NCERT Solutions for Class 12 Subject-wise...7/4/2018 ...

https://play.google.com/store/apps/details?id=haygot.togyah.app&referrer=utm_source%3Dguides%26utm_medium%3Dlanding_app_fragment

https://apps.apple.com/in/app/toppr-learning-app-class-5-12/id1043125663

NCERT Solutions for Class 12 Subject-wise ● Class 12 Mathematics

● Class 12 Physics

● Class 12 Chemistry

● Class 12 Biology

● Class 12 Accountancy

● Class 12 Business Studies

● Class 12 English

https://www.toppr.com/guides/ncert-solutions-for-class-12-maths/

https://www.toppr.com/guides/ncert-solutions-for-class-12-physics/

https://www.toppr.com/signup/?utm_medium=guides

https://www.toppr.com/guides/ncert-solutions-class-12-biology/


https://www.toppr.com/guides/ncert-solutions-for-class-12-business-studies/


7/4/2018 https://community.toppr.com/content/questions/print/?show_answer=1&show_topic=1&show_solution=1&page=1&qid=421842%2C+4278…

https://community.toppr.com/content/questions/print/?show_answer=1&show_topic=1&show_solution=1&page=1&qid=421842%2C+427819%2C+42… 1/22

#421228

Topic: Types of Relations

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

(i) Relation in the set defined as

(ii) Relative in the set of natural numbers defined as

(iii) Relation in the set as

(iv) Relative in the set of all integers defined as

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

Solution

(i)

is not reflexive since

Also, is not symmetric as , but

Also, is not transitive as , but .

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

(ii)

It is seen that is not reflexive.

Also .

But, . is not symmetric.

Now, since there is no pair in such that and , then cannot belong to .

is not transitive.


(iii)

We know that any number is divisible by itself.

,

is reflexive.

Now, , [ as is divisible by ]

But,

, [ as is not divisible by ]

is not symmetric.

Let . Then, is divisible by and is divisible by .

is divisible by .

R A = {1, 2, 3, . . . , 13, 14}

R = {(x, y) : 3x− y = 0}

R N

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

R A = {1, 2, 3, 4, 5, 6}

R = {(x, y) : yis divisible byx}

R Z

R = {(x, y) : x− y is an integer}

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

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

(c)R = {(x, y) : xis exactly7cm taller thany}

(d)R = {(x, y) : xis wife ofy}

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

A = {1, 2, 3, . . . . 13, 14}

R = {(x, y) : 3x− y = 0}

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

R (1, 1), (2, 2). . . . . . (14, 14) ∉ R

R (1, 3) ∈ R (3, 1) ∉ R.

R (1, 3), (3, 9) ∈ R (1, 9) ∉ R

R

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

(1, 1) ∉ R ⇒ R

(1, 6) ∈ R

(6, 1) ∉ R ∴ R

R (x, y) (y, z) ∈ R (x, z) R

∴ R

R

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

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

(x)

⇒ (x, x) ∈ R

∴ R

(2, 4) ∈ R 4 2

(4, 2) ∉ R. 2 4

∴ R

(x, y), (y, z) ∈ R y x z y

∴ z x



is transitive.

Hence, is reflexive and transitive but not symmetric.

(iv)

Now, for every as is an integer.

is reflexive.

Now, for every if , then is an integer.

is also an integer.

is an integer.

is symmetric.

Now,

Let and , where .

and are integers.

is an integer.

is transitive.

Hence, is reflexive, symmetric, and transitive.

(v) (a)

is reflexive.

If , then and work at the same place.

and work at the same place.

is symmetric.

Now, let

and work at the same place and and work at the same place.

and work at the same place.

is transitive.

Hence, is reflexive, symmetric and transitive.

(v) (b)

Clearly as and is the same human being.

is reflexive.

If , then and live in the same locality.

and live in the same locality.

is symmetric.

Now, let and .

and live in the same locality and and live in the same locality.

and live in the same locality.

⇒ (x, z) ∈ R

∴ R

R

R = {(x, y) : x− yis an integer}

x ∈ Z, (x, x) ∈ R x− x = 0

∴ R

x, y ∈ Z (x, y) ∈ R x− y

⇒ −(x− y)

⇒ (y − x)

∴ (y, x) ∈ R

⇒ R

(x, y) (y, z) ∈ R x, y, z ∈ Z

⇒ (x− y) (y − z)

⇒ x− z = (x− y) + (y − z)

∴ (x, z) ∈ R

∴ R

R

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

⇒ (x, x) ∈ R

∴ R

(x, y) ∈ R x y

⇒ y x

⇒ (y, x) ∈ R

∴ R

(x, y), (y, z) ∈ R

⇒ x y y z

⇒ x z

⇒ (x, z) ∈ R

∴ R

R

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

(x, y) ∈ R x x

∴ R

(x, y) ∈ R x y

⇒ y x

⇒ (y, x) ∈ R

∴ R

(x, y) ∈ R (y, z) ∈ R

⇒ x y y z

⇒ x z

⇒ (x, z) ∈ R



is transitive.

Hence, is reflexive, symmetric and transitive.

(v)(c)

Now,

Since human being cannot be taller than himself.

is not reflexive.

Now, let .

is exactly cm taller than

Then, is not taller than .

Indeed if is exactly cm taller than , then is exactly cm shorter than .

is not symmetric.

Now, let

is exactly cm taller than and is exactly cm taller than .

is exactly taller than .

is not transitive.


(v)(d)

Now,

Since cannot be the wife of herself.

is not reflexive.

Now, let

is the wife of .

Clearly is not wife of .

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

is not transitive.

Let

is the wife of and is the wife of .

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

is not transitive.


(v)(e)

As cannot be the father of himself.

is not reflexive.

Now, let

is the father of .

cannot be the father of .

Indeed, is the son or the daughter of .

∴ R

R

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

(x, x) ∉ R

x

∴ R

(x, y) ∈ R

⇒ x 7 y

y x

∴ (y, x) ∉ R

x 7 y y 7 x

∴ R

(x, y), (y, z) ∈ R

⇒ x 7 y y 7 z

⇒ x 14 z

∴ (x, z) ∉ R

∴ R

R

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

(x, x) ∉ R

x

∴ R

(x, y) ∈ R

⇒ x y

y x

∴ (y, x) ∉ R

x y y x

∴ R

(x, y), (y, z) ∈ R

⇒ x y y z

x z

∴ (x, z) ∉ R

∴ R

R

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

(x, x) ∉ R

x

∴ R

(x, y) ∈ R

⇒ x y

⇒ y y

y y



is not symmetric.

Now, let and .

is the father of and is the father of .

is not the father of .

Indeed is the grandfather of .

is not transitive.


#421245


Show that the relation in the set of real numbers, defined as is neither reflexive nor symmetric nor transitive.

Solution

It can be observed that , since .

is not reflexive.

Now, as

But, is not less than .

is not symmetric.

Now,

But,

is not transitive.


#421842


Check whether the relation defined in the set as is reflexive, symmetric or transitive.

Solution

Let .

A relative is defined on set as:

We can find , where .

For instance,

is not reflexive.

It can be observed that , but .

is not symmetric.

Now,

But,

is not transitive.


#421846


∴ (y, x) ∉ R

∴ R

(x, y) ∈ R (y, z) ∈ R

⇒ x y y z

⇒ x z

x z

∴ (x, z) ∉ R

∴ R

R

R R R = {(a, b) : a ≤ }b2

R = {(a, b) : a ≤ }b2

( , ) ∉ R1

2

1

2> =

1

2( )

1

2

2 1

4

∴ R

(1, 4) ∈ R 1 < 42

4 12

∴ (4, 1) ∉ R

∴ R

(3, 2), (2, 1.5) ∈ R

(as3 < = 4and2 < (1.5 = 2.25)22 )2

3 > (1.5 = 2.25)2

∴ (3, 1.5) ∉ R

∴ R

R

R {1, 2, 3, 4, 5, 6} R = {(a, b) : b = a+ 1}

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

R A

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), (6, 6) ∉ R

∴ R

(1, 2) ∈ R (2, 1) ∉ R

∴ R

(1, 2), (2, 3) ∈ R

(1, 3) ∉ R

∴ R

R



Show that the relative in defined as , is reflexive and transitive but not symmetric.

Solution

Clearly as .

is reflexive.

Now,

But, as is greater than .

is not symmetric.

Now, let .

Then, and

is transitive.


#421847


Check whether the relative in defined by is reflexive, symmetric or transitive.

Solution

It is observed that as .

is not reflexive.

Now,

But,

is not symmetric.

We have as and .

But as .

is not transitive.


#421848


Show that the relative in the set given by is symmetric but neither reflexive nor transitive.

Solution

Let

A relation on is defined as .

It is seen that .

is not reflexive.

Now, as and , then is symmetric.

Now, and

However,

is not transitive.

Hence, is symmetric but neither reflexive nor transitive.

#421849


R R R = {(a, b) : a ≤ b}

R = {(a, b); a ≤ b}

(a, a) ∈ R a = a

∴ R

(2, 4) ∈ R(as2 < 4)

(4, 2) ∉ R 4 2

∴ R

(a, b), (b, c) ∈ R

a ≤ b b ≤ c

⇒ a ≤ c

⇒ (a, c) ∈ R

∴ R

R

R R R = {(a, b) : a ≤ }b3

R = {(a, b) : a ≤ }b3

( , ) ∉ R1

2

1

2> =

1

2( )

1

2

3 1

8

∴ R

(1, 2) ∈ R (as 1 < = 8)23

(2, 1) ∉ R(as 2 > )13

∴ R

(3, ) ,( , ) ∈ R3

2

3

2

6

53 < ( )

3

2

3

<3

2( )

6

5

3

(3, ) ∉ R6

53 > ( )6

5

3

∴ R

R

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

A = {1, 2, 3}

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

(1, 1), (2, 2), (3, 3) ∉ R

∴ R

(1, 2) ∈ R (2, 1) ∈ R R

(1, 2) (2, 1) ∈ R

(1, 1) ∉ R

∴ R

R



Show that the relation in the set of all the books in a library of a college, given by is an equivalence relation.

Solution

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

Now, is reflexive since as and has the same number of pages.

Let and have the same number of pages.

and have the same number of pages.

is symmetric.

Now, let and

and and have the same number of pages and and have the same number of pages.

and have the same numbers of pages.

is transitive.

Hence, is an equivalence relation.

#421850


Show that the relative in the set given by , is an equivalence relation. Show that all the elements of are related to

each other and all the elements of are related to each other. But no element of is related to any element of .

Solution

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

is reflexive.

Let

is even.

is also even.

is symmetric.

Now, let and .

is even and is even.

is even and is even.

is even. [Sum of two even integers is even]

is even.

is transitive.


Now, all elements of the set 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 are related to each other as all the elements of this subset are even.

Also, no element of the subset can be related to any element of as all elements of are odd and all elements of are even. Thus, the modulus of

the difference between the two elements (from each of these two subsets) will not be even.


#421854


R A R = {(x, y) : x and yhave same number of pages}

A

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

R (x, x) ∈ R x x

(x, y) ∈ R ⇒ x y

⇒ y x

⇒ (y, x) ∈ R

∴ R

(x, y) ∈ R (y, z) ∈ R

⇒ x y y z

⇒ x z

⇒ (x, z) ∈ R

∴ R

R

R A = {1, 2, 3, 4, 5} R = {(a, b) : |a− b| is even} {1, 3, 5}

{2, 4} {1, 3, 5} {2, 4}

A = {1, 2, 3, 4, 5}

R = {(a, b) : |a− b|is even}

a ∈ A |a− a| = 0

∴ R

(a, b) ∈ R

⇒ |a− b|

⇒ | − (a− b)| = |b− a|

⇒ (b, a) ∈ R

∴ R

(a, b) ∈ R (b, c) ∈ R

⇒ |a− b| |b− c|

⇒ (a− c) (b− c)

⇒ (a− c) = (a− b) + (b− c)

⇒ |a− c|

⇒ (a, c) ∈ R

∴ R

R

{1, 2, 3}

{2, 4}

{1, 3, 5} {2, 4} {1, 3, 5} {2, 4}

R



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

(i)

(ii)

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

Solution

(i)

For any element , we have as is a multiple of .

is reflexive.

Now, let is a multiple of .

is a multiple of .

is symmetric.

Now, let

is a multiple of and is a multiple of .

is a multiple of and is a multiple of .

is a multiple of .

is a multiple of .

.

is transitive.


The set of elements related to is since

is a multiple of ,

is a multiple of , and

is a multiple of


(ii)

For any element , we have , since .

is reflexive.

Now, let .

is symmetric.

Now, let and .

and

is transitive.


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

Hence, the set of elements related to is .

#421860


R A = {x ∈ Z : 0 ≤ x ≤ 12}

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

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

1

A = {x ∈ Z : 0 ≤ x ≤ 12} = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

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

a ∈ A (a, a) ∈ R |a− a| = 0 4

∴ R

(a, b) ∈ R ⇒ |a− b| 4

⇒ | − (a− b)| = |b− a| 4

⇒ (b, a) ∈ R

∴

(a, b), (b, c) ∈ R

⇒ |a− b| 4 |b− c| 4

⇒ (a− b) 4 (b− c) 4

⇒ (a− c) = (a− b) + (b− c) 4

⇒ |a− c| 4

⇒ (a, c) ∈ R

∴ R

R

1 {1, 5, 9}

|1 − 1| = 0 4

|5 − 1| = 4 4

|9 − 1| = 8 4

R

A = {x ∈ Z : 0 ≤ x ≤ 12} = {0, 1, 2, 3, 4, . . . . . . , 11, 12}

R = {(a, b) : a = b} = {(0, ), (1, 1), (2, 2), . . . . . . . . , (11, 11), (12, 12)}

a ∈ A (a, a) ∈ R a = a

∴ R

(a, b) ∈ R

⇒ a = b

⇒ b = a

⇒ (b, a) ∈ R

∴ R

(a, b) ∈ R (b, c) ∈ R

⇒ a = b b = c

⇒ a = c

⇒ (a, c) ∈ R

∴ R

R

R 1 A 1

1 {1}



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.

Solution

(i)

Let

Define a relation on as

Relation is not reflexive as .

Now, as and also , is symmetric.

, but

is not transitive.

Hence, relation is symmetric but not reflexive or transitive.

(ii)

Consider a relation in defined as

For any , we have since cannot be strictly less than a itself. In fact, .

is not reflexive.

Now,

But, is not less than .

is not symmetric.

Now, let .

and

is transitive.

Hence, relation is transitive but not reflexive and symmetric.

(iii)

Let

Define a relation on as:

Relation is reflexive since for every

Relation is symmetric since for all .

Relation is not transitive since , but .

Hence, relation is reflexive and symmetric but not transitive.

(iv)

Define a relation in as:

Clearly as .

is reflexive.

Now,

But,

is not symmetric.

A = {5, 6, 7}

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

R (5, 5), (6, 6), (7, 7) ∉ R

(5, 6) ∈ R (6, 5) ∈ R R

⇒ (5, 6), (6, 5) ∈ R (5, 5) ∉ R

∴ R

R

R R

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

a ∈ R (a, a) ∉ R a a = a

∴ R

(1, 2) ∈ R(as1 < 2)

2 1

∴ (2, 1) ∉ R

∴ R

(a, b), (b, c) ∈ R

⇒ a < b b < c

⇒ a < c

⇒ (a, c) ∈ R

∴ R

R

A = {4, 6, 8}

R A

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

R {a ∈ A, (a, a) ∈ Ri. e. , (4, 4), (6, 6), (8, 8)} ∈ R

R (a, b) ∈ R ⇒ (b, a) ∈ R a, b ∈ R

R (4, 6), (6, 8) ∈ R (4, 8) ∉ R

R

R R

R = {(a, b) : ≥ }a3 b3

(a, a) ∈ R =a3 a3

∴ R

(2, 1) ∈ R(as ≥ )23 13

(1, 2) ∉ R(as < )13 23

∴ R



Let

and

is transitive.

Hence, relation is reflexive and transitive but not symmetric.

(v)

Let .

Define a relation on as:

Relation is not reflexive as

Relation is symmetric as and

It is seen that .

Also, .

the relation is transitive.

Hence, relation is symmetric and transitive but not reflexive.

#423434


Show that the relation in the set of points in a plane given by

, is an equivalence relation. Further, show that the

set of all point related to a point is the circle passing through with origin as centre.

Solution

Clearly, since the distance of point from the origin is always the same as the distance of the same point from the origin.

is reflexive.

Now, let .

the distance of point from the origin is the same as the distance of point from the origin.

The distance of point from the origin is the same as the distance of point from the origin.

is symmetric.

Now, let .

The distance of points and from the origin is the same and also, the distance of points and from the origin is the same.

The distance of points and from the origin is the same.

.

is transitive.

Therefore, is an equivalence relation.

The set of all points related to will be those points whose distance from the origin is the same as the distance of point from the origin.

In other words, if is the origin and , then the set of all points related to is at a distance of from the origin.

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

#423437


Show that the relation defined in the set of all triangles as , is equivalence relation. Consider three right angle triangles with sides

with sides and with sides . Which triangles among and are related?

Solution

(a, b), (b, c) ∈ R

⇒ ≥a3 b3 ≥b3 c3

⇒ ≥a3 c3

⇒ (a, c) ∈ R

∴ R

R

A = {−5, −6}

R A

R = {(−5, −6), (−6, −5), (−5, −5)}

R (−6, −6) ∉ R

R (−5, −6) ∈ R (−6, −5) ∈ R

(−5, −6), (−6, −5) ∈ R

(−5, −5) ∈ R

∴ R

R

R A

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

P ≠ (0, 0) P

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

(P ,P ) ∈ R P P

∴ R

(P ,Q) ∈ R

⇒ P Q

⇒ Q P

⇒ (Q,P ) ∈ R

∴ R

(P ,Q), (Q, S) ∈ R

⇒ P Q Q S

⇒ P S

⇒ (P , S) ∈ R

∴ R

R

P ≠ (0, 0) P

O(0, 0) OP = k P k

P

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

3, 4, 5, T2 5, 12, 13 T3 6, 8, 10 ,T1 T2 T3



is reflexive since every triangle is similar to itself.

Further, if , then is similar to .

is similar to .

is symmetric.

Now,

Let .

is similar to and is similar to .

is similar to .

is transitive.

Thus, is an equivalence relation.

Now, we can observe that:

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

Then, triangle is similar to triangle .

Hence, is related to .

#423440


Show that the relation defined in the set of all polygons as , is an equivalence relation. What is the set of all

elements in related to the right angle triangle with sides and ?

Solution

is reflexive since as the same polygon has the same number of sides with itself.

Let .

and have the same number of sides.


is symmetric.

Now,

Let .

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


is transitive.


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

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

#423441


Let be the set of all lines in plane and be the relation in defined as . Show that is an equivalence relation. Find the set of

all lines related to the line .

Solution

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

R

( , ) ∈ RT1 T2 T1 T2

⇒ T2 T1

⇒ ( , ) ∈ RT2 T1

∴ R

( , ), ( , ) ∈ RT1 T2 T2 T3

⇒ T1 T2 T2 T3

⇒ T1 T3

⇒ ( , ) ∈ RT1 T3

∴ R

R

= = (= )3

6

4

8

5

10

1

2

∴ T1 T3

T1 T3

T1 T3

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

A T 3, 4 5

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

R ( , ) ∈ RP1 P1

( , ) ∈ RP1 P2

⇒ P1 P2

⇒ P2 P1

⇒ ( , ) ∈ RP2 P1

∴ R

( , ), ( , ) ∈ RP1 P2 P2 P3

⇒ P1 P2 P2 P3

⇒ P1 P3

⇒ ( , ) ∈ RP1 P3

∴ R

R

A (T ) 3, 4, 5 3 T 3

A T

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

y = 2x+ 4



is reflexive as any line is parallel to itself i.e., .

Now, let .

is parallel to .

is parallel to .

is symmetric.

Now, let .

is parallel to . Also, is parallel to .

is parallel to .

is transitive.


The set of all lines related to the line is the set of all lines that are parallel to the line .

Slope of line is

It is known that parallel lines have the same slopes.

The line parallel to the given line is of the form , where .

Hence, the set of all lines related to the given line is given by , where .

#425308


Let be the relation in the set given by .

Choose the correct answer.

A is reflexive and symmetric but not transitive.

B is reflexive and transitive but not symmetric.

C is symmetric and transitive but not reflexive.

D is an equivalence relation.

Solution

It is seen that , for every .

is reflexive.

It is seen that , but .

is not symmetric.

Also, it is observed that for all

is transitive.


The correct answer is B.

#425309


Let be the relation in the set given by . Choose the correct answer.

A

B

C

D

Solution

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

R L1 ( , ) ∈ RL1 L1

( , ) ∈ RL1 L2

⇒ L1 L2

⇒ L2 L1

⇒ ( , ) ∈ RL2 L1

∴ R

( , ), ( , ) ∈ RL1 L2 L2 L3

⇒ L1 L2 L2 L3

⇒ L1 L3

⇒ R

R

y = 2x+ 4 y = 2x+ 4

y = 2x+ 4 m = 2

y = 2x+ c c ∈ R

y = 2x+ c c ∈ R

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

R

R

R

R

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

(a, a) ∈ R a ∈ {1, 2, 3, 4}

∴ R

(1, 2) ∈ R (2, 1) ∉ R

∴ R

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

∴ R

R

R N R = {(a, b) : a = b− 2, b > 6}

(2, 4) ∈ R

(3, 8) ∈ R

(6, 8) ∈ R

(8, 7) ∈ R



Now, since

Also, as

And, as

Now, consider .

We have and also, .

Hence the correct answer is C.

#425310

Topic: Types of Functions

Show that the function defined by is one-one and onto, where ' ' is the set of all non-zero real numbers. Is the result true, if the domain ' ' is replaced

by with co-domain being same as ' '?

Solution

R = {(a, b) : a = b− 2, b > 6}

b > 6, (2, 4) ∉ R

3 ≠ 8 − 2, (3, 8) ∉ R

8 ≠ 7 − 2

∴ (8, 7) ∉ R

(6, 8)

8 > 6 6 = 8 − 2

∴ (6, 8) ∈ R

f : R⋅ → R⋅ f(x) =1

xR⋅ R⋅

N R⋅



Given defined by

.....(1)

We have to check if is one-one and onto

(i) One-one:

Let for all

(by (1))

Hence, is one-one.

(ii) Onto :

For every , there is an such that

Hence, is onto.

Now, if co-domain of is replaced by N, then the function defined by

We have to check if is one -one and onto.

Let for

Hence, is one-one.

Onto :

is not onto as for every negative real number there is no in such that

Also, for but there is no such that belongs to .

Hence, the result is not true if domain is replaced by

f : R− {0} → R− {0}

f(x) =1

x

f

f(x) = f(y) x, y ∈ R− {0}

⇒ =1

x

1

y

⇒ x = y

f

y ∈ R− {0} x = ∈ R− {0}1

y

f(x) =1

1

y

⇒ f(x) = y

f

f g : N → R− {0}

g(x) =1

x

g

g(x) = g(y)

⇒ =1

x

1

y

⇒ x = y

g

g x N f(x) = y

1.6 ∈ R− {0} x ∈ N1

1.6N

R− {0} N



It is given that is defined by .

One-one.

is one-one.

Onto:

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

is onto.

Thus, the given function is one-one and onto.

Now, consider function $$g: N \rightarrow R\cdot $$ defined by

we have,

is one-one.

Further, it is clear that g is not onto as for there does not exit any in such that

Hence, function g is one-one but onto.

#427806


Check the injectivity and surjectivity of the following functions:

(i) given by

(ii) given by

(iii) given by

(iv) given by

(v) given by

Solution

(i)

It is seen that for ,

is injective

Now, , But, there does not exist any

in such that .

is not surjective.

Hence, function is injective but not surjective.

(ii)

It is seen that , but

is not injective.

Now, . But, there does not exist any

element such that

f : R⋅ → R⋅ f(x) =1

x

f(x) = f(y)

⇒ =1

x

1

y

⇒ x = y

∴ f

y ∈ R⋅ x = ∈ R⋅1

yy ≠= 0 f(x) = = y1

1

y

∴ f

(f)

g(x) =1

x

g( ) = g( ) ⇒ = ⇒ =x1 x21

x1

1

x2

x1 x2

⇒ g

1.2 ∈ R⋅ x N g(x) =1

1.2

f : N → N f(x) = x2

f : Z → Z f(x) = x2

f : R → R f(x) = x2

f : N → N f(x) = x3

f : Z → Z f(x) = x3

f(x) = x2

x, y ∈ N

f(x) = f(y) ⇒ = ⇒ x = yx2 y2

∴ f

2 ∈ N x

N f(x) = = 2x2

∴ f

f

f(x) = x2

f(−1) = f(1) = 1 −1 ≠ 1

∴ f

−2 ∈ Z

x ∈ Z f(x) = = −2x2



is not surjective.

Hence, function is neither injective nor surjective.

(iii)


is not injective.

Now, . But, there does not exist any element such that

is not surjective.

Hence, function is neither injective nor surjective.

(iv)

It is seen that for

is injective.

Now, . But, there does not exist any element

in domain such that

is not surjective.


(v)

It is seen that for

is injective.

Now, . But, there does not exist any element

in domain such that

is not surjective.


#427811


Prove that the Greatest Integer Function given by , is neither one-one nor onto, where denotes the greatest integer less that or equal to

Solution

It is seen that .

, but

is not one-one

Now, consider

It is known that is always an integer. Thus, there does not exist any element such that

is not onto.

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

∴ f

f

f(x) = x2

f(−1) = f(1) = 1 −1 ≠ 1

∴ f

−2 ∈ R x ∈ R f(x) = = −2x2

∴ f

f

f(x) = x3

x, y ∈ N, f(x) = f(y) ⇒ = ⇒ x = yx3 y3

∴ f

2 ∈ N

x N f(x) = = 2x3

∴ f

f

f(x) = x3

x, y ∈ Z, f(x) = f(y) ⇒ = ⇒ x = yx3 y3

∴ f

2 ∈ Z

x Z f(x) = = 2x3

∴ f

f

f : R → R f(x) = [x] [x] x

f(x) = [x]

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

∴ f(1.2) = f(1.9) 1.2 ≠ 1.9

∴ f

0.7 ∈ R

f(x) = [x] x ∈ R f(x) = 0.7

∴ f



#427812


Show that the modulus function given by , is neither one-one nor onto, where is , if is positive or and is , if is negative.

Solution

It is seen that , but

is not one-one

Now, consider .

but it is known that is always non-negative.

Thus, there does not exist any element in domain such that

is not onto.

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

#427813


Show that the Signum function , given by

.

is neither one-one nor onto.

Solution

.


is not one-one.

Now, as takes only values or for the element in co-domain ,

there does not exist any in domain such that

is not onto

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

#427814


Let { }, { } and let be a function from to . Show that is one-one.

Solution

It is given that { }, { }.

is defined as .

It is seen that the images of distinct elements of

under are distinct

Hence, function is one-one

#427817


f : R → R f(x) = |x| |x| x x 0 |x| −x x

f(x) = |x| = { x if x > 0−x if x < 0

f(−1) = |−1| = 1, f(1) = |1| = 1∴ f(−1) = f(1) −1 ≠ 1

∴ f

−1 ∈ R

f(x) = |x|

x R f(x) = |x| = −1

∴ f

f : R → R

f(x) =⎧⎩⎨

1, if x > 00, if x = 0−1, if x < 0

f(x) =

⎧

⎩⎨⎪

⎪

1

0

−1

ifx > 0

ifx = 0

ifx < 0

⎫

⎭⎬⎪

⎪

f(1) = f(2) = 1 1 ≠ 2

∴ f

f(x) 3 (1, 0, −1) −2 R

x R f(x) = −2

∴ f

A = 1, 2, 3 B = 4, 5, 6, 7 f = (1, 4), (2, 5), (3, 6) A B f

A = 1, 2, 3 B = 4, 5, 6, 7

f : A → B f = (1, 4), (2, 5), (3, 6)

∴ f(1) = 4, f(2) = 5, f(3) = 6

A

f

f



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

(i) defined by

(ii) defined by

Solution

(i)

Let such that

Hence f is one-one

Also the range of f is R = codomain of f,

since it is first order polynomial

Therefore f is both one-one and onto function

(ii)

is defined as

Let such that

does not imply that .

For instance,

is not one-one.

Consider an element in co-domain .

It is seen that is positive for all

Thus, there does not exist any in domain such that

is not onto.

Hence, is neither one-one nor onto

#427818


Let and be sets. Show that such that is bijective function.

Solution

is defined as .

Let such that .

and

is one-one.

Now, let be any element.

Then, there exists such that

. By definition of

is onto.

Hence, is bijective.

f : R → R f(x) = 3 − 4x

f : R → R f(x) = 1 + x2

, ∈ Rx1 x2 f( ) = f( )x1 x2

⇒ 3 − 4 = 3 − 4 ⇒ =x1 x2 x1 x2

f : R → R f(x) = 1 + x2

, ∈ Rx1 x2 f( ) = f( )x1 x2

⇒ 1 + = 1 +x21 x2

2

⇒ =x21 x2

2

⇒ = ±x1 x2

∴ f( ) = f( )x1 x2 =x1 x2

f(1) = f(−1) = 2

∴ f

−2 R

f(x) = 1 + x22 x ∈ R

x R f(x) = −2

∴ f

f

A B f : A×B → B×A f(a, b) = (b, a)

f : A×B → B×A f(a, b) = (b, a)

( , ), ( , ) ∈ A×Ba1 b1 a2 b2 f( , ) = f( , )a1 b1 a2 b2

⇒ ( , ) = ( , )b1 a1 b2 a2

⇒ = )b1 b2 ( = )a1 a2

⇒ ( , ) = ( , )a1 b1 a2 b2

∴ f

(b, a) ∈ B×A

(a, b) ∈ A×B

f(a, b) = (b, a) [ f]

∴ f

f



#427819


Let be defined by for all .

State whether the function is bijective. Justify your answer.

Solution

for all

is defined as

It can be observed that:

and

, where

is not one-one.

Consider a natural number in co-domain Case : is odd

for some .


Case : is even

for some .


is onto.

Hence, is not a bijective function

#427820


Let and . Consider the function defined by

. Is one-one and onto? Justify your answer

Solution

f : N → N f(n) = { }, if n is oddn+1

2

, if n is evenn

2

n ∈ N

f

f(n) = { }, if n is oddn+1

2

, if n is evenn

2

n ∈ N

f : N → N

f(1) = = 11 + 1

2f(2) = = 1

2

2

∴ f(1) = f(2) 1 ≠ 2

∴ f

(n) N I n

∴ n = 2r+ 1 r ∈ N

4r+ 1 ∈ N f(4r+ 1) = = 2r+ 14r+ 1 + 1

2

II n

∴ n = 2r r ∈ N

4r ∈ N f(4r) = = 2r4r

2

∴ f

f

A = R− 3 B = R− 1 f : A → B

f(x) = ( )x− 2

x− 3f



is defined as

Let, such that

is one-one.

Let . Then,

The function is onto if there exists

such that

Now,

Thus, for any , there exists

such that

is onto.

Hence, function is one-one and onto.

#427821


Let be defined as .

Choose the correct answer.

A is one-one onto

B is many-one onto

C is one-one but not onto

D is neither one-one nor onto

Solution

A = R− 3,B = R− 1

f(x) = ( )x− 2

x− 3

f : A → B

x, y ∈ A f(x) = f(y)

⇒ =x− 2

x− 3

y − 2

y − 3

⇒ x− 2y − 3 = y − 2x− 3

⇒ xy − 3x− 2y + 6 = xy − 3y − 2x+ 6

⇒ −3x− 2y = −3y − 2x

⇒ 3x− 2x = 3y − 2y

⇒ x = y

∴ f

y ∈ B = R− 1 y ≠ 1

f x ∈ A

f(x) = y

f(x) = y

⇒ = yx− 2

x− 3

⇒ x− 2 = xy − 3y

⇒ x(1 − y) = −3y + 2

⇒ x = ∈ A2 − 3y

1 − y[y ≠ 1]

y ∈ B

∈ A2 − 3y

1 − y

f ( ) =2 − 3y

1 − y

( )− 22 − 3y

1 − y

( )− 32 − 3y

1 − y

∴ f

f

f : R → R f(x) = x4

f

f

f

f



is defined as .

Let such that

does not imply that

For instance

is not one-one.

Consider an element in co-domain .

It is clear that there does not exist any in domain such that

is not onto.

Hence, function is neither one-one nor onto.

The correct answer is .

#427822


Let be defined as . Choose the correct answer.

A is one-one onto

B is many-one onto

C is one-one but not onto

D is neither one-one nor onto

Solution

is defined as .

Let such that

is one-one

Also, for any real number in co-domian ,

there exists in such that

is onto.

Hence, function is one-one and onto

The correct answer is .

#427823

Topic: Composite and Inverse Functions

Let and be given by and . Write down .

Solution

f : R → R f(x) = x4

x, y ∈ R f(x) = f(y)

⇒ =x4 y4

⇒ x = ±y

∴ f( ) = f( )x1 x2 =x1 x2

f(1) = f(−1) = 1

∴ f

−2 R

x R f(x) = −2

∴ f

f

D

f : R → R f(x) = 3x

f

f

f

f

f : R → R f(x) = 3x

x, y ∈ R f(x) = f(y)

⇒ 3x = 3y

⇒ x = y

∴ f

(y) Ry

3R

f ( ) = 3 ( ) = yy

3

y

3

∴ f

f

A

f : 1, 3, 4 → 1, 2, 5 g : 1, 2, 5 → 1, 3 f = (1, 2), (3, 5), (4, 1) g = (1, 3). (2, 3), (5, 1) gof



The function and are defined as

and

as and

as and

as and

#427825


Let and be functions from to . Show that

(i)

(ii)

Solution

(i)

To prove:

Consider:

for all

Hence,

(ii)

To prove:

Consider:

for all

Hence,

#427827


Find and , if and

ii) and

Solution

f : 1, 3, 5 → 1, 2, 5 g : 1, 2, 5 → 1, 3

f = (1, 2), (3, 5), (4, 1) g = (1, 3). (2, 3), (5, 1)

gof(1) = g(f(1)) = g(2) = 3 [f(1) = 2 g(2) = 3]

gof(3) = g(f(3)) = g(5) = 1 [f(3) = 5 g(5) = 1]

gof(4) = g(f(4)) = g(1) = 3 [f(4) = 1 g(1) = 3]

∴ gof = (1, 3), (3, 1), (4, 3)

f, g h R R

(f + g)oh = foh+ goh

(f. g)oh = (foh). (goh)


((f + g)oh)(x)

= (f + g)(h(x))

= f(h(x)) + g(h(x))

= (foh)(x) + (goh)(x)

= (foh) + (goh)(x)

∴ ((f + g)oh)(x) = (foh) + (goh)(x) x ∈ R



((f. g)oh)(x)

= (f. g)(h(x))

= f(h(x)). g(h(x))

= (foh)(x). (goh)(x)

= (foh). (goh)(x)

∴ ((f. g)oh)(x) = (foh). (goh)(x) x ∈ R


gof fog (i) f(x) = |x| g(x) = |5x− 2|

f(x) = 8x3 g(x) = x1

3



and

And

and

And

(i) f(x) = |x| g(x) = |5x− 2|

∴ (gof)(x) = g(f(x)) = g(|x|) = |5|x| − 2|

(fog)(x) = f(g(x)) = f(|5x− 2|) = ||5x− 2|| = |5x− 2|

(ii)f(x) = 8x3 g(x) = x1

3

∴ (gof)(x) = g(f(x)) = g(8 ) = (8 = 2xx3 x3)

1

3

(fog)(x) = f(g(x)) = f ( ) = 8 = 8xx1

3 ( )x1

3

3

NCERT Solutions for Class 12 Subject-wise...7/4/2018 ...

Documents

Transcript of NCERT Solutions for Class 12 Subject-wise...7/4/2018 ...