NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions Miscellaneous Exercise

Miscellaneous Exercise

1. The relation $f$  is defined by $f(x)=\left\{ \begin{matrix} {{x}^{2}},0\le x\le 3  \\ 3x,3\le x\le 10  \\ \end{matrix} \right.$

And the relation $g$ is defined by \[g(x)=\left\{ \begin{matrix} {{x}^{2}},0\le x\le 2  \\ 3x,2\le x\le 10  \\ \end{matrix} \right.\]

Show that $f$ is a function and $g$ is not a function.

Ans: According to the problem, we have the function $f$ as,

$f(x)=\left\{ \begin{matrix} {{x}^{2}},0\le x\le 3  \\ 3x,3\le x\le 10  \\ \end{matrix} \right.$

We can see that, for $x=3$ ,

$f(x)={{3}^{2}}\,=9$ from the first given condition.

And again, $f(x)=3\times 3=9$ from the second condition.

But now, 

\[g(x)=\left\{ \begin{matrix} {{x}^{2}},0\le x\le 2  \\ 3x,2\le x\le 10  \\ \end{matrix} \right.\]

We can see that, for $x=2$ ,

$f(x)={{2}^{2}}\,=4$ from the first given condition.

And again, $f(x)=3\times 2=6$ from the second condition.

Thus, the domain of the relation $g$ is having two different images from a single element.

So, it can be concluded that the relation is not a function.

2. If $f(x)={{x}^{2}}$ , find $\dfrac{f(1.1)-f(1)}{(1.1-1)}$ .

Ans: We have the function, $f(x)={{x}^{2}}$ .

So, we will have, 

\[\dfrac{f(1.1)-f(1)}{(1.1-1)}\] equalling to,  

\[\dfrac{{{(1.1)}^{2}}-{{1}^{2}}}{1.1-1}\] , putting the values.

After further simplification,

$\dfrac{1.21-1}{0.1} $

$ =\dfrac{0.21}{0.1}$

$=2.1 $ 

3. Find the domain of the function $f(x)=\dfrac{{{x}^{2}}+2x+1}{{{x}^{2}}-8x+12}$

Ans: According to the problem, we have the given function as,

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

Let us try to simplify the given function and bring it to a form where we can analyze the problem.

The denominator can be factorized as,

${{x}^{2}}-8x+12$ 

$={{x}^{2}}-6x-2x+12$ 

$=x(x-6)-2(x-6) $ 

$=(x-2)(x-6)$ 

So, we see that the function is defined for every real numbers except $6,2$ .

Thus, the domain of the function will be, $R-\{2,6\}$ 

4. Find the domain and the range of the real function $f$ defined by $f(x)=\sqrt{(x-1)}$.

Ans: We have the given function as, $f(x)=\sqrt{(x-1)}$ .

Clearly, the term inside the root sign must be non-negative.

So, the function is valid for all values of $x\ge 1$ .

Thus, the domain of the function will be, $[1,\infty )$ .

Now, again, for $x\ge 1$, the value of the function will always be greater than or equal to zero.

So, the range of the function is, $[0,\infty )$ .

5. Find the domain and the range of the real function $f$ defined by $f(x)=\left| x-1 \right|$.

Ans: The function which is given is, $f(x)=\left| x-1 \right|$ .

We can clearly see that, the function is well defined for all the real numbers.

Thus, it can be concluded that, the domain of the function is $R$ .

And for every $x\in R$ , the function gives all non-negative real numbers.

So, the range of the function is the set of all non-negative real numbers. i.e, $[0,\infty )$ .

6. Let $f=\left\{ \left( x,\dfrac{{{x}^{2}}}{1+{{x}^{2}}} \right):x\in R \right\}$ be a function from $R$ to $R$. Determine the range of $f$ .

Ans: We have our given function as,$f=\left\{ \left( x,\dfrac{{{x}^{2}}}{1+{{x}^{2}}} \right):x\in R \right\}$

Expressing it by term to term, we are getting, 

$f=\left\{ \left( 0,0 \right),\left( \pm 0.5,\dfrac{1}{5} \right),\left( \pm 1,\dfrac{1}{2} \right),\left( \pm 1.5,\dfrac{9}{13} \right),\left( \pm 2,\dfrac{4}{5} \right),\left( 3,\dfrac{9}{10} \right),\left( 4,\dfrac{16}{17} \right),.... \right\}$ 

And we also know, the range of $f$ is the set of all the second elements. We can also see that the terms are greater than or equal to $0$ but less than $1$ .

So, the range of the function is, $[0,1)$ .

7. Let $f,g:R\to R$ be defined, respectively by $f(x)=x+1,g(x)=2x-3$ . Find $f+g,f-g$ and $\dfrac{f}{g}$ .

Ans: We have the functions defined as, $f,g:R\to R$is defined as, $f(x)=x+1,g(x)=2x-3$ .

Thus, the function

$(f+g)(x)=f(x)+g(x)$ 

$=(x+1)+(2x-3)$ 

$=3x-2$ 

So, the function $(f+g)(x)=3x-2$ .

 Again, the function, 

$(f-g)(x)=f(x)-g(x)$ 

$=(x+1)-(2x-3)$ 

$=-x+4$

So, the function $(f-g)(x)=-x+4$.

Similarly, 

$\left( \dfrac{f}{g} \right)(x)=\dfrac{f(x)}{g(x)}$ where $g\left( x \right)\ne 0$ and also $x\in R$ .

Now, putting the values, 

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

where, 

$2x-3\ne 0$ 

$\Rightarrow x\ne \dfrac{3}{2}$ 

8. Let $f=\left\{ \left( 1,1 \right),\left( 2,3 \right),\left( 0,-1 \right),\left( -1,-3 \right) \right\}$ be a function from $Z$ to $Z$ defined by $f(x)=ax+b$ , for some integers $a,b$ . Determine $a,b$ 

Ans: We have the given function as,  $f=\left\{ \left( 1,1 \right),\left( 2,3 \right),\left( 0,-1 \right),\left( -1,-3 \right) \right\}$ and also $f(x)=ax+b$ .

As, $(1,1)\in f$ , we get, 

$a\times 1+b=1$

$\Rightarrow a+b=1$

And again, $(0,-1)\in f$ , from this we can get,

$a\times 0+b=-1$

$\Rightarrow b=-1$ 

Putting this value in the first equation, we have,

$a-1=1$

$\Rightarrow a=2$

So, the value of $a$ and $b$ are respectively, $2,-1$ .

9. Let $R$ be a relation from $N$ to $N$ defined by $R = \left\{ {\left( {a,b} \right):\,a,b \in \,N\,and\,a = {b^2}} \right\} $ . Are the following true? Justify your answer in each case.

(i) $\left( a,a \right)\in R$ , for all $a\in N$ .

Ans: We are given our relation as, $R=\left\{ (a,b):a,b\in N\,and\,a={{b}^{2}} \right\}$ 

Let us take, $2\in N$ .

But we have, $2\ne {{2}^{2}}=4$ 

So, the statement that $\left( a,a \right)\in R$ , for all $a\in N$is not true.

(ii) $\left( a,b \right)\in R$ , implies $\left( b,a \right)\in R$ 

Ans: We are given our relation as, $R=\left\{ (a,b):a,b\in N\,and\,a={{b}^{2}} \right\}$ 

Let us take, $(9,3)\in N$ . 

We have to check if, $\left( 3,9 \right)\in N$or not.

But, the condition of the relation says,  $R=\left\{ (a,b):a,b\in N\,and\,a={{b}^{2}} \right\}$and ${{9}^{2}}\ne 3$ .

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

(iii) $\left( a,b \right)\in R,\left( b,c \right)\in R$ implies $\left( a,c \right)\in R$ .

Ans: We are given our relation as, $R=\left\{ (a,b):a,b\in N\,and\,a={{b}^{2}} \right\}$ 

Now, let us take, $\left( 9,3 \right)\in R,\left( 16,4 \right)\in R$ .

We have to check if, $\left( 9,4 \right)\in N$or not.

Thus can also easily see, $9\ne {{4}^{2}}=16$ .

So, the given statement $\left( a,b \right)\in R,\left( b,c \right)\in R$ implies $\left( a,c \right)\in R$ .

10. Let $A=\{1,2,3,4\},B=\{1,5,9,11,15,16\}$ and $f=\left\{ \left( 1,5 \right),\left( 2,9 \right),\left( 3,1 \right),\left( 4,5 \right),(2,11) \right\}$ . Are the following true? Justify your answer in each case.

(i) $f$ is a relation from $A$ to $B$ .

Ans: We are provided with two sets, $A=\{1,2,3,4\},B=\{1,5,9,11,15,16\}$ 

Thus, the Cartesian product of these two sets will be,

$A\times B=${ $\left( 1,1 \right),\left( 1,5 \right),\left( 1,9 \right),\left( 1,11 \right),\left( 1,15 \right),\left( 1,16 \right),$ 

$\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) $

And it is also given that, 

$f=\left\{ \left( 1,5 \right),\left( 2,9 \right),\left( 3,1 \right),\left( 4,5 \right),(2,11) \right\}$

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

Thus, it can be easily checked that $f$ is a relation from $A$ to B.

(ii) f is a function from $A$ to $B$ .

Ans: We are provided with two sets, $A=\{1,2,3,4\},B=\{1,5,9,11,15,16\}$ 

Thus, the Cartesian product of these two sets will be,

$A\times B=\{\left( 1,1 \right),\left( 1,5 \right),\left( 1,9 \right),\left( 1,11 \right),\left( 1,15 \right),\left( 1,16 \right),$ 

$\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), $

$\left( 4,1 \right),\left( 4,5 \right),\left( 4,9 \right),\left( 4,11 \right),\left( 4,15 \right),\left( 4,16 \right)$

And it is also given that, 

$f=\left\{ \left( 1,5 \right),\left( 2,9 \right),\left( 3,1 \right),\left( 4,5 \right),(2,11) \right\}$

If we check carefully, we see that the first element $2$ is providing us two different value of the image $9,11$ .

So, it can be concluded that $f$ is not a function from $A$ to B.

11. Let $f$ be the subset of $Z\times Z$ defined by $f=\left\{ \left( ab,a+b \right):a,b\in Z \right\}$ . If $f$ a function from $Z$ to $Z$. Justify your answer.

Ans: Our given relation $f$ is defined as $f=\left\{ \left( ab,a+b \right):a,b\in Z \right\}$.

We also know that a relation will be called a function from $A$ to $B$ if every element of the set $A$ has unique images in set $B$ .

Let us take 4 elements, $2,6,-2,-6\in Z$ .

So, for the first two elements, 

$\left( 2\times 6,2+6 \right)\in f$ 

$\Rightarrow \left( 12,8 \right)\in f$

And for the last two elements,

$\left( -2\times -6,-2+-6 \right)\in f$

$\Rightarrow \left( 12,-8 \right)\in f$

So, it is clearly visible that one single element $12$ having two different images $8,-8$. Thus, the relation is not a function.

12. Let $A=\left\{ 9,10,11,12,13 \right\}$ and let $f:A\to N$ be defined by $f(n)=$ the highest prime factor of $n$ . Find the range of $f$ .

Ans: We have our given set as, $A=\left\{ 9,10,11,12,13 \right\}$ and the relation is given as $f(n)=$ the highest prime factor of $n$.

The prime factor of $9$ is $3$ .

The prime factors of 10 is $2,5$ .

The prime factor of $11$ is $11$ .

The prime factor of 12 is $2,3$ .

The prime factor of $13$ is $13$ .

Thus, it can be said, 

$f(9)=$ 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 $13=13$ .

Now, the range of the function will be, $\left\{ 3,5,11,13 \right\}$.

Conclusion

NCERT Solutions for Maths Miscellaneous Exercise Class 11 Chapter 2 Relations and Functions, by Vedantu, provides clear explanations for every problem. Expert teachers have prepared these solutions, following CBSE guidelines. Focus on understanding different types of relations and functions, as these are important for the chapter. Practising these solutions will help you understand the key concepts and improve your problem-solving skills, which is important for doing well in exams. Download the PDF and practice regularly for the best results.

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