Paper E Sec 3 IP Solutions for students

1 P aper E Sec 3 I P Solutions for students Qn Solutions 1(a)    38 1 1 22 3 2 1 1 2 x xx x x      =         22 1 4 2 1 2 1 2 1 2 3 2 1 1 2 x x x xx x x x        =       2 2 2 1 4 2 1 21 2 3 2 1 4 2 1 x x x x x x x x        = 21 23 x x   1(b) 1 4 5 3x    4 5 1 and 4 5 3xx    31 22 x   -------- (1)   4 5 18xx      2 2 20 18 20 1 2 0 xx xx xx           12 x   ---------- (2) Combining (1) & (2), 11 2x   2 (a) 2 2 3 28 2 x x   2 1 2 6 3 2 82 22 6 3 2 55 1 xx xx xx x x          2 (b)   1 24 7 2 5 x x  Alternatively, 1 4 5 3 6 4 2 31 422 x x x           S3Math10A Tutor

2        2 2 2 2 7 2 5 let 2 2 7 5 0 2 5 1 0 5 or 12 52 or 2 12 5ln 2 or 0ln 2 1.32(3 . ) or 0 xx x xx a aa aa aa xx s f x               [Accept 2 5log 2 ] 3(a) 5log 2log 5 3 xx 5 5 5 log 5 log 2 3 logx x              2 55 5 2 55 2 log 2 3 log Let log 3 2 0 1 2 0 1 or 2 log 1 or log 2 5, or 5 5 xx ax aa aa aa xx xx x              [Accept 2.24 (3 s.f.)] 3(b) 9 4 2 5xx                2 2 9 4 5 2 9 4 2 9 4 5 5 4 10 5 2 9 4 5 100 100 25 4 9 4 5 9 16 80 0 9 20 4 0 24 or 2 NA9 xx x x x x x x x x x x x xx xx x                           

3 Or alternatively,          2 2 9 4 4 5 4 5 3 16 5 9 16 80 0 9 20 4 0 2 4 or 2 NA 9 x x x xx xx xx x                 4(a) 22 6 13xx = 2 1323 2xx  = 2 3 172 24x   2 3 172 22x   4(b)   22Let 2 0x x k k            2 2 2 2 2 discriminant, - 4 0 2 4 1 0 4 4 3 0 3 4 4 0 3 2 2 0 22 or 3 b ac kk kk kk kk kk                5(i) 225 20 125 50 5y y x x    22 22 5 20 125 50 5 0 10 4 25 0 y y x x x y x y           Centre =  10 4, 5, 222   Radius = 225 2 25 2   units [Copy qn wrongly max 1m] 5(ii) Radius of circle 2 2 7 9C    Equation of circle 2C is     22 5 2 81xy    [PP if answer is     22 25 2 9xy    ]

4 6(i) Let           f ( ) 2 3 3 f ( 1) 8 2 1 3 1 3 1 8 16 1 8 1 2 x x x x k k k k                 6(ii)        1f ( ) 2 3 3 2 3 3 2 1 x x x x x x x           Remainder     f (5) 8 2 11 176    [PP if students did not mention ‘remainder’ in answer] 7(a) 2 1 5 4 5 43 1 3 1 1 1 1                P 6 3 21 24 3 9 6 3 27 27 96             7(b)  2 5 3Q         3.30 3.20 2 5 3 1.90 2.00 3.40 3.45 6.60 9.50 10.20 6.40 10.00 10.35 26.30 26.75 [accept 26.3 26.75          QC ] Conclusion: it is cheaper to shop at mini-market A 8(i)   1( ) (3) 6 7 1 Let g 1 3( 1) 12 1 g x f x x x         8(ii) 2 7 3xx   Alternatively, 1 2() 3 2 (3) 13 1 xgx x x fx x       

5        22 2 2 7 3 3 22 40 0 3 10 4 0 10 or 43 xx xx xx xx           9(i) 2x2 + 7x + 14 = 0 7 14 and 722      9(ii)   2 22 7 2 49 4 49 24 49 14 4 7 4               2  + 2  2222 72 4 7 1 2      Product 22 4  The new equation is 2 1 402xx   [or equivalent equation. A0 if an expression is given.] 10 6 32a 360 180 2 b b   1 graph is shifted 1 unit upwardsc

6 11(i) Draw a line from O to T. AOT is a right-angled triangle. 1 tan 8 tan 8 lAOT lAOT     11(ii) Area of shaded part = Area Quad. AOBT – area of sector AOPB 21 1 112 8 2 8 tan2 2 8 8 64 tan 8 lSl lSl                    11(iii) Perimeter = arc AOB + 2l 1 1 8 2 tan 2 m8 16 tan 2 m8 l l l l              11(iv)   2 2288 128 128 8 2 OT OT     Let top of pole be V. 12 3 2tan 482 VTO   Angle of elevation, VTO = 46.7 (1 d.p.) 12(a) 1 tan 31 tan x x           1 tan 3 3 tan 1 3 tan 3 1 3131tan 1 3 3 1 23 xx x x          12(b) (i) 4sin 2 1x  Basic = 14.48 2 194.48 or 345.52 97.2 ,172.8 (1d.p) x x       

7 12(ii) (b)   23cos 4sin cos cos 3cos 4sin 0 cos 0, 3cos - 4sin 0 3, rad or 4sin 3cos22 3 ta n 4 0.644, 3.79 x x x x x x x x x x x x x x        rad [If students divide the equation by cos x, max 2/3: M1 – Using sintan cos xx x . A1 – Correct values for solving 3tan 4x ] [Minus 1 mark if students give the all correct answers in degree for this question.] 13(i) 22 180 56 124 6 10 2 6 10cos124 14.251 426 19 RSQ QR           14.3 km (3s.f) 13(ii) sin sin124 10 14.25142619 10sin124sin 14.25142619 35.57178777 35.6 (1d.p) SQR SQR SQR        13(iii) Let 2N be north of R 2 (alt. ) 38 N RS QPS s    bearing of S from R is 038 . 13(iv) shortest distance sin6 shortest distance 6sin 35.57178777 3.490335191 3.49 km SQR    14(i) Step1:Horizontal translation of 3 units to the left Step2:a stretch with scale factor 2 parallel to y-axis 14(ii) Diagram

8 14(iii)   12 2 2 12 1 2 e 3 e e 3 e e3 ln( 3) 1 2 2ln( 3) 2 x x x x x x xx xx             Hence the graph of 2yx is to be drawn. The x-coordinate of the point of intersection of these two graphs is the solution to the equation. 15(i) At : 2(0) 4 4 (4,0) Bx xB     Eqn : 2 4 2 2 Let , 2 2 BC y x xy xCx       AB = AC       2 2 2 2 6 6 2 6 0 6 4 2 xx          

9        2 2 2 6 8 40 2 16 48 0 12 4 0 12 or 4 ( ) 12 12 and 2 4 2 Hence, 12, 4 xx xx xx x x given xy C                  12 4 4 0, 8, 222M  15(ii)   1grad 2 grad bisector of = 2 Let , be a point on bisector of 2 28 2 18 BC BC x y BC y x yx           15(iii) Reason 1: 1 6 2grad grad 1 2 6 10 diagonal diagonal BC AD BC AD       Reason 2:   6 10 6 ( 2)mid-pt , 22 8, 2 mid-pt diagonal bisects diagonal Diagonals are bisectors. AD BC AD BC        Reason 3: AD = BC = 80 Hence is a squareABDC Alternatively, Reason 1: AC = DC = 40 => Rhombus Reason 2: D lies on perpendicular bisector of BC (can be checked) or another pair of equal sides Reason 3: AB AC => ABDC is a square

10 15(iv)       2 4 10 12 6 41Area 0 2 4 6 02 1 8 40 72 0 0 24 24 242 1 104 242 40 units              16(i) [See Graph Paper] 16(ii) [See Graph Paper] The gradient = 2 b , so 1.00b (accept 0.90 1.10b ) The vertical intercept = 1 ln2 a , so 4.95a (accept 4.48 5.47b ) 16(iii) ln 1.675 5.34yy   (accept 5.21 5.75y )