(Bedok South) AM4 P1 (2025) Students

1 BEDOK SOUTH SECONDARY SCHOOL PRELIMINARY EXAMINATION 2025 4E5N CANDIDATE NAME CLASS REGISTER NUMBER ADDITIONAL MATHEMATICS Paper 1 4049 / 01 2 hours 15 minutes 90 Answer all the questions. 1 Prove that sin 2 cos 2 tan tanx x x x−= . [4] 2 Show that the solution of 3 4 2 1 3( 1)2 5 5 16x x x x+ + + =  is 4 52lg . [4] 3 Giving your answer in the form 2 3 cd+ , solve, without using a calculator, 18 3 32xx =+ . [5] 4 A square of area (11 120)+ cm2 has a length of () ab+ cm, where ab . Without using a calculator, find the values of a and b. [5] 5 The line 3 4 13xy+= intersects the curve 6 10 31 xy x −= − at points P and Q. Calculate the exact length of the line segment PQ. [5] 6 The coefficient of 3x in the expansion of ( )( ) 8 110 2 2kx x+− is zero. Find the value of the constant k. [5] 7 Solve the equation 5cos sin 2 cotcos 2 5sin xx xxx + =+ for 0 180x   . [6] 8 (a) Show that the derivative of 4( 3)xx − with respect to x is ( )( ) 3 5 3 3xx−− . [2] (b) Hence, given that ( ) 4     3 5     3 xxy x −= − and y is decreasing at a constant rate of 70 units/s, calculate the rate of change of x when 1x= . [4]

2 9 (a) Express 25 12 3y x x= − − in the form 2()y a x b c= + + and hence show that y can never be greater than 20. [3] (b) Explain why there are no values of k for which the curve 2( 1) 2( 2) 3y k x k x k= − + + + + is always positive. [4] 10 A rectangular field has sides (3 5)x− m and ( 10)x− m. Its area is at most 200 m2. (a) Find the range of values of x that satisfies the above sides and area conditions. [4] (b) Justify whether a fence of 98 m is enough to enclose the field. [3] 11 (a) State the range of values of x for which the equation below is valid. 2 392log (4 ) log ( 4) 2xx− − − = [1] (b) Express 2 392log (4 ) log ( 4) 2xx− − − = as a quadratic equation 2 0x bx c+ + = and explain why there is only one real solution. [6] 12 A quadratic curve is given by 2 2 9 6y hx x h= − − + , where h is a constant. (a) Show that the equation 0y= has real roots for all values of h except 0h= . [3] (b) State the value of h in the case where 0y= has two real and equal roots. [1] (c) Given that the line 2 12y x h= − − meets the curve 2 2 9 6y hx x h= − − + , find the range of values of h. [5] 13 It is given that 22f ( ) xx x e += . (a) Show that the range of values of x for which f ( )x is a decreasing function is 20 x−   . [4] (b) The gradient with the least value is in the range 20 x−   . Find the value of this gradient, giving your answer in exact form. [5] 14 It is given that 32f ( ) 3 2 16x x x= + + . The remainder when f ( )x is divided by 3xa− , where a is a constant, is the same as the remainder when it is divided by 32x+ . (a) Find the possible values of a. [3] (b) Show, with clear working, that 2x=− is a solution of f ( ) 0x = . [1] (c) Explain why f ( ) 0x = has only one real root. [5] (d) Hence use your answers to parts (b) and (c) to solve the equation 316 2 3 0yy+ − = . [2] END OF PAPER

3 Mathematical Formulae 1. ALGEBRA Quadratic Equation For the equation ax bx c2 0+ + = , x b b ac a= −  −( )2 4 2 Binomial Expansion ( ) ... ...a b a n a b n a b n r a b bn n n n n r r n+ = +     +     +     + +− − − 1 2 1 2 2 , where n is a positive integer and ( ) ( ) ! 1...1 ! ! )( ! r rnnn rrn n r n +−−=−=      2. TRIGONOMETRY Identities BABABA BABABA AAec AA AA sinsincoscos)(cos sincoscossin)(sin cot1cos tan1sec 1cossin 22 22 22 = = += += =+ tan (A  B) = tan tan tan tan A B A B  1  sin sin cos2 2A A A= cos cos sin cos sin2 2 1 1 22 2 2 2A A A A A= − = − = − tan tan tan2 2 1 2A A A= − Formulae for  ABC a A b B c Csin sin sin= = a b c bc A2 2 2 2= + − cos Cabsin2 1=

4 ANSWERS (Bedok South) AM4 P1 (2025) Students 1. Hint: Convert to sin 𝑥 and cos 𝑥 by apply double angle formulae. 2. 2 lg ( 4 5) 3. 8+4√2 3 4. 𝑎 = 5, 𝑏 = 6 5. 𝑃𝑄 = 5 units 6. 𝑘 = 5 7. 30°, 90°, 150° 8. (a) Hint: Apply Product Rule d(𝑢𝑣) d𝑥 = d𝑢 d𝑥 𝑣 + 𝑢 d𝑣 d𝑥 (b) 2.5 units/s 9. (a) 𝑦 = −3(𝑥 + 2)2 + 17 (b) Hint: Apply the two conditions: (i) coefficient of 𝑥2 > 0 (ii) discriminant < 0 10. (a) 10 < 𝑥 ≤ 15 (b) It is enough to enclose the field. 11. (a) 𝑥 < 4 (b) 𝑥 = −5 12. (a) Hint: Apply completing the square to the discriminant. (b) ℎ = 1 3 (c) ℎ ≤ 1 4 or ℎ ≥ 2 and ℎ ≠ 0 13. (a) Hint: For decreasing function, f ′(𝑥) < 0 (b) (2 − 2√2)𝑒√2 14. (a) 𝑎 = 0 or 𝑎 = − 2 9 (b) Hint: Apply Factor Theorem (c) Hint: Show that discriminant < 0 (d) 𝑦 = 1 2

5 MARKING SCHEME (Bedok South) 4E5N AM P1 (Prelim 2025) (w Ans) Qn Solution Marks Testing 1 Trigonometric Identities 𝐿𝐻𝑆 = sin 2𝑥 − cos 2𝑥 tan 𝑥 = 2 sin 𝑥 cos 𝑥 − (2 cos2 𝑥 − 1) sin 𝑥 cos 𝑥 = 2 sin 𝑥 cos 𝑥 − 2 sin 𝑥 cos 𝑥 + sin 𝑥 cos 𝑥 = tan 𝑥 = 𝑅𝐻𝑆 M2 M1 A1 Apply double angle for sine & cosine Expand & simplify 2 Exponential Equations 22(𝑥+2) × 5 = 5𝑥+3 × 8𝑥 24(22𝑥) × 5 = 53(5𝑥) × 23𝑥 22𝑥 (23𝑥)(5𝑥) = 53 24 × 5 1 (2𝑥) (5𝑥) = 25 16 1 10𝑥 = (5 4) 2 10𝑥 = (4 5) 2 𝑥 = lg (4 5) 2 𝑥 = 𝟐 𝐥𝐠 (𝟒 𝟓) M1 M1 M1 A1 Split the powers Separate factors with x and non-x Combine common index Express in logarithm Power law OR 22(𝑥+2) × 5 = 5𝑥+3 × 8𝑥 22𝑥+4 × 5 = 5𝑥+3 × 23𝑥 22𝑥+4 23𝑥 = 5𝑥+3 5 24 2𝑥 = 5𝑥52 10𝑥 = 24 52 10𝑥 = 42 52 𝑥 = lg (4 5) 2 𝑥 = 𝟐 𝐥𝐠 (𝟒 𝟓) M1 M1 M1 A1 Convert to prime bases Separate factors of different bases Combine common index Express in logarithm Power law 3 Surds Expressions 𝑥√18 = 3𝑥 + √32 3𝑥√2 − 3𝑥 = 4√2 3𝑥(√2 − 1) = 4√2 3𝑥 = 4√2 √2 − 1 M1 Grouping Qn Solution Marks Testing = 4√2 (√2 + 1) (√2 − 1)(√2 + 1) = 8 + 4√2 (√2) 2 − (1)2 𝑥 = 𝟖 + 𝟒√𝟐 𝟑 M1 M1 M1 A1 Rationalise denominator with conjugate Expansion Difference of squares OR 𝑥√18 = 3𝑥 + √32 3√2𝑥 = 3𝑥 + 4√2 (3√2 − 3)𝑥 = 4√2 𝑥 = 4√2 3√2 − 3 = 4√2 (3√2 + 3) (3√2 − 3)(3√2 + 3) = 24 + 12√2 (3√2) 2 − (3)2 = 24 + 12√2 9 = 𝟖 + 𝟒√𝟐 𝟑 M1 M1 M1 M1 A1 Grouping Rationalise denominator with conjugate Expansion Difference of squares 4 Surds (Application) (a) Area of square = 11 + √120 (√𝑎 + √𝑏)2 = 11 + 2√30 𝑎 + 𝑏 + 2√𝑎𝑏 = 11 + 2√30 Equating the rational part, 𝑎 + 𝑏 = 11 𝑏 = 11 − 𝑎……(1) Equating the irrational part, 2√𝑎𝑏 = 2√30 𝑎𝑏 = 30 ……(2) M1 M1 Expand Equating rational & irrational parts separately Subst. (1) into (2), 𝑎(11 − 𝑎) = 30 11𝑎 − 𝑎2 = 30 𝑎2 − 11𝑎 + 30 = 0 (𝑎 − 5)(𝑎 − 6) = 0 𝑎 = 5 or 𝑎 = 6 Corresponding b values: 𝑏 = 6 or 𝑏 = 5 Since 𝑎 < 𝑏 𝒂 = 𝟓, 𝒃 = 𝟔 M1 M1 A1 Substitution method Solve quadratic equation Correct set of values

6 Qn Solution Marks Testing 5 Simultaneous Equations 4𝑥 − 𝑦 = 5 … … [1] 𝑦 = 2𝑥2 − 6𝑥 + 7 … … [2] Subst (2) into (1) 4𝑥 − (2𝑥2 − 6𝑥 + 7) = 5 4𝑥 − 2𝑥2 + 6𝑥 − 7 = 5 2𝑥2 − 10𝑥 + 12 = 0 𝑥2 − 5𝑥 + 6 = 0 (𝑥 − 2)(𝑥 − 3) = 0 𝑥 = 2 𝑜𝑟 𝑥 = 3 Subst respective x values into [1], 𝑦 = 4(2) − 5 𝑦 = 3 𝑦 = 4(3) − 5 𝑦 = 7 𝑃(2, 3) and 𝑄(3, 7) 𝑃𝑄 = √(3 − 2)2 + (7 − 3)2 𝑃𝑄 = √𝟏𝟕 M1 M1