2026 EJC Prelim P1 (Soln)

1 Error! Reference source not found. EJC 2026 H2 Mathematics Prelim P1 1 A recycling company wants to buy three types of broken machines and process them to recover reusable materials. Each type of broken machine contains different amounts of copper, steel, and plastic for recovery. The amounts of each material recoverable from 1 unit of each type of broken machine are show n in the following table. Machine A Machine B Machine C Copper (kg) 2 1 4 Steel (kg) 3 4 3 Plastic (kg) 1 5 8 The unit costs for Machine A, Machine B and Machine C are $550, $400 and $350 respectively. The company recovered exactly 40 kg of copper, 61 kg of steel, and 56 kg of plastic. F ind the total amount paid by the recycling company to purchase the machines. [4] Solution Let a, b and c be no. of machines A, B and C respectively. 2 4 40 (1) 3 4 3 61 (2) 5 8 56 (3) ab c abc abc + + = −−−−−− + + = −−−−− + + = −−−−−− Solving, 12, 4, 3a bc= = = The total amount paid is $550 12 $400 4 $350 3 $9250× +× +× =

2 Error! Reference source not found. 2 Do not use a calculator in answering this question. It is given that one of the roots of the equation 432 7 8 3 10 0px x x x− + −+= , where p∈  , is 2i− . Find the value of p and the other roots in exact form. [6] Solution Since all the coefficients of the polynomial are real, 2i+ is another root. A quadratic factor is ( ) ( ) ( ) ( ) ( ) 2 2 2i 2i 2 i 2 i 21 45 xx xx x xx  − −  − + = − +  − −       = −+ =−+ So we have ( )( ) 432 2 27 8 3 10 4 5 2px x x x x x px qx− + −+= −+ ++ 3Comparing coeff of : 714 2 Comparing coeff of : 35 8 1 x pp x qq −=− ⇒ = −= −⇒ = Solve 22 20xx++= : ( ) ( )( ) ( ) 22 1 1 22 1i 1i 44 41 152 0 or 4 16 2 2 15 15 or x x x  ±−⇒ − ++ = =  −  −−    +   ⇒= The other roots are: 2i+ , 1 15 i44−− and 1 15 i44−+

3 Error! Reference source not found. 3 Do not use a calculator in answering this question. The points V and W are represented by the complex numbers v and w respectively, where ( ) ( )3 3i 1 3iv= +− and iwv= . (a) Find the exact values of v and ( )arg v . [4] (b) Plot the points V and W on an Argand diagram, showing clearly the geometrical relationship between them. [2] Solution (a) ( ) ( )3 3i 1 3i 3 3 3i 3i 3 6 2 3iv= + − =− + +=− ( ) 22623 i 6 23 43v = − = +− = ( ) ( ) 11 23 3arg arg 6 2 3i tan t an 6 36v π−−  =−= − = − = −        (b) O V Re(z) W Im(z)

4 Error! Reference source not found. 4 (a) Without using a calculator, solve the inequality 2 11 13 1 41 x x − < − . [4] (b) Hence solve the inequality 2 4 11 13 1 41 x x − < − , giving your answer in exact form. [3] Solution (a) ( ) ( )( ) ( )( ) ( )( ) ( )( ) 2 2 2 2 2 2 11 13 1 41 11 13 10 41 11 13 4 1 0 41 4 13 12 0 41 43 4 02 12 1 43 4 02 12 1 x x x x xx x xx x xx xx xx xx − < − − −< − −− − < − −−+ < − −− + <+− −+ >+− 11 34 or or 22 4x xx<− − < < > (b) Replace x with 2x : 2 22 11 34 or or 22 4 (no sol) x xx<− − < < > Method 1 (graphical): 311 3 or or22 22x xx<− − < < >

5 Error! Reference source not found. Method 2 (algebraic): 2 22 22 2 11 34 or or 22 4 13 or 024 1 33002 22 11 3300 2222 11 3 3or or2222 x xx xx x xx xx xx x xx <− − < < > ⇒ < −>   −< − + >           − +< − +>              − < < > <−

6 Error! Reference source not found. 5 The diagram shows a sketch of the curve 2exyx= for 0x≥ , with n rectangles of equal width 1n , where n +∈  , drawn under the curve for 01 x≤≤ . Let nA denote the total area of the n rectangles. (a) Show that 1 2 3 1 1 e rn n n r Ar n − = = ∑ . [3] (b) Find the exact value of lim n nA→∞ . [4] Solution (a) ( ) ( )22123 2 122 222 2 2 123 2 1 22 2 2 3 1 2 2 3 1 2 2111 2 3e e e ... e e 1 1 e 2 e 3 e ... ( 2) e ( 1) e 1 e nn nnn n n nn nnn n n rn n r n nnA n nnn n n nn n r n −− −− − =  −−= +++ + +    = + + ++− +−  = ∑ (b) 1 2 0 112 0 0 11 0 0 1 0 lim e d e 2ed e 2 e 2e d e 2e 2e e 2e 2e 2 e2 x n xx xx x nA xx x xx xx →∞ = = − = −− = −+  = −+− = − ∫ ∫ ∫ O x ... ... 1 y

7 Error! Reference source not found. 6 The function f is defined by ( ) 1f 2, , 1x x xx + ∈= − ≥− . (a) Show that 1f− does not exist. [2] The function f has an inverse if its domain is restricted to 2x≥ . (b) Find an expression for 1f− and state its domain. [4] (c) Sketch the graphs of ( )fyx= and ( )1fyx −= on the same diagram, showing clearly the relationship between the two graphs. [3] The functions g and h are defined by 2 4g( ) 41 x x = − , x∈  , 12x≠− , 12x≠ , ( )h 1e xx −= + , x∈  . (d) Find the exact range of hg. [2] Solution (a) Since the line 3y= (or any other line 12,k k y= <≤ ) cuts the graph of ( )fyx= at 2 points, f is not one- to-one. Hence, 1f− does not exist. Alternative: Since ( ) ( )f 1 2 f5−== (or ( ) ( )f0 f 3 4= = , or others), f is not one-to-one. Hence, 1f− does not exist. (b) Let 2 1 for 2yx x= −+ ≥ 2 12yx−= − Since 22 0xx≥⇒−≤ , ( ) 2 12 2y xx−= − − =− 2 1xy⇒ = + 1 ffD R [1, )− = = ∞ 12f ( ) 1, 1xx x− = +≥

8 Error! Reference source not found. (c) (d) Method 1: use Rg as input for Dh ( )hD, = −∞ ∞ , g 2D 11 = \ , 2− From the graph: ( ] ( )g , 4 0,R = −∞ − ∪ ∞ ( ] ( ) ) ( ) g h 4 1 11 1, ,,2 22 2 1 e , 1, 2 , 4 0,    −  −∞ − ∪ ∞ ∞− ∪ − ∪ ∞        →  → + ∞ ∪ Alternative: find Rhg directly 2 4 41 2 4hg( ) h 1 e 41 xx x − −= = + − ( ) ) 41, 2 1 e ,∪+ ∞

9 Error! Reference source not found. 7 [It is given that a sphere of radius r has surface area 24πr and volume 34 π3 r .] An industrial engineer is designing a storage tank in the shape of a closed composite solid which consists of three parts: • the curved surface of a hemisphere of radius r m. • the curved surface of a cylinder of radius r m and height h m. • the circular top of radius r m. The total volume of the tank is fixed at k 3m , where k is a positive constant. You can assume that the wall of the tank is of negligible thickness. During the preliminary design phase, the engineer uses a simulation software to study the tank’s proportions. The software models r increasing at a constant rate of 1 ms-1, while the volume remains a t k m3 and the composite structure of the tank is kept intact. (a) Find ddht at the instant when 3 67πkr = . [4] The total external surface area is given as S 2m . (b) Show that 252π3 kSr r= + . [3] (c) Find the value of r, in terms of k, for which S is a minimum value. [4] Solution (a) hemisphere cylinder 32 2 2 ππ3 2 3π VV V k r rh khr r = + = + = − Method 1: Chain rule 3 d 22 d3 π hk r r= −−

10 Error! Reference source not found. ( ) ( ) 3 3 d dd d dd 6dWhen and 17d 22d 22 1 136 3πd3 π 7π h hr t rt krr t khk kt r π = ⋅ = = −− = −− = = −     Method 2: Implicit differentiation ( ) ( ) 2 3 3 2 3π d 2 d 2d d d 3d π 6dWhen and 17πd d2 2 1 136d3 π 7π khr r h kr r t tt r krr t hk kt = −  = −−    = = = − −= −  (b) Let aS , bS and cS be the curved surface areas of the hemisphere, the curved surface area of the cylinder, and the area of the circular top surface respectively 22 2 2π 2π π 3π 2π abcSS S S r rh r r rh =++ =++ = + ( ) 2 2 2 22 2 3π 2π 23π 2π 3π 243ππ 3 52π shown3 S r rh krr r r krr r kr r = + = +−  = +− = + (c) 2 2 2 3 3 52π3 d 10 2 πd3 dFor stationary value of , set 0 d 10 2π03 33 5π 5π kSr r Sk rr r SS r kr r kkrr = + = − = −= = ⇒= Check: 2 2 3 d 10 4 π03d Skrr = +> so 3 35πkr = gives a minimum.