46 PRIME Opt. Maths Book - IX The progression having rule of multiplication with a constant number is called geometric progression (GP). Examples : 2, 6, 18, 54, ..., ..., ..., ..., ..., ..., ..., ..., ..., ... 3, 6, 12, 24, 48, ..., ..., ..., ..., ..., ..., ..., ..., ..., ... 128, 64, 32, 16, 8, ..., ..., ..., ..., ..., ..., ..., ..., ..., ... nth term of the sequence and series: Let us discuss different examples to find the nth term of the sequence. 2, 4, 6, 8, ..., ..., ..., ..., ..., ..., ..., ..., ..., ... 2n 1, 3, 5, 7, .........................., 2n – 1 12 + 22 + 32 + ... + ... + ... + ... + ... + ... + ... + ... + n2 0, 2, 6, 12, ..., ..., ..., ..., ..., ..., ..., ..., ..., ... n2 – n. Here, taking above examples, Terms 1st example 2nd example 3rd example 4th example 1st 2 × 1 2 × 1 – 1 12 12 – 1 2nd 2 × 2 2 × 2 – 1 22 22 – 2 3rd 2 × 3 2 × 3 – 1 33 33 – 3 4th 2 × 4 2 × 4 – 1 44 44 – 4 So on .............. .............. .............. .............. nth term 2n 2n – 1 n2 n2 – n From the table the last term (nth term) can be generalized by comparing the number of terms which is, nth term (tn ) = 2n (1st example) tn = 2n – 1 (2nd example) tn = n2 (3rd example) tn = n2 – n (4th example) • The sequence whose terms are written in order according to the rule of linear equation (y = ax + b) is called linear sequence. • The sequence whose terms are written in order according to the rule of quadratic equation (y = ax2 + bx + c) is called quadratic sequence.
PRIME Opt. Maths Book - IX 47 Alternative way: i. nth term of linear sequence. Taking Example 1 : 2, 2 6, 2 10, 2 4, 2 8, 2 12, Here, constant difference of the terms of sequence is 2. Where nth term is taken as, tn = an + b The constant difference is 2 So, a = 2. Then, tn = an + b t1 = 2 × 1 + b or, 2 = 2 + b \ b = 0 \ The nth term of the sequence is, tn = 2n Taking example 2 : 2 1, 2 5, 2 2 9, 11, 2 3, 2 7, 13, Here, the nth term is taken as, tn = an + b The constant difference is 2, so a = 2 Then, t1 = a × 1 + b or, 1 = 2 + b \ b = –1 \ nth term of the sequence is, tn = an + b = 2n – 1 ii) Quadratic nth term of quadratic sequence Taking an example 4, 2 2 0, 6 2 6, 10 20, 30, 4 2 2, 8 2 12, Here, the nth term of the sequence is taken as, tn = an2 + bn + c
48 PRIME Opt. Maths Book - IX The second difference is the constant difference 2. So, 2a = 2 \ a = 1 Then, tn = an2 + bn + c t1 = 1 × 12 + b × 1 + c or, 0 = 1 + b + c \ b = – 1 – C ...................................... (i) Again, t2 = 1 × 22 + b × 2 + c or, 2 = 4 – 2 – 2c + c or, c = 0 From equation (i) b = – 1 – C = – 1 \ nth term of the sequence is, tn = an2 + bn + c \ tn = n2 – n • Taking another example 6, 11, 18, 27, ..., ..., ..., ..., ..., ..., ..., ..., ..., ... Here, 5 2 6, 9 18, 7 2 11, 27, The nth term of the sequences is, tn = an2 + bn + c The second difference is the constant difference ‘2’. So, 2a = 2 \ a = 1 Then, tn = an2 + bn + c t1 = 1 × 12 + b × 1 + c 6 = 1 + b + c or, b = 5 – C ................................. (i) Again, t2 = 1 × 22 + b × 2 + c or, 11 = 4 + 10 – 2c + c [\ using (i)] or, c = 3 From equation (i), b = 5 – 3 = 2 \ nth term of the sequence is, tn = an2 + bn + c \ tn = n2 + 2n + 3
PRIME Opt. Maths Book - IX 49 Note : i) If the sequence is written in terms of opposite sign as – , +, –, + etc for each terms respectively tn = (–1)n ii) If the sequence is written in terms of opposite sign as +, –, +, –, +, – etc, the nth term tn = (–1)n + 1 Sum of the terms of the sequence : If a1 , a2 , a3 , a4 , a5 are the terms of the finite sequence, the sum of the terms is written as, S5 = a1 + a2 + a3 + a4 + a5 If a1 + a2 + a3 + a4 + ... + ... + ... + ... + ... + ... + ... + an are the ‘n’ terms of the sequence, the sum can be written as, Sn = a1 , a2 , a3 , a4 , ..., ..., ..., ..., ..., ..., ..., ..., ..., ... + an If a1 , a2 , a3 , a4 , ..., ..., ..., ..., ..., ..., ..., ..., ..., ... are the infinite terms of the sequence, the sum can not be defined. Sum of the terms of a sequence using sigma notation: In the sequence having ‘n’ terms a1 , a2 , a3 , a4 , ..., ..., ..., ..., ..., ..., ..., ..., ..., ... an The sum of them can be written as, Sn = a1 + a2 + a3 + ... + ... + ... + ... + ... + ... + ... + ... + an = ak k n =1 / Where ∑ (sigma) is used for symbol of summation of the terms and ‘k’ is the number of terms from 1 to n. To write the sum of the terms in sigma notation, nth term of the sequence should be know. Let us take an example. 1, 5, 9, 11, 15, 19. For nth term of the sequence, 4 1, 4 9, 4 15, 19, 4 5, 4 11, Here, the nth term of this linear sequence is, tn = an + b The constant difference is 4. So a = 4.
50 PRIME Opt. Maths Book - IX Then, tn = an + b t1 = 4 + b or, 1 = 4 + b \ b = – 3 \ nth term of the sequence, tn = 4n – 3. The sum of the terms is Sn = 1 + 5 + 9 + 11 + 15 + 19 = ( ) 4 3 n – n 1 6 = / 1. If nth term of a sequence is tn = 3n – 2, find the first five terms of the sequence. Solution : The nth term of a sequence is, tn = 3n – 2 Taking n = 1, t1 = 3 × 1 – 2 = 1 n = 2, t2 = 3 × 2 – 2 = 4 n = 3, t3 = 3 × 3 – 2 = 7 n = 4, t4 = 3 × 4 – 2 = 10 n = 5, tn = 3 × 5 – 2 = 13 Hence, the first five terms are 1, 4, 7, 10, 13. 2. If a1 , a2 , a3 , ..., ..., ..., ..., ..., ... are the terms of a sequence where an + 1 = 2an – 3 and a0 = 4, find the value of a1 , a2 , a3 and a4 . Solution : In a sequence a1 , a2 , a3 , ..., ..., ..., ..., ..., ..., ..., ..., ..., ... are the terms. an + 1 = 2an – 3 and a0 = 4 Taking, n = 0 an + 1 = 2an – 3 or, a0 + 1 = 2 × a0 – 3 or, a1 = 2 × 4 – 3 = 5 Again, Taking n = 1 Taking n = 2, Taking n = 3, a1 + 1 = 2a1 – 3 a2 + 1 = 2 × a2 – 3 a3 + 1 = 2 × a3 – 3 \ a2 = 2 × 5 – 3 = 7 a3 = 2 × 7 – 3 = 11 a4 = 2 × 11 – 3 = 19 Worked out Examples
PRIME Opt. Maths Book - IX 51 3. Write down in expanded form of ( ) –1 n n 1 5 = / (n2 +3n). Also find the sum. Solution : The given sum of the terms of a sequence in sigma notation is, Sn = ( ) –1 n n 1 5 = / (n2 + 3n) = t1 + t2 + t3 + t4 + t5 [Here, tn = (–1)n (n2 + 3n)] = (–1)1 (12 + 3 × 1) + (–1)2 (22 + 3 × 2) + (–1)3 (32 + 3 × 3) + (–1)4 (42 + 3 × 4) + (–1)5 (52 + 3 × 5) = –4 + 10 – 18 + 28 – 40 = – 24 4. If sum of the ‘n’ terms of the sequence is Sn = n n 2 2 + , find the 3rd and 4th terms of the sequence. Solution : The sum of the ‘n’ terms is, Sn = n n 2 2 + Then, 3rd term (t3 ) = S3 – S2 = 2 3 3 2 + – 2 2 2 2 + = 6 – 3 = 3 4th term (t4 ) = S4 – S3 = 2 4 4 2 + – 2 3 3 2 + = 10 – 6 = 4 5. Find the nth term of the sequence 3, 5, 8, 12, 17. Also write down in sigma notation. Solution : The given sequence is 3, 5, 8, 12, 17, ..., ..., ..., ..., ..., ..., ..., ..., ..., ... Difference between the terms can be seen as follows, 2 1 3, 4 1 8, 17, 3 1 5, 5 (is not equal) (is equal) 12, Here, the constant difference is obtained in second step. So, the nth term is quadratic form as,
52 PRIME Opt. Maths Book - IX tn = an2 + bn + c Then, 2a = 1 [constant difference] \ a = 2 1 So, tn = an2 + bn + c t1 = 2 1 × 12 + b × 1 + c or, b = 3 – 2 1 – c = c 2 5 2– ....................... (i) Again, t2 = 2 1 × 22 + b × 2 + c or, 5 = × 2 1 42 + c 2 2 5 2– c m + c = 2 1 or, 3 = 5 – c \ c = 2 From equation (i) b = b 2 – ×2 2 = 2 1 \ nth term is, tn = 2 1 n2 + 2 1 n + 2 = 2 1 (n2 + n + 4) Again, The sum of the terms of sequence in sigma notation is, Sn = 3 + 5 + 8 + 12 + 17 = 2 1 n 1 5 = / ( ) n n 4 2 + + = 2 1 n 1 5 = / ( ) n n 4 2 + + 6. Add two more patterns and find the nth term of the sequence so formed of the number of dots. Also write down in sigma notation from the given patterns. Solution : Two more patterns in the given patterns.
PRIME Opt. Maths Book - IX 53 The sequence according to number of dots from the diagrams as, 5, 10, 15, 20, 25, ..., ..., ..., ..., ..., ..., ..., ..., ..., ... For the general term (nth term) 5 5, 5 15, 25, 5 10, 5 (is same) 20, The constant difference is 5 which is obtained in 1st step. So, the nth term becomes, tn = an + b. Here, a = 5 (Constant difference) t1 = 5 × 1 + b or 5 = 5 + b \ b = 0 Then, the nth term becomes, tn = 5n Again, the sum of the terms of the sequence in sigma notation, Sn = 5 + 10 + 15 + 20 + 25 = ( ) 5n n 1 5 = /
54 PRIME Opt. Maths Book - IX Exercise 1.6 1. i) What is sequence? Write down one example. ii) What do you mean by series? Write down with an example. iii) Define the terms finite and infinite sequence with examples. iv) What is progression? Write down with an example. v) Write down the sum t1 + t2 + t3 + t4 + t5 + t6 in sigma notation. 2. Which of the followings are the finite or infinite sequence or series? i) 2 + 5 + 8 + 11 + 14 ii) 3, 7, 11, 15, ..., ..., ..., ..., ..., ..., ..., ..., ..., ... 15 terms iii) 1, 4, 9, 16, 25, 36, ..., ..., ..., ..., ..., ..., ..., ..., ..., ... iv) 2 + 6 + 18 + 54 + ... + ... + ... + ... + ... + ... + v) a, a + d, a + 2d, ..., ..., ..., ..., ..., ..., ..., ..., ..., ... , a + (n – 1)d 3. Find the followings. i) Write down the corresponding series for the sequence 5, 12, 19, 26, 33, ..., ..., ..., ..., ..., ... ii) Write down the corresponding sequence for the series 33 – 29 + 25 – 21 + 17 – 13 iii) Write down the next term of the sequence 1, 4, 9, 16, 25, ..., ..., ..., ..., ..., ... Is it finite or infinite? iv) What is the common difference of the sequence a, a + d, a + 2d, a + 3d, ..., ..., ..., ..., ..., ... v) Is a sequence 2, 6, 11, 17, 25, ..., ..., ..., ..., ..., ... a progression? Why? 4. Find the next three terms of the sequences. i) 1, 6, 11, 16, ..., ..., ..., ..., ..., ... ii) 1, 8, 27, 64, ..., ..., ..., ..., ..., ... iii) 2 1 , 3 2 , 4 3 , 5 4 , ..., ..., ..., ..., ..., ... iv) 80, –75, 70, –65, ..., ..., ..., ..., ..., ... v) – 7 3 , 12 7 , – 17 11 , 22 15 , ..., ..., ..., ..., ..., ... 5. Find the first five terms of the sequence whose nth term are given below. i) tn = 2n – 1 ii) an = 4n + 3 iii) an = n2 + 4n – 5 iv) an = (–1)n n(n + 1) v) tn = (–1)n + 1 n n 1 2 2 + a k 6. Find first five terms of the followings where the terms are explained below. i) an = an + 1 – 2 and a1 = 3 ii) un + 1 = un + 3 and u2 = 5 iii) tn – 1 = 2tn – 1 and t2 = 2 iv) an + 2 = an + 1 + an , a1 = 1 a2 = 2 v) tn = t t n n 1 2 – – , t1 = 1 and t2 = 2 7. Find the nth terms of the sequences. i) 1, 4, 7, 10, 13, ..., ..., ..., ..., ..., ... ii) 5, 2, –1, –4, ..., ..., ..., ..., ..., ...
PRIME Opt. Maths Book - IX 55 iii) 10, 16, 22, 28, ..., ..., ..., ..., ..., ... iv) 3 1 , 7 3 , 11 5 , 15 7 , ..., ..., ..., ..., ..., ... v) – 5 2 , 11 4 , – 17 6 , 23 8 , ..., ..., ..., ..., ..., ... 8. Find the general term of the sequences. i) 1, 4, 9, 16, 25, ..., ..., ..., ..., ..., ... ii) 1, 3, 6, 10, 15, ..., ..., ..., ..., ..., ... iii) 6, 11, 18, 27, 38, ..., ..., ..., ..., ..., ... iv) 0, – 2 1 , 3 2 , – 4 3 , 5 4 , – 6 5 , ..., ..., ..., ..., ..., ... v) a, a + d, a + 2d, a + 3d, ..., ..., ..., ..., ..., ... 9. Write down in expanded form and evaluate. i) ( ) 2 3 n n 1 4 + = / ii) ( ) 4 1 n – n 1 5 = / iii) n n( ) 1 n 2 5 + = / iv) ( ) –1 n n 1 4 = / [ ( n n + + 2 3 ) ] v) ( ) –1 n n 1 1 5 + = / n n + 1 ` j 10. If Sn is the sum of ‘n’ terms of the sequence which is given below. i) Sn = 2n2 + 3, find S4 , S5 and t5 . ii) Sn = n(n + 1), find the 4th term. iii) Sn = 2n2 – 1, find the 5th term. iv) Sn = n n( ) 2 + 1 , find the 3rd and 6th terms. v) Sn = n(n + 1)(2n + 1) 6 , find 4th and 5th terms. 11. Write down the following in sigma notation for the given series. i) 2 + 5 + 8 + 11 + 14 + 17 ii) 1 + 3 + 6 + 10 + 15 iii) –4 + 7 –11 + 16 –22 + 29 iv) 3 1 – 5 2 + 8 3 – 12 4 v) a + ar + ar2 + ar3 + ar4 + ar5 . 12. Write down the following sequence of number of dots used in diagrams in sigma notation after calculating the general term by adding one more pattern for each. i) ii)
56 PRIME Opt. Maths Book - IX iii) iv) v) 13. PRIME more creative questions i) Find the nth term and write down in sigma notation of the series 1 × 3 + 2 × 7 + 3 × 11 + 4 × 15 + 5 × 19. ii) If nth term of the sequence is tn = a + (n – 1)d where a = 3, d = 4, find the t5 and t11. iii) If nth term of a sequence is tn = arn–1 where a = 3 and r = 2, find the series. iv) If sum of first ‘n’ terms of a sequence is Sn = n 2 [2a + (n –1)d] where a = 40 and d = – 4, find the value of S5 . v) If sum of the first ‘n’ terms of a series is Sn = ( ) r a r 1 1 – –n where a = 5 and r = 2, find the value of S4 + S5 . 14. Project work Prepare a chart of the way of finding the sigma notation for linear and quadratic form of sequence and present in your classroom.
PRIME Opt. Maths Book - IX 57 Answer 1. Show to your teacher. 2. Show to your teacher. 3. Show to your teacher. 4. i) 21, 26, 31 ii) 125, 216, 243 iii) , , 6 5 7 6 8 7 iv) 60, –55, 50 v) – 27 19 , 32 23 , – 37 27 5. i) 1, 3, 5, 7, 9 ii) 7, 11, 15, 19, 23 iii) 0, 7, 16, 27, 40 iv) –2, 6, –12, 20, –30 v) 1, – 5 4 , 10 6 , – 17 8 , 26 10 6. i) 3, 5, 7, 9, 11 ii) 2, 5, 8, 11, 14 iii) 3, 2, , , 2 3 4 5 8 9 iv) 1, 2, 3 , 5, 8 v) 1, 2, 2 1 , 4, 8 1 7. i) 3n – 2 ii) 8 – 3n iii) 2(3n + 2) iv) n n 4 1 2 1– – v) ( ) n n 1 6 1 2 – – n 8. i) n2 ii) n 2 (n + 1) iii) n2 + 2n + 3 iv) (–1)n+1 n n 1– ` j v) a + (n – 1)d 9. i) 32 ii) 55 iii) 68 iv) 14 v) 60 37 10. i) 35, 53, 18 ii) 8 iii) 18 iv) 3, 6 v) 16, 25 11. i) ( ) 3 1 n – n 1 6 = / ii) n n( ) 2 1 n 1 5 + = / iii) ( ) –1 n n 1 6 = / ( ) n n2 3 4 2 + + iv) ( ) –1 n n 1 1 4 + = / n n n 4 2 2 + + a k v) ( ) arn– n 1 1 6 = / 12. i) n n( ) 2 1 n 1 6 + = / ii) ( ) 4 3 n – n 1 5 = / iii) ( ) 2 1 n n 1 5 + = / iv) n n 2 1 6 = / v) ( ) 3 2 n n 1 5 Σ + = 13. i) n n ( ) 4 1– n 1 5 Σ = ii) 19, 43 iii) 3, 6, 12, 24, ..., ..., ... iv) 160 v) 230
58 PRIME Opt. Maths Book - IX Algebra Unit Test - 1 Time : 30 minutes [1 × 1 + 3 × 2 + 2 × 4 + 1 × 5 = 20] Attempt all the questions: 1. What is Cartesian product? 2. a. If (3x+y, y–2)and (10, 2y–6)are equal ordered pairs, find the value of ‘x’ and y; b. What must be subtracted to 2x3 – 3x2 + 5x – 3 to get the polynomial x3 – 2x + 5? c. Find the nth term of the sequence 3, 5, 8, 12, 17, 23, ................. . 3. a. Divide the polynomial x3 – 5x + 3x + 4 with x + 3. b. If f(x) = 2x – 3, domain = {–2, –1, 0, 1, 2}, find range and show the function in arrow diagram. 4. Add one more pattern in the given pattern and write down the sequence of dots in sigma notation. Unit Test - 2 Time : 30 minutes [1 × 1 + 3 × 2 + 2 × 4 + 1 × 5 = 20] Attempt all the questions: 1. If A × B = {(1, 3), (1, 4), (1, 5) (2, 3), (2, 4), (2, 5)}, find thee sets A and B. 2. a. If f(x + a) = f(x) + f(a), find f(0) and prove that f(–a) = –f(a). b. If p(x) = 3x3 + 2x2 – 5x + 7 and q(x) = x3 –3x2 – 2x + 3, find p(x) + q(x). Also write down its types according to degree. c. Write down in expanded form and find the sum of ( ) –1 n n 1 1 5 + = / (n2 + 3n). 3. a. Find the nth term of the sequence 2 – 5 + 10 – 17 + 26 – 37. Also write down in sigma notation. b. Find the relation R = {(x, y) : x + y ≤ 10, x, y ∈ N}. Also show in arrow diagram. 4. If f(x) = 2sin x + 1, range = {1, 2, 3 + 1, 3}, find the domain. Which type of function is it? Why? Also show in arrow diagram.
PRIME Opt. Maths Book - IX 59 Unit 2 Limit and Continuity Specification Grid Table K(1) U(2) A(4) HA(5) TQ TM Periods No. of Questions 1 – 1 – 3 5 10 Weight 1 – 4 – K = Knowledge, U = Understanding, A = Application, HA = Higher ability, TQ = Total Question, TM = Total Marks Objectives : At the end of the lesson • Students are able to understand rational and irrational numbers and their Sequence. • Students are able to generalize the decimal numbers to the nearest whole number. • Students are able to find the limit value of a function in diagram. • Students are able to know the limit value of infinite series and graph. • Students are able to know the symbol of limit and indeterminate form. Materials Required: • Chart paper. • Set of rational and irrational numbers. • Graph paper. • Model of graph of functions. • Geometrical shapes.
60 PRIME Opt. Maths Book - IX Limit Enjoy the recall • y = f(x) is a function f : A → B where Domain = { Set of element x} = A. Co-domain = main set of range = B. Range = {Set of element y} Image = element of range = y Pre - image = element of domain = x • If f(x) = x2 – 2x + 3, we can find f(1), f(2), f(3), ...etc where f(1) = 12 – 2 × 1 + 3 = 2 and so on ....................................... . • If f = {(1, 5), (2, 6), (3, 7), (4, 8)} is a function where Domain = {1, 2, 3, 4}, Range = {5, 6, 7, 8} • Sequence in increasing order is 2, 5, 8, 11, ..., ..., ..., ..., ..., ... • A sequence in decreasing order is, 20, 17, 14, 11, 8, 5, ..., ..., ..., ..., ..., ... • When a football is dropped on the ground following result can be obtained. • Round numbers of 2.467 is, 2. 467 = 2.47 (round up in hundred) 2. 467 = 2.5 (round up in tens) 2. 467 = 2 (round up in whole numbers. • If value of ‘n’ goes on increasing, the value of n 1 will decrease and be very close to zero. Diagramatic representation of a sequence and absolute numbers. 1. Let us consider a line segment AB which is going on dividing at mid-point P again PA is divided by a mid-point Q and So on R, S, T can be obtained. A Q R STP B If AB is taken as 12cm. Then the parts of AB by dividing at mid point will be. AP = 6cm QP = 3cm RP = 2 3 cm SP = 4 3 cm
PRIME Opt. Maths Book - IX 61 TP = 8 3 cm Where the sequence will be, 12, 6, 3, 2 3 , 4 3 , 8 3 , 16 3 , ..., ..., ..., ..., ..., ... Here, The last mid - point of the line segment will be approaches to P and the length becomes near to zero (0) but not equal to zero. Note : If a real number is taken in a number line, it will approaches to + ∞ in right side and approaches to – ∞ in left side. 2. Let us consider a triangle and the triangles obtained by joining the mid - point of sides of previous triangle and so on. P Q A R B C Q X Y Z P A R B C According to the figures given in the above example, following conclusions can be obtained. • Area of triangle ABC is taken as 40 square units and the area of DPQR becomes 10 square units. (i.e. Area goes on decreasing by 4 times.) • The sequence so formed according to area becomes, 40, 10, 2 5 , , 8 5 32 5 , ............ and so on. • The area of last triangle so formed will be very close to zero but not equal to zero. 3. Let us consider a cylinder which is cutting down at the middle of the cylinder and so on for each pieces. According to the volume of cylinder goes on decreasing by half following conclusion can be obtained where volume of 1st cylinder is taken as 20 cubic units. • Volume of cylinder goes on decreasing by half and becomes 10 cubic units, 5 cubic units, 2 5 cubic and so on. • The sequence so obtained of their volume is, 20, 10, 5, 2 5 , 4 5 , 8 5 , .......... The volume of last cylinder so formed will be very close to zero but not equal to zero.
62 PRIME Opt. Maths Book - IX Sum of infinite series : 1. Let us consider a line segment PQ, having length 2cm. P A C D E F Q PQ is divided by A at mid - point AQ is divided by C at mid -point CQ is divided by D at mid - point, DQ is divided by E at mid - point, EQ is divided by F at mid - point Then, The sequence of length of the line segment so formed becomes, 2, 1, , , , , , 2 1 4 1 8 1 16 1 , ..., ..., ..., ..., ..., ... After the last division the divided point will be nearest to Q where length becomes very near to zero but not equal to zero. Here, Sum of the length of the line segment becomes, S1 = PA = 1cm = 1cm S2 = PC = 1 + 2 1 cm = 1.5cm S3 = PD = 1 + 2 1 4 1 + = 1.75cm S4 = PE = 1 + 2 1 4 1 8 1 + + = 1.875cm. S5 = PF = 1 + 2 1 4 1 8 1 + + 16 1 + = and So on S6 , S8 , ..., ..., ..., ..., ..., ..., Sn can be calculated which will be very close to 2 but not equal to 2. Here, • For the infinite series, sum of the infinite terms can not be defined but partial sum like S1 , S2 , S3 , S4 , .......... S∞ can be defined. • 2 is the limit value of the sum of infinite series so formed. • The common ratio of the series of the above sequence is 2 1 which is less than 1. So, It has the limit value with fixed absolute value which is 2. • The limit value of the sum of infinite series can be defined using formula also, which is S r a 1 – 3 = . • In the above example The common ratio r = 2 1 , first term a = 1. So, S r a 1 – 3 = = 1 2 1 1 – = 2 1 1 = 2 • This series is called convergent series where its limit exists.
PRIME Opt. Maths Book - IX 63 Exercise 2.1 1. i) What is the round number of 5.678 in two decimal system? ii) Write down the round digit of 6.425. iii) Define range and domain in a function y = f(x). iv) Define image and pre-image in a function f(2) = 5. v) Write down 4 terms of a sequence 0.5. 2. i) Draw a line segment of length 12cm and divide it at mid - point respectively 12 times. What conclusion do you get ? Explain. ii) 0.1, 0.01, 0.001, ................ is a sequence. What will be its 10th term. Also write down it’s limit value. iii) Write down the limit value of the sequence 1.1, 1.01, 1.001, 1.0001, ...................... iv) Write down the limit value of 0.9, 0.99, 0.999, 0.9999, ................... v) Draw a triangle having base of length 12cm and draw the line joining the mid - points of other two sides and so on for the triangle obtained after joining the mid - points continuously 5 times. What will be the length of the line segments obtained by joining the mid - points respectively ? Also write down the limit value of the length of such last line. 3. i) How many squares can be drawn in a square by joining the mid - point of sides of a square continuously ? What is the limit value of area of square so formed ? Show with diagram. 2. Let us consider a series 1 + 2 + 4 + 8 + 16 +... + ... + ... + ... + ... + ... + Here, The common ratio of the terms is 2 which is more than 1. So, the sum of the infinite series will not be the fixed absolute value and limit does not exist. • This series is called divergent series where its limit does not exist. 3. Let us consider a series, – 8 + 4 – 2 + 1 – – 2 1 4 1 8 1 + + ... + ... + ... + ... + ... + ... + ... Here, The common ratio is – 2 1 which is less than ‘1’. Hence the series has the limit value which will be a fixed absolute value. 4. Let us consider a series 0.25 The series 0.25 can be expressed as 0.25 + 0.0025 + 0.000025 + 0.00000025 + ... + ... + ... + ... + ... + ... + and so on and can be extended in terms of infinity where the last term will be very close to zero but not equal to zero. Hence, The limit value of nth term will be the absolute value zero.
64 PRIME Opt. Maths Book - IX ii) A cylindrical glass full of water is drank half/half in each times after a interval. How many times the water can drank ? What is the limit value of volume of water in that activity ? Show in diagram. iii) What will be the value of y on increasing the value of ‘x’ in the given graph? What is the limit value of y in it? iv) The graph given in Q.No. (iv) can be used in our daily life. Discuss with a suitable example for it in the classroom by preparing a graph in chart. v) What is the limit value of the sum of the terms of the series, 3 + 2 3 + 4 3 + 8 3 + 16 3 + 32 3 ? 4. Which of the following series has the limit value will be the fixed absolute number? i. 1 + 2 1 + 4 1 + 8 1 + 16 1 , ..., ..., ..., ..., ..., ... ii. 4 – 2 + 1 – 2 1 + 4 1 ... + ... + ... + ... + ... + ... iii. 8 + 40 + 200 + 1000 + ... + ... + ... + ... + ... + ... iv. 0.6 + 0.06 + 0.006 + 0.0006 + ... + ... + ... + ... + ... + ... v. 0.45 5. A square having side 1cm is dividing making half continuously to their parts as shown in diagrams. Answer the followings. i) Write down the sequence of the shaded area of region. ii) What will be the area shaded region of figure. iii) Write down the limit value of the shaded region. 6. i) If an = 1 + (–1)n 2n 12 , find a1 , a2 , a3 , a4 and a5 ii) If f(x) = 2x – 1, find f(2), f(1.99) and f(2.01). iii) If f(x) = 3x + 1, find f(3), f(2.999) and f(3.001). iv) tn = n 1 1 + , find the difference between t3 and t4 . v) Find the value and fill the given table. x 0.1 0.01 0.001 0.0001 0.00001 0 f(x) = 2x + 3 ....... ....... ....... ....... ....... ....... x – 0.1 – 0.01 – 0.001 – 0.0001 – 0.00001 0 f(x) = 2x + 3 ....... ....... ....... ....... ....... ....... X O Y
PRIME Opt. Maths Book - IX 65 Answer 1. i) 5.68 ii) 6 iii) Domain = Set of the elements of n Range = Set of the elements of y iv) 5 is the image of 2 and 2 is the pre-image of 5 v) 0.5 + 0.05 + 0.005 + 0.0005 2. i) Limit value of the length will be zero. ii) 0.0000000001, zero iii) 1 iv) 1 v) 12, 6, 3, , 2 3 4 3 ; zero 3. i) Infinite, zero ii) Infinite, zero iii) Decreasing zero iv) Show to your teacher v) 6 4. i, ii, iv, v 5. i) , 2 1 4 3 , , 8 7 16 15 , ..., ..., ..., ... ii) Nearly equal to 1 cm2 . iii) The limiting value is 1. 6. i) , , , 2 1 8 9 18 17 , 32 33 50 49 ii) 3, 2.98, 3.02 iii) 10, 9.97, 10.03 iv) 20 1 v) Show to your subject teacher. 7. i) 21.3 3 64 a k ii) 0.5 9 5a k iii) 12 iv) 5 3 1 v) – 3 40 vi) 0.4 9 4a k 7. Find the limiting value of sum of the following infinite series. i) 16 + 4 + 1 + 4 1 + ... + ... + ... + ... + ... + ... + ... ii) 0.5 + 0.05 + 0.005 + 0.0005 + ... + ... + ... + ... + ... + ... + ... iii) 6 + 3 + 2 3 + 4 3 + 8 3 + ... + ... + ... + ... + ... + ... iv) 8 – 4 + 2 – 1 + ... + ... + ... + ... + ... + ... v) –20 + 10 – 5 + 2 5 – 4 5 + ... + ... + ... + ... + ... + ... vi) 0.4
66 PRIME Opt. Maths Book - IX 2.2 Limit using graph i. Let us consider a function f(x) = 3x – 4 and range of the function can be taken using domain as follows for x = 3. From left side: x 1 2 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 f(x) –1 2 2.3 2.6 2.9 3.2 3.5 3.8 4.1 4.4 4.7 Here, we conclude that, On increasing the value of x up to nearly equal to 3, the element of range f(x) also will goes on increasing to nearly equal to 5. O 2 3 X’ Y’ Y X 1 1 2 3 4 5 Hence, the limit value of the function f(x) tends to 5 (very close to 5) from left side which is called Left Hand Limit (LHL).
PRIME Opt. Maths Book - IX 67 Also from right side: x 5 4 3.9 3.8 3.7 3.6 3.5 3.4 3.3 3.2 3.1 f(x) 11 8 7.7 7.4 7.1 6.8 6.5 6.2 5.9 5.6 5.3 Here, we conclude that, On decreasing the value of x up to nearly equal to 3, the value of f(x) also will goes on decreasing to nearly equal to 5. O 2 3 4 5 X’ Y’ Y X 1 1 2 3 4 5 6 7 Hence, the limit value of the function f(x) tends to 5 (very close to 5) from right side which is called Right Hand Limit (RHL).
68 PRIME Opt. Maths Book - IX ii. Let us consider a function f(x) = x2 . x 0 –1 1 –2 2 –3 3 –4 4 f(x) 0 1 1 4 4 9 9 16 16 O X’ Y’ Y X y = x2 Taking x = 1 and 4 in the graph. The rectangle ABCD is formed where area A = AB × BC = 8 × 2 = 16cm2 . y = x2 O X’ Y’ Y X B C A D
PRIME Opt. Maths Book - IX 69 Taking x = 1.5 and 2.5 Taking x = 1.9 and 2.1 O X’ Y’ Y X y = x2 B C A D O X’ Y’ Y X y = x2 The area of rectangle ABCD, The area of rectangle ABCD A = AB × BC = 4 × 1 = 4cm2 A = AB × BC = 0.8 × 0.2 = 0.16 cm2 Conclusion: When value of x approaches to 2 form left hand limit as well as from right hand limit the rectangle ABCD will be fixed near 4 of f(x) Limit of a function Limit and Continuity can be defined by understanding the function y = f(x) and values of f(x) in different situations. The term limit and Continuity are the fundamental mathematical terms to introduce calculus can be defined only by studying the concept of function, limit and Continuity one after another in a sequence. If f is a function from non-empty set A to B which is written as y = f(x) where ‘y’ is the image of x and ‘x’ is the pre- image of ‘y’. Here, value of x = a is called the element of domain and y = f(x) = f(a) is called the range of function. • Let us take an example f(x) = 2x + 3 and x = 1 is element of domain. Then, y = f(x) = 2x + 3 y = f(2) = 2 × 1 + 3 = 5 (a finite number) • Taking a function f(x) = x 1 1 – and x = 1, Then, y = f(x) = x 1 1 – f(1) = 1 1 1 – = 0 1 ( It is not a finite number)
70 PRIME Opt. Maths Book - IX Taking a function f(x) = x x 1 1 – –2 and x = 1 Then, y = f(x) = x x 1 1 – –2 f(1) = 1 1 1 2 – –2 = 0 0 (It is indeterminant form) i.e. It can’t be determined. Meaning of x → a in a limit Let us consider an example by taking a variable x having values 1.9, 1.99, 1.999, 1.9999, 1.99999, ..........., approaching very close to 2 but which is not exactly equal to 2. i.e. The difference between the numbers x and 2 is very small. (too much less) Taking another example of the variable x having values 2.1, 2.01, 2.001, 2.0001, 2.000 ...........2 approaching very close to 2 but which is not exactly equal to 2. i.e., The difference between the numbers x and 2 is very small. Here, two examples discussed above tell us the value of x is very close to 2 in both the cases but not exactly equal to 2 which is read as, x approaches to 2 or x tends to 2 or x → 2 Thus For any constant ‘a’ of the variable ‘x’ we say, x → a. (x tends to a). Limit can be understand on discussing the area bounded by the polygons which are inscribed in a circle according to the number of sides of the polygons. As the number of sides of the polygon increased the area bounded by them also goes on increasing which is very close and equal to the area of circle for the large number of sides taken as infinity (∞). It can be observed in diagrams. The relation of number of sides of polygons and area is, f(n) = An for n → ∞ Where ‘n’ is number of sides f(n) is the area of polygons and An is the area of circle. It can be written as, lim n " 3 f(x) = lim n An " 3 From such examples limit of a function f(x) = ‘l’ can be defined
PRIME Opt. Maths Book - IX 71 Where, the limit for x → a can be written as lim x a " f(x) = l A function f(x) is said to be have a limit ‘l’ when x → a, if the numerical difference between the value of f(x) and ‘l’ can be made very small as well as we please by making x sufficiently close to a and we write it as, lim x a " f(x) = l Meaning of infinity in a limit If f(x) be a function of x which is very close to ‘a’ and the value of f(x) is obtained ultimately very large (∞) we say that the limit of f(x) is at infinity as ‘x’ tends to a. i.e. lim x a " f(x) = ∞ Let us consider an example: f(x) = x2 x 1 10 100 1000 10000 100000 f(x) 1 100 10000 100000 1000000 10000000 The value of f(x) goes on increasing as increasing ‘x’ and will be closed to infinity for x → ∞. Limit at infinity: Let us consider an example f(x) = x 1 2 , as the value of ‘x’ increase here in the function, the value of f(x) will be decreased as the reciprocal value. x 1 10 100 1000 f(x) 1 0.01 0.0001 0.000001 On increasing the value of x and tends to ∞, the value of f(x) ultimately decreased and tends to zero. i.e. ( ) lim x f x " 3 = lim x x 1 2 " 3 = 0 Indeterminate form: Let us consider a function f(x) = x x 1 1 – –2 and taking x = 1. The function will be, f(x) = x x 1 1 – – 2 = 0 0 (It can not be determined.) Such type of form like 0 0 , ∞ ∞, 3 + 3 , 3 – 3 etc are called indeterminate form and limit can not be defined for indeterminate form.
72 PRIME Opt. Maths Book - IX To find the limit of a function in indeterminate form should be changed by using algebraic operation. Examples : i) x x 1 1 – –2 = (x + 1)(x – 1) (x – 1) = (x + 1) ii) x x 2 8 – –3 = (x – 2)(x2 + 2x + 4) (x – 2) = x2 + 2x + 4 Properties of limit i. lim x a " k = k (where k is any constant) ii. lim x a " [f(x) + g(x)] = lim x a " f(x) + lim x a " g(x) iii. lim x a " [f(x) – g(x)] = lim x a " f(x) – lim x a " g(x) iv. lim x a " [f(x) . g(x)] = lim x a " f(x) . lim x a " g(x) v. lim x a " ( ) ( ) g x f x = ( ) ( ) lim lim x a g x x a f x " " Working rule to find limit i. For polynomial function, ( ) lim x a f x " = f(a) ii. For rational function, ( ) lim ( ) x a g x f x " = ( ) ( ) g a f a if it is not in indeterminate form. iii. If the function is in indeterminate form, factorize and cancel the common factors and find f(a). 1. Complete the given table. x 0.9 0.99 0.999 1.01 1.001 f(x) = x x 1 1 – –3 .......... .......... .......... .......... .......... Solution : Taking, f(x) = x x 1 1 – –3 = (x – 1)(x2 + x + 1) (x – 1) = x2 + x + 1 f(0.9) = 2.71 f(0.99) = 2.97 f(0.999) = 2.997 f(1.01) = 3.0301 f(1.001) = 3.003001 Worked out Examples
PRIME Opt. Maths Book - IX 73 It conclude that the whole number is near to 3 from both sides. Hence, it can be written as, ( ) lim x f x 1 3 " = x 0.9 0.99 0.999 1.01 1.001 x → 1 f(x) = x x 1 1 – –3 2.71 2.97 2.997 3.03 3.003 f(x) = 3 2. Show that the functional value does not exist but limit exist at x = 2 for the function f(x) = x x 2 4 – – 2 . Solution: Here, f(x) = x x 2 4 – –2 Functional value at x = 2 is, f(2) = 2 2 2 4 – –2 = 2 2 4 4 – – = 0 0 = It can not be determined. Hence, the functional value does not exist. 3. Show that the limit x lim " 2 x x 2 4 – – 2 exists. Solution: lim x x x 2 2 4 – –2 " = ( lim ( ) x x x 2 2 2 – –2 2 " = lim x 2 " ( ) ( )( ) x x x 2 2 2 – + – = (2 + 2) = 4 = It is finite number. = It exists limit. 4. Evaluate ( ) x x lim 2 3 2 – " Solution : lim x 3 " (2x – 3) = 2 × 3 – 3 = 6 – 3 = 3 Again, For limit at x = 2 lim x 2 " f(x) = lim x 2 " x x 2 4 – –2 = lim x 2 " ( ) ( )( ) x x x 2 2 2 – + – = (2 + 2) = 4 = It is finite number. Hence, limit exists.
74 PRIME Opt. Maths Book - IX 5. Find the value of lim x 1 " x x x x 2 2 3 – – 2 2 + + Solution : lim x 1 " x x x x 2 2 3 – – 2 2 + + = lim x 1 " x x x x x x 2 2 3 3 – – – 2 2 + + + = lim x 1 " x(x + 3) –1(x + 3) x(x + 2) –1(x – 2) = lim x 1 " ( )( – ) ( )( – ) x x x x 2 1 3 1 + + = 1 2 1 3 + + = 3 4 6. Find the value of lim x a x a x a – –3 3 Solution : $ lim x a x a x a – –3 3 " = lim x a $ (x – a)(x² + ax + a²) (x – a) = lim x a " ( ) x ax a 2 2 + + = a2 + a2 + a2 = 3a2 7. Calculate : lim x 0 $ x x x 2 3 2 2 + Solution : lim x 0 $ x x x 2 3 2 2 + = lim x 0 $ ( ) x x x 2 3 2 + = lim x 0 $ x 2 3 2 + = 2 3 0 × + 2 = 1 8. Evaluate : lim x $ 3 x x x x 2 3 1 6 2 – – 2 2 + + Solution : lim x $ 3 x x x x 2 3 1 6 2 – – 2 2 + + = lim x $ 3 x x x x x x 2 3 1 6 2 – – 2 2 2 2 + + = lim x $ 3 x x x x 2 3 1 6 1 2 – – 2 2 + +
PRIME Opt. Maths Book - IX 75 Exercise 2.2 1. i) What is limit of a function? Write in symbol also. ii) What is indeterminate form? Is limit define for it? What have to be done to find limit value for it? iii) Write down the round number of 3.002 and 4.999. Also present them as the limit value with symbol for a function f(x). iv) Complete the table given below. x 0.9 0.99 0.999 1.01 1.001 x → .......... f(x) = x x 1 1 – – 2 .......... .......... .......... .......... .......... f(x) .......... v) Complete the table given below. x 0.9 0.99 0.999 1.01 1.001 x → .......... f(x) = x x 1 1 – –2 .......... .......... .......... .......... .......... f(x) .......... 2. Answer the followings: i) Write down in symbols of x approaches to 4. ii) Write down in sentence of lim x 3 " f(x). iii) Write down in sentence of x → a. iv) Taking a function f(x) = x2 , discuss the value of function f(x)taking x nearly equal to 3 using graph. v) Taking a function f(x) = 2x – 1 and discuss left hand and right hand limit by using graph for a element of domain 3. 3. Which of the following functions exists the limit value of x? i) lim x x x 1 1 1 – – 2 " ii) lim x 0 " x x x x 2 2 6 – 2 2 + + + iii) lim x x x x 3 –2 2 " 3 iv) lim x x x x 0 2 3 5 –2 " v) lim x 2 " ( ) x x 1 2 + + = 2 3 1 6 1 2 – – 3 3 3 3 + + = 2 0 0 6 0 0 – – + + = 3
76 PRIME Opt. Maths Book - IX 4. Find the limit value of the followings. i) ( – ) lim x x 3 3 2 " ii) ( ) lim x x 3 2 3 " + iii) lim x 2 " ( – 2 5 x x 6) 2 + iv) lim x 4 " ( – 2 7 x x 3) 2 + v) lim x 0 " ( – 4 8 x x 3) 2 + 5. Find the limiting value of the following functions. i) lim x x x 2 2 1 3 2– " + ii) lim x x x 2 2 5 3– " + iii) lim x 1 " x x x x 5 4 3 2 4 – – 2 2 + + iv) lim x 3 " x x x x 2 4 6 3 12 – – 2 2 + + v) lim x 2 " x x x x 2 3 7 2 1 – – 2 2 + + 6. Compute the following limits. i) lim x 1 " x x x 1 3 4 – – 2 + ii) lim x 2 " x x x 2 5 6 – – 2 + iii) lim x 3 " x x x 3 2 5 3 – – – 2 iv) lim x x x 4 4 16 – – 2 " v) lim x a x a x a – – 2 2 " 7. Compute the following limits. i) lim x x x x x 0 3 2 4 8 – – 2 3 " ii) lim x x x x x 0 3 3 2 – – 2 3 " iii) lim x 2 " x x x x 2 5 6 – – – 2 2 + iv) lim x 0 " x x x x x 3 4 4 2 – 2 3 2 + + v) lim x x x x 0 5 3 2 " + 8. Compute the following limits. i) lim x x x 1 3 2 " 3 + + ii) lim x " 3 x x x x 4 5 3 2 1 – 2 2 + + + iii) lim x x x 3 2 2 2 2 " 3 + iv) lim x " 3 x x x x 5 4 3 4 3 2 – 2 2 + + + v) lim x " 3 x x x x 3 2 1 6 2 – 7 2 2 + + + 9. PRIME more creative questions (Evaluation the limits) i) lim x a x a x a – – 2 3 2 3 " ii) lim x a x a x a – – "
PRIME Opt. Maths Book - IX 77 iii) lim x x x 64 4 2 – – 3 6 " iv) lim x a x a x a – – 3 3 4 4 " v) lim x " 3 x a – x 1 + 10. Project work Take a piece of paper having length 80 cm and breadth 1 cm. Divide the piece of paper making two equal parts and one half again divide making half and go on continue upto 5 times. i) What will be the area of the smallest part? ii) Write down the sequence formed in this condition. iii) Write value with symbols. Answer 1. i) Show to your teacher. ii) Show to your teacher. iii) 3 and 5, lim x 3 " f(x), lim x 5 " f(x) iv) lim x 1 " f(x) = 2 1 v) lim x 1 " f(x) = 2 2. Show to your teacher. 3. i) doesn’t exists ii) exists iii) doesn’t exists iv) doesn’t exists v) exists 4. i) 7 ii) 9 iii) 4 iv) 7 v) 3 5. i) 5 4 ii) 4 7 iii) 2 1 iv) 3 v) 1 6. i) 5 ii) –1 iii) 7 iv) 8 v) 2a 7. i) 4 ii) 3 2 iii) 3 1– iv) 2 1 v) 3 8. i) 3 ii) 4 3 iii) 3 2 iv) 5 4 v) 2 9. i) 3 a 2 3 ii) 2 a 1 iii) 4 1 iv) a 3 4 v) 0
78 PRIME Opt. Maths Book - IX Limit and Continuity Unit Test Time : 30 minutes [1 × 1 + 3 × 2 + 2 × 4 + 1 × 5 = 20] Attempt all the questions: 1. Write down the 10th term of the sequence 5.1, 5.01, 5.001, 5.0001,------------ 2. a. Round of the number 7.368 to the tenth digit, hundred digit and whole number. b. What is the limiting value of the infinite series of 0.78? c. When does f(x) = x x 3 9 – –2 give certain existing value? Does f(x) give limit value for x → 3? 3. a. Complete the table for f(x) = 2x + 3. x 0.1 0.01 0.001 0.0001 0.00001 f(x) .......... .......... .......... .......... .......... b. Evaluate: lim x 0 " – x x x 2 2 3 2 2 + + a k 4. Taking an equilateral triangle of side 16cm. Take the mid-point of sides continuously three times for the triangle so formed and show in diagram. What will be the limiting value for area of triangles so formed.
PRIME Opt. Maths Book - IX 79 Unit 3 Matrices Specification Grid Table K(1) U(2) A(4) HA(5) TQ TM Periods No. of Questions 1 2 1 – 4 9 20 Weight 1 4 4 – K = Knowledge, U = Understanding, A = Application, HA = Higher ability, TQ = Total Question, TM = Total Marks Objectives : At the end of the lesson • Student are able to know the rows and columns of a matrix and its order. • Students are able to use matrix for the data used in daily life. • Student are able to identify the types of matrices. • Student are able to operate the matrices like addition, subtraction, multiplication, transpose, etc. • Students are able to use the difference laws for addition & multiplication of matrix. Materials Required: • Cost chart, mark chart etc which are used in our daily life. • Chart paper. • Flash card. • Sample of matrix of order m × n. • Chart of laws of addition and multiplication of matrices.
80 PRIME Opt. Maths Book - IX 3.1 Matrices The cost per kg of vegetables in Kalimati vegetable market in a particular day is as follows. Potato Cauliflower Cabbage Shop A 15 25 20 Shop B 20 30 15 Shop C 18 24 12 Above cost of the vegetables can be represented using square bracket as, 15 20 18 25 30 24 20 15 12 R T S S S S S S SS V X W W W W W W WW Here, rows represent the shops column represents the type of vegetables. Where the elements denote the cost. The rectangular arrangement of the numbers in rows and columns enclosed by ( ) or [ ] is called a matrix. (pluralmatrices) where horizontal arrangement are called rows and vertical arrangements are called columns. • The matrix is denoted by capital letter A, B, C, ... • The members of the matrix are called elements which are denoted by small letters a, b, c, d, e, f, ........................ • The horizontal arrangements of the matrix are called rows. • The vertical arrangements of the matrix are called columns. • The no. of rows (m) and no. of columns (n) can be written as m × n which is called the order of the matrix. Order of A = R × C • The matrix is denoted by using the position of elements as the no. of rows and columns as, A = a a a a a a a a a 11 21 31 12 22 32 13 23 33 R T S S S S S S SS V X W W W W W W WW Here, a11 = element in first row and first column. a23 = element in second row and third column. Order of the matrix A = 3 by 3 written as 3 × 3.
PRIME Opt. Maths Book - IX 81 General form of matrix of order m × n. According to the number of rows and columns of a matrix, its elements can be denoted by a11, a12, a13 .............. etc. Where aij is taken as the general element of ith row and jth column. Here, A matrix of order m × n can be written using general elements as, A = a11 a12 a13 ............. a1j .............. a1n a21 a22 a23 ............. a2j .............. a2n a31 a32 a33 ............. a3j .............. a3n ai1 ai2 ai3 ............. aij .............. ain am1 am2 am3 .............amj .............. amn aij is the general element of matrix A of ith row and jth column. If aij = 2i + 3j, then the matrix A of order 2 × 2 can be calculated as below. a11 = 2 × 1 + 3 × 1 = 5 a12 = 2 × 1 + 3 × 2 = 8 a21 = 2 × 2 + 3 × 1 = 7 a22 = 2 × 2 + 3 × 2 = 10 \ Matrix of order 2 × 2 is, A = 5 7 8 10 < F Types of matrices 1. Row matrix The matrix having only one row is called row matrix. Ex : A = [a11 a12 a13]1 × 3 B = [2 4]1 × 2 C = [a]1×1 2. Column matrix: The matrix having only one column is called column matrix. Ex : A = a a a 11 21 31 3 1× R T S S S S S S SS V X W W W W W W WW B = 5 3 2 1 4 1× R T S S S S S S S S S SS V X W W W W W W W W W WW C = [0]1×1 3. Null matrix (zero): The matrix having all the elements zero is called null matrix. It is denoted by ‘0’. Ex : 0 = 0 0 0 0 0 0 2 3× < F
82 PRIME Opt. Maths Book - IX 4. Rectangular matrix: The matrix having unequal number of rows and columns is called rectangular matrix. Ex : A = 2 1 3 3 4 – –2 2 3× < F 5. Square matrix: The matrix having equal number of rows and columns is called square matrix. Ex : A = [2]1×1 B = 1 3 2 4 2 2× < F C = 1 4 7 2 5 8 3 6 9 3 3× R T S S S S S S SS V X W W W W W W WW 6. Diagonal matrix The square matrix having the main diagonal elements (from left top to right bottom) non-zero and remaining elements are zero is called diagonal matrix. Ex: A = a 0 b 0 2 2× < F B = p q r 0 0 0 0 0 0 3 3× R T S S S S S S SS V X W W W W W W WW 7. Scalar matrix: The diagonal matrix having all non-zero diagonal elements are equal is called scalar matrix. Ex : A = a a a 0 0 0 0 0 0 3 3× R T S S S S S S SS V X W W W W W W WW B = k k k 0 0 0 0 0 0 3 3× R T S S S S S S SS V X W W W W W W WW 8. Identity matrix (Unit matrix): The square matrix having the main diagonal elements (from left top to right bottom) all one and remaining elements are zero is called identity matrix. It is denoted by ‘I’. Eg : I = 1 0 0 1 2 2× < F I = 1 0 0 0 1 0 0 0 1 3 3× R T S S S S S S SS V X W W W W W W WW 9. Triangular matrix: The square matrix having zero elements towards upper or lower parts of leading diagonal of the matrix is called triangular matrix. Eg : A = 1 0 2 3 2 2× < F L.T.M B = 1 0 0 2 4 0 3 5 6 3 3× R T S S S S S S SS V X W W W W W W WW U.T.M C = 1 2 0 3 2 2× < F D = 1 2 4 0 3 5 0 0 6 3 3× R T S S S S S S SS V X W W W W W W WW Here, Matrices A and B are called upper triangular matrices and C and D are called lower triangular matrices.
PRIME Opt. Maths Book - IX 83 1. The cost of fruits in a fruits shop in different days is given below. Represent the informations in a matrix with appropriate meaning. Apple Orange Banana Day 1 150 80 75 Day 2 120 100 70 Solution : The cost of fruits given in table of two days are represented by matrix as follows where, rows → represents the days columns → represents the type of fruits A = 150 120 80 100 75 70 < F 2. If A = 3 2 2 4 5 – –1 < F i) Find the order of matrix A. ii) Find the elements a12, a21, & a23. Solution, i) No. of rows = 2 No. of columns = 3 \ Order of matrix A is 2 by 3 (2 × 3). ii) a12 = element in 1st row 2nd column = 2 a21 = element in 2nd row 1st column = –2 a23 = element in 2nd row 3rd column = –1 3. If A = B, where A = x y 2 3 1 + = G and B = y 3 6 2 3 – 1 < F , find the value of x & y. Solution : A = x y 2 3 1 + = G, B = y 3 6 2 3 – 1 < F Here, Matrix A & B are equal. Worked out Examples 10. Equality of the matrices Any two matrices having same order and same corresponding elements are called equal matrices. Ex : A = 2 4 3 1 < F, B = 2 4 3 1 < F Here, A = B (equal matrices)
84 PRIME Opt. Maths Book - IX By equating the corresponding elements x + 2 = 3 and y = 6 – 2y or, x = 3 – 2 y + 2y = 6 or, x = 1 3y = 6 \ x = 1 y = 3 6 \ y = 2 \ x = 1, y = 2 4. If aij = 3i – 2j is the general element of a matrix, find the matrix of order 2 × 3. Also write down its type. Solution : aij = 3i – 2j (general element) Then, 2 × 3 matrix A = a a a a a a 11 21 12 22 13 23 < F where, a11 = 3 × 1 – 2 × 1 = 1 a12 = 3 × 1 – 2 × 2 = –1 a13 = 3 × 1 – 2 × 3 = – 3 a21 = 3 × 2 – 2 × 1 = 4 a22 = 3 × 2 – 2 × 2 = 2 a23 = 3 × 2 – 3 × 2 = 0 \ A = 1 4 1 2 3 0 – – < F Here, Number of rows and column of the matrix are different. So, it is rectangular matrix.
PRIME Opt. Maths Book - IX 85 Exercise 3.1 1. i) What is matrix? Write down one example. ii) What do you mean by order of the matrix? iii) What is scalar matrix? Write down its example. iv) What is identify matrix? Write down its example. v) Write down the standard form of the matrix. 2. Write down the following informations in matrix form with appropriate meaning. Also write down the order of the matrices. i) The number student in a class are given below. 1st column 2nd column 1st row 4 3 2nd row 5 2 3rd row 6 4 4th row 3 5 ii) The cost of clothes in three shops are given below. Shirt Pants Vest 1st shop 500 700 300 2nd shop 450 600 350 3rd shop 400 550 250 iii) The production of crops in Jhapa district in different years in metric tonne is given below. Rice Wheat Maize others 2072 BS 300 150 100 150 2073 BS 350 170 120 130 2074 BS 475 180 200 300 iv) What is matrix? Write short note on order of the matrix. v) In what conditions equality of the matrices takes place? Write down with example. 3. Write down the types of the matrices from the followings. Also write down the order of the matrices. i) 3 2 1 1 4 5 R T S S S S S S SS V X W W W W W W WW ii) 0 0 0 0 iii) 1 4 1 2 5 0 3 2 – 1 R T S S S S S S SS V X W W W W W W WW iv) 2 0 0 2 < F v) a b c 0 0 0 0 0 0 R T S S S S S S SS V X W W W W W W WW vi) 1 0 0 0 1 0 0 0 1 R T S S S S S S SS V X W W W W W W WW
86 PRIME Opt. Maths Book - IX vii) 2 2 3 2 1 – –2 < F viii) 2 3 4 R T S S S S S S SS V X W W W W W W WW ix) 6 @ 3 2– 1 4. Answer the following questions from the matrix 2 1 3 3 0 2 1 5 1 – R T S S S S S S SS V X W W W W W W W i) Element in first row and second column. W ii) The element a22. iii) a31 + a23 + a33 iv) Element aij where i = 3 & j = 2. v) Write down the order of the matrix. 5. If the general element of a matrix is aij = 2i – 3j, what will be the matrices of following order. i) 2 × 2 matrix ii) 2 × 3 matrix iii) 3 × 3 matrix iv) 3 × 2 matrix v) 3 × 1 matrix 6. Find the value of ‘x’ and ‘y’ from the following equal matrices. i) A = 3 x 5 1 6 – < F & B = y 3 2 1 7 + 6 < F ii) A = x y 2 3 2 5 1 + + = G, B = x y 7 2 2 5 3 5 – – < F iii) P = x y y 2 3 2 5 + 7 = G, B = y 7 3 2 5 7 2 3– < F iv) M = x y 4 x y 2 1 3 5 7 2 – – + R T S S S S S S SS V X W W W W W W WW , N = 4 2 1 3 1 5 7 7 –2 R T S S S S S S SS V X W W W W W W WW v) A = x y 5 2 7 + 1 < F= 5 x y 7 3 2 1 – < F Answer 1. Show to your teacher. 2. Show to your teacher. 3. Show to your teacher. 4. i) 3 ii) 0 iii) 9 iv) 2 v) 3 × 3 5. i) 1 1 4 2 – – – < F ii) 1 1 4 2 7 5 – – – – – < F iii) 1 1 3 4 2 0 7 5 3 – – – – – R T S S S S S S SS V X W W W W W W WW iv) 1 1 3 4 2 0 – – – R T S S S S S S SS V X W W W W W W WW v) 1 1 3 – R T S S S S S S SS V X W W W W W W WW 6. i) x = 8, y = 2 ii) x = 1, y = 3 iii) x = 2, y = 3 iv) x = 4, y = 3 v) x = 3, y = 1
PRIME Opt. Maths Book - IX 87 3.2 Operation on matrices The simpli�ication of two or more matrices in to a single matrix by using any kind of mathematical operations indicates the operation on matrices. 3.2.1. Addition of matrices Marks obtained by three students Sita, Pranav and Pranisha in two test in optional maths in two successive months are as follows. 1st month 2nd month 1st 2nd 1st 2nd Sita 60 75 Sita 70 75 Pranav 85 90 Pranav 95 80 Pranisha 95 85 Pranisha 80 95 Total marks obtained by them in two tests in two months as, 1st 2nd Sita 60 + 70 = 130 75 + 75 = 150 Pranav 85 + 95 = 180 90 + 80 = 170 Pranisha 95 + 80 = 175 85 + 95 = 180 This information can be expressed in matrix form as, 60 85 95 75 90 85 R T S S S S S S SS V X W W W W W W WW + 70 95 80 75 80 95 R T S S S S S S SS V X W W W W W W WW = 60 70 85 95 95 80 75 75 90 80 85 95 + + + + + + R T S S S S S S SS V X W W W W W W WW = 130 180 175 150 170 180 R T S S S S S S SS V X W W W W W W WW The sum of any two matrices having same order is called a new single matrix obtained by adding the corresponding elements of the matrices where the single matrix also has the order same as the given matrices.
88 PRIME Opt. Maths Book - IX Points to remember: • Order of the matrices should be same. • Corresponding elements of the matrices should be added respectively. • New matrix obtained after addition has the order same as the given matrices. Eg. If A = 2 1 3 – 5 < F, B = 3 4 2 – –1 < F Then, A + B = 2 1 3 – 5 < F + 3 4 2 – –1 < F = 2 3 1 4 3 2 – – 5 – 1 + + = G = 5 5 5 – 4 < F 3.2.2. Difference of the matrices (Subtraction) Out of three persons in the above example, who has got more marks in which month can be found out by subtraction. Where, Difference of marks of them 1st month 2nd month 1st 2nd 1st 2nd Sita 60 75 Sita 70 75 Pranav 85 90 Pranav 95 80 Pranisha 95 85 Pranisha 80 95 Difference marks obtained by them in two tests in two months as, 1st 2nd Sita 60 – 70 = –10 75 – 75 = 0 Pranav 85 – 95 = –10 90 – 80 = 10 Pranish 95 – 80 = 15 85 – 95 = –15 This information can be expressed in matrix form as, 60 85 95 75 90 85 R T S S S S S S SS V X W W W W W W WW – 70 95 80 75 80 95 R T S S S S S S SS V X W W W W W W WW = 60 70 85 90 95 80 75 75 90 80 85 95 – – – – – – R T S S S S S S SS V X W W W W W W WW = 10 10 15 0 10 15 – – – R T S S S S S S SS V X W W W W W W WW
PRIME Opt. Maths Book - IX 89 The difference of any two matrices having same order is called a new single matrix obtained by subtracting the corresponding elements of the matrices where the single matrix also has the order same as the given matrices. Eg. If P = 2 5 3 3 1 – 2 < F, Q = 1 2 1 1 5 3 – < F Then, P – Q = 2 5 3 3 1 – 2 < F – 1 2 1 1 5 3 – < F = 2 1 5 2 3 1 3 1 1 5 – 2 3 – – – – – + = G = 3 3 2 4 4 – 1 – – < F 3.2.3. Multiplication of a matrix with a scalar. The new matrix formed by multiplying the each elements of a given matrix with a given scalar quantity is called the multiplication of a matrix with a scalar. Eg. If A = 1 3 2 1 – – < F, �ind 3A. Solution : A = 1 3 2 1 – – < F 3A = 3 1 3 2 1 – – < F = 3 9 6 3 – – < F 3.2.4. Transpose of the matrix. In the example, cost of apple is given. 1st shop 2nd shop Sunday 150 140 Monday 160 145 Tuesday 170 150
90 PRIME Opt. Maths Book - IX It can be written in such a way by changing the information of rows and column as, Sunday Monday Tuesday 1st Shop 150 160 170 2nd Shop 140 145 150 Such examples in matrix form. A = 150 160 170 140 145 150 R T S S S S S S SS V X W W W W W W WW A = after changing row and columns = 150 140 160 145 170 150 < F It is called transpose of the matrix. The new matrix obtained by interchanging the rows and columns of a matrix is called transpose of the matrix. Transpose of A is denoted by AT or A`. Order of transpose matrix will be different from matrix A from rectangular matrix but same for the square matrix. Eg. If A = 3 2 4 6 5 9 < F then. AT = 3 4 5 2 6 9 R T S S S S S S SS V X W W W W W W WW 3.2.5. Symmetrical matrix: The square matrix where same matrix is formed by interchanging the rows and columns is called symmetrical matrix. (i.e. A = AT = symmetric) Eg : A = a b b c < F, B = 5 2 3 2 4 8 3 8 1 R T S S S S S S SS V X W W W W W W WW 3.2.6 Properties of matrix addition : 1. Closer property : A + B is hold for matrices A and B of same order. If A = 2 3 1 1 < F, B = 1 4 2 3 < F Then, A + B = 2 3 1 1 < F + 1 4 2 3 < F = 2 1 3 1 1 2 1 3 + + + + = G = 3 7 3 4 < F It has same order of A and B.
PRIME Opt. Maths Book - IX 91 2. Commutative property A + B = B + A If A = 2 3 1 –1 < F, B = 3 2 1 1 – < F Then, A + B = 2 3 1 –1 < F + 3 2 1 1 – < F = 2 3 3 2 1 1 1 1 – – + + + = G = 5 5 0 0 < F B + A = 3 2 1 1 – < F + 2 3 1 –1 < F = 3 2 2 3 1 1 1 1 – – + + + = G = 5 5 0 0 < F \ A + B = B + A(Hence proved) 3. Associative property : (A + B) + C = A + (B + C) If A = 1 3 2 4 < F, B = 3 1 1 – 2 < F, C = 1 2 3 –2 < F Then, L.H.S. = (A + B) + C = 1 3 2 4 < F + 3 1 1 – 2 < F + 1 2 3 –2 < F = 1 3 3 1 2 1 – 4 2 + + + = G + 1 2 3 –2 < F = 4 2 3 6 < F + 1 2 3 –2 < F = 4 1 2 2 3 3 6 2– + + + = G = 5 4 6 4 < F R.H.S. = 1 3 2 4 < F + 3 1 1 – 2 < F + 1 2 3 –2 < F = 3 1 1 2 1 3 – – 2 2 + + + = G = 4 1 4 0 < F = 1 4 3 1 2 4 4 0 + + + + = G = 5 4 6 4 < F \ L.H.S. = R.H.S. proved
92 PRIME Opt. Maths Book - IX 4. Additive inverse: For any matrix A ∃ a matrix – A such that Here – A is called the additive in verse of A. A + (–A) = (A) – A = 0 If A = 2 3 1 2 – – < F ∴ – A = 2 3 1 2 – – < F Then, A + (–A) = 2 3 1 2 – – < F + 2 3 1 2 – – < F = 2 2 3 3 1 1 2 2 – – – – + + = G = 0 0 0 0 < F (–A) + A = 2 3 1 2 – – < F + 2 3 1 2 – – < F = 2 2 3 3 1 1 2 2 – – – – + + = G = 0 0 0 0 < F ∴ A + (–A) = (–A) + A = 0 proved 5. Additive identity : For any matrix A ∃ a matrix ‘O’ of same order such that ‘O’ is called the additive identity. A + O = O + A = A If A = 2 2 3 – 4 < F, O = 0 0 0 0 < F Then, A + O = 2 2 3 – 4 < F + 0 0 0 0 < F = 2 0 2 0 3 0 – 4 0 + + + + = G = 2 2 3 – 4 < F = A O + A = 0 0 0 0 < F + 2 2 3 – 4 < F = 0 2 0 2 0 3 – 0 4 + + + = G = 2 2 3 – 4 < F = A ∴ A + O = O + A = A proved. 6. Distributive property over scalar. K(A + B) = KA + KB where k is a scalar. If A = 2 4 3 1 < F, B = 1 2 1 3 – < F Then, K(A + B) = K 2 4 3 1 1 2 1 3 – (< F + < F2 = K 2 1 4 2 3 1 1 3 – + + + = G = K K K 6 K 4 4 < F
PRIME Opt. Maths Book - IX 93 KA + KB = K 2 4 3 1 < F + K 1 2 1 3 – < F = K K K k 2 4 3 < F + K K K 2 3K – < F = K K K K K K K K 2 4 2 3 3 – + + + = G = K K K 6 K 4 4 < F ∴ L.H.S. = R.H.S. is proved 7. Property of the matrices over transpose: i) (AT)T = A If A = 2 4 3 1 < F Then, AT = 2 4 3 1 T < F = 2 3 4 1 < F (AT)T = 2 3 4 1 T < F = 2 4 3 1 < F = A ∴ (AT)T = A = 2 4 3 1 < F ii) (A + B)T = AT + BT If A = 1 3 2 4 – < F, B = 2 1 4 –1 < F Then, (A + B)T = 1 3 2 4 2 1 4 1 – – T (< F + < F2 = 1 2 3 1 2 4 4 1 – – T + + + = G = 3 4 2 3 T < F = 3 2 4 3 < F AT + BT = 1 3 2 4 – T < F + 2 1 4 –1 T < F = 1 2 3 – 4 < F + 2 4 1 –1 < F = 1 2 2 4 3 1 – 4 – 1 + + + = G = 3 2 4 3 < F ∴ (A + B)T = AT + BT proved
94 PRIME Opt. Maths Book - IX 1. If A = 1 1 2 – 3 < F, B = 3 2 2 1 – < F, prove that (A + B)T = BT + AT. Solution : A + B = 1 1 2 – 3 < F + 3 2 2 1 – < F = 4 1 0 4 < F ∴ L.H.S. = (A + B)T = 4 1 0 4 T < F = 4 0 1 4 < F ∴ R.H.S. = BT + AT = 3 2 2 – 1 < F + 1 2 1 3 – < F = 4 0 1 4 < F ∴ L.H.S. = R.H.S. proved. 2. If A = 1 1 2 – 3 < F, B = 2 1 1 2 – < F and C = 3 2 1 1 – < F, prove that A + (B + C) = (A + B) + C. Solution : L.H.S. = A + (B + C)= 1 1 2 – 3 < F + 2 1 1 2 3 2 1 1 – – (< F + < F2 = 1 1 2 – 3 < F + 5 3 0 1 < F = 6 2 2 4 < F R.H.S. = (A + B) + C= 1 1 2 3 2 1 1 – 2 – (< F + < F2 + 3 2 1 –1 < F = 3 0 1 5 < F + 3 2 1 –1 < F = 6 2 2 4 < F ∴ L.H.S. = R.H.S. proved. Worked out Examples
PRIME Opt. Maths Book - IX 95 3. If A + B = 4 3 1 4 < F and A – B = 2 5 3 2 – – < F, find the matrices A and B. Also prove the closer property of A + B. Solution : A + B = 4 3 1 4 < F ................ (i) A – B = 2 5 3 2 – – < F ................ (ii) Adding (i) and (ii), we get, A + B = 4 3 1 4 < F A – B = 2 5 3 2 – – < F 2A = 2 2 4 – 6 < F A = 1 1 2 – 3 \ < F Putting the value of ‘A’ in equation (i) B = 4 3 1 4 < F – 1 1 2 – 3 < F = 3 4 1 1 – < F \ A = 1 1 2 – 3 < F & B = 3 4 1 1 – < F Again, B + A = 3 4 1 1 – < F + 1 1 2 – 3 < F = 3 1 4 1 1 2 – 1 3 + + – + = G = 4 3 1 4 < F \ (A + B) = (B + A) i.e. closer property is proved.