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Class 12th Relations and Functions

44. For each binary operation * defined below, determine whether * is commutative or associative:

(i) On Z, define a * b = a − b

(ii) On Q, define a * b = ab + 1

(iii) On Q, define a * b =ab/2

(iv) On Z⁺, define a * b = 2^ab

(v) On Z⁺, define a * b = a^b

(vi) On R − {−1}, define a.b = a/b+1

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Vishal Baghel

Contributor-Level 10

(i) On Z, * is defined by a * b = a − b.

It can be observed that 1 * 2 = 1 − 2 = 1 and 2 * 1 = 2 − 1 = 1.

∴1 * 2 ≠ 2 * 1; where 1, 2 ∈ Z

Hence, the operation * is not commutative.

Also we have:

(1 * 2) * 3 = (1 − 2) * 3 = −1 * 3 = −1 − 3 = −4

1 * (2 * 3) = 1 * (2 − 3) = 1 * −1 = 1 − (−1) = 2

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

Hence, the operation * is not associative.

(ii) On Q, * is defined by a * b = ab + 1.

It is known that:

ab = ba & mn For E; a, b ∈ Q

⇒ ab + 1 = ba&nb

...more

(i) On Z, * is defined by a * b = a − b.

It can be observed that 1 * 2 = 1 − 2 = 1 and 2 * 1 = 2 − 1 = 1.

∴1 * 2 ≠ 2 * 1; where 1, 2 ∈ Z

Hence, the operation * is not commutative.

Also we have:

(1 * 2) * 3 = (1 − 2) * 3 = −1 * 3 = −1 − 3 = −4

1 * (2 * 3) = 1 * (2 − 3) = 1 * −1 = 1 − (−1) = 2

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

Hence, the operation * is not associative.

(ii) On Q, * is defined by a * b = ab + 1.

It is known that:

ab = ba & mn For E; a, b ∈ Q

⇒ ab + 1 = ba + 1 & mn For E; a, b ∈ Q

⇒ a * b = a * b &mn For E; a, b ∈ Q

Therefore, the operation * is commutative.

It can be observed that:

(1 * 2) * 3 = (1 * 2 + 1) * 3 = 3 * 3 = 3 * 3 + 1 = 10

1 * (2 * 3) = 1 * (2 * 3 + 1) = 1 * 7 = 1 * 7 + 1 = 8

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

Therefore, the operation * is not associative.

(iii) On Q, * is defined by a * b = ab/2

It is known that:

ab = ba &mnForE; a, b ∈ Q

⇒ab/2 = ba/2 &mnForE; a, b ∈ Q

⇒ a * b = b * a &mnForE; a, b ∈ Q

Therefore, the operation * is commutative.

For all a, b, c ∈ Q, we have:

$\begin{array}{l} (a * b) * c = (\frac{a b}{2}) * c = \frac{(\frac{a b}{2}) c}{2} = \frac{a b c}{4} \\ a * (b * c) = a * (\frac{b c}{2}) = \frac{a (\frac{b c}{2})}{2} = \frac{a b c}{4} \\ ∴ (a * b) * c = a * (b * c) \end{array}$

Therefore, the operation * is associative.

(iv) On Z⁺, * is defined by a * b = 2^ab.

It is known that:

ab = ba &mnForE; a, b ∈ Z⁺

⇒ 2^ab = 2^ba &mnForE; a, b ∈ Z⁺

⇒ a * b = b * a &mnForE; a, b ∈ Z⁺

Therefore, the operation * is commutative.

It can be observed that:

$\begin{array}{l} (1 * 2) * 3 = 2^{(1 * 2)} * 3 = 4 * 3 = 2^{4 * 3} = 2^{12} \\ 1 * (2 * 3) = 1 * 2^{2 * 3} = 1 * 2^{6} = 1 * 64 = 2^{64} \end{array}$

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

Therefore, the operation * is not associative.

(v) On Z⁺, * is defined by a * b = a^b.

It can be observed that:

1 x2 = 1²=1 and 2x 1 =2¹ = 2

∴ 1 * 2 ≠ 2 * 1 ; where 1, 2 ∈ Z⁺

Therefore, the operation * is not commutative.

It can also be observed that:

$\begin{array}{l} (2 * 3) * 4 = 2^{3} * 4 = 8 * 4 = 8^{4} = {(2^{3})}^{4} = 2^{12} \\ 2 * (3 * 4) = 2 * 3^{4} = 2 * 81 = 2^{81} \end{array}$

∴(2 * 3) * 4 ≠ 2 * (3 * 4) ; where 2, 3, 4 ∈ Z+

Therefore, the operation * is not associative.

(vi) On R, * − {−1} is defined by a.b = a/b+1

It can be observed that 1 x 2 = 1/2+1 = 1/2 and 2 x 1 = 2/1+1 = 2/2

∴1 * 2 ≠ 2 * 1 ; where 1, 2 ∈ R − {−1}

Therefore, the operation * is not commutative.

It can also be observed that:

$\begin{array}{l} (1 * 2) * 3 = \frac{1}{3} * 3 = \frac{\frac{1}{3}}{3 + 1} = \frac{1}{12} \\ 1 * (2 * 3) = 1 * \frac{2}{3 + 1} = 1 * \frac{2}{4} = 1 * \frac{1}{2} = \frac{1}{\frac{1}{2} + 1} = \frac{1}{\frac{3}{2}} = \frac{2}{3} \end{array}$

∴ (1 * 2) * 3 ≠ 1 * (2 * 3) ; where 1, 2, 3 ∈ R − {−1}

Therefore, the operation * is not associative.

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New answer posted

a year ago

Class 12th Relations and Functions

Exercise-1.4

43. Determine whether or not each of the definition of given below gives a binary operation. In the event that * is not a binary operation, give justification for this.

(i) On Z⁺ , define * by a * b = a − b

(ii) On Z⁺, define * by a * b = ab

(iii) On R, define * by a * b = ab²

(iv) On Z⁺, define * by a * b = |a − b|

(v) On Z⁺, define * by a * b = a

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Vishal Baghel

Contributor-Level 10

(i) On Z⁺, * is defined by a * b = a − b.

It is not a binary operation as the image of (1, 2) under * is 1 * 2 = 1 − 2= −1 ∉ Z⁺.

(ii) On Z⁺, * is defined by a * b = ab.

It is seen that for each a, b ∈ Z⁺, there is a unique element ab in Z⁺.

This means that * carries each pair (a, b) to a unique element a * b = ab in Z⁺.

Therefore, * is a binary operation.

(iii) On R, * is defined by a * b = ab².

It is seen that for each a, b ∈ R, there is a unique element ab² in R.

This means that * carries each pair (a, b) to a unique element a * b = ab² in R.

Therefore, * is a binary operation.

(iv) On Z⁺, * is defined by a * b = |a −

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Class 12th Relations and Functions

42. Let $f : R - {- \frac{4}{3}} \to R$ be a function defined as $f (x) = \frac{4 x}{3 x + 4}$ . The inverse of f is map g Range $f \to R - {- \frac{4}{3}}$

$g (y) = \frac{3 y}{3 - 4 y}$

$g (y) = \frac{4 y}{4 - 3 y}$

$g (y) = \frac{4 y}{3 - 4 y}$

$g (y) = \frac{3 y}{4 - 3 y}$

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Vishal Baghel

Contributor-Level 10

It is given that $f : R - {- \frac{4}{3}} \to R$ is defined as $f (x) = \frac{4 x}{3 x - 4}$

Let y be an arbitrary element of Range f.

Then, there exists $x \in R - {- \frac{4}{3}}$ such that $y = f (x)$

$\Rightarrow y = \frac{4 x}{3 x + 4}$

$\Rightarrow 3 x y + 4 y = 4 x$ let us define g: Range $f \to R - {- \frac{4}{3}} a s, g (y) = \frac{4 y}{4 - 3 y}$

$\begin{array}{l} \Rightarrow x (4 - 3 y) = 4 y \\ N o w, (g o f) (x) = g (f (x)) = g (\frac{4 x}{3 x + 4}) \\ \Rightarrow x = \frac{4 y}{4 - 3 y} \\ = \frac{4 (4 \frac{x}{3 x + 4})}{4 - 3 (\frac{4 x}{3 x + 4})} = \frac{16 x}{12 x + 16 - 12 x} = \frac{16 x}{16} = x \\ A n d, (f o g) (y) = f (g (y)) = f (\frac{4 y}{4 - 3 y}) \\ = \frac{4 (\frac{4 y}{4 - 3 y})}{3 (\frac{4 y}{4 - 3 y}) + 4} = \frac{16 y}{12 y + 16 - 12 y} = \frac{16 y}{16} = y \\ ∴ g o f = I_{R - (- \frac{4}{3})} a n d, f o g = I_" R a n g e, f " \end{array}$

Thus, g is the inverse of f i.e., $f^{- 1} = g .$

Hence, the inverse of f is the map g: Range $f \to R - {- \frac{4}{3}}$ which is given by

$g (y) = \frac{4 y}{4 - 3 y}$

The correct answer is B.

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New answer posted

a year ago

Class 12th Relations and Functions

41. If $f : R \to R$ be given by $f (x) = {(3 - x^{3})}^{\frac{1}{3}}$ , then f of(x) is

(A) 1/x³

(B) x³

(C) x

(D) (3 − x³)

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Vishal Baghel

Contributor-Level 10

$f : R \to R$ is given as $f (x) = {(3 - x^{\land} 3)}^{\land} (1 / 3)$

$\begin{array}{l} f (x) = {(3 - x^{3})}^{\frac{1}{3}} \\ ∴ f o f (x) = f (f (x)) = f ({(3 - x^{3})}^{\frac{1}{3}}) = {[3 - {({(3 - x^{3})}^{\frac{1}{3}})}^{3}]}^{\frac{1}{3}} \\ = {[3 - (3 - x^{3})]}^{\frac{1}{3}} = {(x^{3})}^{\frac{1}{3}} = x \\ ∴ f o f (x) = x \end{array}$

The correct answer is C.

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New answer posted

a year ago

Class 12th Relations and Functions

40. Let f: X → Y be an invertible function. Show that the inverse of f−1 is f, i.e., (f⁻¹)⁻¹ = f.

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Vishal Baghel

Contributor-Level 10

Let f: X → Y be an invertible function.

Then, there exists a function g: Y → X such that gof = IX and fog = IY.

Here, f⁻¹ = g.

Now, gof = IX and fog = IY

⇒ f⁻¹ of = IX and fof⁻¹ = IY

Hence, f⁻¹ : Y → X is invertible and f is the inverse of f⁻¹

i.e., (f−1)⁻¹ = f.

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New answer posted

a year ago

Class 12th Relations and Functions

39. Consider f: {1, 2, 3} → {a, b, c} given by f(1) = a, f(2) = b and f(3) = c. Find f⁻¹ and show that (f⁻¹⁾⁻¹ = f.

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Vishal Baghel

Contributor-Level 10

$f (1) = a, f (2) b, a n d, f (3) = c$

If we define $g : {a, b, c} \to {1, 2, 3} a s, g (a) = 1, g (b) = 2, g (c) = 3,$ then we have:

$\begin{array}{l} (f o g) (a) = f (g (a)) = f (1) = a \\ (f o g) (b) = f (g (b)) = f (2) = b \\ (f o g) (c) = f (g (c)) = f (3) = c \\ A n d \\ (g o f) (1) = g (f (1)) = f (a) = 1 \\ (g o f) (2) = g (f (2)) = f (b) = 2 \\ (g o f) (3) = g (f (3)) = f (c) = 3 \\ ∴ g o f = I_{X} a n d, f o g = I_{Y} \\ W h e r e, X = {1, 2, 3}, a n d, Y = {a, b, c} . \end{array}$

Thus, the inverse of f exists and $f^{- 1} = g .$

$∴ f^{- 1} : {a, b, c} \to {1, 2, 3}$ is given by,

$\begin{array}{l} f^{- 1} (a) = 1, f^{- 1} (b) = 2, f^{- 1} (c) = 3 \end{array}$

Let us now find the inverse of $f^{- 1}$ i.e., find the inverse of g.

If we define $h : {1, 2, 3} \to {a, b, c} a s$

$h (1) = a, h (2) = b, h (3) = c$ , then we have

$\begin{array}{l} (g o h) (1) = g (h (1)) = g (a) = 1 \\ (g o h) (2) = g (h (2)) = g (b) = 2 \\ (g o h) (3) = g (h (3)) = g (c) = 3 \\ A n d \\ (h o g) (a) = h (g (a)) = h (1) = a \\ (h o g) (b) = h (g (b)) = h (2) = b \\ (h o g) (c) = h (g (c)) = h (3) = c \\ ∴ g o h = I_{X} a n d, h o g = I_{Y} \\ W h e r e, X = {1, 2, 3}, a n d, Y = {a, b, c} . \end{array}$

Thus, the inverse of g exists and $g^{- 1} = h \Rightarrow {(f^{- 1})}^{- 1} = h .$

It can be noted that h=f.

Hence, ${(f^{- 1})}^{- 1} = f .$

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New answer posted

a year ago

Class 12th Relations and Functions

38. Let f: X → Y be an invertible function. Show that f has unique inverse.

(Hint: suppose g₁ and g₂ are two inverses of f. Then for all y ∈ Y, fog₁(y) = IY(y) = fog₂(y). Use one-one ness of f).

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Vishal Baghel

Contributor-Level 10

Let $f : X \to Y$ be an invertible function.

Also, suppose f has two inverses (say g₁ and g₂ ).

Then, for all y ∈ Y, we have:

$\begin{array}{l} f o g_{1} (y) = I_{y} (y) = f o g_{2} (y) \\ \Rightarrow f (g_{1} (y)) = f (g_{2} (y)) \end{array}$ [f is invertible => f is one-one]

$\Rightarrow g_{1} = g_{2}$ [g is one-one]

Hence, f has a unique inverse.

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New answer posted

a year ago

Class 12th Relations and Functions

36. Consider f: R+ → [4, ∞) given by f(x) = x² + 4 Show that f is invertible with the inverse f⁻¹ of f given by f⁻¹(y) = √y - 4 where R+ is the set of all non-negative real numbers.

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Vishal Baghel

Contributor-Level 10

$f : R_{+} \to [4, \infty)$ is given as $f (x) = x^{2} + 4$ .

One-one:

$\begin{array}{l} L e t, f (x) = f (y) . \\ \Rightarrow x^{2} + 4 = y^{2} + 4 \\ \Rightarrow x^{2} = y^{2} \\ \Rightarrow x = y [a s, x = y \in R] \end{array}$

$∴ f$ is a one-one function.

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New answer posted

a year ago

Class 12th Relations and Functions

35. Consider f : R→ R given by f(x) = 4x + 3 Show that f is invertible. Find the inverse of f.

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Vishal Baghel

Contributor-Level 10

$f : R \to R$ is given by,

$\begin{array}{l} f (x) = 4 x + 3 \\ O n e - o n e : \\ L e t, f (x) = f (y) . \\ \Rightarrow 4 x + 3 = 4 y + 3 \\ \Rightarrow 4 x = 4 y \\ \Rightarrow x = y \end{array}$

$∴$ f is a one-one function.

Onto:

$\begin{array}{l} F o r, y \in R, l e t, y = 4 x + 3 . \\ \Rightarrow x = \frac{y - 3}{4} \in R \end{array}$

Therefore, for any $y \in R$ , there exists $x = \frac{y - 3}{4} \in R$ such that

$f (x) = f (\frac{y - 3}{4}) = 4 (\frac{y - 3}{4}) + 3 = y$

$∴$ f is onto.

Thus, f is one-one and onto and therefore, $f^{- 1}$ exists.

Let us define $g : R \to R$ by $g (x) = \frac{y - 3}{4}$

$\begin{array}{l} N o w, (g o f) (x) = g (f (x)) = g (4 x + 3) = \frac{(4 x + 3) - 3}{4} = x \\ (f o g) (y) = f (g (y)) = f (\frac{y - 3}{4}) = 4 (\frac{y - 3}{4}) + 3 = y - 3 + 3 = y \\ ∴ g o f = f o g = I_{R} \end{array}$

Hence, f is invertible and the inverse of f is given by

$f^{- 1} = g (y) = \frac{y - 3}{4}$

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New answer posted

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Class 12th Relations and Functions

34. Show that f: [-1, 1] → R given by f(x) = x/x+2 is one-one. Find the inverse of the function f: [-1, 1] → Range f

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Vishal Baghel

Contributor-Level 10

$f : [- 1, 1] \to R$ is given as $f (x) = \frac{x}{x + 2}$

$\begin{array}{l} L e t, f (x) = f (y) . \\ \Rightarrow \frac{x}{x + 2} = \frac{y}{y + 2} \\ \Rightarrow x y + 2 x = x y + 2 y \\ \Rightarrow 2 x = 2 y \\ \Rightarrow x = y \end{array}$

$∴$ f is a one-one function.

It is clear that $f : [- 1, 1] \to$ Range f is onto.

$∴$ $f : [- 1, 1] \to$ Range f is one-one onto and therefore, the inverse of the function:

$f : [- 1, 1] \to$ Range f exists.

Let g: Range $f \to [- 1, 1]$ be the inverse of f.

Let y be an arbitrary element of range f.

Since $f : [- 1, 1] \to$ Range f is onto, we have:

$\begin{array}{l} \Rightarrow y = \frac{x}{x + 2} \\ \Rightarrow x y + 2 y = x \\ \Rightarrow x (1 - y) = 2 y \\ \Rightarrow x = \frac{2 y}{1 - y}, y \neq 1 \\ g (y) = \frac{2 y}{1 - y}, y \neq 1 \\ N o w, (g o f) (x) = g (f (x)) = g (\frac{x}{x + 2}) = \frac{2 (\frac{x}{x + 2})}{1 - \frac{x}{x + 2}} = \frac{2 x}{x + 2 - x} = 2 \frac{x}{2} = x \\ (f o g) (y) = f (g (y)) = f (\frac{2 y}{1 - y}) = \frac{\frac{2 y}{(1 - y)}}{(\frac{2 y}{1 - y}) + 2} = \frac{2 y}{2 y + 2 - 2 y} = \frac{2 y}{2} = y \\ ∴ g o f = I_{- 1, 1}, a n d, f o g - I_{R a n g e, f} \\ ∴ f^{- 1} = g \\ ∴ f^{- 1} (y) = \frac{2 y}{1 - y}, y \neq 1 \end{array}$

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