ASRJC Prelims 2025 H2 Math P1 Solutions

Paper 1 Solutions 1 A cubic curve passes through the points (2, 3) and ( 3, 22)−− . Find the equation of the curve if it has a stationary point at (1, 6)− . [4] [Solution] Let 32y ax bx cx d= + ++ 32(2) (2) (2) 3a b cd+ + += ⇒ 8a + 4 b + 2c + d = 3 …Eq(1) 32( 3) ( 3) ( 3) 22a b cd− +− +−+= − ⇒ –27a + 9b – 3c + d = –22 …Eq(2) 32(1) (1) (1) 6a b cd+ + += − ⇒ a + b + c + d = –6 …Eq(3) 2d 32d y ax bx cx = ++ 23 (1) 2 (1) 0a bc + += ⇒ 3a + 2b + c = 0 …Eq(4) From GC, 2, 1, 8, 1a bc d=== −= − ∴ Equation of curve is 322 81y xx x= +−−

2 (a) Show that 3cos3 4cos 3cosθ θθ= − . [3] (b) Hence, or otherwise, evaluate 6 0 sin 2 cos3 d π θ θθ∫ exactly. [4] Solutions (i) ( )cos3 cos 2θ θθ= + cos 2 cos sin 2 sinθθ θθ= − ( ) ( ) 22cos 1 cos 2sin cos sinθ θ θθθ= −− ( ) 32 32 33 3 2cos cos 2sin cos 2cos cos 2 1 cos cos 2cos cos 2cos 2cos 4cos 3cos (shown) θ θ θθ θ θ θθ θθ θ θ θθ = −− = − −− = −− + = − (ii) ( ) ( ) ( ) 6 0 36 0 426 0 sin2 cos3 d 2sin cos 4cos 3cos d 8sin cos 6sin cos d π π π θ θθ θ θ θ θθ θθ θθ θ = − = − ∫ ∫ ∫ 53 6 0 8cos 6cos 53 π θθ= −+ 5 3 53 53 88cos 2cos cos 0 2cos 056 6 5 83 3 2 252 2 5 89 3 3 3 2 25 1 62 425 ππ  =− + −− +       = −+ −        = − +−         33 2 10 5= −

B O C A r O B C A r 3 (a) Given that θ is small, show that 21sec 1 2θθ≈+ [2] (b) The diagram below shows a circle, centre O and radius r , with points A and B on the circumference such that radians AOB θ∠= , where θ is small. AC is a tangent to the circle at A and OBC is a straight line. (i) Show that the length of chord AB can be approximated by rθ . [2] (ii) H ence show that the perimeter of triangle ABC can be approximated by ( )r abθθ + , where a and b are constants to be determined. [4] [Solution] (a) 12 1sec cos 1 since is small2 θ θ θ θ − = ≈− = ( ) 2 11 2 θ+− − +  211 2θ≈+ (shown) (b)(i) ( )( ) 222 22 2 cos 2 2 cos AB r r r r AB r r θ θ =+− ⇒= − Since θ is a small angle,

( ) 1 22 22 2 cosAB r r θ= − 1 2 2 21 1 2r θ ≈ −−   rθ= b(ii) ( ) 222 22 tan sec OC r r r θ θ = + = secBC OC OB r r θ=−= − 2 2 111 2 1 2 r r θ θ ≈+ − = Perimeter of triangle ABC = AB + BC + AC 2 2 1 tan2 1 2 r rr r rr θθ θ θ θθ = ++ ≈ ++ = 12 2rθθ + 12, 2ab∴= =

4 The Folium of Descartes is a curve, defined by the equation x³ + y³ = 3axy where a is a real constant. It is given that a ≠ 0 for this question. (a) Show that 2 2 d d y ay x x y ax −= − . [2] (b) The point 33 ,22 aa  lies on this curve. Show that the equation of the normal at this point on the curve is independent of the value of a. [2] (c) Given that the curve has a stationary point at ( ) ,pa qa , where p and q are positive constants, find the exact values of p and q. You need not determine the nature of this stationary point. [4] [Solution] (i) Differentiating wrt x: 22 dd33 33 dd yyx y ay axxx+= + ( ) 22d d y y ax ay xx −= − (*) 2 2 d d y ay x x y ax −= − (Shown) (ii) Gradient of normal at 33,22 aa  = 2 22 2 3 1 4 1333 422 33 22 a aaaa aa a − = −=   −−     −    ∴ Equation of normal is y – 3 2 a = 1 3 2 ax − y = x which is independent of the value of a. (Shown) (iii) Let d 0d y x = ⇒ 2 2 0ay x y ax − =− y = 2x a ⇒ x³ + 32x a   = 2 3 xax a   3 3 31 30xx a +−= ( ) 3 33 3 20x xaa −= x = 0 (rejected, 0 and 0ap≠> ) or 1 32xa=

When x = 1 32 a , y = 21 3 2 3 2 2 a aa    = ( ) 0a≠ ⇒ 12 332 ,2aa   is a stationary point of the curve, where 1 32p= and 2 32q= Method 2 Let d 0d y x = ⇒ 2 2 0ay x y ax − =− At ( , ),pa qa 2 22qa p a= 2 qp⇒= At ( , ),pa qa ( ) ( ) ( )( ) 33 3pa qa a pa qa+= ( ) 33 3 0p q pq a+= ≠ 36 3 3pp p+= ( ) 33 20pp −= ( ) 3 2 0pp= > 1/32p= and ( ) 21/3 2/322q= =

5 The curve C has equation given by y = 21 3 18 93 xx x +− + . (a) Without the use of a calculator, find the range of values that y can take. [4] (b) Sketch the graph of C indicating clearly its asymptotes and all stationary points. [3] (c) State a sequence of transformations that will transform the curve C to the curve with equation y = ( ) ( ) ( ) 2 1 3 211 8 21 92 1 3 xx x + −− − −+ . [2] [Solution] (i) 9xy + 3 y = –18x2 + 3x + 1 18x2 + (9y – 3)x + 3y – 1 = 0 For the equation to have real roots, (9y – 3)2 – 4(18)(3y – 1) ≥ 0 9(3y – 1)2 – 8(9)(3y – 1) ≥ 0 (3y – 1)(3y – 1 – 8) ≥ 0 (3y – 1)(y – 3) ≥ 0 ⇒ y ≤ 1 3 or y ≥ 3 Hence the set required is {y ∈ℝ : y ≤ 1 3 or y ≥ 3} (ii) 21 3 18 93 xx x +− + = 221 93x x− +− + ∴ the asymptotes are y = 21x−+ and x = 1 3− 1/3 3 y x x = –1/3 y = –2x + 1 (0, 1/3) (–2/3, 3) –1/6 1/3 2 2 21 9 3 18 3 1 18 6 91 93 2 x x xx xx x x −+ +− + + −− + + −

(iii) Let y = f(x) = 21 3 18 93 xx x +− + y = f(x) → y = ( )f1 x− = ( ) ( ) ( ) 2 1 3 1 18 1 9 13 xx x + −− − −+ → y = ( )f2 1x− = ( ) ( ) ( ) 2 1 3 211 8 21 92 1 3 xx x + −− − −+ The transformations are (1) Translation of the curve in the positive x direction by 1 unit. (2) Scaling of scale factor 1 2 along the direction of the x-axis. OR: (1) Scaling of scale factor 1 2 along the direction of the x-axis. (2) Translation of the curve in the positive x direction by 1 2 unit.

6 It is given that 1 74 9 2 5 ( 1)( 2) 2 1 2 n r r rr r n n= + = −−++ + +∑ (a) Find 2 3 7 73n r r rr= − −∑ giving your answers in terms of n. [4] (b) Show algebraically that 3( 1) ( 1)( 2)r rr r+>+ + for all positive integers r. [2] (c) Hence show that 3 1 749 ( 1) 2 n r r r= + <+∑ . [3] Solution (a) 2 3 7 73n r r rr= − −∑ 2 7 73 ( 1)( )( 1) n r r r rr= −= −+∑ 21 6 7( 1) 3 ( )( 1)( 2) n r r rr r − = +−= ++∑ 21 5 11 74 74 ( )( 1)( 2) ( )( 1)( 2) n rr rr rr r rr r − = = ++= −++ ++∑∑ 9 2 5 92 5 2 (2 1) 1 (2 1) 2 2 5 1 5 2nn = − − −− − −+ −+ + +  25 2 5 672 2 1 nn=+− − + 22 1 5 21 2 1nn= −− + (b) To prove 3( 1) ( )( 1)( 2)r rr r+> + + Method 1 LHS 3( 1)r= + 32 3 31rrr=+ ++ RHS ( )( 1)( 2)rr r= ++ 32 32rrr= ++ Clearly, 32 323 3 1 3 2 since 0rrr rrr r++ + > ++ > . Therefore, 3( 1) ( )( 1)( 2)r rr r+> + +

Method 2 22( 1) 2 1r rr+=++ 2( 2) 2rr r r+=+ Clearly, 2( 1) ( 2)r rr+>+ 3( 1) ( 1)( 2) since 0r rr r r+>+ + > (shown) (c) 3( 1) ( )( 1)( 2)r rr r⇒+ > + + 3 11 ( 1) ( )( 1)( 2)r rr r⇒< + ++ 3 11 11 ( 1) ( )( 1)( 2) nn rr r rr r= = ⇒< + ++∑∑ 3 11 74 74 ( 1) ( )( 1)( 2) nn rr rr r rr r= = ++⇒< + ++∑∑ 92 5 212nn= −− ++ 9 2< since 2 01n >+ and 5 02n >+