Secondary 1 Mathematics Quiz - Algebra Functions
Answer Key
Section A: Short Answer Questions (20 marks)
1. f(x)=3x−5
f(4)=3(4)−5=12−5=7
[2 marks]
Marking: 1 mark for correct substitution, 1 mark for correct answer.
2. g(x)=x2+2x−1
g(−3)=(−3)2+2(−3)−1=9−6−1=2
[2 marks]
Marking: 1 mark for correct substitution, 1 mark for correct answer.
Common mistake: (−3)2=−9 (incorrect). Correct: (−3)2=9.
3. h(x)=7−2x
7−2x=1
−2x=1−7=−6
x=−2−6=3
[2 marks]
Marking: 1 mark for setting up equation, 1 mark for correct answer.
4. f(x)=4x+3, g(x)=x−2
f(2)=4(2)+3=8+3=11
g(5)=5−2=3
f(2)+g(5)=11+3=14
[2 marks]
Marking: 1 mark for each correct evaluation, final answer must be correct for full marks.
5. p(x)=x2−4x+4
p(0)=02−4(0)+4=4
p(4)=42−4(4)+4=16−16+4=4
[2 marks]
Marking: 1 mark for each correct value.
6. f(x)=32x+6
f(6)=32(6)+6=312+6=318=6
[2 marks]
Marking: 1 mark for correct substitution, 1 mark for correct simplification.
7. f(x)=5x−a and f(3)=7
f(3)=5(3)−a=7
15−a=7
a=15−7=8
[2 marks]
Marking: 1 mark for setting up equation, 1 mark for correct answer.
8. g(x)=x2−9
x2−9=0
x2=9
x=3 or x=−3
[2 marks]
Marking: 1 mark for factorising or rearranging, 1 mark for both correct values.
Common mistake: giving only x=3 and omitting x=−3.
9. f(x)=2x+1 and f(k)=11
2k+1=11
2k=10
k=5
[2 marks]
Marking: 1 mark for setting up equation, 1 mark for correct answer.
10. h(x)=3x2−x+5
h(−1)=3(−1)2−(−1)+5=3(1)+1+5=3+1+5=9
[2 marks]
Marking: 1 mark for correct substitution, 1 mark for correct answer.
Common mistake: −(−1)=−1 (incorrect). Correct: −(−1)=+1.
Section B: Structured Response Questions (20 marks)
11. f(x)=2x2−3x+1
(a) f(3)=2(3)2−3(3)+1=2(9)−9+1=18−9+1=10
[1 mark]
(b) f(−2)=2(−2)2−3(−2)+1=2(4)+6+1=8+6+1=15
[1 mark]
(c) f(x)=1
2x2−3x+1=1
2x2−3x=0
x(2x−3)=0
x=0 or 2x−3=0⇒x=23
x=0 or x=1.5
[2 marks]
Marking: 1 mark for setting up and simplifying equation, 1 mark for both correct solutions.
12. g(x)=ax+b
(a) g(1)=a(1)+b=5⇒a+b=5
g(3)=a(3)+b=13⇒3a+b=13
[1 mark]
Mark for both equations correct.
(b) Subtracting: (3a+b)−(a+b)=13−5
2a=8
a=4
Substitute into a+b=5: 4+b=5⇒b=1
a=4, b=1
[2 marks]
Marking: 1 mark for correct method, 1 mark for both correct values.
(c) g(0)=4(0)+1=1
[1 mark]
13. f(x)=x2−6x+8
(a) f(1)=(1)2−6(1)+8=1−6+8=3
[1 mark]
(b) x2−6x+8=0
(x−2)(x−4)=0
x=2 or x=4
[2 marks]
Marking: 1 mark for correct factorisation, 1 mark for both correct values.
(c) Completing the square: f(x)=(x−3)2−1
Least value occurs at x=3, and the least value is −1
[1 mark]
Accept: least value is −1 when x=3.
14. h(x)=x−24x−8, x=2
(a) h(5)=5−24(5)−8=320−8=312=4
[1 mark]
(b) h(100)=100−24(100)−8=98400−8=98392=4
[1 mark]
(c) h(x)=x−24x−8=x−24(x−2)=4 (for x=2)
As x gets very large, h(x)=4.
The value of h(x) is always 4 (for all x=2), so as x gets very large, h(x) remains 4.
[2 marks]
Marking: 1 mark for simplifying/factorising, 1 mark for stating the value is 4 with explanation.
15. f(x)=3x−2, g(x)=x2+1
(a) f(1)=3(1)−2=1
g(1)=(1)2+1=2
f(1)×g(1)=1×2=2
[1 mark]
(b) g(2)=(2)2+1=4+1=5
f(g(2))=f(5)=3(5)−2=15−2=13
[2 marks]
Marking: 1 mark for finding g(2) correctly, 1 mark for correct final answer.
(c) f(2)=3(2)−2=6−2=4
g(f(2))=g(4)=(4)2+1=16+1=17
[1 mark]
16. C(n)=2.5n+20
(a) C(40) = 2.5(40) + 20 = 100 + 20 = \boxed{\120}$
[1 mark]
(b) 2.5n+20=120
2.5n=100
n=2.5100=40 booklets
[2 marks]
Marking: 1 mark for setting up equation, 1 mark for correct answer.
(c) The value 20 represents the fixed cost (or setup cost) of printing, i.e., the cost when zero booklets are printed.
[1 mark]
Accept: fixed charge / base cost / setup fee.
17. f(x)=x2+bx+c, with f(1)=0 and f(4)=0
(a) f(1)=1+b+c=0⇒b+c=−1
f(4)=16+4b+c=0⇒4b+c=−16
[1 mark]
Mark for both equations correct.
(b) Subtracting: (4b+c)−(b+c)=−16−(−1)
3b=−15
b=−5
Substitute: −5+c=−1⇒c=4
b=−5, c=4
[2 marks]
Marking: 1 mark for correct method, 1 mark for both correct values.
(c) f(x)=x2−5x+4=(x−1)(x−4)
[1 mark]
18. f(x)=2x2−8x+6
(a) f(0)=2(0)2−8(0)+6=6
f(4)=2(16)−8(4)+6=32−32+6=6
[1 mark]
(b) 2x2−8x+6=0
2(x2−4x+3)=0
2(x−1)(x−3)=0
x=1 or x=3
[2 marks]
Marking: 1 mark for correct factorisation, 1 mark for both correct values.
(c) f(x)=2x2−8x+6
=2(x2−4x)+6
=2(x2−4x+4−4)+6
=2[(x−2)2−4]+6
=2(x−2)2−8+6
=2(x−2)2−2
[1 mark]
19. F(d)=3.20+0.85d
(a) F(12) = 3.20 + 0.85(12) = 3.20 + 10.20 = \boxed{\13.40}$
[1 mark]
(b) 3.20+0.85d=13.40
0.85d=13.40−3.20=10.20
d=0.8510.20=12 km
[2 marks]
Marking: 1 mark for setting up equation, 1 mark for correct answer.
(c) The fixed starting fare is \boxed{\3.20}$.
[1 mark]
20. f(x)=ax2+bx+3, with f(1)=8 and f(−1)=6
(a) f(1)=a+b+3=8⇒a+b=5
f(−1)=a−b+3=6⇒a−b=3
[1 mark]
Mark for both equations correct.
(b) Adding: (a+b)+(a−b)=5+3
2a=8⇒a=4
Substitute: 4+b=5⇒b=1
a=4, b=1
[2 marks]
Marking: 1 mark for correct method, 1 mark for both correct values.
(c) f(x)=4x2+x+3
f(2)=4(4)+2+3=16+2+3=21
[1 mark]