6.3 SPM Practice (Long Questions)

Posted on May 28, 2020 by Myhometuition

Question 7:

The diagram above shows the speed-time graph of a moving object for 15 seconds. Find

(a) the speed of the object at t = 9s.

(b) the distance travel by the object for the first 12 seconds.

Solution:
(a)

The speed of the object at t = 9s is 6 ms^-1

(b)

Distance travel by the object for the first 12 seconds

= Area under the speed-time graph

= Area of triangle

= ½ × 12 × 8

= 48 m

Question 8:

The diagram above shows the speed-time graph of a moving object for 15 seconds.

(a) Find the time interval when the object moves with constant speed.

(b) Find the acceleration from t = 6 s to t = 12 s.

(c) State the time when the object is stationary.

Solution:
(a)

Time interval when the object moves with constant speed

= 12 – 6

= 6 s

(b)

\begin{array}{l} Acceleration from t = 6 s to t = 12 s \\ = Gradient of speed-time graph \\ = \frac{6 - 6}{12 - 6} \\ = 0 {ms}^{- 2} \leftarrow \begin{array}{l} The moving object is moving \\ at a uniform speed \end{array} \end{array}

(c)

Time when the object is stationary is at 15 s.

6.2 Quantity Represented by the Area under a Graph (Part 2)

Posted on May 28, 2020 by Myhometuition

Combination of Graphs

Example 2:

The diagram above shows the speed-time graph of a moving object for 15 seconds.

(a) State the length of time, in s, that the particle moves with constant speed.

(b) Calculate the rate of change of speed, in ms^-2, in the first 3 seconds.

(c) Calculate the average speed of the object in 15 seconds.

Solution:

(a)

Length of time that the particle moves with constant speed

= 9 – 3 = 6 s

(b)

Rate of change of speed in the first 3 seconds

= acceleration = gradient

\begin{array}{l} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \\ = \frac{6 - 3}{3 - 0} \\ = 1 m s^{- 2} \end{array}

(c)

Total distance travelled of the object in 15 seconds

= Area under the graph in the 15 seconds

= Area P + Area Q + Area R

\begin{array}{l} = [\frac{1}{2} (3 + 6) \times 3] + [(9 - 3) \times 6] + [\frac{1}{2} (15 - 9) \times 6] \\ = 13.5 + 36 + 18 \\ = 67.5 m \end{array}

Average speed of the object in 15 seconds

\begin{array}{l} = \frac{Total distance travelled}{Total time taken} \\ = \frac{67.5}{15} = 4.5 m s^{- 1} \end{array}

5.1 Direct Variation (Part 3)

Posted on May 28, 2020 by Myhometuition

(D) Solving problems involving direct variations

1. If

y \propto x^{n}, where n = \frac{1}{2}, 2, 3,

then the equation is

y = k x^{n}

where k is a constant.

2. The graph of y against xⁿ is a straight line passing through the origin.

3. If

y \propto x^{n}

and sufficient information is given, the values of variable x or variable y can be determined.

Example

y varies directly to x³and if y = 54 when x = 3, find

(a) The value of x when y = 16

(b) The value of y when x = 4

Solution:

Given y α x³, y = kx³

When y = 54, x = 3,

54 = k(3)³

54 = 27k

k = 2

Therefore y = 2x³

(a) When y = 16

16 = 2x³

x³ = 8

x = 2

(b) When x = 4

y = 2(4)³ = 128

where k is a constant.

SPM Practice (Long Questions)

Posted on May 28, 2020 by Myhometuition

Question 8:
(a) Diagram 8.1 shows point A and point B marked on a Cartesian plane.

Diagram 8.1

Transformation R is a rotation of 90^o, clockwise about the centre B.

Transformation T is a translation

(\begin{array}{l} 5 \\ - 2 \end{array})

State the coordinates of the image of point A under each of the following transformations:
(i) RT,
(ii) R².

(b) Diagram 8.2 shows three trapeziums ABCD, PQRS and TUVS, drawn on a Cartesian plane.

Diagram 8.2

(i) Trapezium PQRS is the image of trapezium ABCD under the combined transformation MN.
Describe in full, the transformation:
(a) N,
(b) M.

(ii) It is given that trapezium ABCD represents a region of area 30 m².
Calculate the area, in m², of the shaded region.

Solution:
(a)

(i) A (–1, 6) → T → (4, 4 ) → R → (3, 1)
(ii) A (–1, 6) → R → (5, 6) → R → (5, 0)

(b)(i)(a)
N is a reflection in the line y = 4.

(b)(ii)(b)
M is enlargement of scale factor 2 with centre S (1, 5).

(b)(ii)
Area of PQRS = (scale factor)² x Area of object ABCD
= 2² x 30
= 120 m²Therefore,
Area of shaded region
= Area PQRS – area ABCD
= 120 – 30
= 90 m²

SPM Practice (Long Questions)

Posted on May 28, 2020 by Myhometuition

Question 7:
Diagram shows the point J(1, 2) and quadrilaterals ABCD and EFGH, drawn on a Cartesian plane.

(a) Transformation U is a rotation of 90^o, clockwise about the centre O.
Transformation T is a translation

(\begin{array}{l} 2 \\ 3 \end{array})

Transformation R is a reflection at the line x = 3.

State the coordinates of the image of point J under each of the following transformations:
(i) RU,
(ii) TR.

(b) EFGH is the image of ABCD under the combined transformation MN.
Describe in full, the transformation:
(i) N,
(ii) M.

(c) It is given that quadrilateral ABCD represents a region of area 18 m².
Calculate the area, in m², of the shaded region.

Solution:

(a)
(i) J (1, 2) → U → (2, –1 ) → R → (4, –1)
(ii) J (1, 2) → R → (5, 2) → T → (7, 5)

(b)(i)
N is a reflection in the line x = 6.

(b)(ii)
M is enlargement of scale factor 3 with centre (8, 7).

(c)
Area EFGH = (scale factor)² x Area of object ABCD
= 3² x 18
= 162 m²
Therefore,
Area of shaded region
= Area of EFGH – Area of ABCD
= 162 – 18
= 144 m²

SPM Practice (Long Questions)

Posted on May 28, 2020 by Myhometuition

Question 6:
(a) Diagram below shows point M marked on a Cartesian plane.

Transformation T is a translation

(\begin{array}{l} 2 \\ 3 \end{array})

and transformation R is an anticlockwise rotation of 90^o about the centre O.
State the coordinates of the image of point M under each of the following transformations:
(i) RT,
(ii) TR,

(b) Diagram below shows two hexagons, ABCDEF and JKLANO, drawn on square grids.

(i) JKLANO is the image of ABCDEF under the combined transformation VW.
Describe in full the transformation:
(a) W (b) V

(ii) It is given that quadrilateral ABCDEF represents a region of area 45 m².
Calculate the area, in m², of the region represented by the shaded region.

Solution:
(a)

(b)

(i)(a)
V: A reflection in the line EC.

(i)(b)

\begin{array}{l} Scale factor = \frac{K L}{B C} = \frac{6}{2} = 3 \\ W: An enlargement of scale factor 3 at centre J . \end{array}

(ii)
Area of JKLANO
= 3² x area of ABCDEF
= 9 x 45
= 405 m²Area of the coloured region
= area of JKLANO – area of ABCDEF
= 405 – 45
= 360 m²

SPM Practice (Long Questions)

Posted on May 28, 2020 by Myhometuition

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

Transformation T is a translation

(\begin{array}{l} 3 \\ - 1 \end{array})

and transformation R is an anticlockwise rotation of 90^o 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 m²Area of C = (Scale factor)² x Area of object
= 2² x area of B
= 2² x 12
= 48 m²

SPM Practice (Long Questions)

Posted on May 28, 2020 by Myhometuition

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

(a) Transformation R is a rotation of 90^o, clockwise about the centre O.
Transformation T is a translation

(\begin{array}{l} 2 \\ 3 \end{array})

.
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)

\begin{array}{l} Scale factor = \frac{U S}{R V} = \frac{6}{2} = 3 \\ V: An enlargement of scale factor 3 at centre (- 4, 2) \end{array}

(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)

Posted on May 28, 2020 by Myhometuition

Transformation III, Long Questions (Question 3)

Question 3:

(a) Transformation P is a reflection in the line x = m.

Transformation T is a translation

(\begin{array}{l} 4 \\ - 2 \end{array})

Transformation R is a clockwise rotation of 90^o 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)

\begin{array}{l} (6, 4) \to P \to (- 2, 4) \\ m = \frac{6 + (- 2)}{2} = 2 \end{array}

(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 90^oabout the centre A (3, 3).

(b)(i)(b)

Scale factor = \frac{H E}{C D} = \frac{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)

Posted on May 28, 2020 by Myhometuition

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.