doc 1104b b.p.s. xi maths chapterwise topicwise worksheets with solution 2014 15

BRILLIANT PUBLIC SCHOOL , SITAMARHI

(Affiliated up to +2 level to C.B.S.E., New Delhi) Affiliation No. - 330419

XI - Maths Chapterwise Topicwise Worksheets with

Solution

Session : 2014-15

Office: Rajopatti, Dumra Road, Sitamarhi(Bihar), Pin-843301 Website: www.brilliantpublicschool.com; E-mail: [email protected]

Ph.06226-252314, Mobile: 9431636758, 9931610902

MATHEMATICS (Class 11)

Index

Chapters page

1. Sets 01

2. Relations and Functions 20

3. Trigonometric Functions 47

4. Principle of Mathematical Induction 69

5. Complex Numbers and Quadratic Equations 100

6. Linear Inequalities 124

7. Permutations and Combinations 136

8. Binomial Theorem 161

9. Sequences and Series 184

10. Straight Lines 210

11. Conic Sections 234

12. Introduction to Three Dimensional Geometry 258

13. Limits and Derivatives 276

14. Mathematical Reasoning 305

15. Statistics 309

16. Probability 343

CBSE TEST PAPER-01

CLASS - XI MATHEMATICS (Sets)

Topic: -Sets

1. Describe the set in Roster form

}{ : is a two digit number such that the sum of its digit is 8x x[1]

2. Are the following pair of sets equal? Give reasons.

A = { x:x is a letter in the word FOLLOW}

B = { y:y is a letter in the word WOLF}

[1]

3. Write down all the subsets of the set {1,2,3} [1]

4. Let A = { 1,2,{3,4,},5} is { }{ }3,4 A is incorrect. Give reason. [1] 5.

Draw venn diagram for ( )A B [1] 6. In a survey of 400 students in a school, 100 were listed as taking apple juice, 150 as

taking orange juice and 75 were listed as taking both apple as well as orange juice.

Find how many students were taking neither apple juice nor orange juice.

[4]

7. A survey shows that 73% of the Indians like apples, whereas 65% like oranges.

What % Indians like both apples and oranges?

[4]

8. In a school there are 20 teachers who teach mathematics or physics. Of these 12

teach mathematics and 4 teach both physics and mathematics. How many teach

physics?

[4]

9. There are 200 individuals with a skin disorder, 120 had been exposed to the

chemical C1, 50 to chemical C2, and 30 to both the chemicals C1 and C2. Find the

number of individuals exposed to

(1) chemical C1 but not chemical C2

(2) chemical C2 but not chemical C1

(4) chemical C1 or chemical C2

[6]

10. In a survey it was found that 21 peoples liked product A, 26 liked product B and 29

liked product C. If 14 people liked products A and B, 12 people like C and A, 14

people like B and C and 8 liked all the three products. Find now many liked product

C only.

[6]

CBSE TEST PAPER-01


Topic: -Sets [ANSWERS]

Ans 01. { 17, 26, 35, 44, 53, 62, 71, 80 }

Ans 02. A= {F, O, L, W}

B = {W, O, L, F }

Hence A=B

Ans 03. , {1},{2},{3},{1,2},{1,3},{2,3},{1,2,3}

Ans 04. {3,4} is an elements of sets A, therefore { }{ }3,4 is a set containing element {3,4}which is belongs to A

Hence { }{ }3,4 A is correct

Ans 05. (A B ) = U- (AB)

Ans 06. Let A denote the set of students taking apple juice and B denote the set of students taking orange juice n( ) = 400, n(A) = 100, n (B) = 150 n(A B) =75 n ( A B ) = n(A B ) =n( ) n(A B) = n() [n(A) + n(B) n(A B)] =400-100-150+75=225

Ans 07. Let A=set of Indian who like apples B= set of Indian who like oranges n (A)= 73, n (B) = 65 n(AUB)=100 n(A B) = n(A) + n(B) - n(AB) =73+65-100 =38 38% like both

Ans 08. n(MP) = 20, n(M)=12 n(M P) =4 n(MP) = n(M)+n(P)- n(M P) n(P)=12

Ans.09. A denote the set of individuals exposed to the chemical C1 and B denote the set of individuals exposed to the chemical C2 n(U) = 200, n(A) = 120, n(B) = 50, n(A B) = 30

(i) n(A-B) = n(A) n(A B) =120-30=90

(ii) n(B-A) = n(B) n(A B)

=50-30 = 20

(iii) n(AB) =n(A)+ n(B)-n(A B)

=120+50-30

=140

Ans.10. a + b + c + d = 21 b + c + e + f = 26 c + d + f + g = 29 b + c = 14, c + f =15, c + d = 12 c = 8 d = 4, c = 8, f = 7, b = 6 g = 10, e = 5, a = 3 like product c only = g = 10

ab

c

d

e

f

g

A B

C

CBSE TEST PAPER-02


Topic: -Sets

1. Write the set in roster form A = The set of all letters in the word T R I G N O M E

T R Y

[1]

2. Ate the following pair of sets equal? Give reasons

A, the set of letters in ALLOY and B, the set of letters in LOYAL.

[1]

3. Write down the power set of A

A = {1, 2, 3}

[1]

4. A = {1, 2, {3, 4}, 5} which is incorrect and why. (i) {3, 4} A (ii) {3, 4} A [1]

5. Fill in the blanks.

(i) A A = --------

(ii) ( )A = ---------(iii) A A = --------

[1]

6. Let U = {1, 2, 3, 4, 5, 6} A = {2, 3} and B = {3, 4, 5}

Find A B , A B and hence show that (A B) = A B .[4]

7. For any two sets A and B prove by using properties of sets that:

( ) ( )A B A - B = A [4]

8. If A, B, and C, are three sets and U is the universe set such that n(U) = 1000,

n(A) = 300, n (B) 300 and n(A B) = 200 find n ( )A B .[4]

9. A college awarded 38 medals in football, 15 in basketball and 20 in cricket. If

these medals went to a total of 58 men and only three men got medal in all the

three sports, how many received medals in exactly two of the three sports?

[6]

10. In a survey of 60 people, it was found that 25 people read news paper H, 26

read newspaper T, 26 read newspaper I, 9 read both H and I, 11 read both H

and T, 8 read both T and I, 3 read all three newspaper. Find

(i) The no. of people who read at least one of the newspapers.

(ii) The no. of people who read exactly one news paper.

[6]

CBSE TEST PAPER-02


[ANSWERS]

Topic: - Sets

Ans 01. A = {T, R, I, G, N, O, M, E, Y}

Ans 02. A = {A, L, O, Y}

B = {L, O, Y, A}

Hence A = B

Ans 03. { } { } { } { } { } { } { }{ }P (A) = 1 , 2 , 3 , 1, 2 , 1,3 , 2,3 , 1, 2,3Ans 04. {3, 4} is an element of set A.

Hence { }3, 4 A is correct and{ }3, 4 A is incorrect.

Ans 05. (i)

(ii) A(iii)

Ans 06. A = U A= { 1, 4, 5, 6}

B = U B={1, 2, 6}

A U B = {2, 3, 4, 5}

(A B ) = U (A B) = {1, 6}

A B = {1, 6}Hence proved.

Ans 07. L. H. S. = ( A B) (A B)

= (A B) (A )B= X (A B) [ X = A B= (X A) (X B) = A (A B )= A

Ans 08. n ( A B ) = n ( )A B = n(U) n (A B)

= n (U) [ n(A) + n (B) n (A B)]

= 1000 [300+300-200]

= 1000 400

= 600

Ans 09. Let A, B and C denotes the set of men who received medals in football,

basketball and cricket respectively.

n (A) = 38, n (B) = 15, n (C) = 20

n (A B C) = 58 and n (A B C) = 3

n (A B C) = n (A) + n (B) + n(C) n (A B) n (B C) n (C A) + n (A

B C)

58 = 38 + 15 + 20 (a + d ) (d + c) (b + d) + 3

18 = a + d + c + b + d

18 = a + b + c + 3d

18 = a + b + c + 3 3

Ans 10. a + b + c + d = 25

b + c + e + f = 26

c + d + f +g = 26

c + d = 9

b + c = 11

c + f = 8

c = 3

f = 5, b = 8, d = 6, c = 3, g = 12

e = 10, a = 8

(i) a + b + c + d + e + f + g = 52

(ii) a + e + g = 30

9 = a + b + c

b

d

a

c

A B

C

b

d

a

c

H T

I

f

g

e

CBSE TEST PAPER-03


Topic: -Sets

1. Write the set

1 2 3 4 5 6, , , , ,

2 3 4 5 6 7

in the set builder form. [1]

2. Is set C = { x : x 5 = 0} and E = {x : x is an integral positive root of the equation x2

2x 15 = 0}

[1]

3. Write down all possible proper subsets of the set {1, {2}}. [1]

4. State whether each of the following statement is true or false.

(i) {2, 3, 4, 5} and {3, 6} are disjoint

(ii) {2, 6, 10} and {3, 7, 11} are disjoint sets

[1]

5. Fill in the blanks

(i) (A B) = --------- (ii) (A B) = ---------

[1]

6. There are 210 members in a club. 100 of them drink tea and 65 drink tea but not

coffee, each member drinks tea or coffee.

Find how many drink coffee, How many drink coffee, but not tea.

[4]

7. If P (A) = P (B), Show that A = B [4]

8. In a class of 25 students, 12 have taken mathematics, 8 have taken mathematics but

not biology. Find the no. of students who have taken both mathematics and biology

and the no. of those who have taken biology but not mathematics each student has

taken either mathematics or biology or both.

[4]

9. These are 20 students in a chemistry class and 30 students in a physics class. Find

the number of students which are either in physics class or chemistry class in the

following cases.

(i) Two classes meet at the same hour

(ii) The two classes met at different hours and ten students are enrolled in both the

courses.

[6]

10. In a survey of 25 students, it was found that 15 had taken mathematics, 12had taken

physics and 11 had taken chemistry, 5 had taken mathematics and chemistry, 9 had

taken mathematics and physics, 4 had taken physics and chemistry and 3 had taken

all three subjects.

Find the no. of students that had taken

(i) only chemistry (ii) only mathematics (iii) only physics

(iv) physics and chemistry but mathematics (v) mathematics and physics but not

chemistry (vi) only one of the subjects (vii) at least one of three subjects

(viii) None of three subjects.

[6]

CBSE TEST PAPER-03


[ANSWERS]

Topic: - Sets

Ans 01. : , where n is a nature no. and 1 n 61

nx x

n

=

+

Ans 02. C = {5}

[ ]

{ }

2

2

2 15 05 3 15 0

( 5) 3( 5) 0( 5)( 3) 0

53 3 reject

55

Henc C = E.

x x

x x x

x x x

x x

x

x x

x

E

=

+ =

+ =

+ =

=

= =

=

=

Ans 03. { } { }{ } { }{ }, 1 , 2 , 1, 2

Ans 04. (i) { } { } { }2,3, 4,5 3,6 = 3 Hence false

(ii) { } { }2,6,10 3,7,11 = true

Ans 05. (A B ) = A B(A B ) = A B

Ans 06. n (T) = 100

n (T C) = 65

n (T C) = 210

n (T C) = n (T) n (T C)

65 = 100 n (T C)

n (T C) = 35

n (T C) = n (T) + n (C) n (T C)

210 = 100 + n (C) 35

n (C) = 145.

Ans 07. a A { }{ }{ } [{ }

{ }{ }{ }{ }

A P(A) P(B) P(A) = P(B) B

a BA B

for all b Bb Bb P (B) [ P (A) = P (B)b P (A)b A

AB A

Thus A B and B A

A=B

a

a

a

a

b

Ans 08. n (M) = 12, n (M B) = 8

(M B) = 25n ( M B) = n(M) + n (B - M)25 = 12 + n (B - M) 13 = n (B-M)

n

n (M B) = n (M-B) + n (M B) + n(B-M)25 = 8+n (M B) +13n(M B) = 4

Ans 09. Let C be the set of students in chemistry class and P be the set of students in physics

class.

n (C) = 20, n (P) = 30

(i) C P = n (C P) = n (C) + n (P)

= 20 + 30 = 50

(ii) n (C P) = 10n (C F) = n(C) + n(F)-n(C P)

= 20 + 30 - 10 = 40

Ans 10. n(M)=a+b+d+e=15 ( ) b + c + e + f = 12

n(C) = d + e+f + g = 11n ( M P) = b + e = 9 n (M C) = d + e = 5 n (P C) = e + f = 4 e = 3 so b = 6, d = 2, f = 1 a = 4, g = 5, c = 2 (i) g = 5, (ii) a = 4, (iii) c = 2 (iv)

n P =

f = 1, (v) b = 6, (vi) g + a + c = 11(vii) a + b + c + d + e + f + g + = 23 (viii) 25 - (a + b + c + d + e + f + g) = 25 - 23 = 2

ab c

de f

g

MP

C

CBSE TEST PAPER-04


Topic: -Sets

1. Write the set of all vowels in the English alphabet which precede k. [1]

2. Is pair of sets equal? Give reasons.

A = {2, 3} B = x : x is solution of x2 + 5x + 6 = 0}

[1]

3. Write the following intervals in set builder form:

(-3, 0) and [6, 12]

[1]

4. If X = {a, b, c, d}

Y = {f, b, d, g}

Find X Y and Y - X

[1]

5. If A and B are two given sets, Then represent the set (A B ) , using Venndiagram.

[1]

6. A and B are two sets such that n (A B) = 20 + x, n (B A) = 3x and n (A B) = x

+ 1. Draw a Venn diagram to illustrate this information. If n (A) = n (B), Find (i)

the value of x (ii) n (A B)

[4]

7. If A and B are two sets such that A B= A B, then prove that A = B [4]

8. Prove that if A B = C and A B = then A = C B [4] 9. In a survey of 100 students, the no. of students studying the various languages

were found to be English only 18, English but not Hindi 23, English and Sanskrit

8, English 26, Sanskrit 48, Sanskrit and Hindi 8, no language 24. Find

(i) How many students were studying Hindi?

(ii) How many students were studying English and Hindi?

[6]

10. In a class of 50 students, 30 students like Hindi, 25 like science and 16 like both.

Find the no. of students who like

(i) Either Hindi or science

(ii) Neither Hindi nor science.

[6]

CBSE TEST PAPER-04


[ANSWERS]

Ans 01. A = {b, c, d, f, g, h, j}

Ans 02. A = {2, 3}

B = {- 2, - 3}

A B

[ x2 + 5x + 6 = 0

x2 + 3x + 2x + 6 = 0

x = - 2 3

Ans 03. (- 3, 0) = {x : x R, - 3 < x < 0}

[6, 12] = {x : x R, 6 x 12}

Ans 04. X Y = {a, b, c, d} {f, b, d, g}

= {a, c}

Y X = {f, b, d, g} {a, b, c, d}

= {f, g}

Ans 05. (A B ) = - (A B)

Ans 06. (i) n (A) = n (A B) + n (A B)

= 14 + x + x

= 14 + 2x

N (B) = n (B A) + n (A B)

= 3x + x

= 4x

but n (A) = n (B) (Given)

14 + 2x = 4x

x = 7

(ii) n (A B) = n (A B) + n (B A) + n (A B)

= 14 + x+ 3x + x

= 14 + 5x = 14 + 5 7 = 49

x14+x

A B

3x

Ans 07. Let a A, then a A B

Since A B = A B

a A B . So a B

Therefore A B

Similarly if b B,

Then b A B. Since

A B = A B, b A B

So b A

Therefore, B A

Thus

Ans 08. C B = A

( )A B B= = (A B) B= B (A B)= ( B A) ( B B)= ( B A) = B A= A B= A B

= A (Proved )

Ans 09. = 100, a = 18

a + e = 23, e + g = 8

a + e + g + d = 26

e + g + f + c = 48

g + f = 8

so, e = 5, g = 3, d = 0, f = 5, c = 35

(i) d + g + f + b = 0 + 3 + 5 + 10 = 18

(ii) d + g = 0 + 3 = 3

Ans 10. Let = all the students of the class , H = students who like Hindi

S = Students who like Science

(i) n ( H S) = n (H) + n (S) n (H S)

= 30 + 25 16

= 39

(ii) n ( H S ) = n (H S )= - n (H S)

= 50 39

= 11

A = B

E Ha

bd

ef

g

s

CBSE TEST PAPER-05


Topic: -Sets

1. List all the element of the set A = { x : x is an integer x2 4} [1]

2. From the sets given below pair the equivalent sets.

A = { 1, 2, 3}, B = {x, y, z, t}, C = {a, b, c} D = {0, a}

[1]

3. Write the following as interval

(i) {x : x R, - 4 < x 6}

(ii) {x : x R, 3 x 4}

[1]

4. If A = {3, 5, 7, 9, 11}, B = {7, 9, 11, 13}, C = {11, 13, 15}

Find (A B) ( B C)

[1]

5. Write the set

1 3 5 7 9 11, , , , ,

3 5 7 9 11 13

in set builder form. [1]

6. In a group of 65 people, 40 like cricket, 10 like both cricket and tennis. How many

like tennis only and not cricket? How many like tennis?

[4]

7. Let A, B and C be three sets A B = A C and A B = A C show that B = C [4]

8. If = {a, e, i. o. u}

A = {a, e, i}

And B = {e, o, u}

C = {a, i, u}

Then verity that A (B C) = (A B) (A C)

[4]

9. In a town of 10,000 families, it was found that 40% families buy newspaper A,

20% families buy newspaper B and 10% families buy newspaper C. 5% families

buy A and B, 3% buy B and C and 4% buy A and C. If 2% families buy all the three

papers. Find the no. of families which buy

(i) A only (ii) B only (iii) none of A, B, and C.

[6]

10. Two finite sets have m and n elements. The total no. of subsets of the first set is

56 more than the total no. of subsets of second set. Find the value of m and n.

[6]

CBSE TEST PAPER-05


[ANSWERS]

Ans 01. {-2, -1, 0, 1, 2}

Ans 02. A = {1, 2, 3} B = {a, b, c} are equivalent sets [ ]( ) ( )n A n B=

Ans 03. (i) (-4, 6]

(ii) [3, 4]

Ans 04. A B = {7, 9, 11}

B C = {7, 9, 11, 13, 15}

(A B) (B C) = {7, 9, 11}

Ans 05. 2 1

: n is a natural no. less than 72 1

n

n

+

Ans 06. Let C = the set of people who like cricket and

T = the set of people who like tennis.

n ( C T) = 56, n (C) = 40

n (C T) = 10

n (C T) = n (C) + n (T) n (C T)

65 = 40 + n (T) 10

n (T) = 35

Ans 07. Let b B b A B b A C [ A B = A C b A or b c

if b c then B Cif b A, then b A B [ A B = A C

b A C b C B C

thus in both cases B CSimiularly C BHence B = C

Ans 08. B C = {e, o}

A (B C) = e

A B = {e}

A C = {a}

(A B) (A C) = e

Hence proved.

Ans 09. x + a + c + d = 4000

y + a + d + b = 2000

z + b + c + d = 1000

a + d = 500, b + d = 300, C + d = 400 d = 200

On Solving a = 300, b = 100, c = 200

(i) x = 4000 300 200 200 = 3300

(ii) y = 2000 300 200 100 = 1400

(iii) z = 1000 100 200 200 = 500

None of these = 10,000 (3300 + 1400 + 500 + 300+ 100 + 200 + 200)

= 10,000 6000

= 4000

Ans 10. Let A and B be two sets having m and n elements respectively

no of subsets of A = 2m

no of subsets of B = 2n

According to question

2m = 56 + 2n

2m - 2n = 56

2n (2m-n 1) = 56

2n (2m-n 1) = 23 (23 1)

2n = 23

n = 3

m n = 3

m 3 = 3

m = 6

A B

a

b

dc

x y

z

C

TEST PAPER-01

CLASS - XI MATHEMATICS (Relations and functions)

1. If the ordered Pairs ( )1, 3x y + and ( )2, 4x + are equal, find x and y( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 3,3 3,4 1, 4 1,0i ii iii iv

[1]

2. ( ) ( )If, 3, 2,n A n B A= = And B are two sets Then no. of relations of A B have.( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 6 12 32 64i ii iii iv

[1]

3. Let ( ) | |f x x= then Range of function( ) ( ) ( ) ( ) ( ) ( ) ( ) 0, , ,0 none of therei ii iii iv

[1]

4. A real function f is defined by ( ) 2 5.f x x= Then the Value of ( )3f ( ) ( ) ( ) ( ) -11 1 0 none of therei ii iii iv

[1]

5. Let ( ){ }, : , , 2 8R x y x y W x y= = + = then(i) Find the domain and the range of R (ii) Write R as a set of ordered pairs.

[4]

6. Let R be a relation from Q to Q defined by ( ){ }, : , and ,R a b a b Q a b z= show that ( )( ) ( )( ) ( ), for all , implies that ,i a a R a Q ii a b R b a R ( )( ) ( ) ( ), and , implies that ,iii a b R b c R a c R

[4]

7. ( ) ( )2 3 1 1If , find 2

1 3x xf x f f

x

+ = +

[4]

8. Find the domain and the range of the function ( ) 23 5.f x x= Also find ( )3f and the numbers which are associated with the number 43 m its range.

[4]

9. If ( ) ( ) ( )2 3 , find such that 2f x x x x f x f x= + = [4] 10. Find the domain and the range of the function ( ) 1f x x= [4] 11. Draw the graphs of the following real functions and hence find their range

( ) 1 , , 0f x x R xx

=

[6]

12. If ( ) 1 ,f x x

x= Prove that ( ) ( )3 3 13f x f x f

x

= +

[6]

TEST PAPER-01


[ANSWERS]

Ans1 ( )3, 4Ans2 64

Ans3 ( ),0Ans4 11

Ans5 (i) Given 2 8x y+ = and .x y w

Put

0,2 0 8 8,1, 2 1 8 6,2, 2 2 8 4,3,2 3 8 2,4, 2 4 8 0

x y yx y yx y yx y yx y y

= + = =

= + = =

= + = =

= + = =

= + = =

for all other values of ,x w we do not get y w

Domain of { }0,1, 2,3, 4,R = and range of { }8,6,4, 2,0R =

(ii) R as a set of ordered pairs can be written as ( ) ( ) ( ) ( ) ( ){ }0,8 , 1,6 , 2, 4 , 3,2 , 4,0R =

Ans6 ( ), : , and R a b a b Q a b z= (i) ( )For all , 0 and 0 , it implies that , .a Q a a z a a R = (ii) ( ) ( )Given ,a b R a b z a b z

( ), .b a z b a R (iii) ( ) ( ) ( ) ( )Given , and , and a b R b c R a b z b c z a b b c z +

( ) .a c z a c R

Ans7 ( ) { }2 3 1

, 11

x xf x Df Rx

+= =

Given

( ) ( ) ( )22 3 2 1 4 6 1 12 1 and

2 1 3 3f + + + = =

( )

21 1 1 13 11 1 3 13 3 9 1 1 9

1 2 23 9 2 613 3 31 11 1 22 1 23 52 3 .3 3 6 6 6 6

f

f f

+

+

= = = = =

+ = = = =

Ans8 ( ) 2Given 3 5f x x= For ( ),Df f x must be real number

23 5x must be a real numberWhich is a real number for every x R

( )..........Df R i = ,for Rf let ( ) 23 5y f x x= =

We know that for all 2 2, 0 3 0x R x x

[ ]23 5 5 5 5,x y Rf = ( ) ( )

( )( )

2

2 2 2

Funthes, as 3 , 3 exists is and 3

3 3 5 22.As 43 on putting 43 is weget

3 -5=43 3 =48 x =16 4, 4.

Df f f

Rf y ix x x

= =

=

=

There fore 4 and 4 are number ( )is Df which are associated with the number 43 in Rf

Ans9 ( ) 2Given 3 1,f x x x Df R= + =( ) ( ) ( )( ) ( ) ( )

2 22 2 3 2 1 4 6 1As 2 Given

f x x x x xf x f x

= + = +

=

2 24 6 1 3 1x x x x + = +( )2 23 3 0 0 1 0

0,1.x x x x x x

x

= = =

=

Ans10 ( )Given 1,f x x= ( )for , must be a real number

1must be a real number1 0 1

Df f xx

x x

[ ]( )

[ ]

1,

for , let 1

1 0 00,

DfRf y f x x

x yRf

=

= =

=

Ans11 Given ( ) 1 , , 0f x x R xx

=

Let ( ) 1 , , 0y f x i y x R xx

= = =

(Fig for Answer 11)

x 4 2 1 0.5 0.25 0.5 1 2 41yx

=

0.25 0.5 1 2 4 2 1 0.5 0.25

Plot the points shown is the above table and join there points by a free hand drawing.

Portion of the graph are shown the right margin

From the graph, it is clear that [ ]0Rf R= This function is called reciprocal function.

Ans12 If ( ) 1 ,f x xx

= prove that ( ) 3f x ( )3 1f x f x = + ( ) [ ]1 , 0f x x Df R

x= = Given

( ) ( )

( )

( ) ( )

3 33

33 3

3

33

33

3

1 1 1 1 1 and ........1

1 1 1 13 .

1 13

1 13

13 using

f x x f x ix x x x

x

f x x x x xx x x x

x xx x

x xx x

f x f ix

= = =

= =

=

= +

= +

TEST PAPER-02


1. If { }, ,P a b c= and { } ,Q d= form the sets P Q andQ P are these twoCartesian products equal?

[1]

2. If A and B are finite sets such that ( )n A m= and ( )n B k= find the number ofrelations from A to B

[1]

3. Let ( ) ( ) ( ) ( ){ }1,1 , 2,3 , 0, 1 , 1,3 ,......f = be a function from z to z defined by( ) ,f x ax b= + for same integers a and b determine a and b.

[1]

4. Express ( ){ }, : 2 5,x y y x xy w+ = as the set of ordered pairs [1] 5. Let a relation ( ) ( ) ( ) ( ) ( ){ }0,0 , 2,4 , 1, 2 , 3,6 , 1,2R = then

(i) write domain of R (ii) write range of R

(iii) write R the set builder form (iv) represent R by an arrow diagram

[4]

6. Let { } { }1,2,3 , 1,2,3,4A B= = and ( ) ( ){ }, : , , 1R x y x y A B y x= = +(i) find A B (ii) write R in roster form(iii) write domain & range of R (iv) represent R by an arrow diagram

[4]

7. The cartesian product A A has a elements among which are found( )1,0 and ( )0,1 . find the set and the remaining elements of A A

[4]

8. Find the domain and the range of the following functions ( ) 1

5f x

x=

[4]

9. Let ( ) 1f x x= + and ( ) 2 3g x x= be two real functions. Find the followingfunctions ( ) ( ) ( ) ( ) ( ) 2 3fi f g ii f g iii fg iv v f g

g+

[4]

10. Find the domain and the range of the following functions

( ) ( ) ( ) ( ) ( ) ( )2

2 23 1

2 1 1 1x xi f x ii f x iii f xx x x

= = =

+ +

[4]

11. Draw the graphs of the following real functions and hence find their range

( ) ( ) ( ) ( )2 12 1

1xi f x x ii f xx

= =

[6]

12. Let f be a function defined by 2: 5 2,F x x x R + (i) find the image of 3 under f (ii) find ( ) ( )3 2f f+(iii) find x such that ( ) 22f x =

[6]

TEST PAPER-02


[ANSWERS]

Ans1. Given { }, ,P a b c= and { } ,Q d= by definition of cartesion product, we set( ) ( ) ( ), , , , ,P Q a d b d c d = and ( ) ( ) ( ), , , , ,Q P d a d b d c =

By definition of equality of ordered pains the pair ( ),a d is not equal to the pair( ),d a therefore .p Q Q P

Ans2. Linen ( ) ( ) and n A n n B k= =( ) ( ) ( )n A B A B MK = =

the number of subsets of 2MKA B =

( ) ,n A m= then the number of subsets of 2mA =Since every subset of A B is a relation from A to B therefore the number ofrelations from A to B = 2mk

Ans3. Given fx ax b= +Since ( ) ( ) ( )11,1 1 1 1......f f a b i = + =( ) ( ) ( )2,3 . 2 3 2 3........f f a b ii = + =Subtracting (i) from(ii) we set a=2

Substituting a=2 is (ii) we get 2+b=1

b = -1

Hence a = 2, b = -1

Ans4. Since 2 5y x+ = and , ,x y w

Put 0, 0 5 5x y y= + = =

1, 2 1 5 32, 2 2 5 1

x y yx y y

= + = =

= + = =

For anther values of ,x w we do not get .y w

Hence the required set of ordered peutes is ( ) ( ) ( ){ }0 '5 , 1,3 , 2,1

Ans5. Given ( ) ( ) ( ) ( ) ( )0,0 , 2, 4 , 1, 2, , 3,6 , 1,2R = (i) Domain of [ ]0, 2, 1,3,1R = (ii) Rang of [ ]0, 4, 2,6,2R = (iii) R in the builder from can be written as

( ), : , 1 3, 2R x y x I x y x= = (iv) The reaction R can be represented by the arrow diagram are shown.

Ans6 (i) ( ) ( ) ( ) ( ){ 1,1 , 1,2 , 1,3 , 1,4( ) ( ) ( ) ( )( ) ( ) ( ) ( )2,1 , 2, 2 , 2,3 , 2,43,1 , 3,2 , 3,3 , 3,4 }

(ii) ( ) ( ) ( )1,2 , 2,3 , 3, 4R = (iii) Domain of { }1,2,3R = and range of { }2,3,4R =(iv) The relation R can be represented by the are arrow diagram are shown.

Ans7. Let ( )n A m=Given ( ) ( ) ( )9 . 9n A A n A n A = =

( )2. 9 9 3 0m m m m m = = = >Given ( )1,0 1A A A and 0 AAlso ( )0,1 0A A A and 1 AThis 1,0,1 A but ( ) 3n A =Therefore [ ]1,0,1A = The remaining elements of A A are( ) ( ) ( ) ( ) ( ) ( ) ( )1, 1 , 1,1 , 0, 1 , 0,0 , 1, 1 , 1,0 , 1,1

Ans8. Given ( ) 15

f xx

=

For ( ),FD f x must be a real number 1

5 x

Must be a real number

5 0 5 5x x x > > >1 05x

>1 0 if and only if 0aa

> >

0y >( )0,FR =

Ans9 Given ( ) 1f x x= + and ( ) 2 3g x x= we note that FD R= and gD R= so therefunctions have the same Domain R (i) ( ) ( ) ( ) ( ) ( ) ( )1 2 3f g x f x g x x x+ = + = + + 3 2,x= for x R(ii) ( )( ) ( ) ( ) ( ) ( )1 2 3f g x f x g x x x = = + 4,x + for all x R(iii) ( ) ( ) ( ) ( ) ( )1 2 3fg x f x x x= = + 22 3,x x= for all x R(iv) ( ) ( )( )

1 3, ,

2 3 2f xf x

x x x Rg g x s

+= =

(v) ( ) ( ) ( )( ) ( )( ) ( )( ) ( )22 23 3 3f g x f x f x f x g x = = ( ) ( )2 21 3 2 3 2 1 6 9x x x x x= + = + + +

2 4 10,x x= + for all x R

Ans10 (i) Given ( ) 32 1xf xx

=

+For ( ),FD f x must be a real number

32 1x

x

+

must be a real number

12 1 02

x x +

FD = set of all real number except1 1

. .

2 2i e R

For 3

, let 2 32 1FxR y xy y xx

= + = +

( ) 32 1 3 but 1 2yy x y x x R

y+

= =

31 2y

y+

Must be a real number 11 2 02

y y

FR = Set of all real number except 1 1

2 2R

(ii) Given ( )2

21xf x

x=

+

For ( ),FD f x must be a real number 2

21x

x

+


( )2 1 0 for all FD R x x R = + For FR let

22 2

21xy x y y x

x= + =

+

( ) 2 21 , 11

yy x y x yy

= =

But 2 0x for all 0, 11

yx R y

y

Multiply both sides by ( )21 ,y a positive real number( )

( ) ( ) ( )

( )

1 01 0 0 1 0

0 1 but 10 1

0,1F

y y

y y y yy yy

R

< =

(iii) Given ( ) 211f x x= For ( )

,FD f x must be a real number

21

1 x


21 0 1,1x x

FD = Set of all real number except [ ]1,1 . 1,1Fi eD R = For FR let 2

1, 0

1y y

x=

2 21 11 1 0x xy y

= =

But 2 0x for all 11 0FD y

But 2 0, 0y y> Multicity bath sides by 2y a positive real number

( ) ( )( )12 1 0 . 1 0 0 1 0y y y y yy

Either 0y or 1y but 0y ( ) ( ),0 1, .FR =

Ans11 (i) Given ( ) . . 1f x i e y x= , which is first degree equation in ,x y and hence itrepresents a straight line. Two points are sufficient to determine

straight lint uniquely

Table of values

x 0 1

y -1 1

A portion of the graph is shown in the figure from the

graph, it is clear that y takes all real values. It therefore

that FR R=

(ii) Given ( ) ( )2 1 1

1 Fxf x D Rx

= =

Let ( ) ( )2 1 1 1

1xy f x x xx

= = = +

i.e 1,y x= + which is a first degree equation is,x y and hence it represents a straight line. Two

points are sufficient to determine a straight line

uniquely

Table of values

x -1 0

y 0 1

A portion of the graph is shown is the figure from the graph it is clear that y

takes all real values except 2. It fallows that [ ]2 .FR R=

Ans12. Given ( ) 25 2,f x x x R= + (i) ( ) 23 5 3 2 5 9 2 47f = + = + =(ii) ( ) 22 5 2 2 5 4 2 22f = + = + =

( ) ( )3 2 47 22 1034f f = =(iii) ( ) 22f x =

2

2

2

5 2 225 20

42, 2

x

x

x

x

+ =

=

=

=

TEST PAPER-03


1. If { }1,2 ,A = find ( )A A A [1] 2. If A and B are two sets containing m and n elements respectively how many

different relations can be defined from A to B ?

[1]

3. A Function f is defined by ( ) 2 3f x x= find ( )5f [1] 4. Let ( ) ( ) ( ) ( ) ( ){ }0, 5 , 1, 2 , 2,1 , 3,4 , 4,7f = be a linear function from z into z find f [1] 5. If { } { }1,2,3 3, 4A B= = and { }4,5,6c =

find ( ) ( ) ( ) ( ) ( ) ( ) ( ) i A B C ii A B C iii A B B C [4]

6. For non empty sets A and B prove that ( ) ( )A B B A A B = = [4] 7. Let m be a given fixed positive integer.

let ( ) ( ), : , and is divisible by R a b a b z a b m= show that R is an equivalence relation on Z.

[4]

8. Let { }1, 2,3,4,5A = and { }1,3,4B = let R be the relation, is greater than from A to .B Write R as a a set of ordered

pairs. find domain ( )R and range ( )R

[4]

9. Define modulus function Draw graph. [4]

10. Let ( ) ( )

2 2, when 0 3 ,0 3

3 , when 3 10 2 ,3 10x x x xf x g xx x x x

=

Show that f is a function, while g is not a function.

[4]

11. The function ( ) 9 32

5xf x = + is the formula to connect 0x C to Fahrenheit

units find ( ) ( ) ( ) ( ) ( )0 10 the value of i f ii f iii x( ) 212f x = interpret the result is each case

[6]

12. Draw the graph of the greatest integer function, ( ) [ ].f x x= [6]

TEST PAPER-03


[ANSWERS]

Ans01. We have ( ) ( ) ( ) ( ) ( ) ( ){ }1,1,1 , 1,1,2 , 1, 2,1 , 2,1,1 , 2,2,1 , 2, 2, 2A A A =

Ans02. 2m n+

Ans03. ( ) 2 3Here f x x= ( ) ( )2 5 3 7f x = =

Ans04. ( ) 3 5f x x=

Ans05. We have ( ) ( ) { } { } { }

( ){ } { }( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( )

3,4 4,5,6 3, 4,5,6

1,2,3 3,4,5,6{ 1,3 , 1, 4 , 1,5 , 1,6 , 2,3 , 2,4 ,

2,5 , 2,6 , 3,3 , 3,4 , 3,5 , 3,6 }

i B CA B C

= =

=

=

( ) ( ) { } { } { }( ) { } { } ( ) ( ) ( ){ }

3, 4 4,5,6 4

1, 2,3 4 1,4 , 2, 4 , 3, 4

ii B C

A B C

= =

= =

( ) ( ) { } { }( ) ( ) ( ) ( ) ( ) ( ){ }

( ) { } { }( ) ( ) ( ) ( ) ( ) ( ){ }

( ) ( ) ( ){ }

1, 2,3 3, 4

1,3 , 1, 4 , 2,3 , 2,4 , 3,3 , 3,4

3, 4 4,5,6

3, 4 , 3,5 , 3,6 , 4,4 , 4,5 , 4,6

3, 4

iii A B

B C

A B B C

=

=

=

=

=

Ans06. First we assume that A B=Then ( ) ( )A B A A = and ( ) ( )B A A A =

( ) ( )A B B A =

This, when ,A B= then ( ) ( )A B B A = Conversely, Let ( ) ( ) ,A B B A = and let be .x AThen, ( ),x A x b A B for same b B

( ) [ ], .

similarly,Hence,

x b B A A B B Ax B

A BB A

A B

=

=

Ans07. ( ) ( ){ }, : , and is divisible by R a b a b Z a b m= (i) Let .Then,a Z

( ) =0, which is divisible by

, for all so is refleseive

a a m

a a R a ZR

(ii) Let ( ),a b R Then( ) ( ),a b R a b is divisible by m

( )a b is divisible by m( )b a is divisible m( ),b a R

Then ( ) ( ), , .a b R b a R So R is symmetric.

(iii) Let ( ),a b R and ( ),b c R( )a b is divisible by m and ( )b c is divisible by m

( ) ( )a b b c + is divisible by m ( )a c is divisible by m

( ),a c R ( ),a b R and ( ) ( ), , .b c R a c R

So, R is transitive this R is reflexive symmetric and transitive Hence, R is an equivalence relation and Z .

Ans08. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) { 2,1 , 3,1 , 3, 2 , 4,1 , 4, 2 , 4,3 , 5,1 , 5,1 , 5,2 , 5,3 , 5,4 }R = Domain of R { }2,3, 4,5= Range of R { }1,2,3, 4=

Ans09. let ( ): : | |f R R f x x = for each .x R then ( ) , when 0| |, when 0

x xf x xx x

= =

for all xdom ( )f R= and range ( )f = set of non negative real numberDrawing the graph of modulus function defined by

( ) when 0: : | | when 0

x xf R R f x xx x

= =

Ans12. Clearly, we have

( )2, when [ 2, 1)1, when [ 1,0)

0, when [0,1)1, when [1,2)

x

xf xx

x

=

x 2 1x < 1 0x < 0 1x < 1 2x < 2 3x < ... y 2 1 0 1 2 ...

TEST PAPER-04


1. If the ordered pairs ( )2,2 1x y + and ( )1, 2y x + are equal, find x & y [1]2. Let { } { }1, 2,5,8 , 0,1,3,6,7A B= = and R be the relation, is one less than from

A to B then find domain and Range of R

[1]

3. Let R be a relation from N to N define by ( ) 2, : , and R a b a b N a b = = .Is the following true ,a b R implies ( ),b a R

[1]

4. Let N be the set of natural numbers and the relation R be define in N byR = ( ), : 2 , , .x y y x x y N= what is the domain, co domain and range of R ? Isthis relation a function?

[1]

5. Let { }1, 2A = and { }3,4b = write A B how many subsets will A B have? Listthem.

[4]

6. Let { } { } { }1,2 , 1,2,3, 4 , 5,6A B C= = = and { }5,6,7,8D = verify that( ) ( ) ( ) ( )i A B C A B A C = ( ) is subset of ii A C B D

[4]

7. Find the domain and the range of the relation R defined by

( ) ( )1, 3 : 0,1, 2,3,4,5R x x x= + + [4]

8. Find the linear relation between the components of the ordered pairs of the

relation R where R ( ) ( ) ( ){ }2,1 , 4,7 , 1, 2 ,.......= [4]

9. Let { }1,2,3, 4,5,6A = define a relation R from A to A by( ){ }, : 1, ,R x y y x x y A= = +

(i) write R in the roaster form (ii) write down the domain co domain and range of R(iii) Represent R by an arrow diagram

[4]

10. A relation ' 'f is defined by 2: 2f x x where { }1, 2,0, 2x (i) list the elements of f(ii) is f a function?

[4]

11. Find the domain and the range of the following functions:

( ) ( ) ( ) ( ) ( ) ( )2 22

14 16 9

i f x x ii f x x iii f xx

= = =

[6]

12. Draw the graphs of the following real functions and hence find range:

( ) 2f x x=[6]

TEST PAPER-04


[ANSWERS]

Ans.1 3, 2x y= =

Ans.2 Given { } { }1, 2,5,8 , 0,1,3,6,7 ,A B== = and R is the relation is one less than fromA to B therefore ( ) ( ) ( )1,0 , 2,3 , 5,6R =

Domain of { }1,2,5R = and range of { }0,3.6R =

Ans.3 No; let 4, 2.a b= = As 24 2 ,= so ( )4,2 R but 22 4 . so ( )2,4 R

Ans.4 Given ( ), : 2 , ,R x y y x x y N= = Domain of .R N= co domain of .R N= and Range of R is the set of even naturalnumbers.

Since every natural number x has a unique image 2x therefore, the relation R is a function.

Ans.5 ( ) ( ) ( ) ( ){ }1,3 , 1,4 , 2,3 , 2, 4 ;A B = 16 Subsets of A B have( ){ } ( ){ } ( ){ } ( ){ }

( ){ } ( ) ( ){ } ( ) ( ){ }( ) ( ){ } ( ) ( ){ } ( ){ }( ){ } ( ) ( ) ( ){ } ( ) ( ){ }( ){ } ( ) ( ) ( ){ } ( ) ( ){ }( ){ }

Subsets , 1,3 , 1,4 , 2,3 , 2,4 ,

1,4 , 1,3 , 2,3 , 1,3 , 2,4 ,

1,4 , 2,3 , 1,4 , 2, 4 , 2,3 ,

2, 4 , 1,3 , 1,4 , 2,3 , 1,3 , 1, 4 ,

2, 4 , , 1,3 , 2,3 , 2, 4 , 1,4 , 2,3 ,

2, 4 ;

=

Ans6. . . . L H S B C = Part-I

( )( ) ( ) ( ) ( )( ) ( ) ( ) ( )

( ) ( ) ( ) ( ){ }

L.H.S

1,1 , 1, 2 , 1,3 , 1, 4. . .

2,1 , 2,2 , 2,3 , 2, 4

1,5 , 1,6 , 2,5 , 2,6

A B C

R H S A B

A C

=

=

=

( ) ( ) . . . .A B A C L H S R H S = =

Part-II

( ) ( ) ( ) ( )( ) ( ) ( ) ( )

( ) ( )

1,5 , 1,6 , 1,7 , 1,82,5 , 2,6 , 2,7 , 2,8

B D

A C B D

=

Ans7. { }Given 0,1, 2,3,4,5x

( ) ( ) ( ) ( ) ( ) ( )

put 0, 1 0 1 1 and 3 0 3 31, 1 1 1 2 and 3 1 3 4,2, 1 2 1 3 and 3 2 3 5,3, 1 3 1 4 and +3=3+3=6,4, 1 4 1 5 and 3 4 3 75, 1 6 and 3 5 3 8

Hence 1,3 , 2,4 , 3,5 , 4,6 , 5,7 , 6,8

x x x

x x x

x x x

x x x

x x x

x x x

R

= + = + = + = + =

= + = + = + = + =

= + = + = + = + =

= + = + =

= + = + = + = + =

= + = + = + =

=

[ ] [ ] Domain of 1, 2,3,4,5,6 and range of 3,4,5,6,7,8R R = =

Ans8. Given ( ) ( ) ( ){ }2,1 , 4,7 , 1, 2 ,....R = Let y ax b= + be the linear relation between the components of RSince ( ) ( )2,1 , 1 2 ........R y ax b a b i = + = +Also ( ) ( )4,7 , 7 4 ........R y ax b a b ii = + = +Subtracting ( )i from ( )ii , we get 2 6 3a a= =Subtracting 3a = is ( )i , we get 1 6 5b b= + = Subtracting there values of a and b in ,y ax b= + we get

3 5,y x= which is the required linear relation between the components of the givenrelation.

Ans9 (i) ( ) ( ) ( ) ( ) ( ){ }1,2 , 2,3 , 3,4 , 4,5 , 5,6(ii) Domain { }1,2,3, 4,5= co domain ,A= range { }2,3,4,5,6=(iii)

.1

1. .2

2. .3

3. .4

4. .5

5. .6

6.

Ans10. 2Re letion is defined by : 2f f x x (i) ( ) { }2is 2 whene 1, 2,0, 2f x x x=

( ) ( )( ) ( )( )( )

( ) ( ) ( ) ( ){ }

2

2

2

2

1 1 2 1 2 1

2 2 2 4 2 2

0 0 2 0 2 2

2 2 2 4 2 2

1, 1 , 2, 2 , 0, 2 , 2 '2

ffff

f

= = =

= = =

= = =

= = =

=

(ii) We note that each element of the domain of f has a unique image; therefore, the relation f is a function.

Ans11. (i) Given ( ) 2 4f x x= For ( )1Df f x must be a real number

2 4x Must be a real number( )( )2 4 0 2 2 0x x x +

either 2 or 2x x ( , 2] [2, ).FD =

For ,FR let ( )2 4.........y x i= As square root of a real number is always non-negative, 0y On squaring (i), we get 2 2 4y x=

2 2 4x y = + but 2 0x for all Fx D2 24 0 4,y y + which is true for all .y R also 0y

[0, )FR =

(ii) Given ( ) 216f x x= For ( ),FD f x must be a real number

216 x must be a real number

( )( )( )

[ ]

2 2

2

16 0 16 0

16 04 4 0 4 4

4, 4 .F

x x

x

x x x

D

+

=

For ,FR let ( )216 ........y x i=

As square root of real number is always non-negative, 0y Squaring ( )i we get

2 216y x= 2 216x y = but 2 0x for all fx D

( )2 2 216 0 16 0 16 0y y y ( ) ( )4 4 0 4 4y y y + but 0y

[ ]0,4FR =

(iii) Given ( )2

19

f xx

=

For ( ),FD f x must be a real number

2

19 x

must be a real number

( )( )( ) ( )

2 29 2 0 9 0 9 0

3 3 0 3 3 3,3F

x x x

x x x D

> > on squaring ( )i we get

2 22 2 2

1 1 12 9 99

y x xx y y

= = =

But 2 0x for all 219 0Fx D y

2 0y >(Multiply bath sides by 2 ,y a positive real number)

2 2 19 1 0 09

y y

1 1 03 3

y y +

either13

y or 13

y

103

1[ , ).3F

y y

R

>

=

Ans12. Given ( ) 2 Ff x x D R= =Let ( ) 2 ,y f x x x R= = x -4 -3 -2 -1 0 1 2 3 4

2y x= 16 9 4 1 0 1 4 9 16

Plot the points

( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( )4,16 , 3,9 , 2, 4 , 1,1 , 0,0 , 1,1 2,4 , 3,9 , 4,16 ....., And join these points by a free hand drawing. A portion of the graph is shown in

sigma (next)

From the graph, it is clear that y takes all non-negative real values, if follows that

[ )0,FR =

TEST PAPER-05


1. Let ( ){ }, : 1R x y y x= = + and { } 0,1, 2,3,4,5y list the element of R [1] 2. Let f be the subset of Q Z defined by

, : . , 0mf m m n Z nn

=

. Is f a function from to ?Q Z Justify your answer

[1]

3. The function `f which maps temperature in Celsius into temperature inFahrenheit is defined by ( ) ( )9 32 find 0

5f c c f= +

[1]

4. If 3 3

1f x xx

=

Prove that ( ) 1 0f x f

x

+ =

[1]

5. If

6 5.

5 6xyx

=

Prove that ( ) 6, 5

f y x x= [4]

6. Let :f X Y be defined by ( ) 2f x x= for all x X where { }2, 1,0,1, 2,3X = and { }0,1, 4,7,9,10y = write the relation f in the roster farm. It f a function?

[4]

7. Determine a quadratic function `f defined by( ) ( ) ( ) ( )2 if 0 6, 2 11 and 3 6f x ax bx c f f f= + + = = =

[4]

8. Find the domain and the range of the function f defied by ( ) 2| 2 |

xf xx

+=

+

[4]

9. Find the domain and the range of ( )

2

21xf x

x=

+

[4]

10.

If

{ } { }( ) ( ){ }

( ) ( )

1,2,3 , 1, 2,3,4 . and

, : , , 1 then

find write domain and Range

A B

R x y x y A B y x

i A B ii

= =

= = +

[4]

11. Define polynomial function. Draw the graph of ( ) 3f x x= find domain andrange

[6]

12. (a) If ,A B are two sets such that ( ) 6n A B = and some elements of A B are( ) ( ) ( )1,2 , 2,3 , 4,3 , than find and A B B A (b) Find domain of the function ( ) [ ]

1f xx x

=

+

[6]

TEST PAPER-05


[ANSWERS]

Ans1 ( ) ( ) ( ) ( ) ( ) ( ){ }1,0 , 0,1 , 1,2 , 2,3 , 3,4 , 4,5R = Ans2 f Is not a function from Q to Z

1 21 22 4

1 2

2 4

f and f

But

= =

=

One element 12

have two images

f is not function

Ans3 ( ) 90 0 325

f = +

( )0 32f =

Ans4 ( ) 3 31f x x x=

( )

33

3

1 1

1

f xx x

f x f xx

=

+ =

3

1x

31x

+ 3x

0=

Ans5 6 55 6

xyx

=

( )

( )

6 55 6

6 56 55 66 55 65 6

xy f xx

x

xf yx

x

= =

=

( ) 36 30xf y = 25 30x +5 6x

30 25 30 365 6

x x

x

+

( ) 11 6,11 5

xf y x x= =

Ans6 : defined byf X Y( )

{ }{ }

( ) ( )( ) ( )( )( )( )( )

( ) ( ) ( ) ( ) ( ) ( ){ }

2

2

2

2

2

2

2

,

2, 1,0,1,2,3 0,1, 4,7,9,10

2 2 4

1 1 1

0 0 0

1 1 1

2 2 4

3 3 9

2, 4 , 1,1 , 0,0 , 1,1 , 2, 4 , 3,9

f x x x Xand X

y

ffffff

f

=

=

=

= =

= =

= =

= =

= =

= =

=

f is a function because different elements of X have different imager in y

Ans7 ( ) 2f x ax bx c= + +( )

( )

[ ] ( )( )

( ) ( )

[ ] ( )

2

2

2

0 60 0 662 11

2 2 114 2 114 2 6 114 2 11 64 2 5

3 6

3 3 09 3 6 09 3 6

fa b cc

fa b c

a b ca ba ba b i

a b ca ba b ii

=

+ + =

=

=

+ + =

+ + =

+ + =

+ =

+ =

=

+ + =

+ =

=

Multiplying eq. (i) by 3 and eq. (ii) by 2

12 6a b+ 1518 6a b

=

2

5

1230 33 1

30 101410

a

a

=

=

= =

( ) 2

2 5

22 55

25 2 2325 5

2310

1 23 610 10

b

b

b

b

f x x x

+ =

=

= =

=

= + +

Ans8 ( ) 2| 2 |xf xx

+=

+

For Df , ( )f x must be a real no.| 2 | 0 2 0 2x x x + +

Domain of f = set of all real numbers{ }

( )( )

( )

2 . . 2

2 0 | 2 | 22 1| 2 |

2 0,| 2 | 222

except i e Df Rfor RfcaseI if x then x x

xf xx

caseII if x x xxf xx

=

+ > + = +

+ = =

+

+ < + = +

+ =

+

{ }1

1,1Range of f

=

=

Ans9 ( )2

21xf x

x=

+

( ) .

Domain of f all real no Rfor Range let f x y

= =

=

2

21xy

x=

+

( )

( )

2 2

2 2

2 2

2

2

1

1

1

1

0 1 01

1 0 1 0

y 1 [0,1).

y x x

y yx x

y x yx

y y xy

xy

yx

yy y

yyalso y and y

Range of f

+ =

+ =

=

=

=

=

> >

+ = =

+ <

=

CBSE TEST PAPER-01

CLASS - XI MATHEMATICS (Trigonometric Functions)

1. Convert into radian measures. 470 30 [1]

2. Evaluate tan 750. [1]

3. Prove that Sin (40 + ). Cos (10 + ) Cos (40 + ). Sin (10 + ) =

12

[1]

4. Find the principal solution of the eq. Sin x =

32

[1]

5. Prove that Cos + Cos = 2 Cos x

4 4x x

pi pi +

[1]

6. The minute hand of a watch is 1.5 cm long. How far does it tip move in 40 minute? [4]

7. Show that tan 3x. tan 2x. tan x = tan 3x tan 2x tan x [4]

8. Find the value of tan

8pi

.[4]

9. If Sin + Sin = a and Cos + Cos = b show that Cos ( + ) =

2 2

2 2b ab a

+

[6]

10. Prove that Cos + Cos + Cos + Cos ( + + )

4 Cos . Cos . Cos 2 2 2

+ + + =

[6]

CBSE TEST PAPER-01


[ANSWERS]

Ans 01. - 470 30 = -

0304760

+

0147

295

radians2 180

19= - radians.

72

pi

pi

=

=

Ans 02. tan 75 = tan (45 + 30)

tan 45 + tan 301 - tan 45 tan 30

3 1

3 1

=

+=

Ans 03. L. H. S = Sin (40 + ). Cos (10 + ) Cos (40 + ). Sin (10 +)

= Sin 140 10 = Sin 30 = 2

+

Ans 04. Sin x = 3

2

Sin x = Sin 3

x = 3

pi

pi

x = Sin 3

2 x =

3

Sin pipi

pi

Ans 05. L. H. S = Cos + Cos

4 4x x

pi pi +

[2 Cos . Cos x Cos (A + B) + Cos (A - B) = 2 Cos A. Cos B4pi

12 . Cos x = 2 Cos x2

=

Ans 06. r = 1.5 cm

Angle made in 60 mint = 3600

Angle made in 1 min = 36060

= 600

Angle made in 40 mint = 6 40

= 2400

= lr

4240 180 3 1.5

4 3.14 1.5 5 =

3 10 22 3.146.28

6.28 cm

l

l

ll

l

pi =

=

=

=

Ans 07. Let 3x = 2x + x

tan 3x = tan (2x + x)

tan 3x tan 2x + tan x

1 1 - tan 2x. tan xtan 3x (1 - tan 2x. tan x) = tan 2x + tan xtan 3x - tan 3x. tan 2x. tan x = tan 2x + tan xtan 3x. tan 2x. tan x = tan 3x - tan 2x - tan x

=

Ans 08. Let x = 8pi

2

2

2

2 tan x tan 2x =

1 - tan x2 tan 8

tan 28 4 1 - tan 8

2 tan 811 - tan 8

put tan 8

pi pi

pi

pi

pi

pi

=

=

=

2

2

1 2 =

1 12 1

=

2 + 2 - 1 = 0

- 2 2 2 =

2 1

= -1 2 2 1

2 1 or - 2 1

=

=

tan 8 2 1pi =

Ans 09. 2 2 2 2b + a = (Cos + Cos ) + (Sin + Sin ) 2 2 2 2

= Cos + Cos + 2 Cos . Cos + Sin + Sin 2 . Sin 1 + 1 + 2 (Cos . Cos + Sin . Sin )

= 2 + 2 Cos ( - ) (1)

Sin

+

=

2 2 2 2b - a = (Cos + Cos ) - (Sin + Sin ) 2 2 2 2

(Cos - Sin ) + (Cos - Sin ) + 2 Cos ( + )= Cos ( + ) Cos ( - ) +Cos ( + )Cos ( - ) + 2 Cos ( + ) = 2 Cos ( + ). Cos ( - ) + 2 Cos ( + )

=

[ ]2 2

2 2

2 2

( + ) 2 Cos ( - ) + 2 Cos ( + ). (b + a ) [from (1)

b - a = Cos ( + )

b + a

Cos

=

=

Ans 10. L. H. S.

[ ] Cos + Cos + Cos + Cos ( + + )

= Cos + Cos ) + Cos + Cos ( + + )

2 Cos . Cos 2 Cos . Cos 2 2 2 2

2 Cos . Cos 2 Cos2 2

=

+ + + + + + = +

+ + = +

2. Cos

2 222 Cos Cos Cos

2 2 2

2 22 2 2 22 Cos 2 Cos . Cos

2 2 2

+ +

+ + + = +

+ + + + + + =

2 Cos 2 Cos . Cos 2 2 2

4 Cos . Cos . Cos 2 2 2

+ + + =

+ + + =

CBSE TEST PAPER-02


1. Convert into radian measures. -370 30 (1)

2. Prove Sin (n+1) x Sin (n+2) x + Cos (n+1) x. Cos (n+2) x = Sin (x + y) (1)

3. Find the value of Sin

313pi (1)

4. Find the principal solution of the eq. tan x =

13

(1)

5. Prove that

( ) tan tan( ) tan tan

Sin x y x ySin x y x y

+ +=

(4)

6. If in two circles, arcs of the same length subtend angles 600 and 750 at the

centre find the ratio of their radii.

(4)

7. Prove that Cos 6x= 32 Cos6x 48 Cos4 x + 18 Cos2 x-1 (4)

8. Solve Sin2x-Sin4x+Sin6x=o (4)

9. Prove that Sin3x +Sin2x-Sinx=4Sinx.Cos

2x

. Cos32x (6)

10. Prove that 2Cos

13pi

.Cos 913pi

+ Cos313pi

+ Cos513pi

=0(6)

CBSE TEST PAPER-02


[ANSWERS]

Ans.01. - 370 30 = -

0 1

2303760

+

075

2

=

75= -

2 180524

radienpi

pi

=

Ans.02. L.H.S = Cos (n+1) x Cos (n+2) x + Sin (n+1) x Sin (n+2) x

=Cos ( ) ( )1 2n x n x+ + =Cos x

Ans.03. Sin 31

3pi

= Sin 103pi

pi

+

= sin 2 53pi

pi

+

= Sin 3pi

=3

2

Ans.04. tan x = 13

tan tan6

56

1tan

3

x

x

x

pipi

pi

=

=

=

tan tan 26

116

x

x

pipi

pi

=

=

Ans.05. L.H.S = ( )( )

Sin x ySin x y

+

. .

. .

.

tan tantan tan

Sinx Cosy Cosx SinySinx Cosy Cosx Siny

DividingNandDbyCosx Cosyx yx y

+=

+=

.Ans.06. 1

lr

=

1

1

60183

(1)

lr

lr

pi

pi

=

=

2

2

2

751812

(2)5

lr

lr

lr

pi

pi

=

=

=

(1) ( 2)

1

2

33 5

12 125

5 : 4

lr l

lr lpipi

pipi

= =

=

Ans.7. L.H.S. = Cos 6x

= cos 2(3x) = 2 4

2

3 2

6

6 2 4

6 4 2

cos 2(3 ) 2cos 3 12(4cos 3cos ) 12 16cos 9cos 24cos 1

32cos 18cos 48cos 132cos 48cos 18cos 1

x x

x x

x x x

x x x

x x x

=

=

= +

= +

= +

Ans.08 6 sin 2 sin 4 0Sin x x x+ =

[ ]

6 2 6 22sin .cos sin 4 02 2

2sin 4 .cos 2 sin 4 0sin 4 2cos 2 1 0sin 4 04

4

x x x xx

x x x

x x

x

x n

nx

pi

pi

+ =

=

=

=

=

=

2cos 2 1 0

cos 2 cos3

2 23

6

x

x

x n

x n

pi

pipi

pipi

=

=

=

=

Ans.09

( )

[ ]

sin 3 sin sin 23 32cos .sin sin 2

2 22cos 2 .sin sin 22cos 2 .sin 2sin .cos2sin 2cos 2 cos

32sin 2cos .cos2 2

x x x

x x x xx

x x x

x x x x

x x x

x xx x

+

+ = +

= +

= +

= +

=

Ans10. L.H.S.

9 3 52cos .cos cos cos13 13 13 13pi pi pi pi

= + +

9 9 3 5cos cos cos cos

13 13 13 13 13 13pi pi pi pi pi pi

= + + + +

10 8 3 5cos cos cos cos

13 13 13 133 5 3 5

cos cos cos cos13 13 13 13

3cos

13

pi pi pi pi

pi pi pi pipi pi

pi

= + + +

= + + +

=

5cos

13pi

3cos

13pi

+5

cos13pi

+

0=

CBSE TEST PAPER-03


1. Convert into radian measures. 05 37 30 [1]

2. Prove Cos 700. Cos 100 + Sin 700. Sin 100 =

12

[1]

3. Evaluate 2 Sin .

12pi

[1]

4. Find the solution of Sin x =

32

[1]

5. Prove that

0 00

0 0Cos 9 - Sin 9

tan 36Cos 9 + Sin 9

=

[1]

6. In a circle of diameter 40cm, the length of a chord is 20cm. Find the length of minor

are of the chord.

[4]

7. Prove that tan 4x =

2

2 44 tan x (1 - tan x)

1 - 6 tan x + tan x[4]

8. Prove that (Cos x + Cos y)2 + (Sin x Sin y)2 = 4 Cos2

2x y+

[4]

9. Find the value of tan ( + ) Given that

1 3 5Cot = , , and Sec = - , ,2 2 3 2

pi pi pi pi

[6]

10. Prove that

Sec 8A - 1 tan 8ASec 4A - 1 tan 2A

=

[6]

CBSE TEST PAPER-03


[ANSWERS]

Ans 01. 50 371 3011 = 50 +

1303760

+

1

0

00

00

0

0

7552

75 152 60

558

558

458

45 radian

8 180

= radian.32

pi

pi

= +

= +

= +

=

=

=

Ans 02. L. H. S. = Cos (70 10) = Cos 60 = 12

Ans 03. 2 Sin 2 Sin - 12 4 6pi pi pi

=

2 Sin . Cos - Cos . Sin 4 6 6 4

1 3 1 1=2 -

22 2 2

3 1 3 1=2 =

2 2 2

pi pi pi pi =

Ans 04. Sin x = 3

2

n

Sin x = Sin 3

4Sin x = Sin 3

if Sin = Sin = n + (-1) .

pipi

pi

pi

+

4n ( 1) .

3nx

pipi= +

Ans 05. L. H. S = tan 360 0 0

0

0

0 0

0 0

tan (45 - 9 )1 - tan 9

=

1 + tan 9Cos 9 - Sin 9Cos 9 + Sin 9

=

=

Ans 06.

= 600

60180 3 20

20 cm.

3

lr

l

l

pi

pi

=

=

=

Ans 07. L. H. S = tan 4x

( )( )

2

2

2

2

2

22 2

22

2 tan 2x1 - tan 2x

2 tan2.1 tan

2 tan11 tan

4 tan1 tan

1 tan 4 tan

1 tan

x

x

x

x

x

x

x x

x

=

=

=

20

20

20

20

60

o

A

B

( )( )2

4 2 22

2

2 4

1 tan4 tan

1 tan 2 tan 4 tan1 tan

4 tan (1 tan )1 6 tan tan

xx

x x xx

x x

x x

= +

=

+

Ans 08. L. H. S = (Cos x + Cos y)2 + (Sin x Sin y)2 22

2 Cos . Cos 2 Cos . Sin 2 2 2 2

x y x y x y x y+ + = +

2 2 2 2

2 2 2

2

4 Cos . Cos 4 Cos . Sin 2 2 2 2

4 Cos Cos 2 2 2

4 Cos 2

x y x y x y x y

x y x y x ySin

x y

+ + = +

+ = +

+ =

Ans 09. tan + tan

tan ( + ) = (1)1 - tan . tan

2 2

2

1Cot = ,2

tan = 21 + tan = Sec

5 51 + tan = 2 Sec 2 3

4tan =

34

tan = - 23

put tan , and tan in eq. (1)

x

pi

=

423tan ( + ) =

41 23

2 =

11

Ans 10. L. H. S

=

1 1 8A

1 1Cos 4A

Cos

( )

2

2

2

1 - Cos 8A Cos 4A

1 - Cos 4A Cos 8A2 Sin 4A Cos 4A

.

2 Sin 2A Cos 8A2 Sin 4A. Cos 4A . Sin 4A

2 Sin 2A. Cos 8A

=

=

=

( )2

Sin 8A. 2 Sin 2A . Cos 2A2 Sin 2A. Cos 8A

Sin 8A Cos 2ASin 2A. Cos 8Atan 8Atan 2A

=

=

=

CBSE TEST PAPER-04


1. Find the value of tan

19.

3pi

[1]

2. Prove Cos 4x = 1 3 Sin2 x. Cos2x [1]

3. Prove 2

Cos ( +x) . Cos (-x) = Cot x

Sin ( -x). Cos 2

x

pi

pipi +

[1]

4. Prove that tan 560 =

0 0

0 0Cos 11 + Sin 11Cos 11 - Sin 11

[1]

5. Prove that Cos 1050 + Cos 150 = Sin 750 Sin 150 [1]

6. If Cot x = -

5,

12x lies in second quadrant find the values of other five trigonometric

functions.

[4]

7. Prove that

Sin 5x - 2Sin 3x + Sin x = tan x

Cos 5x - Cos x[4]

8. Prove that Sin x + Sin 3x + Sin 5x + Sin 7x = 4 Cos x. Cos 2x. Sin 4x [4]

9. Prove that 2 2 2

3Cos x + Cos 3 3 2

x Cos xpi pi + + =

[6]

10. Prove that Cos 2x. Cos

2x

- 9 5Cos 3x. Cos Sin 5x Sin 2 2x x

=

[6]

CBSE TEST PAPER-04


[ANSWERS]

Ans 01. 19

tan = tan 63 3pi pi

pi

+

tan 3 23

tan = 33

pipi

pi

= +

=

Ans 02. L. H. S = Cos 4x 2 2

2

2

2 2

2 2

1 - 2 Sin 2x Cos 2 = 1 - 2Sin

1 - 2 ( Sin2x) 1 - 2 (2 Sin x. Cos x) 1 - 2 (4 Sin x . Cos x)

= 1 - 8 Sin x. Cos x.

= =

=

=

Ans 03. L. H. S = 2 Cos x. Cos x

= Cot x. -Sin x. Sin x

Ans 04. L. H. S = tan 560

= tan (45 + 11)

tan 45 + tan 11

1 - tan 45. tan 111 + tan 11

1 - tan 11Cos 11 + Sin 11

= R H S Cos 11 - Sin 11

=

=

=

Ans 05. L. H. S = Cos 1050 + Cos 150 0 0 0 0

0 0

0

= Cos (90 + 15 ) + Cos (90 - 75 ) = - Sin 15 + Sin 75= Sin 75 - Sin 15

Ans 06. Cot x = 5

12

2 2

12tan x = -

5Sec x = 1 + tan x

13Sec x = 5

[13Sec x = - x lies in IInd quad.5 5Cos x = -

13Sin x = tan x. Cos x

-12 5 12 = =

5 13 1313Cosec x = 12

Ans 07. L. H. S = Sin 5x + Sinx - 2 Sin 3x

Cos 5x - Cos x

2

2Sin 3x. Cos 2x - 2 Sin 3x- 2 Sin 3x. Sin 2x

2 Sin 3x (Cos 2x - 1) 2 Sin 3x. Sin 2x

- (1 - Cos 2x)- Sin 2x2 Sin x

2 Sin x. Cos xSin x

tan xCos x

=

=

=

=

= =

Ans 08. L. H. S. = Sin x + Sin 3x + Sin 5x + Sin 7x

= Sin x + Sin 7x + Sin 3x + Sin 5x

7 7 3 5 3 52 Sin . Cos 2 Sin Cos 2 2 2 2

2 Sin 4x. Cos 3x + 2 Sin 4x. Cos x= 2 Sin 4x [Cos 3x + Cos x]

x x x x x x x x+ + = +

=

3 3= 2 Sin 4x 2 Cos . Cos

2 2 2 Sin 4x [ 2 Cos 2x. Cos x]

= 4 Cos x. Cos 2x. Sin 4x

x x x x +

=

Ans 09. L. H. S =

2 21 + Cos 2 1 + Cos 21 + Cos 2x 3 3

+ + 2 2 2

x xpi pi

+

1 2 21+1+1+Cos 2x + Cos 2 Cos 22 3 31 2 23 2 2 22 3 3

2 2 2 22 2 2 21 3 3 3 33 2 2 . Cos 2 2 3

1 3 + Cos 2

x x

Cos x Cos x Cos x

x x x x

Cos x Cos

pi pi

pi pi

pi pi pi pi

= + +

= + + + +

+ + + + = + +

=

2

3

42x + 2 Cos 2x. Cos 6

1 23 + Cos 2x + 2 Cos 2x. Cos 2 31 3 + Cos 2x + 2 Cos 2x. Cos 2 31 13 + Cos 2x + 2 Cos 2x. 2 23

.

2

pi

pi

pipi

=

=

=

=

Ans 10. L. H. S = 1 92 Cos 2x Cos 2Cos 3x. Cos 2 2 2

x x

1 9 9Cos 2 2 3 32 2 2 2 21 5 3 15 32 2 2 2 21 5 152 2 2

5 15 5 151 2 2 2 22 .2 2 2

- Sin

x x x xx Cos x Cos x Cos x

x x x xCos Cos Cos Cos

x xCos Cos

x x x x

Sin Sin

= + + +

= +

=

+ =

=

5 5x. Sin

25

Sin 5x. Sin .2

x

x

=

CBSE TEST PAPER-05


1. Find the value of Cos (- 17100). [1]

2. A wheel makes 360 revolutions in 1 minute. Through how many radians does it

turn in 1 second.

[1]

3. Prove Sin2 6x Sin2 4x = Sin2 x. Sin 10 x. [1]

4. Prove that

tan 69 + tan 66 11 - tan 69. tan 66

= [1]

5. Prove that

Sin x = tan

1 + Cos x 2x [1]

6. Find the angle between the minute hand and hour hand of a clock when the time is

7. 20.

[4]

7. Prove that Cot 4x (Sin 5x + Sin 3x) = Cot x (Sin 5x Sin 3x) [4]

8. Show that 2 2 + 2 Cos 4 = 2 Cos + [4]

9. Prove that Cos 200. Cos 400. Cos 600 Cos 800 =

116

[6]

10. If tan x =

3 3, x < , Find the value of Sin , Cos and tan .

4 2 2 2 2x x xpi

pi < [6]

CBSE TEST PAPER-05


[ANSWERS]

Ans 01. Cos (-17100) = Cos (1800-90) [Cos (-) = Cos

= Cos [5 360 +90]

= Cos 2pi

= 0

Ans 02. N. of revolutions made in 60 sec. = 360

N. of revolutions made in 1 sec = 360 660

=

Angle moved in 6 revolutions = 2 6 = 12

Ans 03. L. H. S = Sin2 6x Sin2 4x

= Sin (6x + 4x). Sin (6x 4x)

= Sin 10x . Sin 2x

Ans 04. L. H. S = tan (69 + 66)

= tan (135)

= tan (90 + 45)

= - tan 45

= -1

Ans 05. L. H. S = 2 .

2 22

2x

x xSin Cos

xCos= tan

2x

Ans 06. Angle made by mint hand in 15 mint= 15 6 = 900

Angle made by hour hand in 1 hr = 300

in 60 minute = 3060

= 12

[ 0 Angle Traled by hr hand in 12 hr = 360in 20 minute =

1 202

= 010

Angle made = 90 + 10 = 1000

Ans 07. L. H. S = Cot 4x (Sin 5x + Sin 3x)

Cos 4x 5 3 5 3 2 Sin . Cos

Sin 4x 2 2Cos 4x

2 Sin 4x. Cos xSin 4x

= 2 Cos 4x. Cos x

x x x x+ =

=

R. H. S = Cot x (Sin 5x Sin 3x)

Cos x 5 3 5 32 Cos . Sin Sin x 2 2Cos x 2 Cos 4x. Sin xSin x2 Cos 4x. Cos x

L. H. S = R. H. S

x x x x+ =

=

=

Ans 08. L H. S = 2 2 2Cos 4+ +

( )

( )

2

2

2 2 1 Cos 4

2+ 2.2 Cos 22 2 Cos 2

2 1 Cos 2

2.2 Cos2 Cos

= + +

=

= +

= +

=

=

Ans 09. L. H. S = Cos 200. Cos 400. Cos 600. Cos 800.

= Cos 60. Cos 200. Cos 400. Cos 80.

[ ]

( )

0

0

0 0

0

0 0

1 1. Cos40 ( 2 Cos20. Cos80)

2 21

= Cos40 Cos (80 + 20) + Cos (80 - 20)41 Cos40 Cos 100 + Cos 6041 1Cos40 Cos 100+4 21 12 Cos 100. Cos 40 Cos 408 8

=

=

=

= +

0 0 01 1Cos (100 + 40 ) + Cos (100 - 40 ) 408 8

Cos = +

0 0

0

0 0

0

0 0

1 1Cos 140 + Cos 60 + Cos 408 81 1 1Cos 140 + + Cos 408 2 81 1 1

Cos 140 + Cos 408 16 81 1 1

Cos (180 - 40) + Cos 408 16 8

1 1 1 Cos 40 + Cos 40

8 16 81

16

=

=

= +

= +

= +

=

Ans 10.

< x < 3

[Given2pi

Cos x is tive

2 2

2

32 2 4

Sin is + tive and Cos is - tive. 2 2

1 + tan x = Sec x

31 + = Sec2x4

5Sec x = 4

x

x x

pi pi < 3

3n

for n = 1

12 > 13

which is true

Let P (K) be true

P (K) : 12 + 22 + 32 + -- + K2 > 3

(1)3

K

we want to prove that P (K + 1) is true

P (K+1) : 12 + 22 + --- + (K+1)2

2 2 2 2

32

3 2

3 2

3

3

=1 +2 +--+K +(K+1)K

> +(K+1)3

1= K +3(K+1)

31

= K +3K +3+6K31

= (K+1) +(3K+2)31

> (K+1)3P(K+1)is true

Hence by PMI P (n) is true n N

Ans 07. Let P (n) : (ab)n = an bn

for n = 1

ab = ab which is true

Let P (K) be true

(ab)K = aK bK (1)

we want to prove that P (K+1) is true

(ab)K+1 = aK+1. bK+1

L.H.S = (ab)K+1

1

K K 1

K+1 K+1

( ) . (ab)=a b . (ab) [from (1)= a . b

( 1)is true.

Kab

P K

=

+

Ans 08. Let P (n) : a + ar + ar2 + -- + arn-1 = ( 1

1

na r

r

for n = 1

P (1) = a = a which is true

Let P (K) be true

P (K) : a + ar + ar2 + -- + arK-1 = ( 1)

(1)1

Ka r

r

we want to prove that

P (K+1) : a + ar + ar2 + -- + arK = 1( 1)1

Ka r

r

+

L.H.S 2 1K Ka ar ar ar ar= + + + + +

( ) [1 from (1)1

KK

a rar

r

= +

( )( )

K 1 K

K+1K k+1 K

a r -1 +ar -arr-1

a r -1ar -a+ar -ar= =

r-1 r-1

k +

=

Thus P (K+1) is true whenever P(K) is true

Hence by PMI P(n) is true for all n N

CBSE TEST PAPER-03

CLASS - XI MATHEMATICS (Principle of mathematical Induction)

Topic: - Principle of mathematical Induction

1. Prove that x2n y2n is divisible by x + y. [4]

2. Prove that n (n + 1) (2n + 1) is divisible by 6. [4]

3. Show that 23n 1 is divisible by 7 . [4]

4. Prove by P M I.

1. 2. 3 + 2. 3. 4 + --- + n (n + 1) (n + 2) = ( 1)( 2)( 3)

4n n n n+ + +

[4]

5. Prove that

1 1 1 1. +

1.2 2.3 3.4 ( 1) 1n

n n n+ + + =

+ +

[4]

6. Show that the sum of the first n odd natural no is n2. [4]

7. Prove by P M I

23 3 3 3 ( 1)1 2 3

2n n

n+

+ + + + =

[4]

8. Prove. 22

3 5 7 (2n+1)1+ 1+ 1+ ----+ 1+ = (n+1)1 4 9 n

[4]

CBSE TEST PAPER-03


[ANSWERS]

Ans 01. P (n) : x2n y2n is divisible by x + y

for n = 1

p (1) : x2 y2 = (x y) (x + y), which is divisible by x + y

Hence result is true for n = 1

Let P (K) be true

p (K) : x2K y2K is divisible by x + y

( )

2K 2K

2(K+1) 2(K+1) 2K+2 2K+2

2K 2 2K 2

2 2

x - y = (x+y) , where N (i)we want to prove the result is true for n = K + 1 x - y = x - y

= x . x - y . y

= ( ) .Kx y y x

+ + 2 2

2 2 2K 2K 2

2 2K 2 2

2 2K

2

. ( ) = (x+y) x +x y - y . y = (x+y) x + y (x - y ) = (x+y) x +y (x+y) (x-y) = (x+y) x +

Ky y from i

2K y (x-y) is divisible by (x+y)

p (K+1) is true whenever p (K) is true Hence by P.M.I, p (n) is true n N

Ans 02. P (n) : n (n+1) (2n+1) is divisible by 6 for n = 1

P (1) : (1) (2) (3) = 6 is divisible by 6


Let P (K) be true

P (K) : K (K+1) (2K+1) is divisible by 6

[ ]

( 1)(2 1) 6 where N (i)we want to prove that result is true for n = K+1 (K+1) (K+2) (2k+3) = (K+1) (K+2) (2 1) 2

= (K+1) (K+2)(2K+1) + 2 (K+1) (K+2)

K K K

K

+ + =

+ +

[ ]( 2) ( 1)(2 1) + 2 (K+1) (K+2)= K (K+1) (2K+1) + 2 (K+1) (2K+1) + 2 (K+1) (K+2)

K K K= + + +

[ ]

[ ]

6 2( 1)(2 1) 2( 1)( 2) (by i) 6 + 2(K+1) 2 1 26 2( 1)(3 3)6 6( 1)( 1)6 ( 1)( 1)

is divisible by 6.

K K K KK K

K KK KK K

= + + + + + +

= + + +

= + + +

= + + +

= + + +

Ans 03. P (n) : 23n 1 is divisible by 7

for n = 1

P (1) : 23 1 = 7 which is divisible by 7

Let P (K) be true

P (K) : 23K 1 is divisible by 7

3K

3(K+1) 3K+3

3K 3

2 -1 = 7 where N (i)we want to prove that P(K+1) is true whenever P(K) is true 2 - 1 = 2 - 1

= 2 . 2 - 1 = (7 +1). 8 - 1( from i) = 56 +

8 - 1 = 56 + 7 = 7(8 +1) which is divisible by 7

Thus P (K+1) is true

Hence by P.M.I P (n) is true n N

Ans 04. Let P (n) : 1. 2. 3 + 2. 3. 4 + --- + n (n+1) (n+2) = ( 1)( 2)( 3)

4n n n n+ + +

For n = 1

P (1) = 1 (2) (3) = (1)(2)(3)(4)

4

P (1) = 6 = 6 which is true

Let P (K) be true

P (K) : 1. 2. 3 + 2. 3. 4 + -- + K (K+1) (K+2) = ( 1)( 2)( 3)

(1)4

K K K K+ + +

we want to prove that

P (K+1) n: 1. 2. 3 + 2. 3. 4 + -- + (K+1) (K+2) (K+3) = ( 1)( 2)( 3)( 4)

4K K K K+ + + +

L.H.S = 1. 2. 3 + 2. 3. 4 + -- + K (K+1) (K+2) + (K+1) (K+2) (K+3)

[( 1)( 2)( 3) ( 1)( 2)( 3) from (1)4 1

K K K K K K K+ + + + + += +

( 1)( 2)( 3)[ 4],

4K K K K+ + + +

=

Thus P (K+1) is true whenever P(K) is true.

Ans 05. P(n) : 1 1 1 1

. +1.2 2.3 3.4 ( 1) 1

n

n n n+ + + =

+ +

For n = 1

P (1) = 1 1

which is true 2 2

=

Let P (K) be true

1 1 1( ) : --- + (1)1.2 2.3 ( 1) 1

we want to prove that P (K+1) is true 1 1 1 1P (K+1) : --- +

1.2 2.3 ( 1)( 2) 21 1 1 1L.H.S = --- +

1.2 2.3 ( 1) ( 1)( 2)

KP KK K K

KK K K

K K K K

+ + =+ +

++ + =

+ + +

+ + ++ + +

[

2 2

1 from (1)1 ( 1)( 2)

( 2) 1( 1)( 2)

2 1 ( 1)( 1)( 2) ( 1)( 2)

1,

2

KK K KK KK KK K KK K K K

KK

= ++ + +

+ +=

+ +

+ + += =

+ + + +

+=

+

Thus P (K+1) is for whenever P (K) is true.

Ans 06. Let P (n) : 1 + 3 + 5 + --- + (2n-1) = n2

For n = 1

P (1) = 1 = 1 which is true

Let P (K) be true

P (K) : 1 + 3 + 5 + --- + (2K-1) = K2 (1)

we want to prove that P (K+1) is true

P (K+1) : 1 + 3 + 5 + --- + (2K+1) = (K+1)2

L.H.S = 1 3 5 (2 1) (2 1)K K+ + + + + +

[22

2 1 From (1)( 1)K K

K

= + +

= +

Thus P (K+_1) is true whenever P(K) is true.

Hence by PMI, P(n) is true for all n N.

Ans 07. P (n) :

23 3 3 3 ( 1)1 2 3

2n n

n+

+ + + + =

For n = 1

P (1) : 13 = 13 which is true

Let P (K) be true

23 3 3

23 3 3

3 3 3 3

23

( 1)( ) :1 2 (1)2

we want to prove that P (K+1) is true ( 1)( 2)P (K+1) : 1 +2 + -- + (K+1) =

2. . 1 2 ( 1)

( 1) = ( 1)

2

K KP K K

K K

L H S K K

K K K

+ + + + =

+ +

= + + + + +

+ + +

[2 2 3

2 2 3

2 2

from (1)

( 1) ( 1) =

4 1( 1) 4( 1)

=

4( 1) 4( 1)

=

4

K K K

K K K

K K K

+ ++

+ + +

+ + +

2 2

2 2

( 1) ( 4 4) =

4( 1) ( 2)

=

4( 1)( 2)

=

2

K K K

K K

K K

+ + +

+ +

+ +

Thus P (K+1) is true whenever P (K) is true.

Ans 08. P (n) : 223 5 (2n+1)1+ 1+ +----+ 1+ = (n+1)1 4 n

For n = 1

P (1) : 4 = 4 which is true

Let P (K) be true

P (K) : 223 5 (2 1)1 1 1 ( 1) (1)1 4

K KK

+ + + + + + = +

We want to power that P (K+1) is true

[

22

2 2

22

2

3 5 (2 3)( 1) : 1 1 1 ( 2)1 4 ( 1)

3 5 (2 1) (2 3. . 1 1 1 1

1 4 ( 1)(2 3

=(K+1) + 1 from (1)( 1)(

=(K+1)

KP K KK

K KL H SK K

KK

+ + + + + + + = + +

+ + = + + + + + + + +

++ +

2

2

2 2

2

2 2

2

2

1) 2 3( 1)

( 1) ( 4 4) = ( 1)

( 1) ( 2) = ( 1) =(K+2)

K KK

K K KK

K KK

+ + + +

+ + +

+

+ +

+

Thus P (K+1) is true whenever P (K) is true.

CBSE TEST PAPER-04


1. Prove that 32n+2 8n 9 is divisible by 8 [4]

2. Prove by PMI.

xn-yn is divisible by (x-y) whenever x-y 0

[4]

3. Prove (x2n-1) is divisible by (x-1). [4]

4. Prove ( ) ( ) ( ) ( )

1 1 1 211 2 1 2 3 1 2 1

n

n n+ + + + =

+ + + + + + +

[4]

5. Prove 1.3 + 3.5 + 5.7 + -- + (2n 1) (2n + 1) =

( )24 6 13

n n n+ [4]

6. Prove by PMI

3.22 + 32.23 + 33.24 + -- + 3n. 2n+1 = ( )12 6 15 n n N .

[4]

7. Prove 1.3 + 2.32 + 3.33 + --- + n.3n =

( ) 12 1 3 34

nn

+ + [4]

8. Prove

1 1 1 13.5 5.7 7.9 (2 1)(2 3) 3(2 3)

n

n n n+ + + + =

+ + +

[4]

CBSE TEST PAPER-04


[ANSWERS]

Ans 01. P(n) : 32n+2 8n 9 is divisible by 8

For n = 1

P (1) : 32+2 8 9 = 64

which is divisible by 8


Let P (K) be true

P (K) : 32K+2 8K 9 is divisible by 8 2 2

2(K+1)+2 2K+2+2

3 8 9 8 , where N ( i)we want to prove that result is true for n = K+13 -8(K+1)-9=3 -8K-8-9

K K + =

2 4

2 4

2 2 2

3 8 173 .3 8 173 .3 8 17(8 8 9).9 8 1772 72 81 8 1764 64 648(8 8 8)

K

K

k

KK

KK K

K KK

K

+

+

=

=

=

= + +

= + +

= + +

= + +

( from i)

which is divisible by 8 Hence P(K+1) is true whwnever P (K) is true. Hence by P.M.I P (n) is true n N

Ans 02. P (n) : xn-yn is divisible by (x-y)

For n = 1

P (1) : x y is divisible by (x y)

Let P (K) be true

P (K) : xK yK is divisible by (x y)

( )K K

K+1 K+1 K K

x -y = x-y ( )we want to prove that P (K+1) is true whenever P (K) is true x -y = x .x-y .y

i

( )K K(x-y)+y .x-y .y= ( from i)

( )( ) ( )

. .

.

( )

K K

K

K

x y x y x y y

x y x y x y

x y x y

= +

= +

= +

which is divisible by x-y Hence P (K+1) is true

Ans 03. P (n) : (x2n-1) is divisible by (x-1).

For n = 1

P (1) : (x2 1) = (x 1) (x + 1)

which is divisible by (x 1)

Let P (K) be true

( )22K

2(K+1)

( ) : 1 is divisible by x-1 (i) x -1= (x-1)

we want to prove that P (K+1) is true P(K+1) : x -1L.H.S

KP K x

( )

( ) ( )( )( )

2K +2

2K 2

2

2 2

2

2

=x -1=x .x -1= (x-1)+1 .x -1(from i )=(x-1).x +x -1= x-1 .x + x-1 x+1

=(x-1) x + x+1 which is divisible by (x-1) Hence p(K+1) is true whenever p(k) is true

Ans 04. P (n) : ( ) ( ) ( ) ( )1 1 1 21

1 2 1 2 3 1 2 1n

n n+ + + + =

+ + + + + + +

( ) ( ) ( )

( ) ( ) ( )

1 1 1 2111 2 1 2 3 1

21 1 2 21

1 2 1 2 3 1 1for n = 1

2 2P (1) : 1 2 2

n

n n n

n

n n n

+ + + + =++ + + +

+ + + + =+ + + + +

= =

( ) ( ) ( )

which is true Let p(k) be true

1 1 2 2p(k) : 1 (1) 1 2 1 2 3 1 1

kk k k

+ + + + =+ + + + +

we want to prove that p (k + 1) is true

( ) ( ) ( )

( ) ( ) ( ) ( )[

2

1 1 2 2( 1)p(k+1) : 11 2 1 2 3 1 ( 2) 2

1 1 2 2. . 1

1 2 1 2 3 1 1 ( 2)2 2

= from (1)1 ( 1)( 2)

2 ( 2) 2 = ( 1)( 2)

2 4 2 = ( 1)( 2)

kk k k

L H Sk k k k

kk k k

k kk kk k

k k

++ + + + =

+ + + + + +

= + + + + =+ + + + + +

++ + +

+ +

+ +

+ +

+ +2

2

2( 2 1) = ( 1)( 2)

2( 1) = ( 1)( 2)

2( 1) = ( 2)

thus p (k+1) is true whenever p(k) is true Hence by PMI p(n) is true n N.

k kk k

kk k

kk

+ +

+ +

+

+ +

+

+

Ans 05. Let p (n) : 1.3 + 3.5 + -- + (2n-1) (2n+1) = ( )24 6 1

3n n n+

For n = 1

P (1) = (1) (3) = 1(4 6 1)

3+

P(1) = 3 = 3 Hence p (1) is true

Let (k) be true

P(k) : 1.3 + 3.5 + -- + (2k - 1) (2K + 1) = 2(4 6 1)

(1)3

k k k+

we want to prove that p (k+1) is true

p (k+1) : 1.3 + 3.5 + -- + (2k+1) (2k+3) = ( ) 21 4( 1) 6( 1) 1

3k k k + + + +

L. H. S

( )( ) [( ) ( ) ( )

[ ]( )( )

2

2

3 2

2

1.3 3.5 (2 1)(2 1) (2 1)(2 3)2 1 2 3(4 6 1)

from (1)3 1

4 6 1 3 2 1 2 33

4 18 23 9 put k = -1 (k+1) is one fator3

1 4 14 93

k k k kk kk k k

k k k k k

k k k

k k k

+ + + + + + +

+ ++ = +

+ + + +=

+ + +=

+ + +=

Thus p (k+1) is true whenever p (k) is true.

Ans 06. Let p (n) : 3.22 + 32.23 + 33.24 + --- + 3n. 2n+1 = ( )12 6 15 n For n = 1

( )

( )

( )

1 2 1

2 2 3 k k+1

2 2 3 k+1 k+2 1

2 2 3

12(1) : 3 .2 6 15

(1) 12 12p(1) is true Let p(k) be true

12p(k):3.2 +3 .2 +--+3 . 2 = 6 1 (1)5

we want to prove that p (k+1) is true 12p(k+1): 3.2 +3 .2 +---+ 3 . 2 = 6 15

. . 3.2 3 .2

k

k

p

p

L H S

+

=

= =

= + +

( )1 1 2

1 2

1 1 1

1 1

k+1

k+1 1

3 .2 3 .212

= 6 1 3 .2 [from (1)52 12

= .6.6 3 .2 .25 5

2 12 = 6 6 .2

5 52 12

=6 25 512 12 12

=6 6 15 5 5

k k k k

k k k

k k k

k k

k

+ + +

+ +

+ +

+ +

+

+ +

+

+

+

+

=

Thus p(k+1) is truewhenever p(k) is true.

Ans 07. P (n) : 1.3 + 2.32 + 3.33 + --- + n.3n = ( ) 12 1 3 3

4

nn

+ +

For n = 1

P (1) : 1. 31 = 2(2 1).3 3

4 +

( ) 12 3 k

22 3 k+1

2

12 3(1) : 34

hence p(1) is true Let p(k) be true

2 1 3 3p(k) : 1.3+2.3 +3.3 +--+k.3 = (1)

4we want to prove that p (k+1) is true

(2 1)3 3p(k+1) : 1.3+2.3 +3.3 +--+(k+1).3 =4

. . = 1.3+2.3 +

k

k

p

k

k

L H S

+

+

=

+

+ +

3 1

1 1

1 1

1

1

1

3.3 +--+ .3 ( 1).3(2 1).3 3 ( 1)3

= [from (1)4 1

(2 1).3 3 4( 1)3 =

4(2 1 4 4).3 3

=

4(6 3).3 3

=

43(2 1).3 3

=

4

k k

k k

k k

k

k

k

k kk k

k k

k k

k

k

+

+ +

+ +

+

+

+

+ +

+ ++

+ + +

+ + +

+ +

+ +

2(2 1)3 3 =

4

kk ++ +

Thus p (k+1) is true whenever p(k) is true.

Ans 8. P(n) : 1 1 1 1

3.5 5.7 7.9 (2 1)(2 3) 3(2 3)n

n n n+ + + + =

+ + +

For n = 1

1 1(1) : (2 1)(2 3) 3(2 3)1 1(1) Hence p (1) is true

.15 15Let p (k) be true

1 1 1 1p(k) : (1)3.5 5.7 7.9 (2 1)(2 3) 3(2 3)

p

p

kk k k

=

+ + +

= =

+ + + + =+ + +

we want to prove that p (k+1) is true 1 1 1 1 ( 1)p(k+1):

3.5 5.7 7.9 (2 3)(2 5) 3(2 5)k

k k k+

+ + + + =+ + +

1 1 1 1 1. .

3.5 5.7 7.9 (2 1)(2 3) (2 3)(2 5)L H S k k k k= + + + + ++ + + +

( ) ( )1 1

[from (1)3(2 3) 2 3 2 5

(2 5) 3=

3(2 3)(2 5)1

3(2 5)

kk k k

k kk k

kk

= + + + +

+ +

+ +

+=

+

Thus p (k+1) is true whenever p (k) is true

Hence p (n) is true for all n N.

CBSE TEST PAPER-05


1. The sum of the cubes of three consecutive natural no. is divisible by 9. [4]

2. Prove that 12n + 25n-1 is divisible by 13 [4]

3. Prove 11n+2 + 122n+1 is divisible by 133. [4]

4. Prove 13 + 23 + 33 + --- + n3 =

2 2( 1)4

n n + [4]

5. Prove (a)+ (a + d) + (a + 2d) + -- + [a + (n 1)d] = [ ]n 2a+(n-1)d

2[4]

6. Prove that 2n > n positive integers n. [4]

7. Prove ( )

1 1 1 11.2 2.3 3.4 1 1

n

n n n+ + + + =

+ +

[4]

8. Prove

1 1 1 1.

3.6 6.9 9.12 3 (3 3) 9( 1)n

n n n+ + + + =

+ +

[4]

CBSE TEST PAPER-05


[ANSWERS]

Ans 01. P(n) 3 3 3( 1) ( 2)k k k + + + + is divisible by 9For n = 1

P (1) : 1 + 8 + 9 = 18


Let p (k) be true 3 3 3

3 3 3

3 3 3

3 3 3

( ) : ( 1) ( 2) is divisible by 9 ( 1) ( 2) 9 ( )

we want to prove that p (k+1) is true p(k+1) : (k+1) + (k+2) +(k+3)L.H.S = (k+1) + (k+2) + (k+3)

p k k k k

k k k i + + + +

+ + + + =

3 3 3 2

3 3 3 2

2

2

( 1) ( 2) 9 27 27( 1) ( 2) 9( 3 3)

9 9( 3 3) ( from i)9 ( 3 3) which is by 9.

k k k k kk k k k k

k kk k

= + + + + + + +

= + + + + + + +

= + + +

= + + +

Ans 02. P(n) : 12n + 25n-1 is divisible by 13

For n = 1

P(1) : 12 + (25)0 = 13


Let p (k) be true

P(k) : 12k + 25k-1 is divisible by 13 1

(k+1) k+1-1 k 1 k

12 25 13 ( )we want to prove that result is true for n = k+112 + 25 =12 .12 +25

k k i + =

( )

( )

1

1

1

13 25 .12 25 ( )13 12 25 .12 2513 12 25 12 25

k k

k k

k

from i

= +

= +

= + +

( )113 12 25k = +which is divisible by 13.

Ans 03. P(n) : 11n+2 + 122n+1 is divisible by 133.

For n = 1

P(1) : 113 + 123 = 3059


Let p (k) be true 2 2 1

2 2 1

k+1+2 2(k+1)+1

k+3 2k+2+1

k+3 2k+3

( ) :11 12 is divisible by 13311 12 133 ( i)

we want to prove that result is true for n = k+1L.H.S = 11 + 12

=11 +12 =11 +12 =11

k k

k k

p k

+ +

+ +

+

+ =

( )

k 3 2k 3

k+2 2k 3

2 1 2 3

2k+1 2k 3

k 3

.11 +12 .12 =11 .11+12 .12

= 133 12 .11 12 .12 ( from i) =133 11 -12 .11+12 .12 =133 11 +12 (-12 11+12 ) =133 (11 12 (1596)

which

k k

k

+ +

+

is 133.

Ans 04. P(n) : 13 + 23 + 33 + --- + n3 = 2 2( 1)

4n n +

For n = 1 2 2

3

2 23 3 3

2 23 3 3

3 3 3

1 (2)(1) :1 14

which is true Let p(k) be true

( 1)p(k) : 1 + 2 + --- + k = (1)4

we want to prove that p (k+1) is true ( 1) ( 2)p(k+1) : 1 + 2 + --- + (k+1) =

4L.H.S = 1 +2 +---+k +(

p

k k

k k

= =

+

+ +

3k+1) [ from (1)

2 2 3( 1) ( 1) =

4 1k k k+ +

+

2 2 3

2 2

2 2

( 1) 4( 1)4

( 1) [ 4( 1)4

( 1) ( 2)4

k k k

k k k

k k

+ + +=

+ + +=

+ +=

Thus p(k+1) is true whenever p(k) is true.

Ans 05. P(n) : (a)+ (a + d) + (a + 2d) + -- + [a + (n 1)d] = [ ]n 2a+(n-1)d2

For n = 1

p(1) : a + (1-1) d = 12

2a + (1-1) d = a

which is true

Let p (k) be true

( )( ) [ ]

[ ]

kp(k):(a)+(a+d)+(a+2d)+--+ a+ k-1 d = 2a+(k-1)d (1)2

we want to prove that p (k+1) is true k+1p(k+1) : (a) + (a+d) + -- + (a+kd) = 2a+kd2

L.H.S = a + (a+d)+--+ a+kd = a + (a + d) + -- + a + (k-1)d

[ ]

2

+ a + kd k

= 2a+(k-1)d +a+kd [from (1)2

k = ka + (k-1)d+a+kd

22ak+k d-kd+2a+2kd

=

22 ( 1) ( 1) ( 1)(2 )

=

2 2 proved.

a k kd k k a kd+ + + + +=

Ans 06. Let p (n) : 2n > n

For n = 1

P (1) : 21 >1

Which is true

Let p (k) be true

P (k) : 2k > k (1)

we want to prove that p (k+1) is true

2k > k by (1)

kk+1

k+1

2 .2 > 2k 2 > 2k 2 > 2k = k+k > k+1

Hence provtd.

Ans 07. P(n) : ( )1 1 1 1

1.2 2.3 3.4 1 1n

n n n+ + + + =

+ +

For n = 1

1 1(1) which is true 2 2

Let p (k) be true 1 1 1p(k) : (1)

1.2 2.3 ( 1) 1we want to prove that p (k+1) is true

1 1 1 1p(k+1) : 1.2 2.3 ( 1)( 2) 21 1 1 1L.H.S =

1.2 2.3 ( 1) ( 1)(

p

kk k k

kk k k

k k k k

= =

+ + + =+ +

++ + + =

+ + +

+ + + ++ +

[from (1)2)+

2 2

11 ( 1)( 2)

( 2) 1( 1)( 2)

2 1 ( 1)( 1)( 2) ( 1)( 2)

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