· Exercise 1. 3 -3 Exercise 1. 3 -3 er r s er e e si s 3 -i e ice e si s - 1 1 16. stru t 1- 1...
Transcript of · Exercise 1. 3 -3 Exercise 1. 3 -3 er r s er e e si s 3 -i e ice e si s - 1 1 16. stru t 1- 1...
CONTENTS
1. Number System 1-16
Exercise 1.1 1-3
Exercise 1.2 3-5
Exercise 1.3 5-6
Exercise 1.4 6-10
Exercise 1.5 10-13
Exercise 1.6 13
Multiple Choice Questions 13-16
2. Exponents of Real Numbers 17-48
Exercise 2.1 17-24
Exercise 2.2 24-38
Very Short Answer Type Questions 38-39
Multiple Choice Questions 39-48
3. Rationalisation 49-66
Exercise 3.1 49-50
Exercise 3.2 50-59
Very Short Answer Type Questions 59-61
Multiple Choice Questions 61-66
4. Algebraic Identities 67-91
Exercise 4.1 67-70
Exercise 4.2 71-73
Exercise 4.3 73-79
Exercise 4.4 79-83
Exercise 4.5 83-84
Very Short Answer Type Questions 84-86
Multiple Choice Questions 86-91
5. Factorization of Algebraic Expressions 92-105
Exercise 5.1 92-96
Exercise 5.2 96-98
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Exercise 5.3 98
Exercise 5.4 99-101
Very Short Answer Type Questions 101-102
Multiple Choice Questions 102-105
6. Factorization of Polynomials 106-133
Exercise 6.1 108-109
Exercise 6.2 110-112
Exercise 6.3 112-117
Exercise 6.4 117-123
Exercise 6.5 124-129
Very Short Answer Type Questions 129
Multiple Choice Questions 130-133
7. Linear Equations in Two Variables 134-153
Exercise 7.1 134
Exercise 7.2 135-137
Exercise 7.3 137-148
Exercise 7.4 148-151
Very Short Answer Type Questions 151-152
Multiple Choice Questions 152-153
8. Co-ordinate Geometry 154-157
Exercise 8.1 155-156
Multiple Choice Questions 156-157
9. Introduction to Euclid’s Geometry 158-164
Exercise 9.1 162-163
Very Short Answer Type Questions 163-164
10. Lines and Angles 165-204
Exercise 10.1 173-175
Exercise 10.2 175-181
Exercise 10.3 181-185
Exercise 10.4 185-194
Very Short Answer Type Questions 194
Multiple Choice Questions 194-204
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11. Triangle and its Angles 205-234
Exercise 11.1 210-213
Exercise 11.2 213-220
Very Short Answer Type Questions 220-224
Multiple Choice Questions 224-234
12. Congruent Triangles 235-269
Exercise 12.1 245-249
Exercise 12.2 249-250
Exercise 12.3 250-254
Exercise 12.4 254
Exercise 12.5 254-257
Exercise 12.6 257-261
Very Short Answer Type Questions 261-263
Multiple Choice Questions 263-269
13. Quadrilaterals 270-309
Exercise 13.1 281-282
Exercise 13.2 282-285
Exercise 13.3 285-288
Exercise 13.4 288-297
Very Short Answer Type Questions 297-303
Multiple Choice Questions 303-309
14. Areas of Parallelograms and Triangles 310-347
Exercise 14.1 316-317
Exercise 14.2 317-318
Exercise 14.3 319-337
Very Short Answer Type Questions 337-341
Multiple Choice Questions 341-347
15. Circles 348-411
Exercise 15.1 373
Exercise 15.2 373-379
Exercise 15.3 379-380
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Exercise 15.4 380-387
Exercise 15.5 387-398
Very Short Answer Type Questions 398-402
Multiple Choice Questions 402-411
16. Construction 412-421
Exercise 16.1 412-414
Exercise 16.2 414-418
Exercise 16.3 418-421
17. Heron’s Formula 422-440
Exercise 17.1 423-427
Exercise 17.2 427-433
Very Short Answer Type Questions 433-435
Multiple Choice Questions 435-440
18. Surface Area and Volume of a Cuboid and Cube 441-459
Exercise 18.1 441-446
Exercise 18.2 446-453
Very Short Answer Type Questions 453-454
Multiple Choice Questions 454-459
19. Surface Area and Volume of A Right Circular Cylinder 460-478
Exercise 19.1 460-464
Exercise 19.2 464-473
Very Short Answer Type Questions 473-474
Multiple Choice Questions 474-478
20. Surface Area and Volume of A Right Circular Cone 479-496
Exercise 20.1 479-484
Exercise 20.2 485-490
Very Short Answer Type Questions 490-491
Multiple Choice Questions 491-496
21. Surface Area and Volume of A Sphere 497-517
Exercise 21.1 497-501
Exercise 21.2 501-511
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Very Short Answer Type Questions 511-513
Multiple Choice Questions 513-517
22. Tabular Representation of Statistical Data 518-532
Exercise 22.1 520-527
Exercise 22.2 527-531
Multiple Choice Questions 531-532
23. Graphical Representation of Statistical Data 533-569
Exercise 23.1 535-549
Exercise 23.2 549-563
Exercise 23.3 563-568
Multiple Choice Questions 569
24. Measures of Central Tendency 570-598
Exercise 24.1 575-582
Exercise 24.2 582-589
Exercise 24.3 589-592
Exercise 24.4 592-593
Very Short Answer Type Questions 594-595
Multiple Choice Questions 595-598
25. Probability 599-610
Exercise 25.1 599-607
Very Short Answer Type Questions 607-608
Multiple Choice Questions 608-610
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1 Arundeep’s Mathematics (R.D.) 9th
NUMBER SYSTEM
Points to Remember :
1. Natural number : Numbers like 1, 2, 3, 4,
5, ..... are called natural numbers. There
are also called counting numbers. The set
of natural number is denoted by N.
2. Whole numbers : Numbers like 0, 1, 2, 3,
4, ..... are called whole numbers. The set
of whole numbers is denoted by W.
3. Integers : The negative integers together
whole numbers are called integers : ....., –
4, –3, –2, –1, 0, 1, 2, 3, 4, ..... are integers.
The set of integers is denoted by Z.
4. Rational numbers : The number in the
form of q
p where p and q are integers and
q 0, is called rational numbers like 5
2,
7
1, 0,
4
3,
8
7,
4
9 etc.
The set of rational numbers is denoted by
Q.
Q = q
p : p and q z and q 0.
5. Irrational numbers : The numbers which
are not rationals, are called irrational
numbers. e.g. 2 , 3 , 5 etc.
6. Real numbers : Rational numbers and
irrational numbers together are called real
numbers.
7. Number line and representation of
numbers on a number line : A line on
which numbers are shown, is called a
number line.
We can represent real numbers on the
number line.
8. Decimal expansion of rational numbers:
The decimal of rational numbers are either
terminating or non-terminating recurring
decimals and decimal of irrational numbers
are non-terminating non-recurring.
9. If r is a rational number and s is irrational
then r + s, r – s and s
r are irrational
numbers.
10. Finding rational numbers between two
numbers : We can find an infinite number
of rational numbers between two numbers.
Let a and b are two numbers, then a rational
number between then = 2
ba .
11. Some useful results on irrational
numbers :
(i) Negative of an irrational number is an
irrational number.
(ii) The sum of a rational number and an
irrational number is an irrational number.
(iii) The product of a non-zero rational number
and an irrational number is an irrational
number.
(iv) The sum, difference, product and quotient
of two irrational numbers need not be an
irrational number.
EXERCISE 1.1
1. Is zero a rational number? Can you write it
in the form q
p, where p and q are integers
and q 0? [NCERT]
Sol. Yes, zero is a rational number e.g.
11
0,
12
0,
31
0, etc.
2. Find five rational numbers between 1 and
2. [NCERT]
Sol. We know that one rational number between
two numbers a and b = 2
ba
1
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2 Arundeep’s Mathematics (R.D.) 9th
Therefore one rational number between 1
and 2
= 2
21 =
2
3
Similarly, rational number between 1 and
2
3
= 2
1
2
3
1
1 =
2
1
2
32 =
2
1 ×
2
5 =
4
5
Rational number between 2
3 and 2
= 2
1
1
2
2
3 =
2
1
2
43
= 2
1 ×
2
7 =
4
7
Three rational number between 1 and 2
are 4
5,
2
3,
4
7 or
4
5,
4
6,
4
7
and rational number between 1 and 4
5
= 2
1
4
51
= 2
1
4
54 =
2
1 ×
4
9 =
8
9
and rational number between 4
7 and 2
= 2
1
1
2
4
7 =
2
1
4
87
= 2
1 ×
4
15 =
8
15
Hence five rational numbers are 8
9,
4
5,
2
3,
4
7,
8
15
3. Find six rational numbers between 3 and 4.
[NCERT]
Sol. One rational number between 3 and 4
= 2
1(3 + 4) =
2
7
One rational number between 3 and 2
7
= 2
1
2
73 =
2
1
2
76
= 2
1 ×
2
13 =
4
13
One rational number between 2
7 and 4
= 2
1
1
4
2
7 =
2
1
2
87
= 2
1
2
15 =
4
15
One rational number between 3 and 4
13
= 2
1
4
133 =
2
1
4
1312
= 2
1 ×
4
25 =
8
25
One rational number between 4
15 and 4
= 2
1
1
4
4
15 =
2
1
4
1615
= 2
1 ×
4
31 =
8
31
One rational number between 3 and 4
25
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3 Arundeep’s Mathematics (R.D.) 9th
= 2
1
8
253
= 2
1
8
2524 =
2
1 ×
8
49 =
16
49
Six rational numbers between 3 and 4 are
16
49,
8
25,
4
13,
2
7,
4
15,
8
31
4. Find five rational numbers between 5
3 and
5
4.
Sol. _ Rational number between a and b
= 2
1(a + b)
Rational number between 5
3 and
5
4
= 2
1
5
4
5
3 =
2
1 ×
5
7 =
10
7
Rational number between 5
3 and
10
7
= 2
1
10
7
5
3 =
2
1
10
76
= 2
1 ×
10
13 =
20
13
Rational number between 10
7 and
5
4
= 2
1
5
4
10
7 =
2
1
10
87
= 2
1 ×
10
15 =
20
15 =
4
3
Rational number between 5
3 and
20
13
= 2
1
20
13
5
3 =
2
1
20
1312
= 2
1 ×
20
25 =
40
25 =
8
5
and rational number between 4
3 and
5
4
= 2
1
5
4
4
3 =
2
1
20
1615
= 2
1 ×
20
31 =
40
31
Required rational numbers are
8
5,
20
13,
10
7,
4
3,
40
31
5. Are the following statements true or false?
Give reason for your answer.
(i) Every whole number is a natural number.
[NCERT]
(ii) Every integer is a rational number.
(iii) Every rational number is an integer.
(iv) Every natural number is a whole number.
(v) Every integer is a whole number.
(vi) Every rational number is a whole number.
Sol. (i) False, as 0 is not a natural number.
(ii) True.
(iii) False, as 2
1,
3
1 etc. are not integers.
(iv) True.
(v) False, _ negative natural numbers are not
whole numbers.
(vi) False, _ proper fraction are not whole
numbers.
EXERCISE 1.2
1. Express the following rational numbers as
decimals :
(i)100
42(ii)
500
327
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4 Arundeep’s Mathematics (R.D.) 9th
(iii)4
15
Sol. (i) 100
42 = 0.42
(ii)500
327 =
2500
2327
=
1000
654 = 0.654
(iii)4
15 =
254
2515
=
100
375 = 3.75
2. (i) 3
2(ii) –
9
4
(iii)15
2(iv) –
13
22
(v)999
437(vi)
26
33
Sol. (i) 3
2 = 0.66..... = 6.0
0.66.....3)2.000( 18 20 18 20 18 2
(ii) –9
4 = –(0.444.....) = 4.0
0.444.....9)4.000( 36 40 36 40 36 4
(iii)15
2 = –(0.133.....) = 31.0
0.133.....15)2.000( 15 50 45 50 45 5
(iv) –13
22 = –1
13
9
= –1.692307692307..... = 692307.1
0.692307.692307.....13)9.000000000000( 78 120 117 30 26 40 39 100 91 90 78 120 117 30 26 40 39 1
(v)999
437 = 0.437437..... = 437.0
0.437437.....999)437000000( 3996 3740 2997 7430 6993 4370 3996 3740 2997 7430 6993 437
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5 Arundeep’s Mathematics (R.D.) 9th
(vi)26
33 = 1
26
7 = 1.2692307692307.....
= 6923072.1
0.2692307692307.....26)7.000000000000( 52 180 156 240 234 60 52 80 78 200 182 180 156 240 234 60 52 80 78 200 182 18
3. Look at several examples of rational
numbers in the form q
p(q 0), where p
and q are integers with no common factors
other than 1 and having terminating decimal
representations. Can you guess what
property q must satisfy?
Sol. The property is if the denominators have
factors 2 or 5 or both, the decimal
representation will be terminating e.g.
2
1 = 0.5,
4
1 = 0.25,
5
1 = 0.2,
10
1 = 0.1,
20
1 = 0.5,
8
3 = 0.125,
40
3 = 0.075 etc.
EXERCISE 1.3
1. Express each of the following decimals in
the form q
p:
(i) 0.39 (ii) 0.750
(iii) 2.15 (iv) 7.010
(v) 9.90 (vi) 1.0001
Sol. (i) 0.39 = 100
39
(ii) 0.750 = 1000
750 =
4
3
(iii) 2.15 = 100
215 =
20
43
(iv) 7.010 = 71000
10 = 7
100
1 =
100
701
(v) 9.90 = 100
990 =
10
99
(vi) 1.0001 = 10000
10001
2. Express each of the following decimals in
the form q
p:
(i) 4.0 (ii) 37.0
(iii) 54.0 (iv) 621.0
(v) 3.125 (vi) 7.4
(vii) 74.0 [NCERT]
Sol. (i) 4.0
Let x = 4.0 = 0.4444... ...(i)
10x = 4.4444... ...(ii)
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6 Arundeep’s Mathematics (R.D.) 9th
Subtracting (i) from (ii)
9x = 4 x = 9
4
4.0 = 9
4
(ii) 37.0
Let x = 37.0 = 0.373737... ...(i)
100x = 37.373737... ...(ii)
Subtracting (i) from (ii)
99x = 37 x = 99
37
37.0 = 99
37
(iii) 54.0
Let x = 54.0 = 0.545454... ...(i)
100x = 54.545454... ...(ii)
Subtracting (i) from (ii)
99x = 54 x = 99
54
x = 11
6
54.0 = 11
6
(iv) 621.0
Let x = 621.0 = 0.621621621... ...(i)
1000x = 621.621621621... ...(ii)
Subtracting (i) from (ii)
999x = 621 x = 999
621
x = 37
23
(Dividing the numerator and denominator
by 27)
(v) 3.125
Let x = 3.125 = 125.333... ...(i)
10x = 1253.333... ...(ii)
Subtracting (i) from (ii)
9x = 1253 – 125 = 1128
x = 9
1128 =
3
376
3.125 = 3
376
(vi) 7.4
Let x = 7.4 = 4.777... ...(i)
10x = 47.777... ...(ii)
Subtracting (i) from (ii)
9x = 47 – 4 = 43
x = 9
43
7.4 = 9
43
(vii) 74.0
Let x = 74.0
10x = 7.4 = 4.777... ...(i)
and 100x = 47.777... ...(ii)
Subtracting (i) from (ii)
90x = 43 x = 90
43
74.0 = 90
43
EXERCISE 1.4
1. Define an irrational number.
Sol. A number which cannot be expressed in
the form of q
p where p and q are integers
and q 0 is called an irrational number.
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7 Arundeep’s Mathematics (R.D.) 9th
2. Explain, how irrational numbers differ from
rational numbers?
Sol. A rational number can be expressed in either
terminating decimal or non-terminating
recurring decimals but an irrational number
is expressed in non-terminating non-
recurring decimals.
3. Examine, whether the following numbers
are rational or irrational:
(i) 7 (ii) 4
(iii) 2 + 3 (iv) 3 + 2
(v) 3 + 5 (vi) 2)22(
(vii) )22( )22( (viii) 2)32(
(ix) 5 – 2 (x) 23 [NCERT]
(xi) 225 [NCERT] (xii) 0.3796 [NCERT]
(xiii) 7.478478..... [NCERT]
(xiv) 1.101001000100001..... [NCERT]
Sol. (i) 7
It is an irrational as 7 is not a perfect square.
(ii) 4
It is a rational number as 4 is a perfect
square of 2.
(iii) 2 + 3
It is an irrational number as sum of a rational
number and an irrational number is also an
irrational number.
(iv) 3 + 2
Irrational as sum of two irrational numbers
is also an irrational number.
(v) 3 + 5 is an irrational number as sum of
two irrational numbers is also an irrational.
(vi) 2)22( = 2 + 4 + 2 2 × 2 = 6 + 4 2
{_ (a – b)2 = a2 + b2 – 2ab)}
It is an irrational number as sum of a rational
and an irrational number is an irrational
number.
(vii) )22( )22( = (2)2 – ( 2 )2
{_ (a + b) (a – b) = a2 – b2}
= 4 – 2 = 2
Which a rational number
(viii) 2)32( = 2 + 3 + 2 2 3
{_ (a + b)2 = a2 + b2 + 2ab}
= 5 + 2 6
Which is an irrational number as sum of a
rational and an irrational number is an
irrational number.
(ix) 5 – 2 is an irrational number as difference
of an irrational number and a rational
number is also an irrational number.
(x) 23 is an irrational number as 23 is not
a perfect square.
(xi) 225 = 15
Which is a rational number.
(xii) 0.3796 is a rational number its decimal is
terminating.
(xiii) 7.478478..... = 478.7
Which is non-terminating recurring decimal.
Therefore it is a rational number.
(xiv) 1.101001000100001.....
It is an irrational number as its decimal is
non-terminating non-recurring decimal.
4. Identify the following as rational or irrational
numbers. Give the decimal representation
of rational numbers:
(i) 4 (ii) 3 18
(iii) 44.1 (iv)27
9
(v) – 64 (vi) 100
Sol. (i) 4 = 2
It is a rational number.
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8 Arundeep’s Mathematics (R.D.) 9th
(ii) 3 18 = 3 × 29 = 3 × 3 2 = 9 2
It is an irrational number.
(iii) 44.1 = 2.12.1 = 1.2
It is a rational number.
(iv)27
9 =
3
1 =
3
1
= 33
31
=
3
3
It is an irrational number.
(v) – 64 = – 88 = –8
It is a rational number.
(vi) 100 = 1010 = 10
It is a rational number.
5. In the following equations, find which
variables x, y, z etc. represent rational or
irrational numbers:
(i) x2 = 5 (ii) y2 = 9
(iii) z2 = 0.04 (iv) u2 = 4
17
(v) v2 = 3 (vi) w2 = 27
(vii) t2 = 0.4
Sol. (i) x2 = 5 x2 = (+ 5 )2
x = 5
Which is an irrational number.
(ii) y2 = 9 y2 = (3)2
y = 3
Which is a rational number.
(iii) z2 = 0.04 = (0.2)2
z = 0.2
Which is a rational number.
(iv) u2 = 4
17 =
2
2
17
u = 2
17
Which is an irrational number.
(v) v2 = 3 = ( 3 )2
v2 = 3
Which is an irrational number.
(vi) w2 = 27 = 9 × 3 = (3 3 )2
w = 3 3
Which is an irrational number.
(vii) t2 = 0.4 t2 = 10
4 =
2
10
2
t = 10
2 =
1010
102
= 10
102 =
5
10
Which is an irrational number.
6. Given two rational numbers lying between
0.232332333233332... and
0.212112111211112.
Sol. Two rational numbers lying between
0.232332333233332... and
0.212112111211112...
will be 0.232 and 0.212
7. Give two rational numbers lying between
0.515115111511115... and 0.5353353335...
Sol. Two rational numbers lying between
0.515115111511115... and
0.535335333533335...
will be 0.515, 0.535
8. Find one irrational numbers between 0.2101
and 0.2222... = 2.0 .
Sol. One irrational number lying between 0.2101
and 0.2222... = 2.0 will be 2201.0010001...
9. Find a rational number and also an irrational
number lying between the numbers,
0.3030030003... and 0.3010010001...
Sol. Between two numbers 0.3030030003... and
0.3010010001..., a rational will be 0.301 and
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9 Arundeep’s Mathematics (R.D.) 9th
irrational number will be 0.3020020002...
10. Find three different irrational numbers
between the rational numbers 7
5 and
11
9.
[NCERT]
Sol. Let a = 7
5 = 0.714285714285
= 714285.0
and b = 11
9 = 0.8181... = 81.0
0.714285717)5.00000000( 49 10 7 30 28 20 14 60 56 40 35 50 49 10 7 3
0.818111)9.0000( 88 20 11 90 88 20 11 9
We see that a has six digits and b has two
digits, i.e. 0.71 < 0.81
7
5 <
11
9
First irrational number
will be 0.72020020002...
and 0.74020020002...
and 0.76020020002...
11. Give an example of each, of two irrational
numbers whose:
(i) difference is a rational number.
(ii) difference is an irrational number.
(iii) sum is a rational number.
(iv) sum is an irrational number.
(v) product is a rational number.
(vi) product is an irrational number.
(vii) quotient is a rational number.
(viii)quotient is an irrational number.
Sol. (i) Two numbers whose difference is also
a rational number. e.g. 2 , 2 which are
irrational numbers.
Difference = 2 – 2 = 0 which is also a
rational number.
(ii) Two numbers whose difference is an
irrational number.
e.g. 3 and 2 which are irrational
numbers.
Now difference = 3 – 2 which is also
an irrational number.
(iii) Let two irrational numbers be 3 and
– 3 which are irrational numbers.
Now sum = 3 + (– 3 ) = 3 – 3 = 0
Which is a rational number.
(iv) Let two numbers be 5 , 3 which are
irrational numbers.
Now sum = 5 + 3 which is an irrational
number.
(v) Let numbers be 3 + 2 and 3 – 2
which are irrational numbers.
Now product = ( 3 + 2 ) ( 3 – 2 )
= 3 – 2 = 1 which is a rational number.
(vi) Let numbers be 3 and 5 , which are
irrational number.
Now product = 3 × 5 = 53 = 15
which is an irrational number.
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10 Arundeep’s Mathematics (R.D.) 9th
(vii) Let numbers be 6 2 and 2 2 which are
irrational numbers.
Quotient = 22
26 = 3 which is a rational
number.
(viii)Let numbers be 3 and 5 which are
irrational numbers.
Now quotient = 5
3 =
5
3 which is an
irrational number.
12. Find two irrational numbers between 0.5
and 0.55.
Sol. Two irrational numbers between 0.5 and
0.55 will be 0.51010010001... and
0.52020020002...
13. Find two irrational numbers lying between
0.1 and 0.12.
Sol. Two irrational numbers lying between 0.1
and 0.12 will be 0.1010010001... and
0.1020020002...
14. Prove that 3 + 5 is an irrational number..
Sol. Let 3 + 5 be a rational number,,
and Let x = 3 + 5
Squaring both sides,
x2 = ( 3 + 5 )2 = 3 + 5 + 2 × 3 × 5
x2 = 8 + 2 15
x2 – 8 = 2 15
2
82 x
= 15
_ x is rational
2
82 x
will be rational
15 will be rational
But it is not possible as 15 is an irrational
number.
Our supposition is wrong
Hence 3 + 5 is an irrational number..
EXERCISE 1.5
1. Complete the following sentences:
(i) Every point on the number line corresponds
to a ... number which many be either ... or
...
(ii) The decimal form of an irrational number
is neither ... nor ...
(iii) The decimal representation of a rational
number is either ... or ...
(iv) Every real number is either ... number or
... number.
Sol. (i) Every point on the number line
corresponds to a real number which many
be either rational or irrational.
(ii) The decimal form of an irrational number
is neither terminating nor repeating.
(iii) The decimal representation of a rational
number is either terminating or non-
terminating, recurring.
(iv) Every real number is either rational number
or an irrational number.
2. Find whether the following statements are
true or false:
(i) Every real number is either rational or
irrational .
(ii) is an irrational number.
(iii) Irrational numbers cannot be represented
by points on the number line.
Sol. (i) True.
(ii) True. (Value of = 3.14)
(iii) False : we can represent irrational number
also.
3. Represent 6 , 7 , 8 on the number
line.
Sol. (i) 6 =
22
2
16
2
16
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11 Arundeep’s Mathematics (R.D.) 9th
=
22
2
5
2
7
= 22 )5.2()5.3(
Steps of construction :
(a) Take BC = 2.5 cm.
(b) At C, draw a ray BX making an angle of
90º.
(c) With centre B and radius 3.5 cm draw an
arc which intersects CX at A.
(d) Join AB.
Now, AC = 22 BCAB
= 22 )5.2()5.3(
= 25.625.12 = 6
X
A
CB
3.5
cm
2.5 cm
6 c
m
(ii) 7 =
22
2
17
2
17
=
22
2
6
2
8
= (4)2 – (3)2
Steps of construction :
(a) Draw BC = 3 cm.
(b) At C, draw a ray CX making an angle of
90º.
(c) With centre B and radius 4 cm, draw an
arc which intersects CX at A.
(d) Join AB.
Now AC = 22 BCAB
= 22 34
= 916 = 7
X
A
CB
4 cm
5 cm
7 c
m
(iii) 8 =
22
2
18
2
18
=
22
2
7
2
9
= 22 )5.3()5.4(
Steps of construction :
(a) Draw a line segment BC = 3.5 cm.
(b) At C, draw a ray CX making an angle of
90º.
X
A
CB
4.5
cm
3.5 cm
8 c
m
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12 Arundeep’s Mathematics (R.D.) 9th
(c) With centre B and radius 4.5 cm, draw an
arc which intersects CX at A.
(d) Join AB.
AC = 22 BCAB = 22 )5.3()5.4(
= 5.1225.20 = 8
4. Represent 5.3 , 4.9 and 5.10 on the
real number line.
Sol. (i) 5.3
=
22
2
15.3
2
15.3
=
22
2
5.2
2
5.4
= 22 )25.1()25.2(
Steps of construction :
(a) Draw a line segment BC = 1.25 cm.
(b) At C, draw a ray CX making an angle of
90º
(c) With centre B and radius 2.25 cm, draw an
arc which intersects CX at A.
(d) Join AB.
AC = 22 BCAB = 22 )25.1()25.2(
= 5625.10625.5 = 5.3
X
A
CB
2.25 cm
1.25 cm
3.5
(ii) 4.9
=
22
2
14.9
2
14.9
=
22
2
4.8
2
4.10
= 22 )2.4()2.5(
Steps of construction :
(a) Draw a line segment BC = 4.2 cm.
(b) At C, draw a ray CX making an angle of
90º.
(c) With centre B and radius 5.2 cm draw an
arc which intersects CX at A.
(d) Join AB.
AC = 22 BCAB = 22 )2.4()2.5(
= 64.1704.27 = 40.9 = 4.9
X
A
CB
5.2
cm
4.2 cm
9.4
(iii) 5.10
=
22
2
15.10
2
15.10
=
22
2
5.9
2
5.11
= 22 )75.4()75.5(
Steps of construction :
(a) Draw a line segment BC = 5.75 cm.
(b) At C, draw a ray CX making an angle of
90º.
(c) With centre B and radius 5.75 cm, draw an
arc intersecting CX at A.
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13 Arundeep’s Mathematics (R.D.) 9th
(d) Join AB.
AB = 22 )75.4()75.5(
= 5625.220625.33
= 5000.10 = 5.10
X
A
CB
5.75
cm
4.75 cm10.5
EXERCISE 1.6
1. Visualise 2.665 on the number line, using
successive magnification.
Sol. 2.665
_ It lies between 2 and 3
0 1 2 3 4 5 61
2.660
A A1
2.6 2.7
B B1
2.61
2.61
2.62
2.63
2.64
2.65
2.66
2.67
2.68
2.69
2.67
2.6702.6692.660 2.661 2.662 2.663 2.664 2.665 2.666 2.667 2.668
C C1
2. Visualise the representation of 73.5 on the
number line upto 5 decimal places, that is
upto 5.37777. [NCERT]
Sol. 5.37777
0 1 2 3 4 5 61
5.1
7
65
5.2 5.3 5.4 5.5 5.6 5.7 5.8 5.9
8
Multiple Choice Questions (MCQs)
Mark the correct alternative in each of the
following:
1. Which one of the following is a correct
statement?
(a) Decimal expansion of a rational number is
terminating
(b) Decimal expansion of a rational number is
non-terminating
(c) Decimal expansion of an irrational number
is terminating
(d) Decimal expansion of an irrational number
is non-terminating and non-repeating
Sol. Decimal expansion of an irrational number
is non-terminating and non-repeating (d)
2. Which one of the following statements is
true?
(a) The sum of two irrational numbers is always
an irrational number
(b) The sum of two irrational numbers is always
a rational number
(c) The sum of two irrational numbers may be
a rational number or an irrational number
(d) The sum of two irrational numbers is always
an integer
Sol. The sum of two irrational numbers may be
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14 Arundeep’s Mathematics (R.D.) 9th
a rational number or an irrational number
(c)
3. Which of the following is a correct
statement?
(a) Sum of two irrational numbers is always
irrational
(b) Sum of a rational and irrational number is
always an irrational number
(c) Square of an irrational number is always a
rational number
(d) Sum of two rational numbers can never be
an integer
Sol. Sum of a rational and irrational number is
always an irrational number (b)
4. Which of the following statements is true?
(a) Product of two irrational numbers is always
irrational
(b) Product of a rational and an irrational
number is always irrational
(c) Sum of two irrational numbers can never
be irrational
(d) Sum of an integer and a rational number
can never be an integer
Sol. Product of a rational and an irrational
number is always irrational (b)
5. Which of the following is irrational?
(a)9
4(b)
5
4
(c) 7 (d) 81
Sol. 7 is irrational number
_
9
4 =
3
2 and 81 = 9 (c)
6. Which of the following is irrational?
(a) 0.14 (b) 1614.0
(c) 1416.0 (d) 0.1014001400014
Sol. 0.1014001400014..... is irrational as it is
non-terminating nor repeating decimal. (d)
7. Which of the following is rational?
(a) 3 (b)
(c)0
4(d)
4
0
Sol. _ 3 , are irrational and 0
4 is meaningless
4
0 is a rational which is equal to 0. (d)
8. The number 0.318564318564318564..... is:
(a) a natural number
(b) an integer
(c) a rational number
(d) an irrational number
Sol. The number = 0.318564318564318564.....
= 318564.0
_ The decimal is non-terminating and recurring
It is rational number. (c)
9. If n is a natural number, then n is
(a) always a natural number
(b) always a rational number
(c) always an irrational number
(d) sometimes a natural number and sometimes
an irrational number
Sol. If n is a natural number then n may
sometimes a natural number and sometime
an irrational number e.g.
If n = 2 then n = 2 which is are
irrational and if n = 4, then n = 4 = 2
which is a rational number. (d)
10. Which of the following numbers can be
represented as non-terminating, repeating
decimals?
(a)24
39(b)
16
3
(c)11
3(d)
25
137
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15 Arundeep’s Mathematics (R.D.) 9th
Sol.11
3 as its denominator has no factor of 2
or 5 or both, so it has non-terminating nor
repeating.
In 24
39 =
8
13,
16
3,
25
137, all have
terminating decimals. (c)
11. Every point on a number line represents
(a) a unique real number
(b) a natural number
(c) a rational number
(d) an irrational number
Sol. Every point on a number line represents a
unique real number. (a)
12. Which of the following is irrational?
(a) 0.15 (b) 0.01516
(c) 1516.0 (d) 0.5015001500015..
Sol. As it is non-terminating non-repeating
decimals while others are terminating or
non-terminating repeating decimals. (d)
13. The number 27.1 in the form q
p, where p
and q are integers and q 0, is
(a)9
14(b)
11
14
(c)13
14(d)
15
14
Sol. Let x = 27.1 = 1.272727.....
100x = 127.272727.....
Subtracting, 99x = 126 x = 99
126 =
11
14
(b)
14. The number 3.0 in the form q
p, where p
and q are integers and q 0, is
(a)100
33(b)
10
3
(c)3
1(d)
100
3
Sol. 3.0 = 9
3 =
3
1(c)
15. 23.0 when expressed in the form q
p (p, q
are integers q 0), is
(a)25
8(b)
90
29
(c)99
32(d)
199
32
Sol. 23.0
Let x = 23.0 = 0.3222.....
10x = 3.222.....
100x = 32.2222.....
Subtracting,
90x = 29 x = 90
29
23.0 = 90
29(b)
16. 43.23 when expressed in the form q
p(p, q
are integers q 0), is
(a)99
2320(b)
100
2343
(c)999
2343(d)
199
2320
Sol. 43.23
Let x = 43.23 = 23.434343.....
100x = 2343.434343.....
Subtracting,
99x = 2320 x = 99
2320(a)
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16 Arundeep’s Mathematics (R.D.) 9th
17. 001.0 when expressed in the form q
p(p, q
are integers, q 0), is
(a)1000
1(b)
100
1
(c)1999
1(d)
999
1
Sol. 001.0
Let x = 001.0 = 0.001001001......
1000x = 1.001001001.....
Subtracting,
999x = 1
x = 999
1(d)
18. The value of 23.0 + 22.0 is
(a) 45.0 (b) 43.0
(c) 45.0 (d) 0.45
Sol. 23.0 + 22.0
0.232323..... + 0.222222.....
= 0.454545..... = 45.0 (a)
19. An irrational number between 2 and 2.5 is
(a) 11 (b) 5
(c) 5.22 (d) 5.12
Sol. An irrational number between 2 and 2.5 is
5 as it has approximate value 2.236...
(b)
20. The number of consecutive zeros in 23 × 34
× 54 × 7, is
(a) 3 (b) 2
(c) 4 (d) 5
Sol. In 23 × 34 × 54 × 7, number of consecutive
zero will be 3 as 23 × 54 = 2 × 2 × 2 × 5 ×
5 × 5 × 5 = 5000 (a)
21. The smallest rational number by which 3
1
should be multiplied so that its decimal
expansion terminates after one place of
decimal, is
(a)10
1(b)
10
3
(c) 3 (d) 30
Sol. If 3
1 be multiplied by
10
3, then we get
3
1 ×
10
3 =
10
1 = 0.1 (a)
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17 Arundeep’s Mathematics (R.D.) 9th
Points to Remember :
For any real number a and na positive integer n we define an as
a × a × a × a ..... × a (n times), then
an is called nth power of a.
a is called the base and n is called index, or exponent of nth power of a.
Note = a0 = 1
(i) am × an = am + n (ii) am an = am – n, m > n
(iii) a–m = ma
1(iv) (am)n = amn
(v) (ab)m = am.bm (vi)
m
b
a
= m
m
b
a
(vii) a0 = 1
If a is a positive real number and n is positive integer, then nth root of a is denoted by na
1
and
is written as n a
Note : n
m
a = n ma
EXERCISE 2.1
1. Simplify the following:
(i) 3 (a4b3)10 × 5(a2b2)3 (ii) (2x–2y3)3 (iii) 4
57
108
)106()104(
(iv) 22
32
10
)5(4
ba
abab (v)
n
ba
yx
32
22
(vi) 42
693 )(
n
n
a
a
Sol. (i) 3 (a4b3)10 × 5(a2b2)3
= 3a4 × 10 b3 × 10 × 5a2 × 3 × b2 × 3
nmnm
mnnm
aaa
aa )(
= 3a40b30 × 5a6b6
= 3 × 5 × a40 + 6 × b30 + 6
= 15a46b36
(ii) (2x–2y3)3 = 23 × x–2 × 3 × y3 × 3
nmnm
mnnm
aaa
aa )(
EXPONENTS OF REAL NUMBERS2
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18 Arundeep’s Mathematics (R.D.) 9th
= 8 × x–6 × y9 = 8x–6y9
(iii) 4
57
108
)106()104(
= 8
64 107 – 5 – 4
10a
aaa nmnm
= 3 10–2 = 210
3 =
1010
3
= 100
3
(iv) 22
32
10
)5(4
ba
abab =
10
)5(4 × a1 + 1 – 2 b2 + 3 – 2
10a
aaa nmnm
= –2 × a0b3 = –2 × 1 × b3
= –2b3
(v)
n
ba
yx
32
22
= nn
nn
ba
yx32
22 = nn
nn
ba
yx32
22 ·{_ (am)n = amn}
(vi) 42
693 )(
n
n
a
a = 42
6)93(
n
n
a
a
mnnm
mnnm
aa
aaa
)(
= 42
5418
n
n
a
a = a18n – 54 – 2n + 4 = a16n – 50
2. If a = 3 and b = –2, find the values of:
(i) aa + bb (ii) ab + ba (iii) (a + b)ab
Sol. a = 3 and b = –2
(i) aa + bb = a3 + b–2
= (3)3 + (–2)–2
= 27 + 2)2(
1
= 27 + )2(2
1
= 27 + 4
1 =
4
1108 =
4
109
(ii) ab + ba = (3)–2 + (–2)3
= 23
1 + (–2) × (–2) × (–2)
= 33
1
+ (–8) =
9
1 – 8 =
9
721 =
9
71
(iii) (a + b)ab = [3 + (–2)]3 × (–2)
= (3 – 2)–6 = (1)–6 = 6)1(
1 =
1
1 = 1
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19 Arundeep’s Mathematics (R.D.) 9th
3. Prove that:
(i)
22 baba
b
a
x
x
×
22 cbcb
c
b
x
x
×
22 acac
a
c
x
x
= 1
(ii)
c
b
a
x
x
×
a
c
b
x
x
×
b
a
c
x
x
= 1
(iii)
22 baba
b
a
x
x
×
22 cbcb
c
b
x
x
×
22 acac
a
c
x
x
= 1
Sol. (i)
22 baba
b
a
x
x
×
22 cbcb
c
b
x
x
×
22 acac
a
c
x
x
= 1
LHS =
22 baba
b
a
x
x
×
22 cbcb
c
b
x
x
×
22 acac
a
c
x
x
= 22
)( bababax × 22
)( cbcbcbx × 22
)( acacacx
= )()(22
bababax × )()(22
cbcbcbx × )()(22
acacacx
= 33
bax · 33
cbx · 33
acx
= 333333
accbbax = x0 = 1 = RHS
(ii)
c
b
a
x
x
×
a
c
b
x
x
×
b
a
c
x
x
= 1
LHS =
c
b
a
x
x
×
a
c
b
x
x
×
b
a
c
x
x
= (xa – b)c × (xb – c)a × (xc – a)b
= x(a – b)c × x(b – c)a × x(c – a)b
= xac – bc × xab – ac × xbc – ab
= xac – bc + ab – ac + bc – ab = x0 = 1 = RHS
(iii)
22 baba
b
a
x
x
×
22 cbcb
c
b
x
x
×
22 acac
a
c
x
x
= 1
LHS =
22 baba
b
a
x
x
×
22 cbcb
c
b
x
x
×
22 acac
a
c
x
x
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20 Arundeep’s Mathematics (R.D.) 9th
= 22
)( bababax × 22
)( cbcbcbx × 22
)( acacacx
= )()(22
bababax × )()(22
cbcbcbx × )()(22
acacacx
= 33
bax · 33
cbx · 33
acx
= 333333
accbbax = )(2333
cbax
4. Prove that:
(i) bax 1
1 + abx 1
1 = 1
(ii) acab xx 1
1 + bcba xx 1
1 + cacb xx 1
1 = 1
Sol. (i) bax 1
1 + abx 1
1 = 1
LHS = bax 1
1 + abx 1
1
= babb xx
1 + abaa xx
1{Put 1 = xb – b and 1 = xa – a}
= )(
1abb xxx +
)(
1baa xxx
= ba
b
xx
x
+ ba
a
xx
x
= )( ba
ab
xx
xx
=
)( ba
ba
xx
xx
= 1 = RHS
(ii) acab xx 1
1 + bcba xx 1
1 + cacb xx 1
1 = 1
LHS = acab xx 1
1 + bcba xx 1
1 + cacb xx 1
1
= acabaa xxx
1 + bcbabb xxx
1 + cacbcc xxx
1
{Put 1 = xa – a, 1 = xb – b and 1 = xc – c}
= )(
1cbaa xxxx +
)(
1cabb xxxx +
)(
1abcc xxxx
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21 Arundeep’s Mathematics (R.D.) 9th
= cba
a
xxx
x
+ cba
b
xxx
x
+ cba
c
xxx
x
= cba
cba
xxx
xxx
= 1 = RHS
5. Prove that:
(i) 111111
accbba
cba = abc (ii) (a–1 + b–1)–1 =
ba
ab
Sol. (i) 111111
accbba
cba = abc
LHS = 111111
accbba
cba
m
m
aa
1
=
cabcab
cba
111
=
abc
bac
cba
= )(
)(
cba
abccba
= abc = RHS
(ii) (a–1 + b–1)–1 = ba
ab
LHS = (a–1 + b–1)–1
=
111
ba =
1
ab
ab
m
m
aa
1
= ba
ab
= RHS
6. If abc = 1, show that 11
1 ba
+ 11
1 cb
+ 11
1 ac
= 1
Sol. abc = 1
11
1 ba
+ 11
1 cb
+ 11
1 ac
= 1
LHS = 11
1 ba
+ 11
1 cb
+ 11
1 ac
= b
a1
1
1
+ c
b1
1
1
+ a
c1
1
1
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22 Arundeep’s Mathematics (R.D.) 9th
=
b
abb 1
1
+ abb 1
1 +
aab
111
1
abc
abc
abc
1and
1
1
= 1 abb
b +
abb 1
1 +
bab
ab
1
= abb
abb
1
1 = 1
7. Simplify the following:
(i) 11
1
93
93
nn
nn(ii) 132
21
)25(55
525255
nn
nn
(iii) nn
nn
5259
5652
13
(iv) nn
nn
)8(7)2(10
)2(16)8(613
231
Sol. (i) 11
1
93
93
nn
nn = 121
12
)3(3
)3(3
nn
nn
= 221
22
33
33
nn
nn
= 32n + 2 + n – n + 1 – 2n + 2
= 33n – 3n + 2 + 3 = 35 = 3 × 3 × 3 × 3 × 3 = 243
(ii) 132
21
)25(55
525255
nn
nn
= 1232
2212
)5(55
55)5(5
nn
nn
= 2232
2222
55555
55555
nn
nn
= )555(5
)555(5232
222
n
n
= 24
23
55
55
=
25625
25125
= 600
100 =
6
1
(iii) nn
nn
5259
5652
13
= )29(5
)565(52
13
n
n
= 49
30125
= 5
95 = 19
(iv) nn
nn
)8(7)2(10
)2(16)8(613
231
= nn
nn
)2(7)2(10
)2(16)2(6313
2313
= nn
nn
313
2333
27210
21626
= )7210(2
)21626(213
233
n
n
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23 Arundeep’s Mathematics (R.D.) 9th
= 720
4
11648
=
13
448 =
13
52 = 4
8. Solve the following equations for x:
(i) 72x + 3 = 1 (ii) 2x + 1 = 4x – 3
(iii) 25x + 3 = 8x + 3 (iv) 42x = 32
1
(v) 4x – 1 × (0.5)3 – 2x =
x
8
1
(vi) 23x – 7 = 256
Sol. (i) 72x + 3 = 1 = 70
(_ a0 = 1)
Comparing, we get
2x + 3 = 0 2x = –3
x = 2
3
(ii) 2x + 1 = 4x – 3
2x + 1 = (22)x – 3
2x + 1 = 22x – 6
Comparing, we get
x + 1 = 2x – 6
1 + 6 = 2x – x x = 7
x = 7
(iii) 25x + 3 = 8x + 3
25x + 3 = (23)x + 3
25x + 3 = 23x + 9
Comparing, we get
5x + 3 = 3x + 9
5x – 3x = 9 – 3
3x = 6 x = 3
6 = 2
x = 2
(iv) 42x = 32
1 (22)2x = 52
1
24x = 2–5
Comparing, we get
4x = –5 x = 4
5
x = 4
5
(v) 4x – 1 × (0.5)3 – 2x =
x
8
1
(22)x – 1 ×
x23
2
1
=
x
32
1
22x – 2 × 2–3 + 2x = 2–3x
22x – 2 – 3 + 2x = 2–3x
24x – 5 = 2–3x
Comparing, we get
4x – 5 = –3x 4x + 3x = 5
7x = 5 x = 7
5
x = 7
5
(vi) 23x – 7 = 256
23x – 7 = 28
Comparing, we get
3x – 7 = 8
3x = 8 + 7 = 15
x = 3
15 = 5
x = 5
9. Solve the following equations for x:
(i) 22x – 2x + 3 + 24 = 0 (ii) 32x + 4 +
1 = 2 · 3x + 2
Sol. (i) 22x – 2x + 3 + 24 = 0
(2x)2 – 23 × (2)x + 24 = 0
(2x)2 – 8(2)x + 16 = 0
(2x)2 – 2 × 2x × 4 + (4)2 = 0
(2x – 4)2 = 0 2x – 4 = 0
2x = 4 = 22
Comparing, we get
x = 2
(ii) 32x + 4 + 1 = 2 · 3x + 2
32x × 34 + 1 – 2 · 3x × 32 = 0
81 · 32x – 2 × 9(3x) + 1 = 0
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24 Arundeep’s Mathematics (R.D.) 9th
81(3x)2 – 18(3x) + 1 = 0
[9(3x)]2 – 2 × 9 × 3x + (1)2 = 0
[9(3x) – 1]2 = 0
9(3)x – 1 = 0 9(3x) = 1
3x = 9
1 = 23
1 = 3–2
Comparing, we get
x = –2
10. If 49392 = a4b2c3, find the values of a, b
and c, where a, b and c are different positive
primes.
Sol. 49392 = 2a × 3b × 7c
24 × 32 × 73 = a4b2c3
2 493922 246962 123482 61743 30873 10297 3437 497 7
1
Comparing, we get
a = 4, b = 2, c = 7
11. If 1176 = 2a × 3b × 7c, find a, b and c.
Sol. 1176 = 2a × 3b × 7c
23 × 31 × 72 = 2a × 3b × 7c
2 11762 5882 2943 1477 497 7
1
Comparing, we get
a = 3, b = 1 = c = 2
12. Given 4725 = 3a 5b 7c, find:
(i) the integral values of a, b and c
(ii) the value of 2–a 3b 7c
Sol. 4725 = 3a · 5b · 7c
(i) 33 · 52 · 71 = 3a · 5b · 7c
Comparing, we get
a = 3, b = 2, c = 1
(ii) 2–a 3b 7c = 2–3 · 32 · 71
= 32
1 × 32 × 71
= 222
733
= 8
63
13. If a = xyp – 1, b = xyq – 1 and c = xyr – 1, prove
that aq – r br – p cp – q = 1.
Sol. a = xyp – 1, b = xyq – 1 and c = xyr – 1
aq – r br – p cp – q = 1
LHS = aq – r br – p cp – q
= (xyp – 1)q – r · (xyq – 1)r – p · (xyr – 1)p – q
= xq – r · y(p – 1) (q – r) · xr – p · y(q – 1) (r – p) · xp –
q · y(r – 1) (p – q)
= xq – r + r – p + p – q · ypq – pr – q – pr · yqr – pq – r + p ·
yrp – rq – p + q
= x0 · ypq – pr – q + r + qr – pq – r + p + rp – rq – p + q
= x0 · y0 = 1 × 1 = 1 = RHS
nmnm
mnnm
aaa
aa
a
)(
10
EXERCISE 2.2
1. Assuming that x, y, z are positive real
numbers, simplify each of the following:
(i)5
3
x (ii) 23 yx
(iii) (x–2/3y–1/2)2
(iv) 3/2)( x 4y 2/1xy
(v) 5 10510243 zyx
(vi)
4/5
10
4
y
x(vii)
5
3
2
2
7
6
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25 Arundeep’s Mathematics (R.D.) 9th
Sol. (i) 5
3
x =
5
2
3
x = 5
2
3
x
= 2
15
x =
2
15
1
x
m
m
mnnm
xx
xx
1
)(
(ii) 23 yx = 2
123 yx
= 2
3
x. 2
2
y = 2
3
x.y–1
m
m
mnnm
xx
xx
1and
)(
= y
x 2
3
(iii)
2
2
1
3
2
yx = 23
2
x.
22
1
y
= 3
4
x.y–1 =
yx 3
4
1
m
m
mnnm
xx
xx
1
)(
(iv) 3/2)( x 4y 2/1xy
=
3
2
2
1
x .
4
2
1
y 2
1
x
2
1
2
1
y
=
3
2
2
1
x.
42
1
y 2
1
x. 2
1
2
1
y
= 3
1
x.y2 2
1
x. 4
1
y
=
2
1
3
1
x.
4
12
y
= 6
32
x. 4
12
y = 6
5
x. 4
9
y
=
6
5
4
9
x
y
(v) 5 10510243 zyx = 5
110510 ..243 zyx
= 5
1105105 ...3 zyx
= 5
15
3 . 5
110
x. 5
15
y . 5
110
z = 3.x2yz2
(vi)
4/5
10
4
y
x =
4
5
4
10
x
y
m
m
xx
1
=
4
54
4
510
x
y = 5
2
25
x
y
(vii)
5
3
2
2
7
6
=
5
3
2
×
2
7
6
=
5
3
2
× 2
2
)7(
)6(
= 2
5
3
×
49
36
=
2
3
2
×
3
2 ×
49
36
= 9
11 ×
3
2 ×
49
36
= 349
216 =
49493
16162
= 7203
512 =
3
1
7203
512
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26 Arundeep’s Mathematics (R.D.) 9th
2. Simplify:
(i) (16–1/5)5/2 (ii) 5 3)32(
(iii) 3 2)343( (iv) (0.001)1/3
(v) 3/44/5
5/32/3
)8()16(
)243()25(
(vi)
8
5
2
13
5
2
(vii)2
7
42
21
75
75
× 2
5
53
32
75
75
Sol. (i) (16–1/5)5/2 = 2
5
5
142
= 2
5
5
14
2
m
m
mnnm
aa
aa
1
)(
= 2–2 = 22
1
= 22
1
=
4
1
(ii) 5 3)32( = 5332
1 =
5
3
32
1 =
5
3
5 )2(
1
=
5
35
2
1
= 32
1 =
222
1
=
8
1
(iii) 3 2)343( = 32343
1 =
3
2
343
1
=
3
1
3 )7(
1 =
3
23
7
1
= 27
1 =
77
1
=
49
1
(iv) (0.001)1/3 = 3
13)1.0( = 3
13
)1.0(
= (0.1)1 = 0.1
(v) 3/44/5
5/32/3
)8()16(
)243()25(
=
3
4
34
5
4
5
3
52
3
2
)2()2(
)3()5(
=
3
43
4
54
5
35
2
32
22
35
= 45
33
22
35
= 222222222
333555
= 512
3375
(vi)
8
5
2
13
5
2
=
138
5
2
=
5
5
2
= 5
5
5
)2(
= 5
5
)2(
5
= 2
5
5
2
5
= 22222
55555
= 222
3125
=
24
3125
(vi)2
7
42
21
75
75
× 2
5
53
32
75
75
=
2
74
2
72
2
72
2
71
75
75
×
2
55
2
53
2
53
2
52
75
75
= 147
72
7
75
75
=
2
25
2
15
2
15
5
75
75
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27 Arundeep’s Mathematics (R.D.) 9th
= 2
1575
2
7
5
× 2
2514
2
157
7
= 2
21
2
25
5 . 2
4021
7 = 2
4
5. 2
1
7
= 52 × 71 = 25 × 7 = 175
3. Prove that:
(i) 353 3 13 5 × 6 653 = 5
3
(ii) 2
3
9 – 3 × 50 – 2
1
81
1
= 15
(iii)
2
4
1
– 3 × 3
2
8 × 40 + 2
1
16
9
= 3
16
(iv)
5
3
5
1
4
1
3
1
2
1
510
432
64
53
5
3
5
7
3
4
= 10
(v)4
1 + (0.01)–1/2 – (27)2/3 =
2
3
(vi) nn
nn
22
221
1
= 2
3
(vii)3
2
125
64
+ 4
1
625
256
1
+
0
3 64
25
=
16
61
(viii) 3/13/432
23
3)15(25/15
9863
= 28 2
(ix)131
10
3
1
2
3
8
3
)1.0()6.0(
= –
2
3
Sol. (i) 353 3 13 5 × 6 653 = 5
3
L.H.S. = 353 3 13 5 × 6 653
= 2
1353 3
113 2
1
5 × 6
1653
=
2
3
2
1
53
2
1
3
1
53 ×
6
16
6
1
53
=
2
3
2
1
53
2
1
3
1
53 × 16
1
53
= 6
1
3
1
2
1
3
. 12
1
2
3
5
= 6
1
3
1
2
1
3 . 2
213
5
= 6
123
3
. 2
24
5
= 6
6
3 × 2
2
5
= 31.5–1
= 5
3 = R.H.S.
(ii) 2
3
9 – 3 × 50 – 2
1
81
1
= 15
L.H.S. = 2
3
9 – 3 × 50 – 2
1
81
1
= 2
323 – 3 × 1 –
2
1
29
1
{_ a0 = 1}
= 2
32
3
– 3 –
2
12
9
1
= 33 – 3 –
1
9
1
= 27 – 3 –
1
9
1
= 27 – 3 – 9 = 15 = R.H.S.
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28 Arundeep’s Mathematics (R.D.) 9th
(iii)
2
4
1
– 3 × 3
2
8 × 40 + 2
1
16
9
= 3
16
L.H.S. =
2
4
1
– 3 × 3
2
8 × 40 + 2
1
16
9
=
22
2
1
– 3 × 3
232 × 1 +
2
12
4
3
=
)2(2
2
1
– 3 × 3
23
2
× 1 +
2
12
4
3
=
4
2
1
– 3 × 22 × 1 +
1
4
3
=
4
1
2
– 3
× 4 + 3
4
m
m
aa
1
=
4
1
2
– 12 + 3
4 = 16 – 12 +
3
4
= 4 + 3
4 =
3
16 = R.H.S.
(iv)
5
3
5
1
4
1
3
1
2
1
510
432
64
53
5
3
5
7
3
4
= 10
L.H.S. =
5
3
5
1
4
1
3
1
2
1
510
432
64
53
5
3
5
7
3
4
=
5
3
5
1
4
1
23
1
2
1
5)52(
)2(32
115
3
2
5
7
3
4
32)2(
53
=
5
3
5
1
5
1
4
12
3
1
2
1
552
232
115
6
5
7
3
4
322
53
=
5
3
5
1
5
1
2
1
3
1
2
1
552
232
×
5
7
3
4
115
6
53
322
= 5
11
5
6
2
1
2
1
2 . 3
41
3
1
3 . 5
7
5
3
5
1
5
= 10
2101255
2
. 3
431
3
. 5
731
5
= 10
1222
2
.34 – 4. 5
38
5
= 10
10
2.30. 5
5
5
= 21.30.51 = 2 × 1 × 5 = 10
= R.H.S.
(v)4
1 + 2
1
)01.0(
– 3
2
)27( = 2
3
L.H.S. = 4
1 + 2
1
)01.0(
– 3
2
)27(
= 2
1
22
1
+
2
12
)1.0( – 3
2
3 )3(
= 2
1
2 )2(
1
+
2
12
)1.0( – 3
23
3
= 12
1 + (0.1)–1 – 32
= 2
1 +
1
1
10
– 32
= 2
1 + 10 – 9 = 1
2
1 =
2
3 = R.H.S.
(vi) nn
nn
22
221
1
= 2
3
L.H.S. = nn
nn
22
221
1
= nn
nn
22.2
2221
1
= )12(2
)21(21
1
n
n
= 12
2
11
= 1
2
3
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29 Arundeep’s Mathematics (R.D.) 9th
= 2
3 = R.H.S.
(vii)3
2
125
64
+ 4
1
625
256
1
+
0
3 64
25
=
16
61
L.H.S. = 3
2
125
64
+ 4
1
625
256
1
+
3 64
25
= 3
2
3
3
5
4
+
4
1
4
4
5
4
1
+
3 3
2
4
5
=
3
25
4
3
3
23
+
4
14
4
14
5
4
1
+
3
1
3
2
1
2
)4(
)5(
= 2
2
5
4
+
5
4
1 +
4
5
= 2
2
4
5 +
4
5 +
4
5
= 16
25 +
4
10 =
16
25 +
16
40
= 16
65 = R.H.S.
(viii)
3
1
3
4
32
23
3)15(25
15
9863
= 28 2
L.H.S. =
3
1
3
4
32
2
1
23
3)15(25
15
)98(63
=
3
1
3
4
3
4
3
1
2
2
1
223
353)25(5
)492()32(3
=
3
1
3
4
3
4
3
1
22
2
1
22
1
223
353)5(5
)7(2323
= 22. 2
1
2. 3
1
3
423
3 . 3
42
3
2
5 .71
= 4 2 × 30.50 × 71
= 4 2 × 1 × 1 × 7
= 28 2 = RHS
(ix)131
10
3
1
2
3
8
3
)1.0()6.0(
= –
2
3
L.H.S. = 131
10
3
1
2
3
8
3
)1.0()6.0(
= 131
1
1
3
2
3
3
8
10
11
=
1
3
8
27
3
8
101 1
=
1
39
9
= 6
9 =
2
3 = RHS
4. Show that:
(i) bax 1
1 + abx 1
1 = 1
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30 Arundeep’s Mathematics (R.D.) 9th
(ii)
ba
abb
abb
baa
baa
x
x
x
x
)(
)(
)(
)(
= 1
(iii)
ca
bax
1
1 ab
cbx
1
1 bc
acx
1
1
= 1
(iv)
ba
ab
ba
x
x
22 cb
bc
cb
x
x
22 ca
ac
ac
x
x
22
=
)(2333
cbax
(v) (xa – b)a + b (xb – c)b + c (xc – a)c + a = 1
(vi) 1
1
11
a
a
aaax = x
(vii)
yx
y
x
a
a
1
1zy
z
y
a
a
2
2xz
x
z
a
a
3
3
= 1
(viii)
ba
b
a
3
3cb
c
b
3
3ac
a
c
3
3 = 1
Sol. (i) bax 1
1 + abx 1
1 = 1
LHS = bax 1
1 + abx 1
1
= babb xx
1 + abaa xx
1
= )(
1abb xxx +
)(
1baa xxx
= ba
b
xx
x
+ ba
a
xx
x
= ba
ab
xx
xx
= ba
ba
xx
xx
= 1 = RHS.
(ii)
ba
abb
abb
baa
baa
x
x
x
x
)(
)(
)(
)(
= 1
LHS =
ba
abb
abb
baa
baa
x
x
x
x
)(
)(
)(
)(
=
ba
abb
abb
aba
aba
x
x
x
x
2
2
2
2
= baabbabbabaaba
xx
2222
= [x–2ab ÷ x–2ab]a + b
= [x–2ab – (–2a)]a + b
= (x–2ab + 2ab)a + b = (x0)a + b = (x)0 = 1 = RHS
(iii)
ca
bax
1
1 ab
cbx
1
1 bc
acx
1
1
= 1
LHS =
ca
bax
1
1 ab
cbx
1
1 bc
acx
1
1
= )(
1
)(
1
cabax
× abcbx
11
× bcacx
11
= ))((
1
)()(
1
)()(
1
bcacabcbcabax
= ))((
1
)()(
1
)()(
1
cbacbacbacbax
= )()()( accbba
baaccb
x
= )()()(
0
accbbax = x0 =
1 = RHS
(iv)
ba
ab
ba
x
x
22 cb
bc
cb
x
x
22 ca
ac
ac
x
x
22
= )(2333
cbax
LHS =
ba
ab
ba
x
x
22 cb
bc
cb
x
x
22 ca
ac
ac
x
x
22
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31 Arundeep’s Mathematics (R.D.) 9th
= baabbax
22 cb
bccbx
22 cacaacx
22
= )()( 22 bababax ·
)()( 22 cbcbcbx
·)()(
22cacaacx
= 33
bax ·
33cbx
·33
acx
= 333333
accbbax =
)(2333
cbax = RHS
(v) (xa – b)a + b (xb – c)b + c (xc – a)c + a = 1
LHS = (xa – b)a + b (xb – c)b + c (xc – a)c + a
= x(a – b) (a + b) · x(b – c) (b + c) · x(c – a) (c + a)
= 22
bax ·22
cbx ·22
acx
= 222222
accbbax = x0 = 1 = RHS
(vi) 1
1
11
a
a
aaax = x
LHS = 1
1
11
a
a
aaax
=
1
1
1
1
a
a
aa
a
x = 11
112
a
a
aa
a
x
= 11
1)1()1(
a
a
aa
aa
x = x1 = x = RHS
(vii)
yx
y
x
a
a
1
1zy
z
y
a
a
2
2xz
x
z
a
a
3
3
= 1
LHS =
yx
y
x
a
a
1
1zy
z
y
a
a
2
2xz
x
z
a
a
3
3
= (ax + 1 – y – 1)x + y × (ay + 2 – z – 2)y + z
.(az + 3 – x – 3)z + x
= (ax – y)x + y · (ay – z)y + z · (az – x)z + x
= a(x – y) (x + y) · a(y – z) (y + z) · a(z – x) (z + x)
= 22
yxa ·22
zya ·22
xza
= 222222
xzzyyxa = a0 = 1 = RHS
(viii)
ba
b
a
3
3cb
c
b
3
3ac
a
c
3
3 = 1
LHS =
ba
b
a
3
3cb
c
b
3
3ac
a
c
3
3
= (3a – b)a + b · (3b – c)b + c · (3c – a)c + a
= 22
3 ba ·22
3 cb ·22
3 ac
= 222222
3 accbba = 30 = 1 = RHS
5. If 2x = 3y = 12z, show that z
1 =
y
1 +
x
2.
Sol. 2x = 3y = 12z, z
1 =
y
1 +
x
2
Let 2x = 3y = 12z = k
Then, 2 = xk
1, 3 = yk
1
, 12 = zk
1
22 × 3 = 3
1
k
2
1
)( xk × )(
1
yk = zk
1
xk
2
· yk
1
= 3
1
k
yxk
12
= zk
1
Comparing both sides,
z
1 =
y
1 +
x
2Hence Proved.
6. If 2x = 3y = 6–z, show that x
1 +
y
1 +
z
1 =
0.
Sol. Let 2x = 3y = 6–z = k, then
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32 Arundeep’s Mathematics (R.D.) 9th
2 = xk
1
, 3 = yk
1
and 6 = zk
1
2 × 3 = zk
1
xk
1
× yk
1
= zk
1
yxk
11
= zk
1
x
1 +
y
1 =
z
1
x
1 +
y
1 +
z
1 = 0 Hence Proved.
7. If ax = by = cz and b2 = ac, then show that
y = xz
zx
2
.
Sol. ax = by = cz = k, then
a = xk
1
, b = yk
1
and c = zk
1
Now, b2 = ac
2
1
)( yk = xk
1
· zk
1
yk
2
= zxk
11
y
2 =
x
1 +
z
1
y
2 =
xy
xz
y = xz
xy
2
8. If 3x = 5y = (75)z, show that z = yx
xy
2.
Sol. Let 3x = 5y = 75z = k
3 = xk
1
, 5 = yk
1
and 75 = zk
1
75 = zk
1
3 × 52 = zk
1
xk
1
× yk
2
= zk
1
yxk
21
= zk
1
x
1 +
y
2 =
z
1
xy
xy 2 =
z
1
z = yx
xy
2
9. If 27x = x3
9, find x.
Sol. 27x = x3
9 (33)x = x
3
32
33x × 3x = 32 33x + x = 32
34x = 32
Comparing, we get
4x = 2 x = 4
2 =
2
1
x = 2
1
10. Find the values of x in each of the following:
(i) 25x 2x = 5 202 (ii) (23)4 = (22)x
(iii)27
125
3
5
5
32
xx
(iv) 5x – 2 × 32x – 3 = 135
(v) 2x – 7 × 5x – 4 = 1250 (vi) 2
12
3 )4(x
= 32
1
(vii) 52x + 3 = 1 (viii)x)13( = 44 – 34 – 6
(ix)
1
5
3
x
= 27
125
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33 Arundeep’s Mathematics (R.D.) 9th
Sol. (i) 25x 2x = 5 202
x
x
2
25
= 5
1
20 )2( 25x – x = 5
20
2
24x = 5
20
2
Comparing, we get
4x = 5
20 x =
45
20
= 1
x = 1
(ii) (23)4 = (22)x
23 × 4 = 22 × x 212 = 22x
Comparing, we get
2x = 12 x = 2
12 = 6
x = 6
(iii)27
125
3
5.
5
32
xx
x
3
5.
x2
3
5
= 3
3
3
5
xx 2
3
5
=
3
3
5
Comparing, we get
–x + 2x = 3 x = 3
x = 3
(iv) 5x – 2 × 32x – 3 = 135
5x.5–2.32x.3–3 = 135
32
2
35
3.5
xx
= 135
5x.32x = 135 × 52 × 33
5x × 3x × 3x = 135 × 25 × 27
(5 × 3 × 3)x = 3 × 3 × 3 × 5 × 5 × 5 × 3 ×
3 × 3
(45)x = (3 × 5 × 3)3 = (45)3
Comparing, we get
x = 3
(v) 2x – 7 × 5x – 4 = 1250
2x.2–7.5x.5–4 = 2 × 5 × 5 × 5 × 5
4752
52
xx
= 2 × 5 × 5 × 5 × 5
(10)x = 2 × 5 × 5 × 5 × 5 × 2 × 2 × 2 × 2
× 2 × 2 × 2 × 5 × 5 × 5 × 5
(10)x = 28 × 58 = (10)8
Comparing, we get
x = 8
(vi) 2
12
3 )4(x
= 32
1
2
12
3
1
2 )2(
x
= 52
1
2
12
3
2
2x
= 2–5
3
2
2
12x = –5
3
4x +
6
2 = –5 8x + 2 = –5 × 6
8x = –30 – 2 x = 8
32 = –4
(vii) 52x + 3 = 1 = 50
(_ a0 = 1)
2x + 3 = 0 2x = –3
x = 2
3
(viii)x)13( = 44 – 34 – 6
= 256 – 81 – 6 = 169 = (13)2
x = 2
Squaring, x = 4
(ix)
1
5
3
x
= 27
125
1
2
1
5
3
x
=
3
3
5
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34 Arundeep’s Mathematics (R.D.) 9th
2
1
5
3
x
=
3
5
3
m
m
aa
1
2
1x = –3
x + 1 = –6 x = –6 – 1 = –7
x = –7
11. If x = 21/3 + 22/3, show that x3 – 6x = 6.
Sol. x = 3
1
2 + 3
2
2 , x3 – 6x = 6
Cubing both sides,
x3 =
3
3
2
3
1
22
= 2 + 22 + 3 × 3
1
2 ×
3
2
2
3
2
3
1
22
x3 = 2 + 4 + 3 × 3
2
3
1
2
× x
x3 = 6 + 3 × 2 × x
x3 = 6x + 6
x3 – 6x = 6 Hence Proved.
12. Determine (8x)x, if 9x + 2 = 240 + 9x.
Sol. (8x)x, if 9x + 2 = 240 + 9x
9x + 2 = 240 + 9x
9x × 92 = 240 + 9x 9x × 81 = 9x + 240
81 × 9x – 9x = 240 80 × 9x = 240
9x = 80
240 = 3 32x = 31
2x = 1 x = 2
1
Now, (8x)x = 2
1
2
18
= 2
1
4 = 2
1
2 )2( =
21 = 2
13. If 3x + 1 = 9x – 2, find the value of 21 + x.
Sol. 3x + 1 = 9x – 2
3x + 1 = (32)x – 2 3x + 1 = 32x – 4
x + 1 = 2x – 4 1 + 4 = 2x – x
x = 5
Now, 21 + x = 21 + 5 = 26 = 64
14. If 34x = (81)–1 and 101/y = 0.0001, find the
value of 2–x + 4y.
Sol. 34x = (81)–1, y
1
10 = 0.0001
34x = (34)–1 34x = 3–4
4x = –4 x = 4
4 = –1, and
y
1
10 = 0.0001 = 10000
1 = 4)10(
1 = (10)–4
y
1 = –4 y =
4
1
Now, 2–x + 4y =
4
14)1(
2
= 21 – 1 = 20 = 1
15. If 53x = 125 and 10y = 0.001 find x and y.
Sol. 53x = 125 and 10y = 0.001
53x = 125 = (5)3
3x = 3 x = 3
3 = 1
and 10y = 0.001 = 1000
1 = 310
1 = 10–3
y = –3
and x = 1, y = –3
16. Solve the following equations:
(i) 3x + 1 = 27 × 34
(ii) 42x = y/63 )16( = 2)8(
(iii) 3x – 1 × 52y – 3 = 225
(iv) 8x + 1 = 16y + 2 and,
x
3
2
1 =
y3
4
1
(v) 4x – 1 × (0.5)3 – 2x =
x
8
1
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35 Arundeep’s Mathematics (R.D.) 9th
(vi)b
a =
x
a
b21
, where a, b are distinct
positive primes.
Sol. (i) 3x + 1 = 27 × 34
3x + 1 = 33 + 4 = 37
x + 1 = 7 x = 7 – 1 = 6
x = 6
(ii) 42x = y/63 )16( = 2)8(
42x = 2)8( = 22
1
8 = 8 = (2)3
(22)2x = 23 22 × 2x = 23
24x = 23
Comparing, we get
4x = 3 x = 4
3
and y
6
3 )16(
= 2)8( = 23
y
6
3 4 )2(
= 23
y
6
3
4
2
= 23
y
6
3
4
2 = 23
y
8
2
= 23
Comparing, we get
y
8 = 3 y =
3
8
(iii) 3x – 1 × 52y – 3 = 225
3x – 1 × 52y – 3 = (15)2 = (3 × 5)2
3x – 1 × 52y – 3 = 32 × 52
Comparing,
3x – 1 = 32 x – 1 = 2 x = 2 + 1 = 3
and,
52y – 3 = 52 2y – 3 = 2 2y = 2 + 3 = 5
y = 2
5
x = 3, y = 2
5
(iv) 8x + 1 = 16y + 2 and,
x
3
2
1 =
y3
4
1
(23)x + 1 = (24)y + 2
23x + 3 = 24y + 8
3x + 3 = 4y + 8
3x – 4y = 8 – 3 = 5 ...(i)
and
x
3
2
1 =
y3
4
1
x
3
2
1 =
y32
2
1
=
y6
2
1
3 + x = 6y x = 6y – 3 ...(ii)
From (i),
3(6y – 3) – 4y = 5 18y – 9 – 4y = 5
14y = 5 + 9 = 14 y = 14
14 = 1
and x = 6 × 1 – 3 = 6 – 3 = 3
x = 3, y = 1
(v) 4x – 1 × (0.5)3 – 2x =
x
8
1
(22)x – 1 ×
x23
2
1
=
x
32
1
22x – 2 × 2–3 + 2x = 2–3x
22x – 2 – 3 + 2x = 2–3x 24x – 5 = 2–3x
4x – 5 = –3x 4x + 3x = 5
7x = 5 x = 7
5
(vi)b
a =
x
a
b21
2
1
b
a =
x
b
a21
2
1 = –1 + 2x 2x = 1 +
2
1 =
2
3
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36 Arundeep’s Mathematics (R.D.) 9th
x = 22
3
=
4
3
x = 4
3
17. If a and b are distinct positive primes such
that 3 46 ba = axb2y, find x and y.
Sol. 3 46 ba = axb2y
3
1
46 )( ba = ax·b2y
3
6
a · 3
4
b = ax·b2y a2· 3
4
b = ax · b2y
Comparing, we get
ax = az x = 2
and 3
4
b = b2y 2y = 3
4
y = 23
4
= 3
2
x = 2, y = 3
2
18. If a and b are different positive primes
such that
(i)
7
42
21
ba
ba ÷
32
53
ba
ba = axby, find x and y.
(ii) (a + b)–1 (a–1 + b–1) = axby, find x + y + 2.
Sol. (i)
7
42
21
ba
ba ÷
32
53
ba
ba = axby
2814
147
·
ba
ba ÷ 32
53
ba
ba
= axby
2814
147
·
ba
ba × 53
32
ba
ba = axby
a–7 – 14 – 2 – 3 · b14 + 28 + 3 + 5 = axby
a–26 × b50 = axby
Comparing, we get
x = –26, y = 50
(ii) (a + b)–1 (a–1 + b–1) = axby, find x + y + 2
(a + b)–1
ba
11 = axby
ba 1
× ab
ab = axby
ab
1 = axby a–1b–1 = ax·by
Comparing,
x = –1, y = –1
Now put the value of x and y in x + y + 2,
x + y + 2 = –1 – 1 + 2 = 0
19. If 2x × 3y × 5z = 2160, find x, y and z.
Hence, compute the value of 3x × 2–y × 5–z.
Sol. 2x × 3y × 5z = 2160
2 21602 10802 5402 2703 1353 453 155 5
1
2x × 3y × 5z = 24 × 33 × 51
Comparing, we get
x = 4, y = 3, z = 1
Now, 3x × 2–y × 5–z
= 34 × 2–3 × 5–1 = 58
81
=
40
81
20. If 1176 = 2a × 3b × 7c, find the values of a,
b and c. Hence, compute the value of 2a ×
3b × 7–c as a fraction.
Sol. 1176 = 2a × 3b × 7c
2 11762 5882 2943 1477 497 7
1
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37 Arundeep’s Mathematics (R.D.) 9th
2a × 3b × 7c = 23 × 31 × 72
Comparing, we get
a = 3, b = 1, c = 2
2a × 3b × 7–c
= 23 × 31 × 7–2 = 8 × 3 × 49
1 =
49
24
21. Simplify:
(i)
ba
c
ba
x
x
cb
a
cb
x
x
ac
b
ac
x
x
(ii)lm
m
l
x
x ×
mn
n
m
x
x ×
nl
l
n
x
x
Sol. (i)
ba
c
ba
x
x
cb
a
cb
x
x
ac
b
ac
x
x
= (xa + b – c)a – b·(xb + c – a)b – c·(xc + a – b)c – a
= x(a – b) (a + b – c)·x(b – c) (b + c – a)·x(c – a) (c + a – b)
= bcacbax 22
·acabcbx 22
·abbcacx 22
= abbcacacabcbbcacbax 222222
= x0 = 1
(ii)lm
m
l
x
x ×
mn
n
m
x
x ×
nl
l
n
x
x
= lmmlx
1
)( × mnnmx
1
)( × nllnx
1
)(
= lm
ml
x
· mn
nm
x
· nl
ln
x
= nl
ln
mn
nm
lm
ml
x
= lmn
lmmnlnlmmnln
x
= lmnx
0
= x0 = 1
22. Show that:
nm
nm
ab
ab
ba
ba
11
11
=
nm
b
a
Sol. nm
nm
ab
ab
ba
ba
11
11
=
nm
b
a
LHS = nm
nm
ab
ab
ba
ba
11
11
= nm
nm
a
ab
a
ab
b
ab
b
ab
11
11
=
n
n
m
m
nm
nm
a
ab
a
ab
bb
abab
)1()1(
·
)1()1(
= nmnm
nmnm
ababbb
aaabab
)1()1(·
)1()1(
= nm
nm
bb
ca
.
. = nm
nm
b
a
=
nm
b
a
= RHS
23. (i) If a = xm + ny1, b = xn + l ym and c = xl + m
yn, prove that am – n bn – 1 cl – m = 1.
(ii) If x = am + n, y = an + 1 and z = al + m, prove
that xm yn zl = xn yl zm.
Sol. (i) a = xm + n·yl, b = xn + lym, c = xl + myn
am – n·bn – l·cl – m = 1
LHS = am – n·bn – l·cl – m
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38 Arundeep’s Mathematics (R.D.) 9th
= (xm + n·yl)m – n·(xn + l·ym)n – l·(xl + m·yn)l – m
= x(m + n) (m – n)·yl(m – n) × x(n + l) (n – l)·ym(n – l)
× x(l + m) (l – m)·yn(l – m)
= 22
nmx · ylm – ln · 22
lnx · ymn – ml · 22
mlx
· ynl – nm
= 222222
mllnnmx · ylm – ln + mn – ml + nl – mn
= x0 · y0 = 1 × 1 = 1 = RHS
(ii) x = am + n · y = an + 1 · z = al + m,
xm yn zl = xn yl zm
LHS = xm · yn · zl = a(m + n)m · a(n + l)n · a(l + m)l
= mnma 2
·nlna 2
·nlla 2
= lmnlmnlnma 222
RHS = xn · yl · zm
= (am + n)n · (yn + l)l · (al + m)m
= 2
nmna ·
2lnla
·2
mlma
= 222 mlmlnlnmna
= nlmnlmnmla 222
LHS = RHS
VERY SHORT ANSWER TYPE QUESTIONS
(VSAQs)
1. Write (625)–1/4 in decimal form.
Sol. 4
1
)625(
= 4
1
4 )5(
= 4
14
5
= 5–1
= 5
1 = 0.2
2. State the product law of exponents:
Sol. xm × xn = xm + n
3. State the quotient law of exponents.
Sol. xm xn = xm – n
4. State the power law of exponents.
Sol. (xm)n = xm × n = xmn
5. If 24 × 42 = 16x, then find the value of x.
Sol. 24 × 42 = 16x
24 × (22)2 = [(2)4]x
24 × 24 = 24x
24 + 4 = 24x 28 = 24x
Comparing, we get
4x = 8 x = 4
8 = 2
x = 2
6. If 3x – 1 = 9 and 4y + 2 = 64, what is the value
of y
x?
Sol. 3x – 1 = 9 3x – 1 = 32
Comparing, we get
x – 1 = 2 x = 2 + 1 = 3
x = 3
and 4y + 2 = 64 (4)y + 2 = (4)3
Comparing, we get
y + 2 = 3 y = 3 – 2 = 1
y = 1
Now y
x =
1
3 = 3
7. Write the value of 3 7 × 3 49 .
Sol. 3 7 × 3 49 = 3 497 = 3
1
)777(
= 3
13
7
= 71 = 7
8. Write 2
1
9
1
× 3
1
)64(
as a rational number..
Sol.2
1
9
1
× 3
1
)64(
=
2
12
3
× 3
1
3)4(
=
2
12
3
1 ×
3
13
4
=
1
3
1
× 4–1
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39 Arundeep’s Mathematics (R.D.) 9th
= 3 × 4
1 =
4
3
9. Write the value of 3 27125 .
Sol. 3 27125 = 3 33 35
= 31
3)35( = 3
13
)15(
= (15)1 = 15
10. For any positive real number x, find the
value of
ba
b
a
x
x
×
cb
c
b
x
x
×
ac
a
c
x
x
Sol.
ba
b
a
x
x
×
cb
c
b
x
x
×
ac
a
c
x
x
= (xa – b)a + b.(xb – c)b + c.(xc – a)c + a
= x(a – b) (a + b).x(b – c) (b + c).x(c – a) (c + a)
= 22
bax .
22cbx
.22
acx
= 222222
accbbax = x0 = 1
11. Write the value of {5(81/3 + 271/3)3}1/4.
Sol.
4
1
3
3
1
3
1
2785
= 4
13
3
1
3
1
4
1
2785
= 4
3
3
1
3
1
4
1
2785
= 4
3
3
1
33
1
34
1
)3()2(5
= 4
34
1
325 = 4
1
5 × 4
3
5
= 4
3
4
1
5
= 51 = 5
12. Simplify :
2
4
1
2
1
)625(
.
Sol.
2
4
1
2
1
)625(
= 2
4
1
2
1
)625(
= 4
1
)625(
= 4
1
4 )5( = 4
14
5
= 51 = 5
13. For any positive real number x, write the
value of
abbax1
)( bccbx1
)( caacx1
)(
Sol. abbax1
)( . bccbx1
)( . caacx1
)(
= ababx1
. bcbcx1
. cacax1
= ab
ab
x. bc
bc
x. ca
ca
x
= x1 × x1 × x1 = x × x × x = x3
14. If (x – 1)3 = 8, what is the value of (x + 1)2?
Sol. (x – 1)3 = 8 (x – 1)3 = 23
Comparing, we get
x – 1 = 2 x = 2 + 1 = 3
x = 3
Now (x + 1)2 = (3 + 1)2 = (4)2 = 16
MULTIPLE CHOICE QUESTIONS (MCQs)
Mark the correct alternative in each of the
following:
1. The value of {2 – 3 (2 – 3)3}3 is
(a) 5 (b) 125
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40 Arundeep’s Mathematics (R.D.) 9th
(c)5
1(d) –125
Sol. {2 – 3 (2 – 3)3}3 = {2 – 3 (–1)3}3
= {2 – 3 × (–1)}3
= (2 + 3)3 = (5)3
= 125 (b)
2. The value of x – yx – y when x = 2 and
y = –2 is
(a) 18 (b) –18
(c) 14 (d) –14
Sol. x = 2, y = –2
x – yx – y = 2 – (–2)2 – (–2)
= 2 – (–2)2 + 2 = 2 – (–2)4
= 2 – (+16) = 2 – 16 = –14 (d)
3. The product of the square root of x with
the cube root of x, is
(a) cube root of the square root of x
(b) sixth root of the fifth power of x
(c) fifth root of the sixth power of x
(d) sixth root of x
Sol. Product of sixth root of = 6 x × 3 x
= 6
1
x × 3
1
x = 3
1
2
1
x
= 6
23
x = 6
5
x = 6 5x (b)
4. The seventh root of x divided by the eighth
root of x is
(a) x (b) x
(c) 56 x (d) 56
1
x
Sol. Seventh root of x = 7 x = 7
1
x
Eighth root of x = 8 x = 8
1
x
8
1
7
1
x
x = 8
1
7
1
x = 56
78
x = 56
1
x = 56 x (c)
5. The square root of 64 divided by the cube
root of 64 is
(a) 64 (b) 2
(c)2
1(d) 3
2
64
Sol. 64 3 64
= 28 3 34 2
1
2 )8( 3
1
3)4(
= 2
12
8
3
13
4
= 8 4
= 4
8 = 2 (b)
6. Which of the following is (are) not equal to
6
1
5
1
6
5
?
(a)6
1
5
1
6
5
(b)
6
1
5
1
6
5
1
(c)30
1
5
6
(d)30
1
6
5
Sol.
6
1
5
1
6
5
=
6
1
5
1
6
5 =
30
1
6
5
(d)
7. When simplified (x–1 + y–1)–1 is equal to
(a) xy (b) x + y
(c)yx
xy
(d)xy
yx
Sol. (x–1 + y–1)–1 = x
1 +
y
1
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41 Arundeep’s Mathematics (R.D.) 9th
= xy
xy =
xy
yx (d)
8. If 8x + 1 = 64, what is the value of 32x + 1?
(a) 1 (b) 3
(c) 9 (d) 27
Sol. 8x + 1 = 64 = 82
Comparing, we get
x + 1 = 2 x = 2 – 1 = 1
Now 32x + 1 = 32 × 1 + 1 = 32 + 1
= 33 = 3 × 3 × 3 = 27 (d)
9. If (23)2 = 4x, then 3x =
(a) 3 (b) 6
(c) 9 (d) 27
Sol. (23)2 = 4x 23 × 2 = (22)x
26 = 22x
Comparing,
2x = 6 x = 2
6 = 3
Now 3x = (3)3 = 3 × 3 × 3 = 27 (d)
10. If x–2 = 64, then 3
1
x + x0 =
(a) 2 (b) 3
(c)2
3(d)
3
2
Sol. x–2 = 64 2
1
x = 64
21
x = (8)2
x
1 = 8 x =
8
1
3
1
x + x0 = 3
1
8
1
+ 1
=
3
1
3
2
1
+ 1 = 3
13
2
1
+ 1
= 2
1 + 1 =
2
3(c)
11. When simplified 3
2
27
1
is
(a) 9 (b) –9
(c)9
1(d) –
9
1
Sol.3
2
27
1
=
3
2
3
3
1
= 3
23
3
1
=
2
3
1
= (–3)2 = (–3) × (–3) = 9 (a)
12. Which one of the following is not equal to
2
13 8
?
(a) 2
13 2
(b) 6
1
8
(c)
2
13 8
1(d)
2
1
Sol. 2
13 8
=
2
1
3
1
3)2(
=
2
1
3
13
2 = 2
1
2
=
2
1
(a) 2
13 2
=
2
1
3
1
)2(
=
2
1
3
1
2
= 6
1
2
(b) 6
1
8
= 6
1
3)2(
= 6
13
2
= 2
1
2
= 2
1
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42 Arundeep’s Mathematics (R.D.) 9th
(c)
2
13 8
1 =
2
1
3
1
3 )2(
1
=
2
1
3
13
2
1
=
2
1
2
1 =
2
1
(d)2
1
We see that (a) is not equal (a)
13. Which one of the following is not equal to
2
3
9
100
?
(a)2
3
100
9
(b)
2
3
9
100
1
(c)10
3 ×
10
3 ×
10
3
(d)9
100
9
100
9
100
Sol.2
3
9
100
=
2
3
2
3
10
=
2
32
3
10 =
3
3
10
=
3
10
3
(a)2
3
100
9
= 2
32
10
3
=
3
10
3
(b)
2
3
9
100
1
=
2
32
3
10
1
= 3
3
10
1
=
3
10
3
(c)10
3 ×
10
3 ×
10
3 =
3
10
3
(d)9
100
9
100
9
100 =
3
9
100
= 2
3
9
100
= 2
32
3
10
=
3
3
10
= 9
100
9
100
9
100
(d)
14. If a, b, c are positive real numbers, then
ba 1 × cb 1 × ac 1 is equal to
(a) 1 (b) abc
(c) abc (d)abc
1
Sol. ba 1 × cb 1 × ac 1
= a
b ×
b
c ×
c
a
= c
a
b
c
a
b = 1 = 1 (a)
15. If
x
3
2x2
2
3
= 16
81, then x =
(a) 2 (b) 3
(c) 4 (d) 1
Sol.
x
3
2×
x2
2
3
= 16
81
x
2
3×
x2
2
3
= 4
4
2
3 =
4
2
3
xx
2
2
3=
4
2
3
x
2
3=
4
2
3
Comparing, we get
x = 4 (c)
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43 Arundeep’s Mathematics (R.D.) 9th
16. The value of
2
1
23
4
28
is
(a)2
1(b) 2
(c)4
1(d) 4
Sol.
2
1
23
4
28
=
2
1
23
4
32)2(
=
2
1
23
43
22
=
2
1
23
43
22
= 2
1
2
4
2
2
= 2
1
4
2
2
2
= 2
1
242
1
= 2
1
22
1
=
2
12
2
1
=
2
1(a)
17. If a, b, c are positive real numbers, then
5 105103125 cba is equal to
(a) 5a2bc2 (b) 25ab2c
(c) 5a3bc3 (d) 125a2bc2
Sol. 5 105103125 cba = 5 1051055 cba
5 31255 6255 1255 255 5
1
= 5
11051055 cba
= 5
15
5
. 5
110
a . 5
15
b . 5
110
c
= 5a2bc2 (a)
18. If a, m, n are positive integers, then
mnm n a is equal to
(a) amn (b) a
(c) n
m
a (d) 1
Sol. mnm n a =
mn
m
na
1
1
=
mn
mna
1
= mn
mn
a = a (b)
19. If x = 2 and y = 4, then
yx
y
x
+
xy
x
y
=
(a) 4 (b) 8
(c) 12 (d) 2
Sol. x = 2, y = 4
yx
y
x
+
xy
x
y
=
42
8
4
+
24
2
4
=
2
2
1
+ (2)2
= 22 + 22 = 4 + 4 = 8 (b)
20. The value of m for which
4
1
3
12
27
1
= 7m, is
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44 Arundeep’s Mathematics (R.D.) 9th
(a) –3
1(b)
4
1
(c) –3 (d) 2
Sol.
4
1
3
12
27
1
= 7m
4
1
3
122
7
= 7m
4
1
3
1)2(27
= 7m
4
1
3
14
7
= 7m
4
1
3
4
7 = 7m
3
1
7
= 7m
Comparing both sides,
m = –3
1(a)
21. The value of {(23 + 22)2/3 + (150 – 29)1/2}2,
is
(a) 196 (b) 289
(c) 324 (d) 400
Sol. {(23 + 22)2/3 + (150 – 29)1/2}2
= 22
1
3
2
])29150()423[(
= 22
1
3
2
])121()27[(
= 22
1
23
2
3 ])11()3[( = (9 + 11)2
= (20)2 = 400 (d)
22. (256)0.16 × (256)0.09
(a) 4 (b) 16
(c) 64 (d) 256.25
Sol. (256)0.16 × (256)0.09 = (256)0.16 + 0.09
= (256)0.25
= 100
25
8 )2( = 4
1
8 )2(
= 4
18
2
= 22 = 4 (a)
23. If 102y = 25, then 10–y equals
(a)5
1 (b)
50
1
(c)625
1(d)
5
1
Sol. 102y = 25 = 52
(10y)2 = (5)2 10y = 5
Now 10–y = y10
1 =
5
1(d)
24. If 9x + 2 = 240 + 9x, then x =
(a) 0.5 (b) 0.2
(c) 0.4 (d) 0.1
Sol. 9x + 2 = 240 + 9x 9x + 2 – 9x = 240
9x.92 – 9x = 240 9x(92 – 1) = 240
9x(81 – 1) = 240 9x × 80 = 240
9x = 80
40 = 3 (32)x = 31
32x = 31
2x = 1 x = 2
1 = 0.5 (a)
25. If x is a positive real number and x2 = 2,
then x3 =
(a) 2 (b) 2 2
(c) 3 2 (d) 4
Sol. x2 = 2 x = 2
x3 = ( 2 )3 = 2 × 2 × 2
= 2 × 2 = 2 2 (b)
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45 Arundeep’s Mathematics (R.D.) 9th
26. If 5.1x
x = 8x–1 and x > 0, then x =
(a)4
2(b) 2 2
(c) 4 (d) 64
Sol. 5.1x
x = 8x–1
5.1x
x =
x
8 5.1
2
x
x = 8
x2 – 1.5 = 8 x0.5 = 8
2
1
x = 8
Squaring,
x = (8)2 = 64
x = 64 (d)
27. If 3
2
tg + 2
1
4
t , what is the value of g
when t = 64?
(a)2
31(b)
2
33
(c) 16 (d)16
257
Sol. 3
2
tg + 2
1
4
t
= 3
2
t + 4 ×
2
1
1
t
= 3
2
)64( + 4 ×
2
1
64
1
= 3
2
3)4( + 4 ×
2
1
2 )8(
1
= 3
3
2
4
+ 4 ×
2
12
8
1
= 42 +
8
4
= 16 + 2
1 =
2
33(b)
28. If 4x – 4x – 1 = 24, then (2x)x equals
(a) 5 5 (b) 5
(c) 25 5 (d) 125
Sol. 4x – 4x – 1 = 24 4x – 14
4x
= 24
4x
4
11 = 24 4x ×
4
3 = 24
4x = 3
424 = 32
(22)x = 32 = (2)5
22x = 25
Comparing, we get
2x = 5 x = 2
5
(2x)x = 2
5
2
52
= 2
5
)5(
= 55 = 55555
= 5 × 5 5 = 25 5 (c)
29. When simplifed )4( 2/3
256 is
(a) 8 (b)8
1
(c) 2 (d)2
1
Sol. )4( 2/3
256
= – 2/34
14
4
= – 2
3
)4(
= – 2
3
2 )2(
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46 Arundeep’s Mathematics (R.D.) 9th
= –
2
32
)2( = –(2)–3
= – 3)2(
1 =
8
1
Now,
8
1
)256(
1 =
8
1
8 )2(
1
=
8
18
2
1
= 12
1 =
2
1(d)
30. If 225
382 x
= x5
53
, then x =
(a) 2 (b) 3
(c) 5 (d) 4
Sol.225
382 x
= x5
53
32x – 8 × 5x = 53 × 225
8
2
3
3x
× 5x = 5 × 5 × 5 × 5 × 5 × 3 × 3
32x × 5x = 38 × 32 × 55
(32)x × 5x = 310 × 55
9x × 5x = 35 × 35 × 55
(45)x = (45)5
Comparing, we get
x = 5 (c)
31. The value of 3
1
64
3
2
3
1
6464 , is
(a) 1 (b)3
1
(c) –3 (d) –2
Sol. 3
1
64
3
2
3
1
6464
= 3
1
3)4(
3
23
3
13
)4(4
=
3
13
4
3
23
3
13
44
= 4–1[41 – 42] = 4
1[4 – 16]
= 4
1(–12) = –3 (c)
32. If n5 = 125, then n 645 =
(a) 25 (b)125
1
(c) 625 (d)5
1
Sol. n5 = 125 25
n = 53
Comparing, we get
2
n = 3 n = 6
Now n 645 =
n
1
645
= 6
1
645 = 6
1
6 )2(5 = 6
16
25
= 52 = 25 (a)
33. If (16)2x + 3 = (64)x + 3, then 42x – 2 =
(a) 64 (b) 256
(c) 32 (d) 512
Sol. (16)2x + 3 = (64)x + 3 (24)2x + 3 = (26)x + 3
28x + 12 = 26x + 18
Comparing, we get
8x + 12 = 6x + 18 8x – 6x = 18 – 12
2x = 6 x = 2
6 = 3
Now 42x – 2 = 42 × (3) – 2 = 4+6 – 2 = 44
= 4 × 4 × 4 × 4 = 256 (b)
34. If 2–m × m2
1 =
4
1,
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47 Arundeep’s Mathematics (R.D.) 9th
then 14
1
1
2
1
5
1)4(
m
m
is equal to
(a)2
1(b) 2
(c) 4 (d) –4
1
Sol. 2–m × m2
1 =
4
1
2–m – m =
2
2
1
2–2m = 2–2
Comparing, we get
–2m = –2 m = 2
2
= 1
Now, 14
1
1
2
1
5
1)4(
m
m
= 14
1
12
12 )5()2( mm
= 14
1
)1(2
12
52m
m
= 14
1{2m + 5+m}
= 14
1{21 + 51} =
14
1 × 7 =
2
1(a)
35. If mn
nm
2
2 = 16, n
p
3
3 = 81 and a = 10
1
2 , then
122
2
)(
pnm
pnm
a
a =
(a) 2 (b)4
1
(c) 9 (d)8
1
Sol.mn
nm
2
2 = 16, n
p
3
3 = 81 and a = 10
1
2 , then
122
2
)(
pnm
pnm
a
a
2m + n – n + m = 24 22m = 24
Comparing, we get
2m = 4 m = 2
4 = 2
Given, n
p
3
3 = 81
3p – n = (3)4
p – n = 4
122
2
)(
pnm
pnm
a
a = a2m + n – p × am – 2n + 2p
= a2m + n – p + m – 2n + 2p = a3m – n + p
_ a = 10
1
2 =
pnm
3
10
1
2
= 10
3
2
pnm
= 10
23
2
pn
= 10
6
2
pn
= 10
46
2
= 10
10
2 = 2 (a)
36. If x
x
2
25
3
6561813 = 37, then x =
(a) 3 (b) –3
(c)3
1(d) –
3
1
Sol. x
x
2
25
3
6561813 = 37
35x × 812 × 6561 = 37 × 32x
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48 Arundeep’s Mathematics (R.D.) 9th
3 65613 21873 7293 2433 813 273 93 3
1
x
x
2
5
3
3 =
656181
32
7
= 824
7
3)3(
3
35x – 2x = 88
7
33
3
= 37 – 8 – 8 = 3–9
33x = 3–9
Comparing,
3x = –9 x = 3
9 = –3
x = –3 (b)
37. If 0 < y < x, which statement must be
true?
(a) x – y = yx
(b) x + x = x2
(c) x y = y x
(d) xy = x y
Sol. 0 < y < x
Only xy = x y is true (d)
38. If 10x = 64, what is the value of 1
210
x
?
(a) 18 (b) 42
(c) 80 (d) 81
Sol. 10x = 64 x10 = 64
2
1
)10( x = 8 210
x
= 8
Now, 1
210
x
= 210
x
× 101
= 8 × 10 = 80 (c)
39. 1
12
52513
565
nn
nn
is equal to
(a)3
5(b) –
3
5
(c)5
3(d) –
5
3
Sol. 1
12
52513
565
nn
nn
= )5213(5
)565(51
12
n
n
= 5213
5652
= 1013
3025
= 3
5(b)
40. If n2 = 1024, then
44
2
3
n
=
(a) 3 (b) 9
(c) 27 (d) 81
Sol. n2 = 1024 2
1
)2( n = 1024 = 210
2 10242 5122 2562 1282 642 322 162 82 42 2
1
22
n
= 210
Comparing,
2
n = 10 n = 2 × 10 = 20
Now,
44
2
3
n
=
44
202
3
= 32(5 – 4) = 32 × 1
= 32 = 9 (b)
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49 Arundeep’s Mathematics (R.D.) 9th
Points to Remember :
1. Some Identies :
(i) 2a = a
(ii) a . b = ab
(iii)b
a =
b
a
(iv) ba ba = a – b
(v) ba ba = a2 – b
(vi) 2ba = a + b + 2 ab
(vii) ba dc = ac + ad +
bc + bd
2. Rationalisation Factors : Rationalisation
factor
(i) Of a is a
(ii) Of ba is ba
and of ba is ba
(iii) Of a
1 is a and if a is
a
1
EXERCISE 3.1
1. Simplify each of the following:
(i) 3 4 × 3 16 (ii) 4
4
2
1250
Sol. (i) 3 4 × 3 16 = 3 164
(_ a × b = ab )
= 3 64 = 3 444
= 3
1
3)4( = 3
13
4 = 41 = 4
(ii) 4
4
2
1250 = 4
2
1250
b
a
b
a
= 4 625
= 4 5555
= 4
1
4 )5( = 4
14
5
= 51 = 5
2. Simplify the following expressions:
(i) (4 + 7 ) (3 + 2 )
(ii) (3 + 3 ) (5 – 2 )
(iii) ( 5 – 2) ( 3 – 5 )
Sol. (i) (4 + 7 ) (3 + 2 )
= 4 × 3 + 4 2 + 3 7 + 7 × 2
= 12 + 4 2 + 3 7 + 14
(ii) (3 + 3 ) (5 – 2 )
= 3 × 5 – 3 2 + 5 3 – 3 × 2
= 15 – 3 2 + 5 3 – 6
(iii) ( 5 – 2) ( 3 – 5 )
= 5 × 3 – 5 × 5 – 2 3 + 2 5
= 15 – 25 – 2 3 + 2 5
= 15 – 5 – 2 3 + 2 5
= 15 – 2 3 + 2 5 – 5
3. Simplify the following expressions:
(i) (11 + 11 ) (11 – 11 )
(ii) (5 + 7 ) (5 – 7 )
(iii) ( 8 – 2 ) ( 8 + 2 )
RATIONALISATION3
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50 Arundeep’s Mathematics (R.D.) 9th
(iv) (3 + 3 ) (3 – 3 ) [NCERT]
(v) ( 5 – 2 ) ( 5 + 2 ) [NCERT]
Sol. (i) (11 + 11 ) (11 – 11 )
= (11)2 – ( 11 )2
{_ (a + b) (a – b) = a2 – b2}
= 121 – 11 = 110
(ii) (5 + 7 ) (5 – 7 )
= (5)2 – ( 7 )2
{_ (a + b) (a – b) = a2 – b2}
= 25 – 7 = 18
(iii) ( 8 – 2 ) ( 8 + 2 )
( 8 )2 – ( 2 )2
{_ (a + b) (a – b) = a2 – b2}
= 8 – 2 = 6
(iv) (3 + 3 ) (3 – 3 )
= (3)2 – ( 3 )2
{_ (a + b) (a – b) = a2 – b2}
= 9 – 3 = 6
(v) ( 5 – 2 ) ( 5 + 2 )
= ( 5 )2 – ( 2 )2
{_ (a + b) (a – b) = a2 – b2}
= 5 – 2 = 3
4. Simplify the following expressions:
(i) ( 3 + 7 )2 (ii) ( 5 – 3 )2
(iii) (2 5 + 3 2 )2
Sol. ( 3 + 7 )2
= ( 3 )2 + ( 7 )2 + 2 × 3 × 7
{_ (a + b)2 = a2 + b2 + 2ab}
= 3 + 7 + 2 21 = 10 + 2 21
(ii) ( 5 – 3 )2
= ( 5 )2 + ( 3 )2 – 2 × 5 × 3
{_ (a – b)2 = a2 + b2 – 2ab}
= 5 + 3 – 2 15 = 8 – 2 15
(iii) (2 5 + 3 2 )2
= (2 5 )2 + (3 2 )2 + 2 × 2 5 × 3 2
{_ (a + b)2 = a2 + b2 + 2ab}
= 4 × 5 + 9 × 2 + 2 × 2 × 3 × 5 × 2
= 20 + 18 + 12 10
= 38 + 12 10
EXERCISE 3.2
1. Rationalise the denominators of each of the
following (i – vii):
(i)5
3(ii)
52
3
(iii)12
1(iv)
5
2
(v)2
13 (vi)
3
52
(vii)5
23
Sol. (i) 5
3 =
55
53
=
5
53 =
5
35
(ii)52
3 =
552
53
= 52
53
=
10
53 =
10
35
(iii)12
1 =
34
1
=
32
1
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51 Arundeep’s Mathematics (R.D.) 9th
= 332
31
=
32
3
=
6
3
(iv)5
2 =
55
52
=
5
10 =
5
110
(v)2
13 =
22
2)13(
=
2
26
(vi)3
52 =
33
3)52(
= 3
156
(vii)5
23 =
55
523
=
5
103
= 5
310
2. Find the value to three places of decimals
of each of the following. It is given that
2 = 1.414, 3 = 1.732, 5 = 2.236
and 10 = 3.162.
(i)3
2(ii)
10
3
(iii)2
15 (iv)
2
1510
(v)3
32 (vi)
5
12
Sol. 2 = 1.414, 3 = 1.732, 5 = 2.236
and 10 = 3.162
(i)3
2 =
33
32
(Rationalising the denominator)
= 3
32 =
3
732.12 =
3
464.3 = 1.154
(ii)10
3 =
1010
103
=
10
103
(Rationalising the denominator)
= 10
)162.3(3 =
10
486.9 = 0.9486
(iii)2
15 =
22
2)15(
(Rationalising the denominator)
= 2
210 =
2
414.1162.3
= 2
576.4 = 2.288
(iv)2
1510 =
22
2)1510(
(Rationalising the denominator)
= 2
3020 =
2
31054
= 2
31052
= 2
732.1162.3)236.2(2
= 2
476.5472.4 =
2
948.9 = 4.974
(v)3
32 =
3
732.12 =
3
732.3 = 1.244
(vi)5
12 =
55
5)12(
(Rationalising the denominator)
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52 Arundeep’s Mathematics (R.D.) 9th
= 5
510 =
5
236.2162.3
= 5
926.0 = 0.185
3. Express each one of the following with
rational denominator:
(i)23
1
(ii)
56
1
(iii)541
16
(iv)
5335
30
(v)352
1
(vi)
322
13
(vii)246
246
(viii)
352
123
(ix)aba
b
2
2
Sol. (i) 23
1
=
)23)(23(
)23(1
{Multiplying and dividing by (3 – 2 )}
= 22)2()3(
23
{_ (a + b) (a – b) = a2 – b2}
= 29
23
=
7
23
(ii)56
1
=
)56)(56(
)56(1
(Multiplying and dividing by 6 + 5 )
=
)5()6(
562
{_ (a + b) (a – b) = a2 – b2}
= 56
56
=
1
56
= 6 + 5
(iii)541
16
=
)541)(541(
)541(16
{Multiplying and dividing by ( 41 + 5)}
= 22)5()41(
)541(16
{_ (a + b) (a – b) = a2 – b2}
= 2541
)541(16
= 16
)541(16
= 41 + 5
(iv)5335
30
=
)5335)(5335(
)5335(30
{Multiplying and dividing by (5 3 +
3 5 )}
= 22 )53()35(
)5335(30
= 59325
)5335(30
= 4575
)5335(30
= 30
)5335(30 = 5 3 + 3 5
(v)352
1
=
)352)(352(
)352(1
{Multiplying and dividing by (2 5 + 3 )}
= 22)3()52(
352
=
320
352
= 17
352
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53 Arundeep’s Mathematics (R.D.) 9th
(vi)322
13
=
)322)(322(
)322)(13(
{Multiplying and dividing by (2 2 + 3 )}
= 22)3()22(
32233223
= 38
322362
= 5
332262
(vii)246
246
=
)246)(246(
)246)(246(
{Multiplying and dividing by (6 – 4 2 )}
= 22
2
)24()6(
)246(
= 3236
246221636
= 4
2483236 =
4
24868
= 17 – 12 2
(viii)352
123
=
)352)(352(
)352)(123(
{Multiplying and dividing by (2 5 + 3)}
= 22)3()52(
3522335223
= 920
35229106
= 11
35229106
(ix)aba
b
2
2
= ))((
)(
2222
222
abaaba
abab
{Dividing and multiplying by ( 22 ba –a)}
= 2222
222
)(
)(
aba
abab
=
222
222)(
aba
abab
= 2
222)(
b
abab = 22 ba – a
4. Rationales the denominator and simplify:
(i)23
23
(ii)
347
325
(iii)223
21
(iv)
6253
562
(v)1848
2534
(vi)
3322
532
Sol. (i) 23
23
=
)23)(23(
)23)(23(
(Rationalising the denominator)
= 22
2
)2()3(
)23(
=
23
23223
= 1
625 = 5 – 2 6
(ii)347
325
=
)347)(347(
)347)(325(
(Rationalising the denominator)
= 22 )34()7(
33831432035
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54 Arundeep’s Mathematics (R.D.) 9th
= 4849
243635
=
1
3611
= 11 – 6 3
(iii)223
21
=
)223)(223(
)223)(21(
(Rationalising the denominator)
= 22)22()3(
22223223
= 89
4253
= 1
257
= 7 + 5 2
(iv)6253
562
=
)6253)(6253(
)6253)(562(
(Rationalising the denominator)
= 22)62()53(
302553664306
= 6459
5364304
= 2445
3041524
= 21
3049
(v)1848
2534
=
)1848)(1848(
)1848)(2534(
(Rationalising the denominator)
= 22)18()48(
3659655441444
= 1848
656165694124
= 30
3064563448
= 30
6206123048 =
30
6818
= 15
649 (Dividing by 2)
(vi)3322
532
=
)3322)(3322(
)3322)(532(
(Rationalising the denominator)
= 22)33()22(
15310233664
= 278
1531021864
= 19
1536410218
= 19
1536410218
5. Simplify:
(i)35
35
+
35
35
(iii)32
1
+
35
2
+
52
1
(v)35
2
+
23
1
–
25
3
Sol. (i) 35
35
+
35
35
= )35)(35(
)35()35( 22
= 22
22
)3()5(
])3()5[(2
{_ (a + b)2 + (a – b)2 = 2(a2 + b2)}
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55 Arundeep’s Mathematics (R.D.) 9th
= 35
)35(2
= 2
82 = 8
(iii)32
1
+
35
2
+
52
1
= 32
1
=
)32)(32(
)32(1
= 22 )3()2(
)32(1
= 34
32
= 1
32 = 2 – 3
= 35
2
=
)35)(35(
)35(2
= 22 )3()5(
)35(2
= 35
)35(2
=
2
)35(2 = 5 + 3
= 52
1
=
)52)(52(
)52(1
= 22)5()2(
52
=
54
52
= 1
52
= –(–2 – 5 )
Now, 32
1
+
35
2
+
52
1
= (2 – 3 ) + ( 5 + 3 ) + (–2 – 5 )
= 2 – 3 + 5 + 3 – 2 – 5 = 0
(v)35
2
+
23
1
–
25
3
= 35
2
=
)35)(35(
)35(2
= 22)3()5(
)35(2
=
35
)35(2
= 2
)35(2 = 5 – 3
= 23
1
=
)23)(23(
)23(1
= 22)2()3(
23
=
23
23
= 1
23 = 3 – 2
= 25
3
=
)25)(25(
)25(3
= 22 )2()5(
)25(3
=
25
)25(3
= 3
)25(3 = 5 – 2
Now, 35
2
+
23
1
–
25
3
= 5 – 3 + 3 – 2 – ( 5 – 2 )
= 5 – 3 + 3 – 2 – 5 + 2 = 0
6. In each of the following determine rational
numbers a and b:
(i)13
13
= a – b 3
(ii)22
24
= a – b
(iii)23
23
= a + b 2
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56 Arundeep’s Mathematics (R.D.) 9th
(iv)347
335
= a + b 3
(v)711
711
= a – b 77
(vi)534
534
= a + b 5
Sol. (i) 13
13
= a – b 3
= 13
13
=
)13)(13(
)13)(13(
(Rationalising the denominator)
= 22
2
)1()3(
)13(
=
13
3213
= 2
324
= 2 – 3
Now, 2 – 3 = a – b 3
Comparing, we get
a = 2, b = 1
(ii)22
24
= a – b
= 22
24
=
)22)(22(
)22)(24(
(Rationalising the denominator)
= 22 )2()2(
222248
=
24
226
= 2
226 = 3 – 2
Now, 3 – 2 = a – b
Comparing, we get,
a = 3, b = 2
(iii)23
23
= a + b 2
= 23
23
=
)23)(23(
)23)(23(
(Rationalising the denominator)
= 22
2
)2()3(
)23(
=
29
23229
= 7
2611 =
7
11 +
7
62
Now, 7
11 +
7
62 = a + b 2
Comparing, we get
a = 7
11, b =
7
6
(iv)347
335
= a + b 3
= 347
335
=
)347)(347(
)347)(335(
(Rationalising the denominator)
= 22)34()7(
31232132035
= 4849
36335
=
1
13 = –1 + 3
Now, –1 + 3 = a + b 3
Comparing, we get
a = –1, b = 1
(v)711
711
= a – b 77
= 711
711
=
)711)(711(
)711)(711(
(Rationalising the denominator)
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57 Arundeep’s Mathematics (R.D.) 9th
= 22
2
)7()11(
)711(
= 711
7112711
= 4
77218 =
2
779
2
9 –
2
177 = a – b 77
Comparing, we get
a = 2
9, b =
2
1
(vi)534
534
= a + b 5
= 534
534
=
)534)(534(
)534)(534(
(Rationalising the denominator)
= 22
2
)53()4(
)534(
=
4516
5244516
= 29
52461
29
52461
= a + b 5
29
61 –
29
245 = a + b 5
Comparing, we get
a = 29
61, b =
29
24
7. Find the value of 35
6
, it being given
that 3 = 1.732 and 5 = 2.236.
Sol. 3 = 1.732, 5 = 2.236
Now, 35
6
=
)35)(35(
)35(6
(Rationalising the denominator)
= 22)3()5(
)35(6
=
35
)35(6
= 2
)35(6 = 3( 5 + 3 )
= 3[2.236 + 1.732] = 3(3.968)
= 11.904
8. Find the values of each of the following
correct to three places of decimals, it being
given that 2 = 1.4142, 3 = 1.732,
5 = 2.2360, 6 = 2.4495 and 10 =
3.162.
(i)523
53
(ii)
223
21
Sol. 2 = 1.4142, 3 = 1.732, 5 = 2.2360,
6 = 2.4495 and 10 = 3.162
(i)523
53
=
)523)(523(
)523)(53(
(Rationalising the denominator)
= 22 )52()3(
5253569
=
209
59109
= 11
5919
=
11
2360.2919
= 11
124.2019
= 11
124.1
= 0.102
(ii)223
21
=
)223)(223(
)223)(21(
(Rationalising the denominator)
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58 Arundeep’s Mathematics (R.D.) 9th
= 22)22()3(
2223223
=
89
2543
= 1
257 = 7 + 5(1.4142)
= 7 + 7.0710 = 14.071
9. Simplify:
(i)3223
3223
+
23
12
(ii)53
537
–
53
537
Sol. (i) 3223
3223
+
23
12
= 3223
3223
=
)3223)(3223(
)3223)(3223(
(Rationalising the denominator)
= 22
2
)32()23(
)3223(
= 3429
322323429
= 1218
6121218
=
6
61230
= 5 – 2 6
and 23
12
=
)23)(23(
)23)(34(
= 22)2()3(
)23(32
=
23
6232
= 1
626 = 6 + 2 6
3223
3223
+
23
12
= 5 – 2 6 + 6 + 2 6 = 111
(ii)53
537
–
53
537
= )53)(53(
)53)(537()53)(537(
= 22)5()3(
)553595721()53595721(
= 59
)155221()155221(
= 4
)526()526(
= 4
526526 =
4
54 = 5
10. If x = 2 + 3 , find the value of x3 + 3
1
x.
Sol. x = 2 + 3
x
1 =
32
1
=
)32)(32(
)32(1
(Rationalising the denominator)
= 22)3()2(
32
=
34
32
= 2 – 3
x + x
1 = 2 + 3 + 2 – 3 = 4
Cubing both sides,
31
xx = (4)3
x3 + 3
1
x + 3
xx
1 = 64
x3 + 3
1
x + 3 × 4 = 64
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59 Arundeep’s Mathematics (R.D.) 9th
x3 + 3
1
x + 12 = 64
x3 + 3
1
x = 64 – 12 = 52
Hence x3 + 3
1
x = 52
11. If x = 3 + 8 , find the value of x2 + 2
1
x.
Sol. x = 3 + 8
x
1 =
83
1
=
)83)(83(
)83(1
(Rationalising the denominator)
= 22 )8()3(
83
=
89
83
=
1
83
= 3 – 8
Now, x + x
1 = 3 + 8 + 3 – 8 = 6
Squaring both sides
21
xx = (6)2
x2 + 2
1
x + 2 = 36
x2 + 2
1
x = 36 – 2 = 34
12. If x = 2
13 , find the value of 4x3 + 2x2 –
8x + 7.
Sol. x = 2
13 2x = 3 + 1
2x – 1 = 3
Squaring both sides,
(2x – 1)2 = ( 3 )2 4x2 – 4x + 1 = 3
4x2 – 4x + 1 – 3 = 0 4x2 – 4x – 2 = 0
2x2 – 2x – 1 = 0
Now, 4x3 + 2x2 – 8x + 7
= (4x3 – 4x2 – 2x) + (6x2 – 6x – 3) + 10
= 2x(2x2 – 2x – 1) + 3(2x2 – 2x – 1) + 10
= 2x × 0 + 3 × 0 + 10
= 0 + 0 + 10 = 10
VERY SHORT ANSWER TYPE QUESTIONS
(VSAQs)
1. Write the value of (2 + 3 ) (2 – 3 ).
Sol. (2 + 3 ) (2 – 3 ) = (2)2 – ( 3 )2
{_ (a + b) (a – b) = a2 – b2}
= 4 – 3 = 1
2. Write the reciprocal of 5 + 2 .
Sol. Reciprocal of 5 + 2 = 25
1
(Rationalising its denominator)
= )25)(25(
)25(1
= 22 )2()5(
25
=
225
25
=
23
25
3. Write the rationalisation factor of 7 – 3 5 .
Sol. Rationalising factor of 7 – 3 5 is 7 +
3 5
{_ ( a + b ) ( a – b ) = a – b}
4. If 13
13
= x + y 3 , find the value of x
and y.
Sol.13
13
=
)13)(13(
)13)(13(
(Rationalising the denominator)
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60 Arundeep’s Mathematics (R.D.) 9th
= 22
2
)1()3(
)13(
=
13
3213
= 2
324
= 2 – 3
2 – 3 = x + y 3
Comparing, we get
x = 2, y = –1
5. If x = 2 – 1, then write the value of x
1.
Sol. x = 2 – 1
x
1 =
12
1
=
)12)(12(
)12(1
(Rationalising the denominator)
= 22)1()2(
12
=
12
12
= 1
12 = 2 + 1
6. If a = 2 + 1, then find the value of
a – a
1.
Sol. a = 2 + 1
Then, a
1 =
12
1
=
)12)(12(
)12(1
(Rationalising the denominator)
= 22 )1()2(
12
=
12
12
= 2 – 1
a – a
1 = ( 2 + 1) – ( 2 – 1)
= 2 + 1 – 2 + 1 = 2
7. If x = 2 + 3 , find the value of x + x
1.
Sol. x = 2 + 3
Then, x
1 =
32
1
=
)32)(32(
)32(1
= 22)3()2(
32
=
34
32
= 1
32
x + x
1 = 2 + 3 + 2 – 3 = 4
8. Write the rationalisation factor of 5 – 2.
Sol. Rationalisation factor of 5 – 2 is 5 + 2
as ( a + b ) ( a – b ) = a – b
9. Simplify : 223 .
Sol. 223
= 2212
= 122)1()2( 22
{Converting into (a + b)2 = a2 + b2 + 2ab}
= 2)12( = 2 + 1
10. Simplify : 223 .
Sol. 223
= 2212
{Converting into (a – b)2 = a2 + b2 – 2ab}
= 122)1()2( 22
= 2)12( = 2 – 1
11. If x = 3 + 2 2 , then find the value of x
– x
1.
Sol. x = 3 + 2 2
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61 Arundeep’s Mathematics (R.D.) 9th
Then, x
1 =
223
1
=
)223)(223(
)223(1
= 22 )22()3(
223
=
89
223
=
1
223
x + x
1 = 3 + 2 2 + 3 – 2 2 = 6
Now,
21
xx = x +
x
1 – 2
= 6 – 2 = 4 = (2)2
x – x
1 = 2
MULTIPLE CHOICE QUESTIONS (MCQs)
Mark the correct alternative in each of the
following:
1. 10 × 15 is equal to
(a) 5 6 (b) 6 5
(c) 30 (d) 25
Sol. 10 × 15 = 1510
= 3552 = 5 6 (a)
2. 5 6 × 5 6 is equal to
(a) 5 36 (b) 5 06
(c) 5 6 (d) 5 12
Sol. 5 6 × 5 6 = 5 66 = 5 36 (a)
3. The rationalisation factor of 3 is
(a) – 3 (b)3
1
(c) 2 3 (d) –2 3
Sol. Rationalisation factor of 3 is 3
1
1
1
aa (b)
4. The rationalisation factor of 2 + 3 is
(a) 2 – 3 (b) 2 + 3
(c) 2 – 3 (d) 3 – 2
Sol. Rationalisation factor of 2 + 3 is 2 – 3
{_ ( a + b ) ( a – b ) = a – b} (a)
5. If x = 5 + 2, then x – x
1 equals
(a) 2 5 (b) 4
(c) 2 (d) 5
Sol. x = 5 + 2
x
1 =
25
1
=
)25)(25(
)25(1
= 22 )2()5(
25
=
45
25
=
1
25
= 5 – 2
x – x
1 = ( 5 + 2) – ( 5 – 2)
= 5 + 2 – 5 + 2 = 4 (b)
6. If 13
13
= a – b 3 , then
(a) a = 2, b = 1 (b) a = 2, b = –1
(c) a = –2, b = 1 (d) a = b = 1
Sol.13
13
= a – b 3
13
13
=
)13)(13(
)13)(13(
(Rationalising the denominator)
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62 Arundeep’s Mathematics (R.D.) 9th
= 22
2
)1()3(
)13(
=
13
3213
= 2
324
= 2 – 3
Now, 2 – 3 = a – b 3
Comparing, we get
a = 2, b = 1 (a)
7. The simplest rationalising factor of 3 500
is
(a) 3 2 (b) 3 5
(c) 3 (d) none of these
Sol. 3 500 = 3 55522
Making triplet, we find 2 is required to
complete the triplet of 2
Required factor = 3 2 (a)
8. The simplest rationalising factor of 3 +
5 is
(a) 3 – 5 (b) 3 – 5
(c) 3 – 5 (d) 3 + 5
Sol. The simplest rationalising factor of
3 + 5 is 3 – 5 as ( a + b )
( a – b ) = a – b (c)
9. The simplest rationalising factor of 2 5 –
3 is
(a) 2 5 + 3 (b) 2 5 + 3
(c) 5 + 3 (d) 5 – 3
Sol. The simplest rationalising factor if
2 5 – 3 is 2 5 + 3 as ( a + b )
( a – b ) = a – b (b)
10. If x = 73
2
, then (x – 3)2 =
(a) 1 (b) 3
(c) 6 (d) 7
Sol. x = 73
2
=
)73)(73(
)73(2
(Rationalising the denominator)
= 22)7()3(
)73(2
=
79
)73(2
= 2
)73(2 = 3 – 7
Now, x – 3 = 3 – 7 – 3 = – 7
(x – 3)2 = (– 7 )2 = 7 (d)
11. If x = 7 + 4 3 and xy = 1, then 2
1
x +
2
1
y =
(a) 64 (b) 134
(c) 194 (d)49
1
Sol. x = 7 + 4 3 and xy = 1
y = x
1 =
347
1
y = x
1 =
)347)(347(
)347(1
(Rationalising the denominator)
= 22)34()7(
347
=
4849
347
= 1
347
= 7 – 4 3
Now, 2
1
x + 2
1
y
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63 Arundeep’s Mathematics (R.D.) 9th
_ y = x
1
y
1 = x
2
1
x + 2
1
y = 2
1
x + x2
= x2 + 2
1
x =
21
xx – 2
= [(7 + 4 3 ) + 7 – 4 3 ]2 – 2
= (14)2 – 2 = 196 – 2 = 194 (c)
12. If x + 15 = 4, then x + x
1 =
(a) 2 (b) 4
(c) 8 (d) 1
Sol. x + 15 = 4 x = 4 – 15
and x
1 =
154
1
=
)154)(154(
)154(1
(Rationalising the denominator)
= 22)15()4(
154
=
1516
154
= 1
154 = 4 + 15
Now, x + x
1 = 4 – 15 + 4 + 15
= 8 (c)
13. If x = 35
35
and y =
35
35
, then x
+ y + xy =
(a) 9 (b) 5
(c) 17 (d) 7
Sol. x = 35
35
=
)35)(35(
)35)(35(
(Rationalising the denominator)
= 22
2
)3()5(
)35(
=
35
35235
= 2
1528 = 4 + 15
Similarly,
y = 35
35
=
)35)(35(
)35)(35(
= 22
2
)3()5(
)35(
=
35
35235
= 2
1528 = 4 – 15
Now, x + y + xy = 4 + 15 + 4 – 15 +
(4 + 15 ) (4 – 15 )
= 8 + [(4)2 – ( 15 )2] = 8 + (16 – 15)
= 8 + 1 = 9 (a)
14. If x = 23
23
and y =
23
23
, then x2
+ xy + y2 =
(a) 101 (b) 99
(c) 98 (d) 102
Sol. x = 23
23
=
)23)(23(
)23)(23(
(Rationalising the denominator)
= 22
2
)2()3(
)23(
=
23
23223
= 1
625 = 5 – 2 6
Similarly,
y = 23
23
=
)23)(23(
)23)(23(
(Rationalising the denominator)
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64 Arundeep’s Mathematics (R.D.) 9th
= 22
2
)2()3(
)23(
=
23
23223
= 1
625 = 5 + 2 6
x2 + y2 + xy = (5 – 2 6 )2 + (5 + 2 6 )2 +
(5 – 2 6 ) (5 + 2 6 )
= 25 + 24 – 20 6 + 25 + 24 + 20 6 +
(5)2 – (2 6 )2
= 49 + 49 + 25 – 24 = 99 (b)
15.89
1
is equal to
(a) 3 + 2 2 (b)223
1
(c) 3 – 2 2 (d)2
3 – 2
Sol.89
1
=
243
1
=
223
1
= )223)(223(
)223(1
= 22 )22()3(
223
= 89
223
= 1
223 = 3 + 2 2 (a)
16. The value of 1827
3248
is
(a)3
4(b) 4
(c) 3 (d)4
3
Sol.1827
3248
=
2939
216316
= 2333
2434
=
)23(3
)23(4
=
3
4(a)
17. If 32
35
= x + y 3 , then
(a) x = 13, y = –7 (b) x = –13, y = 7
(c) x = –13, y = –7 (d) x = 13, y = 7
Sol.32
35
= x + y 3
32
35
=
)32)(32(
)32)(35(
(Rationalising the denominator)
= 22)3()2(
3323510
=
34
3713
= 1
3713 = 13 – 7 3
13 – 7 3 = x + y 3
Comparing, we get
x = 13, y = –7 (a)
18. If x = 3 32 , then x3 + 3
1
x =
(a) 2 (b) 4
(c) 8 (d) 9
Sol. x = 31
32 x3 = 2 + 3
3
1
x =
32
1
=
)32)(32(
)32(1
= 22 )3()2(
32
=
34
32
(Rationalising the denominator)
= 1
32 = 2 – 3
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65 Arundeep’s Mathematics (R.D.) 9th
x3 + 3
1
x = 2 + 3 + 2 – 3 = 4 (b)
19. The value of 223 is
(a) 2 – 1 (b) 2 + 1
(c) 3 – 2 (d) 3 + 2
Sol. 223 = 2212
= 122)1()2( 22
= 2)12( = 2 – 1 (a)
20. The value of 625 is
(a) 3 – 2 (b) 3 + 2
(c) 5 + 6 (d) none of these
Sol. 625 = 23223
= 232)2()3( 22
= 2)23( = 3 + 2 (b)
21. If 2 = 1.4142, then 12
12
is equal to
(a) 0.1718 (b) 5.8282
(c) 0.4142 (d) 2.4142
Sol. 2 = 1.4142
12
12
=
)12)(12(
)12)(12(
(Rationalising the denominator)
= 22
2
)1()2(
)12(
=
12
)12( 2
= 1
)12( 2 = 2 – 1 = 1.4142 – 1
= 0.4142 (c)
22. If 2 = 1.414, then the value of 6 – 3
upto three places of decimal is
(a) 0.235 (b) 0.707
(c) 1.414 (d) 0.471
Sol. 2 = 1.414
6 – 3 = 32 – 3
= 3 × 2 – 3
= 3 ( 2 – 1)
= 1.732 (1.414 – 1)
= 1.732 × 0.414 = 0.717 (b)
23. The positive square root of 7 + 48 is
(a) 7 + 2 3 (b) 7 + 3
(c) 2 + 3 (d) 3 + 2
Sol. 7 + 48 = 7 + 4 × 12
= 7 + 2 12
= 4 + 3 + 2 34
= ( 4 )2 + ( 3 )2 + 2 × ( 4 × 3 )
= ( 4 + 3 )2
Square root = 4 + 3
= 2 + 3 (c)
24. If x = 6 + 5 , then x2 + 2
1
x – 2 =
(a) 2 6 (b) 2 5
(c) 24 (d) 20
Sol. x = 6 + 5
Then x
1=
56
1
=
)56)(56(
56
= 22)5()6(
56
=
56
56
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66 Arundeep’s Mathematics (R.D.) 9th
= 1
56 = 6 – 5
Now, x2 + 2
1
x – 2 =
21
xx
= [( 6 + 5 ) – ( 6 – 5 )]2
= ( 6 + 5 – 6 + 5 )2 = (2 5 )2
= 4 × 5 = 20 (d)
25. If 1013 a = 8 + 5 , then a =
(a) –5 (b) –6
(c) –4 (d) –2
Sol. 1013 a = 8 + 5
1058 a = 8 + 5
Squaring both sides,
( 1058 a )2 = ( 8 + 5 )2
8 + 5 – a 10 = 8 + 5 + 2 × 8 × 5
–a 10 = 2 40 = 2 × 104
–a 10 = 2 × 2 × 10
–a 10 = 4 10
Comparing, we get
–a = 4 a = –4 (c)
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67 Arundeep’s Mathematics (R.D.) 9th
Points to Remember :
1. (a + b)2 = a2 + 2ab + b2
2. (a – b)2 = a2 – 2ab + b2
3. (a + b) (a – b) = a2 – b2
4. (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc +
2ca
= a2 + b2 + c2 + 2(ab + bc + ca)
5. (x + a) (x + b) = x2 + (a + b)x + ab
6. (a + b)3 = a3 + b3 + 3ab(a + b)
= a3 + 3a2b + 3ab2 + b3
7. (a – b)3 = a3 – b3 – 3ab (a – b)
= a3 – 3a2b + 3ab2 – b3
8. a3 + b3 = (a + b)3 – 3ab(a + b)
= (a + b) (a2 – ab + b2)
9. a3 – b3 = (a – b)3 + 3ab(a – b)
= (a – b) (a2 + ab + b2)
10. a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 +
c2 – ab – bc – ca)
11. a3 + b3 + c3 = 3abc if (a + b + c) = 0
12. (a + b)3 – (a – b)3 = 2(b3 + 3a2b)
13. (a + b)3 + (a – b)3 = 2(b3 + 3ab2)
EXERCISE 4.1
1. Evaluate each of the following using
identities:
(i)
21
2
x
x (ii) (2x + y) (2x – y)
(iii) (a2b – b2a)2 (iv) (a – 0.1) (a + 0.1)
(v) (1.5x2 – 0.3y2) (1.5x2 + 0.3y2)
Sol. (i)
21
2
x
x = (2x)2 +
21
x– 2 × 2x ×
x
1
{_ (a – b)2 = a2 + b2 – 2ab}
= 4x2 + 2
1
x – 4 = 4x2 – 4 + 2
1
x
(ii) (2x + y) (2x – y)
= (2x + y) (2x – y)
= (2x)2 – (y)2
{_ (a + b) (a – b) = a2 – b2)
= 4x2 – y2
(iii) (a2b – b2a)2 = (a2b)2 + (b2a)2 – 2 × a2b ×
b2a
= a4b2 + b4a2 – 2a3b3
{_ (a – b)2 = a2 + b2 – 2ab}
= a4b2 – 2a3b3 + b4a2
(iv) (a – 0.1) (a + 0.1) = (a)2 – (0.1)2
{_ (a + b) (a – b) = a2 – b2}
= a2 – 0.01
(v) (1.5x2 – 0.3y2) (1.5x2 + 0.3y2)
{_ (a + b) (a – b) = a2 – b2}
= (1.5x2)2 – (0.3y2)2
= 2.25x4 – 0.09y4
2. Evaluate each of the following using
identities:
(i) (399)2 (ii) (0.98)2
(iii) 991 × 1009 (iv) 117 × 83
Sol. (i) (399)2 = (400 – 1)2
= (400)2 – 2 × 400 × 1 + (1)2
{_ (a – b)2 = a2 – 2ab + b2}
= 160000 – 800 + 1
= 160001 – 800 = 159201
(ii) (0.98)2 = (1 – 0.02)2
{_ (a – b)2 = a2 – 2ab + b2}
= (1)2 – 2 × 1 × 0.02 + (0.02)2
= 1 – 0.04 + 0.0004
= 1.0004 – 0.04 = 1.0004 – 0.0400
= 0.9604
(iii) 991 × 1009 = (1000 – 9) (1000 + 9)
{_ (a + b) (a – b) = a2 – b2}
= (1000)2 – (9)2
= 1000000 – 81 = 999919
(iv) 117 × 83 = (100 + 17) (100 – 17)
{_ (a + b) (a – b) = a2 – b2}
= (100)2 – (17)2
= 10000 – 289 = 9711
3. Simplify each of the following:
(i) 175 × 175 + 2 × 175 × 25 + 25 × 25
(ii) 322 × 322 – 2 × 322 × 22 + 22 × 22
ALGEBRAIC IDENTITIES4
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68 Arundeep’s Mathematics (R.D.) 9th
(iii) 0.76 × 0.76 + 2 × 0.76 × 0.24 + 0.24 ×
0.24
(iv)66.6
17.117.183.783.7
Sol. (i) 175 × 175 + 2 × 175 × 25 + 25 × 25
Let a = 175 and b = 25, then
a2 + 2ab + b2 = (a + b)2
= (175 + 25)2 = (200)2 = 40000
(ii) 322 × 322 – 2 × 322 × 22 + 22 × 22
Let a = 322 and b = 22, then
a2 – 2ab + b2 = (a – b)2
= (322 – 22)2 = (300)2 = 90000
(iii) 0.76 × 0.76 + 2 × 0.76 × 0.24 + 0.24 ×
0.24
Let a = 0.76 and b = 0.24, then
a2 + 2ab + b2 = (a + b)2
= (0.76 + 0.24)2 = (1.00)2 = 1
(iv)66.6
17.117.183.783.7
Let a = 7.83, b = 1.17, a – b = 7.83 – 1.17
= 6.66
ba
bbaa
=
ba
ba
22
= ba
baba
)()( = a + b
= 7.83 + 1.17 = 9.00 = 9
4. If x + x
1 = 11, find the value of x2 + 2
1
x.
Sol. x + x
1 = 111
Squaring both sides,
21
xx = (11)2
x2 + 2
1
x + 2 = 121
x2 + 2
1
x = 121 – 2 = 119
x2 + 2
1
x = 119
5. If x – x
1 = –1, find the value of x2 + 2
1
x.
Sol. x – x
1 = –1
Squaring both sides,
21
xx = (–1)2 x2 + 2
1
x – 2 = 1
x2 + 2
1
x = 1 + 2 = 3
6. If x + x
1 = 5 , find the values of x2 +
2
1
x and x4 + 4
1
x.
Sol. x + x
1 = 5
Squaring both sides,
2
1
xx = ( 5 )2 x2 + 2
1
x + 2 = 5
(i) x2 + 2
1
x = 5 – 2 = 3
Again squaring,
2
2
2 1
x
x = (3)2
x4 + 4
1
x + 2 = 9
x4 + 4
1
x = 9 – 2 = 7
x2 + 2
1
x = 3, x4 + 4
1
x = 7
7. If 9x2 + 25y2 = 181 and xy = –6, find the
value of 3x + 5y.
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69 Arundeep’s Mathematics (R.D.) 9th
Sol. 9x2 + 25y2 = 181, and xy = –6
(3x + 5y)2 = (3x)2 + (5y)2 + 2 × 3x + 5y
9x2 + 25y2 + 30xy
= 181 + 30 × (–6)
= 181 – 180 = 1
= (+1)2
3x + 5y = +1
8. If 2x + 3y = 8 and xy = 2, find the value of
4x2 + 9y2.
Sol. 2x + 3y = 8 and xy = 2
Now, (2x + 3y)2 = (2x)2 + (3y)2 + 2 × 2x × 3y
(8)2 = 4x2 + 9y2 + 12xy
64 = 4x2 + 9y2 + 12 × 2
64 = 4x2 + 9y2 + 24
4x2 + 9y2 = 64 – 24 = 40
4x2 + 9y2 = 40
9. If 3x – 7y = 10 and xy = –1, find the value
of 9x2 + 49y2
Sol. 3x – 7y = 10, xy = –1
3x – 7y = 10
Squaring both sides,
(3x – 7y)2 = (10)2
(3x)2 + (7y)2 – 2 × 3x × 7y = 100
9x2 + 49y2 – 42xy = 100
9x2 + 49y2 – 42(–1) = 100
9x2 + 49y2 + 42 = 100
9x2 + 49y2 = 100 – 42 = 58
10. Simplify each of the following products:
(i)
ba 32
1
ab2
13
22 94
1ba
(ii)
3
7
nm
7
nm
Sol. (i)
ba 32
1
ab2
13
22 94
1ba
=
ba 32
1
ba 32
1
22 94
1ba
=
2
2
)3(2
1ba
22 94
1ba
{_ (a + b) (a – b) = a2 – b2}
=
22 94
1ba
22 94
1ba
=
2
2
4
1
a – (9b2)2
= 16
1a4 – 81b4
(ii)
3
7
nm
7
nm
=
2
7
nm
7
nm
7
nm
=
2
7
nm
2
2
7
nm
{_ (a + b) (a – b) = a2 – b2}
=
2
7
nm
49
22 n
m
11. If x2 + 2
1
x = 66, find the value of x –
x
1.
Sol. x2 + 2
1
x = 66
21
xx = x2 + 2
1
x – 2
= 66 – 2 = 64 = (+8)2
x – x
1 = +8
12. If x2 + 2
1
x = 79, find the value of x +
x
1.
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70 Arundeep’s Mathematics (R.D.) 9th
Sol. x2 + 2
1
x = 79
21
xx = x2 + 2
1
x + 2
= 79 + 2 = 81 = (+9)2
x + x
1 = +9
13. Simplify each of the following products:
(i)
5
2
2
x
25
2 x – x2 + 2x
(ii) (x2 + x – 2) (x2 – x + 2)
(iii) (x3 – 3x2 – x) (x2 – 3x + 1)
(iv) (2x4 – 4x2 + 1) (2x4 – 4x2 – 1)
Sol. (i)
5
2
2
x
25
2 x – x2 + 2x
=
5
2
2
x
5
2
2
x – x2 + 2x
= –
5
2
2
x
5
2
2
x – x2 + 2x
= –
2
5
2
2
x
– x2 + 2x
= –
5
2
22
5
2
2
22xx
– x2 + 2x
= –
5
2
25
4
4
2 xx – x2 + 2x
= 4
2x
– 25
4 +
5
2x – x2 + 2x
= 4
2x
– x2 + 5
2x + 2x –
25
4
= –
4
4 22 xx +
5
102 xx –
25
4
= –4
52
x +
5
12x –
25
4
(ii) (x2 + x – 2) (x2 – x + 2)
= [x2 + (x – 2)] [x2 – (x – 2)]
= (x2)2 – (x – 2)2
{_ (a + b) (a – b) = a2 – b2}
= x4 – [(x)2 – 2 × x × 2 + (2)2]
= x4 – (x2 – 4x + 4)
= x4 – x2 + 4x – 4
(iii) (x3 – 3x2 – x) (x2 – 3x + 1)
= x(x2 – 3x – 1) (x2 – 3x + 1)
= x[(x2 – 3x) – 1] [(x2 – 3x) + 1]
= x[(x2 – 3x)2 – (1)2]
{_ (a + b) (a – b) = a2 – b2}
= x[(x2)2 – 2 × x2 × 3x + (3x)2 – 1]
= x[x4 – 6x3 + 9x2 – 1]
= x5 – 6x4 + 9x3 – x
(iv) (2x4 – 4x2 + 1) (2x4 – 4x2 – 1)
= [(2x4 – 4x2) + 1] [(2x4 – 4x2) – 1]
= (2x4 – 4x2)2 – (1)2
{_ (a + b) (a – b) = a2 + b2}
= (2x4)2 – 2 × 2x4 × 4x2 + (4x2)2 – 1
= 4x8 – 16x6 + 16x4 – 1
14. Prove that a2 + b2 + c2 – ab – bc – ca is
always non-negative for all values of a, b
and c.
Sol. a2 + b2 + c2 – ab – bc – ca
{Multiplying and dividing by 2}
= 2
1[2a2 + 2b2 + 2c2 – 2ab – 2bc – 2ca]
= 2
1[a2 + b2 – 2ab + b2 + c2 – 2bc + c2 +
a2 – 2ca]
= 2
1[(a – b)2 + (b – c)2 + (c – a)2]
_ The given expression is sum of these
squares
Its value is always positive
Hence the given expression is always non-
negative for all values of a, b and c
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71 Arundeep’s Mathematics (R.D.) 9th
EXERCISE 4.2
1. Write the following in the expanded form:
(i) (a + 2b + c)2 (ii) (2a – 3b – c)2
(iii) (–3x + y + z)2 (iv) (m + 2n – 5p)2
(v) (2 + x – 2y)2 (vi) (a2 + b2 + c2)2
(vii) (ab + bc + ca)2 (viii)
2
x
z
z
y
y
x
(ix)
2
ab
c
ca
b
bc
a
(x) (x + 2y + 4z)2 [NCERT]
(xi) (2x – y + z)2 [NCERT]
(xii) (–2x + 3y + 2z)2 [NCERT]
Sol. We know that (a + b + c)2 = a2 + b2 + c2 +
2ab + 2bc + 2ca, Therefore
(i) (a + 2b + c)2 = (a)2 + (2b)2 + (c)2 + 2 × a
× 2b + 2 × 2b × c + 2 × c × a
= a2 + 4b2 + c2 + 4ab + 4bc + 2ca
(ii) (2a – 3b – c)2 = (2a)2 + (3b)2 + (c)2 + 2 ×
2a × –3b + 2 × –3b × –c + 2 × –c × 2a
= 4a2 + 9b2 + c2 – 12ab + 6bc – 4ca
(iii) (–3x + y + z)2 = (–3x)2 + (y)2 + z2 + 2 ×
(–3x) × y + 2 × y × z + 2 × z × (–3x)
= 9x2 + y2 + z2 – 6xy + yz – 6zx
(iv) (m + 2n – 5p)2 = (m)2 + (2n)2 + (–5p)2 + 2 ×
m × 2n + 2 × (2n) × (–5p) + 2 × (–5p) × m
= m2 + 4n2 + 25p2 + 4mn – 20np – 10pm
(v) (2 + x – 2y)2 = (2)2 + (x)2 + (–2y2) + 2 × 2
× x + 2 × x × (–2y) + 2 × (–2y) × 2
= 4 + x2 + 4y2 + 4x – 4xy – 8y
(vi) (a2 + b2 + c2)2 = (a2)2 + (b2)2 + (c2)2 + 2 ×
a2 × b2 + 2b2 × c2 + 2c2 × a2
= a4 + b4 + c4 + 2a2b2 + 2b2c2 + 2c2a2
(vii) (ab + bc + ca)2 = (ab)2 + (bc)2 + (ca)2 +
2ab × bc + 2bc × ca + 2ca × ab
= a2b2 + b2c2 + c2a2 + 2ab2c + 2bc2a + 2ca2b
(viii)
2
x
z
z
y
y
x =
2
y
x +
2
z
y +
2
x
z +
2 × y
x ×
z
y + 2 ×
z
y +
x
z + 2 ×
x
z ×
y
x
= 2
2
y
x + 2
2
z
y +
x
z2
+ 2z
x + 2
x
y + 2
y
z
(ix)
2
ab
c
ca
b
bc
a =
2
bc
a +
2
ca
b +
2
ab
c + 2 ×
bc
a ×
ca
b + 2
ca
b ×
ab
c +
2ab
c ×
bc
a
= 22
2
cb
a + 22
2
ac
b + 22
2
ba
c + 2
2
c + 2
2
a + 2
2
b
= 22
2
cb
a + 22
2
ac
b + 22
2
ba
c + 2
2
a + 2
2
b + 2
2
c
(x) (x + 2y + 4z)2 = (x)2 + (2y)2 + (4z)2 + 2 × x
× 2y + 2 × 2y × 4z + 2 × 4z × x
= x2 + 4y2 + 16z2 + 4xy + 16yz + 8zx
(xi) (2x – y + z)2 = (2x)2 + (–y)2 + (z)2 + 2 ×
(2x) × (–y) + 2 × (–y) × z + 2z × 2x
= 4x2 + y2 + z2 – 4xy – 2yz + 4zx
(xii) (–2x + 3y + 2z)2 = (–2x)2 + (3y)2 + (2z)2 + 2
× (–2x) × 3y + 2 × 3y × 2z + 2 × 2z × (–2x)
= 4x2 + 9y2 + 4z2 – 12xy + 12yz – 8zx
2. If a + b + c = 0 and a2 + b2 + c2 = 16, find
the value of ab + bc + ca.
Sol. a + b + c = 0
Squaring both sides,
(a + b + c)2 = 0 a2 + b2 + c2 + 2(ab + bc
+ ca) = 0
16 + 2(ab + bc + c) = 0
2(ab + bc + ca) = –16
ab + bc + ca = 2
16 = –8
ab + bc + ca = –8
3. If a2 + b2 + c2 = 16 and ab + bc + ca = 10,
find the value of a + b + c.
Sol. (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
= 16 + 2 × 10 = 16 + 20 = 36
= (+6)2
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72 Arundeep’s Mathematics (R.D.) 9th
a + b + c = +6
4. If a + b + c = 9 and ab + bc + ca = 23,
find the value of a2 + b2 + c2.
Sol. (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc +
ca)
(9)2 = a2 + b2 + c2 + 2 × 23
81 = a2 + b2 + c2 + 46
a2 + b2 + c2 = 81 – 46 = 35
a2 + b2 + c2 = 35
5. Find the value of 4x2 + y2 + 25z2 + 4xy –
10yz – 20zx when x = 4, y = 3 and z = 2.
Sol. x = 4, y = 3, z = 2
4x2 + y2 + 25z2 + 4xy – 10yz – 20zx
= (2x)2 + (y)2 + (5z)2 + 2 × 2x × y – 2 × y
× 5z – 2 × 5z × 2x
= (2x + y – 5z)2
= (2 × 4 + 3 – 5 × 2)2 = (8 + 3 – 10)2
= (11 – 10)2 = (1)2 = 1
6. Simplify:
(i) (a + b + c)2 + (a – b + c)2
(ii) (a + b + c)2 – (a – b + c)2
(iii) (a + b + c)2 + (a – b + c)2 + (a + b – c)2
(iv) (2x + p – c)2 – (2x – p + c)2
(v) (x2 + y2 – z2)2 – (x2 – y2 + z2)2
Sol. (i) (a + b + c)2 + (a – b + c)2
= (a2 + b2 + c2 + 2ab + 2bc + 2ca) + (a2 +
b2 + c2 – 2ab – 2bc + 2ca)
= a2 + b2 + c2 + 2ab + 2bc + 2ca + a2 + b2
+ c2 – 2ab – 2bc + 2ca
= 2a2 + 2b2 + 2c2 + 4ca
= 2(a2 + b2 + c2 + 2ca)
(ii) (a + b + c)2 – (a – b + c)2
= (a2 + b2 + c2 + 2ab + 2bc + 2ca) – (a2 +
b2 + c2 – 2ab – 2bc + 2ca)
= a2 + b2 + c2 + 2ab + 2bc + 2ca – a2 – b2
– c2 + 2ab + 2bc – 2ca
= 4ab + 4bc = 4(ab + bc)
(iii) (a + b + c)2 + (a – b + c)2 + (a + b – c)2
= a2 + b2 + c2 + 2ab + 2bc + 2ca + a2 + b2
+ c2 – 2ab – 2bc + 2ca + a2 + b2 + c2 +
2ab – 2bc – 2ca
= 3a2 + 3b2 + 3c2 + 2ab – 2bc + 2ca
= 3(a2 + b2 + c2) + 2(ab – bc + ca)
(iv) (2x + p – c)2 – (2x – p + c)2
= [(2x)2 + (p)2 + (–c)2 + 2 × 2x × p – 2pc
– 2 × c × 2x] – [(2x)2 + (–p)2 + (c)2 – 2 ×
2x × p – 2pc + 2c × 2x]
= (4x2 + p2 + c2 + 4xp – 2pc – 4cx) – (4x2
+ p2 + c2 – 4xp – 2pc + 4cx)
= 4x2 + p2 + c2 + 4xp – 2pc – 4cx – 4x2 – p2
– c2 + 4xp + 2pc – 4cx
= 8xp – 8cx
= 8x(p – c)
(v) (x2 + y2 – z2)2 – (x2 – y2 + z2)2
= [(x2)2 + (y2)2 + (–z2)2 + 2x2y2 – 2y2z2 –
2z2x2] – [(x2)2 + (–y2)2 + (z2)2 – 2x2y2 – 2y2z2
+ 2z2x2]
= x4 + y4 + z4 + 2x2y2 – 2y2z2 – 2z2x2 – x4 –
y4 – z4 + 2x2y2 + 2y2z2 – 2z2x2
= 4x2y2 – 4z2x2
= 4x2(y2 – z2)
7. Simplify each of the following expressions:
(i) (x + y + z)2 +
2
32
zy
x –
2
432
zyx
(ii) (x + y – 2z)2 – x2 – y2 – 3z2 + 4xy
(iii) (x2 – x + 1)2 – (x2 + x + 1)2
Sol. (i) (x + y + z)2 +
2
32
zy
x
–
2
432
zyx
= (x2 + y2 + z2 + 2xy + 2yz + 2zx) +
4
22 y
x + 9
2z
+ 2
2 yx +
2
2 y ×
3
z
+
xz
3
2 –
4
2x +
9
2y
+ 16
2z
+ 2
2x
× 3
y +
3
2y ×
4
z +
4
2z ×
2
x
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73 Arundeep’s Mathematics (R.D.) 9th
= x2 + y2 + z2 + 2xy + 2yz + 2zx + x2 + 4
2y
+ 9
2z
+ xy + 3
yz +
3
2zx –
4
2x
– 9
2y
–
16
2z
– 3
xy –
6
yz –
4
xz
= x2 + x2 – 4
2x
+ y2 + 4
2y
– 9
2y + z2 +
9
2z –
16
2z
+ 2xy + xy – 3
xy + 2yz +
3
yz –
6
yz + 2zx +
3
2zx –
4
zx
= 4
44 222 xxx +
36
4936222
yyy +
144
916144222
zzz +
3
36 xyxyxy +
6
212 yzyzyz +
12
3824 zxzxzx
= 4
72
x +
36
412
y +
14
1512
z +
3
8xy +
6
13yz
+ 12
29zx
(ii) (x + y – 2z)2 – x2 – y2 – 3z2 + 4xy
= x2 + y2 + 4z2 + 2xy – 4yz – 4zx – x2 – y2 –
3z2 + 4xy
= z2 + 6xy – 4yz – 4zx
(iii) (x2 – x + 1)2 – (x2 + x + 1)2
= (x4 + x2 + 1 – 2x3 – 2x + 2x2) – (x4 + x2 +
1 + 2x3 + 2x + 2x2)
= x4 + x2 + 1 – 2x3 – 2x + 2x2 – x4 – x2 – 1
– 2x3 – 2x – 2x2
= –4x3 – 4x = –4x(x2 + 1)
EXERCISE 4.3
1. Find the cube of each of the following
binomial expressions:
(i)x
1 +
3
y(ii)
x
3 – 2
2
x
(iii) 2x + x
3(iv) 4 –
x3
1
Sol. (i)
3
3
1
y
x =
31
x +
3
3
y
+ 3
21
x ×
3
y + 3 ×
x
1 ×
2
3
y
= 3
1
x +
27
3y
+ 3 × 2
1
x ×
3
y + 3 ×
x
1 ×
9
2y
= 3
1
x +
27
3y
+ 2x
y +
x
y
3
2
(ii)
3
2
23
xx
=
33
x –
3
2
2
x – 3 ×
23
x.
2
2
x + 3 ×
x
3 ×
2
2
2
x
= 3
27
x– 6
8
x – 3 × 2
9
x× 2
2
x + 3 ×
x
3× 4
4
x
= 3
27
x – 6
8
x – 4
54
x + 5
36
x
(iii)
33
2
x
x = (2x)3 +
33
x + 3 × (2x)2 ×
x
3 + 3 × 2x ×
23
x
= 8x3 + 3
27
x + 3 × 4x2 ×
x
3 + 3 × 2x × 2
9
x
= 8x3 + 3
27
x + 36x +
x
54
(iv)
3
3
14
x = (4)3 –
3
3
1
x – 3 × (4)2 ×
x3
1
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74 Arundeep’s Mathematics (R.D.) 9th
+ 3 × 4 ×
2
3
1
x
= 64 – 327
1
x – 3 × 16 ×
x3
1 + 3 × 4 × 29
1
x
= 64 – 327
1
x –
x
16 + 23
4
x
2. If a + b = 10 and ab = 21, find the value of
a3 + b3.
Sol. a + b = 10, ab = 21
Cubing both sides, (a + b)3 = (10)3
a3 + b3 + 3ab (a + b) = 1000
a3 + b3 + 3 × 21 × 10 = 1000
a3 + b3 + 630 = 1000
a3 + b3 = 1000 – 630 = 370
a3 + b3 = 370
3. If a – b = 4 and ab = 21, find the value of
a3 – b3.
Sol. a – b = 4, ab = 21
Cubing both sides,
(a – b)3 = (4)3
a3 – b3 – 3ab (a – b) = 64
a3 – b3 – 3 × 21 × 4 = 64
a3 – b3 – 252 = 64
a3 – b3 = 64 + 252 = 316
a3 – b3 = 316
4. If x + x
1 = 5, find the value of x3 + 3
1
x.
Sol. x + x
1 = 5
Cubing both sides,
31
xx = (5)3
x3 + 3
1
x + 3
xx
1 = 125
x3 + 3
1
x + 3 × 5 = 125
x3 + 3
1
x + 15 = 125
x3 + 3
1
x = 125 – 15 = 110
5. If x – x
1 = 7, find the value of x3 – 3
1
x.
Sol. x – x
1 = 7
Cubing both sides,
31
xx = (7)3
x3 – 3
1
x – 3
xx
1 = 343
x3 – 3
1
x – 3 × 7 = 343
x3 – 3
1
x – 21 = 343
x3 – 3
1
x = 343 + 21 = 364
6. If x – x
1 = 5, find the value of x3 – 3
1
x.
Sol. x – x
1 = 5
Cubing both sides,
31
xx = (5)3
x3 – 3
1
x – 3
xx
1 = 125
x3 – 3
1
x – 3 × 5 = 125
x3 – 3
1
x – 15 = 125
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75 Arundeep’s Mathematics (R.D.) 9th
x3 – 3
1
x = 125 + 15 = 140
x3 – 3
1
x = 140
7. If x2 + 2
1
x= 51, find the value of x3 – 3
1
x.
Sol. x2 + 2
1
x = 51
21
xx = x2 + 2
1
x – 2
= 51 – 2 = 49 = (7)2
x – x
1 = 7
Cubing both sides,
31
xx = (7)3
x3 – 3
1
x – 3
xx
1 = 343
x3 – 3
1
x – 3 × 7 = 343
x3 – 3
1
x – 21 = 343
x3 – 3
1
x = 343 + 21 = 364
x3 – 3
1
x = 364
8. If x2+ 2
1
x = 98, find the value of x3 + 3
1
x.
Sol.
21
xx = x2 + 2
1
x + 2
= 98 + 2 = 100 = (10)2
x + x
1 = 10
Cubing both sides,
31
xx = (10)3
x3 + 3
1
x + 3
xx
1 = 1000
x3 + 3
1
x + 3 × 10 = 1000
x3 + 3
1
x + 30 = 1000
x3 + 3
1
x = 1000 – 30 = 970
x3 + 3
1
x = 970
9. If 2x + 3y = 13 and xy = 6, find the value
of 8x3 + 27y3.
Sol. 2x + 3y = 13, xy = 6
Cubing both sides,
(2x + 3y)3 = (13)3
(2x)3 + (3y)3 + 3 × 2x × 3y(2x + 3y) = 2197
8x3 + 27y3 + 18xy(2x + 3y) = 2197
8x3 + 27y3 + 18 × 6 × 13 = 2197
8x3 + 27y3 + 1404 = 2197
8x3 + 27y3 = 2197 – 1404 = 793
8x3 + 27y3 = 793
10. If 3x – 2y = 11 and xy = 12, find the value
of 27x3 – 8y3.
Sol. 3x – 2y = 11 and xy = 12
Cubing both sides,
(3x – 2y)3 = (11)3
(3x)3 – (2y)3 – 3 × 3x × 2y(3x – 2y) = 1331
27x3 – 8y3 – 18xy(3x – 2y) = 1331
27x3 – 8y3 – 18 × 12 × 11 = 1331
27x3 – 8y3 – 2376 = 1331
27x3 – 8y3 = 1331 + 2376 = 3707
2x3 – 8y3 = 3707
11. Evaluate each of the following:
(i) (103)3 (ii) (98)3
(iii) (9.9)3 (iv) (10.4)3
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76 Arundeep’s Mathematics (R.D.) 9th
(v) (598)3 (vi) (99)3
Sol. We know that (a + b)3 = a3 + b3 + 3ab(a +
b) and (a – b)3 = a3 – b3 – 3ab(a – b)
Therefore,
(i) (103)3 = (100 + 3)3
= (100)3 + (3)3 + 3 × 100 × 3(100 + 3)
{_ (a + b)3 = a3 + b3 + 3ab(a + b)}
= 1000000 + 27 + 900 × 103
= 1000000 + 27 + 92700 = 1092727
(ii) (98)3 = (100 – 2)3
= (100)3 – (2)3 – 3 × 100 × 2(100 – 2)
= 1000000 – 8 – 600 × 98
= 1000000 – 8 – 58800
= 1000000 – 58808 = 941192
(iii) (9.9)3 = (10 – 0.1)3
= (10)3 – (0.1)3 – 3 × 10 × 0.1(10 – 0.1)
= 1000 – 0.001 – 3 × 9.9
= 1000 – 0.001 – 29.7
= 1000 – 29.701 = 970.299
(iv) (10.4)3 = (10 + 0.4)3
= (10)3 + (0.4)3 + 3 × 10 × 0.4(10 + 0.4)
= 1000 + 0.064 + 12(10.4)
= 1000 + 0.064 + 124.8 = 1124.864
(v) (598)3 = (600 – 2)3
= (600)3 – (2)3 – 3 × 600 × 2 × (600 – 2)
= 216000000 – 8 – 3600 × 598
= 216000000 – 8 – 2152800
= 216000000 – 2152808 = 213847192
(vi) (99)3 = (100 – 1)3
= (100)3 – (1)3 – 3 × 100 × 1 × (100 – 1)
= 1000000 – 1 – 300 × 99
= 1000000 – 1 – 29700
= 1000000 – 29701 = 970299
12. Evaluate each of the following:
(i) 1113 – 893 (ii) 463 + 343
(iii) 1043 + 963 (iv) 933 – 1073
Sol. We know that a3 + b3 = (a + b)3 – 3ab(a +
b) and a3 – b3 = (a – b)3 + 3ab(a – b)
(i) 1113 – 893
= (111 – 89)3 + 3 × 111 × 89(111 – 89)
= (22)3 + 3 × 111 × 89 × 22
= 10648 + 652014 = 662662
OR
(a + b)3 – (a – b)3 = 2(b3 + 3a2b)
= 1113 – 893 = (100 + 11)3 – (100 – 11)3
= 2[113 + 3 × 1002 × 11]
= 2[1331 + 330000]
= 331331 × 2 = 662662
(ii) (a + b)3 + (a – b)3 = 2(b3 + 3ab2)
463 + 343 = (40 + 6)3 + (40 – 6)3
= 2[(40)3 + 3 × 40 × 62]
= 2[64000 + 3 × 40 × 36] = 2[64000 +
4320]
= 2 × 68320 = 136640
(iii) 1043 + 963 = (100 + 4)3 + (100 – 96)3
= 2[a3 + 3ab2]
= 2[(100)3 + 3 × 100 × (4)2]
= 2[1000000 + 300 × 16]
= 2[1000000 + 4800]
= 1004800 × 2 = 2009600
(iv) 933 – 1073 = –[(107)3 – (93)3]
= –[(100 + 7)3 – (100 – 7)3]
= –2[b3 + 3a2b] = –2[(7)3 + 3(100)2 × 7]
= –2[343 + 3 × 10000 × 7]
= –2[343 + 210000]
= –2[210343] = –420686
13. If x + x
1 = 3, calculate x2 + 2
1
x, x3 + 3
1
x
and x4 + 4
1
x.
Sol. (i) x + x
1 = 3
Squaring both sides,
21
xx = (3)2 x2 + 2
1
x + 2 = 9
x2 + 2
1
x = 9 – 2 = 7
x2 + 2
1
x = 7
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77 Arundeep’s Mathematics (R.D.) 9th
(ii) x + x
1 = 3
Cubing both sides,
31
xx = (3)3
x3 + 3
1
x + 3
xx
1 = 27
x3 + 3
1
x + 3 × 3 = 27 x3 + 3
1
x + 9 = 27
x3 + 3
1
x = 27 – 9 = 18
x3 + 3
1
x = 18
(iii) x2 + 2
1
x = 7
Squaring both sides,
2
2 1
x
x = (7)2
x4 + 4
1
x + 2 = 49
x4 + 4
1
x = 49 – 2 = 47
x4 + 4
1
x = 47
14. Find the value of 27x3 + 8y3, if
(i) 3x + 2y = 14 and xy = 8
(ii) 3x + 2y = 20 and xy = 9
14
Sol. (i) 3x + 2y = 14 and xy = 8
Cubing both sides,
(3x + 2y)3 = (14)3
(3x)3 + (2y)3 + 3 × 3x × 2y(3x + 2y) = 2744
27x3 + 8y3 + 18xy(3x + 2y) = 2744
27x3 + 8y3 + 18 × 8 × 14 = 2744
27x3 + 8y3 + 2016 = 2744
27x3 + 8y3 = 2744 – 2016 = 728
(ii) 3x + 2y = 20 and xy = 9
14
Cubing both sides,
(3x + 2y)3 = (20)3
(3x)3 + (2y)3 + 3 × 3x × 2y(3x + 2y) = 8000
27x3 + 8y3 + 3 × 3 × 2xy(3x + 2y) = 8000
27x3 + 8y3 + 18 × 9
14 × 20 = 8000
27x3 + 8y3 + 560 = 8000
27x3 + 8y3 = 8000 – 560 = 7440
27x3 + 8y3 = 7440
15. Find the value of 64x3 – 125z3, if 4x – 5z =
16 and xz = 12.
Sol. 4x – 5z = 16, xz = 12
Cubing both sides,
(4x – 5z)3 = (16)3
(4x)3 – (5y)3 – 3 × 4x × 5z(4x – 5z) = 4096
64x3 – 125z3 – 3 × 4 × 5 × xz(4x – 5z) = 4096
64x3 – 125z3 – 60 × 12 × 16 = 4096
64x3 – 125z3 – 11520 = 4096
64x3 – 125z3 = 4096 + 11520 = 15616
16. If x – x
1 = 3 + 2 2 , find the value of x3 –
3
1
x.
Sol. x – x
1 = 3 + 2 2
Cubing both sides,
31
xx = (3 + 2 2 )3
x3 – 3
1
x – 3
xx
1 = (3)3 + (2 2 )3 + 3
× 3 × 2 2 (3 + 2 2 )
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78 Arundeep’s Mathematics (R.D.) 9th
x3 – 3
1
x – 3 × (3 + 2 2 ) = 27 + 16 2 +
18 2 (3 + 2 2 )
x3 – 3
1
x – 3 × (3 + 2 2 ) = 27 + 16 2 +
54 2 + 72
x3 + 3
1
x – 3(3 + 2 2 ) = 99 + 70 2
x3 + 3
1
x = 99 + 70 2 + 3(3 + 2 2 )
= 99 + 70 2 + 9 + 6 2
= 108 + 76 2
17. Simplify each of the following:
(i) (x + 3)3 + (x – 3)3
(ii)
3
32
yx
–
3
32
yx
(iii)
32
xx +
32
xx
(iv) (2x – 5y)3 – (2x + 5y)3
Sol. (i) (x + 3)3 + (x – 3)3
= (x)3 + (3)3 + 3x2 × 3 + 3 × x × (3)2 +
[(x)3 – (3)3 – 3x2 × 3 + 3 × x (3)2]
= x3 + 27 + 9x2 + 27x + x3 – 27 – 9x2 + 27x
= 2x3 + 54x
(ii)
3
32
yx
–
3
32
yx
=
3
2
x +
3
3
y + 3 ×
2
2
x
× 3
y +
3
2
x
2
3
y –
3
2
x –
3
3
y – 3
2
2
x
× 3
y + 3
2
x
2
3
y
=
8
3x +
27
3y
+ 3 × 4
2x
× 3
y + 3 ×
2
x ×
9
2y –
8
3x –
27
3y
– 3 × 4
2x
× 3
y + 3
2
x
×
9
2y
= 8
3x
+ 27
3y
+ 4
2yx
+ 6
2xy
– 8
3x
+ 27
3y
+ 4
2yx
– 6
2xy
= 27
23
y +
2
2yx
(iii)
32
xx +
32
xx
=
3
3 8
xx + 3 × x2 ×
x
2 + 3x ×
2
4
x +
3
3 8
xx – 3x2 ×
x
2 + 3x ×
2
4
x
= x3 + 3
8
x + 6x +
x
12 + x3 – 3
8
x – 6x +
x
12
= 2x3 + x
24
(iv) (2x – 5y)3 – (2x + 5y)3
= [(2x)3 – (5y)3 – 3(2x)2 (5y) + 3 × 2x ×
(5y)2] – [(2x)3 + (5y)3 + 3(2x)2 (5y) + 3 ×
2x × (5y)2]
= [8x3 – 125y3 – 60x2y + 150xy2] – [8x3 +
125y3 + 60x2y + 150xy2]
= 8x3 – 125y3 – 60x2y + 150xy2 – 8x3 –
125y3 – 60x2y – 150xy2
= –250y2 – 120x2y
18. If x4 + 4
1
x = 194, find x3 + 3
1
x, x2 + 2
1
x
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79 Arundeep’s Mathematics (R.D.) 9th
and x + x
1.
Sol. x4 + 4
1
x = 194
Adding 2 to both sides,
x4 + 4
1
x + 2 = 194 + 2 = 196
(x2)2 + 22 )(
1
x + 2 = (14)2
2
2
2 1
x
x = (14)2 x2 + 2
1
x = 14
x2 + 2
1
x = 14
Adding 2 to both sides,
x2 + 2
1
x + 2 = 14 + 2 = 16 = (4)2
21
xx = (4)2
x + x
1 = 4
Cubing both sides,
31
xx = x3 + 3
1
x + 3
xx
1
(4)3 = x3 + 3
1
x + 3 × 4
64 = x3 + 3
1
x + 12
x3 + 3
1
x = 64 – 12 = 52
x3 + 3
1
x = 52, x2 + 2
1
x = 14
and x + x
1 = 4
19. If x4 + 4
1
x= 119, find the value of x3 – 3
1
x.
Sol. x4 + 4
1
x = 119
Adding 2 to both sides,
x4 + 4
1
x + 2 = 119 + 2
(x2)2 +
2
2
1
x+ 2 = 121
2
2
2 1
x
x = (11)2 x2 + 2
1
x = 111
x2 + 2
1
x = 111
Subtracting 2 from both sides,
x2 + 2
1
x – 2 = 11 – 2 = 9
2
1
xx = (3)2 x –
x
1 = 3
x – x
1 = 3
Cubing both sides,
31
xx = (3)3
x3 – 3
1
x – 3
xx
1 = 27
x3 – 3
1
x – 3 × 3 = 27 x3 – 3
1
x – 9 = 27
x3 – 3
1
x = 27 + 9 = 36
EXERCISE 4.4
1. Find the following products:
(i) (3x + 2y) (9x2 – 6xy + 4y2)
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80 Arundeep’s Mathematics (R.D.) 9th
(ii) (4x – 5y) (16x2 + 20xy + 25y2)
(iii) (7p4 + q) (49p8 – 7p4q + q2)
(iv)
yx
22
2
2
44
yxyx
(v)
yx
53
xyyx
1525922
(vi)
x
53
2
25159
xx
(vii)
xx
32
694 2
2x
x
(viii)
22
3x
x
xxx
649 4
2
(ix) (1 – x) (1 + x + x2)
(x) (1 + x) (1 – x + x2)
(xi) (x2 – 1) (x4 + x2 + 1)
(xii) (x3 + 1) (x6 – x3 + 1)
Sol. We know that a3 + b3 = (a + b) (a2 – ab +
b2) and a3 – b3 = (a – b) (a2 + ab + b2)
(i) (3x + 2y) (9x2 – 6xy + 4y2)
= (3x + 2y) [(3x)2 – 3x × 2y + (2y)2]
= (3x)3 + (2y)3 = 27x3 + 8y3
(ii) (4x – 5y) (16x2 + 20xy + 25y2)
= (4x – 5y) [(4x)2 + 4x × 5y + (5y)2]
= (4x)3 – (5y)3
= 64x3 – 125y3
(iii) (7p4 + q) (49p8 – 7p4q + q2)
= (7p4 + q) [(7p4)2 – 7p4 × q + (q)2]
= (7p4)3 + (q)3 = 343p12 + q3
(iv)
yx
22
2
2
44
yxyx
=
yx
22
2
2
)2(222
yyxx
=
3
2
x
+ (2y)3 = 8
3x
+ 8y3
(v)
yx
53
xyyx
1525922
=
yx
53
225533
yyxx
=
33
x–
35
y
= 3
27
x – 3
125
y
(vi)
x
53
2
25159
xx
=
x
53
2
2 553)3(
xx
= (3)3 +
35
x= 27 + 2
125
x
(vii)
xx
32
694 2
2x
x
=
xx
32
2
296
4x
x
=
xx
32
2
2
)3(322
xxxx
=
32
x+ (3x)3 = 3
8
x + 27x3
(viii)
222
3x
xxx
649 4
2
=
222
3x
4
246
9xx
x
=
222
3x
222
2
)2(223
xxxx
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81 Arundeep’s Mathematics (R.D.) 9th
=
33
x– (2x2)3 = 3
27
x – 8x6
(ix) (1 – x) (1 + x + x2)
= (1 – x) [(1)2 + 1 × x + (x)2]
= (1)3 – (x)3 = 1 – x3
(x) (1 + x) (1 – x + x2)
= (1 + x) [(1)2 – 1 × x + (x)2]
= (1)3 + (x)3 = 1 + x3
(xi) (x2 – 1) (x4 + x2 + 1)
= (x2 – 1) [(x2)2 + x2 × 1 + (1)2]
= (x2)3 – (1)3
= x6 – 1
(xii) (x3 + 1) (x6 – x3 + 1)
= (x3 + 1) [(x3)2 – x3 × 1 + (1)2]
= (x3)3 + (1)3 = x9 + 1
2. If x = 3 and y = –1, find the values of each
of the following using in identity:
(i) (9y2 – 4x2) (81y4 + 36x2y2 + 16x4)
(ii)
3
3 x
x
1
9
9 2
2
x
x
(iii)
37
yx
21949
22 xyyx
(iv)
34
yx
91216
22 yxyx
(v)
xx
55
2
22525
25x
x
Sol. x = 3, y = –1
(i) (9y2 – 4x2) (81y4 + 36x2y2 + 16x4)
= (9y2 – 4x2) [(9y2)2 + 9y2 × 4x2 + (4x2)2]
= (9y2)3 – (4x2)3 = 729y6 – 64x6
= 729 × (–1)6 – 64(3)6
= 729 × 1 – 64 × 729
= 729 – 46656 = –45927
(ii)
3
3 x
x
1
9
9 2
2
x
x
=
3
3 x
x
22
33
33 xx
xx
=
33
x–
3
3
x
=
3
3
3
–
3
3
3
= 13 – 13 = 0
(iii)
37
yx
21949
22 xyyx
=
37
yx
22
3777
yyxx
=
3
7
x
+
3
3
y
= 343
3x
+ 27
3y
= 343
)3( 3
+ 27
)1( 3
= 343
27 –
27
1
= 9261
343729 =
9261
386
(iv)
34
yx
91216
22 yxyx
=
34
yx
22
3344
yyxx
=
3
4
x
–
3
3
y
= 64
3x
– 27
3y
= 64
)3(3
– 27
)1(3
= 64
27 +
27
1
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82 Arundeep’s Mathematics (R.D.) 9th
= 1728
64729 =
1728
793
(v)
xx
55
2
22525
25x
x
=
xx
55
2
2
)5(555
xxxx
=
35
x+ (5x)3 = 3
125
x + 125x3
= 3)3(
125 + 125 × (3)3 =
9
125 + 125 × 27
= 27
125 + 3375
= 27
91125125 =
27
91250
3. If a + b = 10 and ab = 16, find the value of
a2 – ab + b2 and a2 + ab + b2.
Sol. a + b = 10, ab = 16
Squaring,
(a + b)2 = (10)2
a2 + b2 + 2ab = 100
a2 + b2 + 2 × 16 = 100
a2 + b2 + 32 = 100
a2 + b2 = 100 – 32 = 68
Now, a2 – ab + b2 = a2 + b2 – ab
= 68 – 16 = 52
and a2 + ab + b2 = a2 + b2 + ab
= 68 + 16 = 84
4. If a + b = 8 and ab = 6, find the value of
a3 + b3.
Sol. a + b = 8, ab = 6
Cubing both sides,
(a + b)3 = (8)3
a3 + b3 + 3ab(a + b) = 512
a3 + b3 + 3 × 6 × 8 = 512
a3 + b3 + 144 = 512
a3 + b3 = 512 – 144 = 368
a3 + b3 = 368
5. If a – b = 6 and ab = 20, find the value of
a3 – b3.
Sol. a – b = 6, ab = 20
Cubing both sides,
(a – b)3 = (6)3
a3 – b3 – 3ab(a – b) = 216
a3 – b3 – 3 × 20 × 6 = 216
a3 – b3 – 360 = 216
a3 – b3 = 216 + 360 = 576
a3 – b3 = 576
6. If x = –2 and y = 1, by using an identity
find the value of the following:
(i) (4y2 – 9x2) (16y4 + 36x2y2 + 81x4)
(ii)
2
2 x
x
1
4
4 2
2
x
x
(iii)
yy
155
2
2 2257525
yy
Sol. x = –2, y = 1
(i) (4y2 – 9x2) (16y4 + 36x2y2 + 81x4)
= (4y2 – 9x2) [(4y2)2 + 4y2 × 9x2 + (9x2)2]
= (4y2)3 – (9x2)3 = 64y6 – 729x6
= 64 × 16 – 729 × (–2)6
= 64 × 1 – 729 × (64)
= 64 – 46656 = –46592
(ii)
2
2 x
x
1
4
4 2
2
x
x
=
2
2 x
x
22
22
22
2 xx
xx
=
32
x–
3
2
x
= 3
8
x –
8
3x
= 3)2(
8
– 8
)2(3
= 8
8
–
8
8
= –1 + 1 = 0
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83 Arundeep’s Mathematics (R.D.) 9th
(iii)
yy
155
2
2 2257525
yy
=
yy
155
2
2 151552)5(
yyyy
= (5y)3 +
315
y= 125y3 + 3
3375
y
= 125 × (1) + 3)1(
3375
= 125 + 1
3375 = 125 + 3375 = 3500
EXERCISE 4.5
1. Find the following products:
(i) (3x + 2y + 2z) (9x2 + 4y2 + 4z2 – 6xy – 4yz
– 6zx)
(ii) (4x – 3y + 2z) (16x2 + 9y2 + 4z2 + 12xy +
6yz – 8zx)
(iii) (2a – 3b – 2c) (4a2 + 9b2 + 4c2 + 6ab –
6bc + 4ca)
(iv) (3x – 4y + 5z) (9x2 + 16y2 + 25z2 + 12xy –
15zx + 20yz)
Sol. (i) (3x + 2y + 2z) (9x2 + 4y2 + 4z2 – 6xy –
4yz – 6zx)
= (3x + 2y + 2z) [(3x)2 + (2y)2 + (2z)2 – 3x
× 2y + 2y × 2z + 2z × 3x]
= (3x)3 + (2y)3 + (2z)3 – 3 × 3x × 2y × 2z
= 27x3 + 8y3 + 8z3 – 36xyz
(ii) (4x – 3y + 2z) (16x2 + 9y2 + 4z2 + 12xy +
6yz – 8zx)
= (4x – 3y + 2z) [(4x)2 + (–3y)2 + (2z)2 –
4x × (–3y) + (3y) × (2z) – (2z × 4x)]
= (4x)3 + (–3y)3 + (2z)3 – 3 × 4x × (–3y) × (2z)
= 64x3 – 27y3 + 8z3 + 72xyz
(iii) (2a – 3b – 2c) (4a2 + 9b2 + 4c2 + 6ab –
6bc + 4ca)
= (2a – 3b – 2c) [(2a)2 + (3b)2 + (2c)2 – 2a
× (–3b) – (–3b) × (–2c) – (–2c) × 2a]
= (2a)3 + (3b)3 + (–2c)3 – 3 × 2a × (–3b)
(–2c)
= 8a3 – 27b3 – 8c3 – 36abc
(iv) (3x – 4y + 5z) (9x2 + 16y2 + 25z2 + 12xy –
15zx + 20yz)
= [3x + (–4y) + 5z] [(3x)2 + (–4y)2 + (5z)2
– 3x × (–4y) – (–4y) (5z) – 5z × 3x]
= (3x)3 + (–4y)3 + (5z)3 – 3 × 3x × (–4y) (5z)
= 27x3 – 64y3 + 125z3 + 180xyz
2. Evaluate:
(i) 253 – 753 + 503 (ii) 483 – 303 – 183
(iii)
3
2
1
+
3
3
1
–
3
6
5
(iv) (0.2)3 – (0.3)3 + (0.1)3
Sol. (i) We know that if a + b + c = 0, then
x3 + y3 + z3 = 3xyz. Therefore,
(i) (25)3 – (75)3 + (50)3
Let 25 = a, –75 = b and 50 = c
a + b + c = 25 – 75 + 50 = 0
a3 + b3 + c3 = 3abc
253 – 753 + 503 = 3 × 25 × (–75) × 50
= –3 × 25 × 75 × 50 = –281250
(ii) 483 – 303 – 183
Let a = 48, b = –30, c = –18
a + b + c = 48 – 30 – 18 = 0, then
a3 – b3 – c3 = 3abc
= 3 × 48 × (–30) × (–18)
= 3 × 48 × 30 × 18 = 77760
(iii)
3
2
1
+
3
3
1
–
3
6
5
Let a = 2
1, b =
3
1, c =
6
5
a + b + c = 2
1 +
3
1 –
6
5
= 6
523 =
6
0 = 0
a3 + b3 + c3 = 3abc
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84 Arundeep’s Mathematics (R.D.) 9th
3
2
1
+
3
3
1
–
3
6
5
= 3 × 2
1 ×
3
1 ×
6
5
= –3 × 2
1 ×
3
1 ×
6
5 =
12
5
(iv) (0.2)3 – (0.3)3 + (0.1)3
Let a = 0.2, b = –0.3, c = 0.1
a + b + c = 0.2 – 0.3 + 0.1 = 0
a3 + b3 + c3 = 3abc
(0.2)3 – (0.3)3 + (0.1)3
= 3(0.2) (–0.3) (0.1)
= –3 × 0.006 = –0.018
3. If x + y + z = 8 and xy + yz + zx = 20, find
the value of x3 + y3 + z3 – 3xyz.
Sol. We know that
x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 +
z2 – xy – yz – zx)
Now, x + y + z = 8
Squaring, we get
(x + y + z)2 = (8)2
x2 + y2 + z2 + 2(xy + yz + zx) = 64
x2 + y2 + z2 + 2 × 20 = 64
x2 + y2 + z2 + 40 = 64
x2 + y2 + z2 = 64 – 40 = 24
Now,
x3 + y3 + z3 – 3xyz = (x + y + z) [x2 + y2 +
z2 – (xy + yz + zx)]
= 8(24 – 20) = 8 × 4 = 32
4. If a + b + c = 9 and ab + bc + ca = 26,
find the value of a3 + b3 + c3 – 3abc.
Sol. a + b + c = 9, ab + bc + ca = 26
Squaring, we get
(a + b + c)2 = (9)2
a2 + b2 + c2 + 2(ab + bc + ca) = 81
a2 + b2 + c2 + 2 × 26 = 81
a2 + b2 + c2 + 52 = 81
a2 + b2 + c2 = 81 – 52 = 29
Now, a3 + b3 + c3 – 3abc
= (a + b + c) [(a2 + b2 + c2 – (ab + bc + ca)]
= 9[29 – 26] = 9 × 3 = 27
5. If a + b + c = 9, and a2 + b2 + c2 = 35, find
the value of a3 + b3 + c3 – 3abc.
Sol. a + b + c = 9
Squaring, we get
(a + b + c)2 = (9)2
a2 + b2 + c2 + 2(ab + bc + ca) = 81
35 + 2(ab + bc + ca) = 81
2(ab + bc + ca) = 81 – 35 = 46
ab + bc + ca = 2
46 = 23
Now, a3 + b3 + c3 – 3abc
= (a + b + c) [a2 + b2 + c2 – (ab + bc + ca)]
= 9[35 – 23] = 9 × 12 = 108
VERY SHORT ANSWER TYPE QUESTIONS
(VSAQs)
1. If x + x
1 = 3, then find the value of x2 +
2
1
x.
Sol. x + x
1 = 3
Squaring both sides,
21
xx = (3)2
x2 + 2
1
x + 2 = 9 x2 + 2
1
x = 9 – 2 = 7
x2 + 2
1
x = 7
2. If x + x
1 = 3, find the value of x6 + 6
1
x.
Sol. x + x
1 = 3,
Squaring both sides,
21
xx = (3)2
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85 Arundeep’s Mathematics (R.D.) 9th
x2 + 2
1
x + 2 = 9
x2 + 2
1
x = 9 – 2 = 7
Cubing both sides,
3
2
2 1
x
x = (7)3
x6 + 6
1
x + 3
2
2 1
xx = 343
x6 + 6
1
x + 3 × 7 = 343
x6 + 6
1
x + 21 = 343
x6 + 6
1
x = 343 – 21 = 322
x6 + 6
1
x = 322
3. If a + b = 7 and ab = 12, find the value of
a2 + b2.
Sol. a + b = 7, ab = 12
Squaring both sides,
(a + b)2 = (7)2
a2 + b2 + 2ab = 49
a2 + b2 + 2 × 12 = 49
a2 + b2 + 24 = 49
a2 + b2 = 49 – 24 = 25
a2 + b2 = 25
4. If a – b = 5 and ab = 12, find the value of
a2 + b2.
Sol. a – b = 5, ab = 12
Squaring both sides,
(a – b)2 = (5)2
a2 + b2 – 2ab = 25
a2 + b2 – 2 × 12 = 25
a2 + b2 – 24 = 25
a2 + b2 = 25 + 24 = 49
a2 + b2 = 49
5. If x – x
1 =
2
1, then write the value of 4x2
+ 2
4
x.
Sol. x – x
1 =
2
1
2x – x
2 = 1 (Multiplying by 2)
Squaring both sides,
22
2
x
x = (1)2
4x2 + 2
4
x – 2 × 2x ×
x
2 = 1
4x2 + 2
4
x – 8 = 1
4x2 + 2
4
x = 1 + 8 = 9
4x2 + 2
4
x = 9
6. If a2 + 2
1
a = 102, find the value of a –
a
1.
Sol. a2 + 2
1
a = 102
21
aa = a2 + 2
1
a – 2
= 102 – 2 = 100 = (10)2
a – a
1 = 10
7. If a + b + c = 0, then write the value of
bc
a2
+ ca
b2
+ ab
c2
.
Sol.bc
a2
+ ca
b2
+ ab
c2
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86 Arundeep’s Mathematics (R.D.) 9th
= abc
cba333
a + b + c = 0
Then, a3 + b3 + c3 = 3abc
abc
cba333
= abc
abc3 = 3
MULTIPLE CHOICE QUESTIONS (MCQs)
Mark the correct alternative in each of the
following:
1. If x + x
1 = 5, then x2 + 2
1
x =
(a) 25 (b) 10
(c) 23 (d) 27
Sol. x + x
1 = 5
Squaring, we get
x2 + 2
1
x + 2 = 25
x2 + 2
1
x = 25 – 2 = 23 (c)
2. If x + x
1 = 2, then x3 + 3
1
x =
(a) 64 (b) 14
(c) 8 (d) 2
Sol. x + x
1 = 2,
Cubing, we get
x3 + 3
1
x + 3
2
1x = (2)3
x3 + 3
1
x × 3(2) = 8
x3 + 3
1
x = 8 – 6 = 2 (d)
3. If x + x
1 = 4, then x4 + 4
1
x =
(a) 196 (b) 194
(c) 192 (d) 190
Sol. x + x
1 = 4
Squaring, we get
x2 + 2
1
x + 2 = 16
x2 + 2
1
x = 16 – 2 = 14
Again squaring,
x4 + 4
1
x + 2 = 196
x4 + 4
1
x = 196 – 2 = 194 (b)
4. x + x
1 = 3, then x6 + 6
1
x =
(a) 927 (b) 414
(c) 364 (d) 322
Sol. x + x
1 = 3
Squaring, we get
x2 + 2
1
x + 2 = 9
x2 + 2
1
x = 9 – 2 = 7
Cubing both sides,
3
2
2 1
x
x = (7)3
x6 + 6
1
x + 3
2
2 1
xx = 343
x6 + 6
1
x + 3 × 7 = 343
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87 Arundeep’s Mathematics (R.D.) 9th
x6 + 6
1
x + 21 = 343
x6 + 6
1
x = 343 – 21 = 322 (d)
5. If x2 + 2
1
x = 102, then x –
x
1 =
(a) 8 (b) 10
(c) 12 (d) 13
Sol. x2 + 2
1
x = 102
Subtracting 2 from both sides
x2 + 2
1
x – 2 = 102 – 2 = 100
21
xx = (10)2
x – x
1 = 10 (b)
6. If x3 + 3
1
x = 110, then x +
x
1 =
(a) 5 (b) 10
(c) 15 (d) none of these
Sol. x3 + 3
1
x = 110
x3 + 3
1
x =
31
xx – 3
xx
1
110 =
31
xx – 3
xx
1
Let x + x
1 = a, then
a3 – 3a – 110 = 0
Factors of –110 = +1, +2, +5, +11
Let a = 1, then 1 – 3 – 110 0
Let a = –1, then –1 + 3 – 110 0
Let a = 2, then 8 – 6 – 110 0
Let a = –2, then –8 – 6 – 110 0
Let a = 5, then 125 – 15 – 110 = 0
a = 5 x + x
1 = 5 (a)
7. If x3 – 3
1
x = 14, then x +
x
1 =
(a) 5 (b) 4
(c) 3 (d) 2
Sol.
31
xx = x3 –
x
1 – 3
xx
1
31
xx = 14 – 3
xx
1
Let x – x
1 = a, then
a3 = 14 – 3a a3 + 3a – 14 = 0
Factors of –14 = +1, +2, +7
Let a= +1, then 1 + 3 – 14 0
Let a = –1, then –1 – 3 – 14 0
Let a = 2, then 8 + 6 – 14 = 0
x – x
1 = 2 (d)
8. If a + b + c = 9 and ab + bc + ca = 23,
then a2 + b2 + c2 =
(a) 35 (b) 58
(c) 127 (d) none of these
Sol. a + b + c = 9, ab + bc + ca = 23
Squaring,
(a + b + c)2 = (9)2
a2 + b2 + c2 + 2(ab + bc + ca) = 81
a2 + b2 + c2 + 2 × 23 = 81
a2 + b2 + c2 + 46 = 81
a2 + b2 + c2 = 81 – 46 = 35 (a)
9. (a – b)3 + (b – c)3 + (c – a)3 =
(a) (a + b + c) (a2 + b2 + c2 – ab – bc – ca)
(b) (a – b) (b – c) (c – a)
(c) 3(a – b) (b – c) (c – a)
(d) none of these
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88 Arundeep’s Mathematics (R.D.) 9th
Sol. (a – b)3 + (b – c)3 + (c – a)3
_ a – b + b – c + c – a = 0
(a – b)3 + (b – c)3 + (c – a)3
= 3(a – b) (b – c) (c – a) (c)
10. If b
a +
a
b = 1, then a3 + b3 =
(a) 1 (b) –1
(c)2
1(d) 0
Sol.b
a +
a
b = 1
ab
ba22
= 1 a2 + b2 = ab
Now, a3 + b3 = (a + b) (a2 – ab + b2)
= (a + b) (a2 + b2 – ab)
= (a + b) (ab – ab)
= (a + b) × 0 = 0 (d)
11. If a – b = –8 and ab = –12 then a3 – b3 =
(a) –244 (b) –240
(c) –224 (d) –260
Sol. a – b = –8, ab = –12
(a – b)3 = a3 – b3 – 3ab (a – b)
(–8)3 = a3 – b3 – 3 × (–12) (–8)
–512 = a3 – b3 – 288
a3 – b3 = –512 + 288 = –224 (c)
12. If the volume of a cuboid is 3x2 – 27, then
its possible dimensions are
(a) 3, x2, –27x (b) 3, x – 3, x + 3
(c) 3, x2, 27x (d) 3, 3, 3
Sol. Volume = 3x2 – 27 = 3(x2 – 9)
= 3(x + 3) (x – 3)
Dimensions are = 3, x – 3, x + 3 (b)
13. 75 × 75 + 2 × 75 × 25 + 25 × 25 is equal
to
(a) 10000 (b) 6250
(c) 7500 (d) 3750
Sol. 75 × 75 + 2 × 75 × 25 + 25 × 25
= (75)2 + 2 × 75 × 25 + (25)2
{_ a2 + 2ab + b2 = (a + b)2}
= (75 + 25)2 = (100)2
= 10000 (a)
14. (x – y) (x + y) (x2 + y2) (x4 + y4) is equal to
(a) x16 – y16 (b) x8 – y8
(c) x8 + y8 (d) x16 + y16
Sol. (x – y) (x + y) (x2 + y2) (x4 + y4)
= (x2 – y2) (x2 + y2) (x4 + y4)
{_ (a + b) (a – b) = a2 – b2}
= [(x2)2 – (y2)2] (x4 + y4) = (x4 – y4) (x4 + y4)
= (x4)2 – (y4)2 = x8 – y8 (b)
15. If x4 + 4
1
x = 623, then x +
x
1 =
(a) 27 (b) 25
(c) 3 3 (d) –3 3
Sol. x4 + 4
1
x = 623
Adding 2 to both sides,
x4 + 4
1
x + 2 = 623 + 2 = 625
2
2
2 1
x
x = (25)2
x2 + 2
1
x = 25
Adding 2 to both sides,
x2 + 2
1
x + 2 = 25 + 2 = 27
21
xx = ( 27 )2 = 239
x + x
1 = 3 3 (c)
16. If x4 + 4
1
x = 194, then x3 + 3
1
x =
(a) 76 (b) 52
(c) 64 (d) none of these
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89 Arundeep’s Mathematics (R.D.) 9th
Sol. x4 + 4
1
x = 194
Adding 2 to both sides,
x4 + 4
1
x + 2 = 194 + 2 = 196
2
2
2 1
x
x = (14)2 x2 + 2
1
x = 14
Adding 2 to both sides,
x2 + 2
1
x + 2 = 14 + 2 = 16
2
1
xx = (4)2
x + x
1 = 4
Cubing both sides,
31
xx = (4)3
x3 + 3
1
x + 3
xx
1 = 64
x3 + 3
1
x + 3 × 4 = 64
x3 + 3
1
x + 12 = 64
x3 + 3
1
x = 64 – 12 = 52 (b)
17. If x – x
1 =
4
15, then x +
x
1 =
(a) 4 (b)4
17
(c)4
13(d)
4
1
Sol. x – x
1 =
4
15
21
xx =
21
xx + 4
=
2
4
15
+ 4 = 16
225 + 4 =
16
64225
= 16
289 =
2
4
17
x + x
1 =
4
17(b)
18. If 3x + x
2 = 7, then
2
2 49
xx =
(a) 25 (b) 35
(c) 49 (d) 30
Sol. 3x + x
2 = 7
Squaring, we get
22
3
x
x = (7)2 = 49
9x2 + 2
4
x + 2 × 3x ×
x
2 = 49
9x2 + 2
4
x + 12 = 49
9x2 + 2
4
x = 49 – 12 = 37
22
3
x
x = 9x2 + 2
4
x – 2 × 3x ×
x
2
= 9x2 + 2
4
x – 12 = 37 – 12 = 25 = (5)2
3x – x
2 = 5
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90 Arundeep’s Mathematics (R.D.) 9th
Now,
x
x2
3
x
x2
3 = 7 × 5
(3x)2 –
22
x= 35
9x2 – 2
4
x = 35 (b)
19. If a2 + b2 + c2 – ab – bc – ca = 0, then
(a) a + b = c (b) b + c = a
(c) c + a = b (d) a = b = c
Sol. a2 + b2 + c2 – ab – bc – ca = 0
2(a2 + b2 + c2 – ab – bc – ca) = 0
(Multiplying by 2)
2a2 + 2b2 + 2c2 – 2ab – 2bc – 2ca = 0
a2 + b2 – 2ab + b2 + c2 – 2bc + c2 + a2 –
2ca = 0
(a – b)2 + (b – c)2 + (c – a)2 = 0
(a – b)2 = 0, then a – b = 0 a = b
Similarly, (b – c)2 = 0, then
b – c = 0 b = c
and (c – a)2 = 0, then c – a = 0 c = a
a = b = c (d)
20. If a + b + c = 0, then bc
a2
+ ca
b2
+ ab
c2
=
(a) 0 (b) 1
(c) –1 (d) 3
Sol. a + b + c = 0
bc
a2
+ ca
b2
+ ab
c2
= abc
a3
+ abc
b3
+ abc
c3
= abc
cba333
_ a + b + c = 0
Then, a3 + b3 + c3 = 3abc
= abc
abc3 = 3 (d)
21. If 3
1
a + 3
1
b + 3
1
c = 0, then
(a) a + b + c = 0
(b) (a + b + c)3 = 27abc
(c) a + b + c = 3abc
(d) a3 + b3 + c3 = 0
Sol. _ 3
1
a + 3
1
b + 3
1
c = 0
3
3
1
a +
3
3
1
b +
3
3
1
c = 3 3
1
a 3
1
b 3
1
c
a + b + c = 3 3
1
a 3
1
b 3
1
c
(a + b + c)3 = (3 3
1
a 3
1
b 3
1
c )3
(Cubing both sides)
(a + b + c)3 = 27abc (b)
22. If a + b + c = 9 and ab + bc + ca = 23,
then a3 + b3 + c3 – 3abc =
(a) 108 (b) 207
(c) 669 (d) 729
Sol. a3 + b3 + c3 – 3abc
= (a + b + c) [a2 + b2 + c2 – (ab + bc + ca)]
Now, a + b + c = 9
Squaring,
a2 + b2 + c2 + 2(ab + bc + ca) = 81
a2 + b2 + c2 + 2 × 23 = 81
a2 + b2 + c2 + 46 = 81
a2 + b2 + c2 = 81 – 46 = 35
Now, a3 + b3 + c3 – 3abc = (a + b + c) [(a2
+ b2 + c2) – (ab + bc + ca)]
= 9[35 – 23] = 9 × 12 = 108 (a)
23. 333
322322322
)()()(
)()()(
accbba
accbba
=
(a) 3(a + b) (b + c) (c + a)
(b) 3(a – b) (b – c) (c – a)
(c) (a – b) (b – c) (c – a)
(d) (a + b) (b + c) (c + a)
Sol. a2 – b2 + b2 – c2 + c2 – a2 = 0
(a2 – b2)3 + (b2 – c2)3 + (c2 – a2)3
= 3(a2 – b2) (b2 – c2) (c2 – a2)
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91 Arundeep’s Mathematics (R.D.) 9th
and a – b + b – c + c – a = 0
(a – b)3 + (b – c)3 + (c – a)3
= 3(a – b) (b – c) (c – a)
Now, 333
322322322
)()()(
)()()(
accbba
accbba
= )()()(3
)()()(3 222222
accbba
accbba
= )()()(3
)()()()()()(3
accbba
acaccbcbbaba
= (a + b) (b + c) (c + a) (a)
24. The product (a + b) (a – b) (a2 – ab + b2)
(a2 + ab + b2) is equal to
(a) a6 + b6 (b) a6 – b6
(c) a3 – b3 (d) a3 + b3
Sol. (a + b) (a – b) (a2 – ab + b2) (a2 + ab + b2)
= (a + b) (a2 – ab + b2) (a – b) (a2 + ab + b2)
= (a3 + b3) (a3 – b3) = (a3)2 – (b3)2
= a6 – b6 (b)
25. The product (x2 – 1) (x4 + x2 + 1) is equal
to
(a) x8 – 1 (b) x8 + 1
(c) x6 – 1 (d) x6 + 1
Sol. (x2 – 1) (x4 + x2 + 1)
= (x2)3 – (1)3 = x6 – 1 (c)
26. If b
a +
a
b = 1, then a3 + b3 =
(a) 1 (b) –1
(c)2
1(d) 0
Sol.b
a +
a
b = 1
ab
ba22
= 1
a2 + b2 = –ab
a2 + b2 + ab = 0 a2 + ab + b2 = 0
Now, a3 – b3 = (a – b) (a2 + ab + b2)
= (a – b) × 0 = 0 (d)
27. If 49a2 – b =
2
17a
2
17a , then the
value of b is
(a) 0 (b)4
1
(c)2
1(d)
2
1
Sol. 49a2 – b = (7a)2 – ( b )2
= (7a + b ) (7a – b )
(7a + b ) (7a – b ) =
2
17a
2
17a
Comparing,
b = 2
1 b =
4
1(b)
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92 Arundeep’s Mathematics (R.D.) 9th
Points to Remember :
We can factorize.
1. By taking common
2. By using identities
(a + b)2 = a2 + 2ab + b2
(a – b)2 = a2 – 2ab + b2
a2 – b2 = (a + b) (a – b)
a3 + b3 = (a + b) (a2 – ab + b2)
a3 – b3 = (a – b) (a2 + ab + b2)
(x + a) (x + b) = x2 + (a + b)x + ab
EXERCISE 5.1
Factorize
1. x3 + x – 3x2 – 3
Sol. x3 + x – 3x2 – 3
x3 – 3x2 + x – 3
x2(x – 3) + 1(x – 3)
= (x – 3) (x2 + 1)
2. a(a + b)3 – 3a2b(a + b)
Sol. a(a + b)3 – 3a2b(a + b)
= a(a + b) {(a + b)2 – 3ab}
= a(a + b) {a2 + b2 + 2ab – 3ab}
= a(a + b) (a2 – ab + b2)
3. x(x3 – y3) + 3xy(x – y)
Sol. x(x3 – y3) + 3xy(x – y)
= x(x – y) (x2 + xy + y2) + 3xy(x – y)
= x(x – y) (x2 + xy + y2 + 3y)
= x(x – y) (x2 + xy + y2 + 3y)
4. a2x2 + (ax2 + 1)x + a
Sol. a2x2 + (ax2 + 1)x + a
= a2x2 + a + (ax2 + 1)x
= a(ax2 + 1) + x(ax2 + 1)
= (ax2 + 1) (a + x) = (x + a) (ax2 + 1)
5. x2 + y – xy – x
Sol. x2 + y – xy – x
= x2 – x – xy + y
= x(x – 1) – y(x – 1)
= (x – 1) (x – y)
6. x3 – 2x2y + 3xy2 – 6y3
Sol. x3 – 2x2y + 3xy2 – 6y3
= x2(x – 2y) + 3y2(x – 2y)
= (x – 2y) (x2 + 3y2)
7. 6ab – b2 + 12ac – 2bc
Sol. 6ab – b2 + 12ac – 2bc
= 6ab + 12ac – b2 – 2bc
= 6a(b + 2c) – b(b + 2c)
= (b + 2c) (6a – b)
8. x(x – 2) (x – 4) + 4x – 8
Sol. x(x – 2) (x – 4) + 4x – 8
= x(x – 2) (x – 4) + 4(x – 2)
= (x – 2) [x(x – 4) + 4]
= (x – 2) (x2 – 4x + 4)
= (x – 2) [(x)2 – 2 × x × 2 + (2)2]
= (x – 2) (x – 2)2 = (x – 2)3
9. (a – b + c)2 + (b – c + a)2 + 2(a – b + c)
(b – c + a)
Sol. (a – b + c)2 + (b – c + a)2 + 2(a – b + c)
(b – c + a)
{_ a2 + b2 + 2ab = (a + b)2}
= [a – b + c + b – c + a]2
= (2a)2 = 4a2
10. a2 + 2ab + b2 – c2
Sol. a2 + 2ab + b2 – c2
= (a2 + 2ab + b2) – c2
= (a + b)2 – (c)2
{_ a2 – b2 = (a + b) (a – b)}
= (a + b + c) (a + b – c)
11. a2 + 4b2 – 4ab – 4c2
Sol. a2 + 4b2 – 4ab – 4c2
= (a)2 + (2b)2 – 2 × a × 2b – (2c)2
= (a – 2b)2 – (2c)2
(a – 2b + 2c) (a – 2b – 2c)
22
222
)()(
)(2
bababa
bababa
12. x2 – y2 – 4xz + 4z2
Sol. x2 – y2 – 4xz + 4z2
= x2 – 4xz + 4z2 – y2
FACTORIZATION OF ALGEBRAIC EXPRESSIONS5
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93 Arundeep’s Mathematics (R.D.) 9th
= (x)2 – 2 × x × 2z + (2z)2 – (y)2
= (x – 2z)2 – (y)2
= (x – 2z + y) (x – 2z – y)
= (x +y – 2z) (x – y – 2z)
13. 2x2 – 6
5x +
12
1
Sol. 2x2 – 6
5x +
12
1
2x2 – 2
1x –
3
1x +
12
1
x
2
12x –
6
1
2
12x
3
1
2
1
6
5
3
1
2
1
6
1
6
1
12
2
12
12
2
12x
6
1x
14. x2 + 35
12x +
35
1
Sol. x2 + 35
12x +
35
1
7
1
5
1
35
12
7
1
5
1
35
1
= x2 + 5
1x +
7
1x +
35
1
= x
5
1x +
7
1
5
1x
=
5
1x
7
1x
15. 21x2 – 2x + 21
1
Sol. 21x2 – 2x + 21
1
= 221x – 2 × 21x × 21
1 +
2
21
1
=
2
21
121
x
16. Give possible expression for the length and
breadth of the rectangle having 35y2 + 13y
– 12 as its area.
Sol. Area of a rectangle = 35y2 + 13y – 12
= 35y2 + 28y – 15y – 12
152813
)15(28420
420)12(35
= 7y(5y + 4) – 3(5y + 4)
= (5y + 4) (7y – 3)
(i) If length = 5y + 4, then breadth = 7y – 3
(ii) and if length = 7y – 3, then length = 5y + 4
17. What are the possible expressions for the
dimensions of the cuboid whose volume is
3x2 – 12x.
Sol. Volume 3x2 – 12x
= 3x(x – 4)
Factors are 3, x, and x – 4
Now, if length = 3, breadth = x and height
= x – 4
if length = 3, breadth = x – 4, height = x
if length = x, breadth = 3, height = x – 4
if length = x, breadth = x – 4, height = 3
if length = x – 4, breadth = 3, height = x
if length = x – 4, breadth = x, height = 3
18.
2
2 1
xx – 4
xx
1 + 6
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94 Arundeep’s Mathematics (R.D.) 9th
Sol.
2
2 1
xx – 4
xx
1 + 6
212
2
xx – 4
xx
1 + 4
2
1
xx – 2 × 2
xx
1 + (2)2
=
2
21
x
x {_ a2 – 2ab + b2 = (a – b)2}
19. (x + 2) (x2 + 25) – 10x2 – 20x
Sol. (x + 2) (x2 + 25) – 10x2 – 20x
= (x + 2) (x2 + 25) – 10x(x + 2)
= (x + 2) [x2 + 25 – 10x]
= (x + 2) [(x)2 – 2 × x × 5 + (5)2]
= (x + 2) (x – 5)2
20. 2a2 + 2 6 ab + 3b2
Sol. 2a2 + 2 6 ab + 3b2
= ( 2 a)2 + 2 × 2 a × 3 b + ( 3 b)2
= ( 2 a + 3 b)2
21. a2 + b2 + 2(ab + bc + ca)
Sol. a2 + b2 + 2(ab + bc + ca)
= a2 + b2 + 2ab + 2bc + 2ca
= (a + b)2 + 2c(b + a)
= (a + b)2 + 2c(a + b)
= (a + b) (a + b + 2c)
22. 4(x – y)2 – 12(x – y) (x + y) + 9(x + y)2
Sol. 4(x – y)2 – 12(x – y) (x + y) + 9(x + y)2
= [2(x – y)]2 + 2 × 2(x – y) × 3(x + y) +
[3(x + y]2
{_ a2 + b2 + 2abc = (a + b)2}
= [2(x – y) + 3(x + y)]2
= (2x – 2y + 3x + 3y)2 = (5x + y)2
23. a2 – b2 + 2bc – c2
Sol. a2 – b2 + 2bc – c2
= a2 – (b2 – 2bc + c2)
{_ a2 + b2 – 2abc = (a – b)2}
= a2 – (b – c)2
= (a)2 – (b – c)2
{_ a2 – b2 = (a + b) (a – b)}
= (a + b – c) (a – b + c)
24. xy9 – yx9
Sol. xy9 – yx9 = xy(y8 – x8)
= –xy(x8 – y8)
= –xy[(x4)2 – (y4)2]
= –xy(x4 + y4) (x4 – y4)
{_ a2 – b2 = (a + b) (a – b)}
= –xy (x4 + y4) {(x2)2 – (y2)2}
= –xy(x4 + y4) (x2 + y2) (x2 – y2)
= –xy(x4 + y4) (x2 + y2) (x + y) (x – y)
= –xy(x – y) (x + y) (x2 + y2) (x4 + y4)
25. x4 + x2y2 + y4
Sol. x4 + x2y2 + y4 = (x2)2 + 2x2y2 + y4 – x2y2
(Adding and subtracting x2y2)
= (x2 + y2)2 – (xy)2
{_ a2 – b2 = (a + b) (a – b)}
= (x2 + y2 + xy) (x2 + y2 – xy)
= (x2 + xy + y2) (x2 – xy + y2)
26. x2 + 6 2 x + 10
Sol. x2 + 6 2 x + 10
= x2 + 5 2 x + 2 x + 10
22510
22526
= x(x + 5 2 ) + 2 (x + 5 2 )
= (x + 5 2 ) (x + 2 )
27. x2 – 2 2 x – 30
Sol. x2 – 2 2 x – 30
= x2 – 5 2 x + 3 2 x – 30
= x(x – 5 2 ) + 3 2 (x – 5 2 x)
= (x – 5 2 ) (x + 3 2 )
28. x2 – 3 x – 6
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95 Arundeep’s Mathematics (R.D.) 9th
Sol. x2 – 3 x – 6
= x2 – 2 3 x + 3 x – 6
= x(x – 2 3 ) + 3 (x – 2 3 )
3323
3326
= (x – 2 3 ) (x + 3 )
29. x2 + 5 5 x + 30
Sol. x2 + 5 5 x + 30
= x2 + 3 5 x + 2 5 + 30
525330
525330 xx
= x(x + 3 5 ) + 2 5 (x + 3 5 )
= (x + 3 5 ) (x + 2 5 )
30. x2 + 2 3 x – 24
Sol. x2 + 2 3 x – 24
323432
)32(3424
x2 + 4 3 x – 2 3 x – 24
x(x + 4 3 ) – 2 3 (x + 4 3 )
= (x + 4 3 ) (x – 2 3 )
31. 5 5 x2 + 20x + 3 5
Sol. 5 5 x2 + 20x + 3 5
= 5 5 x2 + 5x + 15x + 3 5
51520
51575
755355
= 5 x(5x + 5 ) + 3(5x + 5 )
= (5x + 5 ) ( 5 x + 3)
32. 2x2 + 3 5 x + 5
Sol. 2x2 + 3 5 x + 5
55253
55210
1052
= 2x2 + 2 5 x + 5 x + 5
= 2x(x + 5 ) + 5 (x + 5 )
= (x + 5 ) (2x + 5 )
33. 9(2a – b)2 – 4(2a – b) – 13
Sol. 9(2a – b)2 – 4(2a – b) – 13
Let 2a – b = x, then
9x2 – 4x – 13
9134
913117
117)13(9
9x2 + 9x – 13x – 13
9x(x + 1) – 13(x + 1)
(x + 1) (9x – 13)
(2a – b + 1) {9 (2a – b) – 13}
(2a – b + 1) (18a – 9b – 13)
34. 7(x – 2y)2 – 25(x – 2y) + 12
Sol. 7(x – 2y)2 – 25(x – 2y) + 12
Let x – 2y = a
7a2 – 25a + 12
42125
)4(2184
84127
= 7a2 – 21a – 4a + 12
= 7a(a – 3) – 4(a – 3)
= (a – 3) (7a – 4)
= (x – 2y – 3) (7x – 14y – 4)
35. 2(x + y)2 – 9(x + y) – 5
Sol. 2(x + y)2 – 9(x + y) – 5
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96 Arundeep’s Mathematics (R.D.) 9th
Let x + y = a, then
2a2 – 9a – 5
1109
11010
10)5(2
= 2a2 – 10a + a – 5
= 2a(a – 5) + 1(a – 5)
= (a – 5) (2a + 1)
= (x + y – 5) (2x + 2y + 1)
EXERCISE 5.2
Factorize each of the following expressions:
1. p3 + 27
Sol. We know that a3 + b3 = (a + b) (a2 – ab +
b2)
a3 – b3 = (a – b) (a2 + ab + b2)
p3 + 27 = (p)3 + (3)3
= (p + 3) (p2 – p × 3 + 32)
= (p + 3) (p2 – 3p + 9)
2. y3 + 125
Sol. y3 + 125 = (p)3 + (5)3
= (p + 5) (p2 – 5y + 52)
= (p + 5) (p2 – 5y + 25)
3. 1 – 27a3
Sol. 1 – 27a3 = (1)3 – (3a)3
= (1 – 3a) [12 + 1 × 3a + (3a)2]
= (1 – 3a) (1 + 3a + 9a2)
4. 8x3y3 + 27a3
Sol. 8x3y3 + 27a3
= (2xy + 3a) [(2xy)2 – 2xy × 3a + (3a)2]
= (2xy + 3a) (4x2y2 – 6xya + 9a2)
5. 64a3 – b3
Sol. 64a3 – b3 = (4a)3 – (b)3
= (4a – b) [(4a)2 + 4a × b + (b)2]
= (4a – b) (16a2 + 4ab + b2)
6.216
3x
– 8y3
Sol.216
3x
– 8y3 =
3
6
x
– (2y)3
=
yx
26
2
2
)2(266
yyxx
=
yx
26
2
2
4336
yxyx
7. 10x4y – 10xy4
Sol. 10x4y – 10xy4 = 10xy(x3 – y3)
= 10xy(x – y) (x2 + xy + y2)
8. 54x6y + 2x3y4
Sol. 54x6y + 2x3y4 = 2x3y(27x3 + y3)
= 2x3y[(3x)3 + (y)3]
= 2x3y(3x + y) [(3x)2 – 3x × y + y2]
= 2x3y(3x + y) (9x2 – 3xy + y2)
9. 32a3 + 108b3
Sol. 32a3 + 108b3
= 4(8a3 + 27b3) = 4[(2a)3 + (3b)3]
= 4(2a + 3b) [(2a)2 – 2a × 3b + (3b)2]
= 4(2a + 3b) (4a2 – 6ab + 9b2)
10. (a – 2b)3 – 512b3
Sol. (a – 2b)3 – 512b3
= (a – 2b)3 – (8b)3
= (a – 2b – 8b) [(a – 2b)2 + (a – 2b) × 8b
+ (8b)2]
= (a – 10b) [a2 + 4b2 – 4ab + 8ab – 16b2 +
64b2]
= (a – 10b) (a2 + 4ab + 52b2)
11. 8x2y3 – x5
Sol. 8x2y3 – x5 = x2(8y3 – x3)
= x2[(2y)3 – (x)3]
= x2[(2y – x) (2y)2 + 2y × x + (x)2]
= x2(2y – x) (4y2 + 2xy + x2)
12. 1029 – 3x3
Sol. 1029 – 3x3 = 3(343 – x3)
= 3[(7)3 – (x)3]
= 3(7 – x) (49x + 7x + x2)
13. x3y3 + 1
Sol. x3y3 + 1 = (xy)3 + (1)3
= (xy + 1) [(xy)2 – xy × 1 + (1)2]
= (xy + 1) (x2y2 – xy + 1)
= (xy + 1) (x2y2 – xy + 1)
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97 Arundeep’s Mathematics (R.D.) 9th
14. x4y4 – xy
Sol. x4y4 – xy = xy(x3y3 – 1)
= xy[(xy)3 – (1)3]
= xy(xy – 1) [x2y2 + 2xy + 1]
15. a3 + b3 + a + b
Sol. a3 + b3 + a + b
= (a + b) (a2 – ab + b2) + 1(a + b)
= (a + b) (a2 – ab + b2 + 1)
16. Simplify:
(i)127127127173173173
127127127173173173
(ii)555555155155155
555555155155155
(iii)2.02.02.02.12.12.1
2.02.02.02.12.12.1
Sol. (i) 127127127173173173
127127127173173173
Let 173 = a and 127 = b, then
= 22
33
baba
ba
= 22
22)()(
baba
bababa
= a + b = 173 + 127 = 300
(ii)555555155155155
555555155155155
Let 155 = a and 55 = b, then
= 22
33
baba
ba
= 22
22)()(
baba
bababa
= a – b = 155 – 55 = 100
(iii)2.02.02.02.12.12.1
2.02.02.02.12.12.1
Let 1.2 = a and 0.2 = b, then
22
33
baba
ba
= 22
22)()(
baba
bababa
= a – b
= 1.2 – 0.2 = 1.0 = 1
17. (a + b)3 – 8(a – b)3
Sol. (a + b)3 – 8(a – b)3
= (a + b)3 – (2a – 2b)3
= (a + b – 2a + 2b) [(a + b)2 + (a + b) (2a
– 2b) + (2a – 2b)2)]
= (3b – a) [a2 + b2 + 2ab + 2a2 – 2ab +
2ab – 2b2 + 4a2 – 8ab + 4b2]
= (3b – a) [7a2 – 6ab + 3b2]
18. (x + 2)3 + (x – 2)3
Sol. (x + 2)3 + (x – 2)3
= (x + 2 + x – 2) [(x + 2)2 – (x + 2) (x – 2)
+ (x – 2)2]
= 2x [x2 + 4x + 4 – (x2 + 2x – 2x – 4) + x2
– 4x + 4]
= 2x[x2 + 4x + 4 – x2 – 2x + 2x + 4 + x2 –
4x + 4]
= 2x[x2 + 12]
19. x6 + y6
Sol. x6 + y6 = (x2)3 + (y2)3
= (x2 + y2) [x4 – x2y2 + y4]
20. a12 + b12
Sol. a12 + b12 = (a4)3 + (b4)3
= (a4 + b4) [(a4)2 – a4b4 + (b4)2]
= (a4 + b4) (a8 – a4b4 + b8)
21. x3 + 6x2 + 12x + 16
Sol. x3 + 6x2 + 12x + 16
= (x)3 + 3.x2.2 + 3.x.4 + (2)3 + 8
{_ a3 + 3a2b + 3ab2 + b3 = (a + b)3}
= (x + 2)3 + 8 = (x + 2)3 + (2)3
= (x + 2 + 2) [(x + 2)2 – (x + 2) × 2 + (2)2]
{_ a3 + b2 = (a + b) (a2 – ab + b2}
= (x + 4) (x2 + 4x + 4 – 2x – 4 + 4)
= (x + 4) (x2 + 2x + 4)
22. a3 – 3
1
a – 2a +
a
2
Sol. a3 – 3
1
a – 2a +
a
2
=
a
a1
2
2 11
aa – 2
a
a1
=
a
a1
21
12
2
aa
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98 Arundeep’s Mathematics (R.D.) 9th
=
a
a1
2
2 11
aa
=
a
a1
11
2
2
aa
23. a3 + 3a2b + 3ab2 + b3 – 8
Sol. a3 + 3a2b + 3ab2 + b3 – 8
= (a + b)3 – (2)3
= (a + b – 2) [(a + b)2 + (a + b) × 2 + (2)2]
= (a + b – 2) (a2 + b2 + 2ab + 2a + 2b + 4)
= (a + b – 2) (a2 + b2 + 2ab + 2(a + b) + 4]
= (a + b – 2) [(a + b)2 + 2(a + b) + 4]
24. 8a3 – b3 – 4ax + 2bx
Sol. 8a3 – b3 – 4ax + 2bx
(2a)3 – (b)3 – 2x(2a – b)
= (2a – b) [(2a)2 + 2a × b + (b)2] – 2x(2a – b)
= (2a – b) [4a2 + 2ab + b2] – 2x(2a – b)
= (2a – b) [4a2 + 2ab + b2 – 2x]
EXERCISE 5.3
Factorize:
1. 64a3 + 125b3 + 240a2b + 300ab2
Sol. 64a3 + 125b3 + 240a2b + 300ab2
= (4a)3 + (5b)3 + 3 × (4a)2 × 5b + 3(4a) +
(5b)2
= (4a + 5b)3 = (4a + 5b) (4a + 5b) (4a +
5b)
2. 125x3 – 27y3 – 225x2y + 135xy2
Sol. 125x3 – 27y3 – 225x2y + 135xy2
= (5x)3 – (3y)3 – 3 × (5x)2 × (3y) + 3 × 5x
× (3y)2
= (5x – 3y)3 = (5x – 3y) (5x – 3y) (5x – 3y)
3.27
8x3 + 1 +
3
4x2 + 2x
Sol.27
8x3 + 1 +
3
4x2 + 2x
=
3
3
2
x + (1)3 + 3 ×
2
3
2
x × 1 + 3 ×
3
2x × (1)2
=
3
13
2
x =
13
2x
13
2x
13
2x
4. 8x3 + 27y3 + 36x2y + 54xy2
Sol. 8x3 + 27y3 + 36x2y + 54xy2
= (2x)3 + (3y)3 + 3 × (2x)2 × 3y + 3 × 2x ×
(3y)2
= (2x + 3y)3 = (2x + 3y) (2x + 3y) (2x + 3y)
5. a3 – 3a2b + 3ab2 – b3 + 8
Sol. a3 – 3a2b + 3ab2 – b3 + 8
= (a – b)3 + (2)3
= (a – b + 2) [(a – b)2 – (a – b) × 2 + (2)2]
= (a – b + 2) (a2 + b2 – 2ab – 2a + 2b + 4)
6. x3 + 8y3 + 6x2y + 12xy2
Sol. x3 + 8y3 + 6x2y + 12xy2
= (x)3 + (2y)3 + 3 × x2 × 2y + 3 × x × (2y)2
= (x + 2y)3 = (x + 2y) (x + 2y) (x + 2y)
7. 8x3 + y3 + 12x2y + 6xy2
Sol. 8x3 + y3 + 12x2y + 6xy2
= (2x)3 + (y)3 + 3 × (2x)2 × y + 3 × 2x × y2
= (2x + y)3 = (2x + y) (2x + y) (2x + y)
8. 8a3 + 27b3 + 36a2b + 54ab2
Sol. 8a3 + 27b3 + 36a2b + 54ab2
= (2a)3 + (3b)3 + 3 × (2a)2 × 3b + 3 × 2a
× (3b)2
= (2a + 3b)3 = (2a + 3b) (2a + 3b) (2a + 3b)
9. 8a3 – 27b3 – 36a2b + 54ab2
Sol. 8a3 – 27b3 – 36a2b + 54ab2
= (2a)3 – (3b)3 – 3 × (2a)2 × 3b + 3 × 2a ×
(3b)2
= (2a – 3b)3 = (2a – 3b) (2a – 3b) (2a – 3b)
10. x3 – 12x(x – 4) – 64
Sol. x3 – 12x(x – 4) – 64
= x3 – 12x2 + 48x – 64
= (x)3 – 3 × x2 × 4 + 3 × x × (4)2 – (4)3
= (x – 4)3 = (x – 4) (x – 4) (x – 4)
11. a3x3 – 3a2bx2 + 3ab2x – b3
Sol. a3x3 – 3a2bx2 + 3ab2x – b3
= (ax)3 – 3 × (ax)2 × b + 3 × ax × (b)2 – (b)3
= (ax – b)3 = (ax – b) (ax – b) (ax – b)
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99 Arundeep’s Mathematics (R.D.) 9th
EXERCISE 5.4
Factorize each of the following expressions:
1. a3 + 8b3 + 64c3 – 24abc
Sol. We know that a3 + b3 + c3 – 3abc = (a + b
+ c) (a2 + b2 + c2 – ab – bc – ca)
a3 + 8b3 + 64c3 – 24abc
= (a)3 + (2b)3 + (4c)3 – 3 × a × 2b × 4c
= (a + 2b + 4c) [(a)2 + (2b)2 + (4c)2 – a ×
2b – 2b × 4c – 4c × a]
= (a + 2b + 4c) (a2 + 4b2 + 16c2 – 2ab –
8bc – 4ca)
2. x3 – 8y3 + 27z3 + 18xyz
Sol. x3 – 8y3 + 27z3 + 18xyz
= (x)3 + (–2y)3 + (3z)3 – 3 × x × (–2y) (3z)
= (x – 2y + 3z) (x2 + 4y2 + 9z2 + 2xy + 6yz
– 3zx)
3. 27x3 – y3 – z3 – 9xyz [NCERT]
Sol. 27x3 – y3 – z3 – 9xyz
= (3x)3 + (–y)3 + (–z)3 – 3 × 3x × (–y) (–z)
= (3x – y – z) [(3x)2 + (–y)2 + (–z)2 – 3x ×
(–y) – (–y) (–z) – (–z × 3x)]
= (3x – y – z) (9x2 + y2 + z2 + 3xy – yz +
3zx)
4.27
1x3 – y3 + 125z3 + 5xyz
Sol.27
1x3 – y3 + 125z3 + 5xyz
=
3
3
1
x + (–y)3 + (5z)3 – 3 × 3
x × (–y) × 5z
=
zyx 53
1
2
3
1x + (–y)2 + (5z)2 –
3
1x × (–y) – (–y) × 5z) – 5z ×
x
3
1
=
zyx 53
1 2
9
1x + y2 + 25z2 +
3
1xy
+ 5yz –
zx3
5
5. 8x3 + 27y3 – 216z3 + 108xyz
Sol. 8x3 + 27y3 – 216z3 + 108xyz
= (2x)3 + (3y)3 + (6z)3 – 3 × (2x) (3y) (–6z)
= (2x + 3y – 6z) [(2x)2 + (3y)2 + (–6z)2 –
2x × 3y – 3y × (–6z) – (–6z) × 2x]
= (2x + 3y – 6z) (4x2 + 9y2 + 36z2 – 6xy +
18yz + 12zx)
6. 125 + 8x3 – 27y3 + 90xy
Sol. 125 + 8x3 – 27y3 + 90xy
= (5)3 + (2x)3 + (–3y)3 – [3 × 5 × 2x × (–3y)]
= (5 + 2x – 3y) [(5)2 + (2x)2 + (–3y)2 – 5 ×
2x – 2x (–3y) – (–3y) × 5]
= (5 + 2x – 3y) (25 + 4x2 + 9y2 – 10x + 6xy
+ 15y)
7. 8x3 – 125y3 + 180xy + 216
Sol. 8x3 – 125y3 + 180xy + 216
= (2x)3 + (–5y)3 + (6)3 – 3 × 2x (–5y) × 6
= (2x – 5y + 6) [(2x)2 + (–5y)2 + (6)2 – 2x
× (–5y) – (–5y) × 6 – 6 × 2x]
= (2x – 5y + 6) (4x2 + 25y2 + 36 + 10xy +
30y – 12x
8. Multiply:
(i) x2 + y2 + z2 – xy + xz + yz by x + y – z
(ii) x2 + 4y2 + z2 + 2xy + xz – 2yz by x – 2y – z
(iii) x2 + 4y2 + 2xy – 3x + 6y + 9 by x – 2y + 3
(iv) 9x2 + 25y2 + 15xy + 12x – 20y + 16 by 3x
– 5y + 4
Sol. (i) (x2 + y2 + z2 – xy + yz + zx)
by (x + y – z)
= x3 + y3 – z3 + 3xyz
(ii) (x2 + 4y2 + z2 + 2xy + xz – 2yz)
by (x – 2y – z)
= (x – 2y – z) [x2 + (–2y)2 + (–z)2 – x × (–
2y) – (–2y) (z) – (–z) (x)]
= x3 + (–2y)3 + (–z)3 – 3x (–2y) (–z)
= x3 – 8y3 – z3 – 6xyz
(iii) x2 + 4y2 + 2xy – 3x + 6y + 9 by x – 2y + 3
= (x – 2y + 3) (x2 + 4y2 + 9 + 2xy + 6y –
3x)
= (x)3 + (–2y)3 + (3)3 – 3 × x × (–2y) × 3
= x3 – 8y3 + 27 + 18xy
(iv) 9x2 + 25y2 + 15xy + 12x – 20y + 16 by 3x
– 5y + 4
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100 Arundeep’s Mathematics (R.D.) 9th
= (3x – 5y + 4) [(3x)2 + (–5y)2 + (4)2 – 3x
× (–5y) (–5y × 4) – (4 × 3x)]
= (3x)3 + (–5y)3 + (4)3 – 3 × 3x (–5y) × 4
= 27x3 – 125y3 + 64 + 180xy
9. (3x – 2y)3 + (2y – 4z)3 + (4z – 3x)3
Sol. (3x – 2y)3 + (2y – 4z)3 + (4z – 3x)3
_ 3x – 2y + 2y – 4z + 4z – 3x = 0
(3x – 2y)3 + (2y – 4z)3 + (4z – 3x)3
= 3(3x – 2y) (2y – 4z) (4z – 3x)
{_ x3 + y3 + z3 = 3xyz if x + y + z = 0}
10. (2x – 3y)3 + (4z – 2x)3 + (3y – 4z)3
Sol. (2x – 3y)3 + (4z – 2x)3 + (3y – 4z)3
_ 2x – 3y + 4z – 2x + 3y – 4z = 0
(2x – 3y)3 + (4z – 2x)3 + (3y – 4z)3
= (2x – 3y) (4z – 2x) (3y – 4z)
{_ x3 + y3 + z3 = 3xyz if x + y + z = 0}
11.
3
32
z
yx
+
3
3
2
3
zyx
+
3
3
4
36
5
zyx
Sol.
3
32
z
yx
+
3
3
2
3
zyx
+
3
3
4
36
5
zyx
_
2
x + y +
3
z +
3
x –
3
2y + z +
6
5x –
3
y
– 3
4z
= 2
x +
3
x –
6
5x + y –
3
2y –
3
y +
3
z + z
– 3
4z
= 6
523 xxx +
3
23 yyy +
3
43 zzz
= 0 + 0 + 0 = 0
_
3
32
z
yx
+
3
3
2
3
zyx
+
3
3
4
36
5
zyx
=
32
zy
x
zyx
3
2
3
3
4
36
5 zyx
12. (a – 3b)3 + (3b – c)3 + (c – a)3
Sol. (a – 3b)3 + (3b – c)3 + (c – a)3
_ a – 3b + 3b – c + c – a = 0
(a – 3b)3 + (3b – c)3 + (c – a)3
= 3(a – 3b) (3b – c) (c – a)
{_ a3 + b3 + c3 = 3abc if a + b + c = 0}
13. 2 2 a3 + 3 3 b3 + c3 – 3 6 abc
Sol. 2 2 a3 + 3 3 b3 + c3 – 3 6 abc
= ( 2 a)3 + ( 3 b)3 + (c)3 – 3 × 2 a ×
3 b × c
= ( 2 a + 3 b + c) [( 2 a)2 + ( 3 b)2 +
c2 – 2 a × 3 b – 3 b × c – c × 2 a]
= ( 2 a + 3 b + c) (2a2 + 3b2 + c2 –
6 ab – 3 bc – 2 ca)
14. 3 3 a3 – b3 – 5 5 c3 – 3 15 abc
Sol. 3 3 a3 – b3 – 5 5 c3 – 3 15 abc
( 3 a)3 + (–b)3 + (– 5 c)3 – 3 3 a × (–b)
× (– 5 c)
= ( 3 a – b – 5 c) [( 3 a)2 + (–b)2 +
(– 5 c)2 – 3 a × (–b) – (–b) (– 5 c) –
(– 5 c) × 3 a
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101 Arundeep’s Mathematics (R.D.) 9th
= ( 3 a – b – 5 c) (3a2 + b2 + 5c2 +
3 ab – 5 bc + 15 ca)
15. 2 2 a3 + 16 2 b3 + c3 – 12abc
Sol. (2 2 a)3 + (2 2 b)3 + (c)3 – 3 × 2 a ×
2 2 b × c
= ( 2 a + 2 2 b + c) [( 2 a)2 + (2 2 b)2
+ c2 – 2 a × 2 2 b – 2 2 b × c – c ×
2 a
= ( 2 a + 2 2 b + c) (2a2 + 8b2 + c2 –
4ab – 2 2 bc – 2 ca)
16. Find the value of x3 + y3 – 12xy + 64, when
x + y = –4
Sol. x3 + y3 – 12xy + 64
x + y = –4
Cubing both sides,
x3 + y3 + 3xy(x + y) = –64
Substitute the value of (x + y)
x3 + y3 + 3xy × (–4) = –64
x3 + y3 – 12xy + 64 = 0
VERY SHORT ANSWER TYPE QUESTIONS
(VSAQs)
1. If a + b + c = 0, then write the value of a3
+ b3 + c3.
Sol. _ a + b + c = 0,
Then a3 + b3 + c3 = 3abc
2. If a2 + b2 + c2 = 20 and a + b + c = 0, find
ab + bc + ca.
Sol. a2 + b2 + c2 = 20, a + b + c = 0
(a + b + c)2 = 0
a2 + b2 + c2 + 2(ab + bc + ca) = 0
20 + 2(ab + bc + ca) = 0
2(ab + bc + ca) = –20
ab + bc + ca = 2
20 = –10
3. If a + b + c = 9 and ab + bc + ca = 40,
find a2 + b2 + c2.
Sol. a + b + c = 9, ab + bc + ca = 40
Squaring both sides,
(a + b + c)2 = (9)2
a2 + b2 + c2 + 2(ab + bc + ca) = 81
a2 + b2 + c2 + 2 × 40 = 81
a2 + b2 + c2 + 80 = 81
a2 + b2 + c2 = 81 – 80 = 1
4. If a2 + b2 + c2 = 250 and ab + bc + ca = 3,
find a + b + c.
Sol. a2 + b2 + c2 = 250, ab + bc + ca = 3
(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
= 250 + 2 × 3 = 250 + 6 = 256
= (+16)2
a + b + c = +16
5. Write the value of 253 – 753 + 503.
Sol. 253 – 753 + 503
Let a = 25, b = –75 and c = 50
_ a + b + c = 25 – 75 + 50 = 0
a3 + b3 + c3 = 3abc
253 + (–75)3 + 503 = 3 × 25 × (–75) × 50
= –281250
6. Write the value of 483 – 303 – 183.
Sol. 483 – 303 – 183
Let a = 48, b = –30, c = –18
_ a + b + c = 48 – 30 – 18 = 0
a3 + b3 + c3 = 3abc
483 – 303 – 183 = 3 × 48 × (–30) (–18)
= 77760
7. Write the value of
3
2
1
+
3
3
1
–
3
6
5
.
Sol.
3
2
1
+
3
3
1
–
3
6
5
Let a = 2
1, b =
3
1 and c =
6
5
_ a + b + c = 2
1 +
3
1 –
6
5
= 6
523 =
6
0 = 0
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102 Arundeep’s Mathematics (R.D.) 9th
a3 + b3 + c3 = 3abc
3
2
1
+
3
3
1
–
3
6
5
= 3 × 2
1 ×
3
1 ×
6
5 =
12
5
8. Write the value of 303 + 203 – 503.
Sol. 303 + 203 – 503
Let a = 30, b = 20, c = –50
_ a + b + c = 30 + 20 – 50 = 50 – 50 = 0
a3 + b3 + c3 = 3abc
303 + 203 – 503 = 3 × 30 × 20 × (–50)
= 90000
9. Factorize: x4 + x2 + 25.
Sol. x4 + x2 + 25
(x2)2 + (5)2 + 2x2 × 5 – 2x2 × 5 + x2
(x2)2 + (5)2 + 10x2 – 10x2 + x2
= (x2)2 + (5)2 + 10x2 – 9x2
= (x2 + 5)2 – (3x)2
{_ a2 – b2 = (a + b) (a – b)}
= (x2 + 5 – 3x) (x2 + 5 + 3x)
= (x2 – 3x + 5) (x2 + 3x + 5)
10. Factorize: x2 – 1 – 2a – a2.
Sol. x2 – 1 – 2a – a2
= x2 – (1 + 2a + a2) = (x)2 – (1 + a)2
{_ a2 – b2 = (a + b) (a – b)}
= (x + 1 + a) (x – 1 – a)
= (x + a + 1) (x – a – 1)
MULTIPLE CHOICE QUESTIONS (MCQs)
Mark the correct alternative in each of the
following:
1. The factors of x3 – x2y – xy2 + y3 are
(a) (x + y) (x2 – xy + y2)
(b) (x + y) (x2 + xy + y2)
(c) (x + y)2 (x – y)
(d) (x – y)2 (x + y)
Sol. x3 – x2y – xy2 + y3
= x3 + y3 – x2y – xy2
= (x + y) (x2 – xy + y2) – xy(x + y)
= (x + y) (x2 – xy + y2 – xy)
= (x + y) (x2 – 2xy + y2) = (x + y) (x – y)2
(d)
2. The factors of x3 – 1 + y3 + 3xy are
(a) (x – 1 + y) (x2 + 1 + y2 + x + y – xy)
(b) (x + y + 1) (x2 + y2 + 1 – xy – x – y)
(c) (x – 1 + y) (x2 – 1 – y2 + x + y + xy)
(d) 3(x + y – 1) (x2 + y2 – 1)
Sol. x3 – 1 + y3 + 3xy
= (x)3 + (–1)3 + (y)3 – 3 × x × (–1) × y
= (x – 1 + y) (x2 + 1 + y2 + x + y – xy)
= (x – 1 + y) (x2 + 1 + y2 + x + y – xy)
(a)
3. The factors of 8a3 + b3 – 6ab + 1 are
(a) (2a + b – 1) (4a2 + b2 + 1 – 3ab – 2a)
(b) (2a – b + 1) (4a2 + b2 – 4ab + 1 – 2a + b)
(c) (2a + b + 1) (4a2 + b2 + 1 – 2ab – b – 2a)
(d) (2a – 1 + b) (4a2 + 1 – 4a – b – 2ab)
Sol. 8a3 + b3 – 6ab + 1
= (2a)3 + (b)3 + (1)3 – 3 × 2a × b × 1
= (2a + b + 1) [(2a)2 + b2 + 1 – 2a × b –
b × 1 – 1 × 2a]
= (2a + b + 1) (4a2 + b2 + 1 – 2ab – b –
2a) (c)
4. (x + y)3 – (x – y)3 can be factorized as
(a) 2y(3x2 + y2) (b) 2x(3x2 + y2)
(c) 2y(3y2 + x2) (d) 2x(x2 + 3y2)
Sol. (x + y)3 – (x – y)3
= (x + y – x + y) [(x + y)2 + (x + y) (x – y)
+ (x – y)2]
= 2y(x2 + y2 + 2xy + x2 – y2 + x2 + y2 – 2xy)
= 2y(3x2 + y2) (a)
5. The expression (a – b)3 + (b – c)3 + (c –
a)3 can be factorized as
(a) (a – b) (b – c) (c – a)
(b) 3(a – b) (b – c) (c – a)
(c) –3(a – b) (b – c) (c – a)
(d) (a + b + c) (a2 + b2 + c2 – ab – bc – ca)
Sol. (a – b)3 + (b – c)3 + (c – a)3
Let a – b = x, b – a = y, c – a = z
x3 + y3 + z3
x + y + z = a – b + b – c + c – a = 0
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103 Arundeep’s Mathematics (R.D.) 9th
x3 + y3 + z3 = 3xyz
(a – b)3 + (b – a)3 + (c – a)3
= 3(a – b) (b – c) (c – a) (b)
6. The value of 09.069.0)3.2(
027.0)3.2(2
3
is
(a) 2 (b) 3 (c) 2.327 (d) 2.273
Sol.09.069.0)3.2(
027.0)3.2(2
3
{_ a3 – b3 = (a – b) (a2 + ab + b2)}
= 22
33
)3.0()3.0()3.2()3.2(
)3.0()3.2(
= 22
22
)3.0()3.0()3.2()3.2(
})3.0()3.0()3.2()3.2{()}3.0()3.2{(
= (2.3) – (0.3) = 2.3 – 0.3 = 2 (a)
7. The value of 22
33
)007.0(007.0013.0)013.0(
)007.0()013.0(
is
(a) 0.006 (b) 0.02 (c) 0.0091 (d) 0.00185
Sol. 22
33
)007.0(007.0013.0)013.0(
)007.0()013.0(
{_ a3 + b3 = (a + b) (a2 – ab + b2)}
= 22
22
)007.0(007.0013.0)013.0(
])007.0(007.0013.0)013.0[()007.0013.0(
= 0.013 + 0.007 = 0.020 = 0.02 (b)
8. The factors of a2 – 1 – 2x – x2 are
(a) (a – x + 1) (a – x – 1) (b) (a + x – 1) (a – x + 1)
(c) (a + x + 1) (a – x – 1) (d) none of these
Sol. a2 – 1 – 2x – x2
a2 – (1 + 2x + x2) = (a)2 – (1 + x)2
= (a + 1 + x) (a – 1 – x) (c)
9. The factors of x4 + x2 + 25 are
(a) (x2 + 3x + 5) (x2 – 3x + 5) (b) (x2 + 3x + 5) (x2 + 3x – 5)
(c) (x2 + x + 5) (x2 – x + 5) (d) none of these
Sol. x4 + x2 + 25 = x4 + 25 + x2
= (x2)2 + (5)2 + 2 × x2 × 5 – 9x2
= (x2 + 5)2 – (3x)2
= (x2 + 5 + 3x) (x2 + 5 – 3x)
= (x2 + 3x + 5) (x2 – 3x + 5) (a)
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104 Arundeep’s Mathematics (R.D.) 9th
10. The factors of x2 + 4y2 + 4y – 4xy – 2x – 8 are
(a) (x – 2y – 4) (x – 2y + 2) (b) (x – y + 2) (x – 4y – 4)
(c) (x + 2y – 4) (x + 2y + 2) (d) none of these
Sol. x2 + 4y2 + 4y – 4xy – 2x – 8
x2 + 4y2 + 4y – 4xy – 2x – 8
= (x)2 + (2y)2 – 2 × x × 2y + 4y – 2x – 8
= (x – 2y)2 – (2x – 4y) – 8
= (x – 2y)2 – 2 (x – 2y) – 8
Let x – 2y = a, then
a2 – 2a – 8 = a2 – 4a + 2a – 8
= a(a – 4) + 2(a – 4)
= (a – 4) (a + 2)
= (x2 – 2y – 4) (x2 – 2y + 2) (a)
11. The factors of x3 – 7x + 6 are
(a) x(x – 6) (x – 1) (b) (x2 – 6) (x – 1)
(c) (x + 1) (x + 2) (x – 3) (d) (x – 1) (x + 3) (x – 2)
Sol. x3 – 7x + 6 = x3 – 1 – 7x + 7
= (x – 1) (x2 + x + 1) – 7(x – 1)
= (x – 1) (x2 + x + 1 – 7) = (x – 1) (x2 + x – 6)
= (x – 1) [x2 + 3x – 2x – 6]
= (x – 1) [x(x + 3) – 2(x + 3)]
= (x – 1) (x + 3) (x – 2) (d)
12. The expression x4 + 4 can be factorized as
(a) (x2 + 2x + 2) (x2 – 2x + 2) (b) (x2 + 2x + 2) (x2 + 2x – 2)
(c) (x2 – 2x – 2) (x2 – 2x + 2) (d) (x2 + 2) (x2 – 2)
Sol. x4 + 4 = x4 + 4 + 4x2 – 4x2 (Adding and subtracting 4x2)
= (x2)2 + (2)2 + 2 × x2 × 2 – (2x)2
= (x2 + 2)2 – (2x)2
= (x2 + 2 + 2x) (x2 + 2 – 2x) {_ a2 – b2 = (a + b) (a – b)}
= (x2 + 2x + 2) (x2 – 2x + 2) (a)
13. If 3x = a + b + c, then the value of (x – a)3 + (x – b)3 + (x – c)3 – 3(x – a) (x – b) (x – c) is
(a) a + b + c (b) (a – b) (b – c) (c – a)
(c) 0 (d) none of these
Sol. 3x = a + b + c
3x – a – b – c = 0
Now, (x – a)3 + (x – b)3 + (x – c)3 – 3(x – a) (x – b) (x – c)
= {(x – a) + (x – b) + (x – c)} {(x – a)2 + (x – b)2 + (x – c)2 – (x – a) (x – b)
– (x – b) (x – c) – (x – c) (x – a)}
= (x – a + x – b + x – c) {(x – a)2 + (x – b)2 + (x – c)2 – (x – a) (x – b)
– (x – b) (x – c) – (x – c) (x – a)}
= (3x – a – b – c) {(x – a)2 + (x – b)2 + (x – c)2 – (x – a) (x – b) – (x – b) (x – c) – (x – c) (x – a)}
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105 Arundeep’s Mathematics (R.D.) 9th
But 3x – a – b – c = 0, then
= 0 × {(x – a)2 + (x – b)2 + (x – c)2 – (x – a) (x – b) – (x – b) (x – c) – (x – c) (x – a)}
= 0 (c)
14. If (x + y)3 – (x – y)3 – 6y(x2 – y2) = ky2, then k =
(a) 1 (b) 2 (c) 4 (d) 8
Sol. (x + y)3 – (x – y)3 – 6y(x2 – y2) = ky2
LHS = (x + y)3 – (x – y)3 – 3 × (x + y) (x – y) [x + y – x + y]
= (x + y – x + y)3 {_ a3 – b3 – 3ab (a – b) = a3 – b3}
= (2y)3 = 8y3
Comparing with ky3, k = 8 (d)
15. If x3 – 3x2 + 3x – 7 = (x + 1) (ax2 + bx + c), then a + b + c =
(a) 4 (b) 12 (c) –10 (d) 3
Sol. x3 – 3x2 + 3x + 7 = (x + 1) (ax2 + bx + c)
= ax3 + bx2 + cx + ax2 + bx + c
x3 – 3x2 + 3x – 7 = ax3 + (b + a)x2 + (c + b)x + c
Comparing the coefficient,
a = 1
b + a = –3 b + 1 = –3 b = –3 – 1 = –4
c + b = 3 c – 4 = 3 c = 3 + 4 = 7
a + b + c = 1 – 4 + 7 = 8 – 4 = 4 (a)
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