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