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 ∞ − = ∞ − = +++  +++  +++   =     += −   + = − +− +  = −−   = ∑ ∑

Page 6 of 13 Qn Solution 4 Applications of Differentiation (a) When 0y= , 21sin 0 2 1sin 0 2 ut gt t u gt θ θ −=  −= 2 sin0 or utt g θ= = Since 0t ≠ , 2 sinut g θ= . 2 sinut g θ= represents the time taken for the projectile to return back to the same height as the origin at which it is launched. (b) When 2 sinut g θ= , ( ) 2 2 sin cos sin 2 Shown uxu g u g θ θ θ =  = (c) 2d2 cos 2d xu g θθ = When d 0d x θ = , 22 cos2 0u g θ = cos2 0 2 2 since 042 θ πθ ππθθ = = = << 22 2 d4 sin2d xu g θθ =− When 4 πθ = , 22 2 d4 0d xu gθ = −< 4 πθ = gives the maximum range of the projectile. 2 2 sin 2 4 ux g u g π =   = Hence the maximum range of the projectile is 2u g .

Page 7 of 13 Qn Solution 5 Probability (a) P( ') 0.51B = P( ' ') 0.15AB∩= P( ') 0.51 0.15 0.36 AB∩= − = (b) 11 P( ) 11P( | ) 29 P( ) 29 BABA A ∩= ⇒= ( )P( ) P ' 11 P( ) 29 P( ) 0.36 11 P( ) 29 29P( ) 10.44 11P( ) 18P( ) 10.44 P( ) 0.58 A AB A A A AA A A −∩ = − = −= = = (c) Note that P( ') 0.51B = . Let ( )P. Cx= Since A and C are independent, ( )P 0.58AC x∩= . Note that B and C are mutually exclusive. 0.15 0.42 0 5 14 x x −≥ ≤ and 0.36 0.58 0 18 29 x x −≥ ≤ Therefore, maximum ( )P C 5 14= A B 0.15 0.36 A B C 0.15 – 0.36 –

Page 8 of 13 Qn Solution 6 Discrete Random Variable (a) Area of target board = ( ) 25 25ππ = Area of region with score 50 ( ) 21ππ= = Probability of dart hitting region with score 50 1 25 25 π π= = Area of region with score 25 ( ) 238π ππ= −= Probability of dart hitting region with score 25 88 25 25 π π= = ( ) ( ) 81P 75 2!25 25 16 shown625 S = =×× = (b) Area of region with score 0 25 9 16ππ π= −= Probability of dart hitting region with score 0 16 16 25 25 π π= = s 0 25 50 75 100 ( )P Ss= 16 16 25 25 256 625 × = 16 8 2!25 25 256 625 ×× = 16 1 2!25 25 88 25 25 96 625 ×× +× = 16 625 11 25 25 1 625 × = (c) From GC, ( )E 20S = and ( )Var 400S = .

Page 9 of 13 Qn Solution 7 Correlation and Regression (a) No, since product moment correlation coefficient is independent of the scale (or unit) of measurement. (b) Using GC, 0.985r =− . Since r is close to −1, there is a strong negative linear correlation between t and ln (100 ) x− . (c) (d) ln (100 ) 5.93932 0.33739 ln (100 ) 5.94 0.337 (3 s.f) xt xt −= − −= − (e) Sub 10t = , 2.56541 ln(100 ) 5.93932 0.33739(10) 100 e 87.0(3 s.f) x x x −= − −= = Since 10t = lies outside of the given data range of t, the linear relation between ln (100 ) x− and t may no longer hold. The estimate is not reliable. 9.1 5.5 t ln (100 ) x− 2.9 4.2

Page 10 of 13 Qn Solution 8 Normal and Sampling Distribution (a) Let A and S be the mass of a randomly chosen apple from Brand A and Brand S respectively. ( ) 2~ N 78.8,3.1S and ( ) 2~ N 82.2, 2.2A Required Probability ( )P 80 84 0.635 (3sf ) A= << = (b) 12 5Let ... .TAA A= + ++ ( ) ( ) ( ) 2 E 5 82.2 411 Var 5 2.2 24.2 ~ N 411, 24.2 T T T = ×= = ×= ( )P 408 0.729 (3 s.f.)T >= (c) Let 0.9DS A= − . ( ) ( ) ( ) ( ) ( ) 2 22 E 78.8 0.9 82.2 4.82 Var 3.1 0.9 2.2 13.5304 ~ N 4.82,13.5304 D D D = −= = += Required Probability ( ) ( ) ( ) P1 1P 1 1P 1 1 0.907 (3sf ) D D D = > = −< =− −< < =