surd to exponential form with base 10 Y simplification ¥ answer (4) Copyright Reserved Please turn over

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Downloaded from hlayiso.com 7: GRADE II Marking Guideline QUESTION 4 somon Test Maroh 2018 WW | pc = (3-(-b)" +(0-(-2)) 20 ¥ substitution into distance formula Y¥ answer 2) 4.1.2 | sp=J20 Y correct answer ia) 4.1.3] By inspection; p=-2 and g=2 (D is 4 units to lett of C: therefore A also 4 units to left of B Dis 2 units below C; therefore A also 2 units below B.) ¥ value of p | Y value ofg 2) OR Midpoint of BD is OR ip a ae 29 wl2 L4 2 = (2° 2) | | L_ p+3 + | 12 P88 gig 1-428 | 2 2 2 Y value of p | p=2 and q=2 ¥_value of g 2), 414 4-21 Y substitution | hay = 242 z Y gradient of AB Lmgy = 2 ¥ gradient of OE The equation of OE is y =-2x Y answer (4) 4.15 | Equation of AB | I Substitute (2 ; Bpxte 45 dere ; 2 c=3 1 ¥ equation of patasd equation of AB Equate equations of AB and OE: Y equating equations of AB and 1 5 OE axa Tx+3 5 oo ol | 2 | 6 yes * . | 5 y¥ value of x-coordinate If _6),,_2 or) re Y value of y-coordinate | 612) B/ -—;= | @) 4 5 5) Copy eserved Please tun over Mathematics 8 GRADE. ne Gui Common Test March 2018 421 _7-4_1 Migs 2. Y gradient of PS | tan ASQ = my = ; Y tan 4SQ=m,, | ASQ = 26,57" Y unswer (3) | = 3 | ey Y gradient of RQ tan TAQ =m TAQ = 180° — 68,20 =11180° Y size of TAQ ROS =111,80" — 68,20" Y subtracting = 450% Y answer maa 4) 4.23 _-6-(-11) Tx = 1-3 ¥ substitution | | 5 aS ¥ value of gradient | | 2 | | eee Y equal gradients | 2 Because my = ry, .N, Rand Q are collinear. Y concluding, Take note: Alternatively the gradients of NQ and NR may also be shown | to be equal. | Or: NQ and QR. (4) (24) | Copyright Reserved TOTAL: 75