DHS 9758 2023 Prelim P2 Solution

1 DHS 2023 Year 6 H2 Mathematics Preliminary Examination 2023 H2 Math Prelim solutions Qn Suggested Solution 1(a) Area of n rectangles under the above curve between 0 and 1,xx= = ( ) 1 0 1 1112 1 1f 0 f f ... f 1 f n r n n nn nn n n r nn − = −   = +++ +        = ∑ Since area of n rectangles is an underestimate of / less than the actual area under curve 1 0 f( ) d ,xx=∫ 1 1 00 1 f f( ) d (shown) n r r xxnn − = ∴< ∑ ∫ (b) Required expression 1 1 f n r r nn= = ∑ (c)(i) Lower limit 9 0 29 0 1 f10 10 1 110 10 1.285 r r r r = = =  = + = ∑ ∑ (c)(ii) From GC : 1 2 0 +1 d 1.3333xx =∫ Difference from 1 2 0 +1 dxx∫ (a) lower limit = 1.3333 1.285 0.0483 3.62%1.3333 1.3333 − = = (b) upper limit = 1.385 1.3333 0.0517 3.88%1.3333 1.3333 − = = The lower limit is a better estimate since it has a smaller % / absolute difference from the exact value. O … 3 n n n y x f( )yx= Upper limit 10 1 210 1 1 f10 10 1 110 10 1.385 r r r r = = =  = + = ∑ ∑

Qn Suggested Solution 2a(i) (A) c = 5 The sequence initially decreases and subsequently alternates , and converges to 2. (B) c = 2 It is a constant with a value of 2. a(ii) ( ) 1 21 32 3 0.5 3 0.5 3 0.5 3 0.5 3 0.5 1.5 0.25 uc uuc uu c c = = −= − = −= −−= + ( ) ( ) 32Given 2 5 2 1.5 0.25 5 3 0.5 3 0.5 15 2.5 2 18 9 uu cc cc c c =− + = −− + = −+ = = b(i) Given 3 21 12 ,v uu+= − ( ) 12 212 11 2 , 2 2 2 3 0.5 2 262 2 vv uu p cc pc cp + −+= − += − − +=− += b(ii) Given 12 , 2,v pv= = ( ) ( ) ( ) 3 12 4 23 5 34 2 12 2 12 1 2 1 2(2) 2 1 1 2 4 2 1 22 1 2 4 1 6 5 6 5 77 12 v vv p p v vv p p v vv p p p p p = + −= +−= + = + −= + + −= + = + −= + + + −= + += ∴=  Qn Suggested Solution

3(a) ( ) 2 22 2 2 22 2 2 2 20 2 2 100 20 20 100 5 (shown)20 xyh yh h yy yh hy = +  −= +  = +−+ = + = + 2 2 20 2 20 2 5 20 10 10 xy h h = − = −+  = − (b) Volume of prism, ( ) ( )( ) ( ) ( ) 2 2 23 43 2 1 2 1 10 20 22 10 100 1010 1000 100 1010 1 10 100 1000 (shown)10 V hxz hhh h hh h h hh hh h h = =−−  = −− = −−+ = −− + ( ) 32d1 4 30 200 1000d 10 V hh hh = −−+ For max.volume, 32d 0 4 30 200 1000 0d V hh hh = ⇒− − + = From GC : 6.40 or 10 (reject 0 10) 3.9039 hh h = − << ∴=  2 2 2 d1 (12 60 200)d 10 V hhh = −− ( ) ( ) 2 2 2 When 3.9039, d1 (12 3.9039 60 3.9039 200) 25.1 0d 10 h V h = = − − = −< 3.9039h∴= gives maximum volume 2 20 20 2 xy xy += = −

Max volume ( ) ( ) ( )( ) 324 3 1 3.9039 10 3.9039 100 3.9039 1000 3.903910 201.71 202cm (3 sf) =−− + = =

Qn Suggested Solution 4(a) Since the lines AB and AC meet, 12 4 44 6 92 ab λµ λµ λµ ++     += +    ++   . 23λµ−= --- (1) abλµ= --- (2) 23λµ−= --- (3) Solving (1) & (3) gives 1λ = & 1µ =− Hence (2) gives ab=− ⇒ 0ab+= (shown) (b) 21 2 3 12 2 ab ab ba