AJC H2Maths 2012Prelim P1 Solution

AJC Preliminary Examination 2012 H2 Mathematics Paper 1 (9740/01) Solution Qn Soln 1  100 2 99 100002 nS a d   2 99 200ad   a, a+d, a+4d are consecutive terms in GP: 4a d a d a a d     (a+d)2 = a(a+4d)  d2 = 2ad  d = 2a since d  0 . Sub d = 2a into 2a + 99 d = 200, get d = 2 and a = 1. 2 22 8Ax By Cy   (2,1)  4A + B + C = 8 -----(1) Diff (*) wrt x : dd2 2 0 dd yyAx By C xx    d2 d2 y Ax x By c   Tangent at (2,1) // y-axis : 2B + C = 0 ------(2) [1] Diff again wrt x : 222 22 d d d2 2 0 d d d y y yA B y C x x x       When y = 0 , d3 d2 y x  and 2 2 d9 d4 y x   392 2 0 24A B C            (3) Solve the 3 eqns : get A = 3, B = 4 and C = -8 3  6 4 (2 1)( 3) 03 x x x x       221 03 xx x    (2 1)( 1) 03 xx x     3 (2 1)( 1) 0x x x     1 or 1 32xx   6 4cosec 1 2sin1 3cosec      6sin 4 1 2sinsin 3     Replace x by sin , 1sin , 1 sin 32    7 11 or6 6 2      4 (i) Reversing the transformations: a. Stretch parallel to y-axis by factor ½ gives 2 1 24 y x   b. Translate 1 unit to the right gives 2 1 2 4 ( 1) y x   c. Reflection in y-axis gives 22 11 = 2 4 ( 1) 2 4 ( 1) y xx       = f(x) 4 (ii) The graphs of y = g(x) and y = g-1(x): y = g(x) y = g-1(x) -2 -2 ½ ½ x y B 1 3 1 A 2 4 6 −1.5 −1 −0.5 0.5 1 1.5 x y

4 (iii) area of the region bounded by y = g-1(x), the x-axis and the line x = 1 = region A = region B = Rectangle - 0 3 dyx   =    0 2 3 11 3 d 4 x x   = 0 1 3 3 sin 2 x      = 30 3       = 3 3  . 5 2 2 2 2 .....1 2 3 2 3 4 3 4 5 ( 1)( 2) nS n n n            1 2 3 1 1 1 5 1 1 9 1 1 = , = , =3 2 2 3 12 2 3 4 20 2 4 5S S S        . (ii) nS = 11 2 ( 1)( 2)nn  by observation. (iii) Let nP be the statement “ nS = 11 2 ( 1)( 2)nn  ” for nZ  P1 is true from (i) Assume that kP is true for some kZ  ie. kS = 11 2 ( 1)( 2)kk  We need to show that 1kP  is true, ie to prove that 1kS  = 11 2 ( 2)( 3)kk  LHS = 1kS  = kS + (k+1)th term = 11 2 ( 1)( 2)kk  + 2 ( 1)( 2)( 3)k k k   = 1 3 2 2 ( 1)( 2)( 3) k k k k     = 11 2 ( 2)( 3)kk  = RHS Therefore 1kP  is true. Since 1P is true and kP is true 1kP  is true,  by MI, nP is true for nZ  6 (i) By pythagoras’ theorem: 24lr and 2 2 2 2 (2 ) 4 4R r R r R      A = πrl  A = π 4 4 4RR 24πA R R   (ii) dA dA dR dV dt dR dV dt   22 2 (2 1) 1 84 dA R dt R RR     2 (4 1) 1 3 3 2 84 (4) 24 2 2 dA dt       7 (i) Since the sequence converges to L, ie as 1, and nnn x L x L    3 22 1 1 12 3 2 1 13L L L L L LLL          (ii) Consider 1nnxx  = 2 11 23 nn n xx x    324 43 dVV R R dR  

Method 1: 1nnxx  = 2 11 23 nn n xx x    =   33 2 1 2 1 33 nn n xxx    3 2 1 13 n n xx Since 1,nxL 310 nx 11 0.n n n nx x x x     Method 2: Use GC, sketch 2 11 23y x x x     From the graph, for 1,nxL 110 0 . n n n ny x x x x       (iii) The sequence is such that 001 x , and from (i) , 1 nnx  . From (ii) , 1 2 31, 1, 1,.....x x x   and 4 3 2 11 .......nx x x x x     the sequence will decrease and converge to the limit 1 from the right for 1.n Since L =1, 1 1 1 1n n nd x L x      11 1nndx 2 11 213 n n x x         2 11 2 1 13 1 n n d d          21 2 2 1 33 nndd            2231 1 2 1 2 ....3 2! n n nd d d          2 nd Range of validity is 1 1 1.nndd    8a 2( 5) 3yx   -----(1) 2( 5) 3 5 3y x y x       10yx -----(2) Points of intersections are 4, -6) and (7, -3) Volume generated       224 7 7 2 3 4 3 5 3 dx + 10 dx 5 3 dx x x x             127.2345 127 (3 s.f.) 8b    dx sin2 1cos 2 1dx cos 222 xexexe xxx 2 2 2 2 1 1 1cos dx cos sin osx dx 2 4 4 x x x xe x e x e x e c       '222 sin 4 1cos 2 1dx cos4 5 Cxexexe xxx    2 2 2 21cos dx cos sin 55 x x xe x e x e x C      At 0, 0xt . At 1, 2xt  2dy dy dt 2 2dx dt dx cos te t    Equation of tangent at x = 0: 21yx  Exact area bounded 1 0 dxy *(Since the area of both triangles are the same) 22 0 cos dttet   222 0 21 cos sin55 tte t e t     1 25 e  1 1 2 2 cos 2xy e x 21yx  1 y 0 x y nx y y 1

9 (i) 22xay xk        22 22 d ( )(4 ) (2 ) 2 4 d y x k x x a x kx a x x k x k       For the curve to have at least 1 tangent parallel to the x -axis, d 0d y x  must have real roots, i.e. 22 4 0x kx a   has real roots 2 2 2(4 ) 4(2)( ) 0 16 8 0 2k a k a k a       Since 22ka , 2 or (rejected >0) 2 2 2 a a ak k k k     (ii) 2222 22x a k ay x k x k x k     When 22ka , 22y x k Thus, the graph is a straight line. (iii) From diagram, 02 b 10 (i) (i) A(0,1,0) lies on p2: 8(0)+a(1) + (0) = 4 hence a = 4. Director vector of L : 2 8 1 1 4 2 0 1 0                              01 : 1 2 , 00 Lr                      10 (ii) (ii) AB L and 1AB n  1 2 0 0 2 1 0 5 0 0 0 5 1 AB                                         . Hence 0 0 1 AB    2nd part: Method 1 Let 00 00 1 AB k k                   0 0 0 1 0 1 0 OB OA AB kk                                  2 2 2 8 44 081 5 1 4 4 45 8 4 1 1 OB k                          45k . 0 1 45 OB   or 0 1 45    Method 2 BC = 5 = length of projection of AB onto 2n = 2 2 2 2 08 1ˆ 04 98 4 1 1 kAB n k                    . Hence 5 459 k k   . 2 2 4( ) 1 ykxk b    -k 2 a 2 a a k y=2x-2k P2 A B C P1 5 n2

0 0 45 AB   or 0 0 45 AB   0 1 0 OB OA AB AB       . 0 1 45 OB   or 0 1 45    10 (iii) Method 1 : Acute angle between line AB and p2 = acute angle between p1 and p2 = 12nn = 1 1 1 12 28 1 1 20cos cos 1 4 cos 6.4 5 64 16 1 9 501 onn                        Method 2 : acute angle between line AB and p2 = 1 1 1 2 08 1 1 45sin sin 0 4 sin 6.4 45 45 9 64 16 145 1 oAB n AB                         10 (iv) p3 : 2x+ y +z = 6 . 3 1 2 1 2 1 2 0 00 n                               for all values of . Hence p3 // L ----(1) 2(0) + 1+(0) = 1 6  A(0,1,0) does not lie on p3 -----(2) Hence line L does not intersect p3. Therefore p1, p2 and p3 do not meet at a common point. When  = 0, p3 : 2x+ y = 6 , p1 : 2x + y = 1 , p2, 8x +4y + z =4 Geometrically, p1 and p3 are parallel with p2 intersecting both p1 and p3.

11 Angle that locus of Z makes with the real axis = 4   .  2i i ic a a a a a     OC OB AB OB BC     Geometrical relationship: AC is the diameter of the circle with centre B. [Or A, B, C are collinear; Or B is the midpt of A and C]