Class 12 Mathematics – Relations and Functions

Questions 1–16: Solutions in English + Hindi

I have solved each question step-by-step and given the final conclusion clearly in both English and Hindi.

1. Determine whether each relation is Reflexive, Symmetric and Transitive

(i) $A=\{1,2,3,\ldots,14\}$, $R=\{(x,y):3x-y=0\}$

Given:

$$3x-y=0\Rightarrow y=3x$$

Reflexive

For reflexive relation, $(x,x)\in R$.

Putting $y=x$:

$$3x-x=2x=0$$

This is true only for $x=0$, but $0\notin A$.

Therefore, R is not reflexive.

हिंदी: $xRx$ के लिए $3x-x=0$ होना चाहिए, अर्थात $x=0$, लेकिन $0\notin A$। अतः R reflexive नहीं है।

Symmetric

Take $(1,3)\in R$, since $3(1)-3=0$.

But:

$$3(3)-1=8\neq0$$

So $(3,1)\notin R$.

Therefore, R is not symmetric.

हिंदी: $(1,3)\in R$, लेकिन $(3,1)\notin R$। अतः R symmetric नहीं है।

Transitive

Suppose $(x,y)\in R$ and $(y,z)\in R$.

Then:

$$y=3x,\qquad z=3y=9x$$

Thus $z=9x$, whereas for $(x,z)\in R$, we need $z=3x$. Not generally true.

For example:

$$(1,3)\in R,\quad (3,9)\in R$$

but

$$(1,9)\notin R.$$

Therefore, R is not transitive.

Final Answer / अंतिम उत्तर:

$$\boxed{\text{Neither reflexive, nor symmetric, nor transitive}}$$

(ii) $R$ on $N$, defined by $y=x+5$ and $x<4$

Possible values of $x$ are:

$$x=1,2,3$$

assuming $N=\{1,2,3,\ldots\}$.

Thus:

$$R=\{(1,6),(2,7),(3,8)\}$$

Reflexive

No pair of the form $(x,x)$ exists.

Not reflexive.

Symmetric

$$(1,6)\in R$$

but

$$(6,1)\notin R.$$

Not symmetric.

Transitive

There are no pairs $(x,y),(y,z)$ because the second components $6,7,8$ cannot act as first components due to $x<4$.

Hence the implication for transitivity is always satisfied.

Transitive.

Final Answer:

$$\boxed{\text{Not reflexive, not symmetric, but transitive}}$$

हिंदी: यह relation reflexive नहीं, symmetric नहीं, लेकिन transitive है।

(iii) $A=\{1,2,3,4,5,6\}$, $y$ is divisible by $x$

Reflexive

Every number divides itself:

$$x\mid x$$

Therefore, $(x,x)\in R$.

Reflexive.

Symmetric

$$2\mid4$$

so $(2,4)\in R$.

But $4\nmid2$, so:

$$(4,2)\notin R.$$

Not symmetric.

Transitive

If $x\mid y$ and $y\mid z$, then:

$$x\mid z.$$

Therefore, transitive.

Final Answer:

$$\boxed{\text{Reflexive and transitive, but not symmetric}}$$

हिंदी: Reflexive और transitive है, लेकिन symmetric नहीं है।

(iv) $R$ on $Z$, where $x-y$ is an integer

Since $x,y\in Z$,

$$x-y\in Z$$

always.

Therefore:

$$R=Z\times Z.$$

Reflexive

$$x-x=0\in Z.$$

Yes.

Symmetric

If $x-y\in Z$, then:

$$y-x=-(x-y)\in Z.$$

Yes.

Transitive

If $x-y\in Z$ and $y-z\in Z$, then:

$$x-z=(x-y)+(y-z)\in Z.$$

Yes.

Final Answer:

$$\boxed{\text{R is reflexive, symmetric and transitive}}$$

हिंदी: R reflexive, symmetric तथा transitive तीनों है।

(v) Relations among human beings

(a) Same workplace

Reflexive

A person works at the same place as himself/herself.

Yes.

Symmetric

If A works at the same place as B, B works at the same place as A.

Yes.

Transitive

If A and B work at the same place, and B and C work at the same place, then A and C work at the same place.

Yes.

$$\boxed{\text{Equivalence relation}}$$

हिंदी: यह reflexive, symmetric और transitive है।

(b) Live in the same locality

Same reasoning applies.

$$\boxed{\text{Reflexive, symmetric and transitive}}$$

हिंदी: Reflexive, symmetric और transitive।

(c) $x$ is exactly 7 cm taller than $y$

Reflexive

A person cannot be 7 cm taller than himself/herself.

Not reflexive.

Symmetric

If A is 7 cm taller than B, B is 7 cm shorter than A.

Not symmetric.

Transitive

Suppose A is 7 cm taller than B and B is 7 cm taller than C.

Then A is 14 cm taller than C, not 7 cm.

Not transitive.

$$\boxed{\text{Neither reflexive, symmetric nor transitive}}$$

हिंदी: न reflexive, न symmetric, न transitive।

(d) $x$ is wife of $y$

Reflexive

A person is not his/her own wife.

No.

Symmetric

If A is wife of B, B is not wife of A.

No.

Transitive

If A is wife of B and B is wife of C, A is not necessarily wife of C.

No.

$$\boxed{\text{Neither reflexive, symmetric nor transitive}}$$

हिंदी: न reflexive, न symmetric, न transitive।

(e) $x$ is father of $y$

A person cannot be his/her own father → not reflexive.

If A is father of B, B is not father of A → not symmetric.

If A is father of B and B is father of C, A is grandfather of C, not father → not transitive.

$$\boxed{\text{Neither reflexive, symmetric nor transitive}}$$

हिंदी: न reflexive, न symmetric, न transitive।

2. $R=\{(a,b):a\le b^2\}$ on $R$

We have to show that R is neither reflexive, symmetric nor transitive.

Reflexive

For reflexivity:

$$a\le a^2$$

Take:

$$a=\frac12$$

Then:

$$\frac12\le\frac14$$

which is false.

Therefore:

$$\boxed{\text{R is not reflexive}}$$

हिंदी: $a=\frac12$ लेने पर $\frac12\le\frac14$ गलत है। अतः R reflexive नहीं है।

Symmetric

Take:

$$a=1,\quad b=2$$

Then:

$$1\le4$$

so $(1,2)\in R$.

But:

$$2\le1$$

is false.

Therefore:

$$(2,1)\notin R.$$ $$\boxed{\text{R is not symmetric}}$$

Transitive

Take:

$$a=\frac12,\quad b=-1,\quad c=0.$$

Then:

$$a\le b^2$$

because

$$\frac12\le1.$$

Also:

$$b\le c^2$$

because

$$-1\le0.$$

But:

$$a\le c^2$$

means

$$\frac12\le0,$$

which is false.

Therefore:

$$\boxed{\text{R is not transitive}}$$

Final Answer:

$$\boxed{\text{R is neither reflexive nor symmetric nor transitive.}}$$

हिंदी: R न तो reflexive है, न symmetric और न ही transitive।

3. $A=\{1,2,3,4,5,6\}$, $R=\{(a,b):b=a+1\}$

Thus:

$$R=\{(1,2),(2,3),(3,4),(4,5),(5,6)\}.$$

Reflexive

No pair $(a,a)$ exists.

$$\boxed{\text{Not reflexive}}$$

Symmetric

$$(1,2)\in R$$

but

$$(2,1)\notin R.$$ $$\boxed{\text{Not symmetric}}$$

Transitive

$$(1,2)\in R,\quad(2,3)\in R$$

but:

$$(1,3)\notin R.$$ $$\boxed{\text{Not transitive}}$$

Final / अंतिम:

$$\boxed{\text{Neither reflexive, symmetric nor transitive}}$$

4. $R=\{(a,b):a\le b\}$ on $R$

Reflexive

For every real number:

$$a\le a.$$

Therefore, R is reflexive.

Symmetric

Take:

$$1\le2.$$

So $(1,2)\in R$.

But:

$$2\le1$$

is false.

Therefore, not symmetric.

Transitive

If:

$$a\le b,\quad b\le c,$$

then:

$$a\le c.$$

Therefore, transitive.

Final Answer:

$$\boxed{\text{Reflexive and transitive, but not symmetric}}$$

हिंदी: यह relation reflexive और transitive है, लेकिन symmetric नहीं है।

5. $R=\{(a,b):a\le b^3\}$ on $R$

Reflexive

For reflexivity:

$$a\le a^3.$$

Take:

$$a=\frac12.$$

Then:

$$\frac12\le\frac18$$

is false.

Therefore, not reflexive.

Symmetric

Take:

$$a=1,\quad b=2.$$ $$1\le8$$

is true.

But:

$$2\le1$$

is false.

Therefore, not symmetric.

Transitive

Take:

$$a=100,\quad b=5,\quad c=2.$$

First:

$$100\le5^3=125$$

is true.

Second:

$$5\le2^3=8$$

is true.

But:

$$100\le2^3=8$$

is false.

Therefore, R is not transitive.

Final Answer:

$$\boxed{\text{Neither reflexive nor symmetric nor transitive}}$$

हिंदी: R न reflexive है, न symmetric और न transitive।

6. $A=\{1,2,3\}$

$$R=\{(1,2),(2,1)\}$$

Reflexive

For reflexivity, we need:

$$(1,1),(2,2),(3,3).$$

These are absent.

Therefore, not reflexive.

Symmetric

$$(1,2)\in R\Rightarrow(2,1)\in R.$$

Both pairs are present.

Therefore, symmetric.

Transitive

$$(1,2)\in R,\quad(2,1)\in R$$

would require:

$$(1,1)\in R.$$

But $(1,1)\notin R$.

Therefore, not transitive.

Final Answer:

$$\boxed{\text{Symmetric but neither reflexive nor transitive}}$$

हिंदी: यह symmetric है, लेकिन reflexive और transitive नहीं है।

7. Books having the same number of pages

Relation:

$$xRy\iff x\text{ and }y\text{ have the same number of pages}.$$

Reflexive

Every book has the same number of pages as itself.

Yes.

Symmetric

If X has the same number of pages as Y, then Y has the same number of pages as X.

Yes.

Transitive

If X and Y have the same number of pages, and Y and Z have the same number, then X and Z have the same number.

Yes.

Therefore:

$$\boxed{\text{R is an equivalence relation}}$$

हिंदी: R reflexive, symmetric तथा transitive है, इसलिए यह equivalence relation है।

8. $A=\{1,2,3,4,5\}$, $|a-b|$ is even

$$aRb\iff |a-b|\text{ is even}.$$

Reflexive

$$|a-a|=0$$

and 0 is even.

So R is reflexive.

Symmetric

$$|a-b|=|b-a|.$$

Therefore, if $aRb$, then $bRa$.

So R is symmetric.

Transitive

Suppose:

$$|a-b|\text{ is even}$$

and

$$|b-c|\text{ is even}.$$

Then:

$$a-b,\ b-c$$

are even.

Hence:

$$a-c=(a-b)+(b-c)$$

is even.

Therefore R is transitive.

$$\boxed{\text{R is an equivalence relation}}$$

Equivalence classes

For 1:

$$[1]=\{1,3,5\}.$$

For 2:

$$[2]=\{2,4\}.$$

Thus:

$$\boxed{\{1,3,5\}\text{ are related to each other}}$$

and

$$\boxed{\{2,4\}\text{ are related to each other}}$$

No member of $\{1,3,5\}$ is related to any member of $\{2,4\}$.

हिंदी: विषम संख्याएँ आपस में related हैं और सम संख्याएँ आपस में related हैं। विषम और सम संख्या के बीच relation नहीं है।

9. $A=\{x\in Z:0\le x\le12\}$

Thus:

$$A=\{0,1,2,\ldots,12\}.$$

(i) $|a-b|$ is a multiple of 4

Reflexive

$$|a-a|=0$$

and 0 is a multiple of 4.

Symmetric

$$|a-b|=|b-a|.$$

Transitive

If:

$$|a-b|=4m,\quad |b-c|=4n,$$

then $a-b$ and $b-c$ are multiples of 4.

Therefore:

$$a-c=(a-b)+(b-c)$$

is also a multiple of 4.

Hence R is an equivalence relation.

Elements related to 1

Numbers whose difference from 1 is a multiple of 4:

$$\boxed{\{1,5,9\}}$$

(ii) $a=b$

Reflexive

$$a=a.$$

Symmetric

If $a=b$, then $b=a$.

Transitive

If $a=b$ and $b=c$, then $a=c$.

Therefore:

$$\boxed{\text{R is an equivalence relation}}$$

Elements related to 1:

$$\boxed{\{1\}}$$

Final Answer:

10. Give examples of relations

We can use:

$$A=\{1,2,3\}.$$

(i) Symmetric but neither reflexive nor transitive

Take:

$$R=\{(1,2),(2,1)\}.$$

It is symmetric, but not reflexive and not transitive.

$$\boxed{R=\{(1,2),(2,1)\}}$$

(ii) Transitive but neither reflexive nor symmetric

Take:

$$R=\{(1,2),(1,3),(2,3)\}.$$

This relation is transitive, but not reflexive or symmetric.

$$\boxed{R=\{(1,2),(1,3),(2,3)\}}$$

(iii) Reflexive and symmetric but not transitive

Take:

$$R=\{(1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(3,2)\}.$$

It is reflexive and symmetric.

But:

$$1R2,\quad2R3$$

while

$$1\not R3.$$

Hence not transitive.

(iv) Reflexive and transitive but not symmetric

Take:

$$R=\{(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)\}.$$

This is reflexive and transitive but not symmetric.

(v) Symmetric and transitive but not reflexive

Take:

$$R=\{(1,1),(2,2),(1,2),(2,1)\}$$

on

$$A=\{1,2,3\}.$$

It is symmetric and transitive, but $(3,3)\notin R$, so it is not reflexive.

11. Points having the same distance from origin

Given:

$$R=\{(P,Q):OP=OQ\}.$$

Reflexive

$$OP=OP.$$

Therefore $P R P$.

Symmetric

If:

$$OP=OQ,$$

then:

$$OQ=OP.$$

Therefore $QRP$.

Transitive

If:

$$OP=OQ$$

and

$$OQ=OR,$$

then:

$$OP=OR.$$

Therefore $PR$.

Hence:

$$\boxed{\text{R is an equivalence relation}}$$

Equivalence class of $P$

Let $P\ne(0,0)$ and:

$$OP=r.$$

All points $Q$ related to $P$ satisfy:

$$OQ=OP=r.$$

Therefore all such points lie on the circle:

$$\boxed{x^2+y^2=r^2}$$

with centre at the origin and passing through $P$.

हिंदी: P से related सभी points की origin से दूरी $OP$ के बराबर होगी। अतः ये सभी points origin को centre मानकर P से गुजरने वाले circle पर स्थित होंगे।

12. Similarity of triangles

Relation:

$$T_1RT_2\iff T_1\text{ is similar to }T_2.$$

Reflexive

Every triangle is similar to itself.

Symmetric

If $T_1\sim T_2$, then:

$$T_2\sim T_1.$$

Transitive

If:

$$T_1\sim T_2$$

and

$$T_2\sim T_3,$$

then:

$$T_1\sim T_3.$$

Therefore:

$$\boxed{\text{R is an equivalence relation}}$$

Given triangles

$$T_1=(3,4,5)$$ $$T_2=(5,12,13)$$ $$T_3=(6,8,10)$$

Compare $T_1$ and $T_3$:

$$\frac63=\frac84=\frac{10}{5}=2.$$

Therefore:

$$T_1\sim T_3.$$

For $T_1$ and $T_2$, ratios are not equal.

Therefore:

$$T_1\not\sim T_2.$$

Similarly:

$$T_2\not\sim T_3.$$

Final Answer:

$$\boxed{T_1\text{ and }T_3\text{ are related}}$$

हिंदी: केवल $T_1$ और $T_3$ similar हैं, क्योंकि उनके corresponding sides का ratio 2 है।

13. Polygons having the same number of sides

$$P_1RP_2\iff P_1\text{ and }P_2\text{ have the same number of sides}.$$

Reflexive

Every polygon has the same number of sides as itself.

Symmetric

If $P_1$ and $P_2$ have the same number of sides, then $P_2$ and $P_1$ also have the same number.

Transitive

If $P_1$ and $P_2$ have the same number of sides and $P_2$ and $P_3$ have the same number, then $P_1$ and $P_3$ have the same number.

Thus:

$$\boxed{\text{R is an equivalence relation}}$$

Triangle $T$ with sides 3, 4, 5

It is a triangle, so all triangles are related to $T$.

Therefore its equivalence class is:

$$\boxed{\{P\in A:P\text{ is a triangle}\}}$$

हिंदी: $3,4,5$ वाला polygon एक triangle है। अतः उससे related सभी elements वे polygons हैं जिनमें 3 sides हैं।

14. Parallel lines

Let:

$$R=\{(L_1,L_2):L_1\parallel L_2\}.$$

Reflexive

Every line is considered parallel to itself in the context of an equivalence relation.

Thus:

$$L\parallel L.$$

Symmetric

If:

$$L_1\parallel L_2,$$

then:

$$L_2\parallel L_1.$$

Transitive

If:

$$L_1\parallel L_2$$

and

$$L_2\parallel L_3,$$

then:

$$L_1\parallel L_3.$$

Hence:

$$\boxed{\text{R is an equivalence relation}}$$

Given line

$$y=2x+4$$

Its slope is:

$$m=2.$$

All lines parallel to it have slope 2.

Therefore:

$$\boxed{y=2x+c,\quad c\in R}$$

are all the lines related to the given line.

हिंदी: $y=2x+4$ की slope $2$ है। अतः उससे related सभी lines की slope भी $2$ होगी:

$$\boxed{y=2x+c}$$

जहाँ $c\in R$ है।

15. Multiple Choice Question

Given:

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

Set:

$$A=\{1,2,3,4\}.$$

Reflexive?

For reflexivity we need:

$$(1,1),(2,2),(3,3),(4,4).$$

All four are present.

So R is reflexive.

Symmetric?

$$(1,2)\in R$$

but

$$(2,1)\notin R.$$

So R is not symmetric.

Transitive?

We have:

$$(1,3)\in R$$

and

$$(3,2)\in R.$$

Therefore, transitivity requires:

$$(1,2)\in R.$$

It is present.

Other combinations also satisfy transitivity.

Therefore R is transitive.

Answer:

$$\boxed{\text{(B) R is reflexive and transitive but not symmetric}}$$

हिंदी उत्तर:

$$\boxed{\text{(B) R reflexive और transitive है, लेकिन symmetric नहीं है।}}$$

16. Multiple Choice Question

Given:

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

Check each option.

(A) $(2,4)$

Here:

$$b=4$$

but $b>6$ is false.

So:

$$(2,4)\notin R.$$

(B) $(3,8)$

Here:

$$a=3,\quad b=8.$$

Required:

$$a=b-2=8-2=6.$$

But $3\ne6$.

So not in R.

(C) $(6,8)$

Here:

$$b-2=8-2=6=a$$

and:

$$8>6.$$

Therefore:

$$(6,8)\in R.$$

(D) $(8,7)$

$$7-2=5\ne8.$$

So not in R.

Answer:

$$\boxed{\text{(C) }(6,8)\in R}$$

हिंदी उत्तर:

केवल $(6,8)$ के लिए:

$$6=8-2,\quad 8>6$$

दोनों conditions satisfy होती हैं।

$$\boxed{\text{सही उत्तर: (C) }(6,8)\in R}$$

Class 12 Mathematics – Relations and Functions

Exercise 1.2 — Solutions in English & Hindi

1. Show that $f:R^*\to R^*$, defined by $f(x)=\frac1x$, is one-one and onto

Here,

$$R^*=R-\{0\}$$

One-One / Injective

Let:

$$f(x_1)=f(x_2)$$

Then:

$$\frac1{x_1}=\frac1{x_2}$$

Since $x_1,x_2\ne0$,

$$x_1=x_2$$

Therefore, $f$ is one-one.

Onto / Surjective

Let $y\in R^*$.

We need to find $x\in R^*$ such that:

$$f(x)=y.$$

Take:

$$x=\frac1y.$$

Since $y\ne0$, $x\ne0$, so $x\in R^*$.

Then:

$$f(x)=\frac1{1/y}=y.$$

Therefore, $f$ is onto.

Hence:

$$\boxed{f\text{ is one-one and onto, i.e. bijective.}}$$

हिंदी

यदि:

$$f(x_1)=f(x_2)$$

तो:

$$\frac1{x_1}=\frac1{x_2} \Rightarrow x_1=x_2.$$

अतः $f$ one-one है।

अब $y\in R^*$ लें और:

$$x=\frac1y$$

लेते हैं। तब:

$$f(x)=\frac1{1/y}=y.$$

अतः $f$ onto भी है।

$$\boxed{f\text{ bijective है।}}$$

If domain is replaced by $N$

Consider:

$$f:N\to R^*,\qquad f(x)=\frac1x.$$

This function is one-one, because:

$$\frac1{x_1}=\frac1{x_2}\Rightarrow x_1=x_2.$$

But it is not onto, because, for example:

$$\frac12\in R^*$$

but there is no natural number $n$ such that:

$$\frac1n=\frac12$$

except $n=2$, actually this example is in the range. Choose instead:

$$\frac23\in R^*.$$

There is no $n\in N$ such that:

$$\frac1n=\frac23.$$

Therefore:

$$\boxed{\text{One-one but not onto.}}$$

हिंदी: Domain $N$ करने पर function one-one लेकिन onto नहीं रहेगा।

2. Check injectivity and surjectivity

(i) $f:N\to N,\quad f(x)=x^2$

Injective?

Take:

$$f(1)=1,\qquad f(2)=4.$$

In general, for natural numbers:

$$x_1^2=x_2^2\Rightarrow x_1=x_2.$$

Therefore, one-one.

Surjective?

For every $y\in N$, we need:

$$x^2=y.$$

But $2\in N$ is not the square of a natural number.

Therefore, not onto.

$$\boxed{\text{One-one but not onto}}$$

हिंदी: $x^2$ natural numbers पर one-one है, लेकिन हर natural number perfect square नहीं है। अतः one-one लेकिन onto नहीं।

(ii) $f:Z\to Z,\quad f(x)=x^2$

Injectivity

$$f(1)=1,\qquad f(-1)=1$$

but:

$$1\ne-1.$$

Hence, not one-one.

Surjectivity

Negative integers can never be obtained because:

$$x^2\ge0.$$

For example, $-1\in Z$, but no integer $x$ satisfies:

$$x^2=-1.$$

Therefore, not onto.

$$\boxed{\text{Neither one-one nor onto}}$$

हिंदी: $1$ और $-1$ दोनों का image 1 है, इसलिए one-one नहीं। Negative integers का image नहीं मिलता, इसलिए onto भी नहीं।

(iii) $f:R\to R,\quad f(x)=x^2$

$$f(1)=f(-1)=1$$

Therefore, not one-one.

Also:

$$x^2\ge0.$$

Negative real numbers are not in the range.

Therefore, not onto.

$$\boxed{\text{Neither one-one nor onto}}$$

(iv) $f:N\to N,\quad f(x)=x^3$

Injective

If:

$$x_1^3=x_2^3,$$

then:

$$x_1=x_2.$$

Therefore, one-one.

Onto

For every $y\in N$, we need $x\in N$ such that:

$$x^3=y.$$

But $2$ is not a perfect cube.

Therefore, not onto.

$$\boxed{\text{One-one but not onto}}$$

हिंदी: Cube function natural numbers पर one-one लेकिन onto नहीं है।

(v) $f:Z\to Z,\quad f(x)=x^3$

Injective

$$x_1^3=x_2^3\Rightarrow x_1=x_2.$$

So, one-one.

Onto

For every $y\in Z$, take:

$$x=\sqrt[3]{y}.$$

But this $x$ need not be an integer. For example, $y=2$ has no integer cube root.

Therefore, not onto.

$$\boxed{\text{One-one but not onto}}$$

Summary

3. Greatest Integer Function $f:R\to R,\ f(x)=[x]$

Recall:

$$[x]=\text{greatest integer}\le x.$$

Not One-One

Take:

$$f(1.2)=1$$

and

$$f(1.8)=1.$$

But:

$$1.2\ne1.8.$$

Thus:

$$f(1.2)=f(1.8)$$

Hence $f$ is not one-one.

Not Onto

The function takes only integer values.

For example:

$$\frac12\in R$$

but there is no $x\in R$ such that:

$$[x]=\frac12.$$

Therefore, $f$ is not onto $R$.

$$\boxed{\text{Neither one-one nor onto}}$$

हिंदी: Greatest Integer Function कई अलग-अलग real numbers को एक ही integer देता है, इसलिए one-one नहीं है। इसका range केवल integers है, इसलिए यह $R$ पर onto नहीं है।

4. Modulus Function $f:R\to R,\ f(x)=|x|$

Not One-One

$$f(1)=1$$

and

$$f(-1)=1.$$

But:

$$1\ne-1.$$

Therefore, not one-one.

Not Onto

For every $x\in R$:

$$|x|\ge0.$$

Thus negative real numbers cannot be obtained.

For example:

$$-2\in R$$

but:

$$|x|\ne-2.$$

Therefore, not onto.

$$\boxed{\text{Neither one-one nor onto}}$$

हिंदी: $|1|=|-1|=1$, इसलिए one-one नहीं। Modulus कभी negative नहीं होता, इसलिए negative real numbers range में नहीं आते और function onto नहीं है।

5. Signum Function

$$f(x)= \begin{cases} 1,&x>0\\ 0,&x=0\\ -1,&x<0 \end{cases}$$

Not One-One

Take:

$$f(2)=1,\qquad f(5)=1.$$

But:

$$2\ne5.$$

Therefore, not one-one.

Not Onto

The range is:

$$\{-1,0,1\}.$$

But codomain is $R$.

For example, $2\in R$, but no $x\in R$ gives:

$$f(x)=2.$$

Therefore, not onto.

$$\boxed{\text{Neither one-one nor onto}}$$

हिंदी: Signum function का range केवल $\{-1,0,1\}$ है। इसलिए यह न one-one है और न onto।

6. $A=\{1,2,3\}$, $B=\{4,5,6,7\}$

$$f=\{(1,4),(2,5),(3,6)\}.$$

To show one-one, we need different elements of A to have different images.

$$f(1)=4,\quad f(2)=5,\quad f(3)=6.$$

All images are different.

Therefore:

$$\boxed{f\text{ is one-one}}$$

It is not onto, because $7\in B$ has no pre-image.

हिंदी

A के अलग-अलग elements के images हैं:

$$4,5,6$$

जो सभी अलग हैं। इसलिए function one-one है।

लेकिन $B$ में 7 का कोई pre-image नहीं है, इसलिए यह onto नहीं है।

$$\boxed{\text{One-one but not onto}}$$

7. State whether one-one, onto or bijective

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

One-One

Suppose:

$$f(x_1)=f(x_2)$$

Then:

$$3-4x_1=3-4x_2$$ $$-4x_1=-4x_2$$ $$x_1=x_2.$$

Thus one-one.

Onto

Let:

$$y=3-4x.$$

Then:

$$4x=3-y$$ $$x=\frac{3-y}{4}.$$

For every $y\in R$, this $x\in R$.

Therefore, onto.

$$\boxed{\text{Bijective}}$$

हिंदी: यह one-one और onto दोनों है, इसलिए bijective है।

(ii) $f:R\to R,\ f(x)=1+x^2$

One-One?

$$f(1)=2$$

and:

$$f(-1)=2.$$

Since:

$$1\ne-1,$$

it is not one-one.

Onto?

Since:

$$x^2\ge0,$$

we have:

$$f(x)=1+x^2\ge1.$$

Thus values less than 1 cannot be obtained.

For example, $0\in R$, but:

$$1+x^2=0$$

has no real solution.

Therefore, not onto.

$$\boxed{\text{Neither one-one nor onto}}$$

8. $f:A\times B\to B\times A$

Defined by:

$$f(a,b)=(b,a).$$

One-One

Suppose:

$$f(a_1,b_1)=f(a_2,b_2).$$

Then:

$$(b_1,a_1)=(b_2,a_2).$$

Therefore:

$$b_1=b_2,\qquad a_1=a_2.$$

Hence:

$$(a_1,b_1)=(a_2,b_2).$$

So $f$ is one-one.

Onto

Take any:

$$(b,a)\in B\times A.$$

Choose:

$$(a,b)\in A\times B.$$

Then:

$$f(a,b)=(b,a).$$

Thus every element of $B\times A$ has a pre-image.

Therefore, onto.

Hence:

$$\boxed{f\text{ is bijective}}$$

Also:

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

हिंदी: $f(a,b)=(b,a)$ में ordered pair के elements की positions interchange होती हैं। इसका inverse भी इसी प्रकार है। इसलिए यह one-one और onto दोनों, अर्थात bijective है।

9. Function $f:N\to N$

The function is:

$$f(n)= \begin{cases} \dfrac{n+1}{2},&n\text{ is odd}\\[4pt] \dfrac n2+1,&n\text{ is even} \end{cases}$$

We need to check whether it is bijective.

Let's calculate:

For odd $n$:

$$1\to1,\quad3\to2,\quad5\to3,\quad7\to4,\ldots$$

For even $n$:

$$2\to2,\quad4\to3,\quad6\to4,\quad8\to5,\ldots$$

Thus:

$$f(3)=2$$

and:

$$f(2)=2.$$

Since:

$$3\ne2,$$

the function is not one-one.

It is also onto because every natural number occurs as an image. However, because it is not one-one, it cannot be bijective.

$$\boxed{\text{f is onto but not one-one; hence not bijective}}$$

हिंदी: $f(2)=2$ और $f(3)=2$, जबकि $2\ne3$। अतः function one-one नहीं है। हर natural number image के रूप में प्राप्त होता है, इसलिए यह onto है। इसलिए:

$$\boxed{\text{f onto है, लेकिन bijective नहीं।}}$$

10. $A=R-\{3\}$, $B=R-\{1\}$

$$f:A\to B$$

given by:

$$f(x)=\frac{x-2}{x-3}.$$

One-One

Suppose:

$$f(x_1)=f(x_2).$$

Then:

$$\frac{x_1-2}{x_1-3} = \frac{x_2-2}{x_2-3}.$$

Cross multiplying:

$$(x_1-2)(x_2-3) = (x_2-2)(x_1-3).$$

Expanding:

$$x_1x_2-3x_1-2x_2+6 = x_1x_2-3x_2-2x_1+6.$$

Therefore:

$$-x_1-2x_2=-x_2-2x_1$$ $$x_1=x_2.$$

Thus $f$ is one-one.

Onto

Let:

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

Then:

$$y(x-3)=x-2$$ $$yx-3y=x-2$$ $$x(y-1)=3y-2$$ $$x=\frac{3y-2}{y-1}.$$

Since $y\ne1$ because $y\in B=R-\{1\}$, this value of $x$ is defined.

Also $x\ne3$, because $x=3$ would imply:

$$3=\frac{3y-2}{y-1}$$ $$3y-3=3y-2,$$

which is impossible.

Therefore $x\in A$.

Hence every $y\in B$ has a pre-image.

Thus $f$ is onto.

$$\boxed{\text{f is one-one and onto, hence bijective}}$$

हिंदी: $f(x_1)=f(x_2)$ से $x_1=x_2$ मिलता है, इसलिए function one-one है। तथा प्रत्येक $y\ne1$ के लिए:

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

मिलता है। इसलिए function onto भी है।

$$\boxed{\text{f bijective है।}}$$

11. $f:R\to R,\quad f(x)=x^4$

One-One?

$$f(1)=1$$

and:

$$f(-1)=1.$$

But:

$$1\ne-1.$$

So not one-one.

Onto?

$$x^4\ge0.$$

Negative real numbers cannot be obtained.

So not onto.

Answer:

$$\boxed{\text{(D) f is neither one-one nor onto}}$$

हिंदी:

$$\boxed{\text{(D) f न one-one है और न onto}}$$

12. $f:R\to R,\quad f(x)=3x$

One-One

Suppose:

$$f(x_1)=f(x_2).$$

Then:

$$3x_1=3x_2$$ $$x_1=x_2.$$

Therefore, one-one.

Onto

Let:

$$y=3x.$$

Then:

$$x=\frac y3.$$

For every $y\in R$,

$$\frac y3\in R.$$

Therefore every real number has a pre-image.

Thus $f$ is onto.

Hence:

$$\boxed{\text{(A) f is one-one and onto}}$$

हिंदी:

$$3x_1=3x_2\Rightarrow x_1=x_2$$

इसलिए one-one।

और किसी भी $y\in R$ के लिए:

$$x=\frac y3$$

लेने पर $f(x)=y$ मिलता है। इसलिए onto भी है।

$$\boxed{\text{(A) f one-one और onto दोनों है।}}$$