2026 DHS Prelim P2 (Soln)

DHS 2026 Year 6 H2 Mathematics Preliminary Examination Paper 2 Suggested Solutions DHS 2026 Prelim H2 Math Paper 2 Suggested Solutions Qn Suggested Solution 1(a) 2 2 22 2 2 2 2 2 35 134 2 3 5 ( 3 4) 034 69 034 ( 3) 0, 4 or 1( 1)( 4) xx xx xx xx xx xx xx x xxx −+ >+− − +− + − >+− −+ >+− − > ≠−−+ Since 2( 3) 0 for all \ {3},xx−> ∈  ( 1)( 4) 0, 3xx x− +> ≠ 4 or 1, 3 [Alt : 1 3 or 3]x xx x x∴ <− > ≠ < < > 1(b) For 0,x ≠ Replace x by 1x in 222 35 134xxxx −+ >+− ( ) ( ) ( ) ( ) 2 2 2 2 2 2 11 11 32 31 2 35 1 34 5 14 23 5 113 4 xx xx xx xx xx xx −+ > +− −+ >+− −+ >+− Replace x by 1x in the previous solution: 1 11 4 or 1 ( 3) 1 1 1 111, Alt : or 14 3 4 33 x xx xx x x <− > ≠ −<< ≠ −<< <<  Note that 0x = is a solution as ( ) ( )( ) ( ) 2 2 2 30 50 2 1. 1 30 40 −+ = > +−

2 DHS 2026 Year 6 H2 Mathematics Preliminary Examination Paper 2 Suggested Solutions Qn Suggested Solution 2(a) Area of a parallelogram formed by sides OA and OB or twice the area of the triangle OAB. 2(b) ( )3 3 // 3 ( ), ( ) 3λλλ ×−= −⇒ = − ∈ ⇒= − r ba 0 rba r ba r ba  Point R lies on the line passing through origin O and parallel to .AB   2(c) ( )0 ⋅=⋅ ⋅−= rb ra rba Point R lies on the plane containing the origin O with normal .AB   2(d) Letθ be the required acute angle. 31 60 20 3 3cos 7749 1θ −   −•  − = = = Using the trigonometric identity, 22 2 1s c cos s 1 i in sn o θθ θθ − += = 2 2 7 31 7 10   = − = Alternatively Letθ be the required acute angle. 31 0 60 2 20 6sin 749 1 2 10 7 θ −     −×    = = =

3 DHS 2026 Year 6 H2 Mathematics Preliminary Examination Paper 2 Suggested Solutions [Turn over Qn Suggested Solution 3(a) Substitute y xk= −+ into ( ) 2 6,x y axy++= ( ) 2 22 22 ( )6 60 (6 ) 0 (shown) .... (1) xxk a xxk k ax akx ax akx k − + + −+ = − + −= − +− = 3(b) For (1) to have real roots, Discriminant 0≥ ( ) 22 2 2 2 46 0 ( 4) 24 0 24 0 ... (2)4 ak a k ak k a − −≥ + −≥ −≥ + 2 2 2 or 2 ( 2)( 2) 0 4 0 ... (3) k kk kk k ≥ ≤− ≥ − +≥ −≥ Compare (2) & (3) : 24 44 24 4 16 2 a a a =+ = + ∴=

4 DHS 2026 Year 6 H2 Mathematics Preliminary Examination Paper 2 Suggested Solutions Qn Suggested Solution 4(a) [ ] ( ) 2 2 sin 3 cos sin( ) sin cos cos sin Comparing, cos 1 --- (1) sin 3 --- (2) π3 1 2 and tan 3 3 π2sin 3 t tR t Rt t R R R xt α αα α α αα += + = + = = ∴= + = = ⇒ = ∴= +  4(b) By observation from (a), maximum 2x = [This occurs when πsin 1 3t += i.e. ππ π32 6tt+ = ⇒= which satisfy 0 πt≤≤ .] If minimum 2,xR= −= − πsin 1. 3t += − Considering the nearest possible solutions to 0 πt≤≤ , we have π π 3π 5π 7π or o r 32 2 6 6tt+ = − ⇒= − , both of which are not within 0 πt≤≤ . Thus, the minimum value of x is not .R− Alternative 430 π si n 13 33 2 3 3 2sin 2 3 tt t t π ππ π π ≤ ≤ ⇒ ≤ + ≤ ⇒− ≤ + ≤  ⇒− ≤ + ≤  Hence, min value of x is 32 R− ≠− = . 4(c) 1 2 11 22 When 0, 3 cos 0 π sin ( π) 3 cos ( π) 1 y tt x = = ⇒= ⇒= + = y x O (1,0)

5 DHS 2026 Year 6 H2 Mathematics Preliminary Examination Paper 2 Suggested Solutions [Turn over When 0, sin 3 cos 0 2πsin 3 cos tan 3 3 133 22 xt t t tt t y = += = − ⇒ = − ⇒= ⇒= − = − 4(d) 2 2 3 cos cos 3 3sin 3 3 , 3 3. 3 yy tt yt yx yy = ⇒= −= −= + − ≤≤ t

6 DHS 2026 Year 6 H2 Mathematics Preliminary Examination Paper 2 Suggested Solutions Qn Suggested Solution 5(a) Rate in 0.075 10 3 g/min4 = × = Rate out 10200 1 g/min20 x x = × = d 31 .d 4 20x xt = − 5(b) d 31 d 4 20 20 d 115 d x xt x xt = − =− Integrate w.r.t t, 20 20 20 20 20 20 d 1d15 20ln 15 15 e 15 e , e 15 e tc tc t xtx x tc x xA A xA −− −− − =− − −= + −= −= = ± = − ⌠⌡ ∫ Remark: If student attempts to remove modulus by reason, they should explain that the concentration of nitrogen in the tank at any one point cannot be greater than the concentration of the nitrogen from the nutrient solution pumped in. 0.075200 15 15 0 x x x < < −> or Remark: Since the tank initially contains pure water and the more concentrated nitrogen from the nutrient solution is being pumped into the tank, the amount of nitrogen in the tank should always be increasing. 3 04 20 3 4 20 15 15 0 x x x x −> > > −>

7 DHS 2026 Year 6 H2 Mathematics Preliminary Examination Paper 2 Suggested Solutions [Turn over 20 Sub 0, 0, 0 15 15 15 15e t tx A A x − = = = − = = − 5(c) The nitrogen mass will stabilize at 15 grams in the long run. O 15x = 2015 15e t x − = −

8 DHS 2026 Year 6 H2 Mathematics Preliminary Examination Paper 2 Suggested Solutions Qn Suggested Solution 6(a) Total number of ornaments that are either white or with a moon pictorial design (or both) but not green 12 2 4 5 11 34= ++++ = Total number of ornaments that are not green = 90 – (13+10+8) = 59 Method 1: Using reduced sample space 34 2 59 2 34 33 561Required Probability (or ) = (or 0.328 (to 3 s.f.)) 59 58 1711CC= × Method 2: Using conditional probability formula 34 2 90 2 59 2 90 2 34 33 90 89 59 58 90 89 561Required Probability (or ) = (or 0.328 (to 3 s.f.)) 1711 C C C C ×= × 6(b) Number of blue stars = 6 Total number of blue ornaments that are not star = 3 + 5 =8 Total number of star ornaments that are not blue = 12 + 9 + 13 =34 Total number of ornaments that are not star and not blue = 90 – 6 – 8 – 34 = 42 8 34 11 90 2 Required Probability P(one blue ornament that is not a star and one star ornament that is not blue) + P(one blue star and one ornament that is not a star and not blue) 8 34 62 (or ) 90 89 90CCC = =×× + 6 42 11 90 2 42 2 (or )89 524= (or 0.131 (to 3sf)) 4005 CC C×× 6(c) Let x be the number of ornaments of his favourite type. 2 90 2 ( 1) 28 (or ) =90 89 4005xCxx C−× Solving (algebraically or using GC), 8x = ⇒ favourite ornament : Green Moon

9 DHS 2026 Year 6 H2 Mathematics Preliminary Examination Paper 2 Suggested Solutions [Turn over Qn Suggested Solution 7(a) Case 1 - 4 “S”: 1 way Case 2 - 3 “S”: 5 1C ways Case 3 – 2 “S” or 2 “E” or both: 25 12 1 21CC× += ways Case 4 – distinct letters: 6 4C ways Number of selections without restrictions 1 5 21 15 42=++ + = 7(b)(i) Number of ways to arrange “AMNT” as one unit and 6 other letters 7! 4!2!= Number of ways to arrange “AMNT” = 4! Number of arrangements required 7! 4! 25204!2!= ×= 7(b)(ii) _ _ _ _ _ _ ↑ ↑ ↑ ↑ ↑ ↑ ↑ Number of ways to arrange letters except “S” 6!2!= Number of ways to choose arrows to slot in the “S” 7 4C= Number of arrangements required 7 46!2! C= ×= 12600 7(b)(iii) _ E E _ _ _ ↑ ↑ ↑ ↑ ↑ ↑ Number of ways to arrange letters except “S” such that the two “E”s are together 5!= Number of ways to choose arrows to slot in the “S” 6 4C= Number of ways “S” to be separated and two “E”s are together 6 45! 1800C= ×= Number of arrangements required 12600 1800 10800= −=

10 DHS 2026 Year 6 H2 Mathematics Preliminary Examination Paper 2 Suggested Solutions Qn Suggested Solution 8(a)(i) 98 0.95xy= + (join 2 points or 1.0526 103.18yx= − ) 8(a)(ii) Since the school wants to estimate the laboratory light meter reading x from the phone app reading y, the error should be measured in the x-direction . Squaring residuals prevents positive and negative residuals from cancelling out. 8(a)(iii) See part (a)(i) 8(a)(iv) Sum of the squares of the residuals = 5349 8(b)(i) 1.0372 72.883xy= + (to 5 s.f.) 1.04 72.9xy= + (to 3 s.f.) Using GC, r = 0.906 8(b)(ii) Sum of the squares of the residuals ≈ 5172.43 < 5349 (part (a)(iv)) 8(b)(iii) Since 300y = is within the data range of [220,400] (or interpolation), and 0.906r = is close to 1, we would expect this value to be reliable. y x 200 250 300 3