VJC 2026 Prelim P1 Ans

2026 VJC H2 Math Prelim Paper 1 (Solution) 1 Solution (a) ( )( ) ( ) ( ) ( )( ) ( ) 2 43 1 4 13 01 4 01 22 01 x xx xx x xx x xx xx xx + ≤ − + −− ≤− − ≤− +− ≤− 2 0 or 1 2xx−≤< <≤ (b) Replace x with x2 22 2 2 0 or 1 2 (NA since 0) 2 1 or 1 2 xx x xx − ≤< <≤ ≥ − ≤< − <≤ 2 Solution (a) ( ) ( ) 2 4i 4i 3 i 3i 3i 3i 4i 3 i 31 i3i 3 i aa a a aa −= × ++− − = + = − = + 22 2 3 i 3 2 2 (since 0) p aa aa a aa = + = + = = −< 1 3tan aaα −= ( ) 1 2πarg π π tan 3 3p α −= −+ = −+ = − -2 0 1 2 + - + - + Im Re α arg(p)

(b) 3 Solution (a) As the area covered by the plant approaches 120 hectares, the rate of growth of area covered by the plant decreases and approaches zero. (b) ( ) ( ) d 120d 1d 120 d 1 d d120 ln120 ln120 120 e 120 e 120 e kt kt kt A kAt A kAt A ktA A kt C A kt C AD AB AB − − − = − =− −−= − − −=+ −= −− −= −= = − ∫∫ At 0, 3tA= = 3 120 117BB= − ⇒= At 4, 18tA= = 418 120 117e 0.0343 (3sf) k k −= − = (c) Im Re T S Q R A t (0, 3) A = 120 Alternatively, Deduced from part (a), A cannot exceed 120, 120 – A > 0. ( )ln 120 120 e 120 e kt kt A kt C AB AB − − − −= + −= = −

4 Solution (a) At time t, XB = 30 – x and BY = y 30tan 30 xXCB −∠= and tan 30yBCY∠= ( )tan tan tan tan 1 tan tan XCB BCY XCB BCY XCB BCY θ = ∠ +∠ ∠ +∠= −∠ ∠ ( ) 30 30 30 301 30 30 30 30 900 30 900 xy xy xy y xy − + = − −  −+ = −− 30(30 ) (shown)900 30 xyy xy−+= −+ (b) Method 1 Given y = 3x, 22 30(30 2 ) 10(30 2 )tan 900 90 3 300 30xxx x xxθ ++= =−+ −+ Differentiate with respect to x, ( ) 2 2 22 2 22 10 2(300 30 ) ( 30 2 )(30 2 )dsec d (300 30 ) 20(750 30 ) (300 30 ) xx x x x xx xx xx θθ − + −− + += −+ −−= −+ At stationary point, d 0dxθ = . 2750 30 0xx− −= From GC, 16.225 (5 s.f.)x= (as 0 30x≤< ) ∴ θ is maximum when 16.2 (3 s.f.)x= Method 2 2 1 2 10(30 2 )tan 300 30 10(30 2 )tan 300 30 x xx x xx θ θ − += −+ +=  −+ 2 2 22 2 d 0d 1 10( 1500 60 2 ) 0(300 30 )10(30 2 )1 300 30 x xx xxx xx θ = −−×= −+++ −+

( ) 2 222 20(750 30 ) 0 (300 30 ) 10(30 2 ) xx xx x −− = −++ + 2750 30 0xx− −= From GC, 16.225 (5 s.f.)x= (as 0 30x≤< ) ∴ θ is maximum when 16.2 (3 s.f.)x= Method 3 Given y = 3x, 22 30(30 2 ) 10(30 2 )tan 0 for 0 30900 90 3 300 30xx xx x xxθ ++= = > ≤<−+ −+ Hence, to maximise θ, it is to maximise tan θ. ( ) ( ) 2 2 22 2 22 d d 10(30 2 )tand d 300 30 10 2(300 30 ) ( 30 2 )(30 2 ) (300 30 ) 20(750 30 ) (300 30 ) x x x xx xx x x xx xx xx θ +=  −+ − + −− + + = −+ −−= −+ ( )d tan 0dx θ = 2 750 30 0xx⇒ − −= From GC, 16.225 (5 s.f.)x= (as 0 30x≤< ) ∴ θ is maximum when 16.2 (3 s.f.)x= 5 Solution (a) ( ) ( )( ) ( )( ) ( ) π 2 sin d π 2 cos 2 cos d 2 π cos 2sin x xx x x xx x x xC − =− − −−− = − −+ ∫ ∫ (b) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) π 0 π π2 π0 2 π π2 π0 2 Area bounded by 1st curve π 2 sin d π 2 sin d π 2 sin d 2 π cos 2sin 2 π cos 2sin 2 π π2 π2 2π 2π4 x xx x xx x xx x xx x xx = − = − +−− = − −  − − −   =− − − −− − −   =−−− = − ∫ ∫∫ π2ux= − and d sindv xx = d 2dux =− cosvx=−

( ) 2π 2 π 2π π 2π π 2π π Area bounded by 2nd curve cos 1 d cos 2 1 1 d2 1 cos 2 1 d2 11 sin 222 1 2ππ2 π 2 xx x x xx xx = −− +=− − = −− = −− =− − − −  = ∫ ∫ ∫ ( ) π 9πRequired Area 2 2 π4 8 22= −+= − 6 Solution (a) ( ) ( ) ( ) ( ) 22 22 22 22 22 2 2 16 16 16 x y ax by xa yb a b xa yb a b += + − − +− −−= − − +− =+− Centre of C : ( ),ab and Radius of C = 3 3 5 ----- (1)ba= −+ 22 2 22 16 3 25 ----- (2) ab ab +−= += Sub (1) into (2): 22 ( 3 5) 25aa+− + = 210 30 0 10 ( 3) 0 aa aa −= −= 3 or 0 (reject since 0)aa a= = ≠ From (1): b = –4 Equation of C : ( ) ( ) ( ) 22 2 23 4 3 4 16xy− + + = +− − ( ) ( ) 22 3 49xy− ++ = (Shown) (b) Method 1: Subt parametric eq into circle ( ) ( ) ( ) ( )( ) ( ) 2 2 2 2 3 7 3 2 64 9 74 1 9 4 50 45 10 5 04 tt tt tt tt tt + − + −+ = +−= −−= − += = ≥  74d dd and 2 d dd 27 tx yyt tx t= = ⇒=

When 54t = , 54d5 4 2d7 7yx = = (Not recommended) Method 2: Subt cartesian eq of D into circle ( ) 2 33 7 7xx tt −=+ ⇒= Curve D: ( ) 2 23 67xy −= − ( )43dd7 xyx −= ( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) 22 2 2 222 242 42 2 233 64 97 23 379 7 49 3 4 3 14 3 49 441 4 3 7 3 245 0 353 7 (NA) or 4 353 since 32 xx xx xxx xx x xx  −− + −+ = −+ −−= −+ −− −+= −−−− = −= − −= > At the intersection point, gradient of D: ( )43d 2 35d7 7 xyx −= = 7 Solution (a) 22 22 ( 4) ( 2) 111yx−− −= Centre of the hyperbola C is (2, 4). Turning points (vertices) of C: ( ) ( ) ( )2, 4 1 2,3 and 2,5±= (b) Equation of tangent is 3y= Equation of normal is x = 2 (c) Differentiate 22( 4) ( 2) 1yx− −− = w.r.t. x d2( 4) 2( 2) 0dyyx x− − −= d2d4yxxy −= − Normal: x = 2 (2, 3) Tangent: y = 3

When 6y= , 222 ( 2) 1x−− = 2( 2) 3 2 3 or 2+ 3 x x −= = − d3 3 or d2 2yx −= Equation of tangents are: ( ) 36 (2 3)2 39 322 yx yx −= − − − = − ++ and ( ) 36 (2 3)2 39 322 yx yx −= − + = −+ 8 Solution (a) As n→∞ , 1 0 42n →+ , ( )( )1 1 2 12 1 1 2 n r rr= ∴ + →−∑ , a constant. Hence ( )( )1 1 2 12 1r rr ∞ = −+∑ converges and ( )( )1 11 2 12 1 2r rr ∞ = =−+∑ . (b) Since 224 4 1,rr >− 22 11 4 41rr < − . ( )( ) 2 1 2 1 1 111 11 ...4 9 16 14 4 114 422 12 1 2 r r r r r rr ∞ ∞ ∞ = = = +++ += = <= =−+  ∑ ∑ ∑

(c) Method 1 ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( ) ( ) 10 1 10 11 1 1 1 11 1 0 1 2 12 3 1 2 12 1 1 2 12 1 11 2 12 1 2 12 1 1 1 11 24 1 2 24 1 0 2 11 42 4 6 r r r rr N rN N N rr rr rr rr rr N N −= = −= = = + + = ++ = −+ = −+ = −−+ −+   =− −−  ++ +     = − + ∑ ∑ ∑ ∑∑ Method 2 ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( ) ( ) 1 1 1 10 10 11 1 1 2 12 3 11 1 ...21 23 23 25 2 1 2 3 1 2 12 1 11 21 21 21 21 1 1 11 24 1 2 24 1 0 2 11 42 4 6 r N N r N r r rr NN rr rr rr N N = = = = + + ++ = + ++ ++ = −+ = −−+ −+    =− −−   ++ +       = − + ∑ ∑ ∑∑ 9 Solution (a) 11 110, a nd 3,2311 1 4xy xy= = = = = =++ 1/3 1/2

Making x the subject, 2 1 11 11 1 1 y x y yx yx y = ++ + +=  −= − ( ) 1 22 22 1 3 11Volume π 3 1 d 10.472 10.53 y yyπ  −= + −= ≈  ∫ (b) Method 1 – using partial fractions 1ux= + When x = 0, u = 1 When x = 3, u 2 d 11 2 d dd2 21u uu xxu x== ⇒=+ Volume = 23 0 1πd 11 x x   ++∫ ( ) ( ) ( ) ( ) ( ) 2 2 1 2 2 1 2 2 1 21 1 1π 2d1 2πd 1 22πd 1 1 π 2ln 1 2 1 (since 0) 22π 2ln 3 2ln 232 13π 2ln 32 uuu u u u uu u uu u − =  + = + = − + + = ++ + >   = +− +     = −+ ∫ ∫ ∫ Method 2 – using splitting of numerator Volume = 23 0 1πd 11 x x   ++∫

= ( ) ( ) ( ) ( ) 2 2 1 2 2 1 2 22 1 212 1 1π 2d1 2πd 21 22 2πd 21 1 π ln 2 1 2 1 (since 0) 22π ln 9 ln 432 13π 2ln 32 uuu u uuu u uuu u uu u u − =  + = ++ += −++ + = + ++ + >   = +− +     = −+ ∫ ∫ ∫ 10 Solution (a) Normal of 1π = ( ) 1 2 32 1 1 1 34 1 2 3 12 1 −     × = −− =     −−   11 11 12 12 122cos c os 164.207 39 33 −− −     −    − −− −   = = ° Required angle = 180 164.207 15.8°− °= ° (b) Method 1 Equation of 1π : ( ) 1 41 1 1 1 121 51xyz −   =  −−  + − =− −−− r Equation of 2π: ( ) 1 2 10 2 2 2 10 2xyz − −= − − + = −−− r Solve (1) and (2) using GC: 0 5 z z  =−+r 1 00 : 5 1 , 01 l αα      = −+ ∈      r 