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

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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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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)
Let e be the identity element.
∴ a * e = a
$$\frac{a e}{4}$$ = a ⇒ e = 4.

Question 15.
a * b = ab ∀a,b ∈ Z. Show that × is not a binary operation on Z by giving a counter example.
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+.
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.
Let L be the set of all lines in a plane and R be the relation in L defined as R = {(L1, L2): L1 is perpendicular to L2}. R is symmetric only ∵ L1 ⊥ L2 ⇒ L1 ⊥ L2

Question 18.
Let * be a binary operation on N defined by a * b = LCM of a and b. Find 20 * 16.
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.
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.
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.
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 = 2ab. Prove that * is Commutative.
a * b = 2ab = 2ba = b * a v a,b ∈ N
∴ * is commutative.

Question 23.
Let * be a Binary operation on N defined by a * b = ab2. Show that * is not commutative by giving a counter example.
2 * 3 = 2.32 = 2.9 =18
3 * 2 = 3.22 = 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.
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.
(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.

∴ (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.
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.
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]
[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
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.
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

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. ,
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.
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.
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.

-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) = x2,∀ x ∈ N is injective but not surjective.
Domain of f = N Codomain of f = N
∀x1, x2 ∈ N, f (x1) = f(x2) ⇒ x12 = x22 ⇒ (x1 – x2 )(x1 + x2) ≠ 0
⇒x12 – x22 = 0 ⇒ x1 = x2 ⇒ f is one-one, x1, x2 ∈ N(x1 + x2) ≠ 0
Range of f = {12, 22, 32…….. } = {1, 4, 9} ≠ N (codomain) ∴ f is not onto.

Question 8.
Show that the function f : Z → Z defined by f(x) = x2 ∀ x ∈ Z is neither injective nor surjective.
∀x1, x2 ∈ Z, f(x1) = f(x2) ⇒ x12 = x22 ⇒ x2 = ±x1 ⇒ f is not one-one.
Range, of f = {02,(±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) = x2 ∀x∈R is neither injective nor subjective.

-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) = x3 ∀x ∈ N is injective but not surjective.
Let f(x) = f(y) ⇒ x3 = y3 ⇒ x = y ∵ x,y ∈ N ∴ f isone-one.
Range of f = {13,23,33,…..} = {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) = x3 ∀ x ∈ Z is injective but not surjective.
∀ x,y ∈ Z, f(x) = f(y) ⇒ x3 = y3 ⇒ x = y f is one-one.
Range of f = {03, (±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.
f(x1) = f(x2) ⇒ 2x1 = 2x2 ⇒ x1 = x2 ∴ 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.
f(x1) = f(x2) ⇒ 3 – 4x1 = 3 – 4x2 ⇒ x1 = x2 ⇒ 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 = a3 + b3, show that * is commutative but not assossiative.
Let a,b,c ∈ N
Given, a * b = a3 + b3 = b3 + a3 = b × a ⇒ *is commutative.
Now, (a * b) * c = (a3 + b3) × c = (a3 + b3)3 + c3
And a * (b * c) = a * (b3 + c3) = a3 + (b3 + c3)3
Since (a3 + b3)3 + c3 ≠ a3 + (b3 + c3)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.

### 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
Let x1, x2 ∈ A = R – {3} such that f(x1) = f(x2) ⇒ $$\frac{x_{1}-2}{x_{1}-3}=\frac{x_{2}-2}{x_{2}-3}$$
⇒ (x1 – 2) (x2 – 3) = (x1 – 3) (x2 – 2)
⇒ x1 x2 – 2x2 – 3x1 + 6
= x1x2 – 2x1 – 3x2 + 6
⇒ x2 = x1
∴ 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.
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.
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.
|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.
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)
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
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.
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 = IA and gof = IB, where, A = R- $$\left\{\frac{3}{5}\right\}$$,B = R – $$\left\{\frac{7}{5}\right\}$$; IA(X) = x, ∀ x ∈ A,IB(x) = x, ∀ X ∈ B are called identity functions on sets A and B, respectively

Thus, gof(x) = x, ∀ x ∈ B and fog(x) = x, ∀ x ∈ A, which implies that gof = IB and fog = IA

Question 10.
Show that if f : A → B and g : B → C are one-one, then gof : A → C is also one-one.
Suppose gof(x1) = gof(x2)
⇒ g(f(x1)) = g(f(x2))
⇒ f(x1) = f(X2), as g is one-one
⇒ x1 = x2, 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.
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.
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.
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-1og-1.
Proof: To show that gof is invertible with (gof)-1 = f-1og-1, it is enough to show that (f-1og-1)o(gof) = Ix and (gof)o(f-1og-1) = Iz
Now, (f-1og-1)o(gof) = ((f-1og-1)og) of, by Theorem 1
= ((f-1o(g-1og) of, by Theorem I
= ((f-1oIY) of, by definition of g-1
= Ix.
Similarly, it can be shown that (gof)o(f-1og-1) = Iz.

Question 15.
If R1 and R2 are equivalence relations in a set A, show that R1 ∩ R2 is also an equivalence relation.
Since R1 and R2 are equivalence relations, (a, a) ∈ R1, and (a, a) ∈ R2 ∀ a ∈ A. This implies that (a,a) ∈ R1 ∩ R2, ∀ a, showing R1 ∩ R2 is reflexive. Further, (a,b) ∈ R1 ∩ R2 ⇒ (a,b) ∈ R1 and (a,b) ∈ R2 ⇒ (b,a) ∈ R2 and (b,a) ∈ R2 ⇒ (b,a) ∈ R1 ∩ R2, hence R1 ∩ R2 is symmetric. Similarly, (a, b) ∈ R1 ∩ R2 and (b, c) ∈ R1 ∩ R2 ⇒ (a, c) ∈ R1 and (a, c) ∈ R2 ⇒ (a, c) ∈ R1 ∩ R2. This shows that R1 ∩ R2 is transitive. Thus, R1 ∩ R2 is and equivalence relation.

Question 16.
Prove that the relation R defined on the set of real numbers R as
R = {(a,b) : a ≤ b2 ∀ a,b ∈ R} is neither reflexive nor symmetric nor transitive.
$$\left(\frac{1}{2}, \frac{1}{2}\right)$$ ∉ R since $$\frac{1}{2} \leq\left(\frac{1}{2}\right)^{2}$$ is not true .
∴ R is not reflexive.
(1,4) ∈ R ∵ 1 ≤ 42 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) = x2, where Y = {y : y = x2, x ∈ N} is invertible. Also write the inverse off (x). (June 2014)
Let f(x1) = f(x2) ⇒ x12 = x22 ⇒ x1 = x2(∵ x1, x2 ∈ N) ∴ f is one – one.
Domain of f = N, codomain of f = Y
Range of f = {12, 22, 32, ……..} {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 x2, for some X ∈ N x ⇒ √y
This gives a function g : Y → N, defined by g(y) = √y
Now gof (x) = g(x2) = $$\sqrt{x^{2}}$$ = x and fog(y) = $$f(\sqrt{y})=(\sqrt{y})^{2}$$ = y.
⇒ gof = IN and fog = Iy Hence, f is invertible with f-1 = g.

Question 2.
Let f : N → R be defined by f(x) = 4x2 + 12x + 15,show that f : N → S, where S is the range of the function, is invertible. Also find the inverse.
Let f(x1) = f(x2)
⇒ 4x12 + 12x1 + 15
⇒ 4x22 + 12x2 + 15
⇒ 4(x12 – x22) + 12(x1 – x2) = 0
⇒ 4(x1 – x2)(x1 + x2) + 12(x1 – x2) = 0
⇒ (x1 – x2) [(x1 + x2) + 12] = 0
⇒ (x1 – x2) = 0 (∵ 4((x1 + x2) +12 ≠ 0)
⇒ x1 = x2 ∴ 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 = 4x2 + 12x + 15, for some x in
N ⇒ y(2x)2 + 2.2x.3 + 32 + 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)
Let f (x1) = f (x2)
⇒ 4x1 + 3 = 4x2 + 3
⇒ 4x1 = 4x2
⇒ x1 = x2
∴ 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) = x2 + 4 is invertible. Also write the inverse off. (March 2015)

Question 5.
If R+ is the set of all non-negative real numbers prove that the f : R+ → [-5,∞) defined byf(x) = 9x2 + 6x – 5 is invertible. Write also f-1 (x).

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.
To Prove that f is one – one
f(x1) = f(x2)
$$\frac{4 x_{1}+3}{6 x_{1}-4}=\frac{4 x_{2}+3}{6 x_{2}-4}$$
(4x1 + 3)(6x2 – 4) = (4x2 + 3)(6x1 – 4)
24x1x2 – 16x1 + 18x2 – 12 = 24x1x2 – 16x2 + 18x1 – 12
-18x1 – 16x1 = -16x2 – 18x2
– 34x1 = – 34x2
∴ x1 = x2
∴ 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)