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AS Mathematics 9709
Common Test 1 2022
Time allowed: 55 minutes
50 Marks
READ THESE INSTRUCTIONS FIRST
Write in dark blue or black pen.
You may use soft pencil for any diagrams or graphs.
Do not use staples, highlighters, glue or correction fluid.
Answer all the questions on this paper.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in
degrees, unless a different level of accuracy is specified in the question.
The use of an electronic calculator is expected, where appropriate.
You are reminded of the need for clear presentation in your answers.
The number of marks is given in brackets [ ] at the end of each question or part question.
The total number of marks for this test is 50.
Name: Solutions
ACG Parnell College: AS Mathematics: Common Test 1: 2022
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1) Determine P and Q such that ( 2+3 √ 3 ) ( 1−P √ 3 )=Q+ √ 3. [2]
2−2 P √ 3+3 √ 3−9 P=Q+ √ 3
2−9 P+ ( 3−2 P ) √ 3=Q+ √ 3
3−2 P=1
P=1
→ 2−9 P=Q
Q=−7
3 3 √3 6 √5
2) Rectangle ABCD has sides AB = cm, BC = . Show that AC= [2]
√5 √5 5
B C
A D
√( )( ) √
2
+ √ = +
2
3 3 3 9 27
AC=
√5 √5 5 5
¿
√ 36 6
=
5 √5
6 √5 6 √ 5
¿ × =
√5 √5 5
ACG Parnell College: AS Mathematics: Common Test 1: 2022
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2
3) Expand and simplify ( √ x+ 1−√ x−1 ) [2]
¿ ( x +1 )−2 √ x+1 × √ x−1+ ( x−1 )
¿ 2 x−2 √ ( x +1 ) ( x−1 )
Or ¿ 2 x−2 √ x 2−1
ACG Parnell College: AS Mathematics: Common Test 1: 2022
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x
3
4) Solve 9 x−2= [2]
3
x−2
( 32 ) =3 3
x −1
2 x− 4 x−1
3 =3
2 x−4=x−1
x=3
5) Solve the following
4 2
(i) x +2 x =99 [3]
2 2 4
let y=x → y =x (oe)
2
y +2 y=99
2
y +2 y−99=0
( y +11 )( y−9 )=0
y=−11, y=9
2 2
x =−11, x =9
No solution x =3 , x=−3
ACG Parnell College: AS Mathematics: Common Test 1: 2022
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2
(ii) 4 x −6 x−14 =16 [2]
2
x −6 x−14=2
2
x −6 x−16=0
( x−8 )( x +2 )=0
x=8 , x=−2
ACG Parnell College: AS Mathematics: Common Test 1: 2022
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6) For y= ( x +1 )2−1 ,
(i) Sketch the curve stating the coordinates of the vertex and the intercepts. [3]
−3 −2 −1 1 x
−1
(vertex) (intercepts) (shape)
ACG Parnell College: AS Mathematics: Common Test 1: 2022
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(ii) What transformation(s) would map the graph of f (x)=( x+1 )2−1 to the graph g ( x )=( x−1 )2−3 ? [3]
Stating translation
Translation by (20)∧translation by (−20 )
7) The points A and B have coordinates ( 2 , h ) and ( 4 h+6 , 5 h ) respectively.
The equation of the perpendicular bisector of AB is 3 x+ 2 y =k . Find the values of the constants h and k . [5]
2 y=−3 x +k
−3 k 2
y= x+ → gradinet of AB=
2 2 3
2 5 h−h 4h
= =
3 4 h+6−2 4 h+4
2 ( 4 h+4 )=12 h
8 h+ 8=12 h
4 h=8→ h=2
midpoint of AB: ( 2+ 42h+ 6 , h+52 h )
as h=2 midpoint : ( 8 , 6 )
midpoint on 3 x+ 2 y =k →3 × 8+2 ×6=k
k =36
ACG Parnell College: AS Mathematics: Common Test 1: 2022
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8) Find the values of k such that y=kx−5 is below y=x 2 +2 x+ 4 [5]
2
x + 2 x + 4=kx−5
2
x + 2 x−kx+ 9=0
( 2−k )2−36< 0
2
k −4 k +4−36< 0
2
k −4 k−32<0
( k −8 ) ( k + 4 )< 0
−4<k < ¿8 (for -4 and 8) (for correct region)
ACG Parnell College: AS Mathematics: Common Test 1: 2022
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9) Function f is defined as
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f : x ⟼ 3 x −6 x−1 for 0 ≤ x ≤ 4
i) Express f (x) in the form a ( x +b )2+ c, where a , b and c are constants. [3]
f ( x )=3 ( x 2−2 x +1 )−3 × 1−1
2
¿ 3 ( x−1 ) −4
ACG Parnell College: AS Mathematics: Common Test 1: 2022
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ii) Find the range of f ( x ) [2]
min at x =1→ f ( 1 )=−4 up ¿ x=4 → f ( 4 )=23 ¿
Range is−4 ≤ f (x )≤ 23
The function h is defined as h : x ⟼ 3 x 2−6 x +5 for k ≤ x ≤ 4where k is a constant.
(iii) Find the minimum value of k so that h has an inverse. [1]
k =1
(iv) For this value of k find an expression for h−1 ( x ) [3]
2
x=3 ( y−1 ) −4+ 6
2
3 ( y−1 ) =x +2
x−2
( y−1 )2 =
3
y−1=
√ x−2
3
ACG Parnell College: AS Mathematics: Common Test 1: 2022
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f −1 ( x )=1+
√ x−2
3
(v) State the domain of this inverse. [1]
2 ≤ x ≤29
10) The coordinates of points A and B are (−1 , 2 ) and ( 7 , 8 ) respectively.
i) Find the equation of the circle C for which AB is the diameter. [4]
centre : ( −1+7
2
,
2 )
2+8
=( 3 , 5 )
radius=√ (7−3 ) + ( 8−5 ) = √ ( 3−−1 ) + ( 5−2 ) =5
2 2 2 2
2 2
Equation: ( x −3 ) + ( y−5 ) =25
ACG Parnell College: AS Mathematics: Common Test 1: 2022
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ii) Find the equation of the tangent T to the circle C at point B. [4]
8−5 3
gradient of centre ¿ B : =
7−3 4
−4
gradient of tangent=
3
−4
equation of tangent : ( y−8 )= ( x−7 )
3
¿ 3 y +4 x =52
−4 52
¿ y= x+
3 3
ACG Parnell College: AS Mathematics: Common Test 1: 2022
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iii) Find the equation of the circle that is the reflection of C in the line T. [3]
new centre at ( 7+4 , 8+3 )= (11 ,11 )
2 2
Equation: ( x −11) + ( y−11 ) =25
ACG Parnell College: AS Mathematics: Common Test 1: 2022
