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Ncert Solutions Maths class 12th

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Ncert Solutions Maths class 12th Relations and Functions

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

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

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Vishal Baghel

Contributor-Level 10

We have,

R= ${(a, b) : a \leq b^{2}}$ is a relation in R.

For $a \in R$ then is $b = a, a \leq a^{2}$ is not true for all real number less than 1.

Hence, R is not reflexive.

Let $(a, b) \in R$ and a=1 and b=2

Then, $a \leq b^{2}$ = $1 \leq 2^{2}$ = $1 \leq 4$ so, $(1, 2) \in R$

But $(b, a) = (2, 1)$

i.e., $2 \leq 1^{2}$ = $2 \leq 1$ is not true

so, $(2, 1) \notin R$

hence, R is not symmetric.

For, $(a, b) = (10, 4) & (b, c) = (4, 2) \in R$

We have, $a = 10 \leq 4^{2} = b^{2}$ => $10 \leq 16$ is true

So, $(10, 4) \in R$

And $4 \leq 2^{2}$ => $4 \leq 4$ So, $(4, 2) \in R$

But $10 \leq 2^{2}$ => $10 \leq 4$ is not true.

So, $(10, 2) \notin R$

Hence, R is not transitive.

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Ncert Solutions Maths class 12th Relations and Functions

Relation and Function Exercise 1.1 Solutions

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

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

R = {(x, y): 3x − y = 0}

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

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

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

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

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

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

(b) R = {(x, y)

...more

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Vishal Baghel

Contributor-Level 10

(i) We have, $R = {(x, y) : 3 x - y = 0}$ a relation in set A= ${1, 2, 3 . . . . . . . . . . 14}$

For $x \in A, y = 3 x$ or $y \neq x$ i.e.,

$(x, x)$ does not exist in R

$∴$ R is not reflexive.

For $(x, y) \in R, y = 3 x$

Then $(y, x) x \neq 3 y$

So $(y, x) \notin R$

$∴$ R is not symmetric

For $(x, y) \in R$ and $(y, z) \in R$ . We have

$y = 3 x$ and $z = 3 y$

Then $z = 3 (3 x) = 9 x$

i.e., $(x, z) \notin R$

$∴$ R is not Transitive

(ii) We have,

R= ${(x, y) : y = x + 5 & x < 4}$ is a relation in N

= ${(1, 1 + 5), (2, 2 + 5), (3, 3 + 5)}$

= ${(1, 6), (2, 7), (3, 8)}$

Clearly, R is not reflexive as $(x, x) \notin R$ and $x < 4 & x \in N$

Also, R is not symmetric as $(1, 6) \in R$ but $(6, 1) \notin R$

And for $(x, y) \in R (y, z) \notin R$ . Hence, R is not Transitive.

(iii) R= ${(x, y); y$ is divisible by x $}$ is a relation in set

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

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

Hence, R is reflexive because $(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) \in R$ i.e., $(x, x) \in R$

R is not sy

...more

(i) We have, $R = {(x, y) : 3 x - y = 0}$ a relation in set A= ${1, 2, 3 . . . . . . . . . . 14}$

For $x \in A, y = 3 x$ or $y \neq x$ i.e.,

$(x, x)$ does not exist in R

$∴$ R is not reflexive.

For $(x, y) \in R, y = 3 x$

Then $(y, x) x \neq 3 y$

So $(y, x) \notin R$

$∴$ R is not symmetric

For $(x, y) \in R$ and $(y, z) \in R$ . We have

$y = 3 x$ and $z = 3 y$

Then $z = 3 (3 x) = 9 x$

i.e., $(x, z) \notin R$

$∴$ R is not Transitive

(ii) We have,

R= ${(x, y) : y = x + 5 & x < 4}$ is a relation in N

= ${(1, 1 + 5), (2, 2 + 5), (3, 3 + 5)}$

= ${(1, 6), (2, 7), (3, 8)}$

Clearly, R is not reflexive as $(x, x) \notin R$ and $x < 4 & x \in N$

Also, R is not symmetric as $(1, 6) \in R$ but $(6, 1) \notin R$

And for $(x, y) \in R (y, z) \notin R$ . Hence, R is not Transitive.

(iii) R= ${(x, y); y$ is divisible by x $}$ is a relation in set

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

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

Hence, R is reflexive because $(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) \in R$ i.e., $(x, x) \in R$

R is not symmetric as $(1, 2) \in R$ but $(2, 1) \notin R$

And for $x, y, z \in A$ and $(x, y) \in R & (y, z) \in R$

$\frac{y}{z} = n$ and $\frac{z}{y} = m$ where $n & m \in N$

Then, $z = m y = m (n x) = m n . x$

$\frac{z}{x} = m . n, m, m n \in N$

Hence, $(x, z) \in R$

$∴$ R is transitive

(iv) R= ${(x, y) : x - y$ is an integer $}$ is a relation in set Z

For $x \in Z$

$x - x = 0$ is an integer

So, $(x, x) \in R$ i.e., R is reflexive

For $x, y \in Z$ and $(x, y) \in R$

$x - y$ is an integer

$- (y - x)$ is an integer

$(y - x)$ is an integer

So, $(y, x) \in R$ i.e., R is symmetric

For $x, y, z \in R$ and $(x, y) \in R & (y, z) \in R$ We have,

$x - y$ is an integer

$y - z$ is an integer

So, $(x - y) + (y - z)$ is also an integer

$x - z$ is an integer

So, $(x, z) \in R$ i.e., R is transitive.

(v) (a) R= ${(x, y) : x$ and $y$ work at same place $}$ in set A of human being.

For $x \in A$ we get

$x & x$ work at same place

$(x, x) \in R$ So, R is reflexive.

For $x, y \in A$ and $(x, y) \in R$ We get,

$x & y$ work at same place

$y & x$ work at same place

$(y, x) \in R$ . So, R is symmetric.

For $x, y, z \in A$ and $(x, y)$ and $(y, z) \in R$ . We have,

$x & y$ work at same place

$y & z$ work at same place

$x & z$ work at same place

i.e., $(x, z) \in R$ . So, R is Transitive.

(b) R= ${(x, y) : x & y$ live in same locality $}$

For $x \in A$ , we get,

$x & x$ live in same locality

$(x, x) \in R$ So, R is reflexive.

For $x, y \in A$ and $(x, y) \in R$ we get,

$x & y$ live in same locality

$y & x$ live in same locality

i.e., $(y, x) \in R$ . So, R is symmetric.

For $x, y, z \in A$ and $(x, y)$ & $(y, z) \in R$ . Then

$x, y$ live in same locality

$y, z$ live in same locality

$x, z$ live in same locality

So, $(x, z) \in R$ i.e., R is transitive.

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

For $x \in A$ ,

Height $(x) \neq 7 +$ height of $(x)$

So, $(x, x) \notin R$ . i.e., R is not reflexive.

For, $x, y \in A$ and $(x, y) \in R$ we have,

Height $(x)$ $= 7 + h e i g h t (y)$

But $h e i g h t (y) \neq 7 + h e i g h t (x)$

So, $(y, x) \in R$ i.e., R is not symmetric.

For $x, y, z \in A$ and $(x, y) \in R$ and $(y, z) \in R$ we have,

Height $(x) = 7 + h e i g h t (y)$

And $h e i g h t (y) = 7 + h e i g h t (z)$

So, $h e i g h t (x) = 7 + 7 + h e i g h t (z) = 14 + h e i g h t (z)$

i.e., $(x, z) \notin R$

So, R is not transitive.

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

For $x \in A,$

X is not wife of x.

So, $(x, x) \notin R$ i.e., R is not reflexive.

For $x, y \in A$ and $(x, y) \in R$ ,

X is wife of y but y is not wife of x

So, $(y, x) \notin R$ . i.e., R is not symmetric.

For $x, y, z \in A$ and $(x, y)$ and $(y, z) \in R$

X is wife of y

Y is wife of z

But y can never be husband & wife simultaneously

So, R is not transitive.

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

For $x \in A$ ,

X is not a father of x

So, $(x, x) \notin R$ i.e., R is not reflexive.

For $x, y \in A$ and $(x, y) \in R$

X is father of y

Y is father of z

But x is not father of z

So, $(x, z) \notin R$ i.e., R is not transitive.

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Ncert Solutions Maths class 12th Inverse Trigonometric Functions

What are the principal values for the inverse trigonometric functions?

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Chandra Pruthi

Beginner-Level 5

students can check the table for the principal values for all ITFs below;

Function	Principal Value Range (in radians)
sin? ¹x	–? /2 to? /2
cos? ¹x	0 to?
tan? ¹x	–? /2 to? /2
cot? ¹x	0 to?
sec? ¹x	0 to? (except? /2)
cosec? ¹x	–? /2 to? /2 (except 0)

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Ncert Solutions Maths class 12th NCERT

What is the principal value of an inverse trigonometric function?

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Himanshi Singh

Beginner-Level 5

To understand this, Assume you have a bucket that has infinite number of apples and if your mother asks "give me the apple". How will you figure out which one is "The Apple", she is asking for.

Similarly any inversre trigonometric functions behaves like a Many-one Function; which means,

For Example $\sin^ {-1}\left (\frac {2} {3}\right)$ can have many solutions, we need to fix one solutions which can be used as standerd value for the function.

A standerd value (Angles) of any inverse trigonometric value lies between a fiexed range is known as principal value.

For Ex; The value of $\sin^ {-1}\left (\frac {2} {3}\right)$ will always lie between –? /2 to? /2.

$y = \sin^{?1} (\frac{2}{3}) ? \sin (y) = \frac{2}{3} ? y ? [? \frac{?}{2}, \frac{?}{2}]$

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11 months ago

Ncert Solutions Maths class 12th NCERT

What are the Type of Functions discussed in the Class 12 Chapter 1 Relation and Function?

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Jaya Sinha

Beginner-Level 5

The Class 12 Relations and Functions explores various types of Functions, Students can check main types dicussed in this chapter below;

One-One Function (Injective)
Onto Function (Surjective)
One-One and Onto Function (Bijective)
Identity Function
Constant Function
Inverse of a Function
Composite Functions

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Ncert Solutions Maths class 12th Maths Determinants

Which websites offer the best NCERT Class 12 Determinants Solutions with detailed explanations?

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Aayush Kumari

Beginner-Level 5

A lot of website offer NCERT Class 12 Determinants Soluitions for students, a lot of them provide direct solutions on the page, many of them provide solutions PDF to download and study. However, Shiksha provides the best NCERT Class 12 Determinants Solutions with detailed and accurate explanation for each and every question. Shiksha's NCERT Solutions are available in both mode directly accessible on the page, and in the downloadable PDF format. Not only this Shiksha also provide weightage, importan topics and formulas for students to solve the exercise on their own.

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Ncert Solutions Maths class 12th NCERT Maths 12th

Where can I download NCERT Solutions for Class 12 Determinants in PDF format?

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Nishtha Datta

Beginner-Level 5

Students can check Shiksha's page to access the NCERT Class 12 Determinants Solutions and download the PDF of solutions for free. We have attached the Solution PDF for each chapter on the NCERT Solution page for the same page, students can directly access the solutions on the page and also download the Class 12 Determinants Solutions PDF for offline study and better practice.

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Ncert Solutions Maths class 12th NCERT

Do NCERT Class 12 Determinants Solutions cover all types of questions asked in board exams?

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nitesh singh

Contributor-Level 10

NCERT Class 12 Determinants Solutions provide complete step-by-step solutions for all the questions of the chapter. Since NCERT Textbooks cover all types of questions commonly asked in board exams, So the NCERT Solution for Determinats also include all types of questions. Questions are asked in many format such including short answer type, long answer type, and application-based problems in class 12 boards.

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Ncert Solutions Maths class 12th NCERT

How much weightage Determinants chapter holds in the class 12 Board Exam?

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Satyendra Dhyani

Beginner-Level 5

Determinats and Matrices are generally combined as a unit in class 12 syllaus, Determinats and Matrices unit carries in general 10-12 marks weightage in the CBSE 12th boards. Specifically Determinant chapter carries weightage of around 6 marks in the actual exams.

Students can use Shiksha's NCERT Solutions for Determinats to prepare and get full marks in the boards.

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Ncert Solutions Maths class 12th NCERT

How can Shiksha's NCERT Solutions for Determinants help in Class 12th board exam preparation?

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Esha Garg

Beginner-Level 5

Determinats holds good weightage in the class 12 boards, Students know that its an easy chapter for scoring well which makes this chapter both scoring and important for not only boards but also upcoming competitve exams in future. Shiksha can help in more than one way for the same, We have compiled NCERT Solutions for all the class 12 chapters including Determinats NCERT solutions. Shiksha's NCERT Solutions of Determinants are accurate, to the point and provide step-by-step explanations for all the questions of the chapters. Students can use these solutions for practice, revision and other form of study.

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