CBSE Class 11 Maths Important Questions - Chapter 2 Relations and Functions

1 Marks Questions

1. Find  $\mathbf{a}$ and $\mathbf{b}$ if $(\mathbf{a}-\mathbf{1},\mathbf{b}+\mathbf{5})=\left(\mathbf{2}\mathbf{,}\mathbf{3}\right)$.

Ans. It is given that, $(a-1,b+5)=(2,3)$.

Therefore, we have $a-1=2$ and $b+5=3$.

Then it implies that, $a=3$ and $b=-2$.

2. If $\mathbf{A}=\{\mathbf{1},\mathbf{3},\mathbf{5}\}$, $\mathbf{B}=\{\mathbf{2},\mathbf{3}\}$, then find: $\mathbf{A}\times \mathbf{B}$, (Question- 3).

Ans. Recall that if $A$ and $B$ are two non-empty sets, the Cartesian product of $A$ and $B$ is the set of all ordered pairs $(a,b)$ so that $a\in A$, $b\in B$, is represented by $A\times B$. 

Therefore, $A\times B=\{(1,2),(1,3),(3,2),(3,3),(5,2),(5,3)\}$.

3. Find $\mathbf{B}\times \mathbf{A}$.

Ans. Since, $B\times A$ is the cartesian product set of $B$ and $A$ such that for all $b\in B$, $a\in A$, $(b,a)\in B\times A$, so we have

$B\times A=\{(2,1),(2,3),(2,5),(3,1)(3,3),(3,5)\}$.

4. If $\mathbf{A}=\left\{\mathbf{1}\mathbf{,}\mathbf{2}\right\}$, $\mathbf{B}=\{\mathbf{2},\mathbf{3},\mathbf{4}\}$, $\mathbf{C}=\left\{\mathbf{4}\mathbf{,}\mathbf{5}\right\}$, then find: $\mathbf{A}\times (\mathbf{B}\cap \mathbf{C})$, (Question- 5).

Ans.  Note that, $$B\cap C=\left\{4\right\}$$.

Also, given that $A=\{1,2\}$.

So, by the definition of cartesian product of sets, we have

$A\times (B\cap C)=\{(1,4),(2,4)\}$.

5. Find $\mathbf{A}\times (\mathbf{B}\cup \mathbf{C})$.

Ans. Given that, $A=\{1,2\}$, $B=\{2,3,4\}$, and $C=\{4,5\}$.

Then, $B\cup C=\{2,3,4,5\}$.

Therefore, by the definition of cartesian product of sets, we have

$A\times (B\cup C)=\{(1,2),(1,3),(1,4),(1,5),(2,2),(2,3),(2,4),(2,5)\}$.

6. If $\mathbf{P}=\left\{\mathbf{1}\mathbf{,}\mathbf{3}\right\}$, $\mathbf{Q}=\{\mathbf{2},\mathbf{3},\mathbf{5}\}$, find the number of relations from $\mathbf{P}$ to $\mathbf{Q}$.

Ans. It is known that, the number of relations from a set $A$ to $B$ having $m$ and $n$ elements respectively, is $2^{mn}$.

Therefore, the number of relations from $P$ to $Q$, 

$=2^{2\times 3}=2^6=64$.

7. If $\mathbf{A}=\left\{\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{3}\mathbf{,}\mathbf{5}\right\}$ and $\mathbf{B}=\left\{\mathbf{4}\mathbf{,}\mathbf{6}\mathbf{,}\mathbf{9}\right\}$, $\mathbf{R}=\{(\mathbf{x},\mathbf{y}):|\mathbf{x}-\mathbf{y}|\mathbf{is}\mathbf{odd},\mathbf{x}\in \mathbf{A},\mathbf{y}\in \mathbf{B}\}$. Write $\mathbf{R}$ in roster form.

Ans. 

The roster form of the set $R$ is given by

$R=\{(1,4),(1,6),(2,9),(3,4),(3,6),(5,4),(5,6)\}$.

• Are the following relations in Question 8, 9, 10 functions? Give a proper reason.

8. $\mathbf{R}=\left\{\left(\mathbf{1}\mathbf{,}\mathbf{1}\right)\mathbf{,}\left(\mathbf{2}\mathbf{,}\mathbf{2}\right)\mathbf{,}\left(\mathbf{3}\mathbf{,}\mathbf{3}\right)\mathbf{,}\left(\mathbf{4}\mathbf{,}\mathbf{4}\right)\mathbf{,}\left(\mathbf{4}\mathbf{,}\mathbf{5}\right)\right\}$.

Ans. The given relation is $R=\{(1,1),(2,2),(3,3),(4,4),(4,5)\}$.

Notice that, the element $4$ possesses two images, such that $(4,4)$ and $(4,5)$.

Therefore, since a function should have unique image for each element, so $R$ cannot be a function.

9. $\mathbf{R}=\{\left(\mathbf{2}\mathbf{,}\mathbf{1}\right)\mathbf{,}\left(\mathbf{2}\mathbf{,}\mathbf{2}\right)\mathbf{,}(\mathbf{2},\mathbf{3}),(\mathbf{2},\mathbf{4})\}$.

Ans. The given relation is $R=\{(2,1),(2,2),(2,3),(2,4)\}$.

Observe that, for the element $2$, there are four different images such that $(2,1),(2,2),(2,3),$ and $(2,4)$.

Therefore, since a function should have unique image for each of the element, so the relation $R$ cannot be a function.

10. $\mathbf{R}=\left\{\left(\mathbf{1}\mathbf{,}\mathbf{2}\right)\mathbf{,}\left(\mathbf{2}\mathbf{,}\mathbf{5}\right)\mathbf{,}\left(\mathbf{3}\mathbf{,}\mathbf{8}\right)\mathbf{,}\left(\mathbf{4}\mathbf{,}\mathbf{10}\right)\mathbf{,}\left(\mathbf{5}\mathbf{,}\mathbf{12}\right)\mathbf{,}\left(\mathbf{6}\mathbf{,}\mathbf{12}\right)\right\}$. 

Ans. The given relation is 

$R=\{(1,2),(2,5),(3,8),(4,10),(5,12),(6,12)\}$.

Observe that, each of the elements in the given relation possesses a unique image.

Therefore, the relation $R$ is a function.

11. Is the following arrow diagram represent a function? Why?

Ans. Notice that, each of the elements of the set  $X=\{a,b,c,d\}$ corresponds to a unique image of the set $Y=\{0,1,2,3,4\}$.

Therefore, the relation given in the diagram is a function.

12. Is the following arrow diagram represent a function? Why?

Ans. Observe that, each of the elements of the set $X=\{2,3,5,7\}$ corresponds to a unique element of the set $Y=\{-1,0,2,4,3\}$, except  the element $2\in X$ which corresponds to two different images such that $(2,0)$ and $(2,3)$.

Therefore, the relation given in the diagrams cannot be a function.

• Let $\mathbf{f}$ and $\mathbf{g}$ be two real valued functions, defined by, $\mathbf{f}\left(\mathbf{x}\right)={\mathbf{x}}^{\mathbf{2}}\mathbf{,}\mathbf{g}\left(\mathbf{x}\right)=\mathbf{3}\mathbf{x}\mathbf{+}\mathbf{2}$.

13. Find the value of $(\mathbf{f}+\mathbf{g})\left(\mathbf{-}\mathbf{2}\right)$.

Ans. The given functions are $f\left(x\right)=x^2$, $g\left(x\right)=3x+2$.

Therefore, $(f+g)\left(x\right)=f\left(x\right)+g\left(x\right)=x^2+3x+2$

So, substituting $x=-2$, we obtain

$(f+g)(-2)={(-2)}^2+3(-2)+2$

$=4-6+2$

$=0.$

Hence, $(f+g)(-2)=0$.

14. Find the value of $\left(\mathbf{f}\mathbf{-}\mathbf{g}\right)\left(\mathbf{1}\right)$.

Ans. The given functions are $f\left(x\right)=x^2$, $g\left(x\right)=3x+2$.

Therefore, 

  $(f-g)\left(x\right)=f\left(x\right)-g\left(x\right)$

 $=x^2-(3x+2)$

 $=x^2-3x-2$

Substituting $x=1$, we get

  $(f-g)\left(1\right)={\left(1\right)}^2-3\left(1\right)-2$

 $=1-3-2$

 $=-2-2$

 $=-4.$

Hence, $(f-g)\left(1\right)=-4$.

15. Find the value of $\left(\mathbf{fg}\right)\left(\mathbf{-}\mathbf{1}\right)$.

Ans. The given functions are $f\left(x\right)=x^2$, $g\left(x\right)=3x+2$.

Therefore, $\left(fg\right)\left(x\right)=f\left(x\right)g\left(x\right)=x^2(3x+2)$.

Substituting $x=-1$, we have

$\left(fg\right)(-1)={(-1)}^2\{3(-1)+2\}$

 $=-3+2$

 $=-1.$

Hence, $\left(fg\right)(-1)=-1$.

16. Find the value of $\left(\frac{\mathbf{f}}{\mathbf{g}}\right)\left(\mathbf{0}\right)$.

Ans. The given functions are $f\left(x\right)=x^2$, $g\left(x\right)=3x+2$.

Then, $\left(\frac{f}{g}\right)\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)}=\frac{x^2}{3x+2}$.

Substituting, $x=0$, we get

$\left(\frac{f}{g}\right)\left(0\right)=\frac{{\left(0\right)}^2}{3\left(0\right)+2}=0$.

That is, $\left(\frac{f}{g}\right)\left(0\right)=0$.

17. If $\mathbf{f}\left(\mathbf{x}\right)={\mathbf{x}}^{\mathbf{3}}$, find the value of $\frac{\mathbf{f}\left(\mathbf{5}\right)\mathbf{-}\mathbf{f}\left(\mathbf{1}\right)}{\mathbf{5}\mathbf{-}\mathbf{1}}$.

Ans. It is given that,

$f\left(x\right)=x^3$.

Substituting $x=1$ we get,

$f\left(1\right)=1^3=1$ and

substituting $x=5$, we obtain

$f\left(5\right)=5^3=125$.

Therefore, $\frac{\text{f}\left(\text{5}\right)\text{-f}\left(\text{1}\right)}{\text{5-1}}=\frac{125-1}{4}=\frac{124}{4}=31$.

Hence, the value of $\frac{\text{f}\left(\text{5}\right)\text{-f}\left(\text{1}\right)}{\text{5-1}}$ is $31$.

18. Find the domain of the real function, $\mathbf{f}\left(\mathbf{x}\right)=\sqrt{{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{4}}$.

Ans. The given function is $f\left(x\right)=\sqrt{x^2-4}$.

Observe that the function exists if $x^2-4\ge 0$, that is, if $x\ge 2$ and $x\le -2$. Thus, the domain of the real function $f\left(x\right)$ is $(-\mathrm{\infty },-2]\cup [2,\mathrm{\infty })$.

19. Find the domain of the function, $\mathbf{f}\left(\mathbf{x}\right)=\frac{{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{3}}{{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{5}\mathbf{x}\mathbf{+}\mathbf{6}}$. 

Ans. The given function is $\text{f}\left(\text{x}\right)=\frac{{\text{x}}^{\text{2}}\text{+2x+3}}{{\text{x}}^{\text{2}}\text{-5x+6}}$.

Now, $x^2-5x+6=(x-2)(x-3)$.

Therefore, the given function can be written as

$f\left(x\right)=\frac{x^2+2x+3}{(x-2)(x-3)}$.

So, clearly for $x=2,3$, the function will be unbounded. Therefore, $f\left(x\right)$ exists for all real numbers except at $x=2,3$.

Hence, the domain of the function $f\left(x\right)$ is $\mathbb{R}-\{2,3\}$.

• Find the range of the functions in Question 20, 21.

20. $\mathbf{f}\left(\mathbf{x}\right)=\frac{\mathbf{1}}{\mathbf{1}\mathbf{-}{\mathbf{x}}^{\mathbf{2}}}$

Ans. We know that, range of a function is the set of all possible function values. Observe that, $f\left(x\right)$ can have all the real values except the unbounded values $-\mathrm{\infty },\mathrm{\infty }$. 

Thus, the range of the function $f\left(x\right)$ is $(-\mathrm{\infty },0]\cup [1,\mathrm{\infty })$.

21. $\mathbf{f}\left(\mathbf{x}\right)={\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{2}$.

Ans. The given function is $f\left(x\right)=x^2+2$.

Note that, $x^2\ge 0$. So, $x^2+2\ge 2$, for all real value of $x$.

Therefore, $f\left(x\right)\ge 2$, for all real value of $x$.

Thus, the range of the function $f\left(x\right)$ is $[2,\mathrm{\infty })$.

22. Find the domain of the relation, $\mathbf{R}=\{\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)\mathbf{:}\mathbf{x}\mathbf{,}\mathbf{y}\in \mathbf{Z}\mathbf{,}\mathbf{xy}=\mathbf{4}\}$.

Ans. The given relation is $\text{R}=\{\left(\text{x,y}\right)\text{:x,y}\in \mathbb{Z}\text{,}\text{xy}=\text{4}\}$.

Observe that, $xy=4$ for the following pair of values of $(x,y)$.

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

Now, we know that the domain is the set of all $x$-values.

Therefore, the domain of the given relation $R$, that is, 

$R=\{(-4,1),(-2,-2),(-1,-4),(1,4),(2,2),(4,1)\}$ is given by

$\{-4,-2,-1,1,2,4\}$.

• Find the range of the relations in Question 23, 24.

23. $\mathbf{R}=\{\left(\mathbf{a}\mathbf{,}\mathbf{b}\right)\mathbf{:}\mathbf{a}\mathbf{,}\mathbf{b}\in \mathbf{N}\mathbf{and}\mathbf{2}\mathbf{a}\mathbf{+}\mathbf{b}=\mathbf{10}\}$.

Ans. The pair of values of $(a,b)\in \mathbb{N}$ for which the equation $2a+b=10$ is satisfied are given by

$(4,2),(3,4),(2,6),$ and $(1,8)$.

Now, we know that range of a relation is the set of all the images.

Therefore, the range of the given relation $R$, that is, $R=\{(4,2),(3,4),(2,6),(1,8)\}$ is given by $\{2,4,6,8\}$.

24. $\mathbf{R}=\{\left(\mathbf{x}\mathbf{,}\frac{\mathbf{1}}{\mathbf{x}}\right):\mathbf{x}\in \mathbf{z},\mathbf{0}<\mathbf{x}<\mathbf{6}\}$.

Ans. The pair of values of $(x,\frac{1}{x})$, for all $x\in \mathbb{Z}$, such that $0<x<6$ are  $(1,1),(2,\frac{1}{2}),(3,\frac{1}{3}),(4,\frac{1}{4}),$ and $(5,\frac{1}{5})$.

Since, the range of a relation is the set of all images, therefore, the range of the given relation $R$ is given by $\{1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5}\}$.

25. If the ordered Pairs $\left(\mathbf{x}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{+}\mathbf{3}\right)$ and $$\left(\mathbf{2}\mathbf{,}\mathbf{x}\mathbf{+}\mathbf{4}\right)$$ are equal, find $\mathbf{x}$ and $\mathbf{y}$.

(a) $\left(\mathbf{3}\mathbf{,}\mathbf{3}\right)$         

(b) $(\mathbf{3},\mathbf{4})$              

(c) $\left(\mathbf{1}\mathbf{,}\mathbf{4}\right)$             

(d) $\left(\mathbf{1}\mathbf{,}\mathbf{0}\right)$

Ans. Given that the ordered pairs $\left(\text{x-1,y+3}\right)$ and $$\left(\text{2,x+4}\right)$$ are equal.

So, we have 

$x-1=2$

$\Rightarrow x=2+1=3$.

Also, we have 

$y+3=x+4$

$\Rightarrow y=x+4-3=x+1$

$\Rightarrow y=3+1=4$ substituting $x=3$

Thus, the required ordered pair is $(3,4)$.

Hence, option (b) is the correct answer.

26. If $\mathbf{n}\left(\mathbf{A}\right)=\mathbf{3},\mathbf{n}\left(\mathbf{B}\right)=\mathbf{2}$, $\mathbf{A}$ and $\mathbf{B}$ are two sets, then number of relations of $\mathbf{A}\times \mathbf{B}$ have.

(a) $\mathbf{6}$

(b) $\mathbf{12}$

(c) $\mathbf{32}$

(d) $\mathbf{64}$

Ans. Given that, the number of elements in the set $A$ is $n\left(A\right)=3$ and the number of elements of the set $B$ is $n\left(B\right)=2$.

Therefore, the number of relations of $A\times B$

$=2^{n\left(A\right)\times n\left(B\right)}$

$=2^{3\times 2}$

$=2^6$

$=64$

Hence, option (d) is the correct answer.

27. Let $\mathbf{f}\left(\mathbf{x}\right)=\mathbf{-}\left|\mathbf{x}\right|$, then Range of function is

(a) $\left(\mathbf{0}\mathbf{,}\mathrm{\infty }\right)$

(b) $(-\mathrm{\infty },\mathrm{\infty })$

(c) $(-\mathrm{\infty },\mathbf{0}]$

(d) none of the above

Ans. The given function is $f\left(x\right)=-\left|x\right|<0$.

Therefore, the range of the function $f\left(x\right)$ is given by $(-\mathrm{\infty },0]$.

Hence, option (c) is the correct answer.

28. A real function $\mathbf{f}$ is defined by $\mathbf{f}\left(\mathbf{x}\right)=\mathbf{2}\mathbf{x}\mathbf{-}\mathbf{5}$. Then the value of $\mathbf{f}\left(\mathbf{-}\mathbf{3}\right)$ is
(a) $\mathbf{-}\mathbf{11}$

(b) $\mathbf{1}$

(c) $\mathbf{0}$

(d) none of the above

Ans. The given function is $f\left(x\right)=2x-5$.

Substituting $x=-3$, we obtain

 $f(-3)=2(-3)-5$

 $=-6-5$

 $=-11.$

Hence, option (a) is the correct answer.

29. If $\mathbf{P}=\left\{\mathbf{a}\mathbf{,}\mathbf{b}\mathbf{,}\mathbf{c}\right\}$ and $\mathbf{Q}=\left\{\mathbf{d}\right\}$, form the sets $\mathbf{P}\times \mathbf{Q}$ and $\mathbf{Q}\times \mathbf{P}$ are these sets equal?

Ans. The cartesian product $P\times Q$ is such that

$P\times Q=\{(a,d),(b,d),(c,d)\}$.

The cartesian product $Q\times P$ is such that

$Q\times P=\{(d,a),(d,b),(d,c)\}$.

The elements of the two sets $P\times Q$ and $Q\times P$ are not equal, since the ordered pairs are not commutative, namely $(a,d)\ne (d,a)$.

Hence, the sets $P\times Q$ and $Q\times P$ are not equal.

30. If $\mathbf{A}$ and $\mathbf{B}$ are finite sets such that $\mathbf{n}\left(\mathbf{A}\right)=\mathbf{m}$ and $\mathbf{n}\left(\mathbf{B}\right)=\mathbf{k}$, find the number of relations from $\mathbf{A}$ to $\mathbf{B}$.

Ans. It is provided that, the number of elements in the set $A$ and $B$ is $n\left(A\right)=m$ and $n\left(B\right)=k$ respectively.

Therefore, the number of relations from the set $A$ to set $B$ is 

$2^{n\left(A\right)\times \text{n}\left(B\right)}=2^{mk}$.

31. Let $\mathbf{f}=\{\left(\mathbf{1}\mathbf{,}\mathbf{1}\right)\mathbf{,}\left(\mathbf{2}\mathbf{,}\mathbf{3}\right)\mathbf{,}\left(\mathbf{0}\mathbf{,}\mathbf{-}\mathbf{1}\right)\mathbf{,}\left(\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{3}\right)\mathbf{,}...\}$ be a function from $\mathbf{z}$ to $\mathbf{z}$ defined by $\mathbf{f}\left(\mathbf{x}\right)=\mathbf{ax}+\mathbf{b}$, for some integers $\mathbf{a}$ and $\mathbf{b}$. Determine the values of $\mathbf{a}$ and $\mathbf{b}$.

Ans. The function provided is $f\left(x\right)=ax+b$.

$(1,1)\in f$ implies that $f\left(1\right)=1$.

So, substituting $x=1$ into the function, we obtain

$a+b=1$                                       …… (i)

Again, $(0,-1)\in f$ implies that, $f\left(0\right)=-1$.

Therefore, substituting $x=0$ into the function, we get

$a\left(0\right)+b=-1$                              …… (ii)

Substituting the value of $b$ from the equation (ii) into the equation (i), yields

$a-1=1$

$\Rightarrow a=2$

Thus, $a=2$ and $b=-1$.

32. Express $\{\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)\mathbf{:}\mathbf{y}\mathbf{+}\mathbf{2}\mathbf{x}=\mathbf{5},\mathbf{x},\mathbf{y}\in \mathbf{w}\}$ as the set of ordered pairs.

Ans. The values of the ordered pairs for which the equation $y+2x=5$ is satisfied are given by $(0,5),(1,3),$ and $(2,1)$.

There are other values of $x,y$ for which the given equation may satisfy, but these are not whole numbers.

Thus, the set in the form of ordered pairs is given by $\{(0,5),(1,3),(2,1)\}$.

33. If $\mathbf{A}=\left\{\mathbf{1}\mathbf{,}\mathbf{2}\right\}$, find $\mathbf{A}\times \mathbf{A}\times \mathbf{A}$.

Ans. The given set is $A=\{1,2\}$.

The cartesian product $A\times A=\{(1,1),(1,2),(2,1),(2,2)\}$.

Therefore, the cartesian product$A\times A\times A=\{(1,1,1),(1,1,2),(1,2,1),(1,2,2),(2,1,1),(2,1,2),(2,2,1),(2,2,2)\}$.

34. A function $\mathbf{f}$ is defined by $\mathbf{f}\left(\mathbf{x}\right)=\mathbf{2}\mathbf{x}\mathbf{-}\mathbf{3}$. Find $\mathbf{f}\left(\mathbf{5}\right)$.

Ans. The given function is $f\left(x\right)=2x-3$.

Then, substituting $x=5$ into the given function, yields

$f\left(5\right)=2\left(5\right)-3$

$=10-3$

$=7.$

Hence, $f\left(5\right)=7$.

35. Let $\mathbf{f}=\left\{\left(\mathbf{0}\mathbf{,}\mathbf{-}\mathbf{5}\right)\mathbf{,}\left(\mathbf{1}\mathbf{,}\mathbf{-}\mathbf{2}\right)\mathbf{,}\left(\mathbf{2}\mathbf{,}\mathbf{1}\right)\mathbf{,}\left(\mathbf{3}\mathbf{,}\mathbf{4}\right)\mathbf{,}\left(\mathbf{4}\mathbf{,}\mathbf{7}\right)\right\}$ be a linear function from $\mathbf{Z}$ into $\mathbf{Z}$. Find $\mathbf{f}$.

Ans.  The required function is $f\left(x\right)=3x-5$.

36. If the ordered pairs $(\mathbf{x}-\mathbf{2},\mathbf{2}\mathbf{y}+\mathbf{1})$ and $\left(\mathbf{y}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{x}\mathbf{+}\mathbf{2}\right)$ are equal, then find the values of $\mathbf{x}$ and $\mathbf{y}$.

Ans. Since, the ordered pairs $(x-2,2y+1)$ and $(y-1,x+2)$ are equal, so we have $x-2=y-1$ and $2y+1=x+2$. That is,

$x-y=1$                                         …… (i)

$x-2y=-1$                                    …… (ii)

Solving the equations (i) and (ii), we obtain

$x=3$, and $y=2$.

37. Let $\mathbf{A}=\left\{\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{5}\mathbf{,}\mathbf{8}\right\}$, $\mathbf{B}=\left\{\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{3}\mathbf{,}\mathbf{6}\mathbf{,}\mathbf{7}\right\}$ and $\mathbf{R}$ be the relation, is one less than from $\mathbf{A}$ to $\mathbf{B}$. Then, find the domain and Range of $\mathbf{R}$.

Ans. The relation $R:A\to B$ such that $A$ is one less than $B$ is given by

$R=\{(-1,0),(2,3),(5,6)\}$.

Therefore, domain of the relation $R=\{-1,2,5\}$ and range of the relation$R=\{0,3,6\}$.

38. Let $\mathbf{R}$ be a relation from $\mathbf{N}$ to $\mathbf{N}$ define by $\mathbf{R}=[\left(\mathbf{a}\mathbf{,}\mathbf{b}\right)\mathbf{:}\mathbf{a}\mathbf{,}\mathbf{b}\in \mathbf{N}\mathbf{and}\mathbf{a}={\mathbf{b}}^{\mathbf{2}}]$. Is the following statement true?

$\mathbf{a},\mathbf{b}\in \mathbf{R}$ implies $(\mathbf{b},\mathbf{a})\in \mathbf{R}$.

Ans. Given statement: $a,b\in R$ implies $(b,a)\in R$.

Now, suppose $a=4$ and $b=2$ so that, $(4,2)\in R$ and $4=2^2$.

Although, $2\ne 4^2$. This implies $(2,4)\notin R$.

Thus, the given statement is not true.

39. Let $\mathbf{N}$ be the set of natural numbers and the relation $\mathbf{R}$ be defined in $\mathbf{N}$ by $\mathbf{R}=\{\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)\mathbf{:}\mathbf{y}=\mathbf{2}\mathbf{x}\mathbf{,}\mathbf{x}\mathbf{,}\mathbf{y}\in \mathbf{N}\}$. What is the domain, co-domain and range of $\mathbf{R}$? Is this relation a function?

Ans. The given relation is $R=\{(x,y):y=2x,x,y\in \mathbb{N}\}$.

The domain of the relation $R$ is the natural number set $\mathbb{N}$, co-domain of the relation $R$ is also the natural number set $\mathbb{N}$ and Range of the relation $R$is the set of even natural numbers.

The relation $R$ is a function because each natural number $x$ has a distinct image $2x$.

40. Let $\mathbf{R}=\{(\mathbf{x},\mathbf{y}):\mathbf{y}=\mathbf{x}+\mathbf{1}\}$ and $\mathbf{y}\in \{\mathbf{0},\mathbf{1},\mathbf{2},\mathbf{3},\mathbf{4},\mathbf{5}\}$. List the element of $\mathbf{R}$.

Ans. The given relation is $R=\{(x,y):y=x+1\}$.

Now, for $y\in \left\{\text{0,1,2,3,4,5}\right\}$, the $x$-values are obtained as

$-1,0,1,2,3,$ and $4$.

Therefore, the list of the elements of the given relation is $R=\{(-1,0),(0,1),(1,2),(2,3),(3,4),(4,5)\}$.

41. Let $\mathbf{f}$ be the subset of $\mathbf{Q}\times \mathbf{Z}$ defined by $\mathbf{f}=\{(\frac{\mathbf{m}}{\mathbf{n}},\mathbf{m}):\mathbf{m},\mathbf{n}\in \mathbf{Z},\mathbf{n}\ne \mathbf{0}\}$. Is $\mathbf{f}$ a function from $\mathbf{Q}$ to $\mathbf{Z}$? Justify your answer.

Ans. The given set is $\text{f}=\{\left(\frac{\text{m}}{\text{n}}\text{,m}\right)\text{:m,n}\in \text{Z,}\text{n}\ne \text{0}\}$.

Let $m=1$, $n=2$, then $(\frac{m}{n},m)\in f$ implies $(\frac{1}{2},1)\in f$.

Again, if $m=2$, $n=4$, then $(\frac{2}{4},2)\in f$.

But, since $\frac{2}{4}=\frac{1}{2}$, so we are getting two different images for $\frac{1}{2}$.

Therefore, $f$ cannot be a function from $\mathbb{Q}$ to $\mathbb{Z}$.

42. The function $\mathbf{f}$ which maps temperature in Celsius into temperature in Fahrenheit is defined by $\mathbf{f}\left(\mathbf{c}\right)=\frac{\mathbf{9}}{\mathbf{5}}\mathbf{c}\mathbf{+}\mathbf{32}$. Find $\mathbf{f}\left(\mathbf{0}\right)$.

Ans. The given function is $f\left(c\right)=\frac{9}{5}c+32$.

Substituting $c=0$ into the function, we have

$f\left(0\right)=\frac{9}{5}\left(0\right)+32$

$\Rightarrow f\left(0\right)=32$.

43. If $\mathbf{f}\left(\mathbf{x}\right)={\mathbf{x}}^{\mathbf{3}}\mathbf{-}\frac{\mathbf{1}}{{\mathbf{x}}^{\mathbf{3}}}$, then prove that $\mathbf{f}\left(\mathbf{x}\right)\mathbf{+}\mathbf{f}\left(\frac{\mathbf{1}}{\mathbf{x}}\right)=\mathbf{0}$.

Ans. The given function is $f\left(x\right)=x^3-\frac{1}{x^3}$.

Replacing $x$ by $\frac{1}{x}$ into the given function we obtain,

$f\left(\frac{1}{x}\right)=\frac{1}{x^3}-x^3$.

Therefore, adding these two functions, we get

  $f\left(x\right)+f\left(\frac{1}{x}\right)=x^3-\frac{1}{x^3}+\frac{1}{x^3}-x^3$

 $=0$

44. If $\mathbf{A}$ and $\mathbf{B}$ are two sets containing $\mathbf{m}$ and $\mathbf{n}$ elements respectively, how many different relations can be defined from to?

Ans. The number of relations from the set $A$ to $B$ is $2^{mn}$.

4 Marks Questions

1. Let $$\mathbf{A}=\{\mathbf{1},\mathbf{2},\mathbf{3},\mathbf{4}\},\mathbf{B}=\{\mathbf{1},\mathbf{4},\mathbf{9},\mathbf{16},\mathbf{25}\}$$ and $\mathbf{R}$ be a relation defined from $\mathbf{A}$ to $\mathbf{B}$ as, $$\mathbf{R}=\{(\mathbf{x},\mathbf{y}):\mathbf{x}\in \mathbf{A},\mathbf{y}\in \mathbf{B}\mathbf{and}\mathbf{y}={\mathbf{x}}^{\mathbf{2}}\}$$. Then do as directed.

(a) Depict this relation using arrow diagram.

Ans. The relation $$\text{R}=\{\left(\text{x,y}\right)\text{:x}\in \text{A,}\text{y}\in \text{B and y}={\text{x}}^{\text{2}}\}$$ can be represented by the following diagram.

(b) Find domain of R.

Ans. The domain of the given relation $R$ is $$\left\{\text{1,2,3,4}\right\}$$.

(c) Find range of R.

Ans. The range of the given relation $R$ is $$\left\{\text{1,4,9,16}\right\}$$.

(d) Write co-domain of R.

Ans. The co-domain of the relation $R$ is $$\left\{\text{1,4,9,16,25}\right\}$$.

2. Let $$\mathbf{R}=\{(\mathbf{x},\mathbf{y}):\mathbf{x},\mathbf{y}\in \mathbf{N}\mathbf{and}\mathbf{y}=\mathbf{2}\mathbf{x}\}$$ be a relation on $\mathbf{N}$. Find: Domain, Co-domain and Range.

Is this relation a function from $\mathbf{N}$ to $\mathbf{N}$?

Ans.  The given relation is $R=\{(x,y):x,y\in \mathbb{N}\text{ and y}=\text{2x}\}$ on the natural number set $\mathbb{N}$.

Therefore, the domain of $R$ is the set of natural numbers $\mathbb{N}$.

The co-domain of $R$ is also the set of natural numbers $\mathbb{N}$.

And the Range of $R$ is the set of even natural numbers.

yes, the relation $R$  is a function from $N$ to $N$.

3. Find the domain and range of, $$\mathbf{f}\left(\mathbf{x}\right)=\left|\mathbf{2}\mathbf{x}\mathbf{3}\right|\mathbf{3}$$

Ans. The given function is $f\left(x\right)=|2x-3|-3$.

There does not exist any value of $x$ for which $f\left(x\right)$ is unbounded. So, domain of the function $f\left(x\right)$ is the set of all real numbers $\mathbb{R}$.

Observe that, $f\left(x\right)\ge -3$, since $|2x-3|\ge 0$.

Therefore, the range of the function $f\left(x\right)$ is $[-3,\mathrm{\infty })$.

4. Draw the graph of the Constant function,$$\mathbf{f}:\mathbf{R}\in \mathbf{R};\mathbf{f}\left(\mathbf{x}\right)=\mathbf{2}\mathbf{x}\in \mathbf{R}$$. Also find its domain and range.

Ans. 

The domain of the function $f\left(x\right)$ is the set of all real numbers $\mathbb{R}$.

The range of the function is $\left\{2\right\}$.

5.Let $\mathbf{R}=\{\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)\mathbf{:}\mathbf{x}\mathbf{,}\mathbf{y}\in \mathbf{W}\mathbf{,}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{y}=\mathbf{8}\}$, then determine the following results.

(i) Find the domain and the range of R.

Ans. The provided equation is $2x+y=8$ such that $x,y\in W$.

The whole numbers for which the given equation is satisfied are as follows:

$x=0,$ $y=8$ implies $2\left(0\right)+8=8$,

$x=1,$ $y=6$ implies $2\left(1\right)+6=8$,

$x=2,y=4$ implies $2\left(2\right)+4=8$,

$x=3,y=2$ implies $2\left(3\right)+2=8$,

$x=4,y=0$ implies $2\left(4\right)+0=8$.

There does not have any other values of $x,y$ belong to the whole number set for which the given equation is satisfied.

Thus, the domain of the relation $R$ is $\{0,1,2,3,4\}$ and the range of the relation $R$ is $\{8,6,4,2,0\}$.

(ii) Write R as a set of ordered pairs.

Ans. The relation $R$ as the set of ordered pairs is given by

$R=\{(0,8),(1,6),(2,4),(3,2),(4,0)\}$.

6. Let $\mathbf{R}$ be a relation from $\mathbf{Q}$ to $\mathbf{Q}$ defined by $\mathbf{R}=\{\left(\mathbf{a}\mathbf{,}\mathbf{b}\right)\mathbf{:}\mathbf{a}\mathbf{,}\mathbf{b}\in \mathbf{Q}\mathbf{and}\mathbf{a}\mathbf{-}\mathbf{b}\in \mathbf{Z}\}$. Then show the following things.

(i) $\left(\mathbf{a}\mathbf{,}\mathbf{a}\right)\in \mathbf{R}$ for all $\mathbf{a}\in \mathbf{Q}$.

Ans. The given relation is $R=\{(a,b):a,b\in \mathbb{Q}\text{and a-b}\in \mathbb{Z}\}$.

Let $a,a\in \mathbb{Q}$. Then $a-a=0\in \mathbb{Z}$.

Thus, $(a,a)\in R$ for all $a\in \mathbb{Q}$.

(ii) $\left(\mathbf{a}\mathbf{,}\mathbf{b}\right)\in \mathbf{R}$ implies that $\left(\mathbf{b}\mathbf{,}\mathbf{a}\right)\in \mathbf{R}$.

Ans. It is provided that $(a,b)\in \mathbb{R}$. Then $a-b\in \mathbb{Z}$, for all $a,b\in \mathbb{Q}$.

Now, $a-b\in \mathbb{Z}$  implies that $-(a-b)\in \mathbb{Z}$, that is $(b-a)\in \mathbb{Z}$.

Therefore, $(b,a)\in R$.

(iii) $\left(\mathbf{a}\mathbf{,}\mathbf{b}\right)\in \mathbf{R}$ and $(\mathbf{b},\mathbf{c})\in \mathbf{R}$ implies that $\left(\mathbf{a}\mathbf{,}\mathbf{c}\right)\in \mathbf{R}$.

Ans. It is provided that $(a,b)\in R$ and $(b,c)\in R$.

Then, we have $a,b,c\in \mathbb{Q}$ , $a-b\in \mathbb{Z}$ and $b-c\in \mathbb{Z}$.

Therefore, $(a-b)+(b-c)\in \mathbb{Z}$

$\Rightarrow a-b+b-c\in \mathbb{Z}$

$\Rightarrow a-c\in \mathbb{Z}$

Hence, $(a,c)\in R$.

7. If $\mathbf{f}\left(\mathbf{x}\right)=\frac{{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{3}\mathbf{x}\mathbf{+}\mathbf{1}}{\mathbf{x}\mathbf{-}\mathbf{1}}$, then find $\mathbf{f}(-\mathbf{2})+\mathbf{f}\left(\frac{\mathbf{1}}{\mathbf{3}}\right)$.

Ans. The provided function is $f\left(x\right)=\frac{x^2-3x+1}{x-1}$.

Substituting $x=-2$ into the function, we obtain

$f(-2)=\frac{{(-2)}^2-3(-2)+1}{-2-1}$