RVHS 9758 2023 Prelim P2 Solution

River Valley High School 1 2023 H2MA Prelim Paper 2 1 Solution [5] Complex Numbers P2 Q1 (i) 12 12 22 Distance between and 4 2 2(4)(2) cos - Using Cosine Rule33 28 zz zz ππ − = = +− +  = (ii) 1 12 2 zz az a z= ⇒= i3 ii33 i3 2i 3 4 2 2 2 eae e e π ππ π π −− − = = = Either: 1z is the scaling of 2z by factor of 2 and rotating 2z 2 3 π anti-clockwise about the Origin, Or: 2z is the scaling of 1z by factor of 1 2 and rotating 1z 2 3 π clockwise about the Origin, 1 z× 2 z× 3 π 3 π 4 2

River Valley High School 2 2 Solution [6] P2 Sequence (i) When 0n= , 00 10 41 2 33 3uu Q  −= − +    00 4 41 233 33 3 Q −= − +    81 333Q=−= − and 11 21 41 12 33 33uu  −= − −    2 41 12 4 2 33 33 3 3u  = − − +=   (ii) 1 41 12 33 33 nn nnuu+  −= − −    1 11 2 12 24433 3 33 3 nn nn nnuu+    −= + ≤ +       1 52 33 n nnuu+ −≤  Thus 1nnuu ε+ −≤ when 52 33 n ε ≤ 23 35 n ε ≤ 3ln 5 2ln 3 n ε≥  0n can be 3ln 5 2ln 3 ε      or celling of 3ln 5 2ln 3 ε  . ___________________________________________

River Valley High School 3 When 0.001ε = , 3ln 5 2ln 3 n ε≥=  18.29 0n can be 19.

River Valley High School 4 3 Solution [7] Transformation of graphs (i) Graph of ( )fyx= for 14 x−≤ ≤ (ii) Graph of 1f1 2yx = − , for 14 x−≤ ≤ The end-points are at ( 1, 1)−− and (4, 0.5)− Note: When 4x= ( )1 13 1f 1 f 2 1 (2) 12 22 2yx = −= −= − + −= − (b) Scale parallel to y-axis by factor 3. Translate 2 units in the negative x-direction. Reflect about the y-axis. OR Reflect about the y-axis. Translate 2 units in the positive x-direction.

River Valley High School 5 4 Solution [10] P2 3D Vectors (a) 1 01 : 0 1, 30 l λλ    = +∈    r  Let 0 0 3 OB  =   . Given that 1 0 4 OA  =    01 1 00 0 34 1 AB OB OA −   =−=−=    −        Let 1 11 1 01 1 10 1 −   =×=    n 101 1 01 1 31 −−     ⋅= ⋅       r 1 1 :r 1 3 1 − π⋅ = 1:3xyzπ −−= − (ii) Q is the foot of perpendicular of P on 1l 01 0 1 for some 30 OQ λλ    = +      0 13 0 14 3 05 PQ OQ OP λ −     =−=+ −              31 41 20 PQ λ      = −+    −     PQ   perpendicular to the line 1l 0PQ⋅=d  

River Valley High School 6 3 11 4 1 10 2 00 λ      −+ ⋅ =    −   12 0λ−+ = 1 2λ = 0 1 0.5 10 1 0.523 03 OQ      = +=         (iii) 2l parallel to 1π 21 is perpendicular to ⇒ dn 21 =0⇒⋅dn 31 110 1 31 0 2 m m m −     ⇒⋅=       ⇒− + + = ⇒= _______________________________________________ h=distance between 2l and 1π h=distance between ( 3,4,5)P − and 1 1 :r 1 3 1 − π⋅ = Let N be the foot of perpendicular of P on 1π . 31 : 4 1, 51 PNl λλ −−   = +∈    r  --- (1) 1 1 :r 1 3 1 − π⋅ = --- (2) To find N, sub (1) into (2): 3 11 4 1 13 5 11 λ − −  −     + ⋅=      

River Valley High School 7 31 11 4 1 1 13 51 11 λ −− −−       ⋅+ ⋅=          12 3 3λ+= 3λ =− 31 4 31 51 ON −− = −   3 13 1 4 31 4 31 5 15 1 h PN − −− −     = =− −= −         11 3 13 13 3 11 PN −−   = −= =    Alternatively Note that (1,0,4)A is a point on the plane Dist of ( 3, 4,5)P − from 1 1 :r 1 3 1 − π⋅ = is h. ˆh AP= ⋅ n   where 31 4 40 4 541 AP −−     = −=         41 141 311 9 3 33 h −−   = ⋅    = =

River Valley High School 8 (iv) : 2:5PQ PR = Case 1: Q is between P and R Distance between 1π and 2π 3 2 h= Case 2: P is between Q and R Distance between 1π and 2π 7 2 h=

River Valley High School 9 5 Solution [12] Integration (i) 2 1 d21 d d1 d2 ux uu x u xu = + = = when 1, 0 when 1, xu xa u a = −= = −= ( )( )( ) ( ) ( ) 1 1 2 0 42 0 53 0 53 53 53 22 1d 1 2d 2d 112 53 11 112 0053 53 22 53 a a a a xx x u u uu u uu uu aa aa − − + = − = − = −   = −−−     = − ∫ ∫ ∫ (ii) C: 1y xx= + Translate curve C, 2 units in the negative y direction, to obtain curve D. D: 12y xx= +− Let S denote the region bounded by curve D, x- axis and 1x=− . Volume obtained by revolving region R , 2π radians about the line 2y= , is the same as the volume obtained by revolving region S , 2π radians about the x-axis.

River Valley High School 10 From GC, x-intercept happens at 1x= . Volume of solid ( ) 21 1 π 1 2dxx x − = +−∫ ( ) ( ) ( ) ( ) 1 2 1 11 32 11 143 53 22 1 π 1 2 2 1 2d π 2d 2 2 π 1d 22π 2 2 2π 2 243 5 3 x x xx x x x x xx x xx x − −− − = +− ++ = ++ − +  = ++ − −  ∫ ∫∫ 11 11 8 4π 2 2 2 2π 2 243 43 5 3 14 16ππ3 15   = ++ − −− − −      = − 318 π units5=