Secondary 3 Additional Mathematics Quiz - Algebra Functions (Answer Key)
Total Marks: 40
Topic: Algebra Functions (syllabus-first, Stage 4/5 inferred; not claimed as past-year derived)
Section A
1. x 2 + 6 x + 5 = ( x + 3 ) 2 − 9 + 5 = ( x + 3 ) 2 − 4 x^2 + 6x + 5 = (x + 3)^2 - 9 + 5 = (x + 3)^2 - 4 x 2 + 6 x + 5 = ( x + 3 ) 2 − 9 + 5 = ( x + 3 ) 2 − 4 .
q = − 4 q = -4 q = − 4 . [1]
Teaching note: Completing square: half of 6 is 3, so ( x + 3 ) 2 = x 2 + 6 x + 9 (x+3)^2 = x^2+6x+9 ( x + 3 ) 2 = x 2 + 6 x + 9 , subtract 9 and add original 5.
2. a = 2 , b = − 3 , c = 1 a=2, b=-3, c=1 a = 2 , b = − 3 , c = 1 ; Δ = ( − 3 ) 2 − 4 ( 2 ) ( 1 ) = 9 − 8 = 1 \Delta = (-3)^2 - 4(2)(1) = 9 - 8 = 1 Δ = ( − 3 ) 2 − 4 ( 2 ) ( 1 ) = 9 − 8 = 1 . [1]
3. Factor ( x − 2 ) (x-2) ( x − 2 ) means P ( 2 ) = 0 P(2)=0 P ( 2 ) = 0 by Factor Theorem. [1]
4. T r + 1 = ( 5 r ) 3 5 − r x r T_{r+1} = \binom{5}{r} 3^{5-r} x^r T r + 1 = ( r 5 ) 3 5 − r x r , r = 0 , … , 5 r=0,\dots,5 r = 0 , … , 5 . [1]
5. x 2 − 4 x + 3 = ( x − 1 ) ( x − 3 ) < 0 ⇒ 1 < x < 3 x^2 - 4x + 3 = (x-1)(x-3) < 0 \Rightarrow 1 < x < 3 x 2 − 4 x + 3 = ( x − 1 ) ( x − 3 ) < 0 ⇒ 1 < x < 3 . [1]
6. 1 2 + 1 × 2 − 1 2 − 1 = 2 − 1 2 − 1 = 2 − 1 \frac{1}{\sqrt{2}+1} \times \frac{\sqrt{2}-1}{\sqrt{2}-1} = \frac{\sqrt{2}-1}{2-1} = \sqrt{2}-1 2 + 1 1 × 2 − 1 2 − 1 = 2 − 1 2 − 1 = 2 − 1 . [1]
7. Sum of roots = − − 5 1 = 5 = -\frac{-5}{1} = 5 = − 1 − 5 = 5 . [1]
8. Remainder Theorem: f ( 1 ) = 1 − 2 + 1 = 0 f(1) = 1 - 2 + 1 = 0 f ( 1 ) = 1 − 2 + 1 = 0 . [1]
9. a > 0 a > 0 a > 0 and b 2 − 4 a c < 0 b^2 - 4ac < 0 b 2 − 4 a c < 0 . [1]
10. ( 1 + 2 x ) 3 = 1 + 3 ( 2 x ) + 3 ( 2 x ) 2 + ( 2 x ) 3 = 1 + 6 x + 12 x 2 + 8 x 3 (1+2x)^3 = 1 + 3(2x) + 3(2x)^2 + (2x)^3 = 1 + 6x + 12x^2 + 8x^3 ( 1 + 2 x ) 3 = 1 + 3 ( 2 x ) + 3 ( 2 x ) 2 + ( 2 x ) 3 = 1 + 6 x + 12 x 2 + 8 x 3 . Coeff of x 2 = 12 x^2 = 12 x 2 = 12 . [1]
Section B
11. (a) 2 x 2 + 8 x − 3 = 2 ( x 2 + 4 x ) − 3 = 2 [ ( x + 2 ) 2 − 4 ] − 3 = 2 ( x + 2 ) 2 − 8 − 3 = 2 ( x + 2 ) 2 − 11 2x^2+8x-3 = 2(x^2+4x) -3 = 2[(x+2)^2 -4] -3 = 2(x+2)^2 -8 -3 = 2(x+2)^2 -11 2 x 2 + 8 x − 3 = 2 ( x 2 + 4 x ) − 3 = 2 [( x + 2 ) 2 − 4 ] − 3 = 2 ( x + 2 ) 2 − 8 − 3 = 2 ( x + 2 ) 2 − 11 . [2]
(b) Min value = − 11 = -11 = − 11 when x = − 2 x=-2 x = − 2 . [1]
12. Set k x + 1 = x 2 − 2 x + 3 ⇒ x 2 − ( k + 2 ) x + 2 = 0 kx+1 = x^2-2x+3 \Rightarrow x^2 -(k+2)x +2 =0 k x + 1 = x 2 − 2 x + 3 ⇒ x 2 − ( k + 2 ) x + 2 = 0 . Tangent ⇒ Δ = 0 \Rightarrow \Delta=0 ⇒ Δ = 0 : ( k + 2 ) 2 − 8 = 0 ⇒ k + 2 = ± 2 2 ⇒ k = − 2 ± 2 2 (k+2)^2 - 8 =0 \Rightarrow k+2 = \pm 2\sqrt{2} \Rightarrow k = -2 \pm 2\sqrt{2} ( k + 2 ) 2 − 8 = 0 ⇒ k + 2 = ± 2 2 ⇒ k = − 2 ± 2 2 . [3]
13. f ( − 1 ) = 0 : − 1 + a + 3 + b = 0 ⇒ a + b = − 2 f(-1)=0: -1 + a +3 + b =0 \Rightarrow a+b = -2 f ( − 1 ) = 0 : − 1 + a + 3 + b = 0 ⇒ a + b = − 2 (1)
f ( 2 ) = 4 : 8 + 4 a − 6 + b = 4 ⇒ 4 a + b = 2 f(2)=4: 8+4a-6+b=4 \Rightarrow 4a+b=2 f ( 2 ) = 4 : 8 + 4 a − 6 + b = 4 ⇒ 4 a + b = 2 (2)
(2)-(1): 3 a = 4 ⇒ a = 4 / 3 , b = − 10 / 3 3a=4 \Rightarrow a=4/3, b=-10/3 3 a = 4 ⇒ a = 4/3 , b = − 10/3 . [3]
14. ( 2 − x ) 4 (2-x)^4 ( 2 − x ) 4 : term x 2 x^2 x 2 is ( 4 2 ) 2 2 ( − x ) 2 = 6 × 4 × x 2 = 24 x 2 \binom{4}{2}2^2(-x)^2 = 6 \times 4 \times x^2 = 24x^2 ( 2 4 ) 2 2 ( − x ) 2 = 6 × 4 × x 2 = 24 x 2 . Coeff = 24. [2]
15. Square: x + 3 = ( x − 1 ) 2 = x 2 − 2 x + 1 ⇒ x 2 − 3 x − 2 = 0 x+3 = (x-1)^2 = x^2 -2x +1 \Rightarrow x^2 -3x -2 =0 x + 3 = ( x − 1 ) 2 = x 2 − 2 x + 1 ⇒ x 2 − 3 x − 2 = 0 .
x = 3 ± 17 2 x = \frac{3 \pm \sqrt{17}}{2} x = 2 3 ± 17 . Check: x ≈ 3.56 x\approx 3.56 x ≈ 3.56 valid; x ≈ − 0.56 x\approx -0.56 x ≈ − 0.56 gives RHS negative. Extraneous rejected. [3]
16. α + β = 3 , α β = 2 \alpha+\beta=3, \alpha\beta=2 α + β = 3 , α β = 2 . New sum = α 2 + β 2 = 9 − 4 = 5 = \alpha^2+\beta^2 = 9-4=5 = α 2 + β 2 = 9 − 4 = 5 ; new prod = 4 =4 = 4 . Eq: x 2 − 5 x + 4 = 0 x^2 -5x +4=0 x 2 − 5 x + 4 = 0 . [3]
Section C
17. (a) Vertex form: f ( x ) = a ( x + 1 ) 2 + 4 f(x)=a(x+1)^2+4 f ( x ) = a ( x + 1 ) 2 + 4 . Through (0,1): a + 4 = 1 ⇒ a = − 3 a+4=1 \Rightarrow a=-3 a + 4 = 1 ⇒ a = − 3 . So f ( x ) = − 3 ( x + 1 ) 2 + 4 = − 3 x 2 − 6 x + 1 f(x)=-3(x+1)^2+4 = -3x^2 -6x +1 f ( x ) = − 3 ( x + 1 ) 2 + 4 = − 3 x 2 − 6 x + 1 . Thus a = − 3 , b = − 6 , c = 1 a=-3,b=-6,c=1 a = − 3 , b = − 6 , c = 1 . [3]
(b) f ( x ) = − 3 ( x + 1 ) 2 + 4 f(x) = -3(x+1)^2 + 4 f ( x ) = − 3 ( x + 1 ) 2 + 4 . [1]
18. (a) 5 x + 1 ( x + 1 ) ( 2 x − 1 ) = A x + 1 + B 2 x − 1 \frac{5x+1}{(x+1)(2x-1)} = \frac{A}{x+1} + \frac{B}{2x-1} ( x + 1 ) ( 2 x − 1 ) 5 x + 1 = x + 1 A + 2 x − 1 B .
5 x + 1 = A ( 2 x − 1 ) + B ( x + 1 ) 5x+1 = A(2x-1)+B(x+1) 5 x + 1 = A ( 2 x − 1 ) + B ( x + 1 ) . x = − 1 : − 4 = − 3 A ⇒ A = 4 / 3 x=-1: -4 = -3A \Rightarrow A=4/3 x = − 1 : − 4 = − 3 A ⇒ A = 4/3 . x = 1 / 2 : 7 / 2 = ( 3 / 2 ) B ⇒ B = 7 / 3 x=1/2: 7/2 = (3/2)B \Rightarrow B=7/3 x = 1/2 : 7/2 = ( 3/2 ) B ⇒ B = 7/3 . [3]
(b) Used in integration of rational functions. [1]
19. ( 1 + x ) 4 (1+x)^4 ( 1 + x ) 4 : terms 1 , 4 x , 6 x 2 , 4 x 3 , x 4 1,4x,6x^2,4x^3, x^4 1 , 4 x , 6 x 2 , 4 x 3 , x 4 .
( 2 − x ) 3 = 8 − 12 x + 6 x 2 − x 3 (2-x)^3 = 8 -12x +6x^2 - x^3 ( 2 − x ) 3 = 8 − 12 x + 6 x 2 − x 3 .
x 3 x^3 x 3 : 1 ( − x 3 ) + 4 x ( 6 x 2 ) + 6 x 2 ( − 12 x ) + 4 x 3 ( 8 ) = − 1 + 24 − 72 + 32 = − 17 1(-x^3) + 4x(6x^2) + 6x^2(-12x) + 4x^3(8) = -1 +24 -72 +32 = -17 1 ( − x 3 ) + 4 x ( 6 x 2 ) + 6 x 2 ( − 12 x ) + 4 x 3 ( 8 ) = − 1 + 24 − 72 + 32 = − 17 . [4]
20. Always positive: p > 0 p>0 p > 0 and Δ = 16 − 4 p ( p + 3 ) < 0 ⇒ 4 p 2 + 12 p − 16 > 0 ⇒ p 2 + 3 p − 4 > 0 ⇒ ( p + 4 ) ( p − 1 ) > 0 \Delta = 16 -4p(p+3) <0 \Rightarrow 4p^2+12p -16 >0 \Rightarrow p^2+3p-4>0 \Rightarrow (p+4)(p-1)>0 Δ = 16 − 4 p ( p + 3 ) < 0 ⇒ 4 p 2 + 12 p − 16 > 0 ⇒ p 2 + 3 p − 4 > 0 ⇒ ( p + 4 ) ( p − 1 ) > 0 . With p > 0 p>0 p > 0 , p > 1 p>1 p > 1 . [4]