Chapter 12
Exercise 12.1
1. (i) 3x ! 5 " x # 3
⇒ 2x " 8
⇒ x"4
(ii) 6x ! 5 $ 2x ! 1
⇒ 4x $ 4
⇒ x$1
(iii) 1 ! 3x " 10
⇒ !3x " 9
⇒
3x % !9
⇒
x % !3
2. (i) _x_ # 2 % 7, x ∈ N
2
⇒ _x_ % 5
2
⇒ x % 10
_1_
(ii)
(x ! 1) & _1_(x ! 4), x ∈ Z
6
3
⇒ 1(x ! 1) & 2(x ! 4)
⇒
x ! 1 ' 2x ! 8
⇒
!x & !7
⇒
x$7
4 ! x " _____
2 ! x, x ∈ R
_____
(iii)
2
3
⇒ 3(4 ! x) " 2(2 ! x)
⇒
⇒
⇒
3.
0
1
2
3
4
5
6
7
8
9 10
!1 0
1
2
3
4
5
6
7
8
12 ! 3x " 4 ! 2x
!x " !8
x% 8
12x ! 3(x ! 3) % 45, x ∈ R
12x ! 3x # 9 % 45
9x % 36
x% 4
(ii)
x(x ! 4) & x2 # 2, x ∈ R
⇒x2 ! 4x & x2 # 2
⇒!4x & 2
⇒ 4x $ !2
⇒ x $ !_1_
2
(iii)
x ! 2(5 # 2x) % 11
⇒ x ! 10 ! 4x % 11
⇒
!3x % 21
⇒
3x " !21
⇒
x " !7
0
1
2
3
4
5
6
7
8
9
(i)
⇒
⇒
⇒
!2 !1 0
1
2
3
4
!5 !4 !3 !2 !1 !1
— 0
2
5
1
2
!8 !7 !6 !5 !4 !3 !2 !1 0
387
Text & Tests 4 Solution
4. (i)
⇒
(ii)
⇒
⇒
⇒
(iii)
⇒
⇒
⇒
!2 $ x # 1 $ 3, x ∈ R
!3 $ x $ 2
13 " 1 ! 3x & 7, x ∈ R
12 " !3x & 6
!12 % 3x $ 6
!4 % x $ 2
3 ' 4x # 1 " !1
2 ' 4x " !2
1 ' x " !__
1
__
2
2
1 % x $ __
1
!__
2
2
!4 !3 !2 !1 0
1
2
!4 !3 !2 !1 0
1
2
−1 0
!3 !2 !1—
3 (x ! 2) " 0
3 " __
5
⇒ 15 " 3x ! 6 " 0
⇒ 21 " 3x " 6
⇒ 7"x"2
⇒ 2%x% 7
2 (1 ! 3x) $ 1
(ii)
!4 $ __
5
⇒!20 $ 2 ! 6x $ 5
⇒!22 $ !6x $ 3
⇒
22 & 6x & !3
2 & x & !__
1
⇒
3__
3
2
1
2
⇒!__ $ x $ 3__
2
3
x %4
(iii)
3 $ 2 ! __
7
⇒21 $ 14 !x $ 28
⇒7 $ !x $ 14
⇒!7 & x & !14
⇒!14 $ x $ !7
5. (i)
6.
3(x ! 2) " x ! 4
⇒ 3x ! 6 " x ! 4
⇒
2x " 2
⇒
x"1
and
4x # 12 " 2x # 17, x ∈ R
⇒ 2x " 5
1
⇒
x " 2__
2
1
ANS : x " 2__
2
7. (i) 2x ! 5 % x ! 1, x ∈ R
⇒ x% 4
(ii)
7(x # 1) " 23 ! x, x ∈ R
⇒ 7x # 7 " 23 ! x
⇒
8x " 16
⇒
x"2
0 1 2 3 4 5
(iii) 2 % x % 4, x ∈ R
⇒
8. (i) 2x ! 3 " 2, x ∈ R
⇒ 2x " 5
1
⇒ x " 2__
2
388
2
1
2
–
1
2
3
3
Chapter 12
3(x # 2) % 12 # x, x ∈ R
⇒ 3x # 6 % 12 # x
⇒
2x % 6
⇒
x% 3
1 % x % 3, x ∈ R
(iii) 2__
2
(ii)
1 1 –1 2 2 –1 3 3 –1 4
2
2
2
15!x % 2(11 ! x), x ∈ Z
15 ! x % 22 ! 2x
x% 7
(ii)
5(3x ! 1) " 12x # 19, x ∈ Z
⇒ 15x ! 5 " 12x # 19
⇒
3x " 24
⇒
x" 8
(iii) ∴ Null set
9. (i)
⇒
⇒
10. (i) 3x # 8 $ 20, x ∈ N
⇒3x $ 12
⇒x $ 4
(ii)
2(3x ! 7) ' x # 6, x ∈ N
⇒ 6x ! 14 & x # 6
⇒
5x & 20
⇒
x& 4
(iii) x ( 4
Width ( x, Length ( x # 1
Perimeter ( 2(x) # 2(x # 1) $ 38
⇒2x # 2x # 2 $ 38
⇒
4x $ 36
⇒
x$9
Width ( 9 m, length ( 10 m
11.
12.
100 % 2n % 200
⇒ 26.645 % 2n % 27.645
⇒ 6.645 % n % 7.645
Hence a ( 6.645, b ( 7.645, n ( 7
13. Example 1:
Example 2:
(i)
(ii)
(iii)
(iv)
14.
4 " 3 " 0 Thus 42 " 32
1 % __
1.
4 " 3 " 0 Thus 4!2 % 3!2, as ___
16 9
if a % b % 0 and n(odd) " 0, then an % bn.
if a % b % 0 and n(even) " 0, then an " bn.
if a % b % 0 and n(odd) % 0, then an " bn.
if a % b % 0 and n(even) % 0, then an % bn.
5 ! 3x % !10 ∩ 4x # 6 % 32, x ∈ Z
⇒ !3x % !15 ∩
4x % 26
1
⇒
3x " 15
∩
x % 6 __
2
1, x∈Z
⇒
x"5
∩
x % 6 __
2
⇒
x(6
389
Text & Tests 4 Solution
Exercise 12.2
1. (i) Let x2 ! x ! 6 ( 0
⇒ (x # 2)(x ! 3) ( 0
⇒
Roots: x ( !2, 3
Hence x2 ! x ! 6 & 0 ⇒
y
x $ !2
!2 O
x& 3
or
(ii) Let x2 # 3x ! 10 ( 0
⇒ (x # 5)(x ! 2) ( 0
⇒
Roots: x ( !5, 2
2
Hence x # 3x ! 10 $ 0 ⇒ !5 $ x $ 2
O
!5
1
–
2
O
2x2 ! 5x # 2 % 0 ⇒
1
___
2
x
2
x
y
!3 $ x $ 2
!3
O
y
1
12–
!4
(iii) Let !2x2 ! 7x ( 0
⇒ 2x2 # 7x ( 0
⇒ x(2x # 7) ( 0
O
x
y
1
!3 2–
O
x
1
__
2
2
Hence !2x ! 7x & 0 ⇒
3. (i) Let 6x2 ! x ! 15 ( 0
⇒ (2x # 3)(3x ! 5) ( 0
1 , 1__
2
⇒ Roots: x ( !1__
2 3
Hence 6x2 ! x ! 15 " 0
390
2
%x%2
(ii) Let 12 ! 5x ! 2x2 ( 0
⇒
(4 # x)(3 ! 2x) ( 0
1
⇒
Roots: x ( !4, 1__
2
1
Hence 12 ! 5x ! 2x2 " 0 ⇒ !4 % x % 1__
2
Roots: x ( 0, !3
x
2
y
2. (i) Let 6 ! x ! x2 ( 0
⇒ (3 # x)(2 ! x) ( 0
⇒
Roots: x ( !3, 2
Hence 6 ! x ! x2 ' 0 ⇒
⇒
x
y
(iii) Let 2x2 ! 5x # 2 ( 0
⇒ (2x ! 1)(x ! 2)(0
1, 2
⇒
Roots: x ( __
2
Hence
3
1$x$0
!3___
2
y
⇒
x % !1
1
___
2
or x " 1
2
__
3
1
!12–
O
2
13–
x
Chapter 12
(ii) Let 16 ! x2 ( 0
⇒ (4 # x)(4 ! x) ( 0
⇒ Roots: x ( !4, 4
Hence 16 ! x2 $ 0 ⇒
y
x $ !4
(iii) Let 2(x2 ! 6) ( 5x
⇒ 2x2 ! 12 ( 5x
⇒ 2x2 ! 5x ! 12 ( 0
⇒ (2x # 3)(x ! 4) ( 0
1, 4
⇒ Roots: x ( !1__
2
2
Hence 2(x ! 6) ' 5x ⇒
or
O
!4
x'4
x
4
y
1
1
x $ !1___
2
4. Let (4 ! x)(1 ! x) ( x # 11
⇒ 4 ! 5x # x2 ! x ! 11 ( 0
⇒ x2 ! 6x ! 7 ( 0
⇒ (x # 1)(x ! 7) ( 0
⇒ Roots: x ( !1, 7
Hence (4 ! x)(1 ! x) % x # 11 ⇒
!1 2–
or
O
4
x
x'4
y
!1 O
7 x
!1 % x % 7
5. Let x2 ! 6x # 2 ( 0
_____________
(!6)2 ! 4(1)(2)
6
±
√
⇒ Roots: x ( _________________
2(1)
______
6 ± √ 36 ! 8
__________
(
2
___
6 ± √ 28
( _______
2 __
6
±
2√ 7
( _______
2 __
__
( 3 ! √ 7, 3 # √ 7
__
__
Hence x2 ! 6x # 2 $ 0 ⇒ 3 ! √ 7 $ x $ 3 # √ 7
6. x2 # (k # 1)x # 1 ( 0
Real Roots ⇒ b2 ! 4ac ' 0
⇒ (k # 1)2 ! 4(1) (1) & 0
⇒ k2 # 2k # 1 ! 4 & 0
⇒ k2 # 2k # 1 ! 3 & 0
Solve k2 # 2k ! 3 ( 0
⇒ (k # 3)(k ! 1) ( 0
⇒ Roots ( !3, 1
Hence k2 # 2k ! 3 & 0 ⇒ k $ !3 or k & 1
7. kx2 # 4x # 3 # k ( 0
Re al Roots ⇒ b2 ! 4ac & 0
⇒ (4)2 ! 4(k)(3 # k) & 0
⇒ 16 ! 12k ! 4 k2 & 0
⇒ k2 # 3k ! 4 $ 0
solve k2 # 3k ! 4 ( 0
⇒ (k # 4)(k ! 1) ( 0
⇒ Roots: k ( !4, 1
Hence k2 # 3k ! 4 $ 0 ⇒ !4 $ k $ 1
y
3! 7
O
3" 7
x
y
!3
O
x
1
y
!4
O
1
x
391
Text & Tests 4 Solution
y
8. px2 # (p # 3)x # p ( 0
Re al Roots ⇒ b2 ! 4ac & 0
⇒ (p # 3)2 ! 4(p)(p) & 0
⇒ p2 # 6p # 9 ! 4p2 & 0
⇒ !3p2 # 6p # 9 & 0
⇒ p2 ! 2p ! 3 $ 0
Solve p2 ! 2p ! 3 ( 0
⇒ (p # 1)(p ! 3) ( 0
⇒ Roots: p ( !1, 3
Hence p2 ! 2p ! 3 $ 0 ⇒ !1 $ p $ 3
x ( !2 ⇒ p (!2)2 # (p # 3) (!2) # p ( 0
⇒ 4p ! 2p ! 6 # p ( 0
⇒ 3p ( 6
⇒ p(2
O
x#3
________
9. (i)
⇒
% 2, x ≠ !2
x#2
x # 3 (x # 2)2 % 2(x # 2)2
________
x#2
⇒ (x # 3)(x # 2) % 2 (x2 # 4x # 4)
⇒ x2 # 5x # 6 % 2x2 # 8x # 8
⇒ x2 # 3x # 2 " 0
Solve x2 # 3x # 2 ( 0
⇒ (x # 2)(x # 1) ( 0
⇒ Roots x ( !2, !1
Hence x2 # 3x # 2 " 0 ⇒ x % !2
y
O
!2
or
x
!1
x " !1
x#5
________
" 1, x ≠ 3
x!3
x # 5 (x ! 3)2 " 1(x ! 3)2
⇒ ________
x!3
⇒ (x # 5)(x ! 3) " 1(x2 ! 6x # 9)
⇒ x2 # 2x ! 15 " x2 ! 6x # 9
⇒
8x " 24
⇒
x"3
2x ! 1 " 3, x ) 3
______
(iii)
x#3
2x
! 1 (x # 3)2 " 3(x # 3)2
______
⇒
x#3
⇒ (2x ! 1)(x # 3) " 3(x2 # 6x # 9)
⇒ 2x2 # 5x ! 3 " 3x2 # 18x # 27
⇒ x2 # 13x # 30 % 0
Solve x2 # 13x # 30 ( 0
⇒ (x # 10)(x # 3) ( 0
⇒ x ( !10, !3
Hence x2 # 13x # 30 % 0 ⇒ !10 % x % !3
(ii)
y
!10
3x # 4
______
10. (i)
⇒
⇒
⇒
⇒
392
3 x
!1
" 2, x ) 5
x!5
3x # 4 (x ! 5)2 " 2(x ! 5)2
______
x!5
(3x # 4)(x ! 5) " 2(x2 ! 10x # 25)
3x2 ! 11x ! 20 " 2x2 ! 20x # 50
x2 # 9x ! 70 " 0
!3 O
x
y
!14
O
5
x
Chapter 12
(ii)
(iii)
11. (i)
(ii)
Solve x2 # 9x ! 70 ( 0
⇒ (x # 14)(x ! 5) ( 0
⇒ x ( !14, 5
Hence x2 # 9x ! 70 " 0 ⇒ x % !14 or x " 5
1 ! 2x " 2, x ) ! __
1
______
2
4x # 2
1!
2x
______
⇒
(4x # 2)2 " 2(4x # 2)2
4x # 2
⇒ (1 ! 2x)(4x # 2) " 2(16x2 # 16x # 4)
⇒ 4x # 2 ! 8x2 ! 4x " 32x2 # 32x # 8
⇒ 40x2# 32x # 6 % 0
⇒ 20x2 # 16x # 3 % 0
Solve 20x2 # 16x # 3 ( 0
⇒ (2x # 1)(10x # 3) ( 0
3
1 , !___
⇒ Roots: x ( !__
2 10
3
1 % x % ! ___
Hence 20x2 # 16x # 3 % 0 ⇒ ! __
2
10
3 # 4x " 3, x ) __
1
______
5
5x ! 1
3
#
4x
______
⇒
(5x ! 1)2 " 3(5x ! 1)2
5x ! 1
⇒ (3 # 4x)(5x ! 1) " 3(25x2 ! 10x # 1)
⇒ 15x ! 3 # 20x2 ! 4x " 75x2 ! 30x # 3
⇒ 55x2 ! 41x # 6 " 0
Solve 55x2 ! 41x # 6 ( 0
⇒ (5x ! 1)(11x ! 6) ( 0
6
1 , ___
⇒ Roots: x ( __
5 11
6
1 % x % ___
Hence 55x2 ! 41x # 6 % 0 ⇒ __
5
11
3
x $ 1, x ) __
______
2
2x ! 3
x
______
⇒
(2x ! 3)2 $ 1(2x ! 3)2
2x ! 3
⇒ x(2x ! 3) $ 1(4x2 ! 12x # 9)
⇒ 2x2 ! 3x $ 4x2 ! 12x # 9
⇒ 2x2 ! 9x # 9 & 0
Solve 2x2 ! 9x # 9 ( 0
⇒ (2x ! 3)(x ! 3) ( 0
1, 3
⇒ Roots: x ( 1__
2
1 or x & 3
Hence 2x2 ! 9x # 9 & 0 ⇒ x $ 1__
2
2x ! 4 % 1, x ) 1
______
x!1
2x ! 4 (x ! 1)2 % 1(x ! 1)2
⇒ _____
x!1
⇒ (2x ! 4)(x ! 1) % 1(x2 ! 2x # 1)
⇒ 2x2 ! 6x # 4 % x2 ! 2x # 1
⇒ x2 ! 4x # 3 % 0
Solve x2 ! 4x # 3 ( 0
⇒ (x ! 1)(x ! 3) ( 0
⇒ x ( 1, 3
Hence x2 ! 4x # 3 % 0 ⇒ 1 % x % 3
y
1
3
!2–
O
––
!10
x
y
O
1
–
5
6
––
11
x
y
O 11–
2
3
x
y
O
1
3
x
393
Text & Tests 4 Solution
(iii)
12. (i)
x!5
_____
$ 3, x ) 1
x!1
x ! 5 (x ! 1)2 $ 3(x ! 1)2
_____
⇒
x!1
⇒
(x ! 5)(x ! 1) $ 3(x2 ! 2x # 1)
⇒
x2 ! 6x # 5 $ 3x2 ! 6x # 3
⇒
2x2 ! 2 ' 0
⇒
x2 ! 1 ' 0
Solve x2 ! 1 ( 0
⇒
(x # 1)(x ! 1) ( 0
⇒
Roots: x ( !1, 1
Hence x2 ! 1 & 0 ⇒ x $ !1 or x & 1
2x ! 7
______
% 1, x ) !3
x#3
2x ! 7 (x # 3)2 % 1(x # 3)2
⇒ _____
x#3
⇒ (2x ! 7)(x # 3) % 1(x2 # 6x # 9)
⇒ 2x2 ! x ! 21 % x2 # 6x # 9
⇒ x2 ! 7x ! 30 % 0
Solve x2 ! 7x ! 30 ( 0
⇒ (x # 3)(x ! 10) ( 0
⇒ Roots: x ( !3, 10
Hence x2 ! 7x ! 30 % 0 ⇒ !3 % x % 10
394
!1 O
x
1
y
!3 O
x
10
2x ! 3
______
3, x ) 5
% __
2
x!5
2x ! 3 (x ! 5)2 % __
3 (x ! 5)2
_____
⇒
2
x!5
3
__
⇒
(2x ! 3)(x ! 5) % (x2 ! 10x # 25)
2
3x2 ! 30x # 75
⇒
2x2 ! 13x # 15 % ______________
2
⇒
4x2 ! 26x # 30 % 3x2 ! 30x # 75
⇒
x2 # 4x ! 45 % 0
Solve x2 # 4x ! 45 ( 0
⇒
(x # 9)(x ! 5) ( 0
⇒
x ( !9, 5
Hence x2 # 4x ! 45 % 0 ⇒ !9 % x % 5
x # 2 $ 3, x ) 1
_____
(iii)
x!1
x # 2 (x ! 1)2 $ 3(x ! 1)2
_____
⇒
x!1
⇒
(x # 2)(x ! 1) $ 3(x2 ! 2x # 1)
⇒
x2 # x ! 2 $ 3x2 ! 6x # 3
⇒
2x2 ! 7x # 5 & 0
Solve 2x2 ! 7x # 5 ( 0
⇒ (x ! 1) (2x ! 5) ( 0
1
⇒
Roots: x ( 1, 2__
2
1
Hence 2x2 ! 7x # 5 & 0 ⇒ x $ 1 or x & 2__
2
(ii)
y
y
O
!9
x
5
y
O
1
1
2 2–
x
Chapter 12
y
13. From graph:
!3 " x " !2
Using inequality ⇒ 2x2 # 4x " x2 ! x ! 6, x ∈ R
⇒ x2 # 5x # 6 " 0
2
Solve x # 5x # 6 ( 0
⇒ (x # 3)(x # 2) ( 0
⇒ Roots: x ( !3, !2
Hence x2 # 5x # 6 " 0 ⇒ x % !3 or x " !2
!2 O
!3
x
14. No real roots if b2 ! 4ac % 0
x2 # x # 1 ( 0 ⇒ b2 ! 4ac ( (1)2 ! 4(1)(1) ( !3 % 0
Hence x2 # x # 1 " 0
3"0
1 # __
⇒ x2 # x # __
4 __4
2
√3
1 2 # ___
⇒ x # __
"0
True
2
2
(
) ( )
15. (i) f(t) $ 4 ⇒ !11 # 13t ! 2t2 $ 4
⇒ 2t2 ! 13t # 15 ' 0
Solve 2t2 ! 13t # 15 ( 0
⇒ (2t ! 3)(t ! 5) ( 0
1, 5
⇒ t ( 1__
2
1 or
Hence 2t2 ! 13t # 15 ' 0 ⇒ t $ 1__
2
(ii) f(t) ' 7 ⇒ !11 # 13t ! 2t2 & 7
⇒ 2t2 ! 13t # 18 $ 0
⇒ (t ! 2)(2t ! 9) ( 0
1
⇒ Roots: t ( 2, 4__
2
1
Hence 2t2 ! 13t # 18 $ 0 ⇒ 2 $ t $ 4__
2
(iii) 4 % f(t) % 7 ⇒
1 % t % 2 and 4 __
1 %t%5
1__
2
2
y
O
x
5
1
12–
t&5
y
O
(i)
0
(iii)
2
(ii)
1 11– 2
2
1
4 2–
x
(iii)
3
4 4 1– 5
2
(i)
6
1
x " !__
2
(ii) x $ 1 or x ' 3
1 $ x $ __
1
(iii) !1__
2
2
(iv) !1 % x % 5
16. (i) x % !3
or
17. Length ( x and Width ( x ! 3
x %5
Ratio % 5 ⇒ _____
x!3
x (x ! 3)2 % 5(x ! 3)2
⇒ _____
x!3
⇒ x(x ! 3) % 5(x2 ! 6x # 9)
⇒ x2 ! 3x % 5x2 ! 30x # 45
⇒ 4x2 ! 27x # 45 " 0
Solve 4x2 ! 27x # 45 ( 0
⇒ (x ! 3)(4x ! 15) ( 0
⇒ Roots: x ( 3, 3.375
Hence 4x2 ! 27x # 45 " 0 ⇒ x % 3 or x " 3.75
Since x % 3 is not valid, hence (i) Length " 3.75
(ii) Width " 0.75
y
O
3
3.75 x
395
Text & Tests 4 Solution
18.
Positive graphs
⇒ No real roots
⇒ b2 ! 4ac % 0
x2 ! 2px # p # 6 ( 0 ⇒ (!2p)2 ! 4(1)(p # 6) % 0
⇒ 4p2 ! 4p ! 24 % 0
⇒ p2 ! p ! 6 % 0
2
Solve
p !p!6(0
⇒ (p # 2)(p ! 3) ( 0
⇒ Roots:
p ( !2, 3
Hence p2 ! p ! 6 % 0 ⇒ !2 % p % 3
19. (i) Perimeter % 50 ⇒ 2(x # 3) # 2(x # 2) % 50
⇒ 2x # 6 # 2x # 4 % 50
⇒ 4x % 50 ! 10
⇒ 4x % 40
⇒
x % 10
(ii) Area " 12 ⇒ (x # 3)(x # 2) " 12
⇒
x2 # 5x # 6 " 12
⇒
x2 # 5x ! 6 " 0
2
Solve x # 5x ! 6 ( 0
⇒ (x # 6)(x ! 1) ( 0
⇒ x ( !6 (Not valid), x ( 1 (valid)
Hence x2 # 5x ! 6 " 0 ⇒ x " 1
(iii) 1 % x % 10, by combining (i) and (ii) above
20. Use Pythagoras
Perimeter " 8
⇒
h2 ( x2 # 32 ( x2 # 9
⇒
h ( √ x2 # 9
______
______
⇒ √ x2 # 9 # x # 3 " 8
______
⇒ √ x2 # 9 " !x # 5
⇒ x2 # 9 " (!x # 5)2
⇒ x2 # 9 " x2 ! 10x # 25
⇒ 10x " 16
⇒ x______
" 1.6
x2 # 9 # x # 3 % 12
Perimeter % 12 ⇒ √______
⇒ √ x2 # 9 % !x # 9
⇒ x2 # 9 % (!x # 9)2
⇒ x2 # 9 % x2 ! 18x # 81
⇒ 18x % 72
⇒ x%4
Hence 1.6 % x % 4
Since x ∈ Z ⇒ x ( 2 m or 3 m
Exercise 12.3
1. (i)
x # 3 ( !1
⇒
x ( !4
(ii)
x ! 2 ( !4
⇒
x ( !2
(iii)
2x ! 1 ( !5
⇒
2x ( !4
⇒
x ( !2
396
OR
OR
OR
OR
OR
OR
OR
x#3(1
x ( !2
x!2(4
x(6
2x ! 1 ( 5
2x ( 6
x(3
y
!2
O
3
x
y
!6
O
1
x
Chapter 12
3x ! 2 ( !x OR
⇒
4x ( 2
OR
1
__
⇒
x(
OR
2
(v) 2(x ! 3) ( !2 OR
⇒ x ! 3 ( !1 OR
⇒
x(2
OR
(vi)
x ! 5 ( !(x # 1)
⇒ x ! 5 ( !x ! 1
⇒
2x ( 4
⇒
x(2
3x ! 2 ( x
2x ( 2
(iv)
x(1
2(x ! 3) ( 2
x!3(1
x(4
OR x ! 5 ( #(x # 1)
OR x ! 5 ( x # 1
OR
!5 ( #1 (Not valid)
2. Copy and complete the following table and hence sketch a graph of f (x) ( |3x ! 2|
x
f (x) ( |3x ! 2|
!3
11
!2
8
!1
5
0
1
2
3
2
1
4
7
Solve |3x ! 2| ( 5.
1
⇒ x ( !1, 2 __
3
y
10
f(x) # |3x ! 2|
f(x) # 5
!3 !2 !1
5
1
2
3
x
3. f (x) ( |x|, g (x) ( |x ! 4|, h(x) ( |x # 3|
f (!2) ( |!2| ( 2 ⇒ point (!2, 2) ∈ f (x)
g (2) ( |2 ! 4| ( |!2| ( 2 ⇒ point (2, 2) ∈ g (x)
h(!5) ( |!5 # 3| ( |!2| ( 2 ⇒ point (!5, 2) ∈ h(x)
4. 2 points on f (x) are (!1, 0) and (0, 1)
f (!1) ( |a(!1) # b| ( 0 and f (0) ( |a(0) # b| ( 1
⇒
!a # b ( 0
⇒ b(1
Hence, !a # 1 ( 0
⇒ a(1
Hence, f (x) ( |x # 1|.
2 points on g (x) are (!1, 0) and (0, 2)
g (!1) ( |a(!1) # b| ( 0 and g (0) ( |a(0) # b| ( 2
⇒
!a # b ( 0 ⇒
b(2
Hence !a # 2 ( 0
⇒
a(2
|
|
Hence, g (x) ( 2x # 2
2 points on h (x) are (!1, 0) and (0, 3)
h (!1) ( |a(!1) # b| ( 0 and h (0) ( |a(0) # b| ( 3
⇒
!a # b ( 0 ⇒
b(3
Hence !a # 3 ( 0
⇒
a(3
Hence, h (x) ( |3x # 3|
x ( !2 ⇒ f (!2) ( |!2 # 1| ( 1
⇒ (!2, 1) ∈ f (x)
|
|
x ( !2 ⇒ g (!2) ( 2(!2) # 2 ( 2 ⇒ (!2, 2) ∈ g (x)
x ( !2 ⇒ h (!2) ( |3(!2) # 3| ( 3 ⇒ (!2, 3) ∈ h (x)
397
Text & Tests 4 Solution
5.
f (x) ( |x ! 2|
⇒f (0) ( |0 ! 2| ( 2 (0, 2) ∈ f (x)
⇒f (2) ( |2 ! 2| ( 0 (2, 0) ∈ f (x)
⇒f (4) ( |4 ! 2| ( 2 (4, 2) ∈ f (x)
g (x) ( |x ! 6|
⇒g(0) ( |0 ! 6| ( 6 (0, 6) ∈ g (x)
⇒g(6) ( |6 ! 6| ( 0 (6, 0) ∈ g (x)
⇒g(8) ( |8 ! 6| ( 2 (8, 2) ∈ g (x)
Hence, |x ! 2| ( |x ! 6| at x ( 4.
Solve |x ! 2| ( |x ! 6|
⇒ x ! 2 ( !(x ! 6) OR x ! 2 ( #(x ! 6)
⇒ x ! 2 ( !x # 6
OR x ! 2 ( x ! 6
⇒
2x ( 8
OR
!2 ( !6 (Not valid)
⇒
x(4
6. (i) |x ! 6| % 2
⇒ !2 % x ! 6 % 2
⇒4%x%8
(ii) |x # 2| $ 4
⇒ !4 $ x # 2 $ 4
⇒ !6 $ x $ 2
(iii) |2x ! 1| & 5
⇒ !5 & 2x ! 1 & 5
⇒ !4 & 2x & 6
⇒ !4 & 2x or 2x & 6
⇒2x $ !4 or x & 3
⇒x $ !2
or x & 3
(iv) |2x ! 1| & 11
⇒ !11 & 2x ! 1 & 11
⇒ !10 & 2x & 12 ⇒ !10 & 2x or 2x & 12
⇒ 2x $ !10 or x & 6
⇒
x $ !5 or x & 6
(v) |3x # 5| % 4
⇒ !4 % 3x # 5 % 4
⇒ !9 % 3x % !1
1
⇒ !3 % x % ! __
3
(vi) |x ! 4| % 3
⇒ !3 % x ! 4 % 3
⇒1%x%7
7. (i) |2x ! 1| & 7
⇒ !7 & 2x ! 1 & 7
⇒ !6 & 2x & 8
⇒ !3 & x & 4
(ii) |3x # 4| $ |x # 2|
⇒ (3x # 4)2 $ (x # 2)2
⇒ 9x2 # 24x # 16 $ x2 # 4x # 4
⇒ 8x2 # 20x # 12 $ 0
398
y
6
f(x) # |x ! 2|
g(x) # |x ! 6|
4
2
2
4
6
8 x
Chapter 12
⇒ 2x2 # 5x # 3 $ 0
Solve 2x2 # 5x # 3 ( 0
⇒ (2x # 3)(x # 1) ( 0
1 , !1
⇒ Roots: x ( !1__
2
1 $ x $ !1
2x2 # 5x # 3 $ 0 ⇒ !1__
2
(iii) 2|x ! 1| $ |x # 3|
⇒ [2(x ! 1)]2 $ (x # 3)2
⇒ 4(x2 ! 2x # 1) $ x2 # 6x # 9
⇒ 4x2 ! 8x # 4 $ x2 # 6x # 9
⇒ 3x2 ! 14x ! 5 $ 0
Solve 3x2 ! 14x ! 5 ( 0
⇒ (3x # 1)(x ! 5) ( 0
1, 5
⇒ Roots: x ( ! __
3
1$x$5
Hence, 3x2 ! 14x ! 5 $ 0 ⇒ ! __
3
Hence,
8.
f (x) ( |x| ! 4
f (!4) ( |!4| ! 4 ( 4 ! 4 ( 0
f (0) ( |0| ! 4 ( !4
f (4) ( |4| ! 4 ( 4 ! 4 ( 0
y
4
(!4, 0) ∈ f (x)
(0, !4) ∈ f (x)
(4, 0) ∈ f (x)
1– x
2
)#
g(x
2
f(x) # |x| ! 4
1x
g(x) ( __
2
__
g(!4) ( 1 (!4) ( !2
2
1 (0) ( 0
g(0) ( __
2
1
__
g(4) ( (4) ( 2
2
(!4, !2) ∈ g(x)
!4
2
!2
(0, 0) ∈ g(x)
(4, 2) ∈ g(x)
4
6
8 x
!2
!4
1x
f (x) $ g(x) ⇒ |x| ! 4 $ __
2
1
__
Solve |x| ( x # 4
2
1 x2 # 4x # 16
⇒ x2 ( __
4
⇒ 4x2 ( x2 # 16x # 64
⇒ 3x2 ! 16x ! 64 ( 0
⇒ (3x # 8)(x ! 8) ( 0
2, 8
⇒ Roots x ( !2 __
3
2$x$8
3x2 ! 16x ! 64 $ 0 ⇒ !2 __
3
1x#3
9.
f (x) ( __
4
1
f (!24) ( __ (!24) # 3 ( |!6 # 3| ( |!3| ( 3 ⇒ (!24, 3) ∈ f (x)
4
1 (!12) # 3 ( |!3 # 3| ( 0 ⇒ (!12, 0) ∈ f (x)
f (!12) ( __
4
1 (0) # 3 ( |0 # 3| ( 3 ⇒ (0, 3) ∈ f (x)
f (0) ( __
4
1 x # 3 & 3 ⇒ x $ !24 or x & 0
Hence, __
4
Hence,
|
|
|
|
|
|
y
4
|
|
|
|
f(x) # 3
2
!24
!12
x
399
Text & Tests 4 Solution
10. |1 # 2x| % |x # 2|
⇒ (1 # 2x)2 % (x # 2)2
⇒ 1 # 4x # 4x2 % x2 # 4x # 4
⇒ 3x2 ! 3 % 0
⇒ x2 ! 1 % 0
Solve x2 ! 1 ( 0
⇒ (x # 1)(x ! 1) ( 0
⇒ Roots: x ( !1, 1
Hence, x2 ! 1 % 0 ⇒ !1 % x % 1
|
(
|
)
1
1 2 ( (1)2
11. %______
( 1 ⇒ ______
1 # 2x
1 # 2x
1
1
⇒ _____________
( __
1 # 4x # 4x2 1
⇒ 1 # 4x # 4x2 ( 1
⇒ 4x2 # 4x ( 0
⇒ x2 # x ( 0
⇒ x(x # 1) ( 0
⇒ Roots: x ( 0, !1
1
______
Solve
%1
1 # 2x
1
Hence, ____________
%1
1 # 4x # 4x2
⇒ 1 # 4x # 4x2 " 1
⇒ 4x2 # 4x " 0
⇒ x2 # x " 0
⇒ !1 " x " 0
|
|
(i) !4 % x % 2
(ii) x % !4 or x " 2
1 % x % 3 __
1
(iii) 1 __
4
2
(iv) 2 % x % 3
1%x%2
(v) 1 __
4
(vi) 2 % x % 3
1
(vii) 3 % x % 3 __
2
x % !2
13. (i) ______
2x ! 1
x
⇒ # ______
% !2
2x ! 1
x (2x ! 1)2 % !2(2x ! 1)2
⇒ ______
2x ! 1
⇒ x(2x ! 1) % !2(4x2 ! 4x # 1)
⇒ 2x2 ! x % !8x2 # 8x ! 2
⇒ 10x2 ! 9x # 2 % 0
Solve 10x2 ! 9x # 2 ( 0
⇒ (5x ! 2)(2x ! 1) ( 0
2 , __
1
⇒ Roots: x ( __
5 2
2 % x __
1
Hence, 10x2 ! 9x # 2 % 0 ⇒ __
5
2
12.
(
400
)
Chapter 12
(ii) |x ! 3| " 2|x ! 1|
⇒ (x ! 3)2 " [2(x ! 1)]2
⇒ x2 ! 6x # 9 " 4(x2 ! 2x # 1)
⇒ x2 ! 6x # 9 " 4x2 ! 8x # 4
⇒ 3x2 ! 2x ! 5 " 0
⇒ (x # 1)(3x ! 5) " 0
2
⇒ Roots: x " !1, 1 __
3
(iii) |x ! 1| ! |2x # 1| $ 0
⇒ |x ! 1| $ |2x # 1|
⇒ (x ! 1)2 $ (2x # 1)2
⇒ x2 ! 2x # 1 $ 4x2 # 4x # 1
⇒ 3x2 # 6x % 0
⇒ x2 # 2x % 0
Solve x2 # 2x " 0
⇒ x(x # 2) " 0
⇒ Roots: x " !2, 0
Hence, x2 # 2x % 0 ⇒ !2 % x % 0
Exercise 12.4
1. Proof: Assume x and y are both positive integers
and x2 ! y2 " (x # y)(x ! y).
If x $ y, then (x # y) and (x ! y) are positive integers;
hence, (x # y)(x ! y) is a positive integer & 1.
If x % y, then (x # y) is a positive integer
and (x ! y) is a negative integer;
hence, (x # y)(x ! y) is a negative integer & 1.
Hence, there is a contradiction in both cases.
∴ There are no positive integer solutions to x2 ! y2 " 1.
2. Proof: Assume (a # b) is a rational number.
p
Hence, (a # b) can be written as __ where p, q ∈ Ζ, q & 0.
q
m
__
Since “a” is a rational number " where m, n ∈ Ζ, n & 0
n
p
Then, a # b " __
q
p
m # b " __
⇒ __
n
q
p
m
⇒ b " __ ! __
q n
pn ! mq
" ________ which is rational.
qn
This is a contradiction as b is a irrational number;
hence, a # b is an irrational number.
401
Text & Tests 4 Solution
3. Proof: Assume x and y are both positive integers
and x2 ! y2 " (x # y)(x ! y).
If x $ y, then (x # y) and (x ! y) they are positive integers;
hence, (x # y)(x ! y) is a positive integer & 10.
Note: x # y " 5 and x # y " 10
x!y"2
x!y"1
2x " 7
2x " 11
1
1∈Ζ
__
⇒ x "3 ∉ Ζ
⇒ x " 5 __
2
2
If x % y, then (x # y) is a positive integer
and (x ! y) is a negative integer;
hence, (x # y)(x ! y) is a negative integer & 10.
Hence, there is a contradiction in both cases.
∴ There are no positive integer solutions to x2 ! y2 " 10.
4. Proof: a divides b
b divides c
⇒
⇒
⇒
⇒
Then a divides c.
b " k(a), k ∈ Ν
c " m(b), m ∈ Ν
c " m[k(a)]
c " mk(a), m, k ∈ Ν
5. Proof: a divides b ⇒ b " k(a), k ∈ Ν
a divides c ⇒ c " m(a), m ∈ Ν
hence, (b # c)
" k(a) # m(a)
" (k # m)(a), k, m ∈ Ν
Then a divides (b # c).
6. Proof: a2 # b2 ' 2ab
⇒ a2 ! 2ab # b2 ' 0
⇒ (a ! b) (a ! b) ' 0
⇒ (a ! b)2 ' 0, true for a, b ∈ R
Hence, a2 # b2 ' 2ab.
7. Proof: a is a rational number
p
Hence, a can be written as __ where p, q ∈ Ζ, q & 0.
q
b is a rational number.
m where m, n ∈ Ζ, n & 0.
Hence, b can be written as " __
n
p
pn
#
mq
m " ________
Hence, a # b " __ # __
q n
qn
" a rational number, as p, q, m, n ∈ Ζ
qn & 0.
8. Proof: x is an odd number " 2a # 1, where a ∈ Ν.
y is an odd number " 2b # 1, where b ∈ Ν.
Hence, x # y " 2a # 1 # 2b # 1
" 2a # 2b # 2
" 2(a # b # 1)
Hence, x # y has a factor of 2.
⇒ x # y must be even.
Hence, the sum of 2 odd numbers is always even.
402
Chapter 12
Exercise 12.5
1. (i)
a2 # 2ab # b2 ( 0
⇒ (a # b)(a # b) ( 0
⇒ (a # b)2 ( 0, true for a, b ∈ R
(ii)
a2 # 2ab # b2 ( 0
⇒ a2 # 2ab # b2 # b2 ( 0
⇒ (a # b)2 # (b)2 ( 0, true for a, b ∈ R
2.
(a # b)2 ( 4ab
⇒ a2 # 2ab # b2 ! 4ab ( 0
⇒ a2 ! 2ab # b2 ( 0
⇒ (a ! b)2 ( 0, true for a, b ∈ R
3.
⇒
⇒
!(a2 # 2ab # b2) ) 0
a2 # 2ab # b2 ( 0
(a # b)2 ( 0, true for a, b ∈ R
1(2
a # __
a
⇒ a2 # 1 ( 2a
⇒ a2 ! 2a # 1 ( 0
⇒ (a ! 1)2 ( 0, true for a $ 0 and a ∈ R
1 # __
1 ( _____
2
__
(ii)
a b a#b
1 # (a)(b)(a # b) * __
1 ( (a)(b)(a # b) * _____
2
(a)(b)(a # b) * __
a
b
a#b
⇒ b(a # b) # a(a # b) ( 2ab
⇒ ab # b2 # a2 # ab ( 2ab
⇒ a2 # b2 ( 0, true for a $ 0, b $ 0 and a, b ∈ R
4. (i)
5.
a2 ! 6a # 9 # b2 ( 0
⇒ (a ! 3)(a ! 3) # b2 ( 0
⇒
(a ! 3)2 # b2 ( 0,
true for a, b ∈ R
6. (i) x2 # 6x # 9 ( 0
⇒ (x # 3)2 ( 0, true for x ∈ R
(ii) x2 ! 10x # 25 ( 0
⇒ (x ! 5)2 ( 0, true for x ∈ R
(iii)
x2 # 4x # 6 ( 0
⇒ x2 # 4x # 4 # 2 ( 0
⇒
(x # 2)2 # 2 ( 0, true for x ∈ R
(iv)
x2 ! 6x # 10 ( 0
2
⇒ x ! 6x # 9 # 1 ( 0
⇒
(x ! 3)2 # 1 ( 0, true for x ∈ R
(v)
4x2 # 12x # 11 ( 0
11 ( 0
⇒
x2 # 3x # ___
4
9 # __
2(0
⇒ x2 # 3x # __
4 4
1 2 # __
1 ( 0, true for x ∈ R
⇒
x # 1__
2
2
(
)
403
Text & Tests 4 Solution
(vi)
4x2 ! 4x # 2 ( 0
1(0
⇒
x2 ! x # __
2
1
1(0
⇒ x2 ! x # __ # __
4 4
1 2 # __
1 ( 0,
⇒
x ! __
2
4
(
)
true for x ∈ R
7. (i)
!x2 # 10x ! 25 ( 0
⇒ x2 ! 10x # 25 ( 0
⇒
(x ! 5)2 ( 0, true for x ∈ R
(ii)
!x2 ! 4x ! 7 ) 0
⇒
x2 # 4x # 7 ( 0
2
⇒ x # 4x # 4 # 3 ( 0
⇒
(x # 2)2 # 3 ( 0, true for x ∈ R
8. (i)
⇒
⇒
⇒
(ii)
⇒
⇒
⇒
⇒
p2 # 4q2 ( 4pq
p ! 4pq # 4q2 ( 0
(p ! 2q)(p ! 2q) ( 0
(p ! 2q)2 ( 0, true for p, q ∈ R
(p # q)2 ) 2(p2 # q2)
p2 # 2pq # q2 ) 2p2 # 2q2
!p2 # 2pq ! q2 ) 0
p2 ! 2pq # q2 ( 0
(p ! q)2 ( 0, true for p, q ∈ R
2
9. a3 # b3 " (a # b)(a2 ! ab # b2)
Proof: a3 # b3 $ a2b # ab2
⇒ a3 # b3 ! a2b ! ab2 $ 0
⇒ (a # b)(a2 ! ab # b2) ! ab(a # b) $ 0
⇒ (a # b)(a2 ! ab # b2 ! ab) $ 0
⇒ (a # b)(a ! b)2 $ 0, true for a $ 0, b $ 0
10. Given a2 # b2 ( 2ab.
(i) a2 # c2 ( 2ac
(ii) b2 # c2 ( 2bc
a2 # b2 ( 2ab
a2 # c2 ( 2ac
b2 # c2 ( 2bc
Add: ⇒ 2a2 # 2b2 # 2c2 ( 2ab # 2bc # 2ac
⇒ a2 # b2 # c2 ( ab # bc # ac
11.
p#q
_____
2
___
⇒
p # q $ 2√ pq
___
⇒
(p # q)2 $ (2√ pq )2
⇒ p2 # 2pq # q2 $ 4pq
⇒ p2 ! 2pq # q2 $ 0
⇒
(p ! q)2 $ 0 … true
12.
⇒
⇒
⇒
⇒
404
___
$ √ pq
(ax # by)2 ) (a2 # b2)(x2 # y2)
a2x2 # 2axby # b2y2 ) a2x2 # a2y2 # b2x2 # b2y2
!a2y2 # 2axby ! b2x2 ) 0
a2y2 ! 2axby # b2x2 ( 0
(ay ! bx)2 ( 0, true for a, b, x, y ∈ R
Chapter 12
13.
a4 # b4 ( 2a2b2
⇒ a4 ! 2a2b2 # b4 ( 0
⇒ (a2 ! b2)(a2 ! b2) ( 0
⇒ (a2 ! b2)2 ( 0, true for a, b ∈ R
(
14.
⇒
⇒
⇒
⇒
⇒
⇒
⇒
)
1 # ___
1 (4
(a # 2b) __
a 2b
a # ___
2b # 2b * ___
1 # ___
1 (4
a * __
a 2a
a
2b
a # ___
2b # 1 ( 4
1 # ___
2b
a
a # ___
2b ( 2
___
2b
a
a
2b ( 2ab . 2 … since a $ 0, b $ 0
___
2ab *
# 2ab * ___
2b
a
a2 # 4b2 ( 4ab
a2 ! 4ab # 4b2 ( 0
(a ! 2b)2 ( 0, true for a, b positive
a
_______
15.
⇒
⇒
⇒
⇒
⇒
⇒
1
) __
(a # 1)2 4
a
1 (a # 1)2
_______
* (a # 1)2 ) __
4
(a # 1)2
a2 # 2a # 1
a ) ___________
4
4a ) a2 # 2a # 1
!a2 # 2a ! 1 ) 0
a2 ! 2a # 1 ( 0
(a ! 1)2 ( 0, true for a ∈ R
16. (i) a4 ! b4 " (a2 # b2)(a2 ! b2)
" (a2 # b2)(a # b)(a ! b)
5
(ii)
a ! a4b ! ab4 # b5
" a4(a ! b) ! b4(a ! b)
" (a ! b)(a4 ! b4)
" (a ! b)(a2 # b2)(a # b)(a ! b)
" (a2 # b2)(a # b)(a ! b)2
(iii)
a5 # b5 $ a4b # ab4
⇒ a5 ! a4b ! ab4 # b5 $ 0
⇒ (a2 # b2)(a # b)(a ! b)2 $ 0,
true if a and b
are positive unequal real numbers.
17. Given:
a2 # b2 " 1 and c2 # d2 " 1.
a2 # b2 ( 2ab ⇒ a2 ! 2ab # b2 ( 0
⇒ (a ! b)2 ( 0
Hence,
(a ! b)2 # (c ! d)2 ( 0
⇒ a2 ! 2ab # b2 # c2 ! 2cd # d2 ( 0
⇒
1 ! 2ab # 1 ! 2cd ( 0
⇒
!2ab ! 2cd ( !2
⇒
2ab # 2cd ) 2
⇒
ab # cd ) 1
405
Text & Tests 4 Solution
___
2ab
√ ab $ _____
a#b
18.
⇒
⇒
⇒
⇒
___
2ab if a and b are positive and unequal
(a # b) . √ ab $ (a # b)_____
a
#b
___
(a # b)√ ab $ 2ab
2ab
___
(a # b) $ ____
√ ab
___
a # b $ 2√ ab
___
⇒
(a # b)2 $ (2√ ab )2
⇒ a2 # 2ab # b2 $ 4ab
⇒ a2 ! 2ab # b2 $ 0
⇒
(a ! b)2 $ 0, true if a and b are positive and unequal.
19.
20.
9 (4
a # _____
a#2
9 (a # 2) ( 4(a # 2), where (a # 2) $ 0
⇒ a(a # 2) # _____
a#2
⇒
a2 # 2a # 9 ( 4a # 8
⇒
a2 ! 2a # 1 ( 0
⇒
(a ! 1)2 ( 0, true where (a # 2) $ 0
a $ __
c,
If __
b d
a (bd) $ __
c (bd) if a, b, c, d are positive numbers
⇒ __
b
d
⇒
ad $ cd
a # c $ __
c
_____
Hence,
b#d d
(a # c)
c (d)(b # d) if a, b, c, d are positive numbers
⇒ ______ * (d)(b # d) $ __
d
(b # d)
⇒
ad # cd $ cb # cd
⇒
ad $ cb, true
21. (a3 ! b3)(a ! b) " (a ! b)(a2 # ab # b2)(a ! b)
" (a ! b)2(a2 # ab # b2)
If a $ b ⇒ (a $ b) $ 0.
Hence, (a ! b)2 is positive,
and a2 # ab # b2 $ a2 ! 2ab # b2 " (a ! b)2 $ 0
Hence, a2 # ab # b2 is a positive, true if a, b, c, d are positive numbers.
Hence,
a4 # b4 ( a3b # ab3
⇒
a4 ! a3b ! ab3 # b4 ( 0
⇒
a3(a ! b) ! b3(a ! b) ( 0
⇒
(a ! b)(a3 ! b3) ( 0
⇒ (a ! b)(a ! b)(a2 # ab # b2) ( 0
⇒
(a ! b)2(a2 # ab # b2) ( 0, true for a, b ∈ R and a $ b
406
Chapter 12
Exercise 12.6
1. (i) a2 + a3 " a2+3 " a5
(ii) x . x . x2 " x1+1+2 " x4
(iii) 2x3 + 3x3 " 6x3+3 " 6x6
x5 " x5!2 " x3
(iv) __
x2
x4 " x4!5 " x!1
(v) __
x5
0
(vi) a___
"1
3
(vii) √ 27 " 3
(viii) (a3)2 " a6
(x3)2 __
x6
6!3
(ix) ____
" x3
3 " 3"x
x
x
(x) (3ab)2 " 9a2b2
2.
(i)
___
3
√
64 " 4
1 " __
1
(ii) 3!2 " __
32 9
1 " 23 " 8
(iii) ___
2!3
!2
32 __
9
__
(iv) 2___
!2 " 2 "
4
3
2
1
__
1 " 42 " 2
(v) ___
1
!__
2
2
__
3
4
3.
2
__
(i) 8 " (23)3 " 22 " 4
3
__
3
__
2
__
2
__
3
__
3
__
(ii) 164 " (24)4 " 23 " 8
(iii) 273 " (33)3 " 32 " 9
(iv) 814 " (34)4 " 33 " 27
2
__
2
__
(v) 1253 " (53)3 " 52 " 25
4.
()
(ii) ( 4 )
9
2
(i) __
3
!2
9
1 " __
1 " __
" ___
_4
_2 2
4
(3)
1
__
9
3
1 " __
1 " __
__ 2
" ___
_1
2
_
4
2
_ 2
!
( )
9
(iii) ___
25
(9)
3
!__
2
3
125
1 " ___
1 " _____
1 " ___
1 " ____
" ____
3
_3
27
___
3 3
9 _2
27
2
__
_
2
_3
(( ) )
(iv) ( 27 ) " ( ( ) ) " ( 3 )
5
125
____
2
!__
3
( 25 )
_3
5
( ) ( )
5
!2
___
3 3
__
(5)
!2
125
25
1 " ___
1 " __
" ___
9
__
9
_3 2
(5)
25
1
__
(27)3 __
3 __13 " ___
27 __13 " _____
"3
(v) 3__
1
__
8
2
8
(8)3
1
__
42 + 162 " ________
42 + 4 " ______
43 " __
43 " 43!5 " 4!2
5. _______
2
2
__
__
2
3
45
643 + 43 (43)3 + 43 4 + 4
407
Text & Tests 4 Solution
1
__
1
__
6.
1
__
34 + 3__+ 36
__________
√3
"
5
1___
1
+1+ __
34 6
______
3 12
____
"
1
__
32
"3
1
__
5 !__
1
1___
12 2 "
11
___
312
32
11
___
11
3P " 312 ⇒ p " ___
12
(xy2)3 + (x2y)!2 x3y6 . x!4y!2 x!1y4
7. (i) _____________ " _______ " _____
xy
xy
x1y1
y4!1
y3
__
" _____
"
x1#1
x2
( )
p2q 4 ______
p8 q4
p8+4 ___
p12
____
(ii) _____
"
"
"
p!1q3
p!4 q12 q12!4 q8
1
__
5
1 __
__
!5
___
(iii) a4 + a 4 " a4
( )
!
4
1
" a!1 " __
a
2
__
2
2
2
__
__
__
y!2 3
!2+3 3
(iv) ___
) " (y1)3 " y3
!3 " (y
y
__
_1
b)
(a√___
_______
_
(a1b2)!3 _______
a!3b!2 " _______
1
1
_______
"
"
" ____
_3 _1
_9 2
3+_32 _12 +_32
3 1 _12
2
2
2
√ a3b
(a b )
ab
a b
ab
(v)
!3
__
(vi)
4 7
√
x
____
___
√ x3
7
__
4
3 "x
__
"
1
!__
1
__
1
__
x2
(ii) (x #
!
"
1
__
x2)(x
√x
1
__
1
__
x2
1
__
x2)
1
__
1
__
x2
1
!__
(x ! 1)2 + (x ! 1) 2
________________
1
__
(x ! 1)2
1
1 __
__
"
"
x2(1 + x)
________
1
__
"1+x
1
__
1
__
x2
1
__
1
__
1
__
1
__
1
n+__
2
+ 3!n " 3
1
__
1
⇒ 3k " 32 ⇒ k " __
2
1 !n
n+__
2
( )
1
___
1
__
11. 220 . 22 . 212 . 212 " 220 . 24
" 261 . 626
" 262 Hz
2
2
,R 1 ___
R1
A1 4_____
___
12.
"
"
A2 4,R 2 R 22
2
V1 3
___
"
V2
( )
( )
Hence,
A1
___
Hence,
A1
___
2
__
408
_4 ,R 3 __23
3
1
_____
_4 ,R 3
3
2
A2
A2
"
"
1
!__
(x ! 1)2(x ! 1)2
(x ! 1)1 + (x ! 1)0 ________
x
" _______________
" x ! 1 + 1 " _____
(x ! 1)1
x!1
x!1
"3
1
___
2
1
__
3
10. √ 32n+1 + √ 3!3n " (32n+1)2 + (3!3n)3
1
__
#
(x ! 1)2(x ! 1)2 + (x ! 1)2(x ! 1) 2
____________________________
___
_____
1
__
1
__
3
__
1
__
"
x #1 " _____
x#1
" ____
1 #__
1
__
x
x2 2
" x2 ! x . x2 + x . x2 ! x2
" x2 ! x
x2 + x2
______
1
__
9.
1
__
" x4
x 2(x # 1)
_________
!
___
__
√ x + √ x3
_______
__
(iii)
3
! __
2
x2
x2 # x 2
_______
8. (i)
7
__
x4
__
3
2
__
_2
_2
3
3
( )
V1
___
2
__
_4 (,) R ( )
R
(
3)
__________
" ___
"
4
( _3 ) (,) R ( ) R
_2
3 33
1
3
2
1
2
2
3 _2
3
2
2
__
3
V2
2
__
2
__
( 384 ) " ( 64 ) " 169
162
____
3
27
___
3
___
1
__
" 32
Chapter 12
13. f(n) " 3n ⇒ (i) f(n + 3) " 3n+3
and
(ii) f(n + 1) " 3n+1
⇒ f(n + 3) ! f(n + 1) " 3n+3 ! 3n+1
" 3n . 33 ! 3n . 31
" 3n(27 ! 3) " 24(3n)
n
⇒ k f(n) " k(3 ) " 24(3n)
⇒ k " 24
14. f(n) " 3n!1 ⇒ f(n + 3) + f(n) " k f(n)
⇒ 3n+3!1 + 3n!1 " k(3n!1)
⇒ 3n!1 . 33 + 3n!1 " k(3n!1)
⇒ 3n!1(27 + 1)
" k(3n!1)
⇒
28 " k
Exercise 12.7
1. (i)
2x " 32
⇒ 2x " 25 ⇒ x " 5
(ii)
16x " 64
⇒ (42)x " 43
3
⇒ 42x " 43 ⇒ 2x " 3 ⇒ x " _
2
x
(iii)
25 " 125
2. (i)
1
9x " ___
27
1
⇒ (32)x " __
33
3
⇒ 32x " 3!3 ⇒ 2x " !3 ⇒ x " !__
2
1
(ii)
4x " ___
32
1
⇒ (22)x " __
25
5
2x
⇒ 2 " 2!5 ⇒ 2x " !5 ⇒ x " !__
2
(iii)
4x!1 " 2x#1
⇒ (52)x " 53
3
⇒ 52x " 53 ⇒ 2x " 3 ⇒ x " _
2
1
__
x
(iv)
3 "
27
1
⇒ 3x " __
33
⇒ 3x " 3!3 ⇒ x " !3
⇒ (22)x!1 " 2x#1
⇒ 22x!2 " 2x#1 ⇒ 2x ! 2 " x # 1
⇒x"3
1
__
(iv)
" 27
9x
1 " 33
⇒ ____
(32)x
1 " 33
⇒ ___
32x
⇒ 3!2x " 33 ⇒ !2x " 3
3
⇒
x " !__
2
__
3. (i)
2 "
x
√2
___
2
21
1
__
__
" √2
1 " 22
⇒ ____
(23)x
1
__
1 " 22
⇒ ___
23x
2
2
__
⇒ 2x " 22
8
x
1
__
1
__
⇒ 2x "
1
__
(iii)
!1
1
!__
2
"2
1
__
⇒ 2!3x " 22
1
⇒ x " !__
2
125
__
(ii)
25x " ____
√5
53
⇒ (52)x " __
1
⇒ !3x " __
2
1
__
52
1
3!__
2
⇒ 52x " 5
5
⇒ 2x " __
2
__
⇒
x"5
4
1
x " !__
6
1
___
(iv)
7x " 3 __
√7
1
x
⇒ 7 " __
⇒
5
__
" 52
1
__
73
1
!__
3
⇒ 7x " 7
1
⇒ x " !__
3
409
Text & Tests 4 Solution
___
4.
⇒
5
__
1
__
√ 32 " (25)2 " 22
___
16x!1 " 2√ 32
5
__
⇒ (24)x!1 " 21 . 22
5
1#__
7
__
24x!4 " 2 2 " 22
7
⇒ 4x ! 4 " __
2
15
7 # 4 " ___
⇒ 4x
" __
2
2
15
___
⇒ x
"
8
x
5.
27 " 9
and
3 x
2
⇒ (3 ) " 3
⇒
⇒ 33x " 32
⇒
⇒ 3x " 2
2
⇒
x " __
⇒
3
⇒
⇒
2x–y " 64
2x–y " 26
x–y"6
2
__
3
!y"6
16
2 " ___
!y " 6 ! __
3
3
16
y " ! ___
3
6. (i) 2x#2 " 2x . 22 " 4 . 2x
(ii) 2x # 2x " 2 . 2x
Hence, 2x # 2x " 2x#2(c ! 2)
⇒
2 . 2x " 4 . 2x(c ! 2)
⇒
2 " 4c ! 8
⇒
4c " 2 # 8 " 10
10 " __
5
⇒
c " ___
2
4
7. 3x " y ⇒ 32x " (3x)2 " y2
Solve 32x ! 12(3x) # 27 " 0
Let y " 3x ⇒ y2 ! 12y # 27 " 0
⇒ (y ! 3)(y ! 9) " 0
⇒ y " 3, y " 9
Hence, 3x " 3, 3x " 9
⇒ 3x " 31, 3x " 32
⇒ x " 1, x " 2
8. Solve
22x ! 3(2x) ! 4 " 0
Let y " 2x ⇒ y2 ! 3y ! 4 " 0
⇒ (y ! 4)(y # 1) " 0
⇒ y " 4, y " !1
Hence,
2x " 4 " 22, 2x " !1 (Not valid)
⇒x"2
9. (i) Solve
22x ! 9(2x) # 8 " 0
Let y " 2x ⇒ (y2 ! 9y # 8 " 0
⇒ (y ! 1)(y ! 8) " 0
⇒ y " 1, y " 8
Hence, 2x " 1 " 20, 2x " 8 " 23
⇒ x " 0,
x"3
410
Text & Tests 4 Solution
14. Solve
2x#1 # 2(2!x) ! 5 " 0
1 !5"0
Let 2x " y ⇒ 2y # 2 __
y
⇒ 2y2 # 2 ! 5y " 0
⇒ 2y2 ! 5y # 2 " 0
⇒ (2y ! 1)(y ! 2) " 0
1, y " 2
⇒ y " __
2
1 " 2!1,
Hence, 2x " __
2x " 21
2
⇒ x " !1, x " 1
()
15. Solve
3x # 81(3!x) ! 30 " 0
1 ! 30 " 0
Let y " 3x
⇒ y # 81 __
y
⇒ y2 # 81 ! 30y " 0
⇒ y2 ! 30y # 81 " 0
⇒ (y ! 3)(y ! 27) " 0
⇒ y " 3, y " 27
Hence,
3x " 31,
3x " 27 " 33
⇒
x " 1,
x"3
()
Exercise 12.8
1. (i) B
(ii) A
(iii) D
2. A(n) " 1000 + 20.2n
(i) A(0) " 1000 + 20.2(0)
" 1000 + 20
" 1000 + 1 " 1000 ha
(ii) (a) A(10) " 1000 + 20.2(10)
" 1000 + 22
" 1000 + 4 " 4000 ha
(b) A(12) " 1000 + 20.2(12)
" 1000 + 22.4
" 1000(5.278) " 5278 ha
(iii) n !
0
2
A(n) !
(iv) 5 weeks
1000
1320
(iv) C
4
1741
6
2297
8
3031
A(n)
4000
A(n) ! 1000 " 20.2n
3000
2000
1000
(iv)
2
412
4
6
8
10 n
10
4000
Chapter 12
3. (i)
(ii)
(iii)
(iv)
Decreasing
Decreasing
Increasing
Decreasing
4. (i)
(ii)
(iii)
(iv)
x " 0 ⇒ y " (0.6)20 " (0.6)(1) " 0.6
x " 0 ⇒ y " 3(2!0) " 3(1) " 3
x " 0 ⇒ y " 8(20) " 8(1) " 8
x " 0 ⇒ y " 6(4!0) " 6(1) " 6
5. (i)
x!
!2
!1
0
1
2
3
4
2x !
1
__
1
__
1
2
4
8
16
3x !
1
__
1
__
1
3
9
27
81
4
2
9
3
y
80
y ! 3x
60
40
20
#2
(ii)
y ! 2x
1
#1
x!
2
4 x
3
!2
!1
0
1
2
3
4
2"x !
4
2
1
1
__
1
__
1
__
8
1
___
16
3"x !
9
3
1
1
__
3
1
__
9
1
___
27
1
___
81
2
4
y
10
y ! 3#x
5
y ! 2#x
#2
#1
1
2
3
4 x
413
Text & Tests 4 Solution
(iii)
!2
!1
0
1
2
3
4
3.2x !
3
__
4
1
1__
2
3
6
12
24
48
2.3"x !
18
6
2
2
__
2
__
2
___
2
___
x!
3
9
y
40
y ! 3.2x
y ! 2.3#x
#2
(iv)
(v)
(vi)
(vii)
20
#1
1
2
3
4x
!2 - x % 0
0%x-4
x"0
0%x-4
6. D " 18(0.72)T
(i) Decay: Graph is decreasing
(ii) (a) T " 5°C ⇒ D " 18(0.72)5 " 3.48 " 3 days
(b) T " 2°C ⇒ D " 18(0.72)2 " 9.33 " 9 days
(c) T " 0°C ⇒ D " 18(0.72)0 " 18 days
(iii) 18(0.72)T ' 5
5
⇒ (0.72)T ' ___
18
5
⇒ log (0.72)T ' log ___
18
5
⇒ T( log 0.72) ' log ___
18
5
log __
18
⇒ T ' ________
log (0.72)
⇒ T ' 3.899
⇒ T " 3.9°C
7. P " 100(0.99988)n
(a) (i) n " 200 ⇒ P " 100(0.99988)200 " 97.628 " 97.6%
(ii) n " 500 ⇒ P " 100(0.99988)500 " 94.176 " 94.2%
(b) n " 5000
⇒ P " 100(0.99988)5000 " 54.879 " 54.9%
n " 6000
⇒ P " 100(0.99988)6000 " 48.673 " 48.7%
Hence, 100(0.99988)n " 50
⇒
(0.99988)n " 0.5
⇒
log(0.99988)n " log (0.5)
⇒
n log(0.99988) " log (0.5)
log (0.5)
⇒
n " ___________ " 5775.88 " 5780 years
log (0.99988)
414
27
81
Chapter 12
(c)
⇒
⇒
⇒
⇒
100(0.99988)n " 79
(0.99988)n " 0.79
log (0.99988)n " log(0.79)
n log (0.99988) " log(0.79)
log (0.79)
n " ___________ " 1964.2 " 1964 years
log (0.99988)
8. A(n) " 1000 + 20.2n
(i) A(0) " 1000 + 20.2(0)
" 1000 + 20
" 1000 + 1 " 1000 ha
(ii) (a) A(5) " 1000 + 20.2(5)
" 1000 + 21 " 2000 ha
(b) A(10) " 1000 + 20.2(10)
" 1000 + 22 " 4000 ha
(c) A(12) " 1000 + 20.2(12)
" 1000 + 22.4
" 1000(5.278) " 5278 ha
9. P(t) " 90 + 3!0.25t # 50
(i)
t!
90 # 3"0.25t $ 50
A(n)
5000
4000
A(n) ! 1000 " 20.2n
3000
2000
1000
0
4 5
8
10
0
2
4
6
8
10
140
102
80
67
60
56
(ii) P(0) " 90 + 3!0.25(0) # 50
" 90 + 30 # 50
" 90 # 50 " 140 beats/min
(iii) (a) 90 + 3!0.25t # 50 " 70
⇒
90 + 3!0.25t " 20
20 " __
2
⇒
3!0.25t " ___
90 9
2
⇒
log 3!0.25t " log __
9
2
⇒ !0.25t(log 3) " log __
9
2
_
log
⇒
!0.25t " _____9 " !1.369
log 3
t"
12 n
P(t)
150
100
50
!1.369
_______
2
4
6
8
10 t
" 5.476 " 5.5 minutes
!0.25
(b) 90 + 3!0.25t # 50 " 55
⇒
90 + 3!0.25t " 5
5 " ___
1
⇒
3!0.25t " ___
90 18
1
⇒
log 3!0.25t " log ___
18
1
⇒ !0.25t(log 3) " log ___
18
⇒
⇒
⇒
1
log __
18
!0.25t " _____
" !2.63
log 3
!2.63 " 10.52 " 10.5 minutes
t " ______
!0.25
(iv) 50 beats/min because P(t) " 50 as t gets larger using graph of P(t).
415
Text & Tests 4 Solution
10. E ! 101.5M#4.8
M " 7 ⇒ E " 101.5(7)#4.8 " 1015.3
M " 5 ⇒ E " 101.5(5)#4.8 " 1012.3
1015.3 " 1015.3!12.3 " 103 " 1000
⇒ No. of times " _____
1012.3
11. P(t) " 40bt
(i) t " 0 ⇒
(ii) t " 1 ⇒
P(0) " 40b0 " 40
P(1) " 40b1 " 48
48 " 1.2 $ 1
⇒ b " ___
40
⇒ No. of flies is increasing
(iii)
t!
0
1
2
3
4
5
P(t) ! 40(1.2)t
40
48
57.6
69.1
82.9
99.5
P(t)
100
50
0
1
2
3
4
5
t
Exercise 12.9
1. (i) log2 4 " 2
(ii) log3 81 " 4
(iii) log10 1000 " 3
(iv) log2 64 " 6
2. (i) log8 16 " x ⇒
8x " 16
3 x
⇒ (2 ) " 24
⇒ 23x " 24
⇒
3x " 4 ⇒
(ii) log9 27 " x ⇒
9x " 27
⇒ (32)x " 33
⇒
32x " 33
⇒
2x " 3 ⇒
(iii) log16 32 " x ⇒
⇒
⇒
⇒
416
4
x " __
3
3
x " __
2
16x " 32
(24)x " 25
24x " 25
5
4x " 5 ⇒ x " __
4
Chapter 12
(iv) log_1 8 " x ⇒
2
(2) " 8
1
__
x
!1
⇒ (2 )x " 23
⇒
2!x " 23
⇒
!x " 3 ⇒ x " !3
1 x " 81
__
(v) log_1 81 " x ⇒
3
3
⇒ (3!1)x " 34
!x
⇒
3 " 34
⇒
!x " 4 ⇒ x " !4
()
3. (i) log_1 27 " x ⇒
3
( 13 ) " 27
__
x
⇒ (3!1)x " 33
⇒
3!x " 33
⇒
!x " 3 ⇒
__
(ii) log __ 4 " x ⇒ (√ 2 )x " 4
√2
_1
⇒
(22)x " 22
⇒
22 " 22
1x " 2
__
2
x"4
⇒
⇒
x " !3
_1 x
(iii) log8 x " 2
⇒ 82 " x
⇒ x " 64
1
(iv) log64 x " __
2
_1
2
⇒ 64 " x
⇒ x"8
log2 x " !1
⇒ 2!1 " x
1
⇒
x " __
2
___
√
(ii) log3 27 " x___
_1
_3
⇒
3x " √ 27 " (33)2 " 32
3
⇒
x " __
2
(iii) logx 2 " 2 ⇒ x2 " 2
4. (i)
_1
__
x " 22 " √ 2
1 " 2!1
(iv) log2 (0.5) " x ⇒ 2x " __
2
⇒ x " !1
⇒
5. (i) log4 2 # log4 32 " log4 (2) . (32)
" log4 64 " x
⇒ 4x " 64 " 43
⇒ x"3
(9)(8)
(ii) log6 9 # log6 8 ! log6 2 " log6 _____
(2)
" log6 36 " x
⇒ 6x " 36 " 62
⇒ x"2
417
Text & Tests 4 Solution
(iii) log6 4 # 2 log6 3 " log 6 4 # log6 32
" log6 4 # log6 9 " log6 36 " 2
6. (i) log3 2 # 2 log3 3 ! log3 18
" log3 2 # log3 32 ! log3 18
" log3 2 # log3 9 ! log3 18
(2)(9)
" log3 _____ " log3 1 " 0
18
9
(ii) log8 72 ! log8 &__
8
" log8 72 ! (log8 9 ! log8 8)
" log8 72 ! log8 9 # log8 8
(72)(8)
" log8 ______ " log8 64 " 2
(9)
( )
7. log3 5 " a
(i) log3 15 " log3 5 . 3 " log3 5 # log3 3 " a # 1
5 " log 5 ! log 3 " a ! 1
(ii) log3 __
3
3
3
1
__
25 " log 25 ! log 3
(iii) log3 &8 " log3 &___
3
3
3
3
" log3 52 ! 1
" 2 log3 5 ! 1 " 2a ! 1
25 " log 25 ! log 27
(iv) log3 &___
3
3
27
2
" log3 (5) ! log3 (3)3
" 2 log3 5 ! 3 log3 3 " 2a ! 3
(v) log3 75 " log3 (25) . (3) " log3 25 # log3 3
" log3 52 # 1
" 2 log3 5 # 1
" 2a # 1
(& )
( )
( )
( )
8. (i)
2x
" 200
x
⇒ log 2 " log 200
⇒ x log 2 " log 200
log 200
⇒
x " _______ " 7.643 " 7.64
log 2
x
(ii)
5
" 500
x
⇒ log 5 " log 500
⇒ x log 5 " log 500
log 500
⇒
x " _______ " 3.861 " 3.86
log 5
(iii) 3x#1 " 25
⇒ log 3(x#1) " log 25
⇒ (x # 1)log 3 " log 25
log 25
⇒ x # 1 " _____ " 2.929
log 3
⇒
x " 1.929 " 1.93
2x#3
(iv)
5
" 51
⇒ log 5(2x#3)
" log 51
⇒ (2x # 3) log 5 " log 51
log 51
⇒ 2x # 3 " _____ " 2.4429
log 5
⇒
2x " !0.5571
⇒
x " !0.2785 " !0.279
418
Chapter 12
y " 2x#1 # 3
⇒ 2x!1 " y ! 3
⇒
log (2x!1) " log (y ! 3)
⇒ (x ! 1) log 2 " log (y ! 3)
log (y ! 3)
⇒
x ! 1 " _________
log 2
log
(y ! 3)
⇒
x " _________ # 1
log 2
log (8 ! 3)
log 5
(ii) y " 8 ⇒ x " _________ # 1 " _____ # 1 " 2.321928 # 1
log 2
log 2
" 3.321928
" 3.3219
9. (i)
10.
log10 x " 1 # a ⇒ 101#a " x
log10 y " 1 ! a ⇒ 101!a " y
⇒ xy " 101#a . 101!a " 101#a#1!a
" 102 " 100
21 " log 21 ! log 4
11. p " loga ___
a
a
4
" loga 7.3 ! loga 4
" loga 7 # loga 3 ! loga 4
7 " log 7 ! log 3
q " loga __
a
a
3
⇒
p # q " loga 7 # loga 3 ! loga 4 # loga 7 ! loga 3
" 2 loga 7 ! loga 4
7 " 2[log 7 ! log 2]
2r " 2 loga __
a
a
2
" 2 loga 7 ! 2 loga 2
" 2 loga 7 ! loga 22 " 2 loga 7 ! loga 4
Hence, p + q " 2r
12.
If loga x " 4 and loga y " 5
(i) loga x2y " loga x2 # loga y
" 2 loga x # loga y " 2(4) # 5 " 13
(ii) loga axy " loga a # loga y
" 1 # 4 # 5 " 10
__
__
√x
___
(iii) loga
" loga √ x ! loga y
y
_1
" loga x2 ! loga y
1 log x ! log y
" __
a
a
2
1 (4) ! 5 " !3
" __
2
1 log x
13. Prove log25 x " __
5
2
log5 x
Proof log25 x " ______
log5 25
log5 x __
" _____
" 1 log5 x
2
2
14.
(i)
(ii)
(iii)
(iv)
log10 4 " 0.60205999 " 0.602
log10 27 " 1.43136 " 1.43
log10 356 " 2.5514 " 2.55
log10 5600 " 3.748 " 3.75
419
Text & Tests 4 Solution
(v) log10 29 000 " 4.462 " 4.46
(vi) log10 35 0000 " 5.544 " 5.54
(vii) log10 38 70 000 " 6.5877 " 6.59
15. Minimum " 103 " 1000
Maximum " 104 " 10 000
log10 15 _______
16. log3 15 " _______
" 1.17609 " 2.464977
0.47712
log10 3
5
log
0.69897 " 2.3219
10
log2 5 " ______
" _______
log10 2 0.30103
Hence, log3 15 ! log2 5 " 2.464977 ! 2.3219
" 0.143077 " 0.143
log3 81 __
17. (i) log27 81 " ______
"4
log3 27 3
log2 8
3
(ii) log32 8 " ______
" __
log2 32 5
1
18. Show: logb a " _____
loga b
logaa _____
Proof: logb a " _____
" 1
loga b loga b
1 # _____
1 # _____
1 " log 2 # log 3 # log 5
19. _____
x
x
x
log2 x log3 x log5 x
" logx (2)(3)(5)
" logx 30
1
" ______
log30 x
20.
logr p " logr 2 # 3 logr q
⇒ logr p " logr 2 # logr q3
⇒ logr p " logr 2q3
⇒
p " 2q3
21.
3
log3 a # log9 a " __
4
a
log
3
__
3
⇒ log3 a # _____
"
log3 9 4
log3 a __
3
⇒ log3 a # _____
"
4
2
3
⇒ 2 log3 a # log3 a " __
2
3
⇒
3log3 a " __
2
__
⇒
log3 a " 1
2
⇒
_1
__
32 " a ⇒ a " √ 3
22. 3 ln 41.5 ! ln 250 " 3(3.7256934) ! 5.52146
" 11.17708 ! 5.52146
" 5.6556
" 5.66
420
Chapter 12
23. Solve log2 (x ! 2) # log2 x " 3
⇒
log2 (x ! 2)(x) "3
⇒
x2 ! 2x " 23 ! 8
2
⇒
x ! 2x ! 8 " 0
⇒
(x ! 4)(x # 2) " 0
⇒
x " 4,
x " !2 (Not valid)
24.
⇒
⇒
⇒
⇒
⇒
⇒
log10 (x2 # 6) ! log10 (x2 ! 1) " 1
x2 # 6 " 1
log10 ______
x2 ! 1
x2 # 6 " 101 " 10
______
x2 ! 1
10x2 ! 10 " x2 # 6
9x2 ! 16 " 0
(3x + 4)(3x ! 4) " 0
4 , x " __
4 ⇒ x " . __
4
x " !__
3
3
3
25. log 2x ! log (x ! 7) " log 3
2x " log 3
⇒
log _____
x!7
_____
⇒
& 2x " 3
x!7
⇒
3x ! 21 " 2x
⇒
x " 21
26. log (2x # 3) # log(x ! 2) " 2 log x
⇒
log (2x # 3)(x ! 2) " log x2
⇒
2x2 ! x ! 6 " x2
⇒
x2 ! x ! 6 " 0
⇒
(x ! 3)(x # 2) " 0
⇒
x " 3, x " !2 (Not valid)
27. log10 (17 ! 3x) # log10 x " 1
⇒
log10 (17 ! 3x)(x) " 1
⇒
101 " 17x ! 3x2
⇒
3x2 ! 17x # 10 " 0
2, 5
⇒
(3x ! 2)(x ! 5) " 0 ⇒ x " __
3
2
28. log10 (x ! 4x ! 11) " 0
⇒ 100 " x2 ! 4x ! 11
⇒ x2 ! 4x ! 11 " 1
⇒ x2 ! 4x ! 12 " 0
⇒ (x ! 6)(x # 2) " 0
⇒ x " 6, !2
29.
2 log2 x " y and log2(2x) " y # 4
⇒ log2 x2 " y ⇒
2y#4 " 2x
y
2
y
⇒
2 "x ⇒
2 . 24 " 2x
⇒
x2 . 16 " 2x
2
⇒ 16x ! 2x " 0
⇒
8x2 ! x " 0
⇒ x(8x ! 1) " 0
1 , x " 0 (Not valid)
⇒ x " __
8
421
Text & Tests 4 Solution
30. log6 x # log6 y " 1
⇒ log6 x . y " 1
⇒ 6 " xy
6
⇒ x " __
y
Solve log6 x # log6 y " 1 ∩ 5x # y " 17
6
6 # y " 17
⇒ x " __
⇒ 5 &__
y
y
30
___
⇒
# y " 17
y
⇒ 30 # y 2 " 17y
⇒ y 2 ! 17y # 30 " 0
(y ! 2)(y ! 15) " 0
⇒ y " 2, y " 15
6 " 3, x " ___
6 " __
2
⇒ x " __
2
15 5
()
31. (i) Solve 4 logx 2 ! log2 x ! 3 " 0
logx x
⇒ 4 logx 2 ! _____
!3"0
logx 2
1 !3"0
⇒ 4 logx 2 ! _____
logx 2
1!3"0
Let y " logx 2 ⇒ 4y ! __
y
2
⇒ 4y ! 1 ! 3y " 0
⇒ 4y 2 ! 3y ! 1 " 0
⇒ (4y # 1)(y ! 1) " 0
1
__
⇒ y"! ,y"1
4
1 , log 2 " 1
Hence, logx 2 " !__
x
4
1
_
⇒
x −4 " 2, x1 " 2
1 ,x"2
⇒
x " 2!4 " ___
16
(ii) Solve 2 log4 x # 1 " logx 4
log4 4
⇒ 2 log4 x # 1 " _____
log4 x
1
⇒
2 log4 x # 1 " _____
log4 x
1
Let y " log4 x ⇒
2y # 1 " __
y
2
⇒
2y # y " 1
⇒
2y 2 # y ! 1 " 0
⇒ (2y ! 1)(y # 1) " 0
1 , y " !1
⇒ y " __
2
1
__
Hence, log4 x " ,
log4 x " !1
2
_1
⇒
42 " x,
4!1 " x
__
1
⇒
x " √ 4 " 2, x " __
4
422
Chapter 12
Exercise 12.10
1. Property 1 ⇒ loga 1 " 0
Property 2 ⇒ log2 2 " 1
⇒ loge e " 1
⇒ log10 10 " 1
Property 5 ⇒ loga 0 " y
2.
⇒ a0 " 1, true
⇒ 21 " 2, true
⇒ e1 " e, true
⇒ 101 " 10, true
⇒ ay " 0 ⇒ No solution here
x"
0
1
2
y
100
y " 10x "
1
10
100
80
undef.
0
0.3
60
y " log10 x "
y ! 10x
40
31
20
y ! log10x
1
3. (i)
1
1
__
__
x1.5
"
9
y " 10 " 31, using graph 3
y " log3 x "
!2
!1
y
1
3
9
2
0
1
2
1
(ii) Graph y " log3 x
(iii) log3 2.5 " 0.8, using graph
log10 2.5
(iv) log3 2.5 " ________
log10 3
0.39794 " 0.834
" _______
0.47712
4. (i)
(ii)
4
6
9 x
8
#2
0
1
2
y " 5x "
1
5
25
y " log5 x "
2 2.5 3
1
#1
x"
x"
x
2
1.5
y
25
20
0
1
5
10
25
15
undef.
0
1
1.4
2
10
(iii) One graph is the inverse of the other.
y ! 5x
5
y ! log5 x
5
10
15
20
25 x
423
Text & Tests 4 Solution
5.
0
1
2
4
undef.
0
1
2
x"
y " log2 x "
x"
0
1
2
4
2x "
0
2
4
8
undef.
1
2
3
y " log2 2x "
1
0
3
4
6
x!2"
0
1
2
4
undef.
0
1
2
1
10
20
y
undef.
0
1
1.3
1.5
1.3
x"
0
2
10
20
x
__
0
1
5
10
undef.
0
0.7
1
"
x"
y " log10 __
2
!1
8
18
x#2"
0
1
10
20
y " log10 (x # 2) "
u.
0
1
1.3
3x#2 " y # 5
⇒
log 3x#2 " log (y # 5)
⇒ (x # 2) log 3 " log (y # 5)
⇒
log (y # 5)
x # 2 " _________
log 3
log (y # 5)
x " _________ !2
log 3
or
log3 (y # 5) ! 2
log (y # 5)
(ii) x " _________ !2
log 3
log (30 # 5)
log 35
when y " 30 ⇒ x " __________ !2 " ______ ! 2
log 3
log 3
1.544068 ! 2
" ________
0.47712
" 3.2362 ! 2
x " 1.2362 " 1.236
424
3
4
5
x
6
y ! log10 (x $ 2)
y ! log10 x
x
y ! log10 –
2
0.5
!2
⇒
2
1
#2
x"
7. (i)
1
0
x"
2
y ! log2 x
y ! log2 (x#2)
2
y " log10 x "
y ! log2 2x
2
x"
y " log2 (x ! 2) "
6.
y
3
10
20
x
Chapter 12
8. (i) Graph of y " log10 x # 2
y
3
y ! log10 x $ 2
2
1
10 x
5
(ii) Graph of y " log10(x # 2)
y
2
y ! log10 (x $ 2)
1
10 x
5
(iii) Graph of y " log10 x ! 2
y
5
10 x
y ! log10 x $ 2
#1
#2
(iv) Graph of y " 2 log10 x
y
2
y ! 2 log10 x
1
10 x
5
(v) y " !log10 x
y
1
10 x
5
#1
y ! #log10 x
9. Graphs of (i) y " log10(2x)
x
(ii) y " log10 &__
2
( )
y
3
2
y ! log10(2x)
1
#1 0
#1
y ! log10
1
2
3
4
5
x
–
2
6 x
425
Text & Tests 4 Solution
10. Graphs for:
(i) y " 2x # 1
y
4
y ! 2x $ 1
2
1
#2 #1
2
3 x
(ii) y " 1 ! 2x
y
2
1
#2 #1
3 x
2
y ! 1 # 2x
#2
#4
(iii) y " 2x#1
y
4
y ! 2x $ 1
2
1
#2 #1
2
3 x
( )
1 . 2x
(iv) y " &__
2
1
y ! – 2x
2
4
2
#2 #1
426
0
1
2
3 x
Chapter 12
Exercise 12.11
( )
0.6 1 " 5000(1 # 0.006)
1. (i) (a) A " 5000 # 5000 ____
100
" 5000(1.006) " €5030
2
0.6
(b) A " 5000 # 5000 ____ " 5000(1.006)2 " €5060.18
100
0.6 3 " 5000(1.006)3 " €5090.54
(c) A " 5000 # 5000 ____
100
(ii) 5000(1.006)t
(iii) 5000(1.006)t " 10 000
10 000 " 2
⇒
(1.006)t " ______
5000
⇒ ln(1.006)t " ln 2
⇒ t ln(1.006) " ln 2
ln 2 " 115.87 " 116 months
⇒ t " _______
ln 1.006
( )
( )
2.
y " Ae bt
t " 0 ⇒ 100 " Ae b(0) " Ae0 " A . 1
⇒ A " 100
Hence, y " 100e bt
t " 6 ⇒ 100e b(6) " 450
450 " 4.5
⇒
e 6b " ____
100
⇒ ln e 6b " ln 4.5
⇒ 6b ln e " 6b(1) " ln 4.5
ln 4.5 " 0.2507 " 0.25
⇒
b " _____
6
!0.02t
3.
T " 15 # 30 × 10
(i) t " 0 ⇒ T " 15 # 30 × 10!0.02(0)
" 15 # 30 × 100 " 15 # 30(1) " 45°C
(ii) T " 35°C ⇒ 15 # 30 × 10!0.02t " 35
⇒
30 × 10!0.02t " 35 ! 15 " 20
20 " __
2
⇒
10!0.02t " ___
30 3
2
⇒
log10 10!0.02t " log10 __
3
2
⇒ !0.02t (log10 10) " !0.02t(1) " log __
3
⇒
t"
log _23
______
!0.02
!0.17609
" __________
!0.02
" 8.8 minutes
(iii) As t gets larger, 30 × 10!0.02t gets smaller.
Hence, smallest value for 30 × 10!0.02t " 0.
⇒ Room temperature " 15°C
427
Text & Tests 4 Solution
4.
()
(
)
I
L " 10 log10 __I " 10 log10 _________
1 × 10!12
Io
(i) L " 100 ⇒ 100 " 10[log10 I ! log10 (1 × 10!12)]
⇒ 10 " log10 I # 12
⇒ log10 I " !2
I " 10!2 = 0.01 Wm!2
L " 110 ⇒ 110 " 10[log10 I ! log10(1 × 10!12)]
11 " log10 I # 12
⇒ log10 I " !1 ⇒ I " 10!1 " 0.1 Wm!2
⇒ Range is between 0.01 Wm–2 and 0.1 Wm−2
10
(ii) I " 10 Wm–2 ⇒
L " 10 log10 _________
1 × 10!12
" 10 log101013
" 10 . 13 log1010
" 10 . 13(1) " 130 dB
(
5. A " 10M and
Hence,
6.
)
E " 101.5M#4.8
" 101.5M * 104.8
" (10M)1.5 * 104.8
" A1.5 * 104.8 " 10aAb
a " 4.8 and b " 1.5
Value " €100
in
2000
(
)
4.5
(i) t years later ⇒ Value " 100 1 # ____
100
" 100(1.045)t
(ii) 10 years later ⇒ t " 10 ⇒ Value " 100(1.045)10
" 155.2969 " €155.30
(iii) 5 years earlier ⇒ t " !5 ⇒ Value " 100(1.045)!5
" 80.2451 " €80.25
7.
8.
428
t
W " 0.6 × 1.15t
(i) t " 0 ⇒ W " 0.6 × 1.150
" 0.6 × 1 " 0.6 kg
15
(ii) 1.15 " 1 # 0.15 " 1 # ____
100
⇒ growth constant " 15%
(iii) W " 2(0.6) " 1.2 ⇒ " 0.6 × 1.15t " 1.2
1.2 " 2
⇒ 1.15t " ___
0.6
⇒ ln 1.15t " ln 2
⇒ t ln 1.15 " ln 2
ln 2 " 4.959 " 5 months
⇒ t " ______
ln 1.15
M " M0e!kt
(i) M " 10 when t " 0
⇒
⇒
⇒
M0e!k(0) " 10
M0e0 " 10
M0(1) " M0 " 10
Chapter 12
Hence, M " 10e!kt
M " 5 when t " 140
10 . e!k(140) " 5
5 " 0.5
⇒ e!140k " ___
10
⇒ ln e!140k " ln 0.5
⇒ !140k(ln e) " ln 0.5
⇒ !140k(1) " !0.693147
!0.693147
⇒
k " ___________
!140
" 0.00495
⇒
M " 10e!0.00495t
t " 70 ⇒ M " 10 e!0.00495(70)
" 10 e!0.3465
" 10(0.7071)
" 7.071 " 7g
(iii) M " 2g ⇒
10e!0.00495t " 2
2 " 0.2
⇒
e!0.00495t " ___
10
⇒
ln e!0.00495t " ln 0.2
⇒
!0.00495t(ln e) " ln 0.2
!0.00495t(1) = !1.609438
!1.609438 " 325.1 " 325 days
⇒
t " ___________
!0.00495
(ii)
Exercise 12.12(A)
1. Proof:
(i) n " 1? ⇒ 2(1) " 1(1 # 1) ⇒ 2 " 2, true n " 1
(ii) Assume true for n " k.
⇒ 2 # 4 # 6 # 8 # … 2k " k(k # 1)
(iii) Also true for n " k # 1?
⇒ 2 # 4 # 6 # 8 # … # 2k # 2(k # 1) " k(k # 1) # 2(k # 1)
⇒ 2 # 4 # 6 # 8 # … # 2k # 2(k # 1) " k(k # 1)(k # 2)
⇒ 2 # 4 # 6 # 8 # … # 2k # 2(k # 1) " k(k # 1)[(k # 1) # 1]
∴ It is true for n " k # 1.
(iv) But since it is true for n " 1, it now must be true for n " 1 # 1 " 2.
And if it is true for n " 2, it is true for n " 2 # 1 " 3 … etc.
(v) Therefore, it is true for all positive integer values of n.
2. Proof:
1 [3(1) ! 1]
(i) n " 1? ⇒ 1 " __
2
(ii) Assume true for n " k.
1 (2) " 1, true n " 1,
⇒ 1 " __
2
k (3k ! 1)
⇒ 1 # 4 # 7 # 10 # …(3k ! 2) " __
2
(iii) Also true for n " k # 1?
k (3k ! 1) # (3k # 1)
⇒ 1 # 4 # 7 # 10 # … (3k ! 2) # (3k # 1) " __
2
k (3k ! 1) # 2(3k # 1)
⇒ 1 # 4 # 7 # 10 # … (3k ! 2) # (3k # 1) " __________________
2
429
Text & Tests 4 Solution
3k2 ! k # 6k # 2
⇒ 1 # 4 # 7 # 10 # … (3k ! 2) # (3k # 1) " ______________
2
2
3k # 5k # 2
⇒ 1 # 4 # 7 # 10 # … (3k ! 2) # (3k # 1) " ___________
2
(k # 1)(3k # 2)
_____________
⇒ 1 # 4 # 7 # 10 # … (3k ! 2) # (3k # 1) "
2
(k
#
1)[3(k
# 1) ! 1]
⇒ 1 # 4 # 7 # 10 # … (3k ! 2) # (3k # 1) " _________________
2
∴ It is true for n " k # 1.
(iv) But since it is true for n " 1, it now must be true for n " 1 # 1 " 2.
And if it is true for n " 2, it is true for n " 2 # 1 " 3 … etc.
(v) Therefore, it is true for all positive integer values of n.
3. Proof:
1 (1 # 1)(1 # 2) ⇒ 2 " __
1 (2)(3) ⇒ 2 " 2, true n " 1.
(i) n " 1? ⇒ 1 . 2 " __
3
3
(ii) Assume true for n " k.
k (k # 1) (k # 2)
⇒ 1 . 2 # 2 . 3 # 3 . 4 # 4 . 5 # …k(k # 2) " __
3
(iii) Also true for n " k # 1?
k (k # 1)(k # 2) # (k # 1)(k # 2)
⇒ 1 . 2 # 2 . 3 # 3 . 4 # 4 . 5 # … k(k # 1) # (k # 1)(k # 2) " __
3
k (k # 1)(k # 2) # 3(k # 1)(k # 2)
⇒ 1 . 2 # 2 . 3 # 3 . 4 # 4 . 5 # … k(k # 1) # (k # 1)(k # 2) " ____________________________
3
(k # 1)(k # 2)(k # 3)
⇒ 1 . 2 # 2 . 3 # 3 . 4 # 4 . 5 # … k(k # 1) # (k # 1)(k # 2) " __________________
3
(k # 1)[(k # 1) # 1][(k # 1) # 2]
___________________________
⇒ 1 . 2 # 2 . 3 # 3 . 4 # 4 . 5 # … k(k # 1) # (k # 1)(k # 2) "
3
∴ It is true for n " k # 1.
(iv) But since it is true for n " 1, it now must be true for n " 1 # 1 " 2.
And if it is true for n " 2, it is true for n " 2 # 1 " 3 … etc.
(v) Therefore, it is true for all positive integer values of n.
4. Proof:
1 " _______
1
1 " ___
1
(i) n " 1? ⇒ ___
⇒ ___
⇒ true n " 1,
2 . 3 2(1 # 2)
2.3 2.3
(ii) Assume true for n " k.
1 # ___
1 # ___
1 # ___
1 # … ____________
1
k
⇒ ___
# _______
2.3 3.4 4.5 5.6
(k # 1)(k # 2) 2(k # 2)
(iii) Also true for n " k # 1?
1 # ___
1 # ___
1 # ___
1 # … ____________
1
1
k
1
⇒ ___
# ____________
" _______
# ____________
2.3 3.4 4.5 5.6
(k # 1)(k # 2) (k # 2)(k # 3) 2(k # 2) (k # 2)(k # 3)
k (k # 3) # 2(1)
" _____________
2(k # 2)(k # 3)
k # 3k # 2
" _____________
2(k # 2)(k # 3)
(k # 1)(k # 2)
" _____________
2(k # 2)(k # 3)
2
k # 1 " ____________
k#1
" _______
2(k # 2) 2[(k # 1) # 2]
∴ It is true for n " k # 1
430
Chapter 12
(iv) But since it is true for n " 1, it now must be true for n " 1 # 1 " 2.
And if it is true for n " 2, it is true for n " 2 # 1 " 3 … etc.
(v) Therefore, it is true for all positive integer values of n.
5. Proof:
1 " _______
1
1 " ___
1 , true n " 1
(i) n " 1? ⇒ ___
⇒ ___
4 . 5 4(1 # 4)
4.5 4.5
(ii) Assume true for n " k.
1 # ___
1 # ___
1 # … ____________
1
k
⇒ ___
" _______
4.5 5.6 6.7
(k # 3)(k # 4) 4(k # 4)
(iii) Also true for n " k # 1?
1 # ___
1 # ___
1 # … ____________
1
1
k
1
⇒ ___
# ____________
" _______
# ____________
4.5 5.6 6.7
(k # 3)(k # 4) (k # 4)(k # 5) 4(k # 4) (k # 4)(k # 5)
k (k # 5) # 4(1)
" _____________
4(k # 4)(k # 5)
k # 5k # 4
" _____________
4(k # 4)(k # 5)
(k # 1)(k # 4)
" _____________
4(k # 4)(k # 5)
2
k # 1 " ____________
k#1
" _______
4(k # 5) 4[(k # 1) # 4]
∴ It is true for n " k # 1.
(iv) But since it is true for n " 1, it now must be true for n " 1 # 1 " 2.
And if it is true for n " 2, it is true for n " 2 # 1 " 3 … etc.
(v) Therefore, it is true for all positive integer values of n.
6. Proof:
(1)2
(i) n " 1? ⇒ 13 " ____ (1 # 1)2
4
(ii) Assume true for n " k
1
⇒ 1 " __
4
true n " 1.
⇒ 1 " 1,
k (k # 1)2.
⇒ 13 # 23 # 33 # … k3 " __
4
2
(iii) Also true for n " k # 1?
k (k # 1)2 # (k # 1)3
13 # 23 # 33 # … k3 # (k # 1)3 " __
4
2
(k # 1)2 # 4(k # 1)3
k___________________
"
4
(k # 1)2 [k2 # 4(k # 1)]
___________________
"
4
2
(k
#
2)(k # 2)
(k
#
1
)
" __________________
4
(k # 1)2(k # 2)2 __________________
(k # 1)2 [(k # 1) # 1]2
_____________
"
"
4
4
∴ It is true for n " k # 1.
(iv) But since it is true for n " 1, it now must be true for n " 1 # 1 " 2.
And if it is true for n " 2, it is true for n " 2 # 1 " 3 … etc.
(v) Therefore, it is true for all positive integer values of n.
⇒
n
7.
2
n(n # 1)(2n # 7)
∑n(n # 2) " 1 . 3 # 2 . 4 # 3 . 5 # … n(n # 2) " ______________
6
n"1
Proof:
1(1 # 1)(2 # 7)
2.9
(i) n " 1? ⇒ 1 . 3 " _____________ ⇒ 3 " ___
6
6
⇒ 3 " 3, true n " 1.
431
Text & Tests 4 Solution
k(k # 1)(2k # 7)
(ii) Assume true for n " k ⇒ 1 . 3 # 2 . 4 # 3 . 5 # … k(k # 2) " ______________
6
(iii) Also true for n " k # 1?
k(k # 1)(2k # 7)
1 . 3 # 2 . 4 # 3 . 5 # … k(k # 2) # (k # 1)(k # 3) " ______________ # (k # 1)(k # 3)
6
k(k # 1)(2k # 7) # 6(k # 1)(k # 3)
" _____________________________
6
(k # 1)[2k2 # 7k # 6k # 18]
________________________
"
6
2
#
13k # 18]
(k
#
1)[2
k
" ____________________
6
(k # 1)(k # 2)(2k # 9) ____________________________
(k # 1)[(k # 1) # 1][2(k # 1) # 7]
___________________
"
"
6
6
∴ It is true for n " k # 1
(iv) But since it is true for n " 1, it now must be true for n " 1 # 1 " 2.
And if it is true for n " 2, it is true for n " 2 # 1 " 3 … etc.
(v) Therefore, it is true for all positive integer values of n.
⇒
8. Proof:
x (x1 ! 1)
(i) n " 1? ⇒ x " _______
x!1
⇒ x " x,
true n " 1.
x (x k ! 1)
(ii) Assume true for n " k ⇒ x # x2 # x3 # x4 # … # xk " ________
x!1
(iii) Also true for n " k # 1?
x (x k ! 1)
⇒x # x2 # x3 # … # x k # x k#1 " _______ # x k#1
x!1
x (x k ! 1) # x k#1 (x ! 1)
__________________
"
x!1
k
x . x ! x # x . x k#1 ! x k#1
" ___________________
x!1
k#1
k#1
#
x
.
x
! x ! x k#1
x
____________________
"
x!1
x_________
(x k#1 ! 1)
"
x!1
∴ It is true for n " k # 1.
(iv) But since it is true for n " 1, it now must be true for n " 1 # 1 " 2.
And if it is true for n " 2, it is true for n " 2 # 1 " 3 … etc.
(v) Therefore, it is true for all positive integer values of n.
Exercise 12.12(B)
1. Proof:
(i) True for n " 1? ⇒ 5 is a factor of 61 ! 1 " 5, true n " 1.
(ii) Assume true for n " k ⇒ 5 is a factor of 6k ! 1, k ∈ N.
(iii) Also true for n " k # 1?
⇒ 6k#1 ! 1 " 6k . 61 ! 1
" 6k(5 # 1) ! 1
" 5 . 6k # (6k ! 1)
Since 5 . 6k is divisible by 5 and (6k ! 1) is assumed divisible by 5,
∴ 5 . 6k # (6k ! 1) is divisible by 5.
∴ It is true for n " k # 1.
(iv) But since it is true for n " 1, it now must be true for n " 1 # 1 " 2.
And if it is true for n " 2, it is true for n " 2 # 1 " 3 … etc.
(v) Therefore, it is true for all positive integer values of n.
432
Chapter 12
2. Proof:
(i) True for n " 1? ⇒ 4 is a factor of 51 ! 1 " 4, true n " 1.
(ii) Assume true for n " k ⇒ 4 is a factor of 5k ! 1.
(iii) Also true for n " k # 1?
⇒ 5k#1 ! 1 " 5k . 51 ! 1
" 5k (4 # 1) ! 1
" 4 . 5k # (5k ! 1)
Since 4 . 5k is divisible by 4 and (5k ! 1) is assumed divisible by 4,
∴ 4 . 5k # (5k ! 1) is divisible by 4.
∴ It is true for n " k # 1.
(iv) But since it is true for n " 1, it now must be true for n " 1 # 1 " 2.
And if it is true for n " 2, it is true for n " 2 # 1 " 3 … etc.
(v) Therefore, it is true for all positive integer values of n.
3. Proof:
(i) True for n " 1? ⇒ 4 is a factor of 91 ! 51 " 4, true n " 1.
(ii) Assume true for n " k ⇒ 4 is a factor of 9k ! 5k.
(iii) Also true for n " k # 1?
⇒ 9k#1 ! 5k#1 " 9k . 91 ! 5k . 51
" 9k (8 # 1) ! 5k (4 # 1)
" 8 . 9k # 9k ! 4 . 5k ! 5k
" 8 . 9k ! 4 . 5k # (9k ! 5k)
Since 8 . 9k ! 4 . 5k is divisible by 4 and 9k ! 5k is assumed divisible by 4,
∴ 8 . 9k ! 4 . 5k # (9k ! 5k) is divisible by 4.
∴ It is true for n " k # 1.
(iv) But since it is true for n " 1, it now must be true for n " 1 # 1 " 2.
And if it is true for n " 2, it is true for n " 2 # 1 " 3 … etc.
(v) Therefore, it is true for all positive integer values of n.
4. Proof:
(i) True for n " 1? ⇒ 8 is a factor of 32(1) ! 1 " 9 ! 1 " 8, true n " 1.
(ii) Assume true for n " k ⇒ 8 is a factor of 32k ! 1.
(iii) Also true for n " k # 1?
⇒ 32(k#1) ! 1 " 32k#2 ! 1
" 32k . 32 ! 1
" 32k . 9 ! 1
" 32k (8 # 1) ! 1
" 8 . 32k # (32k ! 1)
Since 8 . 32k is divisible by 8 and (32k ! 1) is assumed divisible by 8,
∴ 8 . 32k # (32k ! 1) is divisible by 8.
(iv) But since it is true for n " 1, it now must be true for n " 1 # 1 " 2.
And if it is true for n " 2, it is true for n " 2 # 1 " 3 … etc.
(v) Therefore, it is true for all positive integer values of n.
5. Proof:
(i) True for n " 1? ⇒ 5 is a factor of 71 ! 21 " 5, true n " 1
(ii) Assume true for n " k ⇒ 5 is a factor of 7k ! 2k.
(iii) Also true for n " k # 1?
⇒ 7k#1 ! 2k#1 " 7k . 71 ! 2k . 21
" 7k(5 # 2) ! 2 . 2k
" 5 . 7k # 2 . 7k ! 2 . 2k
" 5 . 7k # 2(7k ! 2k)
433
Text & Tests 4 Solution
Since 5.7k is divisible by 5 and (7k ! 2k) is assumed divisible by 5,
∴ 5 . 7k # 2(7k ! 2k) is divisible by 5.
∴ It is true for n " k # 1.
(iv) But since it is true for n " 1, it now must be true for n " 1 # 1 " 2.
And if it is true for n " 2, it is true for n " 2 # 1 " 3 … etc.
(v) Therefore, it is true for all positive integer values of n.
6. Proof:
(i) True for n " 1? ⇒ 8 is a factor of 72(1)#1 # 1 " 73 # 1 " 344 " 8 . 43 ⇒ true n " 1.
(ii) Assume true for n " k ⇒ 72k#1 # 1 is divisible by 8.
(iii) Also true for n " k # 1?
⇒ 72(k#1)#1 # 1 " 72k#2#1 # 1
" 72k#1 . 72 # 1
" 72k#1 . 49 # 1
" 72k#1 (48 # 1) # 1
" 48 . 72k#1 # (72k#1 # 1)
2k#1
Since 48 . 7
is divisible by 8 and (72k#1 # 1) is assumed divisible by 8,
∴ 48 . 72k#1 # (72k#1 # 1) is divisible by 8.
∴ It is true for n " k # 1.
(iv) But since it is true for n " 1, it now must be true for n " 1 # 1 " 2.
And if it is true for n " 2, it is true for n " 2 # 1 " 3 … etc.
(v) Therefore, it is true for all positive integer values of n.
7. Proof:
(i) True for n " 1? ⇒ 7 is a factor of 23(1)!1 # 3 " 22 # 3 " 7, true n " 1.
(ii) Assume true for n " k ⇒ 7 is a factor of 23k!1 # 3, k ∈ N.
(iii) Also true for n " k # 1?
⇒ 23(k#1)!1 # 3 " 23k#3!1 # 3
" 23k!1 . 23 # 3
" 23k!1 . 8 # 3
" 23k!1(7 # 1) # 3
" 7 . 23k!1 # (23k!1 # 3)
Since 7 . 23k!1 is divisible by 7 and 23k!1 # 3 is assumed divisible by 7,
∴ 7 . 23k!1 # (23k!1 # 3) is divisible by 7.
∴ It is true for n " k # 1.
(iv) But since it is true for n " 1, it must now be true for n " 1 # 1 " 2.
And if it is true for n " 2, it is true for n " 2 # 1 " 3 … etc.
(v) Therefore, it is true for all positive integer values of n.
8. Proof:
(i) True for n " 1? ⇒ 4 is a factor of 51 ! 4(1) # 3 " 4, true n " 1.
(ii) Assume true for n " k ⇒ 4 is a factor of 5k ! 4k # 3, k ∈ N.
(iii) Also true for n " k # 1?
⇒ 5k#1 ! 4(k # 1) # 3 " 5k . 51 ! 4k ! 4 # 3
" 5k(4 # 1) ! 4k ! 4 # 3
" 4 . 5k # 5k ! 4k ! 4 # 3
" 4 . 5k ! 4 # (5k ! 4k # 3)
Since 4 . 5k ! 4 is divisible by 4 and (5k ! 4k # 3) is assumed divisible by 4,
∴ 4 . 5k ! 4 # (5k ! 4k # 3) is divisible by 4.
∴ It is true for n " k # 1.
434
Chapter 12
(iv) But since it is true for n " 1, it must now be true for n " 1 # 1 " 2.
And if it is true for n " 2, it is true for n " 2 # 1 " 3 … etc.
(v) Therefore, it is true for all positive integer values of n.
9. Proof:
(i) True for n " 1? ⇒ 6 is a factor of 71 # 41 # 1 " 12 ⇒ true n " 1.
(ii) Assume true for n " k ⇒ 6 is a factor of 7k # 4k # 1, k ∈ N.
(iii) Also true for n " k # 1?
⇒ 7k#1 # 4k#1 # 1 " 7k . 71 # 4k . 41 # 1
" 7k(6 # 1) # 4k(3 # 1) # 1
" 6 . 7k # 7k # 3 . 4k # 4k # 1
" 6 . 7k # 3 . 4k # (7k # 4k # 1)
4k is an even number ⇒ 3 . 4k is divisible by 6 and
7k # 4k # 1 is assumed divisible by 6;
∴ 6 . 7k # 3 . 4k # (7k # 4k # 1) is divisible by 6.
∴ It is true for n " k # 1.
(iv) But since it is true for n " 1, it now must be true for n " 1 # 1 " 2.
And if it is true for n " 2, it is true for n " 2 # 1 " 3 … etc.
(v) Therefore, it is true for all positive integer values of n.
10. Proof:
(i) True for n " 1? ⇒ 1(1#1)[2(1) # 1] " (1)(2)(3) " 6 ⇒ true n " 1.
(ii) Assume true for n " k ⇒ 3 is a factor of k(k # 1)(2k # 1) " 2k3 # 3k2 # k, k ∈ N.
(iii) Also true for n " k # 1?
⇒ (k # 1)[(k # 1) # 1][2(k # 1) # 1] " (k # 1)(k # 2)(2k # 3)
" (k # 1)(2k2 # 7k # 6)
" 2k3 # 9k2 # 13k # 6
" 2k3 # 6k2 # 3k2 # 12k # k # 6
" 6k2 # 12k # 6 # (2k3 # 3k2 # k)
Since 6k2 # 12k # 6 is divisible by 3, and (2k3 # 3k2 #k) is assumed divisible by 3,
∴ 6k2 # 12k # 6 # (2k3 # 3k2 # k) is divisible by 3.
(iv) But since it is true for n " 1, it now must be true for n " 1 # 1 " 2.
And if it is true for n " 2, it is true for n " 2 # 1 " 3 … etc.
(v) Therefore, it is true for all positive integer values of n.
11. Prove n3 ! n is divisible by 3 for n ∈ N.
Proof:
(i) True for n " 1? ⇒ 3 is a factor of 13 ! 1 " 1 ! 1 " 0 ⇒ true n " 1.
(ii) Assume true for n " k ⇒ k3 ! k is divisible by 3.
(iii) Also true for n " k # 1?
⇒ (k # 1)3 ! (k # 1) " k3 # 3k2 # 3k # 1 ! k ! 1
" 3k2 # 3k # (k3 ! k)
Since 3k2 # 3k is divisible by 3, and (k3 !k) is assumed divisible by 3,
∴ 3k2 # 3k # (k3 ! k) is divisible by 3.
∴ It is true for n " k # 1.
(iv) But since it is true for n " 1, it now must be true for n " 1 # 1 " 2.
And if it is true for n " 2, it is true for n " 2 # 1 " 3 … etc.
(v) Therefore, it is true for all positive integer values of n.
435
Text & Tests 4 Solution
12. Prove 13n ! 6n!2 is divisible by 7 for n ∈ N.
Proof:
(i) True for n " 2? ⇒ 7 is a factor of 132 ! 62!2 " 169 ! 1 " 168 " 7 . 24 ⇒ true n " 2.
(ii) Assume true for n " k ⇒ 13k ! 6k!2 is divisible by 7.
(iii) Also true for n " k # 1?
⇒ 13k#1 ! 6(k#1)!2 " 13k . 131 ! 6k!2 . 61
" 13k(7 # 6) ! 6k!2 . 6
" 7 . 13k # 6 . 13k ! 6 . 6k!2
" 7 . 13k # 6(13k ! 6k!2)
Since 7 . 13k is divisible by 7, and (13k ! 6k!2) is assumed divisible by 7,
∴ 7 . 13k # 6(13k ! 6k!2) is divisible by 7.
∴ It is true for n " k # 1.
(iv) But since it is true for n " 1, it now must be true for n " 1 # 1 " 2.
And if it is true for n " 2, it is true for n " 2 # 1 " 3 … etc.
(v) Therefore, it is true for all positive integer values of n excluding 1.
Exercise 12.12(C)
1. Proof:
(i) True for n " 3? ⇒ 23 $ 2(3) # 1 ⇒ 8 $ 7 ⇒ true n " 3.
(ii) Assume true for n " k ⇒ 2k $ 2k # 1, k ∈ N, k / 3.
(iii) Also true for n " k # 1?
⇒ 2k#1 " 2 . 2k
$ 2(2k # 1)
since 2k $ 2k # 1 (assumed)
" 4k # 2
" 2k # 2k # 2
$ 2k # 3
since 2k # 2 $ 3, for k / 3
" 2(k # 1) # 1
! It is true for n " k # 1.
(iv) But since it is true for n " 3, it now must be true for n " 3 # 1 " 4.
And if it is true for n " 4, it is true for n " 4 # 1 " 5 … etc.
(v) Therefore, it is true for all values of n, n / 3, n ∈ N.
2. Proof:
(i) True for n " 2? ⇒ 32 $ 22 ⇒ 9 $ 4 ⇒ true n " 2.
(ii) Assume true for n " k ⇒ 3k $ k2, k ∈ N, and k / 2.
(iii) Also true for n " k # 1?
⇒ 3k#1 " 31 . 3k
$ 3 . k2
since 3k $ k2 (assumed)
2
2
2
"k #k #k
$ k2 # 2k # 1
since 2k2 $ 2k # 1, for k / 2
" (k # 1)(k # 1)
" (k # 1)2
! It is true for n " k # 1.
(iv) But since it is true for n " 2, it now must be true for n " 2 # 1 " 3.
And if it is true for n " 3, it is true for n " 3 # 1 " 4 … etc.
(v) Therefore, it is true for all values of n, n / 2, n ∈ N.
436
Chapter 12
3. Proof:
(i) True for n " 2? ⇒ 32 $ 2(2) # 2 ⇒ 9 $ 6 ⇒ true n " 2.
(ii) Assume true for n " k ⇒ 3k $ 2k # 2, k ∈ N, k / 2.
(iii) Also true for n " k # 1?
⇒ 3k#1 " 3 . 3k
$ 3(2k # 2)
since 3k $ 2k #2 (assumed)
" 6k # 6
" 2k # 4k # 2 # 4
" 2k # 4 # 4k # 2
$ 2k # 4
since 4k # 2 $ 0, for k / 2
" 2(k # 1) # 2
! It is true for n " k # 1.
(iv) But since it is true for n " 2, it now must be true for n " 2 # 1 " 3.
And if it is true for n " 3, it is true for n " 3 # 1 " 4 … etc.
(v) Therefore, it is true for all values of n, n / 2, n ∈ N.
4. Proof:
(i) True for n " 3? ⇒ 3! $ 23!1 ⇒ 6 $ 4 ⇒ true n " 3.
(ii) Assume true for n " k ⇒ k! $ 2k!1, k ∈ N, k / 3.
(iii) Also true for n " k # 1?
⇒ (k # 1)! " (k # 1)k!
$ (k # 1) . 2k!1
since k! $ 2k!1 (assumed)
$ (1 # 1) . 2k!1
since k # 1 $ 2, for k / 3
" 2 . 2k!1
" 2k
" 2(k#1)!1
! It is true for n " k # 1.
(iv) But since it is true for n " 3, it now must be true for n " 3 # 1 " 4.
And if it is true for n " 4, it is true for n " 4 # 1 " 5 … etc.
(v) Therefore, it is true for all values of n, n / 3, n ∈ N.
5. Proof:
(i) True for n " 2? ⇒ (2 # 1)! $ 22 ⇒ 6 $ 4 ⇒ true n " 2.
(ii) Assume true for n " k ⇒ (k # 1)! $ 2k.
(iii) Also true for n " k # 1?
⇒ (k # 1 # 1)! " (k # 2)!
" (k # 2)(k # 1)!
$ (k # 2) . 2k
since (k # 1)! $ 2k (assumed)
$ 2 . 2k
since (k # 2) $ 2, for k / 2
" 2 . 2k#1
! It is true for n " k # 1.
(iv) But since it is true for n " 2, it now must be true for n " 2 # 1 " 3.
And if it is true for n " 3, it is true for n " 3 # 1 " 4 … etc.
(v) Therefore, it is true for all values of n, n / 2, n ∈ N.
6. Proof:
(i) True for n " 1? ⇒ (1 # 2x)1 " 1 # 2(1)x ⇒ 1 # 2x " 1 # 2x ⇒ true n " 1.
(ii) Assume true for n " k ⇒ (1 # 2x)k / 1 # 2kx for x $ 0, k ∈ N.
437
Text & Tests 4 Solution
(iii) Also true for n " k # 1?
⇒ (1 # 2x)k#1 " (1 # 2x)(1 # 2x)k
/ (1 # 2x)(1 # 2kx)
since (1 # 2x)k / 1 # 2kx (assumed)
2
" 1 # 2kx # 2x # 4kx
/ 1 # 2kx # 2x
since 4kx2 / 0, for x $ 0, k / 1
" 1 # 2(k#1)x
! It is true for n " k # 1.
(iv) But since it is true for n " 1, it now must be true for n " 1 # 1 " 2.
And if it is true for n " 2, it is true for n " 2 # 1 " 3 … etc.
(v) Therefore, it is true for all values of n, n / 1, n ∈ N.
7. Prove (1 # ax)n / 1 # anx for a $ 0, x $ 0, n ∈ N.
Proof:
(i) True for n " 1? ⇒ (1 # ax)1 / 1 # a(1)x ⇒ 1 # ax / 1 # ax, true n " 1.
(ii) Assume true for n " k ⇒ (1 # ax)k / 1 # akx.
(iii) Also true for n " k #1?
⇒ (1 # ax)k#1 " (1 # ax)(1 # ax)k
/ (1 # ax)(1 # akx)
since (1 # ax)k / 1 # akx (assumed)
2 2
" 1 # akx # ax # a kx
/ 1 # akx # ax
since a2kx2 / 0, for a $ 0, x $ 0
" 1 # a(k # 1)x
! It is true for n " k # 1.
(iv) But since it is true for n " 1, it now must be true for n " 1 # 1 " 2.
And if it is true for n " 2, it is true for n " 2 # 1 " 3 … etc.
(v) Therefore, it is true for all values of n, n / 1, n ∈ N.
Revision Exercise12 (Core)
2x # 4 - 2, x ∈ R
1. !1 - ______
3
⇒ !3 - 2x # 4 - 6
⇒ !7 - 2x - 2
⇒ !3.5 - x - 1
2. (a) (i)
(ii)
(iii)
(iv)
103.5 " 3162.278 " 3162
log10 4.5 " 0.6532 " 0.65
t " 0.04 ⇒ 103t " 103(0.04) " 100.12 " 1.318 " 1.32
n " 100 ⇒ log 5n " log 5(100) " log 500 " 2.69897 " 2.7
(b) (i) e3.4 " 29.964 " 30
(ii) ln 589 " 6.378 " 6.38
(iii) t " 40 ⇒ e!0.02t!4 " e!0.02(40)!4 " e!0.8!4 " e!4.8
" 0.0082297 " 0.00823
10 " ln ______
10 " 0.994 " 0.99
(iv) k " 3.7 ⇒ ln ___
3.7
k
( )
3. (i) f(x) " 3 + 4x
Point (a, 6) ⇒
438
( )
f(a) " 3 + 4a " 6
⇒
4a " 2
2 a
⇒ (2 ) " 2
⇒ 22a " 21
1
⇒
2a " 1 ⇒ a " __
2
Chapter 12
(
) ( )
1
__
1 , b ⇒ f !__
1 " 3 + 4! 2 " b
(ii) Point !__
2
2
1"b
⇒ 3 + __
2
3
⇒ b " __
2
4. x ! 8 " !3
⇒x"5
5. (i)
or x ! 8 " 3
or
x " 11
52n + 252n!1 " 625
⇒
52n + (52)2n!1 " 54
⇒
52n + 54n!2 " 54
2n#4n!2
⇒ 5
" 56n!2 " 54
⇒ 6n ! 2 " 4
⇒ 6n " 6 ⇒ n " 1
(ii)
27n!2 " 93n#2
⇒ (33)n!2 " (32)3n#2
⇒
33n!6 " 36n#4
⇒ 3n ! 6 " 6n # 4
10
⇒
!3n " 10 ⇒ n " ! ___
3
6. y " a2x # b
(i) Point (0, 2.5) ⇒ 2.5 " a . 20 # b " a . 1 # b ⇒
Point (2, 4) ⇒
4 " a . 22 # b " 4a # b ⇒
(ii)
4a # b " 4
a # b " 2.5
1.5 " 0.5
⇒ 3a
" 1.5
⇒ a " ___
3
and 0.5 # b " 2.5
⇒ b " 2.5 ! 0.5 " 2
a # b " 2.5
4a # b " 4
7. (i) Curve C " ln (x) ⇒ Point (1, 0) ⇒ ln 1 " 0, true
(ii) Curve B " ln (x # 1) ⇒ Point (0, 0) ⇒ ln(0 # 1) " ln 1 " 0, true
(iii) Curve A " ln (x) # 1 ⇒ Point (1, 1) ⇒ ln 1 # 1 " 0 # 1 " 1, true
8. Solve ln(x ! 1) # ln(x # 2) " ln(6x ! 8)
⇒
ln(x ! 1)(x # 2) " ln(6x ! 8)
⇒
(x ! 1)(x # 2) " 6x ! 8
⇒
x2 # x ! 2 " 6x ! 8
⇒
x2 ! 5x # 6 " 0
⇒
(x ! 2)(x ! 3) " 0
⇒
x " 2, x " 3
9. y " Aebt
y " 6 when t " 1
y " 8 when t " 2
6
⇒ 6 " Aeb(1) ⇒ Ae b " 6 ⇒ eb " __
A
⇒ 8 " Aeb(2) ⇒ Ae 2b " 8
⇒ A(e b)2 " 8
6 2"8
⇒ A __
A
36 " 8
36 " ___
⇒ A * ____
A
A2
9
⇒ 8A " 36 ⇒ A " __
2
( )
439
Text & Tests 4 Solution
6 " __
4
eb "__
_9
3
2
4
⇒ ln eb " ln __
3
4
⇒ b(ln e) " b " ln __
3
10. y " a log2 (x ! b)
(5, 2) ⇒ 2 " a log2 (5 ! b)
⇒ 2 " log2 (5 ! b)a ⇒ (5 ! b)a " 22 " 4
1
__
⇒ 5 ! b " 4a
1
__
⇒ b " 5 ! 4a
(7, 4) ⇒ 4 " a log2(7 ! b)
⇒ 4 " log2(7 ! b)a ⇒ (7 ! b)a " 24 " 16
1
__
1
__
2
__
⇒ 7 ! b " 16 a " (42) a " 4 a
2
__
⇒ b " 7 ! 4a
1
__
2
__
Hence, 5 ! 4 a " 7 ! 4 a
2
__
1
__
⇒ 4a ! 4a " 2
1
__
1
__
⇒ (4 a )2 ! 4 a ! 2 " 0
1
__
Let y " 4 a ⇒ y2 ! y ! 2 " 0
⇒ (y ! 2)(y # 1) " 0
⇒ y " 2, y " !1
1
__
1
__
⇒ 4 a " 2 or 4 a " !1 (Not valid)
⇒ 4 " 2a
⇒ 22 " 2a ⇒ a " 2
1
__
⇒ b " 5 ! 42 " 5 ! 2 " 3
11. Solve 32x!1 " 28
⇒ ln 32x!1 " ln 28
⇒ (x ! 1) ln 32 " ln 28
ln 28 " 0.96147
⇒ x ! 1 " ____
ln 32
⇒ x " 1.96147
⇒ x " 1.96
12. Proof:
3(1)
(i) n " 1? ⇒ 3 " ____(1 # 1) ⇒ 3 " 3, true n " 1.
2
3k (k # 1).
(ii) Assume true for n " k ⇒ 3 # 6 # 9 # ... # 3k " ___
2
(iii) Also true for n " k # 1?
3k (k # 1) # 3(k # 1)
⇒ 3 # 6 # 9 # ... # 3k # (3k # 3) " ___
2
3k(k
# 1) # 2 . 3(k # 1)
⇒ 3 # 6 # 9 # ... # 3k # 3(k # 1) " ___________________
2
3 (k # 1)(k # 2)
⇒ 3 # 6 # 9 # ... #3k # 3(k # 1) " __
2
3 (k # 1)[(k # 1) # 1]
⇒ 3 # 6 # 9 # ... #3k # 3(k # 1) " __
2
! It is true for n " k # 1.
(iv) But since it is true for n " 1, it now must be true for n " 1 # 1 " 2.
And if it is true for n " 2, it is true for n " 2 # 1 " 3 ... etc.
(v) Therefore, it is true for all positive integer values of n.
440
Chapter 12
13. Proof:
(i) True for n " 1? ⇒ 7 is a factor of 81 # 6 " 14 ⇒ true n " 1.
(ii) Assume true for n " k ⇒ 8k # 6 is divisible by 7.
(iii) Also true for n " k # 1?
⇒ 8k#1 # 6 " 8k . 81 # 6
" 8k (7 # 1) # 6
" 7 . 8k # (8k # 6)
Since 7 . 8k is divisible by 7, and (8k # 6) is assumed divisible by 7,
! 7 . 8k # (8k # 6) is divisible by 7.
(iv) But since it is true for n " 1, it now must be true for n " 1 # 1 " 2.
And if it is true for n " 2, it is true for n " 2 # 1 " 3 ... etc.
(v) Therefore, it is true for all positive integer values of n.
14. Prove by induction that n2 $ 4n # 3 for n / 5, n ∈ N.
Proof:
(i) True for n " 5? ⇒ 52 $ 4(5) # 3 ⇒ 25 $ 23 ⇒ true n " 5.
(ii) Assume true for n " k ⇒ k2 $ 4k # 3 for k / 5, k ∈ N.
(iii) Also true for n " k # 1?
⇒ (k # 1)2 " k2 # 2k # 1
$ 4k # 3 # 2k # 1 since k2 $ 4k # 3 (assumed)
" 4k # 4 # 2k
$ 4(k # 1) # 3
since 2k $ 3, for k ≥ 5
! It is true for n " k # 1.
(iv) But since it is true for n " 5, it now must be true for n " 5 # 1 " 6.
And if it is true for n " 6, it is true for n " 6 # 1 " 7 ... etc.
(v) Therefore, it is true for all values of n, n / 5, n ∈ N.
Revision Exercise12 (Advanced)
1. (i) 3x # 4 % x2 ! 6
x2 ! 3x ! 10 $ 0
Let x2 ! 3x ! 10 " 0
(x ! 5)(x # 2) " 0
x " 5 or x " !2
Solution:
x % !2 or x % 5
(ii) x2 ! 6 % 9 ! 2x
x2 # 2x ! 15 % 0
Let x2 # 2x ! 15 " 0
(x # 5)(x ! 3) " 0
x " !5 or x " 3
Solution:
!5 % x % 3
(iii) The solution of
3x # 4 % x2 ! 6 % 9 ! 2x
is then the intersection of the above solution sets, i.e.
!5 % x % !2.
part (i)
#5
#2
0
3
5
part (ii)
441
Text & Tests 4 Solution
2. M " 30 + 2!0.001t
(i) t " 0
⇒ M " 30 + 2!0.001(0) " 30 + 20 " 30 + 1 " 30 g
(ii) M " 10 ⇒
30 + 2!0.001t " 10
10 " __
1
⇒
2!0.001t " ___
30 3
1
⇒
log 2!0.001t " log __
3
1
⇒
!0.001t . log 2 " log __
3
⇒
!0.001t "
⇒
⇒
log _13
_____
" !1.5849625
log 2
!1.5849625 " 1584.9625
t " ___________
!0.001
t " 1585 years
(iii) 1% of 30 " 0.3
⇒
30 + 2!0.001t " 0.3
0.3 " 0.01
⇒
2!0.001t " ___
30
⇒
log 2!0.001t " log 0.01
⇒
⇒
⇒
!0.001t . log 2 " log 0.01
log 0.01
!0.001t " _______ " !6.643856
log 2
_________
t " !6.643856
" 6643.856 " 6644 years
!0.001
3. I " I0 + 100.1S
(i) S " 30 ⇒ I " I0 + 100.1(30)
" I0 + 103 " 1000 I0 ⇒ Answer " 1000
(ii) S " 28 ⇒ I " I0 + 100.1(28) " I0 + 102.8
S " 15 ⇒ I " I0 + 100.1(15) " I0 + 101.5
I0 + 102.8
" 101.3 " 19.95 " 20 Times
⇒ No. of times " _________________
I0 + 101.5
4. Solve log5 x ! 1 " 6 logX 5.
log5 5
6
1 " _____
⇒ log5 x ! 1 " 6 _____
" 6 . _____
log5 x
log5 x log5 x
6
Let y " log5 x ⇒
y ! 1 " __
y
⇒
y2 ! y " 6
⇒
y2 ! y ! 6 " 0
⇒ (y ! 3)(y # 2) " 0
⇒
y " 3, y " !2
Hence, log5 x " 3 or log5 x " !2
1
⇒ x " 53 " 125 or x " 5!2 " ___
25
5. Solve
(0.7)x / 0.3
⇒ log(0.7)x / log(0.3)
⇒ x . log(0.7) / log(0.3)
⇒ x(!0.1549) / !0.5228787
⇒ x(0.1549) - 0.5228787
0.5228787 " 3.3755
⇒
x - __________
0.1549
⇒
x - 3.38
442
Chapter 12
6. (i)
x"
(ii)
!3 !2 !1
3
2
1
f(x) " | x | "
x#2"
h(x) " | x # 2 |
(iv) f(x) ∩ h(x) ⇒
(v) g(x) > h(x)
7.
⇒
1
2
3
0
1
2
3
5
4
3
2
3
4
5
!1
1
0
1
2
3
4
5
0
1
2
3
4
5
g(x) " | x | # 2
(iii)
0
g(x) ! |x| $ 2
5
4
f(x) ! |x|
3
2
h(x) ! |x $ 2|
x " !1
1
#3 #2 #1
!3 ) x % 0
1
y
2
3 x
9
x"
4
5
6
7
8
9
x!3"
1
2
3
4
5
6
ln(x ! 3) "
0
0.7
1.1
1.4
1.6
1.8
8
x ! ey$ 3
7
6
ln(x ! 3) " loge (x ! 3) " y
⇒ ey " x ! 3
⇒ x " ey # 3
5
y!x
4
3
y ! ln(x # 3)
2
1
y
8.
x"1
!30 !24 !18 !12 !6
0
1x "
__
!7.5 !6 !4.5 !3 !1.5 0
4
1x # 3 "
__
!4.5 !3 !1.5 0 1.5 3
4
1 x # 3 " 4.5 3 1.5 0 1.5 3
f(x) " __
4
1x # 3 / 3
Solve __
4
⇒ !24 / x / 0
|
|
|
2
3
4
5
6
7
8
9 x
f(x)
6
5
1.5
1
4
f(x) ! 4– x $ 3
4.5
4.5
3
|
f(x) ! 3
2
1
#30
3
__
1
#24
#18
#12
#6
6 x
1
__
x2 ! x!2
9. _____________
1
1
__
__
x2 ! x!2
1
__
(x # 1)(x ! 1)
x!2 (x2 ! 1) ____________
" ____________________
"x#1
"
1
__
x!1
x!2 (x ! 1)
10. Proof:
1 0 _______
1
1 0 _____
1 ⇒ true n " 1.
(i) True for n " 1? ⇒ _______
⇒ _____
(1 # r)1 1 # (1)r
1#r 1#r
1 0 ______
1 for k / 1, r $ 0, k ∈ N.
(ii) Assume true for n " k ⇒ _______
(1 # r)k 1 # kr
(iii) Also true for n " k # 1?
1
1
1
⇒ ________
" _____________
0 _____________
(1 # r)k#1 (1 # r)k(1 # r) (1 # kr)(1 # r)
1
" ______________
1 # kr # r # kr 2
1 0 ______
1 (assumed)
since _______
(1 # r)k 1 # kr
443
Text & Tests 4 Solution
1
0 _________
1 # kr # r
1
" ___________
1 # (k # 1)r
since kr 2 $ 0 for k / 1, r $ 0
∴ It is true for n " k # 1.
(iv) But since it is true for n " 1, it now must be true for n " 1 # 1 " 2.
And if it is true for n " 2, it is true for n " 2 # 1 " 3 … etc.
(v) Therefore, it is true for all positive integer values of n.
4x 0 1 for all x ∈ R, x & !1.
11. Prove _______
(x # 1)2
4x * (x # 1)2 0 1(x # 1)2
Proof: _______
(x # 1)2
⇒ 4x 0 x2 # 2x # 1
⇒ !x2 # 2x ! 1 0 0
⇒
x2 ! 2x # 1 0 0
⇒
(x ! 1)2 0 0 true for all x ∈ R, x & !1
4x 0 1.
Hence, _______
(x # 1)2
12. (1 # 2k)x2 ! 10x # (k ! 2) " 0
(i) Real Roots ⇒
b2 ! 4ac / 0
2
⇒ (!10) ! 4(1 # 2k)(k ! 2) / 0
⇒
100 ! 4(2k2 ! 3k ! 2) / 0
⇒
100 ! 8k2 # 12k # 8 / 0
⇒
!8k2 # 12k # 108 / 0
⇒
8k2 ! 12k ! 108 0 0
⇒
2k2 ! 3k ! 27 0 0
Factors: (k # 3)(2k ! 9) " 0
1
Roots: k " !3, k " 4 __
2
1
__
Hence, !3 0 k 0 4 .
2
10 $ 5
_____
(ii) Sum of roots $ 5 ⇒
1 # 2k
10
_____
⇒
(1 # 2k)2 $ 5(1 # 2k)2
1 # 2k
⇒
10(1 # 2k) $ 5(1 # 2k)2
⇒
2(1 # 2k) $ (1 # 2k)2
⇒
2 # 4k $ 1 # 4k # 4k2
⇒
!4k2 # 1 $ 0
⇒
4k2 ! 1 % 0
Factors: (2k # 1)(2k ! 1) " 0
1, k " _
1
Roots:
k " !_
2
2
1%k%_
1.
Hence, !_
2
2
13. Proof:
(i) True for n " 1? ⇒ 1 " (1 ! 1)21 # 1 ⇒ 1 " 0 # 1 ⇒ 1 " 1 ⇒ true n " 1.
(ii) Assume true for n " k ⇒ 1 # 2 . 2 # 3 . 22 # … . . k . 2k!1 " (k ! 1) . 2k # 1.
444
Chapter 12
(iii) Also true for n " k # 1?
⇒ 1 # 2 . 2 # 3 . 22 # … . . # k . 2k!1 # (k # 1) . 2(k#1)!1
" (k ! 1)2k # 1 # (k # 1)2(k#1)!1
" (k ! 1)2k # 1 # (k # 1)2k
" k . 2k ! 2k # 1 # k . 2k # 2k
" 2k . 2k # 1
" k . 2k#1 # 1 " [(k # 1) ! 1] . 2k#1 # 1
∴ It is true for n " k # 1.
(iv) But since it is true for n " 1, it is now must be true for n " 1 # 1 " 2.
And if it is true for n " 2, it is true for n " 2 # 1 " 3 … etc.
(v) Therefore, it is true for all positive integer values of n.
un " (n ! 20) . 2n
⇒ un#1 " (n # 1 ! 20) . 2n#1 " (n ! 19) . 2 . 2n
⇒ un#2 " (n # 2 ! 20) . 2n#2 " (n ! 18) . 22 . 2n " (n ! 18) . 4 . 2n
Hence, un#2 ! 4un#1 # 4un
" (n ! 18) . 4 . 2n ! 4(n ! 19) . 2 . 2n # 4 . (n ! 20)2n
" 2n[4n ! 72 ! 8n # 152 # 4n ! 80]
" 2n[8n ! 8n # 152 ! 152] " 0
15.
2 log y " log 2 # log x and 2y " 4x for y $ 0
⇒
2y " (22)x
⇒ log y2 " log 2x
2
⇒
2y " 22x
⇒
y " 2x
⇒
y " 2x
⇒
y2 " y
⇒ y2 − y " 0
⇒ y(y − 1) " 0
1
⇒ y " 0 (Not Valid) or y " 1 ⇒ 2x " 1 ⇒ x " _
2
n
16. (i) An exponential function : P " 40 000(1.03)
(ii) n " 12 ⇒ P " 40 000(1.03)12
" 40 000(1.42576)
" 57 030.43 " 57 030
(iii) n " 0 ⇒ P " 40 000(1.03)0
" 40 000(1) " 40 000
(iv) P " 80 000 ⇒ 40 000(1.03)n " 80 000
80 000
⇒
(1.03)n " _____ " 2
40 000
⇒
log (1.03)n " log 2
⇒ n log (1.03) " log 2
log 2
⇒
n " ______ " 23.449 " 23.5 years
log 1.03
14.
17. (i) P " Aekt, where A " 8000, P " 15 000, t " 8; find k.
⇒ 15 000 " 8000ek(8)
15 000 " 1.875
⇒
e8k " _____
8000
⇒ ln e8k " ln 1.875
⇒ 8k(ln e) " 8k(1) " 8k " 0.62860865
0.62860865 " 0.078576
⇒ k " __________
8
kt
Hence, P " Ae , where k " 0.078576, t " number of years
and A " 8000.
445
Text & Tests 4 Solution
(ii) t " 10 ⇒ P " 8000 e 0.078576(10)
" 8000 e0.78576 " 8000(2.1940738)
" 17 552.59 " 17 553
0.078576t
(iii) P " 30 000 ⇒ 15 000 e
" 30 000
30 000
⇒
e0.078576t " _____ " 2
15 000
⇒
ln e0.078576t " ln 2
⇒ 0.078576t ln e " 0.69314718
⇒ 0.078576t(1) " 0.69314718
0.69314718 " 8.82 " 9 years
⇒
t " ________
0.078576
Ans " 2016
Revision Exercise 12 (Extended-Response Questions)
1. N " 5000e!0.15t
(i) t " 0 ⇒ N " 5000e!0.15(0) " 5000e0 " 5000(1) " 5000
t " 5 ⇒ N " 5000e!0.15(5) " 5000e!0.75
" 5000(0.47236655)
" 2361.83 " 2362
⇒ Claim is justified.
(ii) t " 10 ⇒ N " 5000e!0.15(10)
" 5000e!1.5
" 5000(0.23313)
" 1115.65
(iii) t " 0 ⇒ N " 5000e!0.15(0) " 5000 . e0 " 5000 . 1 " 5000
(iv) N " 100 ⇒ 5000e!0.15t " 100
100 " 0.02
⇒
e!0.15t " _____
5000
⇒ ln e!0.15t " ln(0.02)
⇒ !0.15t(ln e) " ln(0.02)
⇒ !0.15t(1) " !0.15t " !3.912
!3.912 " 26.08 " 26.1 days
⇒
t " _______
!0.15
(v)
0
2
4
6
8
10
t!
No. of bacteria
N ! 5000!0.15t
5000
3704
2744
2033
N!y
5000
4000
3000
2000
1000
0
2 4 6 8 10 12 14 16 18 20 x ! t
Days
x
___
2. (i) A " 0.02(0.92)10
5 " 1.66667
(ii) Length " __
3
1.66667
______
⇒ A " 0.02(0.92) 10 " 0.02(0.92)0.166667
" 0.01972 " 0.0197
446
1506
1116
12
14
16
18
20
826
612
454
336
249
Chapter 12
(iii) S " (0.92)10!3x
W"S+A
x
___
" 0.02(0.92)10 . (0.92)10!3x
" 0.02(0.92)0.1x#10!3x
" 0.02(0.92)10!2.9x
(iv) W % (0.02)(0.92)2.5
⇒ 0.02(0.92)10!2.9x % (0.02)(0.92)2.5
⇒
10 ! 2.9x % 2.5
⇒
!2.9x % 2.5 ! 10 " !7.5
⇒
2.9x $ 7.5
7.5 " 2.586
⇒
x $ ___
2.9
⇒
x $ 2.59
3. (i) A " (0.83)n I; B " (0.66)(0.89)n I
(ii) (0.83)n I " (0.66)(0.89)n I
⇒
log (0.83)n " log(0.66)(0.89)n
⇒
n log (0.83) " log (0.66) # log (0.89)n
⇒
n log (0.83) " log (0.66) # n log (0.89)
⇒
n[log (0.83) ! log (0.89)] " log (0.66)
⇒ n[!0.0809219 # 0.05061] " !0.180456
⇒
n(!0.0303119) " !0.180456
!0.180456 " 5.9533
⇒
n " ___________
!0.0303119
⇒
n " 6 stations
(
)
t
11 " A(1.11)t
4. (i) Pg " A 1 # ____
100
5 t " 10A(0.95)t
(ii) Pr " 10A 1 ! ____
100
(iii)
A(1.11)t
" 10A (0.95)t
t
(1.11)
⇒ ______t
" 10
(0.95)
1.11 t
⇒ ____
" 10
0.95
⇒ (1.168421)t
" 10
t
⇒ log (1.168421) " log 10
⇒ t log (1.168421) " 1
1
⇒
t " ____________
" 14.793 " 14.8 years
log (1.168421)
(iv) A(1.11)t " 100A(0.95)t .... [Note: Pg " 10Pr when proportions are reversed.]
1.11 t
⇒ ____
" 100
0.95
⇒ (1.168421)t
" 100
t
⇒ log (1.168421) " log 100
⇒ t log (1.168421) " 2
2
⇒
t " ____________
" 29.586
log (1.168421)
⇒
t " 29.6 years
(
)
( )
( )
447
Text & Tests 4 Solution
(v) t !
P ! 10(0.95)t
P ! (1.11)t
0
10
1
5
7.7
1.7
10
6.0
2.8
15
4.6
4.8
20
3.6
8.1
25
2.8
13.6
24
22
20
(i) P ! (1.11)t
17
15
14
12
10
7
5
(ii) P ! 10(0.95)t
2
(iii)
P
5
10
15
(iv)
20
25
30 t
5. n " A(1 ! e!bt)
(i) Growth; as t increases, n also increases.
(ii) t " 2 when n " 10 000 ⇒ 10 000 " A(1 ! e!2b)
10,000
⇒ ______ " 1 ! e!2b
A
000
______
⇒ e!2b " 1 ! 10
A
t " 4 when n " 15 000 ⇒ 15 000 " A(1 ! e!4b)
000 " 1 ! e!4b
______
⇒ 15
A
15 000
⇒ e!4b " 1 ! ______
A
Hence, 2e!4b ! 3e!2b # 1
(
) (
)
15 000 ! 3 1 ! ______
10 000 # 1
" 2 1 ! ______
A
A
30 000 ! 3 # ______
30 000 # 1 " 0
" 2 ! ______
A
A
(iii) a " e!2b ⇒ e!4b " (e!2b)2 " a2
Hence, 2e!4b ! 3e!2b # 1 " 0
becomes
2a2 ! 3a # 1 " 0
(iv) Factors: (a ! 1)(2a ! 1) " 0
1
Roots:
a " 1, a " __
2
1
(v) e!2b " 1
or
e!2b " __
2
448
⇒ e!2b " e0
⇒
⇒ !2b " 0
⇒
b " 0 (Not valid)
⇒
⇒
⇒
1
ln e!2b " ln __
2
!2b(ln e) " ln 1 ! ln 2
!2b " !ln 2
1 ln 2
b " __
2
30
2.1
23.9
Chapter 12
1 ln2
!t*__
(vi) n " A(1 ! e 2 )
1 ln 2
!2*__
t " 2 when n " 10 000 ⇒ 10 000 " A(1 ! e 2 )
⇒ 10 000 " A(1 ! e!ln2)
⇒ 10 000 " A(1 ! 0.5)
⇒ 10 000 " A(0.5)
⇒ 20 000 " A(1)
⇒
A " 20 000
1 ln 2
!t*__
2
(vii) n " 20 000(1 ! e
)
t!
n!
0
0
1
5858
2
10 000
3
12 929
4
15 000
5
16 464
n
15 000
1
n = 20000 1 # e#t⋅2– ln2
10 000
5000
1
2
3
4
5
t
1
__
!t* ln 2
2 )
(viii) n " 18 000 ⇒ 20 000(1 ! e
⇒
⇒
" 18 000
!t* ln 2
18 000 " 0.9
(1 ! e 2 ) " ______
20 000
1 ln 2
!t*__
!e 2 " 0.9 ! 1 " !0.1
1
__
⇒
⇒
⇒
⇒
1 ln 2
!t*__
2
1
__
!t* ln 2
e 2
e
⇒
ln
" 0.1
" ln 0.1
1 ln 2(ln e) " !2.3
!t * __
2
2.3
t " ____
_1 ln 2
2
2.3 " 6.636 " 6.64 hours
t " _______
0.34657
449