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11 months ago

Maths Relations and Functions

55. Consider a binary operation * on N defined as a * b = a³ + b³. Choose the correct answer.

(A) Is * both associative and commutative?

(B) Is * commutative but not associative?

(C) Is * associative but not commutative?

(D) Is * neither commutative nor associative?

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Vishal Baghel

Contributor-Level 10

On N, the operation * is defined as a * b = a³ + b³.

For, a, b, ∈ N, we have:

a * b = a³ + b³ = b³ + a³ = b * a [Addition is commutative in N]

Therefore, the operation * is commutative.

It can be observed that:

(1*2)*3 = (1³+2³)*3 = 9 * 3 = 93 + 33 = 729 + 27 = 756

Also, 1* (2*3) = 1* (2³+3³) = 1* (8 +27) = 1 * 35

= 1³ +35³ = 1 + (35)³ = 1 + 42875 = 42876.

∴ (1 * 2) * 3 ≠ 1 * (2 * 3) ; where 1, 2, 3 ∈ N

Therefore, the operation * is not associative.

Hence, the operation * is commutative, but not associative. Thus, the correct answer is B.

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11 months ago

Maths Relations and Functions

54. State whether the following statements are true or false. Justify.

(i) For an arbitrary binary operation * on a set N, a * a = a ∀ a * N.

(ii) If * is a commutative binary operation on N, then a * (b * c) = (c * b) * a

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Vishal Baghel

Contributor-Level 10

(i) Define an operation * on N as:

a * b = a + b a, b ∈ N

Then, in particular, for b = a = 3, we have:

3 * 3 = 3 + 3 = 6 ≠ 3

Therefore, statement (i) is false.

(ii) R.H.S. = (c * b) * a

= (b * c) * a [* is commutative]

= a * (b * c) [Again, as * is commutative]

= L.H.S.

∴ a * (b * c) = (c * b) * a

Therefore, statement (ii) is true.

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11 months ago

Maths Relations and Functions

53. Let A = N * N and * be the binary operation on A defined by

(a, b) * (c, d) = (a + c, b + d)

Show that * is commutative and associative. Find the identity element for * on A, if any.

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Vishal Baghel

Contributor-Level 10

A = N * N

* is a binary operation on A and is defined by:

(a, b) * (c, d) = (a + c, b + d)

Let (a, b), (c, d) ∈ A

Then, a, b, c, d ∈ N

We have:

(a, b) * (c, d) = (a + c, b + d)

(c, d) * (a, b) = (c + a, d + b) = (a + c, b + d)

[Addition is commutative in the set of natural numbers]

∴(a, b) * (c, d) = (c, d) * (a, b)

Therefore, the operation * is commutative.

Now, let (a, b), (c, d), (e, f) ∈A

Then, a, b, c, d, e, f ∈ N

We have:

$\begin{array}{l} ((a, b) . (c, d)) . (e, f) = (a + c, b + d) . (e, f) = (a + c + e, b + d + f) \\ (a, b) . ((c, d) . (e, f)) = (a, b) . (c + e, d + f) = (a + c + e, b + d + f) \\ ∴ ((a, b) . (c, d)) . (e, f) = (a, b) . ((c, d) . (e, f)) \end{array}$

Therefore, the operation* is associative.

An element $e = (e_{1}, e_{2})$ will be an identity element for the operation* if

$a * e = a = e * a \forall a = (a_{1}, a_{2}) \in A$

$i . e ., (a_{1} + e_{1}, a_{2} + e_{2}) = (a_{1}, a_{2}) = (e_{1} + a_{1}, e_{2} + a_{2})$ which is not true for any element in A.

Therefore, the operation* does not have any identity

...more

A = N * N

* is a binary operation on A and is defined by:

(a, b) * (c, d) = (a + c, b + d)

Let (a, b), (c, d) ∈ A

Then, a, b, c, d ∈ N

We have:

(a, b) * (c, d) = (a + c, b + d)

(c, d) * (a, b) = (c + a, d + b) = (a + c, b + d)

[Addition is commutative in the set of natural numbers]

∴(a, b) * (c, d) = (c, d) * (a, b)

Therefore, the operation * is commutative.

Now, let (a, b), (c, d), (e, f) ∈A

Then, a, b, c, d, e, f ∈ N

We have:

Therefore, the operation* is associative.

An element $e = (e_{1}, e_{2})$ will be an identity element for the operation* if

$a * e = a = e * a \forall a = (a_{1}, a_{2}) \in A$

$i . e ., (a_{1} + e_{1}, a_{2} + e_{2}) = (a_{1}, a_{2}) = (e_{1} + a_{1}, e_{2} + a_{2})$ which is not true for any element in A.

Therefore, the operation* does not have any identity element.

Which is not true for any element in A.

Therefore, the operation * does not have any identity element.

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New answer posted

11 months ago

Maths Relations and Functions

24. Find the range of each of the following functions.

(i) f (x) = 2 – 3x, x $\in$ R, x > 0.

(ii) f (x) = x²+ 2, x is a real number.

(iii) f (x) = x, x is a real number.

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Payal Gupta

Contributor-Level 10

24. (i) f(x)=2 – 3x, x $\in$ R, x>0.

Given, x>0

3x>3 * 0

3x>0

(–1) * 3x<(1) * 0.

–3x<0

2 – 3x<0+2

2 – 3x<2

i.e., f(x) < 2

Hene, range of f(x) = (– ?, 2)

(ii) Given, f(x) = x²+2, x is a real number.

Since, x is a real number,

x² ≥ 0 (x²=0 for x=0)

x²+2 ≥ 0+2

x²+2 ≥ 2

f(x) ≥ 2

?Range of f(x) = [2, ?)

(iii) Given, f(x) = x, x is a real number.

As, f(x) = x, the range of f(x) is also real.

i.e., Range of f(x) = R.

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New answer posted

11 months ago

Maths Straight Lines

59. Find perpendicular distance from the origin to the line joining the points (cosθ, sin θ) and (cos ϕ, sin ϕ).

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alok kumar singh

Contributor-Level 10

Kindly go through the solution

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New answer posted

11 months ago

Maths Relations and Functions

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

0 Follower 2 Views

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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11 months 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.

0 Follower 11 Views

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

...more

(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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11 months 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

11 months 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

11 months 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?

0 Follower 2 Views

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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