Answer :

The relation fis defined as

$ f(x)= \begin{cases}x^{2}, & 0 \leq x \leq 3 \\ 3 x, & 3 \leq x \leq 10\end{cases} $

It is observed that for

$0 \leq x<3, f(x)=x^{2}$

$3<x \leq 10, f(x)=3 x$

Also, at $x=3, f(x)=3^{2}=9$ or $f(x)=3 \times 3=9$

i.e., at $x=3, f(x)=9$

Therefore, for $0 \leq x \leq 10$, the images of $f(x)$ are unique.

Thus, the given relation is a function.

The relation gis defined as $g(x)= \begin{cases}x^{2}, 0 \leq x \leq 2 \\ 3 x, 2 \leq x \leq 10\end{cases}$

It can be observed that for $x=2, g(x)=2^{2}=4$ and $g(x)=3 \times 2=6$

Hence, element 2 of the domain of the relation gcorresponds to two different images i.e., 4 and 6 . Hence, this relation is not a function.

Answer :

The given function is

$ f(x)=\dfrac{x^{2}+2 x+1}{x^{2}-8 x+12} $

$f(x)=\dfrac{x^{2}+2 x+1}{x^{2}-8 x+12}=\dfrac{x^{2}+2 x+1}{(x-6)(x-2)}$

It can be seen that function fis defined for all real numbers except at $x=6$ and $x=2$.

Hence, the domain of fis $\mathbf{R}$ - $\{2,6\}$.

Answer :

The given real function is $f(x)=\sqrt{x-1}$.

It can be seen that $\sqrt{x-1}$ is defined for (xâ “ $.^{\prime} 1) \geq 0$.

i.e., $f(x)=\sqrt{(x-1)}$ is defined for $x \geq 1$.

Therefore, the domain of $f$ is the set of all real numbers greater than or equal to 1 i.e., the domain of $f=[1, \infty)$.

As $x \geq 1 \Rightarrow(x - 1) \geq 0 \Rightarrow \sqrt{x-1} \geq 0$

Therefore, the range of fis the set of all real numbers greater than or equal to 0 i.e., the range of $f=[0, \infty)$.

Answer :

The given real function $is f(x)=|x-1|$.

It is clear that $|x-1|$ is defined for all real numbers.

$\therefore$ Domain of $f=\mathbf{R}$

Also, for $x \in \mathbf{R},|x-1|$ assumes all real numbers.

Hence, the range of fis the set of all non-negative real numbers.

Answer :

$ \begin{aligned} & f=\{(x, \dfrac{x^{2}}{1+x^{2}}): x \in \mathbf{R}\} \\ & =\{(0,0),( \pm 0.5, \dfrac{1}{5}),( \pm 1, \dfrac{1}{2}),( \pm 1.5, \dfrac{9}{13}),( \pm 2, \dfrac{4}{5}),(3, \dfrac{9}{10}),(4, \dfrac{16}{17}), \ldots\} \end{aligned} $

The range of fis the set of all second elements. It can be observed that all these elements are greater than or equal to 0 but less than 1.

[Denominator is greater numerator]

Thus, range of $f=[0,1)$

Answer :

$f, g: \mathbf{R} \rightarrow \mathbf{R}$ is defined as $f(x)=x+1, g(x)=2 x$ - 3

$(f+g)(x)=f(x)+g(x)=(x+1)+(2 x - 3)=3 x$ - 2

$\therefore(f+g)(x)=3 x$ - 2

$(f - g)(x)=f(x) - g(x)=(x+1)$ -$(2 x - 3)=x+1 -2 x+3= -x+4$

$\therefore(f-g)(x)=\hat{a}-x+4$

$ \begin{aligned} & (\dfrac{f}{g})(x)=\dfrac{f(x)}{g(x)}, g(x) \neq 0, x \in \mathbf{R} \\ & \therefore(\dfrac{f}{g})(x)=\dfrac{x+1}{2 x-3}, 2 x-3 \neq 0 \text{ or } 2 x \neq 3 \\ & \therefore(\dfrac{f}{g})(x)=\dfrac{x+1}{2 x-3}, x \neq \dfrac{3}{2} \end{aligned} $

Answer :

$f=\{(1,1),(2,3),(0,-1),(-1,-3)\}$

$f(x)=a x+b$

$(1,1) \in f$

$\Rightarrow f(1)=1$

$\Rightarrow a \times 1+b=1$

$\Rightarrow a+b=1$

$(0,-1) \in f$

$\Rightarrow f(0)=-1$

$\Rightarrow a \times 0+b=-1$

$\Rightarrow b=-1$

On substituting $b=-1$ in $a+b=1$, we obtain $a+(-1)=1 \Rightarrow a=1+1=2$.

Thus, the respective values of aand bare 2 and -1 .

Answer:

$ R=\{(a, b): a, b \in \mathbf{N}.$ and $.a=b^{2} \}$

(i) It can be seen that $2 \in \mathbf{N}$;however, $2 \neq 2^{2}=4$.

Therefore, the statement " $(a, a) \in R$, for all $a \in \mathbf{N}$ " is not true.

(ii) It can be seen that $(9,3) \in \mathbf{N}$ because $9,3 \in \mathbf{N}$ and $9=3^{2}$.

Now, $3 \neq 9^{2}=81$; therefore, $(3,9) \in N$

Therefore, the statement " $(a, b) \in R$, implies $(b, a) \in R$ " is not true.

(iii) It can be seen that $(16,4) \in R,(4,2) \in R$ because $16,4,2 \in \mathbf{N}$ and $16=4^{2}$ and $4=2^{2}$.

Now, $16 \neq 2^{2}=4$; therefore, $(16,2) \in N$

Therefore, the statement " $(a, b) \in R,(b, c) \in R$ implies $(a, c) \in R$ " is not true.

Answer :

$A=\{1,2,3,4\}$ and $B=\{1,5,9,11,15,16\}$

$\therefore A \times B=\{(1,1),(1,5),(1,9),(1,11),(1,15),(1,16),(2,1),(2,5),(2,9),(2,11),(2,15),(2,16),(3,1),(3,5),(3,9)$,

$(3,11),(3,15),(3,16),(4,1),(4,5),(4,9),(4,11),(4,15),(4,16)\}$

It is given that $f=\{(1,5),(2,9),(3,1),(4,5),(2,11)\}$

(i) A relation from a non-empty set $A$ to a non-empty set $B$ is a subset of the Cartesian product $A \times B$.

It is observed that fis a subset of $A \times B$.

Thus, fis a relation from $A$ to $B$.

(ii) Since the same first element i.e., 2 corresponds to two different images i.e., 9 and 11 , relation $f$ is not a function.

Answer :

The relation fis defined as $f=\{(a b, a+b): a, b \in \mathbf{Z}\}$

We know that a relation ffrom a set $A$ to a set $B$ is said to be a function if every element of set $A$ has unique images in set B.

Since 2, 6, -2, -6 $\in \mathbf{Z},(2 \times 6,2+6),(-2 \times-6,-2+(-6)) \in f$

i.e., $(12,8),(12,-8) \in f$

It can be seen that the same first element i.e., 12 corresponds to two different images i.e., 8 and -8. Thus, relation fis not a function.

Answer :

$A=\{9,10,11,12,13\}$

$f: A \to$ Nis defined as

$f(n)=$ The highest prime factor of $n$

Prime factor of $9=3$

Prime factors of $10=2,5$

Prime factor of $11=11$

Prime factors of $12=2,3$

Prime factor of $13=13$

$\therefore f(9)=$ The highest prime factor of $9=3$

$f(10)=$ The highest prime factor of $10=5$

$f(11)=$ The highest prime factor of $11=11$

$f(12)=$ The highest prime factor of $12=3$

$f(13)=$ The highest prime factor of $13=13$

The range of fis the set of all $f(n)$, where $n \in A$.

$\therefore$ Range of $f=\{3,5,11,13\}$