Ncert Solutions Maths class 12th

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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 functions defined by $f (x) = x^{3} + 5$ . Then $f^{- 1} (x)$ is

(A) ${(x + 5)}^{\frac{1}{3}}$

(B) ${(x - 5)}^{\frac{1}{3}}$

(C) ${(5 - x)}^{\frac{1}{3}}$

(D) $5 - x$

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 Objective Type Question as classified in NCERT Exemplar

Sol:

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New answer posted•a year ago

Ncert Solutions Maths class 12th Maths NCERT Exemplar Solutions Class 12th Chapter One

Which of the following functions from $Z$ into $Z$ are bijections?

(A) $f (x) = x^{3}$

(B) $f (x) = x + 2$

(C) $f (x) = 2 x + 1$

(D) $f (x) = x^{2} + 1$

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:

$\begin{array}{l} G i v e n t h a t f : Z \to Z \\ L e t x_{1}, x_{2} \in f (x) \Rightarrow f (x_{1}) = x_{1} + 2, f (x_{2}) = x_{2} + 2 \\ f (x_{1}) = f (x_{2}) \Rightarrow x_{1} + 2 = x_{2} + 2 \Rightarrow x_{1} = x_{2} \\ S o, f (x) i s o n e - o n e f u n c t i o n . \\ N o w, l e t y = x + 2 ∴ x = y - 2 \in Z \forall y \in Z \\ S o, f (x) i s o n t o f u n c t i o n . \\ ∴ f (x) i s b i j e c t i v e f u n c t i o n . \\ H e n c e, t h e c o r r e c t a n s w e r i s (b) . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Relations and Functions

Let $f : R \to R$ be defined by $f (x) = 3 x^{2} - 5$ and $g : R \to R$ by $g (x) = \frac{X}{x^{2} + 1}$ . Then $g ? f$ is

(A) $\frac{3 x^{2} - 5}{9 x^{4} - 30 x^{2} + 26}$

(B) $\frac{3 x^{2} - 5}{9 x^{4} - 6 x^{2} + 26}$

(C) $\frac{3 x^{2}}{x^{4} + 2 x^{2} - 4}$

(D) $\frac{3 x^{2}}{9 x^{4} + 30 x^{2} - 2}$

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:

$\begin{array}{l} H e r e, f (x) = 3 x^{2} - 5 a n d g (x) = \frac{x}{x^{2} + 1} \\ ∴ g o f = g o f (x) = g [3 x^{2} - 5] \\ = \frac{3 x^{2} - 5}{(3 x^{2} - 5)^{2} + 1} = \frac{3 x^{2} - 5}{9 x^{4} + 25 - 30 x^{2} + 1} \\ ∴ g o f = \frac{3 x^{2} - 5}{9 x^{4} + 25 - 30 x^{2} + 26} \\ H e n c e, t h e c o r r e c t a n s w e r i s (a) . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Relations and Functions

Let $f : R \to R$ be defined by $f (x) = \frac{1}{x} \forall x \in R$ . Then $f$ is

(A) One-one

(B) Onto

(C) Bijective

(D) $f$ is not defined

11 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:

$\begin{array}{l} G i v e n t h a t f (x) = \frac{1}{x} \\ P u t t i n g x = 0 ∴ f (x) = \frac{1}{0} = \infty \\ S o, f (x) i s n o t d e f i n e d . \\ H e n c e, t h e c o r r e c t a n s w e r i s (d) . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Relations and Functions

Let $A = {1, 2, 3, \dots, n}$ and $B = {a, b}$ . Then the number of surjections from $A$ into $B$ is

(A) ⁿp₂

(B) $2^{n} - 2$

(C) $2^{n} - 1$

(D) None of these

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:

Upvote

New answer posted•a year ago

Ncert Solutions Maths class 12th Relations and Functions

If the set $A$ contains 5 elements and the set $B$ contains 6 elements, then the number of one-one and onto mappings from $A$ to $B$ is

(A) 720

(B) 120

(C) 0

(D) None of these

2 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:

$\begin{array}{l} I f A a n d B s e t s h a v e m a n d n e l e m e n t s r e s p e c t i v e l y, \\ t h e n t h e n u m b e r o f o n e - o n e a n d o n t o m a p p i n g f r o m A t o B i s \\ n! i f m = n \\ a n d 0 i f m \neq n \\ H e r e, m = 5 a n d n = 6 \\ 5 \neq 6 \\ S o, N u m b e r o f m a p p i n g = 0 \\ H e n c e, t h e c o r r e c t a n s w e r i s (c) \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Maths NCERT Exemplar Solutions Class 12th Chapter One

**The identity element for the binary operation $$ defined on $Q ~ {0}$ as $a b = \frac{a b}{2} \forall a, b \in Q ~ {0}$ is

(A) 1

(B) 0

(C) 2

(D)** None of these

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 Objective Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : a * b = \frac{a b}{2} \forall a, b \in Q - {0} \\ L e t e b e t h e i d e n t i t y e l e m e n t \\ ∴ a * e = \frac{a e}{2} = a \Rightarrow e = 2 \\ H e n c e, t h e c o r r e c t a n s w e r i s (c) . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Relations and Functions

Let $A = {1, 2, 3}$ and consider the relation

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

Then $R$ is

(A) Reflexive but not symmetric

(B) Reflexive but not transitive

(C) Symmetric and transitive

(D) Neither symmetric, nor transitive

3 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:

$\begin{array}{l} G i v e n t h a t : A = {1, 2, 3} \\ a n d R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)} \\ H e r e, 1 R 1, 2 R 2 a n d 3 R 3, S o, R i s r e f l e x i v e . \\ 1 R 2 b u t 2 R 1 o r 2 R 3 b u t 3 R 2, \\ S o, R i s n o t s y m m e t r i c . \\ 1 R 1 a n d 1 R 2 \Rightarrow 1 R 3, S o, R i s t r a n s i t i v e . \\ H e n c e, t h e c o r r e c t a n s w e r i s (a) . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Relations and Functions

Let us define a relation $R$ in $R$ as $a R b$ if $a \geq b$ . Then $R$ is:

(A) An equivalence relation

(B) Reflexive, transitive but not symmetric

(C) Symmetric, transitive but not reflexive

(D) Neither transitive nor reflexive but symmetric

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:

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

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New answer posted•a year ago

Ncert Solutions Maths class 12th Relations and Functions

If a relation $R$ on the set ${1, 2, 3}$ be defined by $R = {(1, 2)}$ , then $R$ is

(A) Reflexive

(B) Transitive

(C) Symmetric

(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:

$\begin{array}{l} G i v e n t h a t : R = {(1, 2)} \\ a R a, ? ? s o i t i s n o t r e f l e x i v e . \\ a R b b u t b R a, s o i t i s n o t s y m m e t y r i c . \\ a R b a n d b R c \Rightarrow a R c w h i c h i s t r u e . \\ S o, R i s t r a n s i t i v e . \\ H e n c e, c o r r e c t a n s w e r i s (b) . \end{array}$

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Ncert Solutions Maths class 12th

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