Answer :

The relation $f$ is defined as $f(x)=\left\lbrace \begin{array}{l}x^2, 0 \leq x \leq 3 \\ \\ 3 x, 3 \leq x \leq 10\end{array}\right.$

It is observed that for $0 \leq x \leq 3$, we have $f(x)=x^2$ and for $3 \leq x \leq 10$, we have $f(x)=3 x$

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

i.e., at $\mathrm{x}=3, \mathrm{f}(\mathrm{x})=9$

Therefore for every $\mathrm{x}, \mathrm{0} \leq \mathrm{x} \leq 10$, we have unique image under $f$

Thus, the relation f is a function.

Also, the relation $g$ is defined as $g(x)=\left\lbrace \begin{array}{l}x^2, 0 \leq x \leq 2 \\ \\ 3 x, 2 \leq x \leq 10\end{array}\right.$

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

Thus, the element $2$ of the domain of the relation $g$ has two different images i.e., $4$ and $6$

Hence, this relation is but not a function.

Answer :

Given, $f(x)=x^2$

$ \therefore \ \ \dfrac{\mathrm{f}(1.1)-\mathrm{f}(1)}{(1.1-1)}=\dfrac{(1.1)^2-(1)^2}{(1.1-1)}$

$\hspace{2.6cm} =\dfrac{1.21-1}{0.1}$

$\hspace{2.6cm} =\dfrac{0.21}{0.1}$

$\hspace{2.6cm} =2.1 $

Answer :

The given function is

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

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

$\hspace{0.8cm} =\dfrac{x^{2}+2 x+1}{\left(x-6\right)\left(x-2\right)}$

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

Hence, the domain of f is $\mathbf{R}$ - $\lbrace 2,6 \rbrace $

Answer :

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

It can be seen that $\sqrt{x-1}$ is defined for $\left(x- 1\right) \geq 0$

i.e., $f\left(x\right)=\sqrt{\left(x-1\right)}$ 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 f is 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 $is f\left(x\right)=|x-1|$

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 f is the set of all non-negative real numbers.

Answer :

$ \text { Let, } \ \mathrm{y}=\dfrac{\mathrm{x}^2}{1+\mathrm{x}^2} $

$ \begin{aligned} & \Rightarrow y+x^2 y=x^2 \\ \\ & \Rightarrow y=x^2(1-y) \\ \\ & \Rightarrow x^2=\dfrac{y}{1-y} \end{aligned} $

$ \Rightarrow x=\sqrt{\dfrac{y}{1-y}} $

Since, $x$ is real

$ \begin{aligned} & \Rightarrow \dfrac{y}{1-y} \geq 0 \\ \\ & \Rightarrow \dfrac{y(1-y)}{(1-y)^2} \geq 0 \\ \\ & \Rightarrow y(1-y) \geq 0 \text { and }(1-y)^2>0 \\ \\ & \Rightarrow 0 \leq y \leq 1 \text { and }-y>-1 \\ \\ & \Rightarrow 0 \leq y \leq 1 \text { and } y<1 \end{aligned} $

Hence, $0 \leq y<1$

Range of $f$ is $[0,1)$

Answer :

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

$\hspace{0.5cm} \left(f+g\right)\left(x\right)=f\left(x\right)+g\left(x\right)=\left(x+1\right)+\left(2 x - 3\right)=3 x - 2$

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

$\hspace{0.5cm}\left(f - g\right)\left(x\right)=f\left(x\right) - g\left(x\right)=\left(x+1\right)-\left(2 x - 3\right)=x+1 -2 x+3= -x+4$

$\therefore \ \ \left(f-g\right)\left(x\right)=-x+4$

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

Answer :

$\hspace{0.5cm} f=\lbrace \left(1,1\right),\left(2,3\right),\left(0,-1\right),\left(-1,-3\right) \rbrace $

$f\left(x\right)=a x+b$

$\hspace{0.5cm}\left(1,1\right) \in f$

$\Rightarrow f\left(1\right)=1$

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

$\Rightarrow a+b=1$

$\hspace{0.5cm}\left(0,-1\right) \in f$

$\Rightarrow f\left(0\right)=-1$

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

$\Rightarrow b=-1$

On substituting $b=-1$ in $a+b=1$,

we obtain $a+\left(-1\right)=1 \Rightarrow a=1+1=2$

Thus, the respective values of $a$ and $b$ are $2$ and $-1 .$

Answer:

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

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

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

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

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

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

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

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

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

Answer :

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

$\therefore \ \ A \times B=\left\lbrace \begin{matrix}\left(1,1\right),\left(1,5\right),\left(1,9\right),\left(1,11\right),\left(1,15\right),\left(1,16\right),\left(2,1\right),\left(2,5\right),\left(2,9\right), \left(2,11\right),\left(2,15\right), \left(2,16\right), \\ \left(3,1\right),\left(3,5\right),\left(3,9\right),\left(3,11\right),\left(3,15\right),\left(3,16\right),\left(4,1\right),\left(4,5\right),\left(4,9\right),\left(4,11\right),\left(4,15\right),\left(4,16\right)\end{matrix}\right\rbrace$

It is given that $f=\lbrace \left(1,5\right),\left(2,9\right),\left(3,1\right),\left(4,5\right),\left(2,11\right)\rbrace$

(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 f is a subset of $A \times B$

Thus, f is 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 f is defined as $f=\lbrace \left(a b, a+b\right): a, b \in \mathbf{Z} \rbrace $

We know that a relation from 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 $\lbrace 2, 6, -2, -6 \rbrace \in \mathbf{Z},\left(2 \times 6,2+6\right),\left(-2 \times-6,-2+\left(-6\right)\right) \in f$

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

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

Answer :

$A=\lbrace 9,10,11,12,13 \rbrace $

$f: A \to N$ is defined as

$f\left(n\right)=$ 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 $13=13$

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

$\quad f\left(10\right)=$ The highest prime factor of $10=5$

$\quad f\left(11\right)=$ The highest prime factor of $11=11$

$\quad f\left(12\right)=$ The highest prime factor of $12=3$

$\quad f\left(13\right)=$ The highest prime factor of $13=13$

The range of f is the set of all $f\left(n\right)$, where $n \in A$

$\therefore \ \ $ Range of $f=\lbrace 3,5,11,13 \rbrace $