Ncert Solutions Maths class 12th

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New answer posted•a year ago

Ncert Solutions Maths class 12th Relations and Functions

The maximum number of equivalence relations on the set $A = {1, 2, 3}$ are

(A) 1

(B) 2

(C) 3

(D) 5

6 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Objective Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l} H e r e, A = {1, 2, 3} \\ T h e n u m b e r o f e q u i v a l e n c e r e l a t i o n s a r e a s f o l l o w s : \\ R_{1} = {(1, 1), (1, 2), (2, 1), (2, 3), (1, 3)} \\ R_{2} = {(2, 2), (1, 3), (3, 1), (3, 2), (1, 2)} \\ R_{3} = {(3, 3), (1, 2), (2, 3), (1, 3), (3, 2)} \\ H e n c e, c o r r e c t a n s w e r i s (d) \end{array}$

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New question posted•a year ago

Ncert Solutions Maths class 12th Relations and Functions

The maximum number of equivalence relations on the set <math><mrow><mi>A</mi><mo>=</mo><mrow><mo>{</mo><mrow><mn>1</mn><mo>,</mo><mn>2</mn><mo>,</mo><mn>3</mn></mrow><mo>}</mo></mrow></mrow></math> are (A) 1 (B) 2 (C) 3 (D) 5

5 Views|Posted a year ago

Asked by Alok Kumar Singh

New answer posted•a year ago

Ncert Solutions Maths class 12th Relations and Functions

Consider the non-empty set consisting of children in a family and a relation $R$ defined as $a R b$ if $a$ is the brother of $b$ . Then $R$ is:

(A) Symmetric but not transitive

(B) Transitive but not symmetric

(C) Neither symmetric nor transitive

(D) Both symmetric and transitive.

7 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Objective Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l} H e r e, a R b \Rightarrow a i s a b r o t h e r o f b . \\ a R a \Rightarrow a i s a b r o t h e r o f a w h i c h i s n o t t r u e . \\ S o, R i s n o t r e f l e x i v e . \\ a R b \Rightarrow a i s a b r o t h e r o f b . \\ b R a \Rightarrow w h i c h i s n o t t r u e b e c a u s e b m a y b e s i s t e r o f a . \\ \Rightarrow a R b \neq b R a \\ S o, R i s n o t s y m m e t r i c . \\ N o w, a R b, b R c \Rightarrow a R c \\ \Rightarrow a i s t h e b r o t h e r o f b a n d b i s t h e b r o t h e r o f c . \\ ∴ a i s a l s o t h e b r o t h e r o f c . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Relations and Functions

Let $T$ be the set of all triangles in the Euclidean plane, and let a relation $R$ on $T$ be defined as $a R b$ if $a$ is congruent to $b \forall a, b \in T$ . Then $R$ is:

(A) Reflexive but not transitive

(B) Transitive but not symmetric

(C) Equivalence

(D) None of these.

5 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Objective Type Question as classified in NCERT Exemplar

Sol: $\begin{array}{l} \end{array}$

$\begin{array}{l} I f a = b \forall a, b \in T \\ t h e n a R a \Rightarrow a ≅ a w h i c h i s t r u e f o r a l l a \in T \\ S o, R i s r e f l e x i v e \\ N o w, a R b a n d b R a . \\ i . e ., a ≅ b a n d b ≅ a w h i c h i s t r u e f o r a l l a \in T \\ R i s s y m m e t r i c . \\ L e t a R b a n d b R c . \\ \Rightarrow a ≅ b a n d b ≅ a \Rightarrow a ≅ c \forall a, b, c \in T \\ S o, R i s t r a n s i t i v e . \\ H e n c e, R i s e q u i v a l e n c e r e l a t i o n . \\ S o, t h e c o r r e c t a n s w e r i s (c) . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Relations and Functions

Let $n$ be a fixed positive integer. Define a relation $R$ in $Z$ as follows: $\forall a, b \in Z, a R b$ if and only if $a - b$ is divisible by $n$ . Show that $R$ is an equivalence relation.

5 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Question as classified in NCERT Exemplar

Sol: $\begin{array}{l} \begin{array}{l} \begin{array}{l} \begin{array}{l} \end{array} \end{array} \end{array} \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Relations and Functions

Let $f : R \to R$ be the function defined by $f (x) = \frac{1}{2 - c o s x}, \forall x \in R$ . Then, find the range of $f$ .

5 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l} f (x) \frac{1}{2 - c o s x}, \forall x \in R \\ R a n g e o f c o s x i s [- 1, 1] \\ L e t f (x) = y = \frac{1}{2 - c o s x} \\ \Rightarrow 2 y - y c o s x = 1 \Rightarrow y c o s x = 2 y - 1 \\ \Rightarrow c o s x = \frac{2 y - 1}{y} = 2 - \frac{1}{y} \\ N o w - 1 \leq c o s x \leq 1 \\ \Rightarrow - 1 \leq 2 - \frac{1}{y} \leq 1 \Rightarrow - 1 - 2 \leq - \frac{1}{y} \leq 1 - 2 \\ \Rightarrow - 3 \leq - \frac{1}{y} \leq - 1 \Rightarrow 3 \geq \frac{1}{y} \geq 1 \Rightarrow \frac{1}{3} \leq y \leq 1 \\ ∴ t h e r a n g e o f f = [\frac{1}{3}, 1] . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Relations and Functions

If functions $f : A \to B$ and $g : B \to A$ satisfy $g ? f = I_{A}$ , then show that $f$ is one-one and $g$ is onto.

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l} \end{array} L e t x_{1}, x_{2} \in g o f$

$g o f {f (x_{1})} = g o f {f (x_{2})}$

$\Rightarrow g (x_{1}) = g (x_{2}) \begin{array}{l} \end{array}$

$∴$
$\begin{array}{l} x_{1} = x_{2} \end{array}$

$\begin{array}{l} H e n c e, f i s o n e - o n e a n d g i s o n t o . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Relations and Functions

Let $X = {1, 2, 3}$ and $Y = {4, 5}$ . Find whether the following subsets of $X \times Y$ are functions from $X$ to $Y$ or not:

$I. f = {(1, 4), (1, 5), (2, 4), (3, 5)}$

$II. g = {(1, 4), (2, 4), (3, 4)}$

$III. h = {(1, 4), (2, 5), (3, 5)}$

$IV. k = {(1, 4), (2, 5)}$ .

7 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Question as classified in NCERT Exemplar

Sol:

$G i v e n t h a t,$

$\begin{array}{l} X = {1, 2, 3} a n d Y = {4, 5} \end{array}$

$\begin{array}{l} ∴ X \times ? Y = {(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)} \end{array}$

$\begin{array}{l} (i) f = {(1, 4), (1, 5), (2, 4), (3, 5)} \end{array}$

$\begin{array}{l} f i s n o t a f u n c t i o n b e c a u s e f h a s n o u n i q u e i m a g e . \end{array}$

$\begin{array}{l} (i i) g = {(1, 4), (2, 4), (3, 4)} \end{array}$

$\begin{array}{l} S i n c e, g i s a f u n c t i o n a s e a c h e l e m e n t o f t h e d o m a i n h a s a u n i q u e i m a g e . \end{array}$

$\begin{array}{l} (i i i) h = {(1, 4), (2, 5), (3, 5)} \end{array}$

$\begin{array}{l} I t i s a f u n c t i o n a s e a c h e l e m e n t o f t h e d o m a i n h a s a u n i q u e i m a g e . \end{array}$

$\begin{array}{l} (i v) k = {(1, 4), (2, 5)} \end{array}$

$\begin{array}{l} k i s n o t a f u n c t i o n a s 3 d o e s n o t h a v e a n y i m a g e u n d e r t h e m a p p i n g . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Relations and Functions

Let the function $f : R \to R$ be defined by $f (x) = c o s x, \forall x \in R .$ Show that $f$ is neither one-one nor onto.

6 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n f u n c t i o n, f (x) = c o s x \forall x \in R \\ L e t [- \frac{π}{2}, \frac{π}{2}] \in f (x) \\ f (- \frac{π}{2}) = c o s (- \frac{π}{2}) = c o s \frac{π}{2} = 0 \\ c o s (\frac{π}{2}) = c o s \frac{π}{2} = 0 \\ B u t - \frac{π}{2} \neq \frac{π}{2} \\ ∴, f (x) i s n o t o n e - o n e . \\ N o w, f (x) = c o s x, \forall x \in R i s n o t o n t o a s t h e r e i s n o p r e - i m a g e f o r a n y r e a l n u m b e r . \\ W h i c h d o e s n o t b e l o n g t o t h e i n t e r v a l s [- 1, 1], t h e r a n g e o f c o s x . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Relations and Functions

Let $C$ be the set of complex numbers. Prove that the mapping $f : C \to R$ given by $f (z) = ∣ z ∣, \forall z \in C$ , is neither one-one nor onto.

7 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l} f (z) = | z | \forall z \in C \\ f (1) = | 1 | = 1 \\ f (- 1) = | - 1 | = 1 \\ f (1) = f (- 1) \\ B u t 1 \neq - 1 \\ T h e r e f o r e, i t i s n o t o n e - o n o . \\ N o w, l e t f (z) = y = | z | . \\ H e r e, t h e r e i s n o; p r e - i m a g e o f n e g a t i v e n u m b e r s . \\ H e n c e, i t i s n o t o n t o . \end{array}$

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Ncert Solutions Maths class 12th

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