TJC 2026 Prelim P1 Ans

Q1 Solution (a) Let w, f and b be the number of bowls of wonton noodles, fishball noodles and beef noodles sold respectively. 0.6w + 0.9f + 1.2b = 222 ...  1.2b − 0.6w = 72 ...  w + f + b = 4f  w − 3f + b = 0 ...  Using GC, w = 80, f = 60, b = 100 Mdm Wang sold 80 bowls of wonton noodles, 60 bowls of fishball noodles and 100 bowls of beef noodles. (b) Total refund = 11 80 $1 100 $0.50 $3045     +   =      

Q2 [Solution] (a) lnxy y= Differentiating w.r.t. x, ( ) d 1 d1 ddyyyx x y x+=  1d d yxyyx −=  2d d1 yy x xy= − Since 0y , d d 0yx  at all points on curve. Thus C has no stationary points. (Shown) (b) When 2ey= , 2 22 ln e 2 eex== ( ) ( ) 22 4 2 2 ed e2d 1e eyx = =−− Equation of tangent to C at point where 2ey= is 24 2 2ee eyx − =− −  42e 3eyx=− + Q3 Solution ( ) ( ) 2 2 4 4 0 4 0 xx x x x x      −− −− − −− −  Multiply by ( ) 2 x − on both sides, ( ) ( ) ( ) ( ) 2 2 2 (to find th 2 e 40 L e -int rceptset 4 0 4 2 ) x x x x x x x      − − − − − = − =  − =  =   For ( ) ( )( ) ( )( )2 2 0x x x− − − − +    −2  +2

From the graph, 2 or 2xx−    +   Alternative Solution 4 xx −−  Multiply by ( ) 2 x − on both sides, ( ) ( ) ( ) ( ) ( ) ( )( ) ( )( )( ) 3 3 2 4 40 40 2 2 0 xx xx xx x x x     −  − − − −  − − −  − + − − +  From the graph, 2 or 2xx−    +   Not so efficient method ( ) ( ) 2 22 22 22 4 4 0 42 0 24 0 24 0 xx x x xx x xx x xx x −− −− − − − + − − + + − − − + − −          22Let 2 4 0xx− + − = ( ) ( )( ) 2 22 2 4 1 4 2 16 222x  − − = = =      For ( ) ( )( ) ( )( )2 2 0x x x− − − − +    From the graph, 2 or 2xx−    +   −2  +2 −2  +2

Q4 Solution (a) 1 (point rep lies in 4th quadrant)arg tan 3 zz  −= =− Method 1 3tan 1 −= ( ) 221 3 10zr= = + − = From diagram, 1 3iz=− Method 2 Let iz a b=+ where a > 0 and b < 0 2 2 2 2 10 10z a b a b= + =  + = --- (1) tan 3 3b baa = =−  =− --- (2) Subst (2) into (1): 22 9 10 aa+= ( ) 1 0aa =   3b=− Thus 1 3iz=− Method 3 From diagram, 110 cos 10 1 10 a  = = =  310 sin 10 3 10 b  = = − =−  Thus 1 3iz=− (b) Let complex numbers represented by B and C be b and c respectively. Since OC = 2OA and OA is perpendicular to OC, c = 2 iz = ( )2i 1 3i− 6 2i=+ By vector addition, b = z + c Re Im O A B C Re Im  1 −3 Re Im  1 −3

= (1 − 3i) + (6 + 2i) = 7 − i Q5 [Solution] (a) f (0) 1= and f (0) 1 = (b) When 0x= , f"(0) 1 2! 2= and f"'(0) 3! q= ( ) 2 2 2 dd 2tan secddyy x p xxx =+ When x = 0, ( )( ) 2 2 22 d 1 0 sec 0d cos 0yp ppx = + = = Thus 1 2! 2 p =  1p= Differentiating w.r.t. x, ( ) ( )( ) 32 2 32 2 dd 2 tan secdd d 2sec 2 sec sec tand yy x p xxx y x p x x xx =+ ++ When 0x= , ( )( ) ( )( ) 3 22 3 d 1 0 sec 0 1 2sec 0 0 1 2 3d y px = + + + = + = Thus 31 3! 2q== (c) 3 1 2211 ...22f ( ) 1 x x xx =+ + + + 2 2 3 23 11 2 11 22 ... 2 11 ... ! 22 11 ...22 x x x x x x =+  − + + +  ++  + ++ 2 2111 .... 824 1x xx=+ −++ 2111 ....28xx= + + + (d) The estimate for 0.2 0 f ( ) dxx will give a smaller percentage error because the Maclaurin series curve is closer to the curve f ( )yx= for x-values closer to zero.

Q6 Solution (a) Let y = 2 1 x x+ , x ∈ ℝ, −1 < x ≤ 0 ( ) ( ) 12 2 2 x y x y x y yx y += − =− = −  1f ( ) 2 xx x− = − , , 0xx  Domain of f −1 = Range of f = (−, 0] (b) (c) Given g : x → 1 − ex x ∈ ℝ, 0 ≤ x < ln 3 ( ) ln3g ln 3 1 e 1 3 2= − = − =− Since Rg = (−2, 0] ⊆ (−, 0] = Df −1 , f⁻¹g exists. x y f −1 0 x = −1 x y f 0 x = −1  y = x y = −1  f−1 x y 0 g

Q7 [Solution] (a) (i) Given: ( ) 21 134nS n n=− 1n n nu S S −=− ( ) ( ) ( )( ) 221113 13 1 144 n n n n= − − − − − ( )( ) ( )( ) 2213 1 1144 n n n n= − − − − − ( )13 1 2144 14 2 7 42 n nn = − − −−== (a) (ii) ( ) 1 717 22 nn nnuu − −−−− = − 1 2=− , a constant independent of n Thus the series is an arithmetic series. (Shown) (b) (i) 33 4 3 24T r x −== , where r is the common ratio  33 8rx −=  12rx −= Since the geometric series is convergent, 1r  i.e., 2 1x  2 2 or 2 x xx   − 

(b) (ii) Required sum ( ) ( ) 11 12 101 11 1 10 1 10 9 ... 32 1 1 2 3072 12 3072 2 3072 2 TT xT rx x x xx x xx − − − − = + + ==−− = − = − = − Alternative solution Required sum ( ) ( ) ( ) 10 10 10110 1 10 1 10 9 313 11 323 1 1 2 3072 12 3072 2 3072 2 SS r rr xr rx x x xx x xx  − − − − =− − =−−− ==−− = − = − = − Q8 Solution (a) 2 1 0     0 1 2    = 2 1 2 1  − A vector normal to 1 is 1 2 1  − 1 : 1 1 1 2 0 2 2 1 1 1 r            − = − =                 Cartesian equation of 1 is x − 2y + z = 2

Q8 Solution (b) sin  = 11 22 1 11 22 1 k k        −              −       1 3 = 2 2 2 2 2 2 14 1 (2) 1 ( 2) 1 k k −+ + + + − + 1 3 = 2 3 56 k k − + Squaring both sides, 9(k − 3)2 = 6(5 + k 2) k 2 − 18k + 17 = 0 (k − 1)(k − 17) = 0  k = 1 or 17 (c) 1 : x − 2y + z = 2 i.e. 1 22 1 r  −=  …  l : 21 12 01 r         =+           , ℝ …  Substitute  into , we have 21 1 2 2 2 1    +       + − =           2 +  −2 − 4 +  = 2  = −1 Hence the coordinates of the point of intersection is (1, −1, −1) (d) The distance from origin to the plane 1 is 2 6 2 : 1 2 1 2 1 2 3 1 3 1 r            − = − =−          −       The distance from origin to the plane 2 is 3 3 66 − = Hence the distance between π1 and π2 = 3 6 + 2 6

Q8 Solution 5 5 6 or 66= Q9 [Solutions] (a) Let ( ) 2 4 2 8 22 x a x aax xx + + + +=++ ( ) 2 4 2 8ax x a x a= + + + + 2 4 2 8 0x x a+ + + = ------- (*) Since 1C and 2C do not intersect, equation (*) has no real roots. Thus discriminant < 0 24 4(1)(2 8) 0a− +  8 16 0a− −  2a− Set of values =  : , 2a a a −  Note that when a = −2, the 2 curves also do not intersect. (b) ( ) ( ) ( ) 2 2 ( 2) 2 4 4 2 8d d 2 x x a x a x ay x x + + + − + + + += + OR: differentiate 42 2y x a x= + + + + Let d 0dyx = ( ) ( ) 2( 2) 2 4 4 2 8 0x x a x a x a + + + − + + + + = ( ) ( ) ( ) 222 4 4 2( 4) 4 2 4 0x a x x a x a x a+ + + + + − − + − + = 2 40xx+= ( )40xx += 0x= or 4x=− When x = 0, 4ya=+ When 4x=− , 4ya=− Coordinates of stationary points = ( )0, 4a+ and ( )4, 4a−− (c) y x y = a O y = x + (a + 2)