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Class 12th Relations and Functions

61. Give examples of two functions f: N → Z and g: Z → Z such that g o f is injective but g is not injective.

(Hint: Consider f(x) = x and g(x) = |x| )

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Vishal Baghel

Contributor-Level 10

Define $f : N \to Z$ as $f (x)$ and $g : Z \to Z$ as $g (x) = | x |$

We first show that g is not injective.

It can be observed that:

$\begin{array}{l} g (- 1) = | - 1 | = 1 \\ g (1) = | 1 | = 1 \\ ∴ g (- 1) = g (1), b u t, - 1 \neq 1 \end{array}$

$∴ g$ is not injective.

Now, $g o f : N \to Z$ is defined as

$g o f (x) = y (f (x)) = y (x) = | x |$

Let $x, y \in N$ such that $g o f (x) - g o f (y)$

$\Rightarrow | x | = | y |$

Since $x, y \in N$ , both are positive.

$∴ | x | = | y | \Rightarrow x = y$

Hence, gof is injective

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11 months ago

Class 12th Relations and Functions

60. Show that the function f: R → R given by f(x) = x³ is injective.

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Vishal Baghel

Contributor-Level 10

f: R → R is given as f (x) = x³.

Suppose f (x) = f (y), where x, y ∈ R.

⇒ x³ = y³ … (1)

Now, we need to show that x = y.

Suppose x ≠ y, their cubes will also not be equal.

⇒ x³ ≠ y³

However, this will be a contradiction to (1).

∴ x = y

Hence, f is injective.

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11 months ago

Class 12th Relations and Functions

59. Show that function f: R → {x ∈ R: −1 < x < 1} defined by f(x) = x/1+|x| x R is one-one and onto function.

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Vishal Baghel

Contributor-Level 10

It is given that $f : R \to {x \in R : - 1 < x < 1}$ is defined as $f (x) = \frac{x}{1 + | x |}, x \in R .$

Suppose $f (x) = f (y)$ , where $x, y \in R .$

$\Rightarrow \frac{x}{1 + x} = \frac{y}{1 - y} \Rightarrow 2 x y = x - y$

Since x is positive and y is negative:

$x > y \Rightarrow x - y > 0$

But, 2xy is negative.

Then, $2 x y \neq x - y$ .

Thus, the case of x being positive and y being negative can be ruled out.

Under a similar argument, x being negative and y being positive can also be ruled out

$∴$ x and y have to be either positive or negative.

When x and y are both positive, we have:

$f (x) = f (y) \Rightarrow \frac{x}{1 + x} = \frac{y}{1 + y} \Rightarrow x + x y = y + x y \Rightarrow x = y$

When x and y are both negative, we have:

$f (x) = f (y) \Rightarrow \frac{x}{1 - x} = \frac{y}{1 - y} \Rightarrow x - x y = y - y x \Rightarrow x = y$

$∴ f$ is one-one.

Now, let $y \in R$ such that $- 1 < y < 1$ .

If x is negative, then there exists $x = \frac{y}{1 + y} \in R$ such that

$f (x) = f (\frac{y}{1 + y}) = \frac{(\frac{y}{1 + y})}{1 + \frac{y}{1 + y}} = \frac{\frac{y}{1 + y}}{1 + \frac{- y}{1 + y}} = \frac{y}{1 + y - y} = y$

If x is positive, then there exists $x = \frac{y}{1 - y} \in R$ such that

$f (x) = f (\frac{y}{1 - y}) = \frac{(\frac{y}{1 - y})}{1 + (\frac{y}{1 - y})} = \frac{\frac{y}{1 - y}}{1 + \frac{y}{1 - y}} = \frac{y}{1 - y + y} = y$

$∴ f$ is onto.

Hence, f is

...more

It is given that $f : R \to {x \in R : - 1 < x < 1}$ is defined as $f (x) = \frac{x}{1 + | x |}, x \in R .$

Suppose $f (x) = f (y)$ , where $x, y \in R .$

$\Rightarrow \frac{x}{1 + x} = \frac{y}{1 - y} \Rightarrow 2 x y = x - y$

Since x is positive and y is negative:

$x > y \Rightarrow x - y > 0$

But, 2xy is negative.

Then, $2 x y \neq x - y$ .

Thus, the case of x being positive and y being negative can be ruled out.

Under a similar argument, x being negative and y being positive can also be ruled out

$∴$ x and y have to be either positive or negative.

When x and y are both positive, we have:

$f (x) = f (y) \Rightarrow \frac{x}{1 + x} = \frac{y}{1 + y} \Rightarrow x + x y = y + x y \Rightarrow x = y$

When x and y are both negative, we have:

$f (x) = f (y) \Rightarrow \frac{x}{1 - x} = \frac{y}{1 - y} \Rightarrow x - x y = y - y x \Rightarrow x = y$

$∴ f$ is one-one.

Now, let $y \in R$ such that $- 1 < y < 1$ .

If x is negative, then there exists $x = \frac{y}{1 + y} \in R$ such that

$f (x) = f (\frac{y}{1 + y}) = \frac{(\frac{y}{1 + y})}{1 + \frac{y}{1 + y}} = \frac{\frac{y}{1 + y}}{1 + \frac{- y}{1 + y}} = \frac{y}{1 + y - y} = y$

If x is positive, then there exists $x = \frac{y}{1 - y} \in R$ such that

$f (x) = f (\frac{y}{1 - y}) = \frac{(\frac{y}{1 - y})}{1 + (\frac{y}{1 - y})} = \frac{\frac{y}{1 - y}}{1 + \frac{y}{1 - y}} = \frac{y}{1 - y + y} = y$

$∴ f$ is onto.

Hence, f is one-one and onto.

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11 months ago

Class 12th Relations and Functions

58. If f: R → R is defined by f(x) = x² − 3x + 2, find f(f(x)).

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Vishal Baghel

Contributor-Level 10

It is given that $f : R \to R$ is defined as $f (x) = x^{2} - 3 x + 2 .$

$\begin{array}{l} f (f (x)) = f (x^{2} - 3 x + 2) \\ = {(x^{2} + 3 x + 2)}^{2} - 3 (x^{2} - 3 x + 2) + 2 \\ = x^{4} + 9 x^{2} + 4 - 6 x^{2} - 12 x + 4 x^{2} - 3 x^{2} + 9 x - 6 + 2 \\ = x^{4} - 6 x^{2} + 10 x^{2} - 3 x \end{array}$

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11 months ago

Class 12th Relations and Functions

57. Let f: W → W be defined as f(n) = n − 1, if is odd and f(n) = n + 1, if n is even. Show that f is invertible. Find the inverse of f. Here, W is the set of all whole numbers.

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Vishal Baghel

Contributor-Level 10

It is given that:

$f : W \to W$ is defined as $f (n) = {\begin{cases} n - 1, i f . n, o d d \\ n + 1, i f . n, e v e n \end{cases}$

One-one:

Let, $f (n) = f (m) .$

It can be observed that if n is odd and m is even, then we will have n-1=m+1.

$\Rightarrow n - m = 2$

However, the possibility of n being even and m being odd can also be ignored under a similar argument.

$∴$ Both n and m must be either odd or even.

Now, if both n and m are odd, then we have:

$f (n) = f (m) \Rightarrow n - 1 = m - 1 \Rightarrow n = m$

Again, if both n and m are even, then we have:

$f (n) = f (m) \Rightarrow n + 1 = m + 1 \Rightarrow n = m$

$∴ f$ is one-one.

It is clear that any odd number 2r+1 in co-domain N is the image of 2r in domain N and any even 2r in co-domain N is the image of 2r+1 in domain N.

$∴ f$ is onto.

Hence, f is an invertible function.

Let us define $g : W \to W$ as:

$g (m) = {\begin{cases} m + 1, i f . n, e v e n \\ m - 1, i f . n, o d d \end{cases}$

Now, when n is odd:

$g o f (n) = g (f (n)) = g (n - 1) = n - 1 + 1 = n$

And, when

...more

It is given that:

$f : W \to W$ is defined as $f (n) = {\begin{cases} n - 1, i f . n, o d d \\ n + 1, i f . n, e v e n \end{cases}$

One-one:

Let, $f (n) = f (m) .$

It can be observed that if n is odd and m is even, then we will have n-1=m+1.

$\Rightarrow n - m = 2$

However, the possibility of n being even and m being odd can also be ignored under a similar argument.

$∴$ Both n and m must be either odd or even.

Now, if both n and m are odd, then we have:

$f (n) = f (m) \Rightarrow n - 1 = m - 1 \Rightarrow n = m$

Again, if both n and m are even, then we have:

$f (n) = f (m) \Rightarrow n + 1 = m + 1 \Rightarrow n = m$

$∴ f$ is one-one.

It is clear that any odd number 2r+1 in co-domain N is the image of 2r in domain N and any even 2r in co-domain N is the image of 2r+1 in domain N.

$∴ f$ is onto.

Hence, f is an invertible function.

Let us define $g : W \to W$ as:

$g (m) = {\begin{cases} m + 1, i f . n, e v e n \\ m - 1, i f . n, o d d \end{cases}$

Now, when n is odd:

$g o f (n) = g (f (n)) = g (n - 1) = n - 1 + 1 = n$

And, when n is even:

$g o f (n) = g (f (n)) = g (n + 1) = n + 1 - 1 = n$

Similarly, when m is odd:

$f o g (m) = f (g (m)) = f (m - 1) = m - 1 + 1 = m$

When m is even:

$\begin{array}{l} f o g (m) = f (g (m)) = f (m + 1) = m + 1 - 1 = m \\ ∴ g o f = I_{W}, a n d, f o g = I_{W} \end{array}$

Thus, f is invertible and the inverse of f is given by $f^{- 1} = g$ , which is the same as f.

Hence, the inverse of f is f itself.

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11 months ago

Class 12th Relations and Functions

MISCELLANEOUS

56. Let f: R → R be defined as f(x) = 10x + 7. Find the function g: R → R such that g o f = f o g = 1R.

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Vishal Baghel

Contributor-Level 10

It is given that $f : R \to R$ is defined as $f (x) = 10 x + 7 .$

One-one:

$\begin{array}{l} L e t, f (x) = f (y), w h e r e, x, y \in R \\ \Rightarrow 10 x + 7 = 10 y + 7 \\ \Rightarrow x = y \end{array}$

$∴ f$ is a one-one function.

Onto:

$\begin{array}{l} F o r, y \in R, l e t, y = 10 x + 7 \\ \Rightarrow x = \frac{y - 7}{10} \in R \end{array}$

Therefore, for any $y \in R$ ,there exists $x = \frac{y - 7}{10} \in R$

Such that

$f (x) = f (\frac{y - 7}{10}) = 10 (\frac{y - 7}{10}) + 7 = y - 7 + 7 = y$

$∴ f$ is onto.

Therefore, f is one-one and onto.

Thus, f is an invertible function.

Let us define $g : R \to R$ as $g (y) = \frac{y - 7}{10}$

Now, we have

$\begin{array}{l} g o f (x) = g (f (x)) = g (10 x + 7) = \frac{(10 x + 7) - 7}{10} = 10 \frac{x}{10} = 10 \\ A n d \\ f o g (y) = f (g (y)) = f (\frac{y - 7}{10}) = 10 (\frac{y - 7}{10}) + 7 = y - 7 + 7 = y \end{array}$

Hence, the required function $g : R \to R$ is defined as $g (y) = \frac{y - 7}{10}$

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11 months ago

Class 12th 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

Class 12th 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

Class 12th 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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Class 12th 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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