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## Karnataka 2nd PUC Maths Question Bank Chapter 1 Relations and Functions

### 2nd PUC Maths Relations and Functions One Mark Questions and Answers

Question 1.

Give an example of a relation which is symmetric and transitive but not reflexive.

Answer:

Let A = {a,b,c} Define a relation R on A by R = {(a,b),(b,a),(a,a),(b,b)} R is a relation which is symmetric and transitive but not reflexive because (c,c) ∉ R.

Question 2.

Give an example of a relation which is symmetric but neither reflexive nor transiitive.

Answer:

Let A= {a,b,c} Define a relation R on A by R = {(a,b),(b,a)} R is a relation which is symmetric but neither reflexive nor transitive because (a, a) ∉ R.

Question 3.

Give an example of a relation which is transitive but neither reflexive nor symmetric.

Answer:

Let A = {a,b,c} Define a relation R on A by R = {(a,b),(b,c),(a,c)} R is a relation which is transitive but neither reflexive nor symmetric because (a,a) ∉ R and (b,a) ∉ R.

Question 4.

Give an example of a relation which is reflexive and symmetric but not transitive.

Answer:

Let A = {a,b,c} Define a relation R on A by R = {(a,a),(b,b),(c,c)(b,c),(c,b),(a,b),(b,a)} R is a relation which is reflexive and symmetric but not transitive because (c, b) ∈ R and (b, a) ∈ R but (c,a) ∉ R.

Question 5.

Give an example of a relation which is reflexive and transitive but not symmetric.

Answer:

Let A = {a,b,c} Define a relation R on A by R = {(a,a),(b,b),(c,c)(b,c),(c,b),(a,b)} R is a relation which is reflexive and transitive but not symmetric because (b, a) ∉ R.

Question 6.

Define an equivalence relation.

Answer:

A relation R on a set A is said to be an equivalence relation if R is reflexive, symmetric and transitive.

Question 7.

Define a reflexive relation.

Answer:

A relation R on a set A is said to be reflexive if (a, a) ∈ R, for every a ∈ A.

Question 8.

Define a symmetric relation.

Answer:

A relation R on a set A is said to be symmetric if (a, b) ∈ R implies (b, a) ∈ R, a,b ∈ A.

Question 9.

Define a transitive relation.

Answer:

A relation R on a set A is said to be transitive if (a,b) ∈ R and (b,c) ∈ R implies (a,c) ∈ R a,b,c ∈ A.

Question 10.

Define an injective function.

Answer:

A function f : A → B is said to be one – one (injective) if distinct elements of A have distinct images in B. i. e. for every x, y ∈ A, f(x) = f(y) implies x = y.

Question 11.

Define a surjective function.

Answer:

A function f : A → B is said to be onto (surjective) if every element of B is the image of some element of A under f. i. e. for every y ∈ B, there exists an element x in A such that f(x) = y.

Question 12.

Define a bijective function.

Answer:

A function f : A → B is said to be bijective if it is both one – one and onto.

Question 13.

If f : {1,3,4} → {1,2,5} and g: {1,2,5} → {1,3} are two functions given by f = {(1,2), (3,5), (4,1)} and g = {(1,3), (2,3), (5,1)}. Write down gof.

Answer:

gof (1) = g (f (1) = g (2) = 3, gof (3) = gof (3)) = g (5) = 1

gof (4) = g (f (4)) = g (1) = 3, gof= {(1, 3) (3, 1), (4, 3)}

Question 14.

Let be a binary operation defined on the set of non-zero rational numbers, by a * b = \(\frac{a b}{4}\). Find the identity element. (March 2015)

Answer:

Let e be the identity element.

∴ a * e = a

\(\frac{a e}{4}\) = a ⇒ e = 4.

Question 15.

a * b = a^{b} ∀a,b ∈ Z. Show that × is not a binary operation on Z by giving a counter example.

Answer:

Taking a = 2, b = – 3 we get a * b = 2 * -3 = 2^{-3} = \(\frac { 1 }{ 8 }\) ∉ Z

∴ * is not a binary operation on Z.

Question 16.

The operation defined by a * b = a. Is * a binary operation on z^{+}.

Answer:

a * b = a ∈ z^{+} ∀ a,b ∈ z^{+}

∴ * is a binary operation on z^{+}

Question 17.

Give an example of a relation which is symmetric only.

Answer:

Let L be the set of all lines in a plane and R be the relation in L defined as R = {(L_{1}, L_{2}): L_{1} is perpendicular to L_{2}}. R is symmetric only ∵ L_{1} ⊥ L_{2} ⇒ L_{1} ⊥ L_{2}

Question 18.

Let * be a binary operation on N defined by a * b = LCM of a and b. Find 20 * 16.

Answer:

20 * 16 = LCM of 20 and 16 = 80

Question 19.

Is * defined on the set A = {1,2,3,4,5} by a * b = LCM of a and b. Binary operation on A. Justify your answer.

Answer:

2 * 3 = LCM of 2 and 3 – 6 ∉ A

∴ * is not a binary operation on A.

Question 20.

Let * be defined on the set on A = {1,2,3,4,5} by a * b = HCF of a and b. Is * a binary operation on A.

Answer:

Yes ∵ a * b = HCF of a and b ∈ A ∀ a,b ∈ A

∴ * is a binary operation on A

Question 21.

Let * be a Binary operation on the set Q of Rational numbers by a * b = (a – b)^{2}. Prove that is commutative.

Answer:

a * b = (a – b)^{2} = (b – a)^{2} = b * a ∈ Q a,b ∈ Q

Question 22.

Let * be a Binary operation defined on N by a * b = 2^{ab}. Prove that * is Commutative.

Answer:

a * b = 2^{ab} = 2^{ba} = b * a v a,b ∈ N

∴ * is commutative.

Question 23.

Let * be a Binary operation on N defined by a * b = ab^{2}. Show that * is not commutative by giving a counter example.

Answer:

2 * 3 = 2.3^{2} = 2.9 =18

3 * 2 = 3.2^{2} = 3.4 = 12

∴ 2 * 3 ≠ 3 * 2

∴ * is not commutative

Question 24.

Show that * : R × R → R defined by a * b = a + 2b is not commutative in R.

Answer:

3 * 4 = 3 + 2 (4) = 3 + 8 = 11

4 * 3 = 4 + 2(3) = 4 + 6 = 10

3 * 4 ≠ 4 * 3

∴ * is not commutative

Question 25.

Show that * :R × R → R defined by a * b = a + 2b is not associative in R.

Answer:

(8 * 5) * 3 = (8 + 10) * 3 = 18 * 3 = 18 + 2(3) = 24 .

8 * (5 * 3) = 8 * (5 + 2(3)) = 8 * (5 + 6)

= 8 * 11 = 8 + 2(11)

= 8 + 22

= 30 * 5) * 3) ≠ 8 * (5 * 3)

∴ * is not associative.

Question 26.

Show that the binary operation * defmed on R by a * b = \(\frac{\mathbf{a}+\mathbf{b}}{\mathbf{2}}\) is not associative by giving a counter example.

Answer:

∴ (1 * (2 * 3) ≠ (1 * 2) * 3)

∴ * is not associative.

Question 27.

Show that the operation * defined on R by a * b = 1 ∀ a, b ∈ R is associative.

Answer:

a * (b * c) = a * 1 = 1

(a * b) * c = 1 * c = 1

a * (b * c) = (a * b) * c

∴ * is associative

Question 28.

If f = {(5,2), (6,3)}, g = {(2,5), (3,6)} Write fog.

Answer:

fog = {(2,2), (3,3)}

Question 29.

Let R be the equivalence relation on z defined by R = {(a,b): 2 divides a – b}. Write the equivalence class [0]

Answer:

[0] = {0, ± 2, ± 4, ± 6,…}

Question 30.

Let f: R → R be a function defined by f(x) = 4x – 3 ∀ x ∈ R. Then Write f^{-1}

Answer:

Let f^{-1}(x) – y ⇒ f(y) = x

⇒ 4(y) – 3 = x

⇒ 4(y) = x + 3.

⇒ y = \(\frac{x+3}{4}\)

⇒ f^{-1} = \(\frac{x+3}{4}\)

### 2nd PUC Maths Relations and Functions Two Marks Questions and Answers

Question 1.

Define binary operation on a set. Verify whether the operation * defined on Z, by a × b = ab + 1 is binary or not.

Answer:

A binary operation * on a set A is defined as a function *: A × A → A.

a * b = ab + 1 ∈ Z, ∀ a, b ∈ Z

∴ * is a binary operation on Z.

Question 2.

Show that the signum function f : R → R given by

Answer:

Domain of f is R, co-domain of f is R and Range of f = {-1, 0, 1}

f(2) = 1, f(3) = 1 ∴ f(2) = f(3) but 2 ≠ 3 ∴ f is not one-one.

Range of f ≠ codomain of f ∴ f is not onto

∴ f is neither one-one nor onto.

Question 3.

A relation R is defined on the set A = {1, 2, 3, 4, 5, 6} by R = {(x, y) : y is divisible by JC}. verify whether R is symmetric and reflexive or not. Give reason. ,

Answer:

R = {(1,1), (2,2), (3,3) (4,4), (5, 5), (6,6) (1,2), (1,3), (1,4), (1, 5), (1, 6), (2, 4), (2, 6), (3, 6)}. As (a, a) ∈ R ∀ a ∈ A ⇒ R is reflexive.

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

Let (a,b) and (b,c) ∈ R ⇒ b is divisible by a and c is divisible by b ⇒ c is divisible by a ⇒ (a, c) ∈ R ⇒ R is transitive.

Question 4.

Prove that the greatest integer function, f : R → R defined by f(x) = [x], where [x] indicates the greatest integer not greater than x, is neither one-one nor onto.

Answer:

Domain of f = R, co-domain of f = R and Range of f = Z

f(1.1) = 1, f(1.2) = 1 ∴ f(1.1) = f(1.2) but 1.1 ≠ 1.2 ∴ f is not one-one.

Range of f ≠ co-domain of f ∴ f is not onto.

∴ f is neither one-one nor onto.

Question 5.

Show that the function f : R → R, defined by f(x) = \(\frac{1}{x}\) is one – one and onto where R is the set of all non-zero real numbers.

Answer:

Let x,y ∈ R, such that f(x) = f(y) ⇒ \(\frac{1}{x}=\frac{1}{y}\) ⇒ x = y ∴ f is one – one.

Let y ∈ R (codomain) then there exists x ∈ R, (domain) such that f (x) = y

⇒ \(\frac{1}{x}\) = y or x = \(\frac{1}{y}\) ∴ \(f\left(\frac{1}{y}\right)\) = y ∀ y ∈ R. (co – domain) ∴ f is onto

f is both one-one and onto.

Question 6.

Show that the Modulus function f : R → R, given by f(x) = |x| is neither one-one nor onto.

Answer:

-1 ∈ R (codomain) then there does not exist x∈R (domain)

such that f(x) = – 1

In fact, all negative integers do not have pre-image

Range of f = [0,∞)

∴ Range off ≠ co-domain of f

∴ f is not onto

∴ f is neither one-one nor onto.

Question 7.

Show that the function f : N → N defined by f(x) = x^{2},∀ x ∈ N is injective but not surjective.

Answer:

Domain of f = N Codomain of f = N

∀x_{1}, x_{2 }∈ N, f (x_{1}) = f(x_{2}) ⇒ x_{1}^{2} = x_{2}^{2} ⇒ (x_{1} – x_{2} )(x_{1} + x_{2}) ≠ 0

⇒x_{1}^{2} – x_{2}^{2} = 0 ⇒ x_{1} = x_{2} ⇒ f is one-one, x_{1}, x_{2} ∈ N(x_{1} + x_{2}) ≠ 0

Range of f = {1^{2}, 2^{2}, 3^{2}…….. } = {1, 4, 9} ≠ N (codomain) ∴ f is not onto.

Question 8.

Show that the function f : Z → Z defined by f(x) = x^{2 }∀ x ∈ Z is neither injective nor surjective.

Answer:

∀x_{1}, x_{2} ∈ Z, f(x_{1}) = f(x_{2}) ⇒ x_{1}^{2} = x_{2}^{2} ⇒ x_{2} = ±x_{1} ⇒ f is not one-one.

Range, of f = {0^{2},(±1)^{2}, (±2)^{2}, ….} = {0,1,4…..} ≠ Z (codomain)

∴ f is not onto.

Question 9.

Show that the function f : R → R defined by f(x) = x^{2} ∀x∈R is neither injective nor subjective.

Answer:

-1 ∈ R (codomain) then there does not exist x ∈ R (domain)

such that f(x) = – 1

In fact, all negative integers do not have pre-image

Range of f = [0, ∞)

∴ Range of f ≠ co-domain of f

∴ f is not onto

∴ f is neither one-one nor onto.

Question 10.

Show that the function f : N → N defined by f(x) = x^{3} ∀x ∈ N is injective but not surjective.

Answer:

Let f(x) = f(y) ⇒ x^{3} = y^{3} ⇒ x = y ∵ x,y ∈ N ∴ f isone-one.

Range of f = {1^{3},2^{3},3^{3},…..} = {1, 8, 27,….} ≠ N (codomain)

∴ f is one-one but not onto.

Question 11.

Show that the function f : Z → Z defined by f(x) = x^{3 }∀ x ∈ Z is injective but not surjective.

Answer:

∀ x,y ∈ Z, f(x) = f(y) ⇒ x^{3} = y^{3} ⇒ x = y f is one-one.

Range of f = {0^{3}, (±1)^{3}, (±2)^{3}, ….} = {0, ±1, ±8, } ≠ Z (codomain)

∴ f is not onto.

Question 12.

Prove that the function f : R → R defined by f(x) = 2x, ∀x ∈ R is bijective.

Answer:

f(x_{1}) = f(x_{2}) ⇒ 2x_{1} = 2x_{2} ⇒ x_{1} = x_{2} ∴ f is one-one.

Given any real number y in R there exists \(\frac{y}{2}\) in R such that \(f\left(\frac{y}{2}\right)=2 \cdot\left(\frac{y}{2}\right)=y\) ∴ f is onto.

∴ f is bijective

Question 13.

Prove that the function f : R → R defined by f(x) = 3 – 4x, ∀ x ∈ R is bijective.

Answer:

f(x_{1}) = f(x_{2}) ⇒ 3 – 4x_{1} = 3 – 4x_{2} ⇒ x_{1} = x_{2} ⇒ f is one-one

Let y ∈ R, there exists x = \(\frac{3-y}{4}\) ∈ R such that f(x) = \(f\left(\frac{3-y}{4}\right)=3-4\left(\frac{3-y}{4}\right)\)

= 3 – 3 + y = y ⇒ f is onto. ∴ f is one-one and onto ⇒ f is bijective.

Question 14.

A binary operation * on N defined as a * b = a^{3} + b^{3}, show that * is commutative but not assossiative.

Answer:

Let a,b,c ∈ N

Given, a * b = a^{3} + b^{3} = b^{3} + a^{3} = b × a ⇒ *is commutative.

Now, (a * b) * c = (a^{3} + b^{3}) × c = (a^{3} + b^{3})^{3} + c^{3}

And a * (b * c) = a * (b^{3} + c^{3}) = a^{3} + (b^{3} + c^{3})^{3}

Since (a^{3} + b^{3})^{3} + c^{3} ≠ a^{3} + (b^{3} + c^{3})^{3} ⇒ (a × b) × c ≠ a × (b × c)

∴ * is commutative but not assossiative.

Question 15.

A binary operation * on N defined as a × b = \(\sqrt{a^{2}+b^{2}}\), show; that * is both commutative and associative.

Answer:

### 2nd PUC Maths Relations and Functions Three Marks Questions and Answers

Question 1.

Verify whether the function, f : A → B, where A = R – {3} and B = R – {1}, defined by f(x) = \(\frac{x-2}{x-3}\) is one – one and on to or not . Give reason

Answer:

Let x_{1}, x_{2} ∈ A = R – {3} such that f(x_{1}) = f(x_{2}) ⇒ \(\frac{x_{1}-2}{x_{1}-3}=\frac{x_{2}-2}{x_{2}-3}\)

⇒ (x_{1} – 2) (x_{2} – 3) = (x_{1} – 3) (x_{2} – 2)

⇒ x_{1} x_{2} – 2x_{2} – 3x_{1} + 6

= x_{1}x_{2} – 2x_{1} – 3x_{2} + 6

⇒ x_{2} = x_{1}

∴ f is one – one.

Question 2.

If * is a binary operation defined on A = N × N, by (a, b) * (c, d) = (a + c, b + d), prove that * is both commutative and associative. Find the identity if it exists.

Answer:

Let {a, b), (c, d), (e, f) be any three elements of A = N × N

(a, b) * {(c, d) * (e, j)} = {a, b) * (c + e, d + f)

= ( a + c + e, b + d + f) = (a + c, b + d) * (e, f) = {(a, b) * (c, d)} * (e, f)}

∴ (A, *) is associative.

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

∴ (A, *) is commutative.

Let (x, y) be the identity element in A

∴ (a, b) * (x, y) = (a, b) ⇒ (a + x, b + y) = (a, b) ⇒ a + x = a, b + y = b

⇒ x = 0, y = 0 and 0 ∉ N ⇒ (x, y) = (0,0) ∉ A

∴ (A, *) does not have an identity element.

Question 3.

Show that the relation R in the set of all integers, Z defined by R = {(a, b) : 2 divides a – b} is an equivalence relation.

Answer:

2 divides 0 ⇒ 2 divides a – a ⇒ (a, a) ∈ R ∀ a ∈ Z ∴ R is reflexive.

Let (a,b) ∈ R ⇒ 2 divides a – b ⇒ 2 divides – (a – b)

⇒ 2 divides b – a ⇒ (b, a) ∈ R

∴ R is symmetric.

Let (a,b)) ∈ R, (b,c) ∈ R

⇒ 2 divides a – b and 2 divides b – c

⇒ 2 divides (a – b) + (b-c)

⇒ 2 divides a – c ⇒ (a, c) ∈ R is transitive.

∴ R is an equivalence relation.

Question 4.

Show that the relation R in the set of all natural numbers, N defined by is an R = {(a, b) : |a – b| is even} is an equivalence relation.

Answer:

|a – a| = |0| = 0 is even ∴ (a, a) ∈ R ∴ R is reflexive.

Let (a,b) ∈ R ⇒ |a – b| is even ⇒ |-(a – b)| is even ⇒ |(b – a)| is even

⇒ (b,a) ∈ R ∴ R is symmetric.

Let (a,b) ∈ R and (b,c) ∈ R ⇒ | a – b| is even and |b – c| is even

⇒ a – b is even ⇒ b – c is even. ⇒ a – b + b – c is even

⇒ a – c is even ⇒ |a – c| is even. ∴ (a,c) ∈ R. ∴ R is transitive

∴ R is an equivalence relation.

Question 5.

Show that the relation R in| the set A = {x ∈ Z : 0 ≤ x ≤ 12} given by R = {{a, b) : |a – b| is a multiple of 4} is an equivalence relation.

Answer:

A = {0, 1, 2, 3, 4, 5, 6 ,7, 8, 9, 10, 11, 12}

|a – a| = 0 divisible by 4 ⇒ (a,a) ∈ R ∀ a ∈ A ⇒ R is reflexive.

(a, b) ∈ R ⇒ | a – b | is divisible by 4 ⇒ |-(b -a)| is divisible by 4

⇒ | b – a| is divisible by 4 ⇒ (b, a) ∈ R ⇒ R is symmetric.

Let(a,b) and (b, c) ∈ R ⇒ |a – b| and |b – c| are divisible by 4.

⇒ a – b and b – c are divisible by 4 ⇒ (a – b) + (b – c) is divisible by 4

⇒ a – c is divisible by 4 ⇒ |a – c| is divisible by 4 ⇒ (a,c) ⇒ R

⇒ R is transitive.

Hence R is an equivalence relation.

Question 6.

Let R be relation on the set A = {1,2,3,… 14} by R = {(x, y): 3x – y = 0}. Verify R is reflexive symmetric and transitive. (March 2015)

Answer:

R = {(1,3), (2,6), (3,9), (4,12)}

(1,1) ∉ R

∴ R is not reflexive (1,3) ∉ R But (3,1) ∉ R

∴ R is not symmetric (1,3) ∉ R, (3,9) ∉ R But (1,9) ∉ R

∴ R is not transitive.

Question 7.

Relation R on Z defined as R = {(x,y): x – y is an integer}. Show that R is an equivalence relation

Answer:

O is an integer

x – x is an integer

∴ (x, x) ∈ R ∀ x ∈ z

∴ R is reflexive

Let (x, y) ∈ R ⇒ x – y is an integer

⇒ -(x – y) is an integer

⇒ y – x is an integer

⇒ (y, x) ∈ R

∴ R is symmetric

Let (x, y) ∈ R and (y, z) ∈ R

⇒ x – y is an integer and y – z is an integer

⇒ (x – y) +(y – z) is an integer

⇒ x – z is an integer

⇒ (x, z) ∈ R

∴ R is transitive

∴ R is an equivalence relation

Question 8.

Show that the relation R in R defined by R = {(a, b): a ≤ b} is reflexive and transitive but not symmetric.

Answer:

a ≤ a is always true

∴ (a, a) R a R

∴ R is reflexive

Let (a, b) ∈ R ⇒ a ≤ b which does not imply b ≤a

∴ (b, a) ∉ R

∴ R is not symmetric

Let (a, b) ∈ R and (b, c) ∈ R

⇒ a ≤ b and b ≤ c

⇒ a ≤ c

⇒ (a, c) ∈ R

∴ R is transitive

Question 9.

Show that if f: R – \(\left\{\frac{7}{5}\right\}\) → R – \(\left\{\frac{3}{5}\right\}\) is defined by f(x) = \(\frac{3 x+4}{5 x-7}\) and g : R- \(\left\{\frac{3}{5}\right\}\) → R – \(\left\{\frac{7}{5}\right\}\) is defined by g(x) = \(\frac{7 x+4}{5 x-3}\) then fog = I_{A} and gof = I_{B}, where, A = R- \(\left\{\frac{3}{5}\right\}\),B = R – \(\left\{\frac{7}{5}\right\}\); I_{A}(X) = x, ∀ x ∈ A,I_{B}(x) = x, ∀ X ∈ B are called identity functions on sets A and B, respectively

Answer:

Thus, gof(x) = x, ∀ x ∈ B and fog(x) = x, ∀ x ∈ A, which implies that gof = I_{B} and fog = I_{A}

Question 10.

Show that if f : A → B and g : B → C are one-one, then gof : A → C is also one-one.

Answer:

Suppose gof(x_{1}) = gof(x_{2})

⇒ g(f(x_{1})) = g(f(x_{2}))

⇒ f(x_{1}) = f(X_{2}), as g is one-one

⇒ x_{1} = x_{2}, as f is one-one.

Hence, gof is one-one.

Question 11.

Show that iff: A → B and g: B → C are onto, then gof : A → C is also onto.

Answer:

Given an arbitrary element z – C, there exists a pre-image y of z under g such that g(y) = z i. e., g is onto. Further, for y ∈ B, there exists an element x in A with f(x) = y, since f is onto. Therefore, gof(x) = g(f(x)) = g(y) = z, showing that gof is onto.

Question 12.

Consider f: N → N,g: N → N and h: N → R defined as f(x) = 2x, g(y) = 3y+4 and h(z) = sin z, ∀ x, y and z in N. Show that ho(gof) = (hog) of.

Answer:

We have ,

ho(gof)(x) = h(gof(x)) = h(g(fx))) = h(g(2x))

= h(3(2x) + 4) = h(6x + 4) = sin(6x + 4) ∀ x ∈ N.

Also, ((hog)of)(x) = (hog) (f(x)) + (hog) (2x) h(g(2x))

= h(3(2x) + 4) = h(6x + 4) = sin(6x + 4), ∀ x ∈ N.

This shows that ho(gof) = (hog) of.

Question 13.

If f: X → y, g: Y → 3 z and h: Z → S are functions, then ho(gof) = (hog)of.

Answer:

Proof: We have

ho(gof) (x) = h(gofx)) = h(g(f(x))), ∀ x in X

and (hog) of (x) = hog(gx)) = h(g(f(x))), ∀ x ¡n X

Hence, ho(gof) = (hog) of.

Question 14.

Let f: X → Y and g: Y → Z be two invertible functions. Then gof is also invertible with (gof)^{-1} = f^{-1}og^{-1}.

Answer:

Proof: To show that gof is invertible with (gof)^{-1} = f^{-1}og^{-1}, it is enough to show that (f^{-1}og^{-1})o(gof) = I_{x} and (gof)o(f^{-1}og^{-1}) = I_{z}

Now, (f^{-1}og^{-1})o(gof) = ((f^{-1}og^{-1})og) of, by Theorem 1

= ((f^{-1}o(g^{-1}og) of, by Theorem I

= ((f^{-1}oI_{Y}) of, by definition of g^{-1}

= I_{x}.

Similarly, it can be shown that (gof)o(f^{-1}og^{-1}) = I_{z}.

Question 15.

If R_{1} and R_{2} are equivalence relations in a set A, show that R_{1} ∩ R_{2} is also an equivalence relation.

Answer:

Since R_{1} and R_{2} are equivalence relations, (a, a) ∈ R_{1}, and (a, a) ∈ R_{2} ∀ a ∈ A. This implies that (a,a) ∈ R_{1} ∩ R_{2}, ∀ a, showing R_{1} ∩ R_{2} is reflexive. Further, (a,b) ∈ R_{1} ∩ R_{2 }⇒ (a,b) ∈ R_{1} and (a,b) ∈ R_{2} ⇒ (b,a) ∈ R_{2} and (b,a) ∈ R_{2} ⇒ (b,a) ∈ R_{1} ∩ R_{2}, hence R_{1} ∩ R_{2} is symmetric. Similarly, (a, b) ∈ R_{1} ∩ R_{2} and (b, c) ∈ R_{1} ∩ R_{2} ⇒ (a, c) ∈ R_{1} and (a, c) ∈ R_{2} ⇒ (a, c) ∈ R_{1} ∩ R_{2}. This shows that R_{1} ∩ R_{2} is transitive. Thus, R_{1} ∩ R_{2} is and equivalence relation.

Question 16.

Prove that the relation R defined on the set of real numbers R as

R = {(a,b) : a ≤ b^{2} ∀ a,b ∈ R} is neither reflexive nor symmetric nor transitive.

Answer:

\(\left(\frac{1}{2}, \frac{1}{2}\right)\) ∉ R since \(\frac{1}{2} \leq\left(\frac{1}{2}\right)^{2}\) is not true .

∴ R is not reflexive.

(1,4) ∈ R ∵ 1 ≤ 4^{2} is true. ∴ R is not symmetric.

But(4,1) ∉ R. (3,2),(2,1.5) ∈ R But(3, 1.5) ∉ R. ∴R is not transitive.

### 2nd PUC Maths Relations and Functions Six Marks Questions and Answers

Question 1.

Prove that the function f : N → Ydefined by f(x) = x^{2}, where Y = {y : y = x^{2}, x ∈ N} is invertible. Also write the inverse off (x). (June 2014)

Answer:

Let f(x_{1}) = f(x_{2}) ⇒ x_{1}^{2} = x_{2}^{2} ⇒ x_{1} = x_{2}(∵ x_{1}, x_{2} ∈ N) ∴ f is one – one.

Domain of f = N, codomain of f = Y

Range of f = {1^{2}, 2^{2}, 3^{2}, ……..} {1, 4,9 ………..} codornain of f

∴ f is onto. ∴ f is bijective. ∴ f^{-1} exists.

An arbitrary element y in Y is of the form x^{2}, for some X ∈ N x ⇒ √y

This gives a function g : Y → N, defined by g(y) = √y

Now gof (x) = g(x^{2}) = \(\sqrt{x^{2}}\) = x and fog(y) = \(f(\sqrt{y})=(\sqrt{y})^{2}\) = y.

⇒ gof = I_{N} and fog = I_{y} Hence, f is invertible with f^{-1} = g.

Question 2.

Let f : N → R be defined by f(x) = 4x^{2} + 12x + 15,show that f : N → S, where S is the range of the function, is invertible. Also find the inverse.

Answer:

Let f(x_{1}) = f(x_{2})

⇒ 4x_{1}^{2} + 12x_{1} + 15

⇒ 4x_{2}^{2} + 12x_{2} + 15

⇒ 4(x_{1}^{2} – x_{2}^{2}) + 12(x_{1} – x_{2}) = 0

⇒ 4(x_{1} – x_{2})(x_{1} + x_{2}) + 12(x_{1} – x_{2}) = 0

⇒ (x_{1} – x_{2}) [(x_{1} + x_{2}) + 12] = 0

⇒ (x_{1} – x_{2}) = 0 (∵ 4((x_{1} + x_{2}) +12 ≠ 0)

⇒ x_{1} = x_{2} ∴ f is one-one

For the function f : N → S, codomain of f = S = Range of f (given)

∴ f is onto. ∴ f is bijective. ∴ f^{-1} exists.

Let y be an arbitrary element of S. Then y = 4x^{2} + 12x + 15, for some x in

N ⇒ y(2x)^{2} + 2.2x.3 + 3^{2} + 6 = (2x + 3)^{2} + 6. = (2x+3)^{2} = y – ó

Question 3.

Verify whether the function f : N → Y defined by f (x) = 4x + 3, where Y = {y : y = 4x + 3, x ∈ N} is invertible or not. Write the inverse of f(x) if exists. (March 2014)

Answer:

Let f (x_{1}) = f (x_{2})

⇒ 4x_{1} + 3 = 4x_{2} + 3

⇒ 4x_{1} = 4x_{2}

⇒ x_{1} = x_{2}

∴ f is one-one.

Let y ∈ Y ∃ X ∈ N such that y = 4x + 3 ∴ f is onto.

∴ f is bijective.

Consider an arbitrary element y of Y. By the definition of Y, y = 4x +3, for some x in the

Question 4.

Let R+ be the set of all non-negative real numbers. Show that the function f : R_{+} → [4,∞) defined by f(x) = x^{2} + 4 is invertible. Also write the inverse off. (March 2015)

Answer:

Question 5.

If R_{+} is the set of all non-negative real numbers prove that the f : R_{+} → [-5,∞) defined byf(x) = 9x^{2} + 6x – 5 is invertible. Write also f^{-1} (x).

Answer:

Question 6.

If A → A where A = R – \(\left\{\frac{2}{3}\right\}\) defined by f(x) = \(\frac{4 x+3}{6 x-4}\) is invertible. Prove that f ‘f.

Answer:

To Prove that f is one – one

f(x_{1}) = f(x_{2})

\(\frac{4 x_{1}+3}{6 x_{1}-4}=\frac{4 x_{2}+3}{6 x_{2}-4}\)

(4x_{1} + 3)(6x_{2} – 4) = (4x_{2} + 3)(6x_{1} – 4)

24x_{1}x_{2} – 16x_{1} + 18x_{2} – 12 = 24x_{1}x_{2} – 16x_{2} + 18x_{1} – 12

-18x_{1} – 16x_{1} = -16x_{2} – 18x_{2}

– 34x_{1} = – 34x_{2}

∴ x_{1} = x_{2}

∴ f is one-one

To prove that f is onto.

Let y ∈ A ∃ x ∈ A such that f(x) = y

\(\frac{4 x+3}{6 x-4}=y\)

4x + 3 = 6xy – 4y

4x – 6xy = -4y – 3

6xy – 4x = 4y + 3

x(6y – 4) = 4y + 3

x = \(\frac{4 y+3}{6 y-4}\) ∈ A

∴ f is onto.

∴ f is invertible.

Question 7.

f: R → R be defmed as f(x) = 4x + 5 ∀ X ∈ R show that fis invertible and find f^{-1} (June 2015)

Answer:

To prove f is one-one

Let f(x_{1}) = f(x_{2})

⇒ 4x_{1} + 5 = 4x_{2} + 5 ,

⇒ 4x_{1} = 4x_{2}

⇒ x_{1} = x_{2}

∴ f is one-one.

To prove f is onto

Let y ∈ R there exists X ∈ R such that

∴ f is onto ∴ f is bijective ∴ f is invertible