ACJC NJC RVHS SAJC 2026 FM Prelim P1 (Ans)

1 Prelim P1 Solutions 1(a) () 2 0 f d 0.463960 (to 6 dp)xx = 1(b) () () ( ) ( ) ( ) () 2 0 1f d f 0 4f 0.5 2f 1 4f 1.5 f 26 0.980280 (to 6 dp) xx ≈++ ++ =  1(c) 0.980280 0.463960% Error 100% 0.463960 111% −=× = 1(d) The graph of ()fyx= is sketched in blue. For each pair of strips, Simpson’s Rule approximates the actual curve by using a quadratic curve passing through the three points, as shown in red. However, for the same strip, the tr apezium rule uses a straight line between two points to approximate the actual curve, as shown in black. From the sketch, there is a greater error when approximating the actual curve using Simpson’s Rule than the trapezium rule. 2(a) ()1 : 1 cos , for 0 πCr a θθ=+ ≤ ≤ ()d sind r a θθ =− Total arc length of 1C

2 () () π 222 0 π 22 0 π 0 π 2 0 π 0 π 0 1c o s s i n d 12 c o s c o s s i n d 22 c o s d 22 2 c o s 1 d2 2c o s d 2 22 s i n 2 4 aa a a a a a a θθ θ θθ θ θ θθ θ θ θ θ θ =+ + − =++ + =+ =+ −  = =  =       2(b) () () π 222 2 0 22 2 π 2 0 1 1c o s d2 1c o s d2 n n n Aa k ak θθ θθ − − =+ =+   The integral () π 2 0 1c o s dθθ+ is finite, positive and independent of n. Alternatively, 22 2 π 2 0 π22 2 0 π22 2 0 π22 2 0 22 2 22 2 12 c o s c o s d2 1c o s 212 c o s d22 31 2cos cos2 d22 2 31 2sin sin222 4 3 π 022 3 π4 n n n n n n n akA ak ak ak ak ak θθ θ θθθ θθ θ θθ θ − − − − − − =+ + +=+ + =+ + =+ +  =−  =    () () 22 π 2 0 21 22 2 π 2 0 1c o s d2 1c o s d2 n n n n ak A kakA θθθθ+ − +== + So nA is a geometric progression.

3 Since 201 0 1kk<<  << , this geometric progression is convergent and hence 1 n n A ∞ =  is finite. 3(a) ii iii 22 2 ii 22 i 2 eee e e e2 R e e 2cos e 2 αβ αβ αβ αβ αβ αβ αβ αβ +− −   −     +−     +   += +    =×  −=  Hence, ()arg 2z α β+= and 2cos 2z α β−=  . Note: ()arg z can be found geometrically by considering the appropriate rhombus in an Argand diagram and using the fact that diagonals bisect interior angles. Then, z can be found by applying the cosine rule. Note that since 1122ππ α−< < and 1122ππ β−< < , we have 11 22ππ 2α β−<< so cos 2α β− is positive. 3(b) From (i), () () 2arg 2argzz α β== + On the other hand, () () () () 22ii i+i2 i2 ee e2 e e cos 2 2cos cos 2 i sin 2 2sin sin 2 z αβ αβαβ αα ββ αα ββ =+ =+ + =++ + ++ + + Since () ()() 2 2 2 Im tan arg Re z z z  = , () ()()sin 2 2sin sin 2tan cos 2 2cos cos 2αα ββαβ αα ββ++ ++= ++ + (proven)

4 3(c) Let ii ie, e, epqr α βγ=== , where 1122π ,, πα βγ−< < . Applying (a) and properties of argument: () ( ) () () () () () () () ii ii i i iii arg arg e e arg e e arg e e arg e arg e arg e 222 0 pqqrr p pqr αβ βγ γα αβγ αβ βγ γα αβγ ++ +   =+ ++ + + −−− ++ +=+ + − − − = Thus, () ( ) ()p qqrr p pqr ++ + is a real number. 4(a) () () () () 13π 4 13π 4 13π 4 2 13π 4 sind d2 d2s i nd 2 d s i n d 4 cos .13π xy xy yyx x yy x x yx c = = = =− +  Sub ()1,1 : () () 2 13π 4 13π 4 41c o s 13π 4 cos 1.13π c c =− + =+ Thus, () () () ()() () 2 13π 13π 44 13π 13π 44 44 cos cos 113π 13π 4 cos cos 1 ( 1 1 0)13π yx yx y =− + + =− + = >  When 54x = , 0.913611 (6 d.p.) 0.9136 (4 d.p.). y = = 4(b) Euler method:

5 () ()13π4sin11.25 1 42 0.911612 (6 d.p.) 0.9116 (4 d.p.). y ≈+   = = Improved Euler method: () () ()()13π 13π 544 454 sin sin111 4 2 2 2 0.911612 0.969181 (6 d.p.) 0.9692 (4 d.p.). y  ×≈+ +    = = 4(c) Comparing absolute errors: Euler: 0.9136 0.9116 0.002−= Improved Euler: 0.9136 0.9692 0.0556 0.002−=> The Euler method provided the better approximation. () () () () 13π 4 1,1 13π 5 44 5, 0.9116124 sind 0.3536 0d2 sind 0.1070 0d 2 0.911612 y x y x  == − < ×== > which is a change in sign of the gradient values. Since the improved Euler uses the average of these two gradient values (which partially cancel each other out) in computing an approximation, it produced a worse approximation compared to the Euler method. OR Stationary points occur when d 0dyx = : 13πsin 4 02 13π π where 4 4 .13 x y xk k kx  = =∈ =  k x 3 4 3 0.9230813 ×= 4 4 41 . 2 3 0 813 ×= 5 4 5 1.538513 ×=

6 Since there is exactly one stationary point in the interval ()1,1.25 at 1.2308x = which is very close to 1.25, the two gradient values used in the improved Euler method have different signs, partially cancelling each other out. This led to the Euler method producing a better approximation than the improved Euler method. 5(a) The three given eigenvectors are linearly independent (since the 3 eigenvalue are distinct). Hence A has three independent eigenvectors (forms a basis of 3 ) and is diagonalisable. 1−=AP D P where 110 1 0 11 a n 0 0d 2 0 0 0 0 101 3    ==   PD 3 11 22 2 511 22 2 11 2 A − =−− (from GC) 5(b) 11 2 2 33 1 2 3 ,, ,.αα α α α α=+ + ∈yv v v  100 11 2 2 3 3 100 100 100 11 100 100 1 2 00 23 3 1 3 2 αα α αα α =+ + +=+ Ay Av Av Av vvv 101 101 101 101 11 2 2 3 3123αα α ++=A y vvv 100 101=Ay Ay 100 100 22 331102 2 3αα α ⋅− − ⋅ =vv v 0 Since v1, v2 and v3 are linearly independent, 312 , 0 , 0αα α =∈= 1 0, 1 tt  =∈y  Alternative: Since A is invertible, A100 is invertible, 100 101 1 = = ⋅ A y A y yA y Ay = y

7 1 0 is an eigenvector correspond to eigenvalue 1 1 1 0, 1 tt  =   ∴= ∈  y y  5(c) Let 11 2 2 33cc c=+ +xv v v () ( ) ( )11 2 2 33 1 2 2 3 3 1 123nn n n n n ncc c ccc=+ + + + =Ax Av Av Av v v v 12 23 31 12 333 nnn n ccc  =+ +  Ax vv v 33 33 =lim 3 0 since n nn c c →∞ =  =≠ Ax v0 v0 1 0 11 0, , 1 αβ α β    =+ ∈   x  6(a) Let i , where ,zxy x y=+ ∈  , () () () ( ) () () () ( ) () ()( ) () () () () 22 22 22 2 2 22 2 2 22 22 22 222 94 i 3 14 i 94 3 14 94 9 1 9 4 18 81 8 16 9 18 9 9 72 144 888 0 5 6 0 10 7 0 51 8 53 2 xy x y xy x y xy x y xx y y xx y y xy y xy y xy xy −+ + = − + − −+ + = − + − −+ += − + − −+ + + + =−+ +− + +−+ = +− + = +− = +− = This is a circle centred at 5i with radius 32 . 6(b) Let the centre of the circle be C, the point 94 i− be A, and the point for which the required argument is minimised be P. ()5i 9 4i 9 9i−− = − +

8 Thus, 99 i 9 2AC =− + = , and () 3πarg 9 9i 4−+ = . 132 1sin sin 392CAP CAP − ∠=  ∠=  Minimum () 13π 1arg 9 4i sin 43z − −+ = −  6(c) If ()arg a θ= , then () *arg a θ=− and () *arg πa θ−= − . * ** * 1wa wa wawa− =  −= ++ , i.e. w lies on the perpendicular bisector of the line segment joining *a and *a− . This perpendicular bisector passes through the origin at an angle of π 2 θ− with respect to the positive real axis. The given property implies that the pe rpendicular bisector is tangent to the locus of P, with w as the point of tangency.

9 So, () 11π 32 32arg sin sin25 5w θ−−  =−  =    , And, () 2253 2 7w =− = . Therefore, iear θ= for all 0r > and 1 32sin 5θ − =  , and the corresponding 11 32 7is i n i s i n25 5 7e or 7ew π −−    −           = . 7(a) 33 2 12 3 15 12 4zz x x yx y+= + −−− Differentiate partially with respect to x, () 22 22 2 2 531 2 331 5 ( 1 ) 4zz x yzx y xx z∂∂ + −+ = + − −−−  =∂∂ + Differentiate partially with respect to y, () 2 2 2( 2)31 2 61 2 ( 2 ) 4zz x yzx y yy z∂∂ −+= − − − −  =∂∂ + For stationary points of S, 0zx∂ =∂ and 0zy∂ =∂ Since 2 4 0 for all zz +> ∈  , 20xy −= and 22 50xy +− = Substitute 2y x= into 22 50xy +− = ,

10 2 2 42 22 4 50 5 4 0(shown) (4 ) (1 ) 0 ( 2)( 2)( 1)( 1) 0 2 or 2 or 1 or 1 1 or 1 or 2 or 2 x x xx xx xxx x x y +− = −+ = −− = +−+ − = =− − =− − At (2 ,1 )−− , 3 12 24 0 1.64 ( 2, 1,1.64)zz z+− =  =∴ − − At (2,1) , 3 12 32 0 2 (2,1, 2)zz z++ =  =− ∴ − At (1 ,2 )−− , 3 12 22 0 1.5331 ( 1, 2,1.53)zz