Ncert Solutions Maths class 12th

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New answer posted•a year ago

Ncert Solutions Maths class 12th Relations and Functions

**Let $*$ be the binary operation defined on $Q$ . Find which of the following binary operations are commutative:

(i)** $a * b = a - b \forall a, b \in Q$

(ii) $a * b = a^{2} + b^{2} \forall a, b \in Q$

(iii) $a * b = a + a b \forall a, b \in Q$

(iv) $a * b = {(a - b)}^{2} \forall a, b \in Q$ .

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 Long Answer Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l} (i) a * b = a - b \in Q a, b \in Q, \\ S o, * i s b i n a r y o p e r a t i o n . \\ a * b = a - b a n d b * a = b - a \forall a, b \in Q \\ a - b \neq b - a \\ S o, * i s n o t c o m m u t a t i v e . \\ (i i) a * b = a^{2} + b^{2} \in Q, S o, * i s b i n a r y o p e r a t i o n . \\ a * b = b * a \\ \Rightarrow a^{2} + b^{2} = b^{2} + a^{2} \forall a, b \in Q \\ W h i c h i s t r u e . S o, * i s c o m m u t a t i v e . \\ (i i i) a * b = a + a b \in Q, S o, * i s b i n a r y o p e r a t i o n . \\ a * b = a + a b a n d b * a = b + b a \\ a + a b \neq b + b a \Rightarrow a * b \neq b * a \forall a, b \in Q . \\ S o, * i s n o t c o m m u t a t i v e . \\ (i v) a * b = {(a - b)}^{2} \in Q, S o, * i s b i n a r y o p e r a t i o n . \\ a * b = {(a - b)}^{2} a n d b * a = {(b - a)}^{2} \\ a * b = b * a \Rightarrow {(a - b)}^{2} = {(b - a)}^{2} \forall a, b \in Q . \\ S o, * i s c o m m u t a t i v e . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Relations and Functions

Functions $f, g : R \to R$ are defined, respectively, by $f (x) = x^{2} + 3 x + 1$ , $g (x) = 2 x - 3$ , find

(i) $f \circ g$

(ii) $g \circ f$

(iii) $f \circ f$

(iv) $g \circ g$

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 Long Answer Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l} (i) f o g \Rightarrow f [(g (x)] = [g] {(x)}^{2} + 3 [(g (x)] + 1 \\ = {(2 x - 3)}^{2} + 3 (2 x - 3) + 1 \\ = 4 x^{2} + 9 - 12 x + 6 x - 9 + 1 = 4 x^{2} - 6 x + 1 \\ (i i) g o f \Rightarrow g [f (x)] = 2 [x^{2} + 3 x + 1] - 3 \\ = 2 x^{2} + 6 x + 2 - 3 = 2 x^{2} + 6 x + 1 \\ (i i i) f o f \Rightarrow f [(f (x)] = {[f (x)]}^{2} + 3 [f (x)] + 1 \\ = ({x^{2} + 3 x + 1)}^{2} + 3 (x^{2} + 3 x + 1) + 1 \\ = x^{4} + 9 x^{2} + 1 + 6 x^{3} + 6 x + 2 x^{2} + 3 x^{2} + 9 x + 3 + 1 \\ = x^{4} + 6 x^{3} + 14 x^{2} + 15 x + 5 \\ (i v) g o g \Rightarrow g [g (x)] = 2 [g (x)] - 3 = 2 (2 x - 3) - 3 = 4 x - 6 - 3 = 4 x - 9 \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Relations and Functions

Using the definition, prove that the function $f : A \to B$ is invertible if and only if $f$ is both one-one and 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 Long Answer Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l} A f u n c t i o n f : X \to Y i s s a i d t o b e i n v e r t i b l e i f t h e r e e x i s t s a \\ f u n c t i o n g : Y \to X s u c h t h a t g o f = I_{X} a n d f o g = I_{Y} a n d t h e n t h e \\ i n v e r s e o f f i s d e n o t e d b y f^{- 1} . \\ A f u n c t i o n f : X \to Y i s s a i d t o b e i n v e r t i b l e i f f f i s a b i j e c t i v e f u n c t i o n . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Relations and Functions

Let $A = {1, 2, 3, \dots, 9}$ and $R$ be the relation in $A \times A$ defined by $(a, b) R (c, d)$ if $a + d = b + c$ for $(a, b), (c, d) i n A \times A$ . Prove that $R$ is an equivalence relation and also obtain the equivalent class $[(2, 5)]$ .

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 Long Answer Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l} H e r e, A = {1, 2, 3, \dots, 9} \\ a n d R \to A \times A d e f i n e d b y (a, b) R (c, d) \Rightarrow a + d = b + c \\ \forall (a, b), (c, d) \in A \times A \\ F o r r e f l e x i v e : (a, b) R (a, b) = a + b = b + a \forall a, b \in A w h i c h i s t r u e . S o, R i s r e f l e x i v e . \\ F o r s y m m e t r i c : (a, b) R (c, d) = (c, d) R (a, b) \\ L . H . S . a + d = b + c \\ R . H . S . c + b = d + a \\ L . H . S . = R . H . S . S o, R i s s y m m e t r i c . \\ F o r t r a n s i t i v e : (a, b) R (c, d) = (c, d) R (e, f) \Leftrightarrow (a, b) R (e, f) \\ \Rightarrow a + d = b + c a n d c + f = d + e \\ \Rightarrow a + d = b + c a n d d + e = c + f \\ \Rightarrow (a + d) - (d + e) = (b + c) - (c + f) \\ \Rightarrow a - e = b - f \\ \Rightarrow a + f = b + e \\ \Rightarrow (a, b) R (e, f) \\ S o, R i s t r a n s i t i v e . \\ H e n c e, R i s a n e q u i v a l e n c e r e l a t i o n . \\ E q u i v a l e n t c l a s s o f {(2, 5)} i s {(1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9)} \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Relations and Functions

Each of the following defines a relation on $N$ :

(i) $x$ is greater than $y, x, y \in N$

(ii) $x + y = 10, x, y \in N$

(iii) $x y, i s s q u a r e o f a n i n t e g e r x, y \in N$

(iv) $x + 4 y = 10, x, y \in N$ .

Determine which of the above relations are reflexive, symmetric, and transitive.

9 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

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New answer posted•a year ago

Ncert Solutions Maths class 12th Relations and Functions

Let $A = [- 1, 1]$ , Then, discuss whether the following functions defined on $A$ are one-one, onto, or bijective:

(i) $f (x) = \frac{x}{2}$

(ii) $g (x) = | x |$

(iii) $h (x) = x | x |$

(iv) $k (x) = x^{2}$ .

9 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} (i) G i v e n t h a t - 1 \leq x \leq 1 \\ L e t x_{1}, x_{2} \in f (x) \\ f (x_{1}) = \frac{1}{x_{1}} a n d f (x_{2}) = \frac{1}{x_{2}} \\ f (x_{1}) = f (x_{2}) \Rightarrow \frac{1}{x_{1}} = \frac{1}{x_{2}} \Rightarrow x_{1} = x_{2} \\ S o, f (x) i s o n e - o n e f u n c t i o n . \\ L e t f (x) = y = \frac{x}{2} \Rightarrow x = 2 y \\ F o r y = 1, x = 2 \notin [- 1, 1] \\ S o, f (x) i s n o t o n t o . H e n c e, f (x) i s n o t b i j e c t i v e f u n c t i o n . \\ (i i) H e r e, g (x) = | x | \\ g (x_{1}) = g (x_{2}) \Rightarrow | x_{1} | = | x_{2} | \Rightarrow x_{1} = \pm x_{2} \\ S o, g (x) i s n o t o n e - o n e f u n c t i o n . \\ L e t g (x) = y = | x | \Rightarrow x = \pm y \notin A \forall y \in A \\ S o, g (x) i s n o t o n t o f u n c t i o n . \\ H e n c e, g (x) i s n o t b i j e c t i v e f u n c t i o n . \\ (i i i) H e r e, h (x) = x | x | \\ h (x_{1}) = h (x_{2}) \\ \Rightarrow x_{1} | x_{1} | = x_{2} | x_{2} | \Rightarrow x_{1} = x_{2} \\ S o, h (x) i s o n e - o n e f u n c t i o n . \\ N o w, l e t h (x) = y = x | x | = x^{2} o r - x^{2} \\ \Rightarrow &thi \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Relations and Functions

Let $A = R - {3}, B = R - {1}$ . Let $f : A \to B$ be defined by $f (x) = \frac{x - 2}{x - 3} \forall x \in A$ . Then Show that $f$ is bijective.

8 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} H e r e, A \in R - {3}, B = R - {1} \\ G i v e n t h a t f : A \to B d e f i n e d b y f (x) = \frac{x - 2}{x - 3} \forall x \in A . \\ L e t x_{1}, x_{2} \in f (x) \\ ∴ f (x_{1}) = f (x_{2}) \\ \Rightarrow \frac{x_{1} - 2}{x_{1} - 3} = \frac{x_{2} - 2}{x_{2} - 3} \\ \Rightarrow (x_{1} - 2) (x_{1} - 3) = (x_{2} - 2) (x_{2} - 3) \\ \Rightarrow - x_{1} = - x_{2} \Rightarrow x_{1} = x_{2} \\ S o, i t i s i n j e c t i v e f u n c t i o n . \\ N o w, L e t y = \frac{x - 2}{x - 3} \\ \Rightarrow x y - 3 y = x - 2 \Rightarrow x y - x = 3 y - 2 \\ \Rightarrow x (y - 1) = 3 y - 2 \Rightarrow x = \frac{3 y - 2}{y - 1} \\ f (x) = \frac{x - 2}{x - 3} = \frac{\frac{3 y - 2}{y - 1} - 2}{\frac{3 y - 2}{y - 1} - 3} \Rightarrow \frac{3 y - 2 - 2 y + 2}{3 y - 2 - 3 y + 3} \Rightarrow y \\ \Rightarrow f (x) = y \in B . \\ S o, f (x) i s s u r j e c t i v e f u n c t i o n . \\ H e n c e, f (x) i s b i j e c t i v e f u n c t i o n . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Relations and Functions

Give an example of a map:

(i) which is one-one but not onto

(ii) which is not one-one but onto

(iii) which 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 Long Answer Type Question as classified in NCERT Exemplar

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

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New answer posted•a year ago

Ncert Solutions Maths class 12th Relations and Functions

Given $A = {2, 3, 4}, B = {2, 5, 6, 7}$ , construct an example of each of the following:

(a) An injective mapping from $A$ to $B$

(b) A mapping from $A$ to $B$ which is not injective

(c) A mapping from $B$ to $A$ .

8 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} H e r e, A = {2, 3, 4} a n d B = {2, 5, 6, 7} \\ (i) L e t f : A \to B b e t h e m a p p i n g f r o m A t o B \\ f = {(x, y) : y = x + 3} \\ ∴ f = {(2, 5), (3, 6), (4, 7)} w h i c h i s a n i n j e c t i v e m a p p i n g . \\ (i i) L e t g : A \to B b e t h e m a p p i n g f r o m A \to B s u c h t h a t \\ ∴ g = {(2, 5), (3, 5), (4, 2)} w h i c h i s n o t a n i n j e c t i v e m a p p i n g . \\ (i i i) L e t h : B \to A b e t h e m a p p i n g f r o m B t o A \\ h = {(y, x) : x = y - 2} \\ ∴ h = {(5, 3), (6, 4), (7, 3)} w h i c h i s t h e m a p p i n g f r o m B t o A . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Relations and Functions

Let $R$ be a relation defined on the set of natural numbers N as follows:

$R = {(x, y) : x \in N, y \in N, 2 x + y = 41}$ . Find the domain and range of the relation $R$ . Also, verify whether $R$ is reflexive, symmetric, and transitive.

8 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} G i v e n ? ? t h a t x \in N, y \in N a n d 2 x + y = 41 \\ ∴ D o m a i n o f R = {1, 2, 3, 4, 5, \dots, 20} \\ a n d R a n g e = {39, 37, 35, 33, 31, \dots, 1} \\ H e r e, (3, 3) \notin R \\ a s 2 \times 3 + 3 \neq 41 \\ S o, R i s n o t r e f l e x i v e . \\ R i s n o t s y m m e t r i c a s (2, 37) \in R b u t (37, 2) \notin R \\ R i s n o t t r a n s i t i v e a s (11, 19) \in R a n d (19, 3) \in R \\ b u t (11, 3) \notin R . \\ H e n c e, R i s n e i t h e r r e f l e x i v e, n o r s y m m e t r i c a n d n o r t r a n s i t i v e . \end{array}$

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Ncert Solutions Maths class 12th

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