CJC SAJC JPJC 2026 H3 Math Prelim (Ans)

1 9820/01/PRELIM/2026 [Turn over 2026 JC2 H3 Math Prelim Solutions Solution Q1 Inequalities (a)(i) [1m] 1 1 1 1 1 1 1 1 1 1 1 13 32 1 (shown) x y z x y z x y z x y z − − −+ + = − + − + − = − + + =− = (a)(ii) [2m] Apply Cauchy-Schwarz inequality, ( ) ( ) 2 2 1 1 1( ) 1 1 1 1 1 1 , by ( )( ) 1 1 1 (shown) x y zx y z x y z x y z x y z x y z a i x y z x y z  − − −+ + + +  − + − + − + +  − + − + − + +  − + − + − (b)(i) [1m] 2 11f ( ) 1 0 1, f (1) 1 and f (1) 1 0 1xx x  = − =  = =− =− =−  f(x) only have one turning point which is a maximum point at (1, 1)− , thus f ( ) ln 1 ln 1x x x x x= − −   − for all real values of x. (b)(ii) [3m] By letting iax A= for i = 1, 2, 3, …, n, we have 11 22 33 ln 1 ln 1 ln 1 ln 1nn aa AA aa AA aa AA aa AA − − − −  adding all the inequalities,

2 9820/01/PRELIM/2026 [Turn over 331 2 1 2 1 2 3 1 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 ln ln ln ln 1 1 1 1 ln ln ln 0 1 1 (shown) nn n i ni n n n n n n n n n n n ni i a a a aa a a a A A A A A A A A a a a a a nAA a a a a nA nAA a a a a A a a a a A a a a a A a a a a a n = = + + + +  + + + − − − −  −  −               Equality holds when 1 2 3 na a a a= = = =  (b)(iii) [5m] Applying (b)(ii) twice as follows, a b c d b c d a+ + + = 2( )22a b c d ac bd ac bdb c d a bd ac abcd+   + + +  + =       2( )( ) 222 , by (b)(ii) 8 a c b d abcd ac bd abcd ++=    = The above inequalities turn into equalities when ac= and bd= . Then 22( )( ) 4a c b d ac bd ab a b+ + = +  = + where solving by quadratic formula gives ( )23ba= . Hence a b c d b c d a+ + + attains value of 8 when 1ac== and 23bd= =  . (There are infinitely many solutions by letting a and c be any real number and solve for b using ( )23ba= and b = d.)

3 9820/01/PRELIM/2026 [Turn over Solution Q2 Sequences and Series (a) [1m] Since ( ) 1 3 1 0 1 nnSS n+ − =  + for 1n , n  , the sequence  nS is strictly increasing. (b) [5m] To show  nS is convergent Need to show (1)  nS is strictly increasing and - proved in part (a) (2)  nS is bounded from above. ( ) ( ) ( ) 3 2 1 1 1 111k k k k k kk==− − + − By partial fractions, ( ) ( ) 3 1 1 1 1 2 1 2 1k k k k k= − +− − + 33 33 22 33 22 33 12 11 since 0 11 1111 11 1 nn kk nn kk nn kk kk k k k k k k k k k k k == == ==  −  − +  + − + −   

4 9820/01/PRELIM/2026 [Turn over ( ) 2 1 1 2 11 2 1 1 1 2 1 1 2 3 1 2 1 2 3 4 1 2 1 3 4 511 2 1 2 1 3 2 1 1 2 1 21 1 2 1 11 1 1 1 11 2 2 1 51 4 2 1 n k RHS k k k n n n n n n n n n nn nn = = + − + −+  −+  + − +  + − + = + + + − + − − − + − + −− + − + −+ = + − + + =− +   As n→ , ( ) 1 021nn →+ , 3 2 15 4 n k kk= →− 3 1 15 4 n k k=  Therefore, the sequence is bounded above by 5 4 . Since  nS is strictly increasing and its limit is 5 4 , By Monotonic Convergence Theorem, the sequence is convergent. (c) [3m] 3 1 1 n kn SS k   =+ −=  Consider the graph of 3 1y x= . For an over-estimation of nS , consider the left end of the Riemann sum (equal intervals of 1 unit): 3 2 2 1 1 1 d 22n nxx x n   = − = For an under-estimation of nS , consider the right end of the Riemann sum

5 9820/01/PRELIM/2026 [Turn over (equal intervals of 1 unit): ( ) 231 11 d 21n xx n  + = + ( ) 3 3 31 1 2 32 1 1 1 1 d d 1 1 1 221 nn kn kn xxx k x knn  + =+  =+     +   ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 11 221 1 221 11 2121 n n n SS nn n n S S n n S S n     −  +  −  +  −  + As n→ , 2 11 0, 0nn→→ , 2 11 21 221 nn →++ By Squeeze Theorem, ( ) 2 1lim 2nn n S S→ −= (d) [4m] ( ) ( ) 3 1 33 11 1 11 2 1 2 (odd terms + even terms) k kk S k kk S O E   =  ==  = =+ − =+   ( ) 3 3 11 1 1 1 1 882kk ES kk   == = = = ( ) 3 1 1 1 7 8821k O S S S k   = = = − = − ( ) 1 3 1 1 71 88 3 4 k k A k OE SS S − =   −= =− =− = 

6 9820/01/PRELIM/2026 [Turn over Since 3 3 5 4 4 4AS  =  , Hence 15 16A

7 9820/01/PRELIM/2026 [Turn over Solution Q3 Functions and Graphs (a) [2m] 0 0 f ( )d f ( )d f ( )d aa aa x x x x x x −− =+   Let d 1 dxux u=−  =− . Applying this to 0 f ( )d a xx −  : 0 0 0 0 f ( )d f ( )( 1)d f ( )d f ( )d a a a a x x u u u u x x − = − − =− − = −    Therefore,   0 0 00 0 f ( )d f ( )d f ( )d f ( )d f ( )d f ( ) f ( ) d aa aa aa a x x x x x x x x x x x x x −− =+ = − + = + −      (b) [3m] H( )x is even function H( ) H( )xx − = . ln(G( ))yx= is odd function ln(G( )) ln(G( ))xx − =− . Thus, ln(G( )) 11G( ) G( ) G( ) G( )xx e x xx−− = =  − = . From (a): 0 0 0 0 0 H( ) H( ) H( ) d d1 G( ) 1 G( ) 1 G( ) H( ) H( ) d11 G( ) 1 G( ) H( ) H( )G( ) d1 G( ) G( ) 1 H( )(1 G( )) d1 G( ) H( )d (proven) aa a a a a a x x x xxx x x xx xx x x x x xxx xx xx xx −  −=++ + + −    =+ + + =+  ++ += + =     

8 9820/01/PRELIM/2026 [Turn over (c)(i) [3m] 2 1 0 cos cos dmmI x x x  −= Let 12 dcos ( 1)cos sindmm uu x m x x x−−=  =− − Let d cos sindv x v xx =  = 1 2 2 22 0 0 222 0 2 2 222 00 2 ) ( cos sin ( 1)cos sin d (0 0) 1) cos (1 cos ) d ( 1 cos d ( 1) cos cos d ( 1) ( 1) mm m m mm mm I x x m x x x m x x x m x x m x x x m I m I    −− − −− − = − − − = − + − − = − − − = − − −    2 ( 1)mmmI m I − = − (ii) [3m] 2 0 0 1d 2 Ix  == Using the reduction formula: 2 2 2 24 26 0 21 2 2 1 2 3 2 2 2 2 1 2 3 2 5 2 2 2 2 4 2 1 2 3 2 5 1 ... 2 2 2 2 4 2 nn n n nII n nn Inn n n n In n n n n n In n n − − − −= −−= − − − − −− − − −    = −= − Multiply numerator and denominator by (2 )(2 2) (2)nn − : 2 2 2 22 (2 )(2 1)(2 2)(2 3) (2)(1) [(2 )(2 2) (2)] 2 (2 )! (2 !) 2 (2 )! (shown)2 ( !) 2 n n n n n n nI nn n n n n    − − − = − = =

9 9820/01/PRELIM/2026 [Turn over (d) [3m] For 62 sin 2 cos d1 xx xe  − + , Let 6( ) cosH x x= and sin() xG x e= . Checking: 66( ) cos ( ) cos ( )H x x x H x− = − = = ()Hx is even function ln( ( )) sinG x x = , and since sin( ) sinxx− =− , ln( ( ))Gx is odd function 62 62 sin 0 2 cos d co s1 xx x xdxe   − =+ [From (b) with 2a = ] 6I= 62 6! 2 (3!) 2 = [From (c)(ii) with 3n= ] 720 64 36 2 5 32   =  =

10 9820/01/PRELIM/2026 [Turn over Solution Q4 Counting / Numbers & Proofs (a)(i) [1m] No of ways = 105 9765625= (a)(ii) [3m] Let 1 2 3 4 5, , , and A A A A A be the set of selection with no red pillar selected, no blue pillar selected, no green pillar selected, no yellow pillar selected and no black pillar selected respectively. 10 10 1 2 3 4 5 10 1 2 1 3 4 5 10 1 2 3 1 2 4 3 4 5 10 1 2 3 4 1 2 3 5 2 3 4 5 1 2 3 4 5 5 4 3 2 1 0 S A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A = = = = = =  =  = =  =   =   = =   =    =    = =    =     =    Thus, total no of ways = 10 10 10 10 105 5 5 55 4 3 2 1 51030001 2 3 4       − + − + =               (b)(i) [1m] 1 5p = and ( ) ( )2 4 5 1 4 24p =  +  = (b)(ii) [2m] Case 1: If the first plank is not red (blue, green, yellow and black), then there are 1np − further choices for the remaining 1n− planks. Case 2: If the first plank is red, then there are four choices (blue, green, yellow and black) for the second plank and 2np − further choices for the remaining 2n− planks. Therefore, 1 2 1 24 4 4( )n n n n np p p p p− − − −= + = + for 3n . (b)(iii) [3m]   ( ) 3 2 3 3 3 4 3 4 3 5 3 4 3 4 3 5 3 4 3 5 3 5 4( ) 4 4( ) 20 16 5 4 3 k k k k k k kk k k k p p