2025 EJC Promo (Soln)

1 EJC H2 Mathematics 2025 Promo Solutions 1 Solution Let x, y and z be the respective price per kilogram for pumpkins, carrots and broccoli respectively. 1.20 2.50 1.25 21.86 1.50 2.70 1.35 25.20 1.30 2.60 1.10 21.90 7.80, 1.90, 6.20 x yz x yz x yz x yz ++= ++= ++= ∴= = = Amount that Denise paid 2.30 7.80 3.10 1.90 2.10 6.20 $36.85=×+×+×= 2 Solution The points of intersection: Point A: ( ) 2 52xx− −= + 2 30xx+−=⇒ 1 1 4(1)( 3) 2x −± − −= 1 13 2x −+= or 1 13 2x −−= (reject since x > 0) Point B: 22 5 2 70x x xx ⇒−=+ −−= 1 29 2x += or 1 29 2x −= (reject since x>0) From the graph: 1 13 2x −+< or 1 29 2x +>

2 3 Solution (a) We can deduce that vectors a and b are parallel or either vector a or b is the zero vector. (b) 2 1 1 14 3  = −  n or 2 1 1 14 3  −−  (c) Let the acute angle between 23−+ij k and the y-axis be .θ 20 1.1 30 1cos 14 1 14 θ      −      = = 4 Solution (a) (b) (c) The point of intersection of the transformed curve with the x-axis is the point that was transformed from point ( ,0)Aa . ( ,0) ( ,0) ( ,0)PQa a ka → −  → − y x O x y (-1,0) (0, 0)

3 5 Solution (a) If the sequence converges, let L be the limit. Then as n→∞, nxL→ and 1nxL+ → . So we have ( ) 2 322 2 0 0 or 1LL L L L L L= − −+ ⇒ =⇒= . (bi) 12 3 40.5, 0.375, 0.2285, 0.0925, ...xx x x= = = = As n→∞ , 0nx → (bii) 1 2341.5, 1.1429, 1.1074, 1.0946, ...xx x x= = = = As n→∞ , 1nx → (biii) 1 23 4 1, 3, 9, 891, ...x xx x= −== −= Sequence diverges OR does not converge OR Sequence oscillates between positive and negative terms with increasing magnitude (c) Observe that 23 1 2n n nnnx x xxx+ −= −− , so 1nnyx x += − . From the graph of 232y xxx= −− , when 01 nx<< , 0y< . So 1 0nnxx+ −< (shown). Now, 1 10 nnnnxx x x+ +⇒−< < , i.e. 12 1... ...nnx x xx + >>> > So the (convergent) sequence in b(i) is (strictly) decreasing.

4 6 Solution (a) 26,x ty t= = d 6 6dx x tt= ⇒= and 2 d 2d y tyt t⇒== , so ddd2 d d d 63 y y x tt xtt=÷== . At the point 2)(6 ,Pp p , tp= , so d d3 yp x = So gradient of normal 3 p=− The equation of the normal at the point 2)(6 ,Pp p is ( ) ( ) ( ) 2 3 3 3 6 3 18 18 3 0 shown yp x p p yp p x p p yp x −= − − − = −+ + − −= (b) Since ( )54, 81Q − lies on the normal, ( ) ( ) 3 3 18 81 3 54 0 63 162 0 From the GC 9 , 3 , 6 pp pp p + − −− = −+= =− Since 9p=− gives the point Q, the possible values of p are 3 and 6. (c)

5 7 Solution Let S be the external surface area of the toy. ( ) 2 2212 4 32 2S r rh r r rhππ π ππ= ++ =+ Let V be the volume of the toy. 2 32 3 14 2 23 3V rh r rh rπ ππ π = −= −  Since Vk= , 23 3 2 2 12 33k rh r h k r rππ ππ = −= + ⇒   Sub 3 2 12 3h kr r ππ = +  into S, 23 2 23 2 22 1232 3 223 3 2 4 13 23 33 Sr r k r r r kr r kkr rr rr ππ π π ππ π ππ  = ++   =++  = ++ = + For S to be minimum, 2 2 3 1 3 d 0d 26 2 03 26 2 3 6 26 3 13 S r kr r rk r kr kr π π π π = −= = = = 2 3 2 3 2 3 2 3 1 3 1 23 3 133 13 13 2 13 13 1 3 15 13 13 335 13 13 35 13 khk k kk k kk kk k π π π π ππ ππ ππ π − − −  = +    = +        =       =     = 

6 8 Solution (a) { }fR \0=  while { }fD \1=  so f fR D⊆/ . Hence 2f does not exist. Alternatively: ( ) ff0 1 R= ∈ but f1D∉ . So f fR D . Hence 2f does not exist. (b) ( ) ( ) 2 111 11g g or 111 11 1 1 xxx x xx x x −−  = = = = −  − −−  − − { } ( ) ( ) ( )2 ggD D \ 0,1 or , 0 0,1 1,= = −∞ ∪ ∪ ∞ So 2 1g : 1 , , 0,1x xx x− ∈≠ By considering the graph of 11y x= − for 0,1x≠ , { } ( ) ( ) ( )2gR \ 0,1 or , 0 0,1 1,= −∞ ∪ ∪ ∞ . Alternative for finding range through mapping: { } { } { } 2g g 2gg gg DD R R \ 0,1 \ 0,1 \ 0,1 =  →  →             (c) Method 1: 32g gg= Method 2: 32g gg= ( ) 3 111g g1 1111 xx x xx = −= = =   −−   ( ) ( ) 32 11g g 1 11 11 1 x xxx x = = − = −−= −   − Since ( ) 3g xx= , ( ) 3g k xx= for all integers k. We have 2025 3 675= × , hence ( ) 2025g xx= . So ( ) ( ) 2025g g0 xx+= is equivalent to 1 01x x+ − = 2 10xx⇒ −−= Solving using G.C., 0.618 or 1.62x=− . Alternatively: ( ) ( ) ( )( ) ( ) 2 2 1 1 41 1 1510 21 2 2x xx −− ± − − − −−= ⇒ = = ±

7 9 Solution (a) ( ) ( ) 22 3 4 1 16xy+− += ( ) ( ) 22 22 31 142 xy++ −=⇒ Equations of oblique asymptotes: ( )113 2yx+= ± + ⇒ 1 22 xy= + , 5 22 xy= −− (b) ( ) ( ) 22 3 4 1 16 (1)xy+ − + = −−−− 1 2xt t= − − and 5yt= −+ ------(2) Sub (2) into (1): ( ) 2 21 3 4 6 16 02tt t − + − −+ − = − From GC, 2.0720 (rej),3.6944 (rej),14.453 (rej)1.7806,t = , since 22 t−≤< The coordinates of the point of intersection ( )6.34, 3.22 (c) y-intercept: 21 0 2 102 0.41421 or 2.4142 (rej 2 2) 5.41 (3s.f.) 0x ttt tt y t⇒ =⇒ − −=− ⇒= − ≤ = − −< ⇒ =  Asymptote: 3 3 3 is the horizontal asymptote 22 57 i.e. 7 tt y y − = ≤< ⇒ <≤ <≤ ⇒ −

8 10 Solution (a) ( )tan ln 1 2yx=+  Method 1 – direct differentiation ( ) ( ) ( ) 22 2 ln 1 2d 2 s 1ln 1 2 2d 12 12 21sec ec 12 xy xx x xx y+ = + + × × = = + ++ ( ) ( ) 2d1 2 2 1 (shown)d yxy x+= + Method 2 – arctan first ( ) 1tan ln 1 2yx− = + Diff w.r.t. x, ( ) ( ) 2 2 1d 2 1 d 12 d1 2 2 1 (shown)d y yx x yxy x =++ += + (b) When 0x= , 0y= (So we need at least term in 3x for 1st 3 non-zero terms) Diff (a) implicitly w.r.t. x, ( ) 2 2 ddd1 2 2 4 (*)ddd yy yxy xx x+ + = −−− Diff (*) implicitly w.r.t. x, ( ) 322 2 223 2 d d d d dd12 2 2 4d d d d dd yyy y y yxy x x x x xx + ++= +   ( ) ( ) 232 23 d dd12 44 4d dd y yyxy x xx ⇒+ +− =  When 0x= , d 2d y x = , 2 2 d 4d y x =− , 3 3d 32d y x = So 23 230 2 ... 2 .4 32 16 2! 23! 3 ..y xxx x x x+− ++ += + = −+ (c) ( ) ( ) ( ) ( ) ( ) 2 22 3 .12. .1 2 1 .. 2 2 1! .2 b bbx ax x b ax x abx bax a x b −+ = + + +=+ + − +  From part (b), (1)1ab − −−= − and 22 2 )1 (23 6ab ab =− −−− Sub (1) into (2): ( ) ( ) 2 3 1 131 3 61 aa− ⇒==−− Sub into (1): ( )13 3 3 131b b − =−= ⇒ 13 3,3 13b a=−= ∴ .

9 11 Solution (a) ( ) 1 11 1 2 1d 1 d , where is an arbitrary constante 2 e 2e 1 2 x xx cx x xx c − − − −= = + − −⌠ ⌠ ⌡⌡ (b) ( ) ( ) 222 2 1 1 t 1 1 12 41 1dd d42 211 6 21 4 1 21tan24 1 21tan ,where is an arbitrary constan84 7 4 xx xx xx x c x c x c − − ++ ++ +⋅+ += + = =++ = ⌠⌠⌠ ⌡ ⌡⌡ Alternative: factorise 4 first 22 2 2 1 1 11dd d 1744 4 1 24 2 1 11 2tan2 1 21tan ,where is an arbitrary constant 111 4 17 4 8 2 4 xx xx xx x c cc x x x − − ++ ++  + +   + + = = = =+ + ⌠⌠⌠   ⌡ ⌡ ⌡ (c) 2 d21 21 d xx ux uu u+⇒ ⇒= = += (or ( ) ( ) 1 2 d 21121d2 u xxu − == + ) Sub into integral: ( ) ( ) ( ) 2 42 5 5 2 3 3 2 112 1d d d22 21 211 ,where is an arbitrary constant2 5 3 10 6 x uuuuxx u u xxu c u u cc −+− ++  = = =  − += − +  ⌠⌡∫∫

10 12 Solution (a) 2 73 : 2 1, . 25 l µµ    = +∈  −−  r  12ll⊥ implies that 3 2. 1 0 3 2 5 0 1 15 b bb       =⇒ +−=⇒=    −   Since the two lines intersect, 17 3 3 22 1 1 12 5 a λµ            −+ = +           −−      for some ,.λµ ∈  So we have 7 3 (1) 3 2 2 (2) 1 2 5 (3) a λµ λµ λµ + = + −−−− − + = + −−− + =− − −−− Solving (2) and (3), 2, 1λµ= =− . Substitute into (1): 2a= Thus ( ) 7 34 2 11 1 2 53 OA        = +− =      −−       Coordinates of point A are ( )4, 1, 3 . (b) Let ( )2,0,0C − be a point on the plane 1p . Then the shortest distance from A to 1p is the length of projection of CA    onto n. So shortest distance from A to 1p 2 62 .0 1.0 1 31 15 35 5 55 CA             = = = =    Alternative method: find foot of perpendicular first (not recommended in this case as F is not requested) Let F be t