NCERT Class 12 Mathematics Chapter 1 Relations and Functions

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NCERT Class 12 Mathematics Chapter 1 Relations and Functions

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Relations and Functions

Chapter – 1

Exercise 1.1

1. Determine whether each of the following relations are reflexive, symmetric and transitive:

(i) Relation R in the set A = {1, 2, 3, …, 13, 14} defined as

R = {(x, y) : 3x – y = 0}

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

R is not reflexive because (1, 1), (2, 2) and (14, 14) ∉ R

R is not symmetric because (1, 3) ∈ R, but (3,1) ∉ R.[since 3(3)≠0].

R is not transitive because (1, 3), (3, 9) ∈ R, but (1, 9) ∉ R.[3(1)-9 ≠ 0].

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Hence, R is neither reflexive nor symmetric nor transitive.

(ii) Relation R in the set N of natural numbers defined as

R = {(x, y) : y = x + 5 and x < 4 }

Ans: R = {(1, 6), (2, 7), (3, 8)}

R is not reflexive because (1, 1) ∉ R

R is not symmetric because (1, 6) ∈ R but (6, 1) ∉ R

R is not transitive because there isn’t any ordered pair in R such that

(x, y), (y, z) ∈ R, so (x, z) ∉ R

Hence, R is neither reflexive nor symmetric nor transitive.

(iii) Relation R in the set A = {1, 2, 3, 4, 5, 6} defined as R = {(x, y): y is divisible by x)

Ans: R = {(x, y) : y is divisible by x}

We any number other than 0 is divisible by itself.

Thus, (x, x) ∈ R

So, R is reflexive.

(2, 4) ∈ R [because 4 is divisible by 2]

But (4,2) ∉ R [since 2 is not divisible by 4]

So, R is not symmetric.

Let (x,y) and (y, z) ∈ R So, y is divisible by x and z is divisible by y.

So, z is divisible by x ⇒ (x, z) ∈ R

So, R is transitive.

So, R is reflexive and transitive but not symmetric.

(iv) Relation R in the set Z of all integers defined as

R = {(x, y) : x – y is an integer}

Ans: R = {(x, y) : x – y is an integer}

For x ∈ Z,(x, x) ∉ R because x – x = 0 is an integer.

So, R is reflexive.

For, x, y ∈ Z , if x, y ∈ R, then x – y is an integer ⇒ (y – x) is an integer.

So, (y, x) ∈ R

So, R is symmetric.

Let (x,y) and (y, z) ∈ R where x, y, z ∈ Z

⇒ (x – y) and (y – z) are integers.

⇒ x – z = (x – y) + (y – z) is an integer.

So, R is transitive.

So, R is reflexive, symmetric and transitive.

(v) Relation R in the set A of human beings in a town at a particular time given by:

(a) R = {(x,y) : x and y work at the same place}

Ans: R = {(x, y) : x and y work at the same place}

R is reflexive because (x, x) ∈ R

R is symmetric because,

If (x, y) ∈ R, then x and y work at the same place and y and x also work at the same place. (y, x) ∈ R

R is transitive because,

Let (x, y), (y, z) ∈ R

x and y work at the same place and y and z work at the same place.

Then, x and z also works at the same place. (x, z) ∈ R

Hence, R is reflexive, symmetric and transitive.

(b) R = {(x, y) : x and y live in the same locality}

Ans: R = {(x, y) : x and y live in the same locality}

R is reflexive because (x, x) ∈ R

R is symmetric because,

If(x, y) ∈ R, then x and y live in the same locality and y and x also live in the same locality (y, x) ∈ R

R is transitive because,

Let (x,y), (y.z) ∈ R

x and y live in the same locality and y and z live in the same locality.

Then x and z also live in the same locality. (x,z) ∈ R.

Hence, R is reflexive, symmetric and transitive.

(c) R = {(x, y) : x is exactly 7 cm taller than y}

Ans: R = {(x,y) : x is exactly 7 cm taller than y}

R is not reflexive because (x,x) ∉ R

R is not symmetric because,

If (x,y) ∈ R, then x is exactly 7 cm taller than y and y is clearly not taller than x (y,x) ∉ R.

R is not transitive because,

Let (x,y), (y,z) ∈ R.

x is exactly 7cm taller than y and y is exactly 7 cm taller thanᶻ.

Then x is exactly 14 cm taller thanᶻ (x,z) ∉ R

Hence, R is neither reflexive nor symmetric nor transitive.

(d) R = {(x, y) : x is wife of y}

Ans: R = {(x,y) : x is wife of y}

R is not reflexive because (x, x) ∉ R

R is not symmetric because,

Let (x, y) ∈ R, x is the wife of y and y is not the wife of x. (y,x) ∉ R.

R is not transitive because,

Let (x, y), (y, z) ∈ R.

x is wife of y and y is wife of z, which is not possible.

(x, z) ∉ R

Hence, R is neither reflexive nor symmetric nor transitive.

(e) R = {(x, y): x is father of y}

Ans: R = {(x,y): x is father of y}

R is not reflexive because (x, x) ∉ R.

R is not symmetric because,

Let (x, y) ∈ R, x is the father of y and y is not the father of x. (y, x) ∉ R.

R is not transitive because,

Let (x, y), (y, z) ∈ R.

x is father of y and y is father of z, x is not father of z. (x, z) ∉ R.

Hence, R is neither reflexive nor symmetric nor transitive.

2. Show that the relation R in the set R of real numbers, defined as R = {(a, b) : a ≤ b²} is neither reflexive nor symmetric nor transitive.

Ans: R = {(a, b) : a ≤ b²}

R is not reflexive.

(1, 4) ∈ R as 1<4. But 4 is not less than 1²

(4, 1) ∉ R

R is not symmetric.

(3, 2)(2, 1.5) ∈ R [Because 3 < 2² = 4 and 2 < (1.5)² = 2.25 ]

3 > (1.5)² = 2.25

∴ (3, 1.5) ∉ R

R is not transitive.

R is neither reflective nor symmetric nor transitive.

3. Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as R = {(a, b) : b = a + 1} is reflexive, symmetric or transitive.

Ans: A = {1, 2, 3, 4, 5, 6}

R = {(a, b) : b = a + 1}

R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}

(a,a) ∉ R, a ∈ A

(1, 1), (2, 2), (3, 3), (4, 4), (5, 5) ∉ R

R is not reflexive.

(1,2) ∈ R, but (2,1) ∉ R

R is not symmetric.

(1, 2), (2, 3) ∈ R

(1, 3) ∉ R

R is not transitive.

R is neither reflective nor symmetric nor transitive.

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

Ans: R = {(a,b): a ≤ b}

(a,a) ∈ R

R is reflexive.

(2,4) ∈ R (as 2 < 4)

(4,2) ∉ R (as 4 > 2)

R is not symmetric.

(a,b),(b,c) ∈ R

a ≤ b and b ≤ c

⇒ a ≤ c

⇒ (a,c) ∈ R

R is transitive.

R is reflexive and transitive but not symmetric.

5. Check whether the relation R in R defined by R = {(a, b) : a ≤ b³} is reflexive, symmetric or transitive.

Ans: R = {(a, b) : a ≤ b³}

R is not reflexive.

(1, 2) ∈ R(as 1 < 2³ = 8)

(2, 1) ∉ R(as 2³ > 1 = 8)

R is not symmetric.

R is not transitive.

R is neither reflexive nor symmetric nor transitive.

6. Show that the relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} is symmetric but neither reflexive nor transitive.

Ans: A = {1, 2, 3}

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

(1, 1), (2, 2), (3, 3) ∉ R

R is not reflexive.

(1, 2) ∈ R and (2, 1) ∈ R

R is symmetric.

(1, 2) ∈ R and (2, 1) ∈ R

(1, 1) ∈ R

R is not transitive.

R is symmetric, but not reflexive or transitive.

7. Show that the relation R in the set A of all the books in a library of a college, given by R = {(x, y) : x and y have same number of pages} is an equivalence relation.

Ans: R = {(x, y) : x and y have same number of pages}

R is reflexive since (x, x) ∈ R as x and x have same number of pages.

R is reflexive.

(x, y) ∈ R

x and y have same number of pages and Y and x have same number of pages (y, x) ∈ R R is symmetric.

(x, y) ∈ R, (y, z) ∈ R

x and y have same number of pages, y and z have same number of pages. Then x and z have same number of pages.

(x, z) ∈ R

R is transitive.

R is an equivalence relation.

8. Show that the relation R in the set A = {1, 2, 3, 4, 5} given by

R = {(a, b) : |a – b| is even}, is an equivalence relation. Show that all the elements of {1, 3, 5} are related to each other and all the elements of {2, 4} are related to each other. But no element of {1, 3, 5} is related to any element of {2,4}.

Ans: a ∈ A

|a – a| =0( which is even)

R is reflective.

(a, b) ∈ R

⇒ |a – b| [is even]

⇒ |- (a – b)| = |b – a| [is even]

(b, a) ∈ R

R is symmetric.

(a, b) ∈ R and (b, c) ∈ R

⇒ |a – b| is even and |b – c| is even

⇒ (a – b) is even and (b – c) is even

⇒ (a – c) = (a + b) + (b – c) is even

⇒ |a – b| is even

⇒ (a, c) ∈ R

R is transitive.

R is an equivalence relation.

All elements of {1, 3, 5} are related to each other because they are all odd. So, the modulus of the difference between any two elements is even.

Similarly, all elements {2, 4} are related to each other because they are all even.

No element of {1, 3, 5} is related to any elements of {2, 4} as all elements of {1, 3, 5} are odd and all elements of {2, 4} are even. So, the modulus of the difference between the two elements will not be even.

9. Show that each of the relation R in the set A = {x ∈ Z : 0 ≤ x ≤ 12} , given by is an equivalence relation. Find the set of all elements related to 1 in each case.

(i) R = {(a, b) : |a – b| is a multiple of 4}

Ans: R= {(a, b) : |a – b| is a multiple of 4}

a ∈ A, (a, a) ∈ R [|a – a| = 0 is a multiple of 4]

R is reflexive.

(a,b) ∈ R ⇒ |a – b| [is a multiple of 4]

⇒ |- (a – b)| = |b – a| [is a multiple of 4]

(b, a) ∈ R

R is symmetric.

(a, b) ∈ R and (b, c) ∈ R

⇒ |a – b| a multiple of 4 and |b – c| is a multiple of 4

⇒ (a – b) is a multiple of 4 and (b – c) is a multiple of 4

⇒ (a – c) = (a – b) + (b – c) is a multiple of 4

⇒ |a – c| is a multiple of 4

⇒ (a, c) ∈ R

R is transitive.

R is an equivalence relation.

The set of elements related to 1 is {1, 5, 9} as

|1 – 1| = 0 is a multiple of 4

|5 – 1| = 4 is a multiple of 4

|9 – 1| = 8 is a multiple of 4

(ii) R = {(a, b) : a = b}

Ans: R = {(a, b) : a = b}

a ∈ A, (a, a) ∈ R. [since a = a]

R is reflective.

(a, b) ∈ R

⇒ a = b

⇒ b = a

⇒ (b, a) ∈ R

R is symmetric.

(a, b) ∈ R and (b,c) ∈ R

⇒ a = b and b = c

⇒ a = c

⇒ (a, c) ∈ R

R is transitive.

R is an equivalence relation.

The set of elements related to 1 is {1}.

10. Give an example of a relation. Which is:

(i) Symmetric but neither reflexive nor transitive.

Ans: A = {5, 6, 7}

R = {(5, 6), (6, 5)}

(5, 5), (6, 6), (7, 7) ∉ R

R is not reflexive as (5, 5), (6, 6), (7, 7) ∉ R

(5, 6), (6, 5) ∈ R (6, 5) ∈ R, R is symmetric.

⇒ (5,6), (6, 5) ∈ R, but (5, 5) ∉ R

R is not transitive.

Relation R is symmetric but not reflexive or transitive.

(ii) Transitive but neither reflexive nor symmetric.

Ans: R = {(a, b) : a < b}

a ∈ R, (a, a) ∉ R [since a cannot be less than itself]

R is not reflexive.

(1, 2) ∈ R (as1 < 2)

But 2 is not less than 1

∴ (2, 1) ∉ R

R is not symmetric.

(a, b), (b, c) ∈ R

⇒ a < b and b < c

⇒ a < c

⇒ (a, c) ∈ R

R is transitive.

Relation R is transitive but not reflexive and symmetric.

(iii) Reflexive and symmetric but not transitive.

Ans: A = {4, 6, 8}

A = {(4, 4), (6, 6), (8, 8), (4, 6), (6, 8), (8, 6)}

R is reflexive since a ∈ A, (a, a) ∈ R

R is symmetric since (a, b) ∈ R

⇒ (b, a) ∈ R for a, b ∈ R

R is not transitive since (4,6), (6,8) ∈ R, but (4,8) ∉ R

R is reflexive and symmetric but not transitive.

(iv) Reflexive and transitive but not symmetric.

Ans: R = {(a, b) : a³ > b³}

(a, a) ∈ R

R is reflexive.

(2, 1) ∈ R

But (1,2) ∉ R

∴ R is not symmetric.

(a, b), (b, c) ∈ R

⇒ a³ ≥ b³ and b³ < c³

⇒ a³ < c³

⇒ (a, c) ∈ R

∴ R is transitive.

R is reflexive and transitive but not symmetric.

(v) Symmetric and transitive but not reflexive.

Ans: Let A = {-5, -6}.

Define a relation R on A as:

R = {(-5, -6), (-6, -5), (-5,-5)}

Relation R is not reflexive as (-6, -6) ∉ R

Relation R is symmetric as (-5, -6) ∈ R and (-6, -5) ∈ R

It is seen that (-5, -6), (-6, -5) ∈ R.

Also, (-5, -5) ∈ R.

∴ the relation R is transitive.

Hence, relation R is symmetric and transitive but not reflexive.

11. Show that the relation R in the set A of points in a plane given by R = {P, Q) : distance of the point P from the origin is same as the distance of the point Q from the origin}, is an equivalence relation. Further, show that the set of all points related to a point P ≠ (0, 0) is the circle passing through P with origin as centre.

Ans: R = {(P, Q)} : Distance of the point P from the origin is same as the distance of the point Q from the origin}

Clearly, (P, P) ∈ R

R is reflexive.

(P, Q) ∈ R

Clearly R is symmetric.

(P, Q), (Q, S) ∈ R

⇒ The distance of P and Q from the origin is the same and also, the distance of Q and S from the origin is the same.

⇒ The distance of P and S from the origin is the same.

(P, S) ∈ R

R is transitive.

R is an equivalence relation.

The set of points related to P ≠ (0, 0) will be those points whose distance from origin is same as distance of P from the origin.

Set of points forms a circle with the centre as origin and this circle passes through P.

12. Show that the relation R defined in the set A of all triangles as R = {(T₁, T₂) : T₁ is similar to T₂}, is equivalence relation. Consider three right angle triangles T₁ with sides 3, 4, 5, T₂ with sides 5, 12, 13 and T₃ with sides 6, 8, 10. Which triangles among T₁,T₂ and T₃ are related?

Ans: R = {(T₁, T₂) : T₁ is similar to T₂}

R is reflexive since every triangle is similar to itself.

Further, If (T₁,T₂) ∈ R, then T₁ is similar to T₂

⇒ T₂ is similar to T₁.

⇒ (T₂,T₁) ∈ R

∴ R is symmetric.

Now,

Let (T₁,T₂),(T₂,T₃) ∈ R

⇒ T₁ is similar to T₂ and T₂ is similar to T3.

⇒ T₁ is similar to T₃.

⇒ (T₁, T₃) ∈ R

∴ R is transitive.

Thus, R is an equivalence relation.

Now, we can observe that:

∴ The corresponding sides of triangles T₁ and T₃ are in the same ratio.

Then, triangle T₁ is similar to triangle T₃.

Hence, T₁ is related to T₃.

13. Show that the relation R defined in the set A of all polygons as R = {(P₁, P₂) : P₁ and P₂ have same number of sides}, is an equivalence relation. What is the set of all elements in A related to the right angle triangle T with sides 3, 4 and 5?

Ans: R = {(P₁,P₂) : P₁ and P₂ have same number of sides}

(P₁, P₂) ∈ R as same polygon has same number of sides.

∴ R is reflexive.

(P₁, P₂) ∈ R

⇒ P₁ and P₂ have same number of sides.

⇒ P₂ and P₁ have same number of sides.

⇒ (P₂, P₁) ∈ R

∴ R is symmetric.

(P₁,P₂),(P₂,P₃) ∈ R

⇒ P₁ and P₂ have same number of sides.

P₂ and P₃ have same number of sides.

⇒ P₁ and P₃ have same number of sides.

⇒ (P₁ , P₃) ∈ R

∴ R is transitive.

R is an equivalence relation.

The elements in A related to right-angled triangle (T) with sides 3,4,5 are those polygons which have three sides.

Set of all elements in a related to triangle T is the set of all triangles.

14. Let L be the set of all lines in XY plane and R be the relation in L defined as R = {(L₁, L₂) : L₁ is parallel to L₂}. Show that R is an equivalence relation. Find the set of all lines related to the line y = 2x + 4.

Ans: R = {(L₁, L₂) : L₁ is parallel to L₂ }

R is reflexive as any line L₁ is parallel to itself i.e., (L₁, L₂) ∈ R

If(L₁, L₂) ∈ R, then

⇒ L₁ is parallel to L₂

⇒ L₂ is parallel to L₁.

⇒ (L₂, L₁) ∈ R

∴ R is symmetric.

(L₁, L₂), (L₂, L₃) ∈ R

⇒ L₁ is parallel to L₂

⇒ L₂ is parallel to L₃

∴ L₁ is parallel to L₃

⇒ (L₁, L₃) ∈ R

∴ R is transitive.

R is an equivalence relation.

Set of all lines related to the line y = 2x + 4 is the set of all lines that are parallel to the line y = 2x + 4

Slope of the line y = 2x + 4 is m = 2

Line parallel to the given line is in the form y = 2x + c, where c ∈ R

Set of all lines related to the given line is given by y = 2x + c where c ∈ R

15. Let R be the relation in the set {1, 2, 3, 4} given by R = {(1, 2), (2, 2), (1, 1), (4, 4) (1, 3), (3, 3), (3, 2)}. Choose the correct answer.

(A) R is reflexive and symmetric but not transitive.

(B) R is reflexive and transitive but not symmetric.

(D) R is an equivalence relation.

Ans: R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}

(a, a) ∈ R for every a ∈ {1, 2, 3, 4}

∴ R is reflexive.

(1, 2) ∈ R but (2, 1) ∉ R

∴ R is not symmetric.

(a, b), (b, c) ∈ R for all a, b, c ∈ {1, 2, 3, 4}

∴ R is not transitive.

R is reflexive and transitive but not symmetric.

The correct answer is B.

16. Let R be the relation in the set N given by R = {(a, b) : a = b – 2, b > 6} . Choose the correct answer.

(A) (2, 4) ∈ R.

(B) (3, 8) ∈ R.

(D) (8, 7) ∈ R.

Ans: R = {(a, b) : a = b – 2, b > 6}

Now,

b > 6, (2, 4) ∉ R

3 ≠ 8 – 2

∴ (3, 8) ∉ R and as 8 ≠ 7 – 2

∴ (8, 7) ∉ R

Consider (6,8)

8 > 6 and 6 = 8 – 2

∴ (6, 8) ∈ R

The correct answer is C.

EXERCISE 1.2

1. Show that the function f : R. → R. defined by f(x) = 1/x is one-one and onto, where R. is the set of all non-zero real numbers. Is the result true, if the domain R. is replaced by N with co-domain being same as R.?

Ans: f : R. → R. is by f(x) = 1/x

For one-one:

x, y ∈ R. such that f(x) = f(y)

⇒ x = y

∴ f is one-one.

For onto:

∴ f is onto.

Given function f is one-one and onto.

Consider function g : N → R.defined by

We have,

∴ g is one-one.

g is not onto as for 1.2 ∈ R, there exist any X in N such that

Function g is one-one but not onto.

2. Check the injectivity and surjectivity of the following functions:

(i) f : N → N given by f(x) = x².

Ans: For f : N → N given by f(x) = x²

x, y ∈ N

f (x) = f (y) ⇒ x² = y² ⇒ x = y

∴ f is injective.

2 ∈ N But, there does not exist any x in N such that f(x) = x² = 2

∴ f is not surjective

Function is injective but not surjective.

(ii) f : Z → Z given by f(x) = x².

Ans: f : Z → Z given by f(x) = x²

f(- 1) = f(1) = 1 but – 1 ≠ 1

∴ f is not injective.

– 2 ∈ Z But, there does not exist any x ∈ Z such that f(x) = -2 ⇒ x² = – 2

∴ f is not surjective.

Function f is neither injective nor surjective.

(iii) f : R → R given by f(x) = x².

Ans: f : R → R given by f(x) = x².

f(- 1) = f(1) = 1 but – 1 ≠ 1

∴ f is not injective.

– 2 ∈ Z But, there does not exist any x ∈ Z such that f(x) = -2 ⇒ x² = – 2

∴ f is not surjective.

Function f is neither injective nor surjective.

(iv) f : N → N given by f(x) = x³.

Ans: f : N → N given by f(x) = x³.

x, y ∈ N

f (x) = f (y) ⇒ x³ = y³ ⇒ x = y

∴ f is injective.

2 ∈ N. But, there does not exist any X in N such that f (x) = x³ = 2

∴ f is not surjective.

Function f is injective but not surjective.

(v) f : Z → Z given by f(x) = x³

Ans: f : Z → Z given by f(x) = x³

x, y ∈ Z

f (x) = f (y) ⇒ x³ = y³ ⇒ x = y

∴ f is injective.

2 ∈ Z But, there does not exist any X in Z such that f(x) = x³ = 2

∴ f is not surjective.

Function f is injective but not surjective.

3. 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 equal to x.

Ans: f : R → R given by f (x) = [x]

f (1.2) = [1.2] = 1, f(1.9) = [1.9] = 1

∴ f (1.2) = f(1.9) but 1.2 ≠ 1.9

∴ f is not one-one.

Consider 0.7 ∈ R

f(x) = [x] is an integer. There does not exist any element x ∈ R such that f (x) = 0.7

∴ f is not onto.

The greatest integer function is neither one-one nor onto.

4. Show that the Modulus Function f : R → R given by f(x) = x, is neither one – one nor onto, where |x| is x, if x is positive or 0 and |x| is-x, if x is negative.

Ans: f : R→ R is

f (-1) = |-1| =1 and f(1) = |1| = 1

∴ f (-1)= f (1) but -1 ≠ 1

∴ f is not one-one.

Consider -1∈ R

f(x) = |x| is non-negative. There exist any element x in domain R such that f(x) = |x| = -1

∴ f is not onto.

The modulus function is neither one-one nor onto.

5. Show that the Signum Function f : R → R, given by is neither one-one nor onto.

Ans: f : R→ R is

f(1) = f (2) = 1, but 1 ≠ 2

∴ f is not one-one.

f (x) takes only 3 values (1,0,-1) for the element -2in co-domain

R, there does not exist any X in domain R such that f(x) = -2

∴ f is not onto.

The signum function is neither one-one nor onto.

6. Let A = {1, 2, 3} B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. Show that f is one-one.

Ans: A = {1, 2, 3} B = {4, 5, 6, 7}

f : A → B is defined as f = {(1, 4), (2, 5), (3, 6)}

∴ f(1) = 4, f (2) = 5, f(3) = 6

It is seen that the images of distinct elements of A under f are distinct.

∴ f is one-one.

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

(i) f : R → R defined by f(x) = 3 – 4x.

Ans: f : R → R defined by f(x) = 3 – 4x.

x₁, x₂ ∈ R such that f(x₁) = f(x₂)

⇒ 3 – 4x₁ = 3 – 4x₂

⇒ – 4x = – 4x₂

⇒ x₁ = x₂

∴ f is one-one.

∴ f is onto.

Hence, f is bijective.

(ii) f : R → R defined by f(x) = 1 + x².

Ans: f : R → R defined by f(x) = 1 + x².

x₁, x₂ ∈ R such that f(x₁) = f(x₂)

⇒ 1 + x² = 1 + x²

⇒ x₁² = x₂²

⇒ x₁ = ±x₂

∴ f(x₁) = f(x₂) does not imply that x₁ = x₂

Consider f(1) = f(-1) = 2

∴ f is not one-one.

Consider an element -2 in co domain R.

It is seen that f(x)= 1 + x² is is positive for all x ∈ R.

∴ f is not onto.

Hence, f is neither one-one nor onto.

8. Let A and B be sets. Show that f : A × B → B × A such that f(a, b) = (b, a) is bijective function.

Ans: f: A × B→ B x A is defined as (a,b) = (b,a).

(a₁,b₁), (a₂,b₂) ∈ A x B such that f(a₁,b₁) = f (a₂,b₂)

⇒ (b₁,a₁) = (b₂,a₂)

⇒ b₁= b₂ and a₁ = a₂

⇒ (a₁,b₁) = (a₂,b₂)

∴ f is one-one.

(b,a) ∈ B x A there exist (a,b) ∈ A x B such that f (a,b) = (b,a)

∴ f is onto.

f is bijective.

State whether the function f is bijective. Justify your answer.

Ans: f : N → N be defined as

f (1) = f (2), where 1 ≠ 2

∴ f is not one-one.

Consider a natural number n in co domain N.

Case l : n is odd

∴ n = 2r + 1 for some r ∈ N there exists 4r + 1 ∈ N such that

Case II : n is even

∴ n = 2r for some r ∈ N there exists 4r ∈ N such that

∴ f is onto.

f is not a bijective function.

10. Let A = R – {3} and B = R – {1}. Consider the function f : A → B defined by

Ans: A = R – {3}, B = R – {1} and f : A → B B defined by

x, y ∈ A such that f (x) = f (y)

⇒ (x – 2)(y – 3) = (y – 2)(x – 3)

⇒ xy – 3x – 2y + 6 = xy – 3y – 2x + 6

⇒ – 3x – 2y = – 3y – 2x

⇒ 3x – 2x = 3y – 2y

⇒ x = y

∴ f is one-one.

Let y ∈ B = R – {1} , then y ≠ 1

The function f is onto if there exists x ∈ A such that f(x) = y

Now,

f(x) = y

∴ f is onto.

Hence, the function is one-one and onto.

11. Let f : R → R be defined as f(x) = x⁴ Choose the correct answer.

(A) f is one-one onto.

(B) f is many-one onto.

(D) f is neither one-one nor onto.

Ans: f : R → R defined as f(x) = x⁴

x,y ∈ R such that f(x) = f(y)

⇒ x⁴ = y⁴

⇒ x = ±y

∴ f (x) = f (y) does not imply that x = y.

For example f(1) = f(- 1) = 1

∴ f is not one-one.

Consider an element 2 in co domain R there does not exist any x in domain R such that f (x) = 2

∴ f is not onto.

Function f is neither one – one nor onto.

The correct answer is D.

12. Let f : R→ R be defined as f(x) = 3x. Choose the correct answer.

(A) f is one-one onto.

(B) f is many-one onto.

(D) f is neither one-one nor onto.

Ans: f : R→ R defined as f(x) = 3x

x,y ∈ R such that f(x) = f(y)

⇒ 3x = 3y

⇒ x = y

∴ f is one-one

For any real number y in co domain R, there exist y/3 in R such that

∴ f is onto.

Hence, function f is one-one and onto.

The correct answer is A.

Miscellaneous Exercise on Chapter 1

1. Show that the function f : R → {x ∈ R : – 1 < x < 1} defined by x ∈ R is one – one and onto function.

Ans: f : R → {x ∈ R : – 1 < x < 1} is defined by x ∈ R.

For one – one:

f(x) = f (y) where x, y ∈ R

If x is positive and y is negative,

⇒ 2xy = x – y

Since, X is positive and y is negative,

x > y ⇒ x – y > 0

2xy is negative.

2xy ≠ x – y

Case of x being positive and y being negative, can be ruled out.

∴ x and y have to be either positive or negative.

If x and y are positive,

f(x) = f(y)

⇒ x – xy = y – xy

⇒ x = y

∴ f is one-one.

For onto:

Let y ∈ R such that – 1 < y < 1

If x is negative, then there exists such that

If x is positive, then there exists such that

∴ f is onto.

Hence, f is one-one and onto.

2. Show that the function f : R → R given by f(x) = x³ is injective.

Ans: f : R → R is defined by f(x) = x³

For one-one:

f (x) = f (y) where x, y ∈ R

x³ = y³……………….(1)

We need to show that x = y

Suppose x ≠ y, their cubes will also not be equal.

⇒ x³ ≠ y³

This will be a contradiction to (1).

∴ x = y . Hence, f is injective.

3. Given a non empty set X, consider P(X) which is the set of all subsets of X. Define the relation R in P(X) as follows:

For subsets A, B in P(X), ARB if and only if A ⊂ B. Is R an equivalence relation on P(X)? Justify your answer.

Ans: Since every set is a subset of itself, ARA for all A ∈ P(X).

∴ R is reflexive.

Let ARB ⇒ A ⊂ B

This cannot be implied to B ⊂ A.

If A = {1,2} and B = {1, 2,3}, then it cannot be implied that B is related to A.

∴ R is not symmetric.

If ARB and BRC, then A ⊂ B and B ⊂ C.

⇒ A ⊂ C

⇒ ARC

∴ R is transitive.

R is not an equivalence relation as it is not symmetric.

4. Find the number of all onto functions from the set {1, 2, 3 ,…..,n} to itself.

Ans: Onto functions from the set {1,2,3,…, n} to itself is simply a permutation on n symbols 1,2,3,…, n.

Thus, the total number of onto nto maps from {1,2,3,…,n} to itself is the same as the total number of permutations on n symbols 1,2,3,…,n, which is n!.

5. Let A = {- 1, 0, 1, 2} B = {- 4, – 2, 0, 2} and f, g : A → B be functions defined by f(x) = x² – x, x ∈ A and g(x)