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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}.

Bab 17 Pilir Atur dan Gabungan

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Soalan 8:
Dalam sebuah kotak terdapat 10 biji gula-gula yang berlainan perisa.
Cari
(a)  Bilangan cara 3 biji gula-gula boleh dipilih dari kotak itu.
(b) Bilangan cara sekurang-kurangnya 8 biji gula-gula boleh dipilih dari kotak itu.

Penyelesaian:
(a)
Bilangan cara memilih 3 daripada 10 biji gula-gula
=   10 C 3 =120  

(b)
Bilangan cara memilih 8 biji gula-gula = 10 C 8  
Bilangan cara memilih 9 biji gula-gula = 10 C 9  
Bilangan cara memilih 10 biji gula-gula = 10 C 10

Oleh itu, bilangan cara sekurang-kurangnya 8 biji gula-gula boleh dipilih dari kotak
=   10 C 8 +  10 C 9 +   10 C 10 =56

 

Soalan 9 (4 markah):
Danya mempunyai sebuah kedai barangan perhiasan rumah. Pada suatu hari, Danya menerima 14 set cawan daripada seorang pembekal. Setiap set mengandungi 6 biji cawan yang berlainan warna.

(a) Danya memilih 3 set cawan secara rawak untuk diperiksa.
Cari bilangan cara yang berlainan yang digunakan oleh Danya untuk memilih set-set cawan itu.

(b) Danya mengambil satu set cawan untuk dipamerkan dengan menyusunnya secara sebaris.
Cari bilangan cara yang berlainan cawan-cawan itu boleh disusun dengan keadaan cawan berwarna biru tidak diletak bersebelahan cawan berwarna merah.

Penyelesaian:
(a)
Bilangan cara yang berlainan yang digunakan oleh Danya untuk memilih 3 set cawan secara rawak untuk diperiksa
= 14C3
 =364

(b)


Bilangan cara (Cawan berwarna biru diletak bersebelahan cawan berwarna merah)
= 5! × 2!
= 240

Bilangan cara yang berlainan cawan berwarna biru tidak diletak bersebelahan cawan berwarna merah
= 6! – 240
= 720 – 240
= 480



Soalan 10 (2 markah):
( a ) Diberi  C 6 n >1, senaraikan semua nilai-nilai yang mungkin bagi n. ( b ) Diberi  C y m = C y n , ungkapkan y  dalam sebutan m dan n.

Penyelesaian:
(a)
n = 1, 2, 3, 4, 5

(b)
y = m + n

7.5.3 Kebarangkalian, SPM Praktis (Kertas 1)

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Soalan 5 (3 markah):
sebiji dadu berbentuk kubus yang tidak adil dilambung. Kebarangkalian mendapat angka ‘4’ ialah  1 16  dan kebarangkalian mendapat selain daripada nombor ‘4’ adalah sama antara satu sama lain.
Jika dadu itu dilambung dua kali, cari kebarangkalian mendapat dua nombor yang berlainan.
Beri jawapan anda dalam bentuk pecahan termudah.

Penyelesaian:

P( 1 )+P( 2 )+P( 3 )+P( 4 )+P( 5 )+P( 6 )=1 Diberi kebarangkalian mendapat nombor lain adalah sama. x+x+x+ 1 16 +x+x=1 5x=1− 1 16 5x= 15 16 x= 3 16 P( Nombor sama ) =P( 1, 1 )+P( 2, 2 )+P( 3, 3 ) +P( 4, 4 )+P( 5, 5 )+P( 6, 6 ) =( x×x )+( x×x )+( x×x )+ ( 1 16 × 1 16 )+( x×x )+( x×x ) =5 x 2 + 1 256 =5 ( 3 16 ) 2 + 1 256 = 23 128 P( Dua nombor yang berlainan ) =1− 23 128 = 105 128

Bab 19 Taburan Kebarangkalian

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Soalan 7 (4 markah):
Sebuah badan sukarela menganjurkan kursus pertolongan cemas 4 kali sebulan, setiap Sabtu dari Mac hingga September.
[Andaikan setiap bulan mempunyai empat hari Sabtu]

Salmah berhasrat untuk menyertai kursus tersebut tetapi dia mungkin perlu meluangkan satu hari Sabtu setiap bulan untuk menemani ibunya ke hospital.
Kebarangkalian bahawa Salmah akan hadir ke kursus tersebut pada setiap Sabtu ialah 0.8. Salmah akan diberi sijil kehadiran bulanan jika dia boleh menghadiri kursus tersebut sekurang-kurangnya 3 kali sebulan.

(a) Cari kebarangkalian bahawa Salmah akan diberi sijil kehadiran bulanan.

(b) Salmah akan layak untuk menduduki ujian pertolongan cemas jika dia memperoleh lebih daripada 5 sijil kehadiran bulanan.
Cari kebarangkalian bahawa Salmah layak untuk menduduki ujian pertolongan cemas itu.

Penyelesaian:
(a)

P( X=r )= C n r p r q n−r p=0.8, q=0.2, n=4, r=3, 4 P( X≥3 ) =P( X=3 )+P( X=4 ) = C 4 3 ( 0.8 ) 3 ( 0.2 ) 1 + C 4 4 ( 0.8 ) 4 ( 0.2 ) 0 =0.4096+0.4096 =0.8192

(b)

P( X=r )= C n r p r q n−r p=0.8192, q=0.1808, n=7, r=6, 7 P( X>5 ) =P( X=6 )+P( X=7 ) = C 7 6 ( 0.8192 ) 6 ( 0.1808 ) 1 + C 7 7 ( 0.8192 ) 7 ( 0.1808 ) 0 =0.3825+0.2476 =0.6301



Soalan 8 (4 markah):
Rajah menunjukkan satu graf taburan normal piawai.

Rajah

Kebarangkalian yang diwakili oleh luas kawasan berlorek ialah 0.2881.
(a) Cari nilai h.
(b) X ialah pemboleh ubah rawak selanjar bertaburan secara normal dengan min, μ dan varians 16.
Cari nilai μ jika skor-z bagi X = 58.8 ialah h.

Penyelesaian:
(a)
P(X < h) = 0.5 – 0.2881
P(X < h) = 0.2119
P(X < –0.8) = 0.2119
h = –0.8

(b)

X=58.8 X−μ σ = 58.8−μ σ    Z= 58.8−μ 4    h= 58.8−μ 4 −0.8= 58.8−μ 4 −3.2=58.8−μ μ=58.8+3.2 μ=62

Bab 19 Taburan Kebarangkalian

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Soalan 5 (2 markah):
Rajah menunjukkan graf taburan kebarangkalian bagi suatu pemboleh ubah rawak X, X ~ N(μ, σ2).

Rajah

Diberi bahawa AB adalah paksi simetri bagi graf itu.
(a) Nyatakan nilai μ.
(b) Jika luas kawasan berlorek ialah 0.38, nyatakan nilai bagi P(5 ≤ X ≤ 15).

Penyelesaian:
(a)
μ = 0

(b)
P(10 ≤ X ≤ 15)
= 0.5 – 0.38
= 0.12

P(5 ≤ X ≤ 10)
= P(10 ≤ X ≤ 15)
= 0.12

Maka P(5 ≤ X ≤ 15)
= 0.12 + 0.12
= 0.24




Soalan 6 (3 markah):
Rajah menunjukkan graf bagi taburan binomial X ~ B(3, p).

Rajah

(a) Ungkapkan P(X = 0) + P(X > 2) dalam sebutan a dan b.
(b) Cari nilai p.

Penyelesaian:
(a)
P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 1
P(X = 0) + a + b + P(X = 3) = 1
P(X = 0) + P(X = 3) = 1 – a – b
P(X = 0) + P(X > 2) = 1 – a – b

(b)

 P( X=0 )= 27 343 3 C 0 ( p 0 ) ( 1−p ) 3 = 27 343 1×1× ( 1−p ) 3 = ( 3 7 ) 3 1−p= 3 7 p= 4 7