Plus Two Maths Chapter Wise Questions and Answers Chapter 1 Relations and Functions are part of Plus Two Maths Chapter Wise Questions and Answers. Here we have given Plus Two Maths Chapter Wise Questions and Answers Chapter 1 Relations and Functions.

Board | SCERT, Kerala |

Text Book | NCERT Based |

Class | Plus Two |

Subject | Maths Chapter Wise Questions |

Chapter | Chapter 1 |

Chapter Name | Relations and Functions |

Number of Questions Solved | 47 |

Category | Plus Two Kerala |

## Kerala Plus Two Maths Chapter Wise Questions and Answers Chapter 1 Relations and Functions

**Short Answer Type Questions**

**(Score 3)**

Question 1.

Let T be the set of all triangles in a plane with R a relation in T given by R = {(T_{1},T_{2}):T_{1} ≅ T_{2}) show that R is an equivalence relation.

Answer.

Given R = {(T_{1}, T_{2}) ∈ T x T / T_{1} = T_{2}}

For reflexive : (T_{1}, T_{1}) ∈ R is true as T_{1} ≅ T

for all T_{1} ∈ T(i.e.triangle is congruent to itself). Hence, R is reflexive.

For symmetric : (T_{1},T_{2})∈ R ⇒ T_{1} ≅ T_{2}

⇒T_{2}≅T_{1}

⇒(T_{2}T_{1})∈R

Hence R is symmetric.

For transitive : Let (T_{1}, T_{2})∈R and (T_{2},T_{3})∈R ⇒ T_{1}≅T_{2}

and T_{2} ≅ T_{3} ⇒ T_{1} ≅ T_{2} => (T_{1,}T_{3}) ∈ R

Hence, R is transitive.

Since R is reflexive, symmetric and transitive.

Hence, R is an equivalence relation.

Question 2.

* be a binary operation on Q, defined as

a. Show that * is commutative.

b. Show that * is associative.

c. Find the identity element of * if any.

Answer.

a. For commutative for a, b ∈ Q

Question 3.

Consider the following functions.

f : Z → Z defined by f(x) = 3x + 7 and g : R → R defined by g(x) = 2x – 3

i. Find the images of 5/2 under the functions f and g, if it exists.

ii. Of the above two functions one is a bijective function and the other is not. Give reasons.

iii. Find the inverse of the bijective function.

Answer.

i. does not exist and

ii. g is bijection and f is not. g is one-one and onto and f is one-one, not onto.

iii. g^{-1} : R→R defined by

Question 4.

Which of the relations R on the set of the real numbers is an equivalence relation?

a. x R y if |x| = |y|

b. x R y if x – y ≥0

Answer.

a. Since |x| = |x| ∀ x, xRx. So R is Reflexive. Since |x| = |y| ⇒ |y| = |x|.

Hence the relation is symmetric.

Also |x| = |y|, |y| = |z| ⇒ |x| = |z|

Therefore, the relation is transitive.

Thus R is an equivalence relation.

b. Since (x – y) ≥ 0 does not imply (y – x) > 0 the relation is not symmetric.

Therefore here R is not an equivalence relation.

Question 5.

In the set of all natural numbers, let a relation defined by R = {(a, b) : a, b ∈ N, a – b is divisible by 5}. Prove that R is an equivalence relation.

Answer.

For a∈N,a – a = 0 which is divisible by 5

(a, a) ∈ R ∀ a e N and R is reflexive

Let (a, b) ∈ R ⇒ a – b is divisible by 5

⇒ -(a – b) is divisible by 5

⇒ b – a is divisible by 5 ⇒ (b, a) ∈ R

R is symmetric; Let (a, b), (b, c)∈R

⇒ a – b and b – c are divisible by 5

⇒a – b + b – c = a – c is divisible by 5

⇒ (a, c) ∈ R

R is transitive.

Hence R is an equivalence relation.

Question 6.

a. If f : R → R is given by/(x) = 3x – 1, g : R → R is given by g(x) = 2x, show that fog – gof = 1.

b. Find the domain, range and inverse of

Answer.

a. (fog)(x) = f(g (x)) = f(2x)

=3(2x) – 1 = 6x – 1

(gof)(x) = g (f(x)) = g (3x – 1)

= 2 (3x – 1) = 6x – 2

∴fog – gof = (6x – 1) – (6x – 2) = 1

Question 7.

Prove that f : N → N defined by f(x) = x² + x + 1 is one to one but not onto.

Answer.

Let x_{1} = x_{2} ∈ N such that f(x_{1}) = f(x_{2})

=>x_{1}^{2} + x_{1} + 1 = x_{2}^{2} + x_{2} + 1

=> (x_{1}^{2} – x_{2}^{2} ) = (x_{2} – x_{1})

=> (x_{1} – x_{2})(x_{1} + x_{2} + 1) = 0 => x_{1} = x_{2},

∴ f is one-one.

Let y ∈ N. Then f(x) = y => x² + x + 1 = y

=> x² + x + (1 – y) = 0

which is not always a natural number.

∴ f is not onto.

Question 8.

Consider the binary operation * on the set {1,2,3,4,5} given by the following table.

i. Is * commutative?

ii. Compute (2 * 3) * 4 and 2*(3 * 4).

iii. Compute (2 * 3) * (4 * 5).

Answer.

i. We observe that a*b = b*a ∀ a, b ∈ {1,2,3,4,5}, ⇒ * is commutative

ii. (2 * 3) * 4 = 1 *4 = 1; 2 * (3 * 4) = 2 * 1 = 1

iii. (2 * 3) * (4 * 5) = 1 * 1 = 1

Question 9.

a. Let A = {-1, 0, 2, 3}, B = {1, 2, 5, 8, 9,10} and f = {(x, y) : y = x² + 1, x ∈ A and y ∈ B}. List the elements of f and state the type of the function.

b. Let * be a binary operation on N, defined by a * b = a^{b}, a, b ∈ N. Is (N, *) associative or commutative?

Answer.

a. Since y = x² + 1 = f(x), therefore,

f(-1) = 2, f(0) = 1, f(2) = 5,

f (3) = 10

f = {(-1, 2), (0, 1), (2, 5), (3, 10)}

It is one to one into function

b. * is neither commutative nor associative.

2 * 3 = 2^{3} = 8, but 3 * 2 = 3^{2} = 9

(3 * 4) * 2 = 34 * 2 = (3^{4})^{2} = 3^{8}

3 * (4 * 2) = 3^{4*2} = 3^{(4)²} = 3^{16}

Question 10.

If the function f : R→R is given by

and g : R→R is given by g(x) = 2x – 3. Find.

(i) fog

(ii) gof

(iii) is f ^{-1} = g

Answer.

i. (fog) x = f(g (x)) = f(2x – 3)

**Long Answer Type Questions**

**(Score 4)**

Question 1.

Consider f : R_{+} → [-5, ∞] given by f(x) = 9x² + 6x – 5. Show that f is invertible with

Answer.

Let y = 9x² + 6x – 5 = (3x +1)² – 6

⇒(3x + 1)² = y + 6

⇒ 3x + 1 = √y + 6

⇒ 3x = √y + 6 – 1

Question 2.

A is a non empty set and let * be a binary operation on P(A) the power set of A defined by X* Y = X∩Y, for X, Y∈P(A).

i. Show that A*B = B*A for A,B∈P(A).

ii. Show that * is associative.

iii. Find the identity element of (P(A), *) and show that A ∈ P(A) is the only invertible element of (P(A), *).

Answer.

i. A*B = A∩B = B∩A = B *A,

A, B∈P(A)

ii. (A * B) * C = A * (B * C),

∴ * is associative

iii. Let E be the identity element in P(A) with respect to *

Then B * E = B = E * B for all B ∈ P(A)

⇒B∩E = B = E∩B ∀BCA ⇒ E = A

Thus A is the identity element with respect to * on P(A)

Let B be an invertible element of P(A) and let S be its inverse. Then, B*S = A = S*B ⇒ B ∩ S = A = S ∩ B

⇒ B = S = A [∵B⊂A S⊂A]

Thus A is the only invertible element of P(A) with respect to * and it is the inverse of itself

Question 3.

a. Let S be the set of all sets and let

R = {(A, B) : A ⊂ B} i.e. A is a proper subset of B. Show that R is

i. Transitive

ii. Not reflexive

iii. Not symmetric

b. Prove that the function f: N → N f(x) = 3x, is one-one and into.

Answer.

a. i. A R B, B R C ⇒A⊂B and B⊂C

⇒A ⊂ C => A R C

∴R is transitive

ii. A ⊂ A is not true R is not reflexive.

iii. {1,2}⊂{1,2,3} But {1,2,3}⊂{1,2} is not true

∴R is not symmetric

∴f is one-one

b. Let 2 ∈ co domain N,

then 2 = 3x; x = 2/3 ∉ N

i. e., 2 does not have pre-image so f is into.

Question 4.

a. If f: R → R be defined by f(x) = 2x – 3 and g : R → R be defined by Show that fog = I_{R} = gof

b. Let f : R→R be defined by f(x) = 3x + 2. Show that f is invertible. Find f^{-1} : R → R. Hence find f^{-1} (3) and f^{-1} (0).

Answer.

Question 5.

Consider the binary operation * : Q → Q, where Q is the set of rational numbers, is defined as a*b = a + b – ab.

i. Is * associative? Justify your answer.

ii. Is identity for * existing? If it exists find the identity element.

iii. Are elements of Q invertible? If so find the inverse of an element in Q.

Answer.

i. (a * b) *c = (a * b) + c – (a * b) c = (a + b – ab) + c – (a + b – ab) c = a + b + c – ab – ac – bc + abc — (1)

a * (b * c) = a + (b * c) – a(b * c)

= a + (b + c – bc) – a(b + c – bc)

= a + b + c – ab – ac – bc + abc — (2)

By (1) & (2), (a * b) * c = a * (b * c) and it is associative.

ii. a*e = a + e – ae = a

⇒ e – ae = 0 , ⇒ e (1 – a) = 0

⇒ e = 0, the identity element (0 ∈ Q )

iii. a * a^{-1} = a + a^{-1} – aa^{-1} = 0

⇒ a^{-1} (1 – a) = -a

Question 6.

A function f: R→R defined by

i. Is f a one-one function? Why?

ii. Prove that f is invertible.

iii. Find f^{-1}.

Answer.

i. Yes, f is a one-one function

f(x_{1}) = f(x_{2}) ⇒

Question 7.

Show that the relation ‘S’ in set

A = {x∈z: 0 ≤ x ≤ 12} given by

S = {(a,b) : a,b∈z, |a – b| is divisible by 4} is an equivalence relation.

Answer.

i. |x – x| = 0 is divisible by 4

∴S is reflexive

ii. |x – y| is divisible by 4

⇒|x – y| is divisible by 4

ie. (x,y) ∈R =>( y,x) ∈R

∴ S is symmetric

iii. (x, y) ∈ S and (y, z) ∈ S

⇒ |x – y| is divisible by 4 and |y – z| is divisible by 4

⇒ |x – y + y – z| is divisible by 4

⇒ |x – z| is divisible by 4

⇒ (x,z) ∈ S

∴‘S’is transitive.

∴S is an equivalence relation.

Question 8.

If A = (1, 2, 3}, B = {a, b}, write all functions from A to B. How many are one- one? How many are into?

Answer.

The possible functions are

f_{1} = {(1, a), (2, a), (3, a)}

f_{2} = {(1, b), (2, b), (3, b)}

f_{3} = {(1, a), (2, b), (3, b)}

f_{4} = {(1, b), (2, a), (3, b)}

f_{5} = {(1, b), (2, b), (3, a)}

f_{6} = {(1. a), (2, a), (3, b)}

f_{7} = {(1, a), (2, b), (3, a)}

f_{8} = {(1, b), (2, a), (3, a)}

As the number of elements of B is less than number of. elements of A, one-one function is not possible. All are many one. f_{1} and f_{2} are into and the others are onto.

Question 9.

Let f: {1,2,3}→{3,5,7} and

g : {3,5,7}→{7,23,47} given by

f(1) = 3,f(2) = 5, f(3) = 7, g(3) = 7, g(5) = 23, g(7) = 47

i. Find gof.

ii. Verify (gof)^{-1} = fog^{-1}

Answer.

i.f = {(1,3), (2,5), (3,7)}

g = {(3,7), (5,23), (7,47)}

(gof)(1) = g(f(1)) = g(3) = 7

(gof) (2) = g(f(2)) = g(5) = 23

(gof)(3) = g(f(3)) = g(7) = 47

∴ gof ={(1,7), (2,23), (3,47)}

From the arrow diagram, we observe that f,g and gof are invertible.

f^{-1}= {(3,1), (5,2), (7,3)}

g^{-1} = {(7,3), (23,5) (47,7)}

(gof)^{-1}= {(7,1), (23,2), (47,3)}

f^{-1}og^{-1}(7) = f^{-1}(g^{-1}(7)) = f^{-1}(3) = 1

f^{-1}og^{-1}(23) = f^{-1}(g^{-1}(23)) = f^{-1}(5) = 2

fog^{-1}(47) = f^{-1}(g^{-1}(47)) = f^{-1}(7) = 3

∴ f^{-1}og^{-1} = {(7,1), (23,2), (47,3))

Hence (gof)^{-1}= f^{-1}og^{-1}

Question 10.

Let Q be the set of rational numbers and * be the binary operation on Q defined by a*b= for a,b ∈ Q

i. What is the identity element of * in Q?

ii. Find the inverse of the element a under * on Q.

Answer.

i. Let e ∈ Q be the identity element under *.

Then a*e = e*a = a for all a ∈ Q

**Very Long Answer Type Questions**

**(Score 6)**

Question 1.

i. Show that the function f:R→R defined by f(x) = 2x – 3 is one-one and onto. Find f^{-1}

ii. Which of the following figures represents the graph of a function on R which is onto but not one-one.

iii. Write a function on R which is onto but not one-one.

Answer.

i. f(x) = 2x – 3

one – one if f(x 1)= f(x) => x1 = x2

For showing onto y = 2x – 3

Question 2.

A = {1,2,3,6}, * is a binary operation on A is defined as a*b = HCF of a and b.

i. Represent * with the help of an operation table.

ii. Find the identity element.

iii. Write a commutative binary operation on A with 3 as the identity element.

(Hint: Operation table may be used)

Answer.

i. A = {1,2,3,6}

a*b = HCF of a and b

Question 3.

a. Give an example of a relation which is

i. reflexive but not symmetric

ii. transitive but not symmetric

b. Let A = {3, 5} and B = {7,11}. Let R = {(a, b) : a ∈ A, b ∈ B, a – b is odd}. Show that R is an empty relation from A into B.

Answer.

a. i. In the set of real numbers x|y (x divides y) is reflexive but not symmetric.

ii. In the set of real numbers, the relation x < y is transitive but not symmetric.

b. Since none of the numbers (3 – 7) (3 – 11) (5 – 7) (5 – 11) is an odd number, R is an empty relation.

Question 4.

Let * be a binary operation on set Q of rational numbers such that

a * b = (2a – b)²: a, b ∈ Q

i. find 3*5

ii. Let * be a binary operation on Z+ defined by a * b = |a – b| for all a, b ∈ Z+. Find 3a * 2b where a = -5 and b = -6.

iii. Let f :R→R; f(x) = (3 – x^{3})^{1/3} find fof

Answer.

i. 3 * 5 = (2 x 3 – 5)² = 1

ii. 3a * 2b = |3a – 2b| = |3 x – 5 – 2 x – 6| = 3

iii. (fof)(x) = f(f(x)) = f((3 – x^{3})^{1/3})

= [3 – (3 – x^{3})]^{1/3} = (x^{3})^{1/3} = x

Question 5.

a. Let * be the binary operation on N given by x*y = LCM of x, y. Find 9*5 ?

b. For each binary operation * defined below, determine whether * is commutative or associative.

i. On Q define a*b =

ii. On Z+, define a*b = a^{b}

Answer.

a. LCM of 9, 5. is 45

b. i. a*b =

Question 6.

Let A = NxN and * be the binary operation on A defined by (a,b) * ( c,d) = (a+c, b+d). Show that * is commutative and associative. Find the identity element for * on A, if any.

Answer.

(a,b)* (c,d) = (a+c, b+d)

(c,d)*(a,b) = (c+a, d+b) = (a+c, b+d) (a,b)* (c,d) = (c,d)*(a,b)

∴* is commutative

Let (a,b), (c,d), (e,f)∈A

(a,b)*[(c,d)*(e,f)] = (a,b)*[(c+e),(d+f)]

= (a+c+e,b+d+f)

[(a,b)*(c,d)]*(e,f) = (a+c, b+d)*(e,f)

= (a+c+e, b+d+f)

i.e, (a,b)* [(c,d)*(e,f)] = [(a,b)*(c,d)]*(e,f)

∴* is associative

Here identify element does not exist

Let (e1, e2) ∈ A be the identify element for * in A

(a,b)*(e1 ,e2) = (e1 ,e2)*(a,b) = (a,b)

⇒(a+e1,b+e2)= ( e1+a, e2+b) = (a,b)

⇒a+e1= a and b+e2 = b

⇒ e1 = 0 and e2 = 0

(e1, e2) = (0,0) ∉ A , since A = NxN

Hence there is no identity element for * in A.

Question 7.

a. Show that the function f in defined as is one-one and onto

b. Hence find f^{-1}.

Answer.

a. Consider the function

For one-one: Let for x1, x2 ∈ A

Question 8.

If R1 and R2 are two equivalence relations in a given set A, show that R1 ∩ R2 is also on equivalence relation.

Answer.

We have R1 ∩ R2 = {(a,b)/(a,b) ∈ R1 and (a,b) ∈ R2}

For reflective: Let a ∈ A , then (a,a) ∈ R, and (a, a)∈R2

⇒ (a, a) ∈R1 ∩ R2

Hence reflexive

For symmetric: Let (a,b) ∈ R1 ∩ R2

⇒ (a – b) ∈ R1 and (a, b) ∈ R1

(As R1 and R2 are equivalnce relations)

⇒ (b, a) ∈ R1 ∩ R2.

Hence symmetric for transitive :

Let (a,b), (b,c) ∈ ( R1 ∩ R2)

⇒ (a,b),(b,c) ∈ R1 and (a,b),(b,c) ∈ R2

⇒ (a,c)∈R1 and (a,c)∈R2

(R1 and R2 are equivalence relations)

⇒ (a,c)∈R1∩R2

As (a,b), (b,c) ∈ R1∩R2

⇒ (a,c) ∈ R1∩R2. So, transitive

Hence, R1∩R2 is an equivalence relation.

Question 9.

Consider the sets A = {1, 2, 3, 4, 5), B= {1, 4, 9, 16, 25} and a function f:A → B defined by f(1) = 1, f(2) = 4, f(3) = 9, f(4) = 16 and f(5) = 25.

a. Show that f is one-to-one.

b. Show that f is onto.

c. Does f^{-1} exist? Explain.

Answer.

a. Since the images of all elements of A are different, f is one – one.

b. Since every element of B is the image of some element of A, f is onto.

c. Yes, f^{-1} exists, since f is one-one and onto.

Question 10.

a. Let f: R→R be given by find fof and show that f is invertible.

b. Find the identify element of the binary operation*on N defined by a*b = ab²

Answer.

**Edumate Questions & Answers**

Question 1.

Let R be a relation on the set A= {1,2,3,4,5,6} defined as R={(x,y) : y = 2x-1}

i. Write R in roster form and find it’s domain and range.

ii. Is R and equivalence relation? Justify.

Answer.

i. R={(1,1), (2,3) (3,5)}

Domain = {1,2,3} ; Range = {1,3,5}

ii. Since (2,2) ∉ R, R is not reflexive

(2,3) ∉ R, but (3,2) ∉ R; R is not symmetric

(2,3) ∉ R, (3,5) ∉ R but (2,5) ∉ R, R is not transitive

∴R is not an equivalence relation.

Question 2.

The relation R defined on the A = {-1,0,1} as R = {(a,b):a² = b}

i. Check whether R is reflexive, symmetric and transitive

ii. Is R and equivalence relation?

Answer.

i. (-1,-1) ∉R, R is not reflexive

(-1,1)∈R but (1,-1)∉R, R is not symmetric

(-1,1)∈R, (1,1) ∈R and (-1,1)∈R, R is transitive.

ii. R is not reflexive, not symmetric and transitive.

So R is not an equivalence relation

Question 3.

Let A = {1,2,3}. Give an example of relation on A which is

i. Symmetric but neither reflexive nor transitive.

ii. Transitive but neither reflexive nor symmetric.

Answer.

i. R = {1,2), (2,1)}

(1,1)∉R ⇒ R is not reflexive.

(1,2)∈R⇒(2,1)∈R, R is symmetric

(1,2)∈R, (2,1)∈R but (1,1)∉R, R is not transitive.

ii. R = {1,2), (1,3), (2,3)}

(1,1)∉R ⇒ R is not reflexive.

(1,2)∈R, but (2,1)∉R, R is not symmetric.

(1,2)∈R, (2,3)∈R⇒( 1,3) ∈R, R is transitive.

Question 4.

i. Let f be a function defined by

f(x) = √x is a function if it defined from

(f: N→ N, f: R → R,f: R →R*,f: R^{+} →R^{+})

ii. Check the injectivity and subjectivity of the following Functions.

a. f: N → N given by f(x) = x^{3}

b. f: R → R given by fix) = [x]

Answer.

i. f:R^{+}→R^{+} For x,y∉N,

ii. a. f(x) = f(y) ⇒ x^{3} = y^{3} ⇒ x = y ⇒ f is injective.

For 2∉N, there does not exist x in the domain N such that f(x) = x^{3 }= 2

∴f is not surjective

b. f:R → R given by f(x)=[x]

It seen that f(1.1) = 1 and f(1.8) = 1 But 1.1 ≠ 1.8

∴f is not injective

There does not exist any element x∈R such that f(x) = 0.7

∴f is not suijective

Question 5.

a. Find fog and gof if

i. f(x) = |x| and g(x) = |3x+4|

ii. f(x) = 16x^{4} and g(x) = x^{1/4}

b. If , Prove that fof(x)=x

Answer.

a. i. f(x) = |x| and g(x) = |3x + 4|

⇒ fog(x) = f(g(x))

= f(|3x + 4|) = |3x + 4| = |3x + 4|

gof(x) = g(f(x)) = g (|x|) = |3|x| + 4|

ii. fog(x) = f(g(x))

= f(x^{1/4}) = 16(x^{1/4})^{4} = 16x

gof(x) = g(f(x))

= g(16x^{4}) = 16(x^{4})^{1/4} = 4x

b. fof(x) = f(f(x))

Question 6.

Let S = {(1,2), (2,3), (3,4)}. Give an example of relation on A which is

i. Find the domain and range of S.

ii. Find S^{-1}

iii. Find the domain and range of S^{-1}

iv. Verify that S^{-1} is a function using the graph of S and S^{-1}

Answer.

i. Domain = {1,2,3} ; Range = {2,3,4}

ii. S^{-1}={(2,1), (3,2), (4,3)} .

iii. Domain {2,3,4} ; Range = {1,2,3}

iv. Yes, S^{-1} is a function because x coordinates do riot intersect

Question 7.

i. Consider f : {3,4,5,6} → {8,10,12,13, 14} and f = {3,8), (4,10), (5,12), (6,14)}. State whether f has inverse? Give reason

ii. Consider f : R→R given by f(x) = 3x+2. Show that f is invertible. Find the inverse of f.

Answer.

i. District elements in set {3,4,56} has distinct images.

Under f ∴ f is one-one

But 13 in the codomain has no pre image, ∴f is not onto.

∴f has no inverse

f(x) = 3x+2; then

Question 8.

Choose the correct answer from the bracket;

If x≠1 and is a real function, then fof (2) = _____(1, 2, 3, 4)

i. What is the inverse of f

ii. Find f(3) + f^{-1}(3)

Answer.

i. 2

ii. Let g: image of f→R-{1} be the inverse f

Let y be any aribtrary element in the range of f then y = f(x)

⇒ xy – y = x + 1

Question 9.

i. Determine whether the following is a binary operation or not? Justify a*b = 2^{a}b defined on Z

ii. Determine whether * is commutative or associative if

a*b = ,a,b∉R

Answer.

i. a*b = 2^{a}b

If a is negative, then 2^{a} becomes a fraction

Question 10.

Consider the binary operation * : Q → Q where Q is the set of rational numbers is defined as a*b=a+b-ab

i. Fine 2*3.

ii. Is identify for * exist? If yes, find the identity element.

iii. Are elements of Q invertible? If yes, find the inverse of an element in Q.

Answer.

i. 2*3 = 2+3-6 = 1

ii. a*e = a+e-ae = a (e-ae = 0 )

⇒ e(1-a) = 0

⇒ e = 0

⇒ e*a = a

⇒ e+a-ea = a

⇒ e-ea = 0

⇒ e = 0 is the identity element

iii. a*a^{-1 }= a+a^{-1 }– aa^{-1} = 0

iv.

**NCERT Questions & Answers**

Question 1.

State the reason for the relation R in the set {1,2,3} given by R ={(1,2), (2,1)} not to be transitive.

Answer.

The given relation R ={(1,2), (2,1)} is not transitive. Because for any relation to be transitive, if (a,b) ∈ R and (b,c) ∈ R

⇒ (a,c) ∈ R

Here (1,2) ∈ R and (2,1)∈R but (1,1)∉R.

Question 2.

Show that the function f defined as is one-one and onto. Hence find f^{-1}

Answer.

To show f is one-one:

Question 3.

Let R be a relation from Q to Q defined by R = {(a, b) : a,b ∈ Q and a – b∈Z}

Show that N

(i) (a,a) ∈R, ∀a∈Q

(ii) (a,b) ∈ R implies that (b, a) ∈ R

(iii) (a,b),(b,c) ∈ R implies that (a,c) ∈ R

(iv) What type of a relation is R?

Answer.

(i) Let a∈Q, Then a – a = 0∈Z,

(a,a) ∈ R, ∀a ∈ Q

(ii) Given (a,b) ∈ R ∴a – b∈R….( 1)

It follows from (1) that b – a∈Z

∴(b,a)∈R

(iii) Given (a,b),(b,c) ∈ R ∴a – b∈Z, and b – c∈Z

∴ a – b + b – c ∈ Z,i.e., a – c∈Z,

∴(a,c)∈R

(iv) R is an equivalence relation.

Question 4.

Let A = {1, 2, 3, 4, 5, 6, 7}. Which of the following two is a partition giving rise to an equivalence relation? Why?

(i) A_{1} = {1,3,5},A_{2} = {2}, A_{3} = {4,7}

(ii) B_{1} = {1, 2, 5, 7}, B_{2} = {3}, B_{3}= {4,6}

Answer.

(i)Since A≠A_{1}∪A_{2}∪A_{3}, therefore, sets A_{1}, A_{2}, A_{3} do not form a partition of A and hence not giving rise to an equivalence relation.

(ii) Since B_{1}∩B_{2} = φ, B_{1}∩B_{3} = φ, B_{2}∩B_{3} = φ and B = B_{1}∪B_{2}∪B_{3}, therefore, sets B_{1}, B_{2}, B_{3} form a partition of A and hence giving rise to an equivalence relation.

Question 5.

In each of the following cases, state whether the function is onto, one to one or bijective. Justify your answer.

i. If f: R→R defined by/(x) = 3 – 4x

ii.If f: R→R defined by/(x) = 1 + x²

iii.If f: N→N defined by

Answer.

i. f(x) = f(y)

⇒ 3 – 4x = 3 – 4y

⇒ x = y

Therefore f is one to one Given

such that

So f is onto. Hence f is a bijective function.

ii. f is not one to one and onto Take 1, -1 ∈ R.

Obviously 1 ≠ -1, but f(1) = f(-1) = 2.

So f is not one to one.

If f is onto, then 0 ∈ R implies that

three exists x ∈ R such that f (x) = 0

So 1 + x² = 0.

Therefore x² = -1, x ∈ R,

which is not true. Thus f is not onto,

iii. Take 3, 4 ∈ N. Obviously 3≠4, but

f (3) = f (4) = 2; So f is not one to one. Therefore f is not bijective.

Note that f(1) = 1, f(3) = 2,…,f

(2n – 1) = n and so on. So range of f = N.

Hence f is onto.

Question 6.

Let

and g(x) = [x]

Find (i) f o g;

ii. gof

iii. Prove that f o g and g o f coincides in (-1, 0).

Answer.

For any x ∈ (-1, 0)

i. (fog)(x) = f(g (x)) = f([x]) = f(-1) = -1

ii. (gof)(x) = g (f(x)) = g (-1) = -1

iii. from (i) & (ii) gof = fog.

Question 7.

Prove that the Greatest Integer Function f:R→R, given by f(x)=[x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x .

Answer.

One-One:- Let x_{1} = 2.1, x_{2} = 2.5

f(x_{1}) = f(2.1) = [2.1] = 2

f(x_{2}) = f(2.5) = [2.5] = 2

i.e., f(x_{1}) = f(x_{2}) for x_{1} ≠ x_{2}

∴ f is not one-one

Onto: Let y = 2.5 ∈ R, the co-domain of f

f(x) = 2.5 ⇒[x] = 2.5, which is not possible. It is not onto

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