ASRJC 9649 2025 Prelim P1 Solutions

1 (a) ff f ff uv x ux vx u v ∂ ∂∂ ∂∂ ∂ ∂ =+= + ∂ ∂∂ ∂∂ ∂ ∂  ff f ff uv y uy vy u v  ∂ ∂∂ ∂∂ ∂∂=+= − ∂ ∂∂ ∂∂ ∂ ∂  Normal vector of tangent plane: f ff f ff 11 x uv y uv ∂ ∂∂   +  ∂ ∂∂  ∂ ∂∂  = −  ∂ ∂∂   −−       Since tangent plane at the point P(a, b, c) is parallel to the plane with equation 1xz−= , ff 1 ff 0 11 uv uv ∂∂ +∂∂ ∂∂  −= ∂∂ − − ff 1uv ∂∂⇒+=∂∂ ff ff 0uv u v ∂∂ ∂∂−= ⇒=∂∂ ∂∂ f f f121 2u uv ∂ ∂∂∴ = ⇒==∂ ∂∂ , shown Alternative Method Since tangent plane at the point P(a, b, c) is parallel to the plane with equation 1xz−= , f 1 f ff 0 1, 0 11 x y xy ∂ ∂ ∂ ∂∂  = ⇒= =∂ ∂∂ − − f f f 1 11 102 22 xy u xu yu ∂ ∂∂ ∂∂     = + =+=     ∂ ∂∂ ∂∂      f f f 1 11 102 22 xy v xv yv ∂ ∂∂ ∂∂      = + = +−=      ∂ ∂∂ ∂∂       f f1 2uv ∂∂⇒==∂∂ , shown (b) xy uzxy zu xy v +=+− ⇒=− −

f1 1 2u uv ∂ = −∂ , ( ) 3 f 2 u v v ∂ =∂ 11 11 22 v uv u − = ⇒= ( ) 2 33 1 1 1 112 12 uu uu u v u =⇒ =⇒ =⇒= ± ⇒=   (reject 1u=− ) 1v∴= ( )11xy+ = −−− ( )12xy− = −−− Solving gives 1, 0xy= = Coordinates of point P is ( )1 ,0 ,0 . 2 (a) Let U be the specified subset. 12 11 1, 1 00 UU      = −∈ = ∈       uu . However, 12 2 0 0 U  += ∉  uu , since 2220≠ . Not closed under vector addition. Therefore, it does not form a linear space. (b) Let W be the specified subset and let 1 11 1 a b c  = w and 2 22 2 a b c  = w be two arbitrary vectors in W, i.e. 111 2 2 22 0 and 2 0abc a bc−+= −+= . For any real constants ,λµ ∈  , 12 1 2 12 12 aa bb cc λµ λ µ λµ λµ + +=+  + ww . Here, ( ) ( ) ( ) ( ) ( )1 2 12 12 1 1 1 2 2 22 2 20a a b b c c abc a bcλµ λµ λµ λ µ+ − + + + = −+ + −+ = . Closed under vector addition and scalar multiplication. W is a subspace. (b) 2 3223 2 3++ = ⇒+ +=M M I0 M M M0 ( ) 32 23 22 3 3 6 (shown) = −− = −− − − = + M MM MI M MI

( ) ( ) ( ) ( ) ( ) ( ) 8 7 6 62 232 2 2 2 4 24 ( 2 3) 6 12 36 2 3 13 36 11 33 11( 3 ) + + += + + + = + + ++ = ++ =+ ++ = −−+ + = + = + M M M MMM M I M M M MIIM M IIM M M IM MI M I MI MI 3 2 23 2 23 d dd d d22d dd d d y wy wywx x xx x x= + ⇒ = +⇒ = ( ) ( ) 2 2 2 dd 4 24 2 8 1edd xww wx xxx −+ −+ − + += 2 2 2 dd 4 4edd xww wxx −+ += Consider Homogeneous Equation: 2 2 dd 4 40dd ww wxx + += Characteristic Equation: 2 4 40mm+ += ( ) 24 4 4(1) 4 22m −± −= =− Complementary Function: ( ) 2e x cw Ax B −= + where A and B are arbitrary constants. Let P.I. be 22e x pw Cx −= 2 22' 2e 2 e xx pw Cx Cx −−= − 222 2 2' ' 2 e 4e 4e 4 exxx x pw C Cx Cx Cx−−− −= −−+ ( ) 2 2 2 22 2 22 22 22 e 4e 4e 4 e 4 2e 2 e 4 e exxx x x x x xC Cx Cx Cx Cx Cx Cx−−− − − − − −−−+ + − + = 2 2 22 2 22 22 22 e 8e 4 e 8e 8 e 4 e ex xx x xx xC Cx Cx Cx Cx Cx− −− −−− −−+ +− + = 222e e xxC −− = Comparing coefficients, 121 2CC= ⇒= 221 e2 x pwx −∴= ( ) 2 22 1ee 2 xxw Ax B x −−= ++ where A and B are arbitrary constants ( ) ( ) 2 22 2 22d 1d 12 ee ee 2d 2d 2 xx xxyy x Ax B x Ax B x xxx −− −−+ = ++ ⇒ = ++ −

22d1 e2d2 xy x Ax B xx −= ++ − 221 e 2 d2 xy x Ax B x x −= ++ −∫ ( ) 22 2 211 1e e d22 2 xx x A xB xA xx−− = − + + −− + − ∫ ( ) 22 2 2 211 1 1e e e d22 4 4 x xx x A xB xA xx− −−= − ++− ++ − ∫ ( ) 2 2 2 2211 1 1e ee22 4 8 x xxx Ax B x A x D− −−= − ++− +− −+ ( ) 22 21 e 2 4 24 218 x x Ax x B A x D−= − + ++++ − + ( ) 22 2 12 1 e28 x x Cx C x D−= − + + −+ where 1C , 2C and D are arbitrary constants. 4 (a) 1 01 10 7 nn+ =− vv (b) ( ) 21det det 7 10 010 7 λλ λλλ −− = = −++=−− AI ( 5)( 2) 0λλ− −= 5 or 2λ = For 5:λ = 5 1 515 10 2 0 0 −− − =  → −   AI 505xy y x−+=⇒= 1 ,5 x xxy  = ∈    An eigenvector is 1 5   . For 2:λ = 21 212 10 5 0 0 −− − =  → −   AI 202xy y x−+=⇒= 1 ,2 x xxy  = ∈    An eigenvector is 1 2   . (c) 2 12 0 1 0 1 0 011 11 11 50 52 52 02 21521 513 52 22(5 ) 5(2 ) 5 21 93 n nn n n n n nn nn n n nn −− − − ++ = = = = =  =     − =−  −  − −+ =−   v Av A v A v PD P v v v  

( ) 1 1 4(5 ) 10(2 ) 9(5 ) 9(2 )3 1 5(5 ) 23 1 523 n nnn n nn nn a + = − − −+ = −− − = + (d) If the matrix associated with the recurrence relation is not diagonalisable (only one unique eigenvalue and the dimension of its eigenspace is 1, e.g. 21 69n nnb bb++ = − ), then it is not possible to adopt the previous approach directly. 5 (a) 11 2 0 0 11 de de e 1 (e ) x xx xxx− = ⋅++ ⌠⌠ ⌡ ⌡ ( ) 11 0 11 tan e tan e tan 1 x− −− = = − Let 1tan eC −= , 1tan 1D −= . tan tan e 1tan( ) 1 tan tan 1 e(1) CDCD CD −−−= = ++ . 1 1 0 1 e1d tan (shown)e e e1xx x − − −∴= ++  ⌠⌡ (Note that in this case, 11tan 1 tan e 2 π−− << ) (b) Simpson’s Rule Step size, 1 6h= . n 0 1 2 3 4 5 6 nx 0 1/6 1/3 1/2 2/3 5/6 1 ny 0.5 0.493135 0.473453 0.443409 0.406314 0.365554 0.324027 ( ) ( ) 1 06 135 24 0 1 d 4 2 0.432886337ee 3xx hx yy yyy yy− ≈ ++ ++ + + ≈+ ⌠⌡ Series Expansion ( ) 122 12 1 2 2 1 11e e 2! 2! 2 11122 11 24 xx xxxx x x x − − − − =+ +++ − +−+  =++ =++ = −+     11 2 0 0 1 11 5dde e 2 4 12xx x xx− ≈− =+  ⌠⌠ ⌡ ⌡ Let the exact answer be A. Percentage error using Simpson’s rule 0.432886337 100% 0.000369%A A −= ×= Percentage error using series expansion 512 100% 3.75%A A −= ×=

(c) Student A (using Simpson’s rule) would obtain a better approximation. With the same step-size, Simpson’s rule performs similarly for a wider interval. As for series expansion, it is only accurate near x = 0 and becomes increasingly inaccurate in approximating the original curve as we move away from this point. 6 (a) Using GP sum, 2 (1 ) 1 n n zzzz z z −+++= − Let iez θ= . So ii i i2 i i e (1 e )ee e 1e n n θθ θθ θ θ −+ ++ = − i i( /2) i( /2) i( /2) i( /2) i( /2) i( /2) i(( 1) /2) e e( e e) e( e e) e 2i sin 2 2i sin 2 ( 1) ( 1)cos isin sin2 22 sin 2 nn n n n n nn θθ θ θ θθ θ θ θ θ θ θθ θ − − + −= − −= − ++  +    = Comparing real and imaginary parts, 1 ( 1)cos cos sin cosec2 22 n r nnr θθ θθ = +=∑ and 1 ( 1)sin sin sin cosec222 n r nnr θθ θθ = +=∑ . (b) Taking modulus and squaring: ( ) ( ) 2 22 sin 2cos cos2 cos sin sin2 sin sin 2 n nn θ θθ θ θθ θ θ  + ++ + + ++ =     (c) Take 2 1n πθ = + . From (b), the expression is equal to 2 22 sin sin sin1 11 1 sin sin sin111 nn n nn nnn π ππ π πππ        −      + ++       = = =          +++      (constant independent of n) 7 (a) 12 1 22 12 sin d cos sin ( cos )( 1)sin cos d cos sin ( 1) (1 sin )sin d sin d cos sin ( 1) sin d nn n nn n nn xx x x xn x xx x x n x xx n xx x x n xx −− −− −− = − −− − = − +− − = − +− ∫∫ ∫ ∫∫ Dividing throughout by n, 1 2cos sin 1sin d 1 sin d n nn xxxx xxnn − −= − +− ∫∫ (shown). (b)(i) Let 2 0 sin dn nI π θθ=∫ . Then 1 2 22 0 1 cos sin 11 n nn n xx nII In nn π − −−  − = −− =   , and 0 2I π= . Then 20 42 64 86 1 33 55 73 5,,,2 4 4 16 6 32 8 256II II II II πππ π= = = = = = = = .

Required area = ( ) 242 80 1 352 sin d2 256I π πθθ×= =∫ . (ii) At intersection, 44sin cotθθ= ( ) 84 42 22 sin cos sin (1 sin ) θθ θθ = = − Let 2sinu θ= . Then 42 (1 )uu= − . Since 0u≥ , 2 511 2u uu −=−⇒= . 2 51 51sin sin22θθ −−= ⇒= (since sine is positive in the first 2 quadrants) Note that ( ) ( ) ( ) 21 51 1 51 215 1 51 5 2 φ − −= = = +