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SPM Practice (Long Questions)

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Question 5:
(a) Diagram below shows two points, M and N, on a Cartesian plane.



Transformation T is a translation (  3 −1 ) and transformation R is an anticlockwise rotation of 90o about the centre (0, 2).
(i) State the coordinates of the image of point M under transformation R.
(ii) State the coordinates of the image of point N under the following transformations:
(a) T2,
(b) TR,

(b) Diagram below shows three pentagons, A, B and C, drawn on a Cartesian plane.


(i) C is the image of A under the combined transformation WV.
Describe in full the transformation:
(a) V    (b) W
(ii) Given A represents a region of area 12 m2, calculate the area, in m2, of the region represented by C.


Solution:
(a)



(b)


(b)(i)(a)
V: A reflection in the line x  = 8

(b)(i)(b)
W: An enlargement of scale factor 2 with centre (14, 0).

(b)(ii)
Area of B = area of A = 12 m2
Area of C = (Scale factor)2 x Area of object
 = 22 x area of B
  = 22 x 12
 = 48 m2

SPM Practice (Long Questions)

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Question 4:
Diagram below shows three triangles RPQ, UST and RVQ, drawn on a Cartesian plane.


(a) Transformation R is a rotation of 90o, clockwise about the centre O.
Transformation T is a translation ( 2 3 ) .
State the coordinates of the image of point B under each of the following transformations:
(i) Translation T2,
(ii) Combined transformation TR.

(b)
(i) Triangle UST is the image of triangle RPQ under the combined transformation VW.
Describe in full the transformation:
(a) W   (b) V

(ii) It is given that quadrilateral RPQ represents a region of area 15 m2.
Calculate the area, in m2, of the region represented by the shaded region.


Solution:


(a)
(i) (–5, 3) → T → (–3, 6) ) → T → (–1, 9)
(ii) (–5, 3) → R → (3, 5) → T → (5, 8)

(b)(i)(a)
W: A reflection in the line URQT.

(b)(i)(b)
Scale factor= US RV = 6 2 =3 V: An enlargement of scale factor 3 at centre ( −4,2 )

(b)(ii)
Area of UST = (Scale factor)2 x Area of RPQ
= 32 x area of RPQ
= 32 x 15
= 135 m2

Therefore,
Area of the shaded region
= Area of UST – area of RPQ
= 135 – 15
= 120 m2

SPM Practice (Long Questions)

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Transformation III, Long Questions (Question 3)

Question 3:
(a) Transformation P is a reflection in the line x = m.
Transformation T is a translation ( 4 − 2 ) .
Transformation R is a clockwise rotation of 90o about the centre (0, 4).

(i) The point (6, 4) is the image of the point ( –2, 4) under the transformation P.
State the value of m. 

(ii) Find the coordinates of the image of point (2, 8) under the following combined transformations:
(a) T2,
(b) TR.

(b) Diagram below shows trapezium CDFE and trapezium HEFG drawn on a Cartesian plane.



(i) HEFG is the image of CDEF under the combined transformation WU.
Describe in full the transformation:
(a) U (b) W

(ii) It is given that CDEF represents a region of area 60 m2.
Calculate the area, in m2, of the region represented by the shaded region.
 
Solution:
(a)(i)
( 6 , 4 ) → P → ( − 2 , 4 ) m = 6 + ( − 2 ) 2 = 2

(a)(ii)



(a) (2, 8) → T → (6, 6) → T → (10, 4)
(b) (2, 8) → R → (4, 2) → T → (8, 0)

(b)(i)(a)
U: An anticlockwise rotation of  90oabout the centre A (3, 3).

(b)(i)(b)
Scale factor = H E C D = 4 2 = 2
W: An enlargement of scale factor 2 with centre B (3, 5).

(b)(ii)
Area of HEFG= (Scale factor)2 × Area of object
= 22 × area of CDEF
= 4 × 60
= 240 m2
Therefore,
Area of the shaded region
= Area of HEFG– area of CDEF
= 240 – 60
= 180 m2

2.5 SPM Practice (Long Questions)

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Question 10:
(a) Complete the table in the answer space for the equation y = x3 – 13x + 18 by writing down the values of y when x = –2 and x = 3.

(b) For this part of the question, use graph paper. You may use a flexible curve rule.
By using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 5 units on the y-axis, draw the graph of y = x3 – 13x + 18 for –4 ≤ x ≤ 4 and 0 ≤ y ≤ 40.

(c) From your graph, find
(i) the value of y when x = –1.5,
(ii) the value of x when y = 25.

(d) Draw a suitable straight line on your graph to find the values of x which satisfy the equation x3 – 11x – 2 = 0 for –4 ≤ x ≤ 4 and 0 ≤ y ≤ 40.

Answer:


Solution:
(a)
y = x3 – 13x + 18   

when x = –2,
y = (–2)3 – 13(–2) + 18
   = –8 + 26 + 18
   = 36

when x = 3,
y = (3)3 – 13(3) + 18
   = 27 – 39 + 18
   = 6


(b)



(c)
(i) From the graph, when x = –1.5, y = 34.
(ii) From the graph, when y = 25, x = –3.25, –0.55 and 3.85.


(d)
y = x3 – 13x + 18 ----- (1)
0 = x3 – 11x – 2 ----- (2)
(1) – (2) : y = –2x + 20

The suitable straight line is y = –2x + 20.
Determine the x-coordinates of the two points of intersection of the curve y = x3 – 13x + 18 and the straight line y = –2x + 20.


From the graph, x = –3.2, –0.2 and 3.4.

2.5 SPM Practice (Long Questions)

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Question 9:
(a) Complete the table in the answer space for the equation y = 8 – 3x – 2x2 by writing down the values of y when x = –4 and x = 2.

(b) For this part of the question, use graph paper. You may use a flexible curve rule.
By using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 5 units on the y-axis, draw the graph of y = 5 – 8x – 2x2 for –5 ≤ x ≤ 3 and –27 ≤ y ≤ 9.

(c) From your graph, find
(i) the value of y when x = –2.5,
(ii) the positive value of x when y = –16.

(d) Draw a suitable straight line on your graph to find the values of x which satisfy the equation 8 – 3x – 2x2 = 0 for –5 ≤ x ≤ 3 and –27 ≤ y ≤ 9.

Answer:

x
–5
–4
–3.5
–2
–1
0
1
2
3
y
–27
r
–6
6
9
8
3
s
–19
Calculate the value of r and s.


Solution:
(a)
y = 8 – 3x – 2x2  
when x = –4,
r = 8 – 3(–4) – 2 (–4)2
= 8 + 12 – 32 = –12

when x = 2,
s = 8 – 3(2) – 2(2)2
= 8 – 6 – 8 = –6


(b)



(c)
(i) From the graph, when x = –2.5, y = 2.5
(ii) From the graph, when y = –16, positive value of x = 2.8


(d)
y = 8 – 3x – 2x2 ----- (1)
0 = 5 – 8x – 2x2  ----- (2)
(1) – (2) : y = 3 + 5x → y = 5x +3
The suitable straight line is y = 5x +3.

Determine the x-coordinates of the two points of intersection of the curve y = 8 – 3x – 2x2 and the straight line y = 5x +3.

x
–5
0
y = 5x + 3
22
3
From the graph, x = –4.5, 0.55.

2.5 SPM Practice (Long Questions)

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Question 8:
(a) Complete the table in the answer space for the equation y = x3 – 4x – 10 by writing down the values of y when x = –1 and x = 3.

(b) For this part of the question, use graph paper. You may use a flexible curve rule.
By using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 10 units on the y-axis, draw the graph of y = x3 – 4x – 10 for –3 ≤ x ≤ 4 and –25 ≤ y ≤ 38.

(c) From your graph, find
(i) the value of y when x = 2.2,
(ii) the value of x when y = 15.

(d) Draw a suitable straight line on your graph to find the values of x which satisfy the equation x3 – 12x – 5 = 0 for –3 ≤ x ≤ 4 and –25 ≤ y ≤ 38.

Answer:

x
–3
–2
–1
0
1
2
3
3.5
4
y
–25
–10
–10
–13
–10
18.9
38


Solution:
(a)
y = x3 – 4x – 10  
when x = –1,
y = (–1)3 – 4(–1) – 10
= –7

when x = 3,
y = (3)3 – 4(3) – 10
= 5

(b)




(c)
(i) From the graph, when x = 2.2, y = –8
(ii) From the graph, when y = 15, x = 3.4


(d)
y = x3 – 4x – 10 ----- (1)
0 = x3 – 12x – 5 ----- (2)
(1) – (2) : y = 8x – 5

The suitable straight line is y = 8x–5. Determine the x-coordinates of the two points of intersection of the curve y = x3 – 4x – 10 and the straight line y = 8x –5.

x
0
2
y = 8x – 5
–5
–11
From the graph, x = –0.45, 3.7.

2.5 SPM Practice (Long Questions)

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Question 7:
(a) The following table shows the corresponding values of x and y for the 
equation y= –x3 + 3x + 1. 
 
x
–3
–2
–1
0
1
2
3
3.5
4
y
19
3
r
1
3
–1
s
–31.4
–51
Calculate the value of r and s.
 
(b) For this part of the question, use graph paper. You may use a flexible curve rule.
By using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 5 units on the y-axis, draw the graph of  y = –x3 + 3x + 1 for –3 ≤ x ≤ 4 and –51 ≤ y ≤ 19.

(c) From your graph, find
(i) The value of y when x = –2.8,
(ii) The value of x when y = –30.

(d) Draw a suitable straight line on your graph to find the values of x which satisfy the equation –x3 + 13x – 9 = 0 for –3 ≤ x ≤ 4 and –51 ≤ y ≤ 19.
 
Solution:
(a)
y= –x3 + 3x + 1 
when x = –1,
r = – (–1)3 + 3(–1) + 1
  = 1 – 3 + 1 = –1
when x = 3,
s = – (3)3 + 3(3) + 1 = –17
 

(b)



(c) 
(i) From the graph, when x = –2.8, y = 15
(ii) From the graph, when y = –30, x = 3.5


(d)
y = –x3 + 3x+ 1 ----- (1)
–x3+ 13x – 9 = 0 ----- (2)
y = –x3 + 3x + 1 ----- (1)
0 = –x3 + 13x – 9 ------ (2) ← (Rearrange (2))
(1)  – (2) : y = –10x + 10
 
The suitable straight line is y = –10x + 10.
 
Determine the x-coordinates of the two points of intersection of the curve 
y = –x3 + 3x + 1 and the straight line y = –10x + 10.
 
x
0
4
y = –10x + 10
10
–30
From the graph, x= 0.7, 3.25.

2.5 SPM Practice (Long Questions)

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Question 3:
On the graph in the answer space, shade the region which satisfies all three inequalities y ≥ x – 2, y > –3x + 6 and y ≤ 4.

Answer:


Solution:



Question 4:
On the graph in the answer space, shade the region which satisfies all three inequalities y > x – 2, y ≤ –x + 6 and x ≥ 1.

Answer:


Solution:



Question 5 (3 marks):
On the graph in the answer space, shade the region which satisfies all three inequalities y ≥ –3x + 6, y > x + 1 and y ≤ 5.

Answer:


Solution:

2.5 SPM Practice (Long Questions)

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Question 1:
On the graph in the answer space, shade the region which satisfies all three inequalities y > –2x + 8, y ≥ x + 1 and y ≤ 7.

Answer:



Solution:





Question 2:
On the graph in the answer space, shade the region which satisfies all three inequalities 2x + y ≥ 2, x + y < 6 and y ≥ 1.

Answer:



Solution: