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Answers of Mathematics NCERT solutions for class 12 Chapter 1 – Relations and Functions

Exercise 1.1:

Question 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 = {(xy): 3x − y = 0}

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

R = {(xy): y = x + 5 and x < 4}

(iii) Relation R in the set A = {1, 2, 3, 4, 5, 6} as

R = {(xy): y is divisible by x}

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

R = {(xy): x − y is as integer}

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

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

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

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

(d) R = {(xy): x is wife of y}

(e) R = {(xy): x is father of y}

Answer:

(i)  A = {1, 2, 3 … 13, 14}

R = {(xy): 3x − y = 0}

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

R is not reflexive since (1, 1), (2, 2) … (14, 14) ∉ R.

Also, R is not symmetric as (1, 3) ∈R, but (3, 1) ∉ R. [3(3) − 1 ≠ 0]

Also, R is not transitive as (1, 3), (3, 9) ∈R, but (1, 9) ∉ R.

[3(1) − 9 ≠ 0]

Hence, R is neither reflexive, nor symmetric, nor transitive.

(ii) R = {(xy): y = x + 5 and x < 4} = {(1, 6), (2, 7), (3, 8)}

It is seen that (1, 1) ∉ R.

∴R is not reflexive.

(1, 6) ∈R

But,

(6, 1) ∉ R.

∴R is not symmetric.

Now, since there is no pair in R such that (xy) and (yz) ∈R, then (xz) cannot belong to R.

∴ R is not transitive.

Hence, R is neither reflexive, nor symmetric, nor transitive.

(iii)  A = {1, 2, 3, 4, 5, 6}

R = {(xy): y is divisible by x}

We know that any number (x) is divisible by itself.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/5971/Exercise2_html_74849e29.gif (xx) ∈R

∴R is reflexive.

Now,

(2, 4) ∈R [as 4 is divisible by 2]

But,

(4, 2) ∉ R. [as 2 is not divisible by 4]

∴R is not symmetric.

Let (xy), (yz) ∈ R. Then, y is divisible by x and z is divisible by y.

z is divisible by x.

⇒ (xz) ∈R

∴R is transitive.

Hence, R is reflexive and transitive but not symmetric.

(iv) R = {(xy): x − y is an integer}

Now, for every x ∈ Z, (xx) ∈R as x − x = 0 is an integer.

∴R is reflexive.

Now, for every xy ∈ Z if (xy) ∈ R, then x − y is an integer.

⇒−(x − y) is also an integer.

⇒ (y − x) is an integer.

∴ (yx) ∈ R

∴R is symmetric.

Now,

Let (xy) and (yz) ∈R, where xyz ∈ Z.

⇒ (x − y) and (y − z) are integers.

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

∴ (xz) ∈R

∴R is transitive.

Hence, R is reflexive, symmetric, and transitive.

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

⇒ (xx) ∈ R

∴ R is reflexive.

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

⇒ y and x work at the same place.

⇒ (yx) ∈ R.

∴R is symmetric.

Now, let (xy), (yz) ∈ R

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

⇒ x and z work at the same place.

⇒ (xz) ∈R

∴ R is transitive.

Hence, R is reflexive, symmetric, and transitive.

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

Clearly (xx) ∈ R as x and x is the same human being.

∴ R is reflexive.

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

⇒ y and x live in the same locality.

⇒ (yx) ∈ R

∴R is symmetric.

Now, let (xy) ∈ R and (yz) ∈ R.

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

⇒ x and z live in the same locality.

⇒ (x, z) ∈ R

∴ R is transitive.

⇒Hence, R is reflexive, symmetric, and transitive.

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

Now,

(xx) ∉ R

Since human being cannot be taller than himself.

∴R is not reflexive.

Now, let (xy) ∈R.

⇒ x is exactly 7 cm taller than y.

Then, y is not taller than x.

∴ (yx) ∉R

Indeed, if x is exactly 7 cm taller than y, then y is exactly 7 cm shorter than x.

∴R is not symmetric.

Now,

Let (xy), (yz) ∈ R.

⇒ x is exactly 7 cm taller thanand y is exactly 7 cm taller than z.

⇒ x is exactly 14 cm taller than z.

∴(xz) ∉R

∴ R is not transitive.

Hence, R is neither reflexive, nor symmetric, nor transitive.

(d) R = {(xy): x is the wife of y}

Now,

(xx) ∉ R

Since x cannot be the wife of herself.

∴R is not reflexive.

Now, let (xy) ∈ R

⇒ x is the wife of y.

Clearly y is not the wife of x.

∴(yx) ∉ R

Indeed, if x is the wife of y, then y is the husband of x.

∴ R is not symmetric.

Let (xy), (yz) ∈ R

⇒ x is the wife of y and y is the wife of z.

This case is not possible. Also, this does not imply that x is the wife of z.

∴(xz) ∉ R

∴R is not transitive.

Hence, R is neither reflexive, nor symmetric, nor transitive.

(e) R = {(xy): x is the father of y}

(xx) ∉ R

As x cannot be the father of himself.

∴R is not reflexive.

Now, let (xy) ∈R.

⇒ x is the father of y.

⇒ y cannot be the father of y.

Indeed, y is the son or the daughter of y.

∴(yx) ∉ R

∴ R is not symmetric.

Now, let (xy) ∈ R and (yz) ∈ R.

⇒ x is the father of y and y is the father of z.

⇒ x is not the father of z.

Indeed, x is the grandfather of z.

∴ (xz) ∉ R

∴R is not transitive.

Hence, R is neither reflexive, nor symmetric, nor transitive.

 

Question 2:

Show that the relation R in the set R of real numbers, defined as

R = {(ab): a ≤ b2} is neither reflexive nor symmetric nor transitive.

Answer:

R = {(ab): a ≤ b2}https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/5979/Chapter%201_html_m28016277.gif

It can be observed that 

∴R is not reflexive.

Now, (1, 4) ∈ R as 1 < 42

But, 4 is not less than 12.

∴(4, 1) ∉ R

∴R is not symmetric.

Now,

(3, 2), (2, 1.5) ∈ R

(as 3 < 22 = 4 and 2 < (1.5)2 = 2.25)

But, 3 > (1.5)2 = 2.25

∴(3, 1.5) ∉ R

∴ R is not transitive.

Hence, R is neither reflexive, nor symmetric, nor transitive.

 

Question 3:

Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as

R = {(ab): b = a + 1}is reflexive, symmetric or transitive.

Answer:

Let A = {1, 2, 3, 4, 5, 6}.

A relation R is defined on set A as:

R = {(ab): b = a + 1}

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

We can find (aa) ∉ R, where ∈ A.

For instance,

(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) ∉ R

∴R is not reflexive.

It can be observed that (1, 2) ∈ R, but (2, 1) ∉ R.

∴R is not symmetric.

Now, (1, 2), (2, 3) ∈ R

But,

(1, 3) ∉ R

∴R is not transitive

Hence, R is neither reflexive, nor symmetric, nor transitive.

 

Question 4:

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

Answer:

R = {(ab); a ≤ b}

Clearly (aa) ∈ R as a.

∴R is reflexive.

Now,

(2, 4) ∈ R (as 2 < 4)

But, (4, 2) ∉ R as 4 is greater than 2.

∴ R is not symmetric.

Now, let (ab), (bc) ∈ R.

Then,

a ≤ b and b ≤ c

⇒ a ≤ c

⇒ (ac) ∈ R

∴R is transitive.

Hence,R is reflexive and transitive but not symmetric.

 

Question 5:

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

Answer:

R = {(ab): Spaj9z V jPhvqG4QDIB2F7DqsBqcUVz9i3RnIs7sjXrXKRALlbLoN469W9jBhCnAJsWdXU4BPnI7MHfB5ebAlEsEVsMyjtRg Mm37 tA1Y7gpvTLkqgHU76z6hisC6Zj4tHEEb3}https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/5993/Chapter%201_html_m17570f67.gif

It is observed that

∴ R is not reflexive.

Now,

(1, 2) ∈ R(as 1 < 23 = 8)

But,

(2, 1) ∉R(as 2 > 13 = 1)(element)

R is not symmetric.

We have

But 

R∈∈is not transitive.

Hence, R is neither reflexive, nor symmetric, nor transitive.

 

Question 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.

Answer:

Let A = {1, 2, 3}.

A relation R on A is defined as R = {(1, 2), (2, 1)}.

It is seen that (1, 1), (2, 2), (3, 3) ∉R.

∴ R is not reflexive.

Now, as (1, 2) ∈ R and (2, 1) ∈ R, then R is symmetric.

Now, (1, 2) and (2, 1) ∈ R

However,

(1, 1) ∉ R

∴ R is not transitive.

Hence, R is symmetric but neither reflexive nor transitive.

 

Question 7:

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

Answer:

Set A is the set of all books in the library of a college.

R = {xy): x and y have the same number of pages}

Now, R is reflexive since (xx) ∈ R as x and x has the same number of pages.

Let (xy) ∈ R ⇒ x and y have the same number of pages.

⇒ y and x have the same number of pages.

⇒ (yx) ∈ R

∴R is symmetric.

Now, let (xy) ∈R and (yz) ∈ R.

⇒ x and y and have the same number of pages and y and z have the same number of pages.

⇒ x and z have the same number of pages.

⇒ (xz) ∈ R

∴R is transitive.

Hence, R is an equivalence relation.

 

Question 8:

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

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}.

Answer:

A = {1, 2, 3, 4, 5}

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6005/Chapter%201_html_m169bdf1e.gif

It is clear that for any element a ∈A, we have (which is even).

∴R is reflexive.

Let (ab) ∈ R.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6005/Chapter%201_html_m33537aa1.gif

∴R is symmetric.

Now, let (ab) ∈ R and (bc) ∈ R.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6005/Chapter%201_html_1d3cb8e0.gif

⇒ (ac) ∈ R

∴R is transitive.

Hence, R is an equivalence relation.

Now, all elements of the set {1, 3, 5} are related to each other as all the elements of this subset are odd. Thus, the modulus of the difference between any two elements will be even.

Similarly, all elements of the set {2, 4} are related to each other as all the elements of this subset are even.

Also, no element of the subset {1, 3, 5} can be related to any element of {2, 4} as all elements of {1, 3, 5} are odd and all elements of {2, 4} are even. Thus, the modulus of the difference between the two elements (from each of these two subsets) will not be even.

 

Question 9:

Show that each of the relation R in the set, given by

(i)  https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6007/Chapter%201_html_1e9a99ef.gif

(ii) https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6007/Chapter%201_html_m7a54ba86.gif

is an equivalence relation. Find the set of all elements related to 1 in each case.

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6007/Chapter%201_html_43efc75.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6007/Chapter%201_html_1e9a99ef.gif

(i) 

For any element a ∈A, we have (aa) ∈ R as is a multiple of 4.

∴R is reflexive.

Now, let (ab) ∈ R ⇒ is a multiple of 4.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6007/Chapter%201_html_6c903344.gif

⇒ (ba) ∈ R

∴R is symmetric.

Now, let (ab), (bc) ∈ R.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6007/Chapter%201_html_2da5cd47.gif

⇒ (ac) ∈R

∴ R is transitive.

Hence, R is an equivalence relation.

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

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6007/Chapter%201_html_m68841b74.gif

(ii) R = {(ab): a = b}

For any element a ∈A, we have (aa) ∈ R, since a = a.

∴R is reflexive.

Now, let (ab) ∈ R.

⇒ a = b

⇒ b = a

⇒ (ba) ∈ R

∴R is symmetric.

Now, let (ab) ∈ R and (bc) ∈ R.

⇒ a = b and b = c

⇒ a = c

⇒ (ac) ∈ R

∴ R is transitive.

Hence, R is an equivalence relation.

The elements in R that are related to 1 will be those elements from set A which are equal to 1.

Hence, the set of elements related to 1 is {1}.

Question 10:

Given an example of a relation. Which is

(i) Symmetric but neither reflexive nor transitive.

(ii) Transitive but neither reflexive nor symmetric.

(iii) Reflexive and symmetric but not transitive.

(iv) Reflexive and transitive but not symmetric.

(v) Symmetric and transitive but not reflexive.

Answer:

(i) Let A = {5, 6, 7}.

Define a relation R on A as R = {(5, 6), (6, 5)}.

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

Now, as (5, 6) ∈ R and also (6, 5) ∈ R, R is symmetric.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6008/Chapter%201_html_74849e29.gif(5, 6), (6, 5) ∈ R, but (5, 5) ∉ R

∴R is not transitive.

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

(ii) Consider a relation R in defined as:

R = {(ab): a < b}

For any ∈ R, we have (aa) ∉ R since a cannot be strictly less than a itself. In fact, a = a.

∴ R is not reflexive.

Now,

(1, 2) ∈ R (as 1 < 2)

But, 2 is not less than 1.

∴ (2, 1) ∉ R

∴ R is not symmetric.

Now, let (ab), (bc) ∈ R.

⇒ a < b and b < c

⇒ a < c

⇒ (ac) ∈ R

∴R is transitive.

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

(iii) Let A = {4, 6, 8}.

Define a relation R on A as:

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

Relation R is reflexive since for every a ∈ A, (aa) ∈R i.e., (4, 4), (6, 6), (8, 8)} ∈ R.

Relation R is symmetric since (ab) ∈ R ⇒ (ba) ∈ R for all ab ∈ R.

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

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

(iv) Define a relation R in R as:

R = {ab): a3 ≥ b3}

Clearly (aa) ∈ R as a3 = a3.

∴R is reflexive.

Now,

(2, 1) ∈ R (as 23 ≥ 13)

But,

(1, 2) ∉ R (as 13 < 23)

∴ R is not symmetric.

Now,

Let (ab), (bc) ∈ R.

⇒ a3 ≥ b3 and b3 ≥ c3

⇒ a3 ≥ c3

⇒ (ac) ∈ R

∴R is transitive.

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

(v) 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.

Question 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 point related to a point P ≠ (0, 0) is the circle passing through P with origin as centre.

Answer:

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

Clearly, (P, P) ∈ R since the distance of point P from the origin is always the same as the distance of the same point P from the origin.

∴R is reflexive.

Now,

Let (P, Q) ∈ R.

⇒ The distance of point P from the origin is the same as the distance of point Q from the origin.

⇒ The distance of point Q from the origin is the same as the distance of point P from the origin.

⇒ (Q, P) ∈ R

∴R is symmetric.

Now,

Let (P, Q), (Q, S) ∈ R.

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

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

⇒ (P, S) ∈ R

∴R is transitive.

Therefore, R is an equivalence relation.

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

In other words, if O (0, 0) is the origin and OP = k, then the set of all points related to P is at a distance of k from the origin.

Hence, this set of points forms a circle with the centre as the origin and this circle passes through point P.

Question 12:

Show that the relation R defined in the set A of all triangles as R = {(T1T2): 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 T1T2 and T3 are related?

Answer:

R = {(T1T2): T1 is similar to T2}

R is reflexive since every triangle is similar to itself.

Further, if (T1T2) ∈ R, then T1 is similar to T2.

⇒ T2 is similar to T1.

⇒ (T2T1) ∈R

∴R is symmetric.

Now,

Let (T1T2), (T2T3) ∈ R.

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

⇒ T1 is similar to T3.

⇒ (T1T3) ∈ R

∴ R is transitive.

Thus, R is an equivalence relation.

Now, we can observe that:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6010/Chapter%201_html_m3cc13ebe.gif

∴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.

Question 13:

Show that the relation R defined in the set A of all polygons as R = {(P1P2): 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?

Answer:

R = {(P1P2): P1 and P2 have same the number of sides}

R is reflexive since (P1P1) ∈ R as the same polygon has the same number of sides with itself.

Let (P1P2) ∈ R.

⇒ P1 and P2 have the same number of sides.

⇒ P2 and P1 have the same number of sides.

⇒ (P2P1) ∈ R

∴R is symmetric.

Now,

Let (P1P2), (P2P3) ∈ R.

⇒ P1 and P2 have the same number of sides. 

Also, P2 and P3 have the same number of sides.

⇒ P1 and P3 have the same number of sides.

⇒ (P1P3) ∈ R

∴R is transitive.

Hence, R is an equivalence relation.

The elements in A related to the right-angled triangle (T) with sides 3, 4, and 5 are those polygons which have 3 sides (since T is a polygon with 3 sides).

Hence, the set of all elements in A related to triangle T is the set of all triangles.

Question 14:

Let L be the set of all lines in XY plane and R be the relation in L defined as 

R = {(L1L2): 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.

Answer:

R = {(L1L2): L1 is parallel to L2}

R is reflexive as any line L1 is parallel to itself i.e., (L1L1) ∈ R.

Now,

Let (L1L2) ∈ R.

⇒ L1 is parallel to L2.

⇒ L2 is parallel to L1.

⇒ (L2L1) ∈ R

∴ R is symmetric.

Now,

Let (L1L2), (L2L3) ∈R.

⇒ L1 is parallel to L2. Also, L2 is parallel to L3.

⇒ L1 is parallel to L3.

∴R is transitive.

Hence, R is an equivalence relation.

The 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 line y = 2x + 4 is m = 2

It is known that parallel lines have the same slopes.

The line parallel to the given line is of the form y = 2x + c, where c ∈R.

Hence, the set of all lines related to the given line is given by y = 2x + c, where c ∈ R.

*Question 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.

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

(D) R is an equivalence relation.

Answer:

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

It is seen that (aa) ∈ R, for every a ∈{1, 2, 3, 4}.

∴ R is reflexive.

It is seen that (1, 2) ∈ R, but (2, 1) ∉ R.

∴R is not symmetric.

Also, it is observed that (ab), (bc) ∈ R ⇒ (ac) ∈ R for all abc ∈ {1, 2, 3, 4}.

∴ R is transitive.

Hence, R is reflexive and transitive but not symmetric.

The correct answer is B.

Question 16:

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

(A) (2, 4) ∈ R (B) (3, 8) ∈R (C) (6, 8) ∈R (D) (8, 7) ∈ R

Answer:

R = {(ab): b − 2, b > 6}

Now, since b > 6, (2, 4) ∉ R

Also, as 3 ≠ 8 − 2, (3, 8) ∉ R

And, as 8 ≠ 7 − 2

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6018/Chapter%201_html_4dd19828.gif (8, 7) ∉ R

Now, consider (6, 8).

We have 8 > 6 and also, 6 = 8 − 2.

∴(6, 8) ∈ R

The correct answer is C.

Exercise 1.2

Question 1:

Show that the function fR* → R* defined byis 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*?

Answer:

It is given that fR* → R* is defined by.

One-one:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6022/Chapter%201_html_4accecb1.gif

f is one-one.

Onto:

It is clear that for y R*, there exists (Exists as )such that

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6022/Chapter%201_html_fe5622f.gif

f is onto.

Thus, the given function (f) is one-one and onto.

Now, consider function g: N → R*defined by

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6022/Chapter%201_html_m15c053d.gif

We have,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6022/Chapter%201_html_1baa05a9.gif

g is one-one.

Further, it is clear that g is not onto as for 1.2 ∈R* there does not exit any x in N such that g(x)

Hence, function g is one-one but not onto.

Question 2:

Check the injectivity and surjectivity of the following functions:

(i)  fN → N given by f(x) = x2

(ii)  fZ → Z given by f(x) = x2

(iii)  fR → R given by f(x) = x2

(iv)  f→ N given by f(x) = x3

(v)  fZ → Z given by f(x) = x3

Answer:

(i)  fN → N is given by,

f(x) = x2

It is seen that for xy ∈Nf(x) = f(y) ⇒ x2 = y2 ⇒ x = y.

f is injective.

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

∴ f is not surjective.

Hence, function f is injective but not surjective.

(ii)  fZ → Z is given by,

f(x) = x2

It is seen that f(−1) = f(1) = 1, but −1 ≠ 1.

∴ is not injective.

Now,−2 ∈ Z. But,there does not exist any element Z such that f(x) = x2 = −2.

∴ f is not surjective.

Hence, function f is neither injective nor surjective.

(iii)  fR → R is given by,

f(x) = x2

It is seen that f(−1) = f(1) = 1, but −1 ≠ 1.

∴ is not injective.

Now,−2 ∈ R. But, there does not exist any element ∈ R such that f(x) = x2 = −2.

∴ f is not surjective.

Hence, function f is neither injective nor surjective.

(iv)  fN → N given by,

f(x) = x3

It is seen that for xy ∈Nf(x) = f(y) ⇒ x3 = y3 ⇒ x = y.

f is injective.

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

∴ f is not surjective

Hence, function f is injective but not surjective.

(v)  f→ Z is given by,

f(x) = x3

It is seen that for xy ∈ Zf(x) = f(y) ⇒ x3 = y3 ⇒ x = y.

∴ f is injective.

Now, 2 ∈ Z. But, there does not exist any element x in domain Z such thatf(x) = x3 = 2.

∴ f is not surjective.

Hence, function f is injective but not surjective.

Question 3:

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

Answer:

fR → R is given by,

f(x) = [x]

It is seen that 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.

Now, consider 0.7 ∈ R.

It is known that f(x) = [x] is always an integer. Thus, there does not exist any element x ∈ R such that f(x) = 0.7.

∴ f is not onto.

Hence, the greatest integer function is neither one-one nor onto.

Question 4:

Show that the Modulus Function f→ R given by, is neither one-one nor onto, whereis x, if x is positive or 0 andis − x, if x is negative.

Answer:

fR → R is given by,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6030/Chapter%201_html_45680a01.gif

It is seen that

f(−1) = f(1), but −1 ≠ 1.

∴ f is not one-one.

Now, consider −1 ∈ R.

It is known that f(x) =  is always non-negative. Thus, there does not exist any element x in domain R such that f(x) = = −1.

∴ f is not onto.

Hence, the modulus function is neither one-one nor onto.

Question 5:

Show that the Signum Function fR → R, given by

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6032/Chapter%201_html_m312b8b2c.gif

is neither one-one nor onto.

Answer:

fR → R is given by,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6032/Chapter%201_html_m312b8b2c.gif

It is seen that f(1) = f(2) = 1, but 1 ≠ 2.

f is not one-one.

Now, as f(x) takes only 3 values (1, 0, or −1) for the element −2 in co-domain R, there does not exist any x in domain R such that f(x) = −2.

∴ f is not onto.

Hence, the signum function is neither one-one nor onto.

Question 6:

Let A = {1, 2, 3}, = {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.

Answer:

It is given that A = {1, 2, 3}, = {4, 5, 6, 7}.

fA → 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.

Hence, function f is one-one.

Question 7:

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

(i)  f→ R defined by f(x) = 3 − 4x

(ii)  f→ R defined by f(x) = 1 + x2

Answer:

(i)  f→ R is defined as f(x) = 3 − 4x.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6034/Chapter%201_html_m267e3da6.gif.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6034/Chapter%201_html_6697f990.gif

∴ f is one-one.

For any real number (y) in R, there existsin R such that

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6034/Chapter%201_html_m16b4e4d6.gif

is onto.

Hence, is bijective.

(ii)  fR → R is defined as

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6034/Chapter%201_html_m57ec1cac.gif.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6034/Chapter%201_html_m267e3da6.gif.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6034/Chapter%201_html_1bbb0484.gif

does not imply that

For instance,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6034/Chapter%201_html_1040cc0f.gif

∴ f is not one-one.

Consider an element −2 in co-domain R.

It is seen thatis positive for all x ∈ R.

Thus, there does not exist any x in domain R such that f(x) = −2.

∴ f is not onto.

Hence, f is neither one-one nor onto.

Question 8:

Let A and B be sets. Show that fA × B → × A such that (ab) = (ba) is bijective function.

Answer:

fA × B → B × A is defined as f(ab) = (ba).

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6035/Chapter%201_html_3bf55b34.gif.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6035/Chapter%201_html_m58ad0bd6.gif

∴ f is one-one.

Now, let (ba) ∈ B × A be any element.

Then, there exists (ab) ∈A × B such that f(ab) = (ba). [By definition of f]

∴ f is onto.

Hence, f is bijective.

*Question 9:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6037/Chapter%201_html_m18543bea.gif

Let fN → N be defined by

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

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6037/Chapter%201_html_m18543bea.gif

fN → N is defined as

It can be observed that:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6037/Chapter%201_html_m5a4452fc.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6037/Chapter%201_html_2fb4345b.gif

∴ f is not one-one.

Consider a natural number (n) in co-domain N.

Case I: n is odd

n = 2r + 1 for some r ∈ N. Then, there exists 4+ 1∈Nsuch that

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6037/Chapter%201_html_1504181f.gif.

Case II: n is even

n = 2r for some r ∈ N. Then,there exists 4r ∈N such that.

∴ f is onto.

Hence, f is not a bijective function.

Question 10:

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

is f one-one and onto? Justify your answer.

Answer:

A = R − {3}, B = R − {1}

f: A → B is defined as.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6039/Chapter%201_html_m38b158f8.gif.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6039/Chapter%201_html_23f1897b.gif

∴ f is one-one.

Let y ∈B = R − {1}. Then, y ≠ 1.

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

Now,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6039/Chapter%201_html_m63b9e71a.gif

Thus, for any ∈ B, there existssuch that

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6039/Chapter%201_html_59a7ba34.gif

Hence, function f is one-one and onto.

*Question 11:

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

(A)  f is one-one onto (B) f is many-one onto

(C)  f is one-one but not onto (D) f is neither one-one nor onto

Answer:

fR → R is defined as

Let x∈ R such that f(x) = f(y).

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6041/Chapter%201_html_4824c8c1.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6041/Chapter%201_html_93d66d8.gifdoes not imply that.

For instance,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6041/Chapter%201_html_6189c965.gif

∴ f is not one-one.

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

∴ f is not onto.

Hence, function f is neither one-one nor onto.

The correct answer is D.

Question12:

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

(A)  f is one-one onto (B) f is many-one onto

(C)  f is one-one but not onto (D) f is neither one-one nor onto

Answer:

fR → R is defined as f(x) = 3x.

Let x∈ R such that f(x) = f(y).

⇒ 3x = 3y

⇒ x = y

is one-one.

Also, for any real number (y) in co-domain R, there exists in R such thathttps://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6043/Chapter%201_html_5269aead.gif.

is onto.

Hence, function f is one-one and onto.

The correct answer is A.

Exercise 1.3

Question 1:

Let f: {1, 3, 4} → {1, 2, 5} and g: {1, 2, 5} → {1, 3} be given by = {(1, 2), (3, 5), (4, 1)} and = {(1, 3), (2, 3), (5, 1)}. Write down gof.

Answer:

The functions f: {1, 3, 4} → {1, 2, 5} and g: {1, 2, 5} → {1, 3} are defined as

= {(1, 2), (3, 5), (4, 1)} and = {(1, 3), (2, 3), (5, 1)}.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6045/Chapter%201_html_57f3b03d.gif

Question 2:

Let fg and h be functions from to R. Show that

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6048/Chapter%201_html_m59307f71.gif

Answer:

To prove:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6048/Chapter%201_html_4bfa0242.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6048/Chapter%201_html_566b6382.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6048/Chapter%201_html_m4f3371b3.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6048/Chapter%201_html_m715a924e.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6048/Chapter%201_html_m377bec3b.gif

*Question 3:

Find goand fog, if

(i) https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6050/Chapter%201_html_5f375722.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6050/Chapter%201_html_m2242302d.gif

(ii) 

Answer:https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6050/Chapter%201_html_5f375722.gif

(i) 

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6050/Chapter%201_html_mc9dd7a9.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6050/Chapter%201_html_m2242302d.gif

(ii) 

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6050/Chapter%201_html_m1498fc9.gif

*Question 4:

If, show that  for all. What is the inverse of f?

Answer:

It is given that.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6054/Chapter%201_html_m196322d8.gif

Hence, the given function f is invertible and the inverse of f is f itself.

Question 5:

State with reason whether following functions have inverse

(i)  f: {1, 2, 3, 4} → {10} with

f = {(1, 10), (2, 10), (3, 10), (4, 10)}

(ii)  g: {5, 6, 7, 8} → {1, 2, 3, 4} with

g = {(5, 4), (6, 3), (7, 4), (8, 2)}

(iii)  h: {2, 3, 4, 5} → {7, 9, 11, 13} with

h = {(2, 7), (3, 9), (4, 11), (5, 13)}

Answer:

(i)  f: {1, 2, 3, 4} → {10}defined as:

f = {(1, 10), (2, 10), (3, 10), (4, 10)}

From the given definition of f, we can see that f is a many one function as: f(1) = f(2) = f(3) = f(4) = 10

is not one-one.

Hence, function does not have an inverse.

(ii)  g: {5, 6, 7, 8} → {1, 2, 3, 4} defined as:

g = {(5, 4), (6, 3), (7, 4), (8, 2)}

From the given definition of g, it is seen that g is a many one function as: g(5) = g(7) = 4.

is not one-one,

Hence, function g does not have an inverse.

(iii)  h: {2, 3, 4, 5} → {7, 9, 11, 13} defined as:

h = {(2, 7), (3, 9), (4, 11), (5, 13)}

It is seen that all distinct elements of the set {2, 3, 4, 5} have distinct images under h.

∴Function h is one-one.

Also, h is onto since for every element y of the set {7, 9, 11, 13}, there exists an element x in the set {2, 3, 4, 5}such that h(x) = y.

Thus, h is a one-one and onto function. Hence, h has an inverse.

*Question 6:

Show that f: [−1, 1] → R, given byis one-one. Find the inverse of the function f: [−1, 1] → Range f.

(Hint: For y ∈Range fy =, for some x in [−1, 1], i.e.,)

Answer:

f: [−1, 1] → R is given as

Let f(x) = f(y).

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6059/Chapter%201_html_2b620184.gif

∴ f is a one-one function.

It is clear that f: [−1, 1] → Range f is onto.

∴ f: [−1, 1] → Range f is one-one and onto and therefore, the inverse of the function:

f: [−1, 1] → Range exists.

Let g: Range f → [−1, 1] be the inverse of f.

Let y be an arbitrary element of range f.

Since f: [−1, 1] → Range f is onto, we have:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6059/Chapter%201_html_m6dada837.gif

Now, let us define g: Range f → [−1, 1] as

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6059/Chapter%201_html_7466b8c7.gif

and 

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6059/Chapter%201_html_4dd19828.giff−1 = g

AOhJ zxoxEiP9N144YvnSg9reoSQzaIiMMcr Jd7Gv rZFXaTqdwGSobPSsYN5tS3lklfgsN47kOONBX MbVF

Question 7:

Consider fR → R given by f(x) = 4x + 3. Show that f is invertible. Find the inverse of f.

Answer:

fR → R is given by,

f(x) = 4x + 3

One-one:

Let f(x) = f(y).

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6060/Chapter%201_html_m761b1178.gif

∴ f is a one-one function.

Onto:

For y ∈ R, let y = 4x + 3.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6060/Chapter%201_html_4195f8f1.gif

Therefore, for any ∈ R, there exists such that

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6060/Chapter%201_html_m6d747ee9.gif

∴ f is onto.

Thus, f is one-one and onto and therefore, f−1 exists.

Let us define gR→ R by

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6060/Chapter%201_html_18a30fb0.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6060/Chapter%201_html_m7e9141ec.gif

Hence, f is invertible and the inverse of f is given by

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6060/Chapter%201_html_2a2d1bb7.gif

Question 8:

Consider fR→ [4, ∞) given by f(x) = x2 + 4. Show that f is invertible with the inverse f−1 of given by, where R+ is the set of all non-negative real numbers.

Answer:

fR+ → [4, ∞) is given as f(x) = x2 + 4.

One-one:

Let f(x) = f(y).

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6061/Chapter%201_html_m7cfa4fd1.gif

∴ f is a one-one function.

Onto:

For y ∈ [4, ∞), let y = x2 + 4.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6061/Chapter%201_html_506595d.gif

Therefore, for any ∈ R, there exists  such that

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6061/Chapter%201_html_1774a55e.gif.

∴ f is onto.

Thus, f is one-one and onto and therefore, f−1 exists.

Let us define g: [4, ∞) → Rby,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6061/Chapter%201_html_2c94539a.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6061/Chapter%201_html_m31641b49.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6061/Chapter%201_html_m1c8142f3.gif

Hence, f is invertible and the inverse of f is given by

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6061/Chapter%201_html_55a4df99.gif

Question 9:

Consider fR+ → [−5, ∞) given by f(x) = 9x2 + 6x − 5. Show that f is invertible withhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6065/Chapter%201_html_m6578c33e.gif.

Answer:

fR+ → [−5, ∞) is given as f(x) = 9x2 + 6x − 5.

Let y be an arbitrary element of [−5, ∞).

Let y = 9x2 + 6– 5.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6065/Chapter%201_html_m61e2d07d.gif

f is onto, thereby range f = [−5, ∞).

Let us define g: [−5, ∞) → R+ as

We now have:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6065/Chapter%201_html_md23d7aa.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6065/Chapter%201_html_46d51344.gif

and

Hence, f is invertible and the inverse of f is given by

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6065/Chapter%201_html_m19f5c246.gif

Question 10:

Let fX → Y be an invertible function. Show that f has unique inverse.

(Hint: suppose g1 and g2 are two inverses of f. Then for all y ∈ Y,

fog1(y) = IY(y) = fog2(y). Use one-one ness of f).

Answer:

Let fX → Y be an invertible function.

Also, suppose f has two inverses (say)

Then, for all y ∈Y, we have:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6068/Chapter%201_html_36b3004b.gif

Hence, f has a unique inverse.

*Question 11:

Consider f: {1, 2, 3} → {abc} given by f(1) = af(2) = b and f(3) = c. Find f−1 and show that (f−1)−1 = f.

Answer:

Function f: {1, 2, 3} → {abc} is given by,

f(1) = af(2) = b, and f(3) = c

If we define g: {abc} → {1, 2, 3} as g(a) = 1, g(b) = 2, g(c) = 3,  then we have:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6072/Chapter%201_html_m6c2426fc.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6072/Chapter%201_html_m2d002022.gifandhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6072/Chapter%201_html_m2a3edc62.gif, where X = {1, 2, 3} and Y= {abc}.

Thus, the inverse of exists and f−1 = g.

f−1: {abc} → {1, 2, 3} is given by,

f−1(a) = 1, f−1(b) = 2, f-1(c) = 3

Let us now find the inverse of f−1 i.e., find the inverse of g.

If we define h: {1, 2, 3} → {abc} as

h(1) = ah(2) = bh(3) = c, then we have:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6072/Chapter%201_html_6a6a33b2.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6072/Chapter%201_html_3c64248.gif, where X = {1, 2, 3} and Y = {abc}.

Thus, the inverse of exists and g−1 = h ⇒ (f−1)−1 = h.

It can be noted that h = f.

Hence, (f−1)−1 = f.

Question 12:

Let fX → Y be an invertible function. Show that the inverse of f−1 is f, i.e.,

(f−1)−1 = f.

Answer:

Let fX → Y be an invertible function.

Then, there exists a function gY → X such that gof = IX and fo= IY.

Here, f−1 = g.

Now, gof = IX and fo= IY

⇒ f−1of = IXand fof−1= IY

Hence, f−1Y → X is invertible and f is the inverse of f−1

i.e., (f−1)−1 = f.

Question 13:

If f→ be given by, then fof(x) is

(A)  (B) x3 (C) x (D) (3 − x3)

Answer:

fR → R is given as.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6075/Chapter%201_html_25362c5e.gif

The correct answer is C.

Question 14:

Letbe a function defined as. The inverse of f is map g: Range

given by

(A)  (B) 

(C)    (D) 

Answer:

It is given that is defined as

Let y be an arbitrary element of Range f.

Then, there exists such that 

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6076/Chapter%201_html_6a72f987.gif

Let us define g: Rangeas

Now,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6076/Chapter%201_html_5420ebc8.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6076/Chapter%201_html_m10d6f44e.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6076/Chapter%201_html_1f2105c2.gif

Thus, g is the inverse of f i.e., f−1 = g.

Hence, the inverse of f is the map g: Range, which is given by

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6076/Chapter%201_html_m59b54d57.gif

The correct answer is B.

Exercise 1.4

Question 1:

Determine whether or not each of the definition of given below gives a binary operation. In the event that * is not a binary operation, give justification for this.

(i) On Z+, define * by − b

(ii) On Z+, define * by ab

(iii) On R, define * by ab2

(iv) On Z+, define * by = |− b|

(v) On Z+, define * by a

Answer:

(i) On Z+, * is defined by * b = a − b.

It is not a binary operation as the image of (1, 2) under * is 1 * 2 = 1 − 2 = −1 ∉ Z+.

(ii) On Z+, * is defined by a * b = ab.

It is seen that for each ab ∈ Z+, there is a unique element ab in Z+.

This means that * carries each pair (ab) to a unique element * b ab in Z+.

Therefore, * is a binary operation.

(iii) On R, * is defined by a * b = ab2.

It is seen that for each ab ∈ R, there is a unique element ab2 in R.

This means that * carries each pair (ab) to a unique element * b abin R.

Therefore, * is a binary operation.

(iv) On Z+, * is defined by * b = |a − b|.

It is seen that for each ab ∈ Z+, there is a unique element |a − b| in Z+.

This means that * carries each pair (ab) to a unique element * b = |a − b| in Z+.

Therefore, * is a binary operation.

(v) On Z+, * is defined by a * b = a.

* carries each pair (ab) to a unique element * b a in Z+.

Therefore, * is a binary operation.

Question 2:

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

(i) On Z, define − b

(ii) On Q, define ab + 1

(iii) On Q, define 

(iv) On Z+, define = 2ab

(v) On Z+, define ab

(vi) On − {−1}, define 

Answer:

(i) On Z, * is defined by a * b = a − b.

It can be observed that 1 * 2 = 1 − 2 = 1and 2 * 1 = 2 − 1 = 1.

∴1 * 2 ≠ 2 * 1; where 1, 2 ∈ Z

Hence, the operation * is not commutative.

Also we have:

(1 * 2) * 3 = (1 − 2) * 3 = −1 * 3 = −1 − 3 = −4

1 * (2 * 3) = 1 * (2 − 3) = 1 * −1 = 1 − (−1) = 2

∴(1 * 2) * 3 ≠ 1 * (2 * 3); where 1, 2, 3 ∈ Z

Hence, the operation * is not associative.

(ii) On Q, * is defined by * b = ab + 1.

It is known that:

ab = ba ∀a, b ∈ Q

⇒ ab + 1 = ba + 1; a, b ∈ Q

⇒ * b = * b ; a, b ∈ Q

Therefore, the operation * is commutative.

It can be observed that:

(1 * 2) * 3 = (1 × 2 + 1) * 3 = 3 * 3 = 3 × 3 + 1 = 10

1 * (2 * 3) = 1 * (2 × 3 + 1) = 1 * 7 = 1 × 7 + 1 = 8

∴(1 * 2) * 3 ≠ 1 * (2 * 3); where 1, 2, 3 ∈ Q

Therefore, the operation * is not associative.

(iii) On Q, * is defined by * b 

It is known that:

ab = ba ; a, b ∈ Q

a, b ∈ Q

⇒ * b = * a ∀a, b ∈ Q

Therefore, the operation * is commutative.

For all a, b, c ∈ Q, we have:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6079/Chapter%201_html_5b28e7e9.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6079/Chapter%201_html_1bd6e76f.gif

Therefore, the operation * is associative.

(iv) On Z+, * is defined by * b = 2ab.

It is known that:

ab = ba ; ∀a, b ∈ Z+

⇒ 2ab = 2ba ∀a, b ∈ Z+

⇒ * b = * a ∀a, b ∈ Z+

Therefore, the operation * is commutative.

It can be observed that:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6079/Chapter%201_html_m2c873722.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6079/Chapter%201_html_m29ceb7e9.gif

∴(1 * 2) * 3 ≠ 1 * (2 * 3); where 1, 2, 3 ∈ Z+

Therefore, the operation * is not associative.

(v) On Z+, * is defined by * b = ab.

It can be observed that:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6079/Chapter%201_html_1e5b10e4.gif and https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6079/Chapter%201_html_2cf8b621.gif

∴ 1 * 2 ≠ 2 * 1; where 1, 2 ∈ Z+

Therefore, the operation * is not commutative.

It can also be observed that:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6079/Chapter%201_html_m4f475604.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6079/Chapter%201_html_4f45b508.gif

∴(2 * 3) * 4 ≠ 2 * (3 * 4) ; where 2, 3, 4 ∈ Z+

Therefore, the operation * is not associative.

(vi) On R, * − {−1} is defined by

It can be observed thatand 

∴1 * 2 ≠ 2 * 1 ; where 1, 2 ∈ − {−1}

Therefore, the operation * is not commutative.

It can also be observed that:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6079/Chapter%201_html_531cf619.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6079/Chapter%201_html_66962d18.gif

∴ (1 * 2) * 3 ≠ 1 * (2 * 3); where 1, 2, 3 ∈ − {−1}

Therefore, the operation * is not associative.

Question 3:

Consider the binary operation ∨ on the set {1, 2, 3, 4, 5} defined by = min {ab}. Write the operation table of the operation∨.

Answer:

The binary operation ∨ on the set {1, 2, 3, 4, 5} is defined as  b = min {ab}

ab ∈ {1, 2, 3, 4, 5}.

Thus, the operation table for the given operation ∨ can be given as:

1 2 3 4 5
1 1 1 1 1 1
2 1 2 2 2 2
3 1 2 3 3 3
4 1 2 3 4 4
5 1 2 3 4 5

Question 4:

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

(i) Compute (2 * 3) * 4 and 2 * (3 * 4)

(ii) Is * commutative?

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

(Hint: use the following table)

* 1 2 3 4 5
1 1 1 1 1 1
2 1 2 1 2 1
3 1 1 3 1 1
4 1 2 1 4 1
5 1 1 1 1 5

Answer:

(i) (2 * 3) * 4 = 1 * 4 = 1

2 * (3 * 4) = 2 * 1 = 1

(ii) For every a, b ∈{1, 2, 3, 4, 5}, we have * b = b * a. Therefore, the operation * is commutative.

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

∴(2 * 3) * (4 * 5) = 1 * 1 = 1

Question 5:

Let*′ be the binary operation on the set {1, 2, 3, 4, 5} defined by *′ = H.C.F. of and b. Is the operation *′ same as the operation * defined in Exercise 4 above? Justify your answer.

Answer:

The binary operation *′ on the set {1, 2, 3 4, 5} is defined as *′ b = H.C.F of a and b.

The operation table for the operation *′ can be given as:

*′ 1 2 3 4 5
1 1 1 1 1 1
2 1 2 1 2 1
3 1 1 3 1 1
4 1 2 1 4 1
5 1 1 1 1 5

We observe that the operation tables for the operations * and *′ are the same.

Thus, the operation *′ is same as the operation*.

Question 6:

Let * be the binary operation on given by a * = L.C.M. of and b. Find

(i) 5 * 7, 20 * 16 (ii) Is * commutative?

(iii) Is * associative? (iv) Find the identity of * in N

(v) Which elements of are invertible for the operation *?

Answer:

The binary operation * on N is defined as * b = L.C.M. of a and b.

(i) 5 * 7 = L.C.M. of 5 and 7 = 35

20 * 16 = L.C.M of 20 and 16 = 80

(ii) It is known that:

L.C.M of a and b = L.C.M of b and a ; a, b ∈ N.

a * b = * a

Thus, the operation * is commutative.

(iii) For a, b∈ N, we have:

(* b) * c = (L.C.M of a and b) * c = LCM of ab, and c

a * (b * c) = a * (LCM of b and c) = L.C.M of ab, and c

∴(* b) * c = a * (* c)

Thus, the operation * is associative.

(iv) It is known that:

L.C.M. of a and 1 = a = L.C.M. 1 and aa ∈ N

⇒ a * 1 = a = 1 * aa ∈ N

Thus, 1 is the identity of * in N.

(v) An element a in N is invertible with respect to the operation * if there exists an element b in N, such that * b = e = b * a.

Here, e = 1

This means that:

L.C.M of a and b = 1 = L.C.M of b and a

This case is possible only when a and b are equal to 1.

Thus, 1 is the only invertible element of N with respect to the operation *.

Question 7:

Is * defined on the set {1, 2, 3, 4, 5} by = L.C.M. of and a binary operation? Justify your answer.

Answer:

The operation * on the set A = {1, 2, 3, 4, 5} is defined as

a * b = L.C.M. of a and b.

Then, the operation table for the given operation * can be given as:

* 1 2 3 4 5
1 1 2 3 4 5
2 2 2 6 4 10
3 3 6 3 12 15
4 4 4 12 4 20
5 5 10 15 20 5

It can be observed from the obtained table that:

3 * 2 = 2 * 3 = 6 ∉ A, 5 * 2 = 2 * 5 = 10 ∉ A, 3 * 4 = 4 * 3 = 12 ∉ A

3 * 5 = 5 * 3 = 15 ∉ A, 4 * 5 = 5 * 4 = 20 ∉ A

Hence, the given operation * is not a binary operation.

Question 8:

Let * be the binary operation on defined by = H.C.F. of and b. Is * commutative? Is * associative? Does there exist identity for this binary operation on N?

Answer:

The binary operation * on N is defined as:

* b = H.C.F. of a and b

It is known that:

H.C.F. of a and b = H.C.F. of b and aa, b ∈ N.

a * b = * a

Thus, the operation * is commutative.

For ab∈ N, we have:

(* b)* c = (H.C.F. of a and b) * c = H.C.F. of ab, and c

*(* c)= *(H.C.F. of b and c) = H.C.F. of ab, and c

∴(* b) * c = a * (* c)

Thus, the operation * is associative.

Now, an element ∈ N will be the identity for the operation * if * e = a = e* a https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6085/Chapter%201_html_m24db17a.gif∈ N.

But this relation is not true for any ∈ N.

Thus, the operation * does not have any identity in N.

Question 9:

Let * be a binary operation on the set of rational numbers as follows:

(i)  −  (ii) a2 + b2

(iii)  ab  (iv) = (− b)2

(v)  (vi) ab2

Find which of the binary operations are commutative and which are associative.

Answer:

(i) On Q, the operation * is defined as * b = a – b.

It can be observed that:

u9wYI9Srk XCSW07EZ60X R69Om9pFD wB4 P5D WO4sre6DIYhlXeiyR4aCUAHm FilnCYbeCWr3DewfWdbzaTl1vF4ROOrPXFkEK3JgLqF70gQ3dcrbvGXyVwXISXlgsWJpXg
OY0CUm1BawMIvmlN6vRx8nXIrUqd07W 9QpjSulR 4RokCesK

Thus, the operation * is not commutative.

It can also be observed that:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6093/Chapter%201_html_m337a2963.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6093/Chapter%201_html_m58a3c011.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6093/Chapter%201_html_m32386eb0.gif

Thus, the operation * is not associative.

(ii) On Q, the operation * is defined as * b = a2 + b2.

For a, b ∈ Q, we have:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6093/Chapter%201_html_m670d7202.gif

* b = b * a

Thus, the operation * is commutative.

It can be observed that:

(1*2)*3 =(12 + 22)*3 = (1+4)*3 = 5*3 = 52+32= 25 + 9 = 34

1*(2*3)=1*(22+32) = 1*(4+9) = 1*13 = 12+ 13= 1 + 169 = 170∴ (1*2)*3 ≠ 1*(2*3) , where 1, 2, 3 ∈ Q

Thus, the operation * is not associative.

(iii) On Q, the operation * is defined as * b = a + ab.

It can be observed that:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6093/Chapter%201_html_m565d07f9.gif

Thus, the operation * is not commutative.

It can also be observed that:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6093/Chapter%201_html_m1179ebba.gif

Thus, the operation * is not associative.

(iv) On Q, the operation * is defined by a * b = (a − b)2.

For ab ∈ Q, we have:

* b = (a − b)2

* a = (b − a)2 = [− (a − b)]2 = (a − b)2

∴ * b = b * a

Thus, the operation * is commutative.

It can be observed that:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6093/Chapter%201_html_m32b45009.gif

Thus, the operation * is not associative.

(v) On Q, the operation * is defined as 

For ab ∈ Q, we have:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6093/Chapter%201_html_m1b52bbe8.gif

∴ * b = * a

Thus, the operation * is commutative.

For a, b, c ∈ Q, we have:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6093/Chapter%201_html_2951557c.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6093/Chapter%201_html_m77554e0b.gif

∴(* b) * c = a * (* c)

Thus, the operation * is associative.

(vi) On Q, the operation * is defined as * b = ab2

It can be observed that:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6093/Chapter%201_html_63ffb90a.gif

Thus, the operation * is not commutative.

It can also be observed that:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6093/Chapter%201_html_3d547ebd.gif

Thus, the operation * is not associative.

Hence, the operations defined in (ii), (iv), (v) are commutative and the operation defined in (v) is associative.

Question 10:

Find which of the operations given above has identity.

Answer:

An element ∈ Q will be the identity element for the operation * if

* e = a = e * ahttps://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6098/Chapter%201_html_m24db17a.gifa ∈ Q.

We are given

* b =ab4

⇒ a*e = a⇒ae4=a⇒ e=4

Similarly, it can be checked for e*a=awe get e=4Thus, e = 4 is the identity.

*Question 11:

Let A = × and * be the binary operation on A defined by

(ab) * (cd) = (cd)

Show that * is commutative and associative. Find the identity element for * on A, if any.

Answer:

A = N × N

* is a binary operation on A and is defined by:

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

Let (a, b), (c, d) ∈ A

Then, a, b, c, d ∈ N

We have:

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

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

[Addition is commutative in the set of natural numbers]

∴(a, b) * (c, d) = (c, d) * (a, b)

Therefore, the operation * is commutative.

Now, let (a, b), (c, d), (e, f) ∈A

Then, a, b, c, d, e∈ N

We have:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6103/Chapter%201_html_ma309ea.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6103/Chapter%201_html_1f4f19c1.gif

Therefore, the operation * is associative.

An element https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6103/Chapter%201_html_m107bb298.gifwill be an identity element for the operation * if

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6103/Chapter%201_html_m78ce6f74.gif, i.e., https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6103/Chapter%201_html_7e9327a.gifwhich is not true for any element in A.

Therefore, the operation * does not have any identity element.

Question 12:

State whether the following statements are true or false. Justify.

(i) For an arbitrary binary operation * on a set N N.

(ii) If * is a commutative binary operation on N, then * (c) = (b) * a

Answer:

(i) Define an operation * on N as:

* b = a + bhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6106/Chapter%201_html_m24db17a.gifa, b ∈ N

Then, in particular, for b = a = 3, we have:

3 * 3 = 3 + 3 = 6 ≠ 3

Therefore, statement (i) is false.

(ii) R.H.S. = (* b) * a

= (* c) * [* is commutative]

a * (* c) [Again, as * is commutative]

= L.H.S.

∴ a * (* c) = (* b) * a

Therefore, statement (ii) is true.

*Question 13:

Consider a binary operation * on defined as a3 + b3. Choose the correct answer.

(A) Is * both associative and commutative?

(B) Is * commutative but not associative?

(C) Is * associative but not commutative?

(D) Is * neither commutative nor associative?

Answer:

On N, the operation * is defined as * b = a3 + b3.

For, ab, ∈ N, we have:

* b = a3 + b3 = b3 + a3 = * a [Addition is commutative in N]

Therefore, the operation * is commutative.

It can be observed that:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6109/Chapter%201_html_6455216b.gif

∴(1 * 2) * 3 ≠ 1 * (2 * 3) ; where 1, 2, 3 ∈ N

Therefore, the operation * is not associative.

Hence, the operation * is commutative, but not associative. Thus, the correct answer is B.

Miscellaneous Exercise

Question 1:

Let fR → be defined as f(x) = 10x + 7. Find the  function gR → R such that  gof  =  fo= 1R.

Answer:

It is given that fR → R is defined as f(x) = 10x + 7.

One-one:

Let f(x) = f(y), where xy ∈R.

⇒ 10x + 7 = 10y + 7

⇒ x = y

∴ is a one-one function.

Onto:

For ∈ R, let y = 10x + 7.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6113/Chapter%201_html_m6cba91ea.gif

Therefore, for any ∈ R, there exists such that

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6113/Chapter%201_html_m3508d959.gif

∴ is onto.

Therefore, is one-one and onto.

Thus, f is an invertible function.

Let us define gR → R as

Now, we have:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6113/Chapter%201_html_m3371d141.gif

Hence, the required function gR → R is defined as

Question 2:

Let f: W → W be defined as f(n) = n – 1, if is odd and f(n) = n + 1, if n is even. Show that f is invertible. Find the inverse of f. Here, W is the set of all whole numbers.

Answer:

It is given that:

f: W → W is defined as

One-one:

Let f(n) = f(m).

It can be observed that if n is odd and m is even, then we will have n − 1 = m + 1.

⇒ n − m = 2

However, this is impossible.

Similarly, the possibility of n being even and m being odd can also be ignored under a similar argument.

∴Both n and m must be either odd or even.

Now, if both n and m are odd, then we have:

f(n) = f(m) ⇒ n − 1 = m − 1 ⇒ n = m

Again, if both n and m are even, then we have:

f(n) = f(m) ⇒ n + 1 = m + 1 ⇒ n = m

f is one-one.

It is clear that any odd number 2+ 1 in co-domain is the image of 2in domain and any even number 2in co-domain is the image of 2+ 1 in domain N.

f is onto.

Hence, is an invertible function.

Let us define g: W → W as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6117/Chapter%201_html_5cc1a749.gif

Now, when n is odd:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6117/Chapter%201_html_5f1e987a.gif

And, when n is even:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6117/Chapter%201_html_7809955c.gif

Similarly, when is odd:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6117/Chapter%201_html_112e0d8.gif

When is even:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6117/Chapter%201_html_74c1f870.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6117/Chapter%201_html_m5d3ea30e.gif

Thus, f is invertible and the inverse of is given by f—1 = g, which is the same as f.

Hence, the inverse of f is f itself.

Question 3:

If f→ R is defined by f(x) = x2 − 3+ 2, find f(f(x)).

Answer:

It is given that fR → R is defined as f(x) = x2 − 3x + 2.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6119/Chapter%201_html_me696a15.gif

Question 4:

Show that function fR → {x ∈ R: −1 < x < 1} defined by f(x) , R is one-one and onto function.

Answer:

It is given that fR → {x ∈ R: −1 < x < 1}is defined as f(x), R.

Suppose f(x) = f(y), where x∈ R.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6122/Chapter%201_html_6e6bb21.gif

It can be observed that if x is positive and y is negative, then we have:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6122/Chapter%201_html_m59e831a3.gif

Since is positive and y is negative:

x > y ⇒ x − y > 0

But, 2xy is negative.

Then, 

Thus, the case of x being positive and y being negative can be ruled out.

Under a similar argument, x being negative and y being positive can also be ruled out

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6122/Chapter%201_html_4dd19828.gif x and y have to be either positive or negative.

When x and y are both positive, we have:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6122/Chapter%201_html_m191a8ff7.gif

When x and y are both negative, we have:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6122/Chapter%201_html_7583119d.gif

∴ f is one-one.

Now, let y ∈ R such that −1 < < 1.

If x is negative, then there exists such that

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6122/Chapter%201_html_3cc4d3b.gif

If x is positive, then there existssuch that

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6122/Chapter%201_html_m783e1055.gif

∴ f is onto.

Hence, f is one-one and onto.

Question 5:

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

Answer:

fR → R is given as f(x) = x3.

Suppose f(x) = f(y), where xy ∈ R.

⇒ x3 = y3 … (1)

Now, we need to show that x = y.

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

⇒ x3 ≠ y3

However, this will be a contradiction to (1).

∴ x = y

Hence, f is injective.

Question 6:

Give examples of two functions fN → Z and gZ → Z such that g o f is injective but g is not injective.

(Hint: Consider f(x) = x and )

Answer:

Define fN → Z as f(x) = x and gZ → Z as 

We first show that g is not injective.

It can be observed that:

DNNWQurRbFUeQdRSIzhBAJDwFeDL9TeDgH6DNP9MlPDIhPLIxmYtJdZA8KCtx 62OplCII8l4TtXifizVuZjrhV8Q 647Lpzwx0gl8FSmTvntX8KRy LAUJuihuZGKfrBUnX2Q
7uvW7VUtWEgJb4DqYh3BBoM0TUGaiVI3je4F1M7PZvNsSZb0lQ3kktMdohZ83r2n0fxy1aavFp4ZNDZhKwpYj8nuHMcESV0mG5UcgSp0KlwPig1Rh 8rf6NI HwR

∴ g(−1) = g(1), but −1 ≠ 1.

∴ g is not injective.

Now, gofN → Z is defined as

Let xy ∈ N such that gof(x) = gof(y).

mq8IN59K4CR3GJcDarUhFCDoBtHdH16LsTHmYfFs2Hq xQyyhAvt KtFvg6br05pCNhJr3uaf 79nW EOwLRgpCXmEEfOiizWttTLEESeUk evOrlEyoMrtIGF40RWvuBlbCQ

Since x and y ∈ N, both are positive.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6127/Chapter%201_html_244b96fa.gif

Hence, gof is injective

Question 7:

Given examples of two functions fN → N and gN → N such that gof is onto but is not onto.

(Hint: Consider f(x) = x + 1 and

Answer:

Define fN → N by,

f(x) = x + 1

And, gN → N by,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6129/Chapter%201_html_m72219cf.gif

We first show that g is not onto.

For this, consider element 1 in co-domain N. It is clear that this element is not an image of any of the elements in domain N.

∴ f is not onto.

Now, gofN → N is defined by,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6129/Chapter%201_html_32937bc5.gif

Then, it is clear that for y ∈ N, there exists ∈ N such that gof(x) = y.

Hence, gof is onto.

Question 8:

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 AB in P(X), ARB if and only if A ⊂ B. Is R an equivalence relation on P(X)? Justify you answer:

Answer:

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.

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

∴ R is not symmetric.

Further, if ARand BRC, then A ⊂ B and ⊂ C.

⇒ A ⊂ C

⇒ ARC

∴ R is transitive.

Hence, R is not an equivalence relation since it is not symmetric.

Question 9:

Given a non-empty set X, consider the binary operation *: P(X) × P(X) → P(X)given by A * B = A ∩ B ; AB in P(X) is the power set of X. Show that is the identity element for this operation and is the only invertible element in P(X) with respect to the operation*.

Answer:

It is given that is defined as

We know that

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6135/Chapter%201_html_m4badca8f.gif

Thus, X is the identity element for the given binary operation *.

Now, an elementis invertible if there existssuch that

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6135/Chapter%201_html_m36354f52.gif

This case is possible only when A = X = B.

Thus, X is the only invertible element in P(X) with respect to the given operation*.

Hence, the given result is proved.

Question 10:

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

Answer:

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

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

Question 11:

Let S = {abc} and T = {1, 2, 3}. Find F−1 of the following functions F from S to T, if it exists.

(i)  F = {(a, 3), (b, 2), (c, 1)} (ii) F = {(a, 2), (b, 1), (c, 1)}

Answer:

S = {abc}, T = {1, 2, 3}

(i) F: S → T is defined as:

F = {(a, 3), (b, 2), (c, 1)}

⇒ F (a) = 3, F (b) = 2, F(c) = 1

Therefore, F−1T → S is given by

F−1 = {(3, a), (2, b), (1, c)}.

(ii) F: S → T is defined as:

F = {(a, 2), (b, 1), (c, 1)}

Since F (b) = F (c) = 1, F is not one-one.

Hence, F is not invertible i.e., F−1 does not exist.

Question 12:

Consider the binary operations*: ×→ and o: R × R → defined as  and a o b = a,;ab ∈ R. Show that * is commutative but not associative, o is associative but not commutative. Further, show that;abc ∈ Ra*(b o c) = (a * b) o (a * c). [If it is so, we say that the operation * distributes over the operation o]. Does o distribute over *? Justify your answer.

Answer:

It is given that *: ×→ and o: R × R → isdefined as

 and a o b = a;ab ∈ R.

For ab ∈ R, we have:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6142/Chapter%201_html_m4474dc4c.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6142/Chapter%201_html_32b9b83e.gif

a * b = b * a

∴ The operation * is commutative.

It can be observed that,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6142/Chapter%201_html_m1d19d376.gif

∴The operation * is not associative.

Now, consider the operation o:

It can be observed that 1 o 2 = 1 and 2 o 1 = 2.

∴1 o 2 ≠ 2 o 1 (where 1, 2 ∈ R)

∴The operation o is not commutative.

Let ab∈ R. Then, we have:

(b) o c = a o c a

a o (b o c) = a o b = a

⇒ b) o c = a o (b o c)

∴ The operation o is associative.

Now, let ab∈ R, then we have:

a * (b o c) = a * b =

(b) o (a * c) =

Hence, * (c) = (b) o (c).

Now,

1 o (2 * 3) =

(1 o 2) * (1 o 3) = 1 * 1 =

∴1 o (2 * 3) ≠ (1 o 2) * (1 o 3) (where 1, 2, 3 ∈ R)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6142/Chapter%201_html_4dd19828.gifThe operation o does not distribute over *.

Question 13:

Given a non-empty set X, let *: P(X) × P(X) → P(X) be defined as A * B = (A − B) ∪ (B − A); AB ∈ P(X). Show that the empty set Φ is the identity for the operation * and all the elements A of P(X) are invertible with A−1 = A. (Hint: (A − Φ) ∪ (Φ − A) = A and (A − A) ∪ (A − A) = A * A = Φ).

Answer:

It is given that *: P(X) × P(X) → P(X) is defined as

A * B = (A − B) ∪ (B − A); AB ∈ P(X).

Let ∈ P(X). Then, we have:

A * Φ = (A − Φ) ∪ (Φ − A) = A ∪ Φ = A

Φ * A = (Φ − A) ∪ (A − Φ) = Φ ∪ A = A

A * Φ = A = Φ * AA ∈ P(X)

Thus, Φ is the identity element for the given operation*.

Now, an element A ∈ P(X) will be invertible if there exists B ∈ P(X) such that

A * B = Φ = B * A. (As Φ is the identity element)

Now, we observed that

Hence, all the elements A of P(X) are invertible with A−1 = A.

Question 14:

Define a binary operation *on the set {0, 1, 2, 3, 4, 5} as

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6145/Chapter%201_html_23b339d7.gif

Show that zero is the identity for this operation and each element a ≠ 0 of the set is invertible with 6 − a being the inverse of a.

Answer:

Let X = {0, 1, 2, 3, 4, 5}.

The operation * on X is defined as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6145/Chapter%201_html_m44f621db.gif

An element e ∈ X is the identity element for the operation *, ifhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6145/Chapter%201_html_m7e4668e9.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6145/Chapter%201_html_74ee7c50.gif

Thus, 0 is the identity element for the given operation *.

An element a ∈ X is invertible if there exists b∈ X such that a * b = 0 = b * a.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6145/Chapter%201_html_3865723c.gif

i.e.,

a = −b or b = 6 − a

But, X = {0, 1, 2, 3, 4, 5} and ab ∈ X. Then, a ≠ −b.

b = 6 − a is the inverse of aa ∈ X.

Hence, the inverse of an element a ∈Xa ≠ 0 is 6 − a i.e., a−1 = 6 − a.

Question 15:

Let A = {−1, 0, 1, 2}, B = {−4, −2, 0, 2} and fgA → B be functions defined by f(x) = x2 − xx ∈ A and. Are f and g equal?

Justify your answer. (Hint: One may note that two function fA → B and g: A → B such that f(a) = g(a);a ∈A, are called equal functions).

Answer:

It is given that A = {−1, 0, 1, 2}, B = {−4, −2, 0, 2}.

Also, it is given that fgA → B are defined by f(x)= x2 − xx ∈ A and

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6146/Chapter%201_html_1989f306.gif.

It is observed that:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6146/Chapter%201_html_m2d9450ff.gif

Hence, the functions and g are equal.

Question 16:

Let A = {1, 2, 3}. Then number of relations containing (1, 2) and (1, 3) which are reflexive and symmetric but not transitive is

(A) 1 (B) 2 (C) 3 (D) 4

Answer:

The given set is A = {1, 2, 3}.

The smallest relation containing (1, 2) and (1, 3) which is reflexive and symmetric, but not transitive is given by:

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

This is because relation R is reflexive as (1, 1), (2, 2), (3, 3) ∈ R.

Relation R is symmetric since (1, 2), (2, 1) ∈R and (1, 3), (3, 1) ∈R.

But relation R is not transitive as (3, 1), (1, 2) ∈ R, but (3, 2) ∉ R.

Now, if we add any two pairs (3, 2) and (2, 3) (or both) to relation R, then relation R will become transitive.

Hence, the total number of desired relations is one.

The correct answer is A.

Question 17:

Let A = {1, 2, 3}. Then number of equivalence relations containing (1, 2) is

(A) 1 (B) 2 (C) 3 (D) 4

Answer:

It is given that A = {1, 2, 3}.

The smallest equivalence relation containing (1, 2) is given by,

R1 = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}

Now, we are left with only four pairs i.e., (2, 3), (3, 2), (1, 3), and (3, 1).

If we odd any one pair [say (2, 3)] to R1, then for symmetry we must add (3, 2). Also, for transitivity we are required to add (1, 3) and (3, 1).

Hence, the only equivalence relation (bigger than R1) is the universal relation.

This shows that the total number of equivalence relations containing (1, 2) is two.

The correct answer is B.

Question 18:

Let fR → R be the Signum Function defined as

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6149/Chapter%201_html_m37007912.gif

and gR → be the Greatest Integer Function given by g(x) = [x], where [x] is greatest integer less than or equal to x. Then does fog and gof coincide in (0, 1]?

Answer:

It is given that,

fR → R is defined as

Also, gR → R is defined as g(x) = [x], where [x] is the greatest integer less than or equal to x.

Now, let x ∈ (0, 1].

Then, we have:

[x] = 1 if x = 1 and [x] = 0 if 0 < x < 1.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6149/Chapter%201_html_m45f8aa1.gif
https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/230/6149/Chapter%201_html_m1d030a9d.gif

Thus, when x ∈ (0, 1), we have fog(x) = 0and gof (x) = 1.

Hence, fog and gof do not coincide in (0, 1].

Question 19:

Number of binary operations on the set {ab} are

(A) 10 (B) 16 (C) 20 (D) 8

Answer:

A binary operation * on {ab} is a function from {ab} × {ab} → {ab}i.e., *is a function from {(aa), (ab), (ba), (bb)} → {ab}.

Hence, the total number of binary operations on the set {ab} is 24 i.e., 16.

The correct answer is B.

 

NCERT Solutions Math Class 12 Chapter 1

Maths is often regarded as one of the most difficult topics for students in all grades, from the lowest to the highest. Class 12 pupils, in particular, have a considerably more difficult time studying for exams, as the level of difficulty of the chapters and topics offered continues to rise.

Functions and Relations  Chapter 1 consists of review of general notation for relations and functions. In class 11, students studied domain, codomain, and range, as well as the several types of specialised real-valued functions and their graphs. Students will study about several forms of connections and functions, as well as the construction of functions, in depth in class 12 Math Chapter 1.

 

Topics to Study in Class 12 Chapter 1 

Section no.

Topics

1.1

Introduction

1.2

Types of Relation

1.3

Types of Function

1.4

Composition of Functions and Invertible Function

1.5

Binary Operations

 

Weightage of Relations And Functions in Term I

Chapter

Marks

Relations And Functions

8 Marks

 

Why opt for Swastik Classes for Math Class 12 Chapter 1?

One of the top IIT JEE coaching institutes is Swastik Classes. Shobhit Bhaiya and Alok Bhaiya, pioneering mentors of IIT JEE Coaching Classes, started Swastik Classes in Anand Vihar. Over the last 15 years, they have educated and sent over 2000+ students to IITs and 5000+ students to different famous universities such as BITS, NITs, DTU, and NSIT. When it comes to coaching programmes for IIT JEE, Swastik Classes is the top IIT JEE Coaching in Delhi, favoured by students from all over India.

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The study process in Swastik courses is separated between pre-class and post-class work, which is one of the most significant aspects. They are precisely created to improve the student’s mental ability and comprehension.

 

Related Links

NCERT Solution for Class XIIth Maths Chapter 1 Relations and Function

NCERT Solution for Class XIIth Maths Chapter 13 Probability

NCERT Solution for Class XIIth Maths Chapter 12 Linear Programming

NCERT Solution for Class XIIth Maths Chapter 11 Three Dimensional Geometry

NCERT Solution for Class XIIth Maths Chapter 10 Vector

NCERT Solution for Class XIIth Maths Chapter 9 Differential Equations

 

What are the advantages of NCERT Solutions?

NCERT Books and answers will become your best friends as you prepare for the CBSE Boards Maths exam. One cannot afford to neglect any topic while studying Maths because it may be connected to other topics and chapters. As a result, have NCERT Solution for CBSE Board Exams in your arsenal of study materials to guarantee that your CBSE Class 12 board exam preparation is on track. NCERT Solutions for CBSE Board Examinations is the only source of information on how to prepare for CBSE board exams, particularly in maths.

 

FAQs on NCERT Solutions for Class 12 Math Chapter 1?

 

What is Relation Maths?

The Cartesian product is a subset of it. Or, to put it another way, a slew of points (ordered pairs). In other words, the collection of the ordered pair is defined as the relationship between the two sets, where the ordered pair is generated by the item from each set.

Example: (-2, 1), (4, 3), (7, -3), generally expressed with curly brackets in set notation form.

 

How do you find r2 in a Relationship?

The coefficient of determination, often known as R 2, is a metric that indicates a model’s quality of fit. It is a statistical measure of how well the regression line approximates the real data in the context of regression. When a statistical model is used to forecast future events or test hypotheses, it is critical to remember this. 

 

The total sum of squares is the sum of the distance the data is away from the mean all squared, and the sum squared regression is the sum of the residuals squared. It will take values between 0 and 1 because it is a percentage.

 

What is a Function Notation?

A function can be expressed using symbols and signs through function notation. Function notation is a less verbose way of explaining a function without having to write it out.

 

f(x), which is read as “f” of “x,” is the most commonly used function notation. The domain set and range set, respectively, are represented by the letter x enclosed in parenthesis and the complete symbol f(x).

 

What does F mean in calculus?

 

It is a function that provides a value within the range for any value of x inside the domain.

 

Typically, we write y=f(x), where x is a domain element and y is the range element that corresponds to it.

 

Why do we use FX instead of Y?

Although the notion that y is a function of x is implicitly accepted in most circumstances, y is commonly employed to explain the relationship between the expression and the coordinate axis. The f(x) notation is far more generic, and it’s used to indicate how an expression is related to its variables.

 

What is the Difference between Function and Equation?

 

There are at least two variables in a function: an output variable and one or more input variables. An equation is a mathematical formula that indicates that two expressions are equal. It can have any number of variables (none, one, or more). Although a function may be expressed as an equation, not all equations are functions.

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NCERT Solutions Class 12 Maths Chapters

  • Chapter 1 Relations and Function
  • Chapter 2 Inverse Trigonomtry
  • Chapter 3 Matrices
  • Chapter 4 Determinants
  • Chapter 5 Continuity and Differentiability
  • Chapter 6 Applications of Derivatives
  • Chapter 7 Integrals
  • Chapter 8 Application of Integrals
  • Chapter 9 Differential Equations
  • Chapter 10 Vector
  • Chapter 11 Three Dimensional Geometry
  • Chapter 12 Linear Programming
  • Chapter 13 Probability

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