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Class 12th Relations and Functions

23. In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.

(i) f: R → R defined by f(x) = 3 − 4x

(ii) f: R → R defined by f(x) = 1 + x²

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Vishal Baghel

Contributor-Level 10

(i) $f : R \to R$ defined as $f (x) = 3 - 4 x$

For $x_{1}, x_{2} \in R$ such that $f (x_{1}) = f (x_{2})$

$\begin{array}{l} \Rightarrow 3 - 4 x_{1} = 3 - 4 x_{2} \\ \Rightarrow 4 x_{1} = 4 x_{2} \\ \Rightarrow x_{1} = x_{2} \end{array}$

So, $f$ is one-one

For $y \in R$ , there exist

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

Hence, $f$ is onto

$∴ f$ is bijective

(ii) Given, $f : R \to R$ defined as $f (x) = 1 + x_{}^{2}$

For $x_{1}, x_{2} \in R$ such that $f (x_{1}) = f (x_{2})$

$\begin{array}{l} \Rightarrow 1 + x_{1}^{2} = 1 + x_{2}^{2} \\ \Rightarrow x_{1}^{2} = x_{2}^{2} \\ \Rightarrow x_{1}^{} = \pm x_{2}^{} \end{array}$

$\Rightarrow x_{1}^{} = x_{2}^{}$ or $x_{1}^{} = - x_{2}^{}$

$∴ f$ is not one-one

The range of $f (x)$ is always a positive real number which is not equal to co-domain $R$

So, $f$ is not onto

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New answer posted

a year ago

Class 12th Relations and Functions

22. Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. Show that f is one-one.

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Vishal Baghel

Contributor-Level 10

Given, $f : A \to B$ and $f = {(1, 4), (2, 5), (3, 6)}$

$\begin{array}{l} ∴ f (1) = 4 \\ f (2) = 5 \\ f (3) = 6 \end{array}$

i.e., the image elements of $A$ under the given $f_{X}^{n}$ $f$ are unique

So, $f$ is one-one

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New answer posted

a year ago

Class 12th Relations and Functions

21. Show that the Signum Function f: R → R , given by $f (x) = {\begin{cases} 1 i f x > 0 \\ 0 i f x = 0 \\ - 1 i f x < 0 \end{cases}$ is neither one-one nor onto.

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Vishal Baghel

Contributor-Level 10

The $f x_{}^{n}$ $f : R ? R$ is given by $f (x) = (\begin{matrix} 1_{} & i f & x > 0 \\ 0_{} & i f & x = 0 \\ ? 1_{} & i f & x < 0 \end{matrix})$

For $x_{1} = 1, x_{2} = 2, ? R$

$f (x_{1}) = f (1) = 1$

$f (x_{2}) = f (2) = 1$ but $1 ? 2$

So, $f$ is not one-one

And the range of $f (x) = {1, 0, ? 1}$ hence it is not equal to the co-domain $R$

So, $f$ is not onto

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New answer posted

a year ago

Class 12th Relations and Functions

20. Show that the Modulus Function f : R → R, given by f(x) = |x| is neither one-one nor onto, where is |x| if x is positive or 0 and |-x| is -x if x is negative.

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Vishal Baghel

Contributor-Level 10

The $f x_{}^{n}$ $f : R \to R$ is given by $f (x) = | x |$

$\Rightarrow f (x) = (\begin{matrix} x_{}, i f & x \geq 0 \\ - x_{}, i f & x < 0 \end{matrix})$

For $x_{1} = - 1$ and $x_{2} = 1$

$f (x_{1}) = f (- 1) = | - 1 | = 1$

$f (x_{2}) = f (1) = | 1 | = 1$

So, $f (x_{1}) = f (x_{2})$ but $x_{1} \neq x_{2}$

i.e., $f$ is not one-one

For $x = - 1 \in R$

$f (x) = | x |$

i.e., $f (- 1) = | - 1 | = 1$

So, range of $f (x)$ is always a positive real number and is not equal to the co-domain $R$

i.e., $f$ is not onto

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New answer posted

a year ago

Class 12th Relations and Functions

19.Prove that the Greatest Integer Function f: R → R given by f(x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.

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Vishal Baghel

Contributor-Level 10

The $f x_{}^{n}$ $f : R \to R$ is given by $f (x) = [x]$

Let $x_{1} = 1.5$ and $x_{2} = 1.2 \in R$ Then,

$f (x_{1}) = f (1.5) = [1.5] = 1$

$f (x_{2}) = f (1.2) = [1.2] = 1$

So, $f (x_{1}) = f (x_{2})$ but $x_{1} \neq x_{2}$

i.e., $f (1.5) = f (1.2)$ but $1.5 \neq 1.2$

So, $f$ is not one-one

The range of $f (x)$ is a set of all integers, $Z$ which is not a co-domain of $R$

$∴ f$ is not onto

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New answer posted

a year ago

Class 12th Relations and Functions

2. Check the injectivity and surjectivity of the following functions:

(i) f: N → N given by f(x) = x²

(ii) f: Z → Z given by f(x) = x²

(iii) f: R → R given by f(x) = x²

(iv) f: N → N given by f(x) = x³

(v) f: Z → Z given by f(x) = x³

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Vishal Baghel

Contributor-Level 10

(i) $f : N \to N$ given by $f (x) = x_{}^{2}$

For, $x_{1}, x_{2} \in N$ , $f (x_{1}) = f (x_{2})$

$\Rightarrow x_{1}^{2} = x_{2}^{2}$

$\Rightarrow x_{1}^{} = x_{2}^{} \notin N$

So, $f$ is one-one/ injective

For $x \in N$ , i.e., $x = 1, 2, 3 . . . .$

Range of $f (x) = {1_{}^{2}, 2_{}^{2}, 3_{}^{2} . . .} = {1, 4, 9 . . .} \neq N$

i.e., co-domain of $N$

So, $f$ is not onto/ subjective

(ii) $f : Z \to Z$ given by $f (x) = x_{}^{2}$

For, $x_{1}, x_{2} \in Z$ , $f (x_{1}) = f (x_{2})$

$\Rightarrow x_{1}^{2} = x_{2}^{2}$

$\Rightarrow x_{1}^{} = \pm x_{2}^{} \notin Z$

i.e., $x_{1} = x_{2}$ and $x_{1} = - x_{2}$

So, $f$ is not one-one/ injective

For $x \in Z$ , $x = 0, \pm 1, \pm 2, \pm 3 . . . .$

Range of $f (x) = {0_{}^{2}, {(\pm 1)}_{}^{2}, {(\pm 2)}_{}^{2}, {(\pm 3)}_{}^{2} . . .}$

${0, 1, 4, 9 . . . .} \neq$ co-domain $Z$

So, $f$ is not onto/ subjective

(iii) $f : R \to R$ given by $f (x) = x_{}^{2}$

For, $x_{1}, x_{2} \in R$ , $f (x_{1}) = f (x_{2})$

$\Rightarrow x_{1}^{2} = x_{2}^{2}$

$\Rightarrow x_{1}^{} = \pm x_{2}^{}$

So, $f$ is not injective

For $x \in R$

Range of $f (x) = {x_{}^{2}, x \in R}$ gives a set of all positive real numbers

Hence, range of $f (x \neq)$ co-domain of $R$

So, $f$ is not subjective

(iv) $f : N \to N$ given by&n

...more

(i) $f : N \to N$ given by $f (x) = x_{}^{2}$

For, $x_{1}, x_{2} \in N$ , $f (x_{1}) = f (x_{2})$

$\Rightarrow x_{1}^{2} = x_{2}^{2}$

$\Rightarrow x_{1}^{} = x_{2}^{} \notin N$

So, $f$ is one-one/ injective

For $x \in N$ , i.e., $x = 1, 2, 3 . . . .$

Range of $f (x) = {1_{}^{2}, 2_{}^{2}, 3_{}^{2} . . .} = {1, 4, 9 . . .} \neq N$

i.e., co-domain of $N$

So, $f$ is not onto/ subjective

(ii) $f : Z \to Z$ given by $f (x) = x_{}^{2}$

For, $x_{1}, x_{2} \in Z$ , $f (x_{1}) = f (x_{2})$

$\Rightarrow x_{1}^{2} = x_{2}^{2}$

$\Rightarrow x_{1}^{} = \pm x_{2}^{} \notin Z$

i.e., $x_{1} = x_{2}$ and $x_{1} = - x_{2}$

So, $f$ is not one-one/ injective

For $x \in Z$ , $x = 0, \pm 1, \pm 2, \pm 3 . . . .$

Range of $f (x) = {0_{}^{2}, {(\pm 1)}_{}^{2}, {(\pm 2)}_{}^{2}, {(\pm 3)}_{}^{2} . . .}$

${0, 1, 4, 9 . . . .} \neq$ co-domain $Z$

So, $f$ is not onto/ subjective

(iii) $f : R \to R$ given by $f (x) = x_{}^{2}$

For, $x_{1}, x_{2} \in R$ , $f (x_{1}) = f (x_{2})$

$\Rightarrow x_{1}^{2} = x_{2}^{2}$

$\Rightarrow x_{1}^{} = \pm x_{2}^{}$

So, $f$ is not injective

For $x \in R$

Range of $f (x) = {x_{}^{2}, x \in R}$ gives a set of all positive real numbers

Hence, range of $f (x \neq)$ co-domain of $R$

So, $f$ is not subjective

(iv) $f : N \to N$ given by $f (x) = x_{}^{3}$

For, $x_{1}, x_{2} \in N$ , $f (x_{1}) = f (x_{2})$

$\Rightarrow x_{1}^{3} = x_{2}^{3}$

$\Rightarrow x_{1}^{} = x_{2}^{} \in N$

So, $f$ is injective

For $x \in N,$

Range of $f (x) = {1_{}^{3}, 2_{}^{3}, 3_{}^{3} . . . .}$

${1_{}^{}, 8, 2 7_{}^{} . . . .} \neq$ co-domain of $N$

So, $f$ is not subjective

(v) $f : R \to R$ given by $f (x) = x_{}^{3}$

For, $x_{1}, x_{2} \in N$ , $f (x_{1}) = f (x_{2})$

$\Rightarrow x_{1}^{3} = x_{2}^{3}$

$\Rightarrow x_{1}^{} = x_{2}^{} \in N$

So, $f$ is injective

For $x \in R,$

Range of $f (x) = {1_{}^{3}, 2_{}^{3}, 3_{}^{3} . . . .}$

${1_{}^{}, 8, 2 7_{}^{} . . . .} \neq$ co-domain of R

So, $f$ is not subjective

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New answer posted

a year ago

Class 12th Relations and Functions

Relation and Function Exercise 1.2 Solutions

17. Show that the function f: R* → R* defined by f(x) = 1/x is one-one and onto, where: R* is the set of all non-zero real numbers. Is the result true, if the domain: R* is replaced by N with the co-domain being same as: R*?

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Vishal Baghel

Contributor-Level 10

The $f x$ n is $f (x) = \frac{1}{x}$ , which is a $f : R_{*}$ $\to$ $R_{*}$ and $R_{*}$ is set of all non-zero real numbers

For, $x_{1}, x_{2} \in R_{*}, f (x_{1}) = f (x_{2})$

$\Rightarrow \frac{1}{x_{1}} = \frac{1}{x_{2}}$

$\Rightarrow x_{1} = x_{2}$ So, $f$ is one-one

For, $y \in R_{*},$ $x = \frac{1}{f (x)} = \frac{1}{y}$ such that

So, $f (x) = y$

So, every element in the co-domain has a pre-image in $f$

So, $f$ is onto

If $f : N \to R_{*}$ such that $f (x) = \frac{1}{x}$

For, $x_{1}, x_{2} \in N,$ $f (x_{1}) = f (x_{2})$

$\Rightarrow \frac{1}{x_{1}} = \frac{1}{x_{2}}$

$\Rightarrow x_{1} = x_{2}$ So, $f$ is one-one

For, $y \in R_{*}$ and $f (x) = y$ we have $x = \frac{1}{y} \notin N$

Eg., $3 \in R_{*}$ so $x = \frac{1}{3} \notin N$

So, $f$ is not onto

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New answer posted

a year ago

Class 12th Relations and Functions

16. Let R be the relation in the set N given by R = {(a, b): a = b − 2, b > 6}. Choose the correct answer.

(A) (2, 4) ∈ R (B) (3, 8) ∈R (C) (6, 8) ∈R (D) (8, 7) ∈ R

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Vishal Baghel

Contributor-Level 10

The given relation in set N defined by

$R = {(a, b) : a = b - 2, b > 6}$

For (2,4), 4>6 is not true

For (3,8), 8>6 but 3= 8-2 ⇒3=6 is not true

For (6,8), 8>6 and 6= 8-2 ⇒6=6 is true

And for (8,7), 7>6 but 8= 7-2 ⇒8=5 is not true

Hence, option (C) is correct

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New answer posted

a year ago

Class 12th Relations and Functions

15. Let R be the relation in the set {1, 2, 3, 4} given by R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}. Choose the correct answer.

(A) R is reflexive and symmetric but not transitive.

(B) R is reflexive and transitive but not symmetric.

(C) R is symmetric and transitive but not reflexive.

(D) R is an equivalence relation.

0 Follower 25 Views

Vishal Baghel

Contributor-Level 10

The set in $A = {1, 2, 3, 4}$

The relation in this set $A$ is given by

$R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}$

$R$ is reflexive as $(1, 1), (2, 2), (3, 3), (4, 4) \in R$

As, $(1, 2) \in R$ but $(2, 1) \notin R$

$R$ is not symmetric

For $(1, 2) \in R$ and $(2, 2) \in R; (1, 2) \in R$

And for $(1, 3) \in R$ and $(3, 2) \in R; (1, 3) \in R$

∴ $R$ is transitive

Hence, option (B) is correct

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New answer posted

a year ago

Class 12th Relations and Functions

14. Let L be the set of all lines in XY plane and R be the relation in L defined as R = {(L₁, L₂): L₁ is parallel to L₂}. Show that R is an equivalence relation. Find the set of all lines related to the line y = 2x + 4.

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Vishal Baghel

Contributor-Level 10

The given relation in the set $L =$ all lines in $X Y -$ plane is defined as

$R = {(L_{1}, L_{2}) : L_{1}$ is parallel to $L_{2}}$

Let $L_{1} \in A$ then as $L_{1}$ is parallel to $L_{1}$ ,

$(L_{1}, L_{1}) \in R$

So, $R$ is reflexive

Let $L_{1}, L_{2} \in A$ and $(L_{1}, L_{1}) \in R$

Then, $L_{1}$ is parallel to $L_{2}$

$L_{2}$ is parallel to $L_{1}$

So, $(L_{2}, L_{1}) \in R$

i.e., $R$ is symmetric

Let $L_{1}, L_{2}, L_{3} \in A$ and $(L_{1}, L_{2})$ and $(L_{2}, L_{3}) \in R$

Then, $L_{1} ? L_{2}$ and $L_{2} ? L_{3}$

So, $L_{1} ? L_{3}$

i.e., $(L_{1}, L_{2}) \in R$

So, $R$ is transitive

Hence, $R$ is an equivalence relation

The set of lines related to $y = 2 x + 4$ is given by the equation $y = 2 x + C$ where $C$ is some constant.

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