NYJC TJC VJC 2026 H3 Math Prelim (Soln)

1 TEMASEK JUNIOR COLLEGE 2026 PRELIMINARY EXAMINATIONS Higher 3 CANDIDATE NAME CG SUBJECT TUTOR’S NAME MATHEMATICS 9820 PRINTED ANSWER BOOKLET 21 September 2026 3 hours Candidates answer on the Printed Answer Booklet. READ THESE INSTRUCTIONS FIRST Write your name, CG and subject tutor’s name on all the work you hand in. Write your answer to each question in the space provided in the Printed Answer Booklet. If additional space is needed, you should use the additional pages at the end of this booklet; the question number or numbers must be clearly shown. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. This document consists of 20 printed pages. [Turn over For Examiner’s Use Question Number Marks Obtained 1 2 3 4 5 6 Total Marks

2 @TJC 2025 9820 1(a)(i) 1(a)(ii) x y −2 −1 0 1 2 2 1 − 1 − 2 (i) y = ⌊x⌋ x y −2 −1 0 1 2 2 1 − 1 −2 (ii) y = ⌈x⌉

3 @TJC 2026 9820 [Turn over 1(a)(iii) 1(b) Given that the function f is defined by f(x) = |x − p| − |x − q|, where 0 < p < q. f ( ) ( ) ( ), ( ) ( ), ( ) ( ), ( ), 2 ( ), ( ), x x p x q x p x q x p x p x q p x q x p x q x q q p x p x p q p x q q p x q = − − − − − + −  = − + −    − − −  − −  = − +    − [M1] x y −2 −1 0 1 2 4 3 2 1 − 1 − 2 − 3 −4 (iii) y = ⌊x⌋ + ⌈x⌉ x y 0 𝑝 + 𝑞 2 p q q+p q−p −( q−p) g f

4 @TJC 2025 9820 2(a) Choose the k non-empty boxes from the 1n+ labelled boxes. Number of choices: 1n k +  . Distributing n identical balls among these k boxes, each containing at least one ball, gives ( ) 11 11 nn k k kk − − + −  =  −−  ways. By multiplication principle, , 11 1nk nnT kk+−  =   −   . 2(b) Suppose exactly k boxes are non -empty. Since each non - empty box contains at least 2 balls, 2kn . Hence 1 2nk   ( 2 nk  as fewer balls than boxes). Choose the 𝑘 non-empty boxes. Number of choices: 1n k +  . Placing 2 balls into each chosen box uses 2k balls. Distribute the remaining 2nk− balls among the same k boxes without restriction. Number of ways: ( ) 121 11 nkn k k kk −− − + −  =  −−  . Therefore, for fixed k, number of distributions: 11 1 n n k kk + − −      −   Summing over all possible k, 2 1 11 1 n n k n n kV kk= + − −  =    −   .

5 @TJC 2026 9820 [Turn over 2(c) Suppose exactly k boxes are non-empty. Choose the 𝑘 non-empty boxes. Number of choices: 1n k +  . Placing 2 balls into each chosen box uses 2k balls. Distribute the remaining 2nk− balls among the chosen boxes, gives ( )2 21 1 11 n nkk kk  − + −  −=  −−  ways. By multiplication principle, 1 12 1 n nnE k k+ − =  − (n even). Summing over all possible k, 2 2 1 1 1 1 n n n k nV k k= +  −=  −  2(d) Suppose • a boxes contain exactly 2 balls, • b boxes contain exactly 1 ball. Then 2a b n+=  2b n a=− . 0 2na   The remaining boxes are empty after selecting the 2-ball and 1- ball boxes. Choose the a boxes containing 2 balls. Number of choices: 1n a +  . From the remaining 1na+− boxes choose the 2b n a=− boxes containing 1 ball. Number of choices: 1 2 na na +−  − . Hence for fixed a, number of distributions: 11 2 n n a a n a + + −      −   . Since 0 2na   , we obtain 2 0 11 2 n n a n n aW a n a= + + −  =    −   .

6 @TJC 2025 9820 3(a) By the Cauchy-Schwarz inequality: ( ) ( )( ) 2 2 2 221 2 1 2 1 2 1 2 1 2 1 2 x x x xy y y y y y y y       +  + +                  ( ) ( ) 22 2 12 1 2 1 2 12 xxx x y y yy+  + + ( ) 222 1212 1 2 1 2 xxxx y y y y ++ + Equality holds if and only if 12 11 12 xx yy yy = , i.e. 12 12 xx yy= . 3(b) Let ( )P n be the statement 2 2 1 1 1 n in ii n i i i i xx y y = = =     , for n ≥ 2, n   +. By part (a), ( )P2 is true. Assume that ( )P k is true for some 2k , i.e. 2 2 1 1 1 k ik ii k i i i i xx y y = = =     . ( ) ( )( ) 2 2 21 1 11 1 2 2 1 1 1 1 2 1 1 1 1 21 1 1 1 by induction hypothesis since P 2 is true kk i i k ii i i k k i i k k k i i k ik i k ik i k i i k i i x x x y y y x x yy xx yy x y + + == + = + + = + = + = + = + = =+  +  + +  =        Since ( )P2 is true and ( ) ( )P P 1kk+ is true, hence ( )P n is true for each 2n , n   +.

7 @TJC 2026 9820 [Turn over 3(b) (cont’d) 3(c)(i) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 0 2 2 0 a b b c c a a b c ab bc ac a b c ab bc ac − + − + −  + + − + +  + +  + + Alternative ( ) 22 22 ,02ab a b ab a b+  =   Similarly, 22 2 bc bc+  and 22 2 ca ca+  . 2 2 2 2 2 2 2 2 2 2 2 2 a b b c c a ab bc ca a b c ab bc ac + + ++ +  + + + +  + + 3(c)(ii) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 2 2 2 by part (b)2 2 2 2 3 2 by part (c)(i)3 1 a b c b c c a a b a b c ab ac bc ba ca cb abc ab ac bc ba ca cb a b c ab bc ca ab bc ca ab bc ca ab bc ca ab bc ca +++ + + = + ++ + + ++ + + + + + + + + + + = ++ + + + + + ++ = Equality holds when abc== .

8 @TJC 2025 9820 4(a)(i) Substitute 1 1yy u=+ into ( ) ( ) ( ) 2d PQdy x x y R x yx = + + : ( ) ( ) ( ) 2 1 1 1 d 1 1 1 P Q Rd y x x y x yx u u u     + = + + + +           ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 2 2 1 11 2 11 12 d 1d dd Q 2 1P Q R R Qd2 1R is a particular solutiond y u x u x x yx x y x y x u u u xyy xyx u u u − = + + + + +  = + + +   ( ) ( ) ( ) ( ) ( ) 1 22 1 Q 21 d 1 Rd d Q 2R Rd x yu xu x u u u u x u x uy xx − = + +  − = + + ( ) ( ) ( )1 d Q 2R R 0du x x y u xx + +  + = 4(a)(ii) 22d 1 2 4 2dy x xy yx = − + − Observe that ( )Q4xx= and ( )R2x =− . By part (a)(i), ( ) ( )d 4 2 2 2 0du x x ux + + −  + − = . Hence, d 2dux = , giving 2u x C=+ . The general solution is given by 1 1yy u=+ , i.e. 1 2yx xC=+ + .

9 @TJC 2026 9820 [Turn over 4(b)(i) ( )( ) ( )( ) ( ) ( ) ( ) ( ) 4 1 3 2 4 2 3 1 4 1 3 2 4 2 3 1 3 2 3 14 1 4 2 4 1 3 2 4 2 3 1 ln ln ln ln ln ' ' ' '' ' ' '' y y y y y y y y y y y y y y y y y y y yy y y y y y y y y y y y     −−= −− = − + − − − − − −−−−= + − −− − − − ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( )( ) ( ) ( ) ( )( )( ) ( ) ( )( ) ( ) 22 22 ' ' P Q R P Q R QR QR '' QR i j i i j j i j i j i j i j ij i j i j ij y y x x y x y x x y x y x y y x y y y y x x y y yy x x y y y yyy − = + + − − − = − + − = − + + − = + + −  ( ) ( )( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )( ) 4 1 3 2 4 2 3 1 ' Q R Q R Q R Q R 0 x x y y x x y y x x y y x x y y   = + + + + + − + + − + +     = Hence, '0 = , which implies that  is a constan