NJC 2025 H3 Mathematics Prelim Solutions

2025 NJC H3 Math Prelim Solutions Qn Solution 1(a) Consider the points ( ),P b a and ( ),Q d c in the x-y plane. Since ,ac bd OQ has a more positive gradient than OP. Also, the point ( ),R b d a c++ is such that OR ⎯⎯ → is the vector sum of OP ⎯⎯ → and .OQ ⎯⎯ → From the above diagram, we see that the gradient of OR is in between the gradients of OP and OQ. Thus, .a a c c b b d d + + 1(b) For π0, 2x 0 cos 1 and sin 0 11 and sin 0cos sinsin cos sin tan xx xx xx x xx         By applying the result from part (a), 2 2 sin 2sin sin 1 1 cos cos 4sin cossin sin 22 1 cos 1 2cos 1 2 4sin cossin sin 22 1 cos 2cos 2 sin 2 tan tan (shown)2 x x x xx xx xx x x xx xx x x xxx +  +−  

Qn Solution 1(c) By AM-GM Inequality, ( ) ( ) ( ) ( ) 22 22 2 2 2 2 2 2 2 2 2 22 2 22 2 22 2 22 2 2 2 22 22 2 2 2 2 22 2 4 2 42 2 4 42 42 4 2 (shown) abab ab a b ab ab a b a b ab a ab b a b ab a b a b abab ab ab a b a b abab a b a b ab abab ab a b a b b a b a + + + + + + + + +  +  + ++ ++    +  +     +  +  

Qn Solution 2(a) Substituting 0m= into the equation, ( ) ( ) ( ) ( ) ( ) ( ) 2 2 f 0 2f f 0 f 0 2f f (1) nn nn + = + += Substituting 1m= into the equation, ( ) ( ) ( ) ( ) ( ) ( ) 2 2 f 2 2f f 1 f 2 2f f 1 (2) nn nn + = + + = + Since n is arbitrary, from equation (1), ( ) ( ) ( ) 2f 1 f 0 2f 1 (3)nn+ = + + Substituting equation (3) into equation (2), ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) f 2 2f f 0 2f 1 f 2 f 0f 1 f , which is a constant.2 nn nn + = + + −+ − = This implies that f is an arithmetic progression. Thus ( )f x ax b=+ for some constants a and b. Hence, ( ) ( ) ( ) ( ) ( ) ( )( ) ( )( ) ( ) 2 22 22 f 2 2f f 22 2 2 2 2 2 2 0 20 0 or 2 0 m n m n a m b an b a a m n b b ma b an b a m a n ab b ma ma na na b ab ma na b a m n a b a + = + + + + = + + + + + + = + + + − + − + − = + + − = + + = − = For ( ) 0m n a b+ + = for all integers m and n, 0.ab== Thus, f is the zero function in this case. For 2 0,a−= 2a= and .b But since f is an integer -valued function with domain , .b Thus either ( )f0 x = or ( )f 2 ,x x b=+ where .b

Qn Solution 2(b) ( ) 2 22 22 (1) 4 16 2 16 16 2 x y x y x xy y x y xy    + = + = + + = + = − ( ) ( ) ( ) 3 3 2 2 3 3 3 2 2 33 33 33 4 64 3 3 64 3 ) 3 64 3 64 3 4 64 6 12 (4 2 x y x y x x y xy y x y x y xy x y xy x y x y xy x y xy + = + = + + + = + + + = + + + = =       + + = +− From equations (1) and (2), ( )( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) 2 2 3 3 5 3 2 2 3 5 5 5 2 2 2 2 2 2 2 2 2 2 2 2 2 2 22 22 16 2 64 12 2 8 4 16 3 8 8 16 3 464 4 8 128 16 24 3 4 116 8 128 40 3 116 256 80 6 5 80 140 0 16 28 0 ( 14)( 2) x y x y xy xy x x y x y y xy xy x y x y x y xy xy x y xy xy x y x y xy x y x y xy x y x y xy x y xy xy xy + + = − − + + + = −  − + + + = − − + = − − + + = − + + = − + − + = − + = − − = 0 14 or 2xy xy== If 14,xy= then 2 2 (4 ) 14 4 14 4 14 0 xx xx xx −= −= − + = 2( 4) 4(14) 40 0D= − − =−  This equation has no real solution for x (and thus y). So it must be that 2.xy= Hence 2 2 2 (4 ) 2 42 4 2 0 4 2 2 or 2 24 4(2 2 ( ) ) xx xx xx x −= −= − + = = = +− −− If 2 2,x=− ( )4 2 2 2 2y= − − = + If 2 2,x=+ ( )4 2 2 2 2y= − + = − Since x  y, 22x=+ and 2 2.y=−

Qn Solution 3(a) Let iS be the total training hours up till the ith day, where 1 140.i Then 1 2 1401 ... 20(13) 260.S S S     = Let 19,iiTS=+ where 1 140.i Then 1 2 14020 ... 279.T T T     Each iS and iT is a value between 1 and 279 (inclusive), and the list  1 2 140 1 2 140, ,..., , , ,...,S S S T T T has 280 elements. Consider slotting the 280 elements into boxes labelled 1 to 279. By Pigeonhole Principle, there exists iS and ( )jT i j  such that iS = jT , since all ’s are distinct and all jT ’s are distinct. Thus, 19 19.i j i jS S S S= +  − = Hence, the player will have trained for exactly 19 hours successively from day 1j+ to day i. Alternative: let iS be the total training hours up till the thi day , where 1 140i and 1 20 13 260.iS mod 19iSr , where 0 18r , r Z . Consider slotting the 140 elements iS into 19 boxes labelled 0 to 18, depending on their corresponding value of r when divided by 19. Notice that 140 19 7 7 . By Pigeonhole Principle, there exists 1 2 8, , .** .. *,S S S such that each of them is congruent to r* modulo 19, where 0 18 *r , *r Z . We can therefore re-write each term in the following form: 1 22 88 119 * 19 * ... 19 * * * * S qr qS S r qr where 18,...,qq Z Notice that since 1 260iS and 260 1319 , we have 1 28 3, ,.01 ..,q q q . iS

Thus, there must be at least two iq ’s which are consecutive values. Let this two iq ’s be jq and kq where kjqq . 19 * 19 * 19 19** kj kj r S q S qr Hence the player would have trained for exactly 19 hours from day 1j to day k. 3(b)(i) Let iiyx= for 1,2,3,4i= and 55 0, where .yy 3(b)(ii) Let 1 1 2 2 3 3 4 43, , 8, 6y x y x y x y x= − = = + = − and 5 0.y  Then 1 2 3 4 50x x x x+ + +  becomes ( ) ( ) ( )1 2 3 4 53 8 6 50y y y y y+ + + − + + + = i.e. 1 2 3 4 5 49,y y y y y+ + + + = where 10 9,y 20 7,y 30 20,y 40 14y and 5 0.y  Let 1 2 3 4, , ,A A A A be the set of integer solutions such that 1 2 3 410, 8, 21, 15y y y y    respectively. 1 2 3 4 49 10 4 1234104 49 8 4 1489954 49 21 4 359604 49 15 4 738154 A A A A −+== −+== −+== −+== 12 13 14 23 24 34 49 18 4 523604 49 31 4 73154 49 25 4 204754 49 29 4 106264 49 23 4 274054 49 36 4 23804 AA AA AA AA AA AA −+ = = −+ = = −+ = = −+ = = −+ = = −+ = =

1 2 3 1 2 4 1 3 4 234 49 39 4 10014 49 33 4 48454 49 46 4 354 49 44 4 1264 A A A A A A A A A A A A −+  = =  −+  = =  −+  = =  −+  = =  1 2 3 4 0A A A A   = By PIE, required number of integer solutions ( ) ( ) ) 4 1 1 2 3 4 49 4 4 25199 i i j i i j i j k i j k A A A A A A A A A A =  += − −  +   −    =  

Qn Solution 4(a) The line passing through point ( )00,xy with gradient m, has an equation ( )00y y m x x− = − , where 0m . For its axial intercepts: When 0x= , 00y y mx=− … y-intercept When 0y= , 0 0 yxx m=− … x-intercept For the y-intercept ( )000, y mx− to be the midpoint of ( )00,xy and the x-intercept 0 0 ,0yx m  − : 0 00 02 yxx m +− = and 0 00 0 2 y y mx+ =− 0 0 2 yx m= and 00 2y mx= 0 02 y mx = For ( ),Q x y on C, d d2 xy yx−= d2 d xxy y−= Integrating both sides with respect to y: 2 2 22 2 d d 2 2 , 2 0 x x y y yxc x y D D c −= − = + + = =−   4(b) ( ) ( ) ( ) ( ) ( ) ( ) 3sin 3sin 3sin 3sin 3sin 3sin 3sin 3sin 3sin sin d d deed d d dde 3e cos 3e cosdd d3e cos e 3e cos d de 3cos 1 3e cos d 1 d 3cos d 3cos 1 xx x x x x x x xx vvx x x vv x v xxx vx x v x vx x v x vx v x x   =     += −= −= = −

Integrating both sides with respect to x: 1 d 1 1d 3cos 1 v v x x=+ − 11d 1 d 3cos 1vxvx =+ −  Let 2 2 1cos 1 tx t −= + . Then ( )( ) ( ) ( ) ( ) 22 22 22 2 2 1 2 1 2dsin d 1 2 d 4 1d 1 d2 d1 t t t txx t t t x t tt t x tt + − − − −= + −−=+ + = + 22 2 22 2 2 2 11dd 3cos 1 12 d11311 12 d3 3 1 1 1 1 d12 1 1 2ln 2 2 1 2 v x xvx xt tt t xt tt t t xt t txc t =+ − =+ + − − + =+ − − − + + =+ − += + + −          1 1 1 1 1 221 1 1 2 tan1 2ln ln 22 1 2 tan 2 1 2 tan1 2ln 22 1 2 tan 2 1 2 tan 2e , e 1 2 tan 2 xc x v x c x x xc x x v A A x − − − − − − + = + + − + = + + − + = = −

Qn Solution 5(i) 1 0 2 1 4 2 8 3 2 1 3 2 1 5 2 1 17 2 1 257 F F F F = + = = + = = + = = + = 5(ii) k 0 1 2 1 ... kF F F F − kF 1 0 3F = 5 2 01 15FF = 17 3 0 1 2 255F F F = 257 Conjecture: 0 1 2 1 ... 2,kkF F F F F k + − = −  5(iii) Let ( )P k be the statement: 0 1 2 1 ..