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

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

Edited By Vishal kumar | Updated on Nov 15, 2023 10:08 AM IST

## NCERT Solutions for Class 11 Maths Chapter 2 - Relations and Functions Miscellaneous Exercise- Download Free PDF

NCERT Solutions for Class 11 Maths Chapter 2: Relations and Functions Miscellaneous Exercise- In the previous exercises of this chapter NCERT Solutions For Class 11 Maths Chapter 2 Relations and Functions, you have learned about the relations, domain, codomain, and range of the relations, functions, domain and range of the functions, different types of function and their graphs, etc. In the NCERT solutions for Class 11 Maths chapter 2 miscellaneous exercise, you will get questions related to relations, functions, domain and range of relations and functions, etc. As the name suggests the miscellaneous exercise consists of mixed kinds of questions from all the topics of this miscellaneous exercise class 11 chapter 2.

If you solved all the previous exercises of this chapter, you can solve all the NCERT problems of class 11 chapter 2 maths miscellaneous solutions by yourself. The miscellaneous exercise Chapter 2 Class 11 is considered to be tougher as compared to the previous exercises of this chapter. So it is not easy to solve these problems on your own at first, but Class 11 Maths chapter 2 miscellaneous exercise solutions are here to help you. You will find all the NCERT problems solved in a descriptive manner. You can NCERT Solutions if you are looking for NCERT solutions for other classes as well.

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## Relations And Functions Class 11 Chapter 2 Miscellaneous Exercise

It is given that
$f (x) = \left\{\begin{matrix} x^2 & 0 \leq x\leq 3 \\ 3x &3 \leq x \leq 10 \end{matrix}\right.$
Now,
$f(x) = x^2 \ for \ 0\leq x\leq 3$
And
$f(x) = 3x \ for \ 3\leq x\leq 10$

At x = 3,$f(x) = x^2 = 3^2 = 9$

Also, at x = 3,$f(x) = 3x = 3\times 3 = 9$

We can see that for $0\leq x\leq 10$, f(x) has unique images.

Therefore, By definition of a function, the given relation is function.

Now,
It is given that
$g(x) = \left\{\begin{matrix} x^2 & 0 \leq x\leq 2 \\ 3x &2 \leq x \leq 10 \end{matrix}\right.$
Now,
$g(x) = x^2 \ for \ 0\leq x\leq 2$
And
$g(x) = 3x \ for \ 2\leq x\leq 10$

At x = 2, $g(x) = x^2 = 2^2 = 4$
Also, at x = 2, $g(x) = 3x = 3\times2 = 6$
We can clearly see that element 2 of the domain of relation g(x) corresponds to two different images i.e. 4 and 6. Thus, f(x) does not have unique images
Therefore, by definition of a function, the given relation is not a function

Hence proved

Given function is
$f(x)= x^2$
Now,
$\frac{f ( 1.1)- f (1)}{(1.1-1)} = \frac{(1.1)^2-1^2}{(1.1-1)} = \frac{1.21-1}{0.1}= \frac{0.21}{0.1}= 2.1$

Therefore, value of $\frac{f ( 1.1)- f (1)}{(1.1-1)}$ is 2.1

Given function is
$f (x) = \frac{x^2 + 2 x +1}{x^2 - 8x + 12 }$
Now, we will simplify it into
$f (x) = \frac{x^2 + 2 x +1}{x^2 - 8x + 12 }$
$= \frac{x^2+2x+1}{x^2-6x-2x+12}$
$= \frac{x^2+2x+1}{x(x-6)-2(x-6)}$
$= \frac{x^2+2x+1}{(x-2)(x-6)}$
Now, we can clearly see that $x \neq 2 , 6$
Therefore, the Domain of f(x) is $(R-\left \{ 2,6 \right \})$

Given function is
$f (x) = \sqrt{(x-1)}$
We can clearly see that f(x) is only defined for the values of x , $x\geq 1$
Therefore,
The domain of the function $f (x) = \sqrt{(x-1)}$ is $[1,\infty)$
Now, as
$\Rightarrow x\geq 1$
$\Rightarrow x-1\geq 1-1$
$\Rightarrow x-1\geq 0$
take square root on both sides
$\Rightarrow \sqrt{x-1}\geq 0$
$\Rightarrow f(x)\geq 0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because f(x)= \sqrt{x-1})$
Therefore,
Range of function $f (x) = \sqrt{(x-1)}$ is $[0,\infty)$

Given function is
$f (x) = |x-1|$
As the given function is defined of all real number
The domain of the function $f (x) = |x-1|$ is R
Now, as we know that the mod function always gives only positive values
Therefore,
Range of function $f (x) = |x-1|$ is all non-negative real numbers i.e. $[0,\infty)$

Given function is
$f = \left \{ \left ( x , \frac{x^2}{1+ x^2} \right ) : x \epsilon R \right \}$
Range of any function is the set of values obtained after the mapping is done in the domain of the function. So every value of the codomain that is being mapped is Range of the function.
Let's take
$y = \frac{x^2}{1+x^2}$
$\Rightarrow y(1+x^2)=x^2$
$\Rightarrow y+yx^2=x^2$
$\Rightarrow y=x^2(1-y)$
$\Rightarrow x^2= \frac{y}{1-y}$
$\Rightarrow x= \pm \sqrt{\frac{y}{1-y}}$
Now, 1 - y should be greater than zero and y should be greater than and equal to zero for x to exist because other than those values the x will be imaginary
Thus, $1 - y > 0 , y < 1 \ and \ y \geq 0$
Therefore,
Range of given function is $[0,1)$

It is given that
$f,g : R \rightarrow R$
$f(x)=x+1 \ \ and \ \ g(x) = 2x - 3$
Now,
$(f+g)x = f(x)+g(x)$
$= (x+1)+(2x-3)$
$= 3x-2$
Therefore,
$(f+g)x= 3x-2$

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

Now,
$\left ( \frac{f}{g} \right )x = \frac{f(x)}{g(x)} , g(x)\neq 0$
$=\frac{x+1}{2x-3} \ , x \neq \frac{3}{2}$
Therefore, values of $(f+g)x,(f-g)x \ and \ \left ( \frac{f}{g} \right )x$ are $(3x-2),(-x+4) \ and \ \frac{x+1}{2x-3}$ respectively

It is given that
$f =\left \{ {(1,1), (2,3), (0,-1), (-1, -3)} \right \}$
And
$f(x) = ax+b$
Now,
At x = 1 , $f(x) = 1$
$\Rightarrow f(1)= a(1)+b$
$\Rightarrow a+b = 1 \ \ \ \ \ \ \ \ \ \ -(i)$
Similarly,
At $x = 0$ , $f(x) = -1$
$\Rightarrow f(0) = a(0)+b$
$\Rightarrow b = -1$
Now, put this value of b in equation (i)
we will get,
$a = 2$
Therefore, values of a and b are 2 and -1 respectively

$( a,a ) \epsilon R ,$ for all $a \epsilon N$

It is given that
$R = \left \{ ( a,b): a,b \epsilon N \: \:and \: \: a = b ^ 2 \right \}$
And
$( a,a ) \epsilon R ,$ for all $a \epsilon N$
Now, it can be seen that $2 \ \epsilon \ N$ But, $2 \neq 2^ 2 = 4$
Therefore, this statement is FALSE

$( a,a ) \epsilon R ,$ implies (b,a) $\epsilon$ R

It is given that
$R = \left \{ ( a,b): a,b \epsilon N \: \:and \: \: a = b ^ 2 \right \}$
And
$( a,b ) \ \epsilon \ R ,$ implies $(b,a) \ \epsilon \ R$
Now , it can be seen that $( 2,4 ) \ \epsilon \ R ,$ and $4 = 2^2 = 4$ , But $2 \neq 4^2 =16$
Therefore, $(2,4) \ \notin \ N$
Therefore, given statement is FALSE

(a,b) $\epsilon$ R, (b,c) $\epsilon$ R implies (a,c) $\epsilon$ R.

It is given that
$R = \left \{ ( a,b): a,b \epsilon N \: \:and \: \: a = b ^ 2 \right \}$
And
$( a,b ) \ \epsilon \ R , (b,c) \ \epsilon \ R$ implies $(a,c) \ \epsilon \ R$
Now, it can be seen that $(16,4) \ \epsilon \ R , ( 4,2 ) \ \epsilon \ R$ because $16 = 4^2 = 16$ and $4 = 2^2 = 4$ , But $16 \neq 2^2 =4$
Therefore, $(16,2) \ \notin \ N$
Therefore, the given statement is FALSE

It is given that
$A =\left \{ {1,2,3,4} \right \}$
$B =\left \{ {1,5,9,11,15,16} \right \}$
and $f =\left \{ {(1,5), (2,9), (3,1), (4,5), (2,11)} \right \}$
Now,
$A \times B =\left \{ (1,1), (1,5), (1,9), (1,11), (1,15), (1,16), (2,1), (2,5), (2,9), (2,11), (2,15), (2,16), (3,1), (3,5), (3,9), (3,11), (3,15), (3,16), (4,1), (4,5), (4,9), (4,11), (4,15), (4,16) \right \}$Now, a relation from a non-empty set A to a non-empty set B is a subset of the Cartesian product A × B
And we can see that f is a subset of $A \times B$
Hence f is a relation from A to B
Therefore, given statement is TRUE

It is given that
$A =\left \{ {1,2,3,4} \right \}$
$B =\left \{ {1,5,9,11,15,16} \right \}$
and $f =\left \{ {(1,5), (2,9), (3,1), (4,5), (2,11)} \right \}$
Now,
$A \times B =\left \{ (1,1), (1,5), (1,9), (1,11), (1,15), (1,16), (2,1), (2,5), (2,9), (2,11), (2,15), (2,16), (3,1), (3,5), (3,9), (3,11), (3,15), (3,16), (4,1), (4,5), (4,9), (4,11), (4,15), (4,16) \right \}$
As we can observe that same first element i.e. 2 corresponds to two different images that is 9 and 11.
Hence f is not a function from A to B
Therefore, given statement is FALSE

It is given that
$f = \{ (ab, a + b) : a, b \epsilon Z \}$
Now, we know that relation f from a set A to a set B is said to be a function only if every element of set A has a unique image in set B
Now, for value 2, 6, -2, -6 $\epsilon \ Z$
$f = \left \{ (2 \times 6, 2 + 6), (-2 \times -6, -2 - 6), (2 \times -6, 2 - 6), (-2 \times 6, -2 + 6) \right \}$
$\Rightarrow f = \left \{ (12, 8), (12, -8), (-12, -4), (-12, 4) \right \}$
Now, we can observe that same first element i.e. 12 corresponds to two different images that are 8 and -8.
Thus, f is not a function

It is given that
A = {9,10,11,12,13}
And
f : A
N be defined by f(n) = the highest prime factor of n.
Now,
Prime factor of 9 = 3

Prime factor of 10 = 2,5

Prime factor of 11 = 11

Prime factor of 12 = 2,3

Prime factor of 13 = 13

f(n) = the highest prime factor of n.

Hence,

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

As the range of f is the set of all f(n), where $n \ \epsilon \ A$

Therefore, the range of f is: {3, 5, 11, 13}.

## More About NCERT Solutions for Class 11 Maths Chapter 2 Miscellaneous Exercise:-

Class 11 Maths chapter 2 miscellaneous solutions consists of questions related to relations, functions, finding the domain and range of the functions, etc. There are five examples related to functions and their properties are given before the miscellaneous exercise chapter 2 Class 11. You can go through these examples in order to get conceptual clarity. The topics covered in this exercise are same as the previous exercises of this chapter, so you can solve the problems of this exercise after solving the previous exercises.

Also Read| Relations And Functions Class 11th Notes

## Benefits of NCERT Solutions for Class 11 Maths Chapter 2 Miscellaneous Exercise:-

• Class 11 Maths chapter 2 miscellaneous exercise solutions are designed by subject matter experts who have expertise in the subject, so you can easily rely upon these solutions.
• NCERT solutions for Class 11 Maths chapter 2 miscellaneous exercise are useful to understand the algebra of functions which is an important concept of function.
• Class 11 Maths chapter 2 miscellaneous solutions are not only important for the CBSE exams but are very useful for competitive exams like JEE Main, SRMJEE, etc.
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## Key Features of NCERT Class 11 Maths Ch 2 Miscellaneous Exercise Solutions

Comprehensive Clarification: Miscellaneous exercise class 11 chapter 2 Solutions provide a thorough explanation of all topics, fostering a strong grasp of relations and functions.

Stepwise Clarity: These class 11 chapter 2 maths miscellaneous solutions break down each problem, aiding students in following a logical progression of concepts, and ensuring a smoother understanding.

Clear Conciseness: Presented in clear and concise language, the class 11 maths miscellaneous exercise chapter 2 solutions simplify complex mathematical concepts, enhancing accessibility for students.

Free PDF Access: Additionally, these class 11 chapter 2 miscellaneous exercise solutions offer the convenience of free PDF access, promoting widespread availability and ease of use for all students.

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## Subject Wise NCERT Exampler Solutions

Happy learning!!!

1. Let f(x) = 2x + 3 than f(3) ?

f(x) = 2x + 3

f(3) = 2(3) + 3 = 6 +3 = 9

2. Let f(x) = 2^x + x + 1 find f(2) ?

f(x) = 2^x + x + 1

f(2) = 2^2 + 2 + 1

f(2) = 4 + 2 + 1 = 7

3. Let f(x) = x^2 and g(x) = x + 2 be two real functions. Find (f+g)(x) ?

f(x) = x^2 and g(x) = x + 2

(f+g)(x) = x^2 + x + 2

4. Let f(x) = x^2 and g(x) = x + 2 be two real functions. Find (f-g)(x) ?

f(x) = x^2 and g(x) = x + 2

(f-g)(x) = x^2 - x - 2

5. Let f(x) = x^2 and g(x) = x + 2 be two real functions. Find (fg)(x) ?

f(x) = x^2 and g(x) = x + 2

(fg)(x) = (x^2)( x + 2)

(fg)(x) = x^3 + 2x^2

6. Let f(x) = x^2 and g(x) = x + 2 be two real functions. Find (f/g)(x) ?

f(x) = x^2 and g(x) = x + 2

(f/g)(x) = (x^2)/( x + 2) where x can't be equal to -2.

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