TMJC 9758 2023 Prelim P2 Solution

Page 1 of 13 2023 H2 MATH (9758/02) JC 2 PRELIMINARY EXAMINATION – SUGGESTED SOLUTIONS Qn Solution 1 Vectors (a) ( ) ( ) 32 33 3 BP OP OB= − =−− = − = − a bb ab ab       and ( ) ( ) 32 22 2 AP OP OA= − =−− = − = − a ba ab ab       3Since 2BP AP=     , points A, B and P are collinear. (Shown) : 3:2BP AP = (b) ( ): , BPl λλ= +− ∈r b ab  Since N lies on line BP, ( ) for some .ON λλ= +− ∈b ab   ( )Since is perpendicular to , is perpendi cular to .BPON l ON −ab   ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) 22 0 1 0 11 0 λ λλ λ λ λλ + − −= − + −= − −− + − = b ab ab b a ab ba b a ab    Given that ( )44 21, , 1 cos3 3 33 π= = = = b a ab  , ( ) ( )( ) ( ) 2 22 421 11 03 33 1 16 210 39 3 1 1 10 033 9 13 1 93 3 13 λ λλλ λ λλ λλ λ λ   − −− + − =     − −+ − = −+ + = = = ( )3 13 3 13 BN ON OB BA = − = + −− = b ab b         Therefore, 3 13 BN BA = .

Page 2 of 13 Alternative Method ( ) ( ) ( ) 2 2 22 2 cos 3 411 32 1 3 2 cos 3 4 411 213 32 13 9 1 33 13 13 9 BO BABN BA BA BA BN BA π π = −−= − −= − + = −+ = −+  = =+−   = +−     = = = bb a b b a bb ba b ba b a ba                

Page 3 of 13 Qn Solution 2 Maclaurin’s Series (a)(i) ( ) 2 2 ln 1 2 3 e 12 3y y xx xx = ++ = ++ Differentiating with respect to x, 22 2 d e 26d dd e e6dd y yy y xx yy xx = + += When 0,x= 2 2 0 d 2d d 2d y y x y x = = = ( ) ( ) 2 22 20 2 ... 2! 2 ... up to y xx xx x = +++ =++ (ii) ( ) ( ) ( ) ( ) ( )( ) 2 22 2 2 2 22 ln 1 2 3 23 2 3 ... using standard series2 22 3 ...2 2 ... up to verified y xx xx xx xxx xx x = ++ + = +− + = +− + =++ (b) ( ) ( ) ( ) 1 2 1 1 2 2 2 23 3 4 4 41 4 13 1 221 ...2 24 2 4 11 3 up to 2 16 256 x xx x xx xx x xx x x − − − = + + = +    −−     = +− + +       = −+

Page 4 of 13 Qn Solution 3 Sequences and Series & Complex Numbers (a) ( ) ( ) ( ) ( ) ( ) ( ) 11 11 22 2 2 11 2 12 1 1 22 22 2 2 2 1 2 22 2 22 2 12 1 12 12 121 2 12 cos isin cos isin cos isin cos isin cos isin cos icos sin isin cos i sin c cos cos i sin sin cos cos sin sin os c isin c os s sin i os i rz zr r r r r r r θθ θθ θθ θθ θθ θθ θθ θθ θ θθ θθ θθ θθ θ θθ θ += + +−= ×+− − − = += + + − − ( ) ( ) ( ) 22 2 121 2 2 2 1 2 co cos sin . n ins sir r θ θ θ θ θ θ θ + = + − − 1 12 2 arg z z θθ = −  (b) ( ) ( ) ( ) ( ) 11 1 12 23 21 0 0 1 arg arg arg arg arg arg arg ... arg arg arg arg arg arg 0 i a arg arg rg i arg i2 n rr r nn n nn ww w ww ww ww w ww w w n nπ − = −− − −= − +− +− + +− + = − = − − − ++ = + ∑ .2k π∴= (c) As ( ), arg i 0,nn→∞ + → hence 1 1 arg n r r r w w − =    ∑ converges. 1 1 arg 2 r r r w w π∞ − =  =  ∑

Page 5 of 13 (d) ( ) ( )( ) 333 1 2 1 1 1i 2i 3iarg arg arg ...2i 3i 4i 3 arg 0i3 arg arg 1i 3 arg 0 i arg 1 i2 3 2 24 3 4 r r r r r r w w w w π π ππ π ∞ − = ∞ − = +++  +++  +++