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Class 12th Relations and Functions

Let $*$ be binary operation defined on $R$ by $a * b = 1 + a b, \forall a, b \in R$ . Then the operation $*$ is:

(i) Commutative but not associative

(ii) Associative but not commutative

(iii) Neither commutative nor associative

(iv) Both commutative and associative.

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alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} (i) : G i v e n t h a t \\ a * b = 1 + a b \forall a, b \in R \\ a n d b * a = 1 + b a \forall a, b \in R \\ a * b = b * a = 1 + a b \\ S o, * i s c o m m u t a t i v e . \\ N o w a * (b * c) = (a * b) * c \forall a, b, c \in R \\ L . H . S . a * (b * c) = a * (1 + b c) = 1 + a (1 + b c) = 1 + a + a b c \\ R . H . S . (a * b) * c = (1 + a b) * c = 1 + (1 + a b) . c = 1 + c + a b c \\ S o, * i s n o t a s s o c i a t i v e . \\ H e n c e, * i s c o m m u t a t i v e b u t n o t a s s o c i a t i v e . \end{array}$

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New answer posted

7 months ago

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

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alok kumar singh

Contributor-Level 10

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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7 months ago

Class 12th Chemistry NCERT Exemplar Solutions Class 12th Chapter Seven

In the preparation of compounds of Xe, Bartlett had taken O₂⁺ PtF₆⁻ as a base compound. This is because

(i) Both O₂ and Xe have the same size.

(ii) Both O₂ and Xe have the same electron gain enthalpy.

(iii) Both O₂ and Xe have almost the same ionisation enthalpy.

(iv) Both O₂ and Xe are gases.

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

Correct option is (iii)

Bartlett noticed that Xenon's first ionisation potential is nearly identical to that of oxygen and used Born-Haber calculations to predict the presence of a stable compound: Xe⁺ Pt F₆^-

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Class 12th Chemistry NCERT Exemplar Solutions Class 12th Chapter Seven

A black compound of manganese reacts with a halogen acid to give greenish yellow gas. When excess of this gas reacts with NH3 an unstable trihalide is formed. In this process the oxidation state of nitrogen changes from _________.

(i) – 3 to +3

(ii) – 3 to 0

(iii) – 3 to +5

(iv) 0 to – 3

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

Correct option is (iv)

MnO₂ ( Black )+4HCl→MnCl₂+2H₂O+Cl₂

Greenish yellow color

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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) $g \circ g$

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alok kumar singh

Contributor-Level 10

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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7 months ago

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.

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alok kumar singh

Contributor-Level 10

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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7 months ago

Class 12th Chemistry NCERT Exemplar Solutions Class 12th Chapter Seven

Hot conc. H₂SO₄ acts as moderately strong oxidising agent. It oxidises both metals and nonmetals. Which of the following element is oxidised by conc. H₂SO₄ into two gaseous products?

(i) Cu

(ii) S

(iii) C

(iv) Zn

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

Correct option is (iii)

C+2H₂SO₄→CO₂+2SO₂+2H₂O

Conc. H₂SO₄ oxidises C to produce two gaseous products.

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7 months ago

Class 12th Relations and Functions

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

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alok kumar singh

Contributor-Level 10

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

Sol:

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