NJC 2026 H3 Math Prelim (Ans)

* 1 Suggested Solutions (a) ( ) ( ) ( ) ( ) ( ) 3 1 22 3 1 22 5 5 2 2 5 2 2 3 3 3 3 3 3 3 3 3 1 22 33 232 2 2 3 d 3 3 3 d 3 3 3 d 3 1 3 d 33 1 1 1 1 1 1 11 2 3 d 1 2 1 x x x x x x xx c x x c x x x x x x x x x x x x x x xxx + = + = + = + = + − + ++ =− +− +− +− + = + − + +            Method 2: ( ) ( ) 3 2 3 2 3 3 33 11 22 3 3 5 2 3 32 33 3 3 3 2 3 2 33 3 1 d 3 d 3d 3 1 1 11 2 1 2 1 2 1 2 42 1 1 3 d x x xxx x x x xx c x x xx x x x x x x c x x x + = + ++= = + −  + + =+ − −+ = + +   +  −        (b) ( ) 2 3 3 23 2 3 3 2 3 3 2 sin2d sin 2sindd cos cos 2sin dd cos sin 2sin 2sin dd cos sin 2sin 2sin dd cos sin xxx x xx xxxx x x x xxx x x x x x x xxx x x x x x x x xxx x x x xx cxx + =+ =− − + =− − − − + =− − − + =− − +     

2 (c) ( ) 22 2 2 22 22 2 22 22 2 2 2 ee dd e e e dd e e e dd e 0 ee 02 e 2e 02 e e 2 02 e2 ln 2 aa xx aa a aax x x aaa a aax x x aaa ax a aa aa aa a xxxx xxx x x xxx x x x aa a a a =  − − =  +=  = −= − = − = = =   

3 2 Suggested Solutions (a) By binomial series: ( ) 2 2 2 2 3 2 2 1/2 2 4 6 1 3 1 3 5 ( )( ) ( )( )( )1 2 2 2 2 21 ( ) ( ) ( ) 2 2! 3! ... 1 . 1 (1 ) 1 1 3 1 3 51 2 2 4 2 ..46 x x x x x x x x − − − − − − + − − + − + − + − =−     +   += + + Integrating both sides: 3 5 7 1 1 1 3 1 3 5sin ... 2 3 2 4 5 2 4 6 7x x xx C x−        = + + + + +             . Substitute 0x= , 1 0 s 0n 0i C C − =+ = 3 5 7 1 1 1 3 1 3 5sin 2 3 2 4 5 2 4 6 7x x xxx−       = + + +          (up to 7x ) (b)(i) From part (a), 1 1 (2 1) 1 21(2 ) n k n n k k a nk= = − = +   1 1 1 2 11 1 1 (2 1) 1 23(2 ) (2 1) .2( 1)(2 3)(2 1) 1 21(2 ) n k n nk n n k n k k nka n a n n k nk + = + += = = − + +== ++− +     (b)(ii) 23 211 21 n nn n nn a x a xa x a +++ + = Since 10 2x , 2 10 4x . Since 2 1 2 2 2 3n n n+  +  + , 1 2 1 2 1 12 2 2 3nna nna n n+ ++  =  ++   . ( ) 23 1 21 111 44nn nnaxax ++ + = .

4 (c) Let 1 2x= in (a). Then 3 5 7 21 3 21 3 1 1 1 1 1 1 3 1 1 1 3 5 1 1 ... 2 2 3 2 2 4 5 2 2 4 6 7 2 s 2 1 2 π 1 1 3 1 3 5 1 1 6 2 48 1 80 2 4 6 2 1 2 1 1 3 5 1 1π 3 6 64 n 8 01 i 9 2 4 6 2 2 r r r r r r − + = + =   + + + +         =             = + + +   +   = + + +   +     2 3 21 3 5 5 1 5 16 ... (from ) 14336 14336 4 14336 4 1 3 5 1 16 2 4 6 2 1 2 5 143366 11 4 5 1792 r r r R r = + = + + +  =   +                =   − =  (b)(ii) Thus 5 1792R .

5 3 Suggested Solutions (a) Differentiate the given equation: 2( 1) d d d( 1) 2d d d dd( 1) 2 dd d1 ( 2 ) d 1d 2d d 2d d 2.d y x p p y p p p x px x x ppp p x p xx pxp x p x p x x xpp x xpp = + + + + + =+ = = ++ − = + − =+ −− + =− (b) dd( e ) e edd p p p xxpp x= + Multiply both sides of (a) by e p , d e e 2 ed p p px xpp + =− Integrate both sides with respect to p, ( ) e 2 e d e e d e 2e 2e 2 2 2 pp pp pp p x p p p C C p p px − = =− =− = +− − − ++ +   (c) ( )1 e2 ppxC −−− += 22 2 2 2 2 22 ( ( 1) [ 2( 1) ]( 1) 2( 1) 1) ( 1) ( 1) ( 1) .2 p p p p y x p p p Ce p p p p C p e p C p e Cp pp ep − − − − = + + = − − + + + =− − + + + + ++ ++ =− + + =− + (d) Assume p is constant. Differentiating the differential equation gives 1 d d 1 0 1y px pp =+ + = = This is a contradiction, so no such solution exists.

6 (e) For stationary point d 0dyp x== ( ) ( ), 2 , 2x y C C= + + Substitute 1x= , 3y= into the parametric equations, For x: ( ) 1 2 e 1 2 e .. 2 .1 p p p p C C − − =− −+ ++ −= For y: ( ) 2 2 3 ( 1) ( 1) ... 2 2 1 p p C p e C p e p p − − =+ = + + − −+ + ( ) ( )( ) ( )( ) ( ) ( )( ) 2 2 2 2 1 1 , 1 1 1 2 1 1 1 2 20 2 1 0 2 or 1 p p p p p p p p pp pp pp −+ − + − =− + + + + = + + − = + − = =− = From (1), 2 2 When 2, 1 4 e 5e p C C − + =− −= =− or 1 When 1, 1 2 e e p C C − = −= +− = The stationary points are ( ) 222 5e , 2 5e−−−− and ( )2 e, 2 e++ .

7 4 Suggested Solutions (a) Total number of selections without restriction is 328= . Selections which satisfy the condition are (1, 1, 1), (1, 2, 2), (2, 1, 2), (2, 2, 1) and (2, 2, 2), with a total of 5. Thus 2 5 8p = . (b) Total number of selections without restriction is 3k . Consider the complement event, i.e. cases in which ( ),,abc cannot form a triangle. When there is a tie for the longest side, ( ),,abc can always form a triangle, thus not in the complement event. Without loss of generality, let c be the longest side. When 1:c= 0 ways When 2:c= ( ) ( ), 1,1ab = … 1 way When 3:c= ( ) ( ) ( ) ( ) , 1,1 1, 2 2,1 ab = … 1 + 2 ways When 4:c= ( ) ( ) ( )( ) ( ) ( ) , 1,1 1, 2 1, 3 2,1 2, 2 (3,1) ab = … 1 + 2 + 3 ways … When :ck= ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) , 1,1 1, 2 1, 3 ... 1, 1 2,1 2, 2 ... 2, 2 (3,1) ... 3, 3 ... 2,1 2, 2 1,1 ab k k k kk k = − − − −− − … ( )1 2 ... 1 ways k+ + + − Total number of ways in the complement event where c is the longest side:

8 ( )( ) ( ) ( )( ) ( ) ( )  ( )( ) 1 1 2 11 3 1 2 ... 1 1 2 1 2 1 2 1 11 2 6 2 1 2 1 312 2 1 1 12 6 k r k r kk rr r rr rr k k k k k kk k k k k kk = = == + + + − −= =−  + + +=−   += + − +−= −=    Total number of ways in the complement event is 33 3 62k k k k −− = . 3 3 3 33 3 2 2 2 2 2 1 2 k kkk p k k k k k k k −− = −+= += (c) Without loss of generality, we assume abc first in the computation then multiply the result by 3! = 6. In counting 22mQ + , the triples which are not included in counting 2mQ have the longest side either 21m+ or 22m+ . When 21cm=+ , the triplets that cannot form triangles are ( ) ( )( )( ) ( ) ( )( ) ( ) ( ) ( ) ( ) , 1, 2 1, 3 1, 4 ... 1, 2 2, 3 2, 4 ... 2, 2 1 3, 4 ... 3, 2 2 ... ,1 a b m m m mm = − − + The number of triplets here is ( ) ( ) ( ) 2 2 1 2 3 ... 1 2 1 1 2mm m m m − + − + + = − + = The number of triplets that can form triangles is ( ) ( ) ( ) ( ) 2 2 2 2 22 2 2 2 13! C 6 2 62 6 m mmmm m m m mm −− = −   = − − =−

9 When 22cm=+ , the triplets that cannot form triangles are ( ) ( )( )( ) ( )( ) ( )( ) ( )( ) ( ) ( )( ) ( )( ) , 1, 2 1, 3 1, 4 ... 1, 2 1, 2 1 2, 3 2, 4 ... 2, 2 1 2, 2 3, 4 ... 3, 2 2 2, 2 1 ... , 1 , 2 a b m m mm mm m m m m =+ − −− ++ The number of triplets here is ( ) ( ) 2 2 2 2 ... 2 2 2 2mm m m mm + − + + = + =+ . The number of triplets that can form triangles is ( ) ( )( ) ( ) 2 1 2 2 2 22 2 2 1 23! C 6 2 62 6 m mmm m m m m m m m m + + − + = − −   = + − − = Therefore ( ) ( ) ( ) 22 2 2 2 2 66 62 6 2 1 mmQ Q m m m mm mm + − = − + =− =− (d) Let Pm be the statement that ( )( )2 1 4 5mQ m m m= − − . When 2m= , we have the permutations of (4, 3, 2) in counting 4Q , 4 3! 6Q == . ( )( ) ( )RHS 2 2 1 4 2 5 6= − − =  . 2P is true. Assume Pk is true for some 1k  , i.e. ( )( )2 1 4 5kQ k k k= − − For 1Pk+ : ( ) ( )( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) 2 2 2 2 2 6 2 1 (from ) 1 4 5 6 2 1 1 4 5 6 2 1 4 9 5 12 6 4 3 1 1 4 1 1 1 1 4 1 5 kkQ Q k k k k k k k k k k k k k k k k k k k k k k k k + = + − = − − + − = − − + − = − + + − = + − = + − = + + − + −     (c) 1Pk+ is true. Since 2P is true and Pk is true implies that 1Pk+ is true, Pm is true for all integer 1m . ( )( )2 1 4 5mQ m m m= − − .

10 5 Suggested Solutions (a) As z, y are non-negative and a  b  c, and .a byy az cz Thus, ( ) (shown) ax ay aza a x y z x by cz  + ++ + + + (b)(i) 1 1 1 2 2 2S ax by cz= + + (b)(ii) From part (i), 2 ax y z S+ + Also, equality is achieved when 2 ,Sx a= 0.yz== Minimum sum 2S a= (c)(i) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ( ) ( ) ( ) ( ) 22 2 2 2 2 bx ay cy bz az cx ax by cz b x baxy a y c y cbyz b z a z aczx c x a x b y c z abxy bcyz acxz b x a y c y b z a z c x a x b y c z a b c x a b c