Mathematics Grade 12 Paper 1 Gauteng Mock 2015 Memo

Downloaded from hlayiso.com INCE PREPARATORY EXAMINATION 2015 MEMORANDUM SUBJECT: MATHEMATICS P1 (10611)

Downloaded -from-hlayise-com————___ 10611/15 GAUTENG DEPARTMENT OF EDUCATION PREPARATORY EXAMINATION -~ 2015 MATHEMATICS (First Paper) MEMORANDUM QUES TION 1 [21 MARKS] Ll 111 x=-2 or x=V2 or x=-J2 Le [3] [| 2-12 [| x*>9 of x?-950 1A x23 or x<-3 1A for critical value of 3 2A for each inequality Max 2/3 if ‘and’ is used Full marks if ‘or’ is omitted | [4] | 1.13 6"*_2?* — 8 6 1A for factorizing bases or gfe 32x = 8.32%,22* or gt 322 = Q3+2x 32x — ar ICA = 22-23 ICA || aS 3 25 ICA are 9 12 (+)? => 1A ————} = —- | 9 +1=4,/— ane fa [x=112 or _x=-3,12 2CA for each solution i | 2x? +4x-7=0 1A standard form — — | ~44f4? —4(2)(-7 x= elf 42) ICA substitution into correct 2(2) formula — | x=112 or x=-3]2 2CA for each solution | =e]

Downloaded from hlayiso.com 10611/15 13 C2xtT+5 _ 1 aa] _ 1A correct substitution. 2(-2x +12) =x—-1 1CA simplification x=5 i ICA Substitutex=5: y=—2(5)+7 a y=-3 ICA OR Jo LE vi 2 aS 1A correct substitution 5 Ce are 2 1CA simplification Lode 1CA substitute y=-3: x= pati — ae ICA {4] 1.4 1-p>0o 1A Ded ICA [2] QUESTION 2 [12 MARKS] | 21 ( sr 1A for —= 2 2 pa 7 5 1CA for multiplication with x (2] 2 | xt+xt+2 TA for x+2 = 2x+2 1CA for adding to x (21. 22 ¢ IM Seda eery iCA x? +6x=0 1CA for simplified form x(x+6)=0 1CA factorizing x=-6 1CA choosing correct x-value [5] 23 The series does NOT converge. ICA 2M _ for any logical explanatior. r=3. toconverge, —l<r<l [3]

Downloaded-from-htayise-com 10611/15 * QUESTION3 [8 MARKS] 3.1 | (250)? + 4(250) 1A substitution = 63500 ICA [2] 3.2 Foo = Sica ~ Soo 1M = 100° + 4(100) —[99? + 4(99)} 1A correct substitution = 203 ICA OR T=5;T,=7; T, =9 1A for first three terms Dooo = 5+ 99(2) 1M for correct formula = 203 1CA [3] 3.3 n? +4n =1440 1A (n+ 40)\(n—36) =0 ICA correct factors or correct substitution into correct formula. n= 36 1CA choice of correct 7. Full marks for answer only {3} QUESTION 4 [6 MARKS] | 4.1 2; 6; 12; 20 1A for first three terms 1A for the 4" term [2] — 4.2 a=], b=1; c=0 3A finding a, b, c using any correct method. T =n? +n or T, =na+l ” Trop = 100(100+1) =10100cm? ICA [4]

Downloaded from hlayiso.com 10611/15 QUESTION 5 [15 MARKS] 5.1 |5.1.1 | 4=6500000-0,3)* 1A substitution into correct formula =R156 065 1CA {2] 5.1.2 | 4 =650000(1+0,15)* 1A substitution into correct formula =R1 136 854,06 ICA | [2] Bye) 1136854, 06 ~ 156065 = 980789, 06 ICA 0,095)" 1M for use of correct formula x (: + oe} -1 1CA us of correct formula 980789, 06 = ———_—____—___+ j 780789 0,095 12 = R16 875,92 1CA [4] 52 | 521 0,092)" 1M wx 11636,02 (: 102) 1] 1A correct substitution oB =1275000/ 1422092) _ ® 12 0,092 12 = R1056 675,39 ICA [3] REELS) 5 0,092 1056675 39(1 +o) 1CA substitution into correct formula = R1097807,15 ICA 0,092) eS x{1-} 1+ Te 1CA substitution into correct formula 1097807,15 =—-—______*_/ 97807,15 0,052 12 = R12297,82 ICA [4] a i Ea | | QUESTION 6 [3 MARKS] oN 1M for correct shape 1A for horizontal asymptote 1A for negative y-value 3]

Downloaded from -hlayiso-com———__ _ 4061115 ‘QUESTION7 [19 MARKS] — a -2x? ~4x+30=0 1M equating to 0 a) (x+5)(x-3) =0 1A correct factors oo A(-5; 0) B(3;0) 2CA for both intercepts -1 if not in coordinate form | [4] oo 7.2 y =-2(x? +2x-15) 1A for taking out -2 | y=-2(x? +2x4+1-15-1 2 , ICA (2) added and deducted y =~2[(x+1)? ~16] ICA incomplete square form | y=-2(x+1)? +32 1CA | Tpt 1:32) ica 2 OR = ooo 1A substitute into correct formula ae) edie ICA y=32 1CA y =-2(x+1)? +32 1CA in correct form Tpt (-1,32) ICA 5] aa y =-2(1)? - 4() +30 1M substitution into equation y=24 - No ICA value and conclusion OR p 2x? —4x4+30=2x+10 IM finding point of intersection x’ +3x-10=0 x=2 or x=-5 -No 1CA for values with conclusion | — [2] a 74 2x? —4x430=mx430 IM for equating | —2x* +(-4—m)x-2=0 1CA for standard form A=0 iM = 7 (-4~m)? —4(-2)(-2) =0 1CA correct substitution m=0 or m=-8 2.CA for each value of m ad [6] ——_—_ TEs) (-29;3)—{-5} or x<3; x¥-5 2A [2]

Downloaded from hlayiso.com 10611/15 | QUESTIONS [il MARKS] | [ x=2 1A 8.1 Vas HA: y=-] 1A - [2] 82 | i y-intercept: ae. IM 3 ICA x-intercept: ye 1=0 1M x=3 ICA [4] iS) 1A intercepts with axes 1A. asymptotes : 1A Shape [3] 8.4 1) i I 1 _ 1A correct substitution nN Pa i 09 | Ba) a | 1CA | [2]

Downloaded from hlayiso.com 0611/15 QUESTION 9 [6 MARKS] : 9.1 9 1A correct substitution a —2=log,| — 4 ae sg 1CA exponential form ea 2 ' 5 ICA E [3] 9.2 y=~log, x or y=log, x 3 2 1A {4 9.3 g(x) >-2 1A {i] 9.4 (—©0} 00) 1A [4 QUESTION 10 [14 MARKS] 10.1 [1011 | fG+A)-f3) h 1M correct formula -23+h) +1-[26" +1] = h 1A correct substitution -12h—2h? it 7 1CA correct simplification _ h(-12—2h) 2 ICA =-12-—2h Max 1/3 if ‘limit’ is used oo OR fath-f@ h 1M. correct formula —2(x+ hy +1-[-2x? +1] = ™ 1A correct substitution _ -4xh—2h? = h : 1CA correct simplification _ h(-4G)—- 2h) _ h 1CA substitution of x=3 =-12-2h Max 1/3 if ‘limit’ is used [4] 10.1.2 | limtl2-2h) 1M using limit h-0 =12 ICA [2] 10.1.3 m=0 1A [1]

Downloaded from hlayiso.com 10611/15 10.2 | 10.2.1 | y=16-16x+4x" 1A for simplification 7 a Ve Ok Fa oe 2CA (-1 for incorrect notation) [3] 10.2.2 as yax~ax? ta 1A for simplification reels yi=l-Sax? 2CA oe 7] a 1CA positive exponents 2x2 [4] QUESTION 11 [11 MARKS] ILA ‘The graph passes through the origin. Or f(x) = (~-0)? (@-a) 1M_any logical explanation {1} 11.2 7112.1 | f(x) =2(x-2)? 1A factorization a=2 ICA [2] 1122 | £0) 0 1M 3x? -8xt+4=0 1A correct differentiation (3x —2)(x-2)=0 1CA factors 2 le ICA [4] 11.3 pO 3(0)? —8(0) +4 1CA substitution into derivative m=4 1CA {2] 114 4x2 44x = p-2 1M taking 2 to RHS p<2 1CA (full marks for answer only) [2]

Downloaded from hlayiso.com—. 10611/15 QUESTION 12 [9 MARKS] | | : x - [+9 -2] <0 1M for inequality x > 120km/ hour 1A for inequality sign ICA for 120km/hour Full marks if units are omitted. [3] | | 3 P=40x? ~— | 3 P(x)=0 1M 80x—x? =0 1CA. ic x(80—x)=0 1CA_ correct factorization : x = 80km/ hour ICA 40 303 Ora ICA ICA Rae ene (aay Re533 el Full marks if units are omitted. [6]

Downloaded from hlayiso.com 10611/15 | QUESTION 13. [15 MARKS] ie i 7 6 IG 2A for both outcome L Ur IM for addition | l9 [73 1CA for simplification — [4] 13.2 | 13.2.1 Ss 3A for each section [3] . 1322) 1,541.) 6 12.3 1M ol w= 75 1A 1 1 P(A) ==4+—== ‘) 6 12°54 1CA [3] 13.3 36 000 1A E y 134 [13.4.1 [291 2A a [2] 1342 | 33) 2A a | [2] 10

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