2022 SCGS AMath Prelims P1 w Ans

SINGAPORE CHINESE GIRLS’ SCHOOL PRELIMINARY EXAMINATION 2022 SECONDARY FOUR O-LEVEL PROGRAMME CANDIDATE NAME Solution CLASS 4 REGISTER NUMBER CENTRE NUMBER INDEX NUMBER ADDITIONAL MATHEMATICS 4049/01 Paper 1 Wednesday 30 August 2022 2 hours 15 minutes Candidates answer on the Question Paper. No Additional Materials are required.

2 1. A quadratic curve is given by the equation 2f ( ) 6 13 3xx x= −− . By expressing f ( x) in the form 2()ax h k−+ , explain why the curve has no real roots. [4] 2f ( ) 6 13 3xx x= −− 2 1332 3xx= − −+ ( ) 2 133 11 3x= − − −+ ( ) 2 1031 3x= − −+ ( ) 2 3 1 10x= − −− Since the curve has a maximum point of (1, –10) which is below the x-axis, the curve doesn’t cut the x-axis and thus it has no real roots. Alternative Method Since ( ) 2 1 10x −≥ , ( ) 2 310x− −≤ ( ) 2 3 1 10 10x− − − ≤− f ( ) 10x∴ ≤− , f( ) 0x ≠ for all values of x Hence, the curve has no real roots. 2. (a) Without the use of a calculator, find the value of 6n , given that 1 3 322 9 n nn n + + −= . [3] (b) The value of an antique is estimated to increase by k % per year. Given that its value at the beginning of 2017 was $20 000. The value V, after t years, can be modelled by (1.11)tVA= . (i) State the value of A and of k. [2] (ii) Calculate the year when its estimated value will first reach $190 000. [2] (a) 1 3 322 9 n nn n + + −= ( ) 3 322 1 3 n n−= 36 7 n = (b) (i) (ii) A = 20 000, k = 11 190000 20000(1.11) t= 1.11 9.5t = lg 9.5 lg1.11t = 21.6t = The year 2038

3 [Turn over 3. The term containing the highest power of x in the polynomial f (x) is 43x . Two of the roots of the equation f (x) = 0 are x = –1 and x = k, where k in an integer. Given that 2 26xx −+ is a quadratic factor of f( x) and f( x) leaves a remainder of – 36 when divided by x, (a) show that k = 2. [3] Hence, (b) determine the number of real roots of the equation f (x) = 0. [3] (a) (b) 2f ( ) 3( 2 6)( 1)( )x x x x xk= −+ + − f (0) 36= − 3(6)( 1)( ) 36k−= − k = 2 (shown) 2 2 60xx − += Discriminant = ( ) ( ) 2 2 46 0−− < ∴ no real roots When f (x) = 0, there are 2 real roots. 4. Given that o 1cos 20 k= , where k > 0, find an expression, in terms of k , for the following trigonometric ratios. (a) osec( 20 )− , [1] (b) osin160 , [2] (c) otan110 . [1] (a) osec( 20 ) k−= (b) osin160 osin 20= 2 1k k −= (c) ootan110 tan 70= − 2 1 1k = − − k 1

4 5. The concentration of a drug, C( t), in a patient’s bloodstream, t hours after an injection of the drug, can be modelled by 3 16() 75 5 ttCt = −+ . Find the length of time in which the drug concentration in the bloodstream is increasing after an injection [4] 2 16'( ) 25 5 tCt = −+ '( ) 0Ct >⇒ 2 16 025 5 t−+> 2 2 80 0 80 0 t t −> −< ( 80)( 80) 0tt+ −< 80 80t− << Since t ≥ 0, 0 80t≤< or 0 45t≤< Duration = 45 h or 8.94 h 6. (a) Express ( ) 2 2 13 2 12 xx xx −− + in partial fractions. [5] (b) Find the value of ( ) 2 2 2 1 13 2 d 12 xx x xx −− +∫ . [4] (a) Let 2 2 13 2 (1 2 ) xx A xx x −− = ++ 2 12 ( 12 ) BC xx+++ 221 3 2 (1 2 ) ( )(1 2 ) ( )x x A x Bx x Cx⇒− − = + + + + 1When , 2x = − 31 11 22 2 C +−= −  4C∴= − When 0,x = 1 (1)A= 1A∴= When 1,x = 1 3 2 1(9) (1)(3) 4(1)B−− = + − 3B∴= − Or Comparing coefficient of x2, 42 2AB+= − 3B∴= − 2 22 13 2 1 3 4 ( 12 ) 12 ( 12 ) xx xx x x x −−∴ = −−+ ++

5 [Turn over (b) ( ) 22 21 13 2 d 12 xx x xx −− +∫ 2 21 13 4 d12 (1 2 ) xxx x = −− + +∫ ( ) 2 1 34ln ln 1 2 2 ( 1)(2)(1 2 )xx x = − +− −+ ( ) 2 1 32ln ln 1 22 12xx x = − ++ + 32 32ln 2 ln 5 ln1 ln 325 23 1.321009 0.98125   = − +− − +     = −+ 0.340= −

6 7. The diagram shows a rectangle ABCD. The vertices of the rectangle are A(1, 0), B(−3 , 3), C and D. The area of the rectangle is 50 square units. (a) Show that x-coordinate of C satisfies the equation 2 6 27 0xx +−= . [6] (b) Find the coordinates of C and D. [3] (a) 30 3 31 4 ABm −= = −−− 4 3 BCm = Equation of BC, 43 ( 3)3yx−= + 4 73 xy = + AB = 2243+ = 5 units ∴ BC = 10 units BC2 = 100 unit2 2 2 4( 3) 7 3 1003 xx + + +− = 2 2 (4 12)6 9 1009 xxx ++ ++ = 229 54 81 16 96 144 900xx xx++ + ++= 225 150 675 0xx + −= 2 6 27 0xx +−= (b) ( 3)( 9) 0xx− += 3, 9 (NA)x = − 4(3) 7 113y = += ∴C(3, 11) y x O A(1, 0) B (−3, 3) C D

7 [Turn over Let D(x, y) Midpoint of AC = Midpoint of BD ( 3) 1 3 722 x x+− + = ⇒= 3 11 0 822 y y++ = ⇒= ∴D(7, 8) 8. The diagram shows the vertical cross-section of a piece of solid conical wood. The solid cone has base radius 13.5 cm and height 36 cm. A solid cylinder of radius r cm and height h cm was to be carved out of the solid cone. (a) Show that the total surface area of the solid cylinder, A cm2, is given by 21072 3Ar r ππ= − . [4] (b) Given that r can vary, find the stationary value of A and determine its nature. [5] (a) By similar triangles, 13.5 36 36 r h =− 8 108 3rh= − 108 8 3 rh −= or 836 3 r− A 222 r rhππ= + 2 108 8= 2 2 3 rrrππ −+  2 2 216 16= 2 33 rrr πππ +− 210= 72 3rrππ − (Shown) r cm 13.5 cm 36 cm h cm

8 (b) d 20 72d3 A rr ππ= − When d 0d A r = , 2072 0 3 rππ−= r = 10.8 2 2 d 20 03d A r π= −< is maximumA⇒ 21072 (10.8) (10.8)3 1221.45 A ππ= − = 21944 = 1220 or cm 5A π∴

9 [Turn over 9. The diagram shows a circle, centre O, with diameter AB. The point C lies on the circle. The tangent to the circle at A meets BC extended at D and CO is perpendicular to AB. (a) Prove that C is the midpoint of BD. [4] (b) Prove that triangle ADC is isosceles. [3] (a) ∠ BAD = 90° (tan ⊥ radius) = ∠BOC (CO ⊥ AB) ∠CBO = ∠DBA (common angle) ∆BOC and ∆BAD are similar. (AA) ∴ 1 2 BO BC BA BD= = Hence, C is the midpoint of BD. Alternative Method ∠ BAD = 90° (tan ⊥ radius) = ∠BOC (CO ⊥ AB) ∴CO is parallel to DA. ∠CBA = ∠DBA (common angle) ∠BOC = ∠BAD (alternate angles) ∆BOC and ∆BAD are similar. (AA) ∴ 1 2 BO BC BA BD= = Hence, C is the midpoint of BD. (b) ∠DAC = ∠CBA (alternate segment theorem) ∠BCO = ∠CBA (base angles of an isosceles ∆) ∠BCO = ∠CDA (corresponding angles of similar ∆) ∴∠CDA = ∠DAC Hence, triangle ADC is isosceles. B A O D C

10 10. (a) Show that 2 1 1 2cos sec 1 sec 1 sin x xx x +=+− . [3] (b) Hence find, for ππ 22 x−≤≤ , the exact solutions of the equation 11 4cossec 1 sec 1 xxx +=+− [4] (a) (b) 11 sec 1 sec 1xx ++− 2 2sec sec 1 x x= − 2 2sec tan x x= 2 2 2 cos cos sin x xx =  2 2cos sin x x= (shown) 2 2cos 4cossin x xx = 22cos 4sin cosx xx= ( ) 22cos 1 2sin 0xx −= 2cos 0x = or 21 2sin 0 x−= 2x π= ± 1sin 2 x = ± 4x π= ±