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Short Questions (Question 5 & 6)

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Question 5:
Diagram below shows a sector QOR of a circle with centre O.

It is given that PS = 8 cm and QP = PO= OS = SR = 5 cm.
Find
(a) the length, in cm, of the arc QR,
(b) the area, in cm2, of the shaded region.

Solution:
(a) Length of arc QR = r θ = 10 (1.75) = 17.5 cm

(b)
Area of the shaded region
= Area of sector QOR – Area of triangle POS
= ½ (10)2 (1.75) – ½ (5) (5) sin 1.75 (change calculator to Rad mode)
= 87.5 – 12.30
= 75.2 cm2



Question 6:
Diagram below shows a circle with centre O and radius 12 cm.

Given that A, B and C are points such that OA = AB and  ∠OAC = 90°, find
(a)   ∠BOC, in radians,
(b)  the area, in cm2, of the shaded region.   

Solution:
(a) For triangle OAC,
  cos  ∠AOC = 6/12 
  ÐAOC = 1.047 rad (change calculator to Rad mode)
  ÐBOC = 1.047 rad

(b) 
Area of the shaded region
= Area of BOC – Area of triangle AOC
= ½ (12)2 (1.047) – ½ (6) (12) sin 1.047 (change calculator to Rad mode)
= 75.38 – 31.17
= 44.21 cm2

Long Questions (Question 7 & 8)

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Question 7:
The diagram shows a straight line PQ which meets a straight line RS at the point Q. The point P lies on the y-axis.
(a) Write down the equation of RS in the intercept form.
(b) Given that 2RQ = QS, find the coordinates of Q.
(c) Given that PQ is perpendicular to RS, find the y-intercept of PQ.


Solution:
(a) 
Equation of RS
x 12 + y − 6 = 1 x 12 − y 6 = 1

(b)
Given  2 R Q = Q S R Q Q S = 1 2 Lets coordinates of  Q = ( x ,   y ) ( ( 0 ) ( 2 ) + ( 12 ) ( 1 ) 1 + 2 , ( − 6 ) ( 2 ) + ( 0 ) ( 1 ) 1 + 2 ) = ( x ,   y ) x = 12 3 = 4 y = − 12 3 = − 4 Q = ( 4 , − 4 )

(c) 
Gradient of  R S ,   m R S = − ( − 6 12 ) = 1 2 m P Q = − 1 m R S = − 1 1 2 = − 2
Point Q = (4, –4), m = –2
Using y = mx+ c
–4 = –2 (4) + c
c = 4
y–intercept of PQ = 4



Question 8:
In the diagram, the equation of FMG is y = – 4. A point P moves such that its distance from E is always half of the distance of E from the straight line FG. Find
(a) The equation of the locus of P,
(b) The x-coordinate of the point of intersection of the locus and the x-axis.


Solution:
(a) 
Gradient of the straight line FMG = 0
EM is perpendicular to FMG, so gradient of EM also = 0, equation of EM is x = 2
Thus, coordinates of point M = (2, –4).

Let coordinates of point P= (x, y).
Given PE = ½ EM
2PE = EM
2 [(x – 2)2+ (y – 4)2]½ = [(2 – 2)2 + (4 – (–4))2]½
4 (x2 – 4x + 4 + y2 – 8y +16) = (0 + 64) → (square for both sides)
4x2 – 16x + 16 + 4y2 – 32y + 64 = 64
4x2 + 4y2 – 16x – 32y + 16 = 0
x2 + y2 – 4x – 8y + 4 = 0

(b) 
x2 + y2 – 4x – 8y + 4 = 0
At x axis, y = 0.
x2 + 0 – 4x – 8(0) + 4 = 0
x2  – 4x+ 4 = 0
(x – 2) (x – 2) = 0
x = 2

The x-coordinate of the point of intersection of the locus and the x-axis is 2.

Long Questions (Question 6)

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Question 7:
Solutions by scale drawing will not be accepted.
Diagram below shows a triangle OPQ. Point S lies on the line PQ.

(a) A point Y moves such that its distance from point S is always 5 uints.
Find the equation of the locus of Y.  

(b) It is given that point P and point Q lie on the locus of Y    .
Calculate
(i) the value of k,
(ii) the coordinates of Q.

(c) Hence, find the area, in uint2, of triangle OPQ.



Solution:
(a)
The equation of the locus  Y   ( x , y )  is given by  Y S = 5  units ( x − 5 ) 2 + ( y − 3 ) 2 = 5 x 2 − 10 x + 25 + y 2 − 6 y + 9 = 25 x 2 + y 2 − 10 x − 6 y + 9 = 0

(b)(i)
Given P (2, k) lies on the locus of Y.
(2)2 + (k)2– 10(2) – 6(k) + 9 = 0  
4 + k2– 20 – 6k + 9 = 0
k2 – 6k – 7 = 0
(k – 7) (k + 1) = 0
k = 7   or   k = – 1
Based on the diagram, k = 7. 
 
(b)(ii) 
As P and Q lie on the locus of Y, S is the midpoint of PQ. P = (2, 7), S = (5, 3).
Let the coordinates of Q = (x, y),
( 2+x 2 , 7+y 2 )=( 5,3 ) 2+x 2 =5    and     7+y 2 =3 2+x=10  and    7+y=6 x=8 and    y=−1
Coordinates of point Q = (8, –1).

(c)
Area of △ OPQ = 1 2 | 0  8  2    0  −1  7   0 0 | = 1 2 |0+( 8 )( 7 )+0−0−( −1 )( 2 )−0| = 1 2 | 58| =29  units 2

Long Questions (Question 5)

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Question 5:
Diagram below shows a quadrilateral ABCD. Point C lies on the y-axis.

The equation of a straight line AD is 2y = 5x – 21
(a) Find
(i) the equation of the straight line AB,
(ii) the coordinates of A,
(b) A point P moves such that its distance from point D is always 5 units.
Find the equation of the locus of P.

Solution:
(a)(i)
2y=5x−21 y= 5 2 x− 21 2 m AD = 5 2 m AB × m AD =−1 m AB × 5 2 =−1 m AB =− 2 5 Equation of AB y− y 1 = m AB ( x− x 1 ) y+1=− 2 5 ( x+2 ) 5y+5=−2x−4 5y=−2x−9

(a)(ii)
2y=5x−21 .......... ( 1 ) 5y=−2x−9 .......... ( 2 ) ( 1 )×5:10y=25x−105 .......... ( 3 ) ( 2 )×2:10y=−4x−18 .......... ( 4 ) ( 2 )−( 4 ):0=29x−87 x=3 From ( 1 ), 2y=15−21 2y=−6 y=−3 A=( 3 , −3 )

(b)
y=2, 4=5x−21 5x=25 x=5 Point D=( 5, 2 ) PD=5 ( x−5 ) 2 + ( y−2 ) 2 =5 ( x−5 ) 2 + ( y−2 ) 2 =25 x 2 −10x+25+( y 2 −4y+4 )=25 x 2 + y 2 −10x−4y+4=0

Quadratic Functions, SPM Practice (Short Questions)

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Question 7:
The diagram below shows the graph of the quadratic function f(x) = (x + 3)2 + 2h – 6, where h is a constant.


(a) State the equation of the axis of symmetry of the curve.
(b) Given the minimum value of the function is 4, find the value of h.

Solution:
(a)
When x + 3 = 0
 x = –3
Therefore, equation of the axis of symmetry of the curve is x = –3.

(b)
When x + 3 = 0, f(x) = 2h – 6
Minimum value of f(x) is 2h – 6.
2h – 6 = 4
2h = 10
h = 5



Question 8 (4 marks):
The quadratic function f is defined by f(x) = x2 + 4x + h, where h is a constant.
(a) Express f(x) in the form (x + m)2 + n, where m and n are constants.

(b) Given the minimum value of f(x) is 8, find the value of h.

Solution:
(a)
f(x) = x2 + 4x + h
   = x2 + 4x + (2)2 – (2)2 + h
   = (x + 2)2 – 4 + h

(b)
Given the minimum value of f(x) = 8
– 4 + h = 8
h = 12



Question 9 (3 marks):
Find the range of values of x such that the quadratic function f(x) = 6 + 5x – x2 is negative.

Solution:
(a)
f(x) < 0
6 + 5x – x2 < 0
(6 – x)(x + 1) < 0
x < –1, x > 6



3.9.4 Quadratic Functions, SPM Practice (Long Question)

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Question 6:
Quadratic function f(x) = x2 – 4px + 5p2 + 1 has a minimum value of m2 + 2p, where m and p are constants.
(a) By using the method of completing the square, shows that m = p – 1.
(b) Hence, find the values of p and of m if the graph of the quadratic function is symmetry at x = m2 – 1.

Solution:
(a)
f( x )= x 2 −4px+5 p 2 +1 = x 2 −4px+ ( −4p 2 ) 2 − ( −4p 2 ) 2 +5 p 2 +1 = ( x−2p ) 2 + p 2 +1 Minimum value, m 2 +2p= p 2 +1 m 2 = p 2 −2p+1 m 2 = ( p−1 ) 2 m=p−1

(b)
x= m 2 −1 2p= m 2 −1 p= m 2 −1 2 Given m=p−1⇒p=m+1 m+1= m 2 −1 2 2m+2= m 2 −1 m 2 −2m−3=0 ( m−3 )( m+1 )=0 m=3 or −1 When m=3, p= 3 2 −1 2 =4 When m=−1, p= ( −1 ) 2 −1 2 =0

Quadratic Functions, SPM Practice (Long Questions)

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Question 4:
(a) Find the range of values of k if the equation x2 – kx + 3k – 5 = 0 does not have real roots.
(b) Show that the quadratic equation hx2 – (h + 3)x + 1 = 0 has real and distinc roots for all values of h.

Solution:
(a)
x 2 −kx+( 3k−5 )=0 If the above equation has no real root, ∴  b 2 −4ac<0. k 2 −4( 3k−5 )<0 k 2 −12k+20<0 ( k−2 )( k−10 )<0

Graph function y = (k – 2)(k – 10) cuts the horizontal line at k = 2 and k = 10 when b2 – 4ac < 0.



The range of values of k that satisfy the inequality above is 2 < k < 10.

(b)
h x 2 −( h+3 )x+1=0 b 2 −4ac= ( h+3 ) 2 −4( h )( 1 ) = h 2 +6h+9−4h = h 2 +2h+9 = ( h+ 2 2 ) 2 − ( 2 2 ) 2 +9 = ( h+1 ) 2 −1+9 = ( h+1 ) 2 +8

The minimum value of (h + 1) + 8 is 8, a positive value. Therefore, b2 – 4ac > 0 for all values of h.
Hence, quadratic equation hx2 – (h + 3)x + 1 = 0 has real and distinc roots for all values of h.

Quadratic Equations, SPM Practice (Paper 2)

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2.10.3 Quadratic Equations, SPM Practice (Paper 2)

Question 5:
Given 3t and (t – 7) are the roots of the quadratic equation 4x2 – 4x + m = 0 where m is a constant.
(a)  Find the values of t and m.
(b)  Hence, form the quadratic equation with roots 4t and 2t + 6.

Solutions:
(a)
Given 3t and (t – 7) are the roots of the quadratic equation 4x2 – 4x + m = 0
a = 4, b = – 4, c = m
Sum of roots = − b a  
3t + (t– 7) = − − 4 4  
3t + t– 7 = 1
4t = 8
t = 2

Product of roots = c a  
3t (t– 7) = m 4  
4 [3(2) (2 – 7)] = m ← (substitute t = 2)
4 [3(2) (2 – 7)] = m
4 (–30) = m
m = –120

(b)
t = 2
4t = 4(2) = 8
2t + 6 = 2(2) + 6 = 10

Sum of roots = 8 + 10 = 18
Product of roots = 8(10) = 80

Using the formula, x2– (sum of roots)x + product of roots = 0
Thus, the quadratic equation is,
x2 – 18x + 80 = 0

1.6.3 Function, SPM Practice (Short Question)

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Question 7:
Diagram below shows the function g : x → x – 2k, where k is a constant.

Find the value of k.

Solution:
Given g( 6 )=12   g( 6 )=6−2k  12=6−2k  2k=6−12    k=−3



Question 8:
Diagram below shows the relation between set M and set N in the arrow diagram.

(a) Represent the relation in the form of ordered pairs.
(b) State the domain of the relation.

Solution: 
(a) Relation in the form of ordered pairs = {(–4, 8), (3, 3), (4, 8)}.
(b) Domain of the relation = {–4, 3, 4}.



Question 9 (3 marks):
Diagram shows the graph of the function f : x → |1 – 2x| for the domain –2 ≤ x ≤ 4.

Diagram

State
(a) the object of 7,
(b) the image of 3,
(c) the domain of 0 ≤ f(x) ≤ 5.


Solution:
(a)
The object of 7 is 4.

(b)
f (x) = |1 – 2x|
f (3) = |1 – 2(3)|
= |1 – 6|
= |–5|
= 5

The image of 3 is 5.

(c)
|1 – 2x| = 5
1 – 2x = ±5
Given when f(x) = 5, x = –2.

When f(x) = –5
1 – 2x = –5
2x = 6
x = 3

Domain: –2 ≤ x ≤ 3.


1.6.2 Function, SPM Practice (Short Question)

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Question 4:
It is given the functions g(x) = 3x and h(x) = m – nx, where m and n are constants.
Express m in terms of n such that hg(1) = 4.

Solution:
hg( x )=h( 3x )  =m−n( 3x )  =m−3nx hg( 1 )=4 m−3n( 1 )=4 m−3n=4 m=4+3n



Question 5:
Diagram below shows the relation between set M and set N in the graph form.
State
(a) the range of the relation,
(b) the type of the relation between set M and set N.

Solution: 
(a) Range of the relation = {p, r, s}.
(b) Type of the relation between set M and set N is many to one relation.



Question 6:
Diagram below shows the relation between set P and set Q.


State
(a) the object of 3,
(b) the range of the relation.

Solution: 
(a) The object of 3 is 7.
(b) The range of the relation is {–3, –1, 1, 3}.