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Maths Relations and Functions

52. Find which of the operations given above has identity.

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Vishal Baghel

Contributor-Level 10

Let the identity be I.

An element $e \in Q$ will be the identity element for the operation* if

$a * e = a = e * a, \forall a \in Q .$

We are given

$\begin{array}{l} a * b = a b 4 \\ \Rightarrow a * e = a \Rightarrow a e 4 = a \Rightarrow e = 4 \end{array}$

Similarly, it can be checked for $e * a = a$ , we get e=4. Thus, e=4 is the identity

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New answer posted

a year ago

Maths Relations and Functions

51. Let * be a binary operation on the set Q of rational numbers as follows:

(i) a * b = a − b

(ii) a * b = a² + b²

(iii) a * b = a + ab

(iv) a * b = (a − b)²

(v) a * b = ab/4 (vi) a * b = ab²

Find which of the binary operations are commutative and which are associative.

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Vishal Baghel

Contributor-Level 10

(i) On Q the operation* is defined as a*b=a-b.

It can be observed that:

$\begin{array}{l} \frac{1}{2} . \frac{1}{3} = \frac{1}{2} - \frac{1}{3} = \frac{3 - 2}{6} = \frac{1}{6} " a n d " \frac{1}{3} . \frac{1}{2} = \frac{1}{3} - \frac{1}{2} = \frac{2 - 3}{6} = \frac{- 1}{6} \\ ∴ \frac{1}{2} . \frac{1}{3} \neq \frac{1}{3} . \frac{1}{2} w h e r e, \frac{1}{2}, \frac{1}{3} \in Q \end{array}$

Thus, the operation* is not commutative.

It can also be observed that:

$\begin{array}{l} (\frac{1}{2} . \frac{1}{3}) . \frac{1}{4} = (\frac{1}{2} - \frac{1}{3}) . \frac{1}{4} = \frac{1}{6} . \frac{1}{4} = \frac{1}{6} - \frac{1}{4} = \frac{2 - 3}{12} = \frac{- 1}{12} \\ \frac{1}{2} . (\frac{1}{3} . \frac{1}{4}) = \frac{1}{2} . (\frac{1}{3} - \frac{1}{4}) = \frac{1}{2} . \frac{1}{12} = \frac{1}{2} - \frac{1}{12} = \frac{6 - 1}{12} = \frac{5}{12} \\ ∴ (\frac{1}{2} . \frac{1}{3}) . \frac{1}{4} \neq \frac{1}{2} . (\frac{1}{3} . \frac{1}{4}) w h e r e, \frac{1}{2}, \frac{1}{3}, \frac{1}{4} \in Q \end{array}$

Thus, the operation* is not associative.

(ii) On Q the operation* is defined as a*b=a²+b.

For $a, b \in Q$ , we have:

$\begin{array}{l} a . b = a^{2} + b^{2} = b^{2} + a^{2} = b . a \\ ∴ a * b = b * a \end{array}$

Thus, the operation* is commutative.

It can be observed that:

$\begin{array}{l} (1 * 2) * 3 = (12 + 22) * 3 = (1 + 4) * 3 = 5 * 3 = 52 + 32 = 25 + 9 = 34 \\ 1 * (2 * 3) = 1 * (22 + 32) = 1 * (4 + 9) = 1 * 13 = 12 + 132 = 1 + 169 = 170 \\ ∴ (1 * 2) * 3 \neq 1 * (2 * 3), w h e r e, 1, 2, 3 \in Q \end{array}$

Thus, the operation* is not commutative.

(iii) On Q the operation* is defined as a*b=a+ab.

It can be observed that:

$\begin{array}{l} 1.2 = 1 + 1 * 2 = 1 + 2 = 3 \\ 2.1 = 2 + 2 * 1 = 2 + 2 = 4 \\ ∴ 1.2 \neq 2.1, w h e r e, 1, 2 \in Q \end{array}$

Thus, the operation* is not commutative.

It can also be observed that:

$\begin{array}{l} (1.2) . 3 = (1 + 1 * 2) . 3 = 3.3 = 3 + 3 * 3 = 3 + 9 = 12 \\ 1 . (2.3) = 1 . (2 + 2 * 3) = 1 + 1 * 8 = 9 \\ ∴ (1.2) . 3 \neq 1 . (2.3), w h e r e, 1, 2, 3 \in Q \end{array}$

Thus, the operation* is not associative.

(iv) On Q the operation* is defined as a*b=(a-b)²

For $a, b \in Q$ , we have:

$\begin{array}{l} a * b = {(a - b)}^{2} \\ b * a = {(b - a)}^{2} = {[- (a - b)]}^{2} = {(a - b)}^{2} \\ ∴ a * b = b * a \end{array}$

Thus, the operation* is commutative

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(i) On Q the operation* is defined as a*b=a-b.

It can be observed that:

Thus, the operation* is not commutative.

It can also be observed that:

Thus, the operation* is not associative.

(ii) On Q the operation* is defined as a*b=a²+b.

For $a, b \in Q$ , we have:

$\begin{array}{l} a . b = a^{2} + b^{2} = b^{2} + a^{2} = b . a \\ ∴ a * b = b * a \end{array}$

Thus, the operation* is commutative.

It can be observed that:

Thus, the operation* is not commutative.

(iii) On Q the operation* is defined as a*b=a+ab.

It can be observed that:

$\begin{array}{l} 1.2 = 1 + 1 * 2 = 1 + 2 = 3 \\ 2.1 = 2 + 2 * 1 = 2 + 2 = 4 \\ ∴ 1.2 \neq 2.1, w h e r e, 1, 2 \in Q \end{array}$

Thus, the operation* is not commutative.

It can also be observed that:

$\begin{array}{l} (1.2) . 3 = (1 + 1 * 2) . 3 = 3.3 = 3 + 3 * 3 = 3 + 9 = 12 \\ 1 . (2.3) = 1 . (2 + 2 * 3) = 1 + 1 * 8 = 9 \\ ∴ (1.2) . 3 \neq 1 . (2.3), w h e r e, 1, 2, 3 \in Q \end{array}$

Thus, the operation* is not associative.

(iv) On Q the operation* is defined as a*b=(a-b)²

For $a, b \in Q$ , we have:

$\begin{array}{l} a * b = {(a - b)}^{2} \\ b * a = {(b - a)}^{2} = {[- (a - b)]}^{2} = {(a - b)}^{2} \\ ∴ a * b = b * a \end{array}$

Thus, the operation* is commutative.

It can be observed that:

$\begin{array}{l} (1.2) . 3 = {(1.2)}^{2} . 3 = {(- 1)}^{2} . 3 = 1.3 = {(1.3)}^{2} = {(- 2)}^{2} = 4 \\ 1 . (2.3) = 1 . {(2.3)}^{2} = 1 . {(- 1)}^{2} = 1.1 = {(1.1)}^{2} = 0 \\ ∴ (1.2) . 3 \neq 1 . (2.3), w h e r e, 1, 2, 3 \in Q \end{array}$

Thus, the operation* is not associative.

(v) On Q the operation* is defined as a.b=ab/4

For $a, b \in Q$ , we have:

$\begin{array}{l} a . b = \frac{a b}{4} = b \frac{a}{4} = b . a \\ ∴ a * b = b * a \end{array}$

Thus, the operation* is commutative.

For $a, b, c \in Q$ , we have:

$\begin{array}{l} (a . b) . c = \frac{a b}{4} . c = \frac{\frac{a b}{4} . c}{4} = \frac{a b c}{16} \\ a . (b . c) = a . b \frac{c}{4} = \frac{a . \frac{b c}{4}}{4} = \frac{a b c}{16} \\ = ∴ (a * b) * c \neq a * (b * c) \end{array}$

Thus, the operation* is associative.

(vi) On Q the operation* is defined as a*b=ab²

It can be observed that:

$\begin{array}{l} \frac{1}{2} . \frac{1}{3} = \frac{1}{2} . {(\frac{1}{3})}^{2} = \frac{1}{2} . \frac{1}{9} = \frac{1}{18} \\ \frac{1}{3} . \frac{1}{2} = \frac{1}{3} . {(\frac{1}{2})}^{2} = \frac{1}{3} . \frac{1}{4} = \frac{1}{12} \\ ∴ \frac{1}{2} . \frac{1}{3} \neq \frac{1}{3} . \frac{1}{2}, w h e r e, \frac{1}{2}, \frac{1}{3} \in Q \end{array}$

Thus, the operation* is not commutative.

It can be observed that:

$\begin{array}{l} (\frac{1}{2} . \frac{1}{3}) . \frac{1}{4} = [\frac{1}{2} . {(\frac{1}{3})}^{2}] . \frac{1}{4} = \frac{1}{18} . \frac{1}{4} = \frac{1}{18} . {(\frac{1}{4})}^{2} = \frac{1}{18 * 16} \\ \frac{1}{2} . (\frac{1}{3} . \frac{1}{4}) = \frac{1}{2} . [\frac{1}{3} . {(\frac{1}{4})}^{2}] = \frac{1}{2} . \frac{1}{48} = \frac{1}{2} . {(\frac{1}{48})}^{2} = \frac{1}{2 * {(48)}^{2}} \\ ∴ (\frac{1}{2} . \frac{1}{3}) . \frac{1}{4} \neq \frac{1}{2} . (\frac{1}{3} . \frac{1}{4}), w h e r e, \frac{1}{2}, \frac{1}{3}, 1.4 \in Q \end{array}$

Thus, the operation* is not associative.

Hence, the operations defined (ii), (iv), (v) are commutative and the operation defined in (v) is associative.

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New question posted

a year ago

Maths Relations and Functions

23. The function ‘t’ which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by t(C) = 9C/5 +32

Find (i) t(0) (ii) t(28) (iii) t(–10) (iv) The value of C, when t(C) = 212.

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New answer posted

a year ago

Maths Straight Lines

58. What are the points on the y-axis whose distance from the line $\frac{x}{3} + \frac{y}{4} = 1$ is 4 units.

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alok kumar singh

Contributor-Level 10

58. Let (0, y) be the point on y-axis which is at a distance 4 unit from the line $\frac{x}{3} + \frac{y}{4} = 1$

Then, the line $\frac{x}{3} + \frac{y}{4} = 1$

4x + 3y = 12.

4x + 3y - 12 = 0

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New answer posted

a year ago

Maths Relations and Functions

50. Let * be the binary operation on N defined by a * b = H.C.F. of a and b. Is * commutative? Is * associative? Does there exist identity for this binary operation on N?

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Vishal Baghel

Contributor-Level 10

The binary operation * on N is defined as:

a * b = H.C.F. of a and b

It is known that:

H.C.F. of a and b = H.C.F. of b and a & mn For E; a, b ∈ N.

∴a * b = b * a

Thus, the operation * is commutative.

For a, b, c ∈ N, we have:

(a * b)* c = (H.C.F. of a and b) * c = H.C.F. of a, b, and c

a * (b * c)= a * (H.C.F. of b and c) = H.C.F. of a, b, and c

∴ (a * b) * c = a * (b * c)

Thus, the operation * is associative.

Now, an element e ∈ N will be the identity for the operation * if a * e = a = e* a ∈ a ∈ N.

But this relation is not true for any a ∈ N.

Thus, the operation * does not have any identity in N.

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New answer posted

a year ago

Maths Straight Lines

57. Find the equations of the lines, which cut-off intercepts on the axes whose sum and product are 1 and – 6, respectively.

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alok kumar singh

Contributor-Level 10

57. Let a and b be the x & y intercept. Then,

$\frac{x}{a} + \frac{y}{b} = 1 .$ _____ (1)

Given, a + b = 1. ______ (2) b = 1 -a _____ (3)

and ab = -6 _____ (4)

Putting eqn (B) in (iii) we get a

a (1- a) = - 6

a - a² = - 6

a² - a - 6 = 0

a² + 2a - 3a - 6 = 0

a (a + 2) - 3 (a + 2) = 0

(a + 2) (a -3) = 0

(a + 2) (a -3) = 0.

a = 3 or a = -2.

When a = 3, b = 1- a = 1 - 3 = - 2

When a = - 2, b = 1 - (-2) = 1 + 2 = 3

So, (a, b) = (3, -2) and (-2, 3)

Hence, eqⁿ (1) becomes,

$\frac{x}{3} + \frac{y}{- 2} = 1$ and $\frac{x}{- 2} + \frac{y}{3} = 1 .$

2x – 3y = 6 and 2y - 3x = 6

Gives the read eqⁿ of lines

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New answer posted

a year ago

Maths Relations and Functions

49. Is * defined on the set {1, 2, 3, 4, 5} by a * b = L.C.M. of a and b a binary operation? Justify your answer.

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Vishal Baghel

Contributor-Level 10

The operation * on the set A = {1, 2, 3, 4, 5} is defined as

a * b = L.C.M. of a and b.

Then, the operation table for the given operation * can be given as:

*	1	2	3	4	5
1	1	2	3	4	5
2	2	2	6	4	10
3	3	6	3	12	15
4	4	4	12	4	20
5	5	10	15	20	5

It can be observed from the obtained table that:

3 * 2 = 2 * 3 = 6 ∉ A, 5 * 2 = 2 * 5 = 10 ∉ A, 3 * 4 = 4 * 3 = 12 ∉ A

3 * 5 = 5 * 3 = 15 ∉ A, 4 * 5 = 5 * 4 = 20 ∉ A

Hence, the given operation * is not a binary operation.

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New answer posted

a year ago

Maths Relations and Functions

22. A function f is defined by f(x) = 2x –5. Write down the values of (i) f (0), (ii) f (7), (iii) f (–3)

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Payal Gupta

Contributor-Level 10

22. Give, f (x) = 2x – 5.

(i) f (0)= (2 * 0) –5=0 – 5= –5

(ii) f (7)= (2 * 7) –5=14 – 5=9

(iii) f (–3)=2 * (–3) –5= –6 – 5= –11.

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New answer posted

a year ago

Maths Straight Lines

56. Find the values of q and p, if the equation x cosØ + y sinØ = p is the normal form of the line √3 $x + y + 2 = 0 .$

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alok kumar singh

Contributor-Level 10

56. The given equation of the line is 3x + y + 2 = 0

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New answer posted

a year ago

Maths Relations and Functions

48. Let * be the binary operation on N given by a * b = L.C.M. of a and b. Find

(i) 5 * 7, 20 * 16 (ii) Is * commutative?

(iii) Is * associative? (iv) Find the identity of * in N

(v) Which elements of N are invertible for the operation *?

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

The binary operation * on N is defined as a * b = L.C.M. of a and b.

(i) 5 * 7 = L.C.M. of 5 and 7 = 35

20 * 16 = L.C.M of 20 and 16 = 80

(ii) It is known that:

L.C.M of a and b = L.C.M of b and a & mnForE; a, b ∈ N.

∴a * b = b * a

Thus, the operation * is commutative.

(iii) For a, b, c ∈ N, we have:

(a * b) * c = (L.C.M of a and b) * c = LCM of a, b, and c

a * (b * c) = a * (LCM of b and c) = L.C.M of a, b, and c

∴ (a * b) * c = a * (b * c)

Thus, the operation * is associative.

(iv) It is known that:

L.C.M. of a and 1 = a = L.C.M. 1 and a &mnForE; a ∈ N

⇒ a * 1 = a = 1 * a &mnForE; a ∈ N

Thus, 1 is the ide

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