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NCERT Exemplar Class 12 Maths Solutions Chapter 1 Relations and Functions

NCERT Exemplar Class 12 Maths Solutions Chapter 1 Relations and Functions

Edited By Ravindra Pindel | Updated on Sep 15, 2022 05:48 PM IST | #CBSE Class 12th

NCERT Exemplar Class 12 Maths solutions chapter 1 is crucial to building a base from ground up for the entire academic session. We here will help you to find the solutions of all NCERT questions of the chapter with complete steps and detailed answers. This chapter of NCERT Class 12 Maths Solutions is divided into two parts; relations and functions. We have solutions for questions from the parts explained and solved in detail. NCERT Exemplar Class 12 Maths solutions chapter 1 PDF download is to be available for students for extending learning of the concepts.

This Story also Contains
  1. NCERT Exemplar Class 12 Maths Solutions Chapter 1: Exercise-1.3
  2. Subtopics Covered in NCERT Exemplar Solutions for Class 12 Maths Chapter 1
  3. What student will learn in NCERT Exemplar Class 12 Maths Solutions Chapter 1?
  4. NCERT Exemplar Class 12 Maths Solutions
  5. Important Topics To Cover From NCERT Exemplar Class 12 Maths Solutions Chapter 1

NCERT Exemplar Class 12 Maths Solutions Chapter 1: Exercise-1.3

Question:1

Let A = {a, b, c} and the relation R be defined on A as follows:

R = {(a, a), (b, c), (a, b)}.

Then, write the minimum number of ordered pairs to be added in R to make R reflexive and transitive.

Answer:

Here, R = {(a, a), (b, c), (a, b)}

The minimum number of ordered pairs to be added to make R as reflexive is (b, b) and (c, c) to R. Whereas, to make R transitive, the minimum number of ordered pairs to be added is (a, c) to R.

Therefore, we need 3 ordered pairs to add with R to make it reflexive and transitive.

Question:2

Let D be the domain of the real-valued function f defined by f(x) = \sqrt {25 - x\textsuperscript{2}} . Then, write D.

Answer:

f(x) = \sqrt {25 - x\textsuperscript{2}}
Therefore, the function can be defined as:
\\25 - x\textsuperscript{2} \geq 0\\ or, x\textsuperscript{2} \leq 25\\ -5 \leq x \leq 5\\

Therefore, the given function has a domain of [-5, 5].

Question:3

Let f, g:R \rightarrow R be defined by f(x) = 2x + 1 and g (x) = x\textsuperscript{2} - 2, \forall x \in R, respectively. Then, find gof.

Answer:

Here,
f(x) = 2x + 1
g (x) = x\textsuperscript{2} -2, \forall x \in R\\
Therefore, gof= g(f(x)) = g (2x + 1) = (2x + 1)\textsuperscript{2} -2 = 4x\textsuperscript{2} + 4x + 1 -2 = 4x\textsuperscript{2} + 4x - 1\\

Question:4

. Let f: R \rightarrow R be the function defined by f (x) = 2x - 3, \forall x \in R. write f\textsuperscript{-1}.

Answer:

Here, f (x) = 2x - 3, \forall x \in R\\
Let’s say,
y = 2x - 3\\ or, x = (y + 3)/ 2

Therefore,
f\textsuperscript{-1}(x) = (x + 3)/ 2\\

Question:5

If A = \{ a, b, c, d \} and the function f = \{ (a, b), (b, d), (c, a), (d, c) \} , write f\textsuperscript{-1}.

Answer:
Here, A = \{ a, b, c, d \} and the function f = \{ (a, b), (b, d), (c, a), (d, c) \} \\
Therefore, f\textsuperscript{-1} = \{ (b, a), (d, b), (a, c), (c, d) \} \\

Question:6

If f:R \rightarrow R is defined by f (x) = x\textsuperscript{2} - 3x + 2, write f (f (x)).

Answer:

Here, f (x) = x\textsuperscript{2} - 3x + 2
Therefore,
\\ f(f(x)) = f(x\textsuperscript{2} - 3x + 2)\\ = (x\textsuperscript{2} - 3x + 2)\textsuperscript{2} - 3(x\textsuperscript{2} - 3x + 2) + 2,\\ = x\textsuperscript{4} + 9x\textsuperscript{2} + 4 - 6x\textsuperscript{3} + 4x\textsuperscript{2} - 12x - 3x\textsuperscript{2} + 9x - 6 + 2\\ = x\textsuperscript{4} - 6x\textsuperscript{3} + 10x\textsuperscript{2} - 3x\\
Similarly,
\\f(f(x)) = x\textsuperscript{4} - 6x\textsuperscript{3} + 10x\textsuperscript{2} - 3x\\

Question:7

Is g = \{ (1, 1), (2, 3), (3, 5), (4, 7) \} a function? If g is described by g (x) = \alpha x + \beta , then what value should be assigned to \alpha\: \: and\: \: \beta .

Answer:

Here,g = \{ (1, 1), (2, 3), (3, 5), (4, 7) \}
Here, each and every element of a domain has a unique image. Therefore, g is a function.
Also, g (x) = \alpha x + \beta
Therefore,
\\g(1) = \alpha (1) + \beta = 1\\ \alpha + \beta = 1 \ldots \ldots .. (1)\\
Similarly,
g(2) = \alpha (2) + \beta = 2 \alpha + \beta = 3 \ldots \ldots .. (2)\\
By solving (1) and (2), we get
\alpha = 2 and \beta = -1\\
Therefore, g (x) = 2x - 1


Question:8

Are the following set of ordered pairs functions? If so, examine whether the mapping is injective or surjective.
(i) \{ (x, y): \text{x is a person, y is the mother of x} \} .\\
(ii) \{ (a, b): \text{a is a person, b is an ancestor of a} \} .\\

Answer:
(i)Here, \{ (x, y): \text{x is a person, y is the mother of x} \} .\\
Therefore, each person (x) has only one biological mother.
Hence, the given set of ordered pairs make a function.
Therefore, there are more than one person who may have the same mother. Hence, the function is many-one and surjective.
(ii) Here, \{ (a, b): \text{a is a person, b is an ancestor of a} \} .\\
It’s seen that any person ‘a’ has more than one ancestor.
Therefore, it is not a function.

Question:9

If the mappings f and g are given by f = \{ (1, 2), (3, 5), (4, 1) \} and g = \{ (2, 3), (5, 1), (1, 3) \} , write fog.\\

Answer:
Here,
f = \{ (1, 2), (3, 5), (4, 1) \} and g = \{ (2, 3), (5, 1), (1, 3) \}
Therefore,
\\fog (2) = f(g(2)) = f(3) = 5\\ fog (5) = f(g(5)) = f(1) = 2\\ fog (1) = f(g(1)) = f(3) = 5\\
So, we can write that:
\\fog = \{ (2, 5), (5, 2), (1, 5) \} \\


Question:10

Let C be the set of complex numbers. Prove that the mapping f: C \rightarrow R is given by f (z) = \vert z \vert , \forall z \in C, is neither one-one nor onto.

Answer:

Here, f: C \rightarrow R is given by f (z) = \vert z \vert , \forall z \in C
If we assume z = 4 + 3i
Then,
f(4 + 3i) = \vert 4 + 3i \vert = \sqrt (4\textsuperscript{2} + 3\textsuperscript{2})= \sqrt 25 = 5\\
Similarly, for z = 4 - 3i
f(4 - 3i) = \vert 4 - 3i \vert = \sqrt (4\textsuperscript{2} + 3\textsuperscript{2}) = \sqrt 25 = 5\\
Therefore, it is clear that f(z) is many-one.
So, \vert z \vert \geq 0, \forall z \in C,\\
However, in the question R is the co-domain given.
Hence, f(z) is not onto. So, f(z) is neither one-one nor onto.

Question:11

Let the function f: R \rightarrow R be defined by f (x) = cos x, \forall x \in R. Show that f is neither one-one nor onto.

Answer:
It is given that, f: R \rightarrow R, f(x) = cos x, \forall x \in R\\
So, we can write:
f(x\textsubscript{1}) = f(x\textsubscript{2})\\ or, cos x\textsubscript{1}= cos x\textsubscript{2}\\
Hence, x\textsubscript{1 }= 2n \pi \pm x\textsubscript{2}, where n \in Z\\
It is understandable that for any value of x\textsubscript{1 }and\: \: x\textsubscript{2}, the above equation has an infinite number of solutions.
Therefore, f(x) is a many one function.
We know the range of cos x is [-1, 1] and it is a subset of the given co-domain R.
Hence, the given function is not onto.

Question:12

Let X = \{ 1, 2, 3 \} and Y = \{ 4, 5 \} . Find whether the following subsets of X \times Y are functions from X to Y or not.
(i) f = \{ (1, 4), (1, 5), (2, 4), (3, 5) \}
(ii) g = \{ (1, 4), (2, 4), (3, 4) \} \\
(iii) h = \{ (1,4), (2, 5), (3, 5) \}
(iv) k = \{ (1,4), (2, 5) \} .\\

Answer:

Here, X = \{ 1, 2, 3 \} and Y = \{ 4, 5 \} \\
Therefore, X \times Y = \{ (1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5) \} \\
(i) f = \{ (1, 4), (1, 5), (2, 4), (3, 5) \} \\
Here, f(1) = 4 and again f(1) = 5.
Therefore, f is not a function here.
As a result, there is no unique of pre- image ‘1’.
(ii) g = \{ (1, 4), (2, 4), (3, 4) \} \\
We can clearly see that g is a function. Here in g, each element of the given domain has a unique image at the given range
(iii) h = \{ (1,4), (2, 5), (3, 5) \} \\
It’s clear that h is a function of each pre-image that has a unique image.
Again, h(2) = h(3) = 5
Therefore, the function h is also many-one.
(iv) k = \{ (1, 4), (2, 5) \} \\
Here, ‘3’ does not have any image under the mapping. Therefore, k is not a function.

Question:13

If functions f: A \rightarrow B \: \: and\: \: g: B \rightarrow A \text{ satisfies } g of = IA, then show that f is one-one and g is onto.

Answer:

Here, it is given:
f: A \rightarrow B \: \: and\: \: g: B \rightarrow A \text{ satisfies } g of = IA
It’s clear here that the function ‘g’ is inverse of ‘f’.
So, ‘f’ has to be both one-one as well as onto.
As a result, ‘g’ is both one-one and onto.

Question:14

Let f: R \rightarrow R be the function defined by f(x) = 1/(2 - cos x) \forall x \in R. Then, find the range off.

Answer:

Here, f(x) = 1/(2 - cos x) \forall x \in R
Let’s say,
\\y = 1/(2 - cos x)\\ \therefore2y - ycos x = 1\\ Or, cos x = (2y - 1)/ y\\ Or, cos x = 2 - 1/y\\
As the range of cos x is, -1 \leq cos x \leq 1\\
Therefore,
\\-1 \leq 2 - 1/y \leq 1\\ -3 \leq - 1/y \leq -1\\ 1 \leq - 1/y \leq 3\\ 1/3 \leq y \leq 1\\
Hence, the range of the given function is [1/3, 1].

Question:15

Let n be a fixed positive integer. Define a relation R in Z as follows: \forall a, b \in Z, aRb if and only if a - b is divisible by n. Show that R is an equivalence relation.

Answer:

Here, we have to a relation R in Z as follows: \forall a, b \in Z, aRb if and only if a - b is divisible by n.
Here, aRa \Rightarrow (a - a) is divisible by n and this is true for all integers.
Therefore, R is reflective.
For aRb, aRb \Rightarrow (a - b) is also divisible by n.
or, - (b - a) is divisible by n.
or, (b - a) is divisible by n
Hence, we can write it as bRa.
Therefore, R is symmetric.
For aRb , (a - b) is divisible by n.
For bRc, (b - c) is divisible by n.
Hence, (a - b) + (b-c) is divisible by n.
Or, (a-c) is divisible by n. This can be expressed as aRc.
Therefore, R is transitive.
So, R is an equivalence relation.

Question:16

If A = {1, 2, 3 }, define relations on A which have properties of being:

(a) reflexive, transitive but not symmetric

(b) symmetric but neither reflexive nor transitive

(c) reflexive, symmetric, and transitive.

Answer:

Here, A = \{ 1, 2, 3 \} .\\
(i) Assume R\textsubscript{1}= \{ (1, 1), (1, 2), (1, 3), (2, 3), (2, 2), (1, 3), (3, 3) \} \\
Here, (1, 1), (2, 2) and (3, 3) \in \ R\textsubscript{1}. R\textsubscript{1} is reflexive.
(1, 2) \in R\textsubscript{1}, (2, 3) \in R\textsubscript{1} \Rightarrow (1, 3) \in R\textsubscript{1}
. Hence,R\textsubscript{1} is transitive.
Now, (1, 2) \in R\textsubscript{1 } \Rightarrow (2, 1) \notin R\textsubscript{1}.
Therefore, R\textsubscript{1} is not symmetric.
(ii) Let say, R\textsubscript{2}= \{ (1, 2), (2, 1) \}
So, (1, 2) \in R\textsubscript{2}, (2, 1) \in R\textsubscript{2}\\
Therefore, R\textsubscript{2 } is symmetric,
(1, 1) \notin R\textsubscript{2}. Therefore, R\textsubscript{2 } is not reflexive.
(1, 2) \in R\textsubscript{2}, (2, 1) \in R\textsubscript{2 } but (1, 1) \notin R\textsubscript{2}. Hence, R\textsubscript{2 } is not transitive.
(iii) Let R\textsubscript{3 } = \{ (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3) \} \\
R\textsubscript{3 } is reflexive as (1, 1) (2, 2) and (3, 3) \in R\textsubscript{3 }\\
R\textsubscript{3 } is symmetric as (1, 2), (1, 3), (2, 3) \in R\textsubscript{3 } \Rightarrow (2, 1), (3, 1), (3, 2) \in R\textsubscript{3 }\\
Therefore, R\textsubscript{3 } is reflexive, symmetric and transitive.

Question:17

Let R be relation defined on the set of natural number N as follows:
R = \{ (x, y): x \in N, y \in N, 2x + y = 41 \} . Find the domain and range of the relation R. Also verify whether R is reflexive, symmetric, and transitive.

Answer:

Here, R = \{ (x, y): x \in N, y \in N, 2x + y = 41 \}
So, the domain D= \{ 1, 2, 3, \ldots .., 20 \} \\
And the Range = \{ 1, 3, 5, \ldots .., 39 \} \\
Here, (2, 2) \notin \: \: R\: \: as\: \: 2 \times 2 + 2 \neq 41. Therefore, R is not reflexive.
Again, (1, 39) \in R \: \: but\: \: (39, 1) \: \: \notin R. So, R is not symmetric.
Again, (11, 19) \notin R, (19, 3) \notin R; but (11, 3) \notin R. So, Further R is not transitive.
Therefore, R is neither reflexive nor symmetric and nor transitive.

Question:18

Given A = {2, 3, 4}, B = {2, 5, 6, 7}. Construct an example of each of the following:

(a) an injective mapping from A to B

(b) a mapping from A to B which is not injective

(c) a mapping from B to A.

Answer:

Here, A = \{ 2, 3, 4 \} , B = \{ 2, 5, 6, 7 \} \\
(i) Assume, f: A \rightarrow B \text{ denote a mapping}
Then, f = \{ (x, y): y = x + 3 \} or\\
f = \{ (2, 5), (3, 6), (4, 7) \} , this is an injective mapping.
(ii) Again assume, g: A \rightarrow B denotes a mapping: g = \{ (2, 2), (3, 2), (4, 5) \} . Hence, it is not an injective mapping.
(iii) Again assume, h: B \rightarrow A denotes a mapping: h = \{ (2, 2), (5, 3), (6, 4), (7, 4) \} . Hence, it is a mapping from B to A.

Question:19

Give an example of a map

(i) which is one-one but not onto

(ii) which is not one-one but onto

(iii) which is neither one-one nor onto.
Answer:
(i) Say, f: N \rightarrow N, be a mapping defined by f (x) = x\textsuperscript{2}\\
If, f(x\textsubscript{1}) = f (x\textsubscript{2})\\
Then, x\textsubscript{1}\textsuperscript{2}= x\textsubscript{2}\textsuperscript{2}\\
So,x\textsubscript{1 }= x\textsubscript{2 }(as, x\textsubscript{1 }+ x\textsubscript{1 }\text{cannot be } 0)
f(x\textsubscript{1 }) = f(x\textsubscript{2 }), hence, f(x) is one-one.
However, ‘f’ is not onto, as for 1 \in N, therefore, there is no existence of x in N : f(x) = 2x + 1.
(ii) Let f: R \rightarrow [0, \infty ), is a mapping that is defined by f(x) = \vert x \vert \\
Then, we can conclude that f(x) is not a one-one as f(2) and f(-2) are the same here.
But \vert x \vert \geq 0, so the range is [0, \infty ].\\
Therefore, f(x) is onto.
(iii) Assume, f: R \rightarrow R , be a mapping which is defined by f(x) = x\textsuperscript{2}\\
Then we can say that f(x) is not one-one as f(1) and f (-1) are the same.
The range of f(x) is [0, \infty ).
Therefore, f (x) is neither one-one nor onto.

Question:20

Let A = R - \{ 3 \} , B = R - \{ 1 \} . Let f : A \rightarrow B be defined by f(x) = x - 2/ x - 3 \forall x \in A . Then show that f is bijective.

Answer:

Here,
A = R - \{ 3 \} , B = R - \{ 1 \} \\
f : A \rightarrow B be defined by f (x) = x - 2/ x - 3 \forall x \in A\\
Hence, f (x) = (x - 3 + 1)/ (x - 3) = 1 + 1/ (x - 3)\\
Say f(x1) = f (x2)\\
1+\frac{1}{x_{1}-3}=1+\frac{1}{x_{2}-3}
\\\frac{1}{x_{1}-3}=\frac{1}{x_{2}-3} \\\\ x_{1}=x_{2}
So, f (x) is an injective function.
\\If, y = (x - 2)/ (x -3)\\ Or, x - 2 = xy - 3y\\ Or, x(1 - y) = 2 - 3y\\ Or, x = (3y - 2)/ (y - 1)\\ Or, y \in R - \{ 1 \} = B\\
So, f (x) is onto or subjective.
Therefore, f(x) is a bijective function.

Question:21

Let A = [-1, 1]. Then, discuss whether the following functions defined on A are one-one, onto, or bijective:
(i) f(x) = x/2 (ii) g(x) = \vert x \vert \\
(iii) h(x) = x \vert x \vert (iv) k(x) = x\textsuperscript{2}\\

Answer:

Here, A = [-1, 1]
(i) f: [-1, 1] \rightarrow [-1, 1], f (x) = x/2\\
If, f(x\textsubscript{1}) = f(x\textsubscript{2})\\
x\textsubscript{1}/2 = x\textsubscript{2}/2\\
So, f(x) is one-one.
Also, x \in [-1, 1]\\
x/2 = f(x) = [-1/2, 1/2]\\
Hence, the range is a subset of co-domain A
So, f(x) is not onto.
Therefore, f (x) is not bijective.
(ii)
\\g(x) = \vert x \vert \\ \text{Let} g(x\textsubscript{1}) = g (x\textsubscript{2})\\ \vert x\textsubscript{1} \vert = \vert x\textsubscript{2} \vert \\ x\textsubscript{1}= \pm x\textsubscript{2}\\
So, g(x) is not one-one.
Also, g(x) = \vert x \vert \geq 0, \text{for all real x}\\
Hence, the range is [0, 1], which is subset of co-domain ‘A’
So, f(x) is not onto.
Therefore, f(x) is not bijective.
(iii)
\\\\h(x) = x \vert x \vert \\ \text{Let } h(x\textsubscript{1}) = h(x\textsubscript{2})\\ x\textsubscript{1} \vert x\textsubscript{1} \vert = x\textsubscript{2} \vert x\textsubscript{2} \vert \\ \text{If }x\textsubscript{1}, x\textsubscript{2}> 0\\ x\textsubscript{1}\textsuperscript{2}= x\textsubscript{2}\textsuperscript{2}\\ x\textsubscript{1}\textsuperscript{2}- x\textsubscript{2}\textsuperscript{2}= 0\\ (x\textsubscript{1} - x\textsubscript{2})(x\textsubscript{1} + x\textsubscript{2}) = 0\\ x\textsubscript{1 }= x\textsubscript{2 }(as x\textsubscript{1}+ x\textsubscript{2} \neq 0)\\ \text{Again }, x\textsubscript{1}, x\textsubscript{2 }< 0, and x\textsubscript{1}= x\textsubscript{2}\\ \text{Therefore }, x\textsubscript{1 }and x\textsubscript{2 }\text{of opposite sign}, x\textsubscript{1 } \neq x\textsubscript{2 }.\\
Hence, f(x) is one-one.
\\\text{For x} \in [0, 1], f(x) = x\textsubscript{2} \in [0, 1]\\ \text{For x} < 0, f(x) = -x\textsubscript{2} \in [-1, 0)\\
Hence, the range is [-1, 1].
So, h(x) is onto.
Therefore, h(x) is bijective.
(iv)
\\k(x) = x^{2}\\ \text{Let }k (x\textsubscript{1}) = k (x\textsubscript{2})\\ x\textsubscript{1}\textsuperscript{2}= x\textsubscript{2}\textsuperscript{2}\\ x\textsubscript{1 }= \pm x\textsubscript{2}\\
Therefore, k(x) is not one-one.

Question:22

Each of the following defines a relation on N:
(i) x is greater than y, x, y \in N\\
(ii) x + y = 10, x, y \in N\\
(iii) x y is square of an integer x, y \in N\\
(iv) x + 4y = 10 x, y \in N\\
Determine which of the above relations are reflexive, symmetric, and transitive.

Answer:

(i)Here, x is greater than y; x, y \in N\\
If (x, x) \in R, then x > x, that does not satisfy for any x \in N.\\
Therefore, R is not reflexive.
Say,(x, y) \in R
\\Or, xRy\\ Or, x > y
\Rightarrow y > x \\ For any x, y \in N, the above condition is not true.
Hence, R is not symmetric.
Again, xRy and yRz
\\Or, x > y and y > z\\ Or, x > z\\ Or, xRz\\
Hence, R is transitive.
(ii) x + y = 10;x, y \in N
Thus,
\\R = \{ (x, y); x + y = 10, x, y \in N \} \\ R = \{ (1, 9), (2, 8), (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1) \} \\
Therefore, (1, 1) \notin R\\
So, R is not reflexive.
Again, (x, y) \in R \Rightarrow (y, x) \in R\\
Therefore, R is symmetric.
And, (1, 9) \in R, (9, 1) \in R, but (1, 1) \notin R\\
Therefore, R is not transitive.
(iii)Here, xy is square of an integer x, y \in N
R = \{ (x, y) : \text{xy is a square of an integer} x, y \in N \} \\
So, (x, x) \in R, \forall x \in N\\
For any x \in N, x\textsuperscript{2 } is an integer.
Thus, R is reflexive.
If (x, y) \in R \Rightarrow (y, x) \in R\\
So, R is symmetric.
Again, if xy and yz both are square of an integer.
Then, xy = m\textsuperscript{2} \: \: and \: \: yz = n\textsuperscript{2} \text{for some m, n} \in Z\\
\\x = m\textsuperscript{2} /y and z = n\textsuperscript{2} /y\\ xz = m\textsuperscript{2} n\textsuperscript{2} / y\textsuperscript{2}, this must be the square of an integer.
Therefore, R is transitive.
(iv) x + 4y = 10; x, y \in N\\
\\R = \{ (x, y): x + 4y = 10; x, y \in N \} \\ R = \{ (2, 2), (6, 1) \} \\ Here, (1, 1) \notin R\\
Hence, R is not symmetric.
\\(x, y) \in R \Rightarrow x + 4y = 10\\ And, (y, z) \in R \Rightarrow y + 4z = 10\\ Or,\ x - 16z = -30\\ Or,\ (x, z) \notin R\\
Therefore, R is not transitive.

Question:23

Let A = {1, 2, 3, … 9} and R be the relation in A \times A defined by (a, b) R (c, d) if a + d = b + c for (a, b), (c, d) in A \times A. Prove that R is an equivalence relation and also obtain the equivalent class [(2, 5)].

Answer:

\\Here, A = \{ 1, 2, 3, \ldots 9 \} \: \: and \: \: (a, b) R (c, d)\: \: if a + d = b + c for (a, b), (c, d) \in A \times A.\\ Say (a, b) R (a, b)\\ Therefore, a + b = b + a, \forall a, b \in A .
This must be true for any a, b \in A.\\
Hence, R is reflexive.
Say, (a, b) R (c, d)
Then,
\\a + d = b + c\\ c + b = d + a\\ (c, d) R (a, b)\\
Therefore, R is symmetric.
Let \\(a, b) R(c, d) ,and, (c, d) R(e, f)
\\ a + d = b + c ,and, c + f = d + e\\ a + d = b + c ,and, d + e = c + f\\ Or, (a + d) - (d + e = (b + c) - (c + f)\\ Or, a - e = b - f\\ Or, a + f = b + e\\ Or, (a, b) R(e, f)\\
So, R is transitive.
Therefore, R is an equivalence relation.

Question:24

Using the definition, prove that the function f: A \rightarrow B is invertible if and only if f is both one-one and onto.

Answer:

Say, f: A \rightarrow B be many-one functions.
If, f(a) = p \text{ and } f(b) = p\\
Then f\textsuperscript{-1}(p) = a \: \: and \: \: f\textsuperscript{-1}(p) = b\\
In this case, we have two images ‘a and b’ for one pre-image ‘p’. This is because the inverse function is not defined here.
However, to be one-one, f must be invertible.
Say, f: A \rightarrow B is not onto function.
B = \{ p, q, r \} and \{ p, q \} is the range of f.
There is no pre-image for the image r, which will have no image in set A.
And, f must be onto to be invertible.
Thus, to be both one-one and onto f must be invertible
If f is a bijective function, then f = X \rightarrow Y is invertible.

Question:25

Functions f, g : R \rightarrow R are defined, respectively, by f(x) = x\textsuperscript{2} + 3x + 1, g(x) = 2x - 3, find
(i) fog (ii) gof (iii) fof (iv) gog

Answer:

Here,
f(x) = x\textsuperscript{2} + 3x + 1, g (x) = 2x - 3\\

\\(i)fog = f(g(x))\\ = f(2x - 3)\\ = (2x - 3)\textsuperscript{2} + 3(2x - 3) + 1\\ = 4x\textsuperscript{2}\ + 9 - 12x + 6x - 9 + 1\\ = 4x\textsuperscript{2}\ - 6x + 1\\ (ii) gof = g(f(x))\\ = g(x\textsuperscript{2}\ + 3x + 1)\\ = 2(x\textsuperscript{2}\ + 3x + 1) - 3\\ = 2x\textsuperscript{2}\ + 6x - 1\\ (iii) fof = f(f(x))\\ = f(x\textsuperscript{2}\ + 3x + 1)\\ = (x\textsuperscript{2}\ + 3x + 1)2 + 3(x\textsuperscript{2} + 3x + 1) + 1\\ = x4 + 9x\textsuperscript{2}\ + 1 + 6x\textsuperscript{3} + 6x + 2x\textsuperscript{2}\ + 3x\textsuperscript{2}\ + 9x + 3 + 1\\ = x4 + 6x\textsuperscript{3} + 14x\textsuperscript{2}\ + 15x + 5\\ (iv) gog = g(g(x))\\ = g(2x - 3)\\ = 2(2x - 3) - 3\\ = 4x - 6 - 3\\ = 4x - 9\\

Question:26

Let \ast be the binary operation defined on Q. Find which of the following binary operations are commutative
\\(i) a \ast b = a - b\: \: \forall a, b \in Q\\ (ii) a \ast b = a^2 + b^2 \: \: \forall a, b \in Q\\ (iii) a \ast b = a + ab \: \: \forall a, b \in Q\\ (iv) a \ast b = (a - b)\textsuperscript{2}\: \: \forall a, b \in Q\\

Answer:

Here, \ast is a binary operation defined on Q.
(i)
\\a \ast b = a - b, \forall a, b \in Q \: \: and \: \: b \ast a = b - a\\ So, a \ast b \neq b \ast a\\
Hence, \ast is not commutative.
(ii)
\\a \ast b = a\textsuperscript{2}+ b\textsuperscript{2}\\ b \ast a = b\textsuperscript{2}+ a\textsuperscript{2}\\
Therefore, \ast is commutative.
(iii)
\\a \ast b = a + ab\\ b \ast a = b + ab\\ Hence, a + ab \neq b + ab\\
Therefore, \ast is not commutative.
(iv)
\\a \ast b = (a - b)\textsuperscript{2}, \forall a, b \in Q\\ b \ast a = (b -a)\textsuperscript{2}\\ As, (a - b)\textsuperscript{2} = (b - a)\textsuperscript{2}\\
Therefore, \ast is commutative.

Question:27

Let \ast be binary operation defined on R by a \ast b = 1 + ab, \forall a, b \in R. Then the operation \ast is
(i) commutative but not associative
(ii) associative but not commutative
(iii) neither commutative nor associative
(iv) both commutative and associative

Answer:

(i) Here,
\ast is a binary operation defined on R by a \ast b = 1 + ab, \forall a, b \in R
So, a \ast b = ab + 1 = b \ast a\\
Therefore, \ast is a commutative binary operation.
\\Now, a \ast (b \ast c) = a \ast (1 + bc) = 1 + a (1 + bc) = 1 + a + abc\\ Again, (a \ast b) \ast c = (1 + ab) \ast c = 1 + (1 + ab) c = 1 + c + abc\\ Therefore, a \ast (b \ast c) \neq (a \ast b) \ast c\\
Hence, \ast is not associative.
Therefore, \ast is commutative but not associative.

Question:28

Let T be the set of all triangles in the Euclidean plane, and let a relation R on T be defined as aRb if a is congruent to b \forall a, b \in T. Then R is
(A) reflexive but not transitive (B) transitive but not symmetric
(C) equivalence (D) none of these

Answer:

(C) equivalence
Here aRb, if a is congruent to b, \forall a, b \in T.\\
So, in aRa, a is congruent to a. This must always be true.
Therefore, R is reflexive.
Say, aRb \Rightarrow a \sim b\\
b \sim a \Rightarrow bRa\\
Therefore, R is symmetric.
Say, aRb and bRc
\\a \sim b \: \: and\: \: b \sim c\\ a \sim c \Rightarrow aRc\\
Therefore, R is transitive.
Therefore, R is an equivalence relation.

Question:29

Consider the non-empty set consisting of children in a family and a relation R defined as aRb if a is brother of b. Then R is

(A) symmetric but not transitive (B) transitive but not symmetric

(C) neither symmetric nor transitive (D) both symmetric and transitive

Answer:

(B) transitive but not symmetric

If, aRb means a is brother of b.

Then, it does not mean b is also a brother of a. Because, b can be a sister of a too.

Therefore, R is not symmetric.

If, aRb implies that a is the brother of b.

and bRc implies that b is the brother of c.

Therefore, a must be the brother of c.

Hence, R is transitive.

Question:30

The maximum number of equivalence relations on the set A = {1, 2, 3} are

(A) 1 (B) 2

(C) 3 (D) 5

Answer:

(D) 5
Given, set A = \{ 1, 2, 3 \} \\
Now, the number of equivalence relations as follows
\\R1 = \{ (1, 1), (2, 2), (3, 3) \} \\ R2 = \{ (1, 1), (2, 2), (3, 3), (1, 2), (2, 1) \} \\ R3 = \{ (1, 1), (2, 2), (3, 3), (1, 3), (3, 1) \} \\ R4 = \{ (1, 1), (2, 2), (3, 3), (2, 3), (3, 2) \} \\ R5 = \{ (1, 2, 3) \Leftrightarrow A \times A = A^2 \} \\
Thus, the maximum number of equivalence relations is ‘5’.

Question:31

If a relation R on the set {1, 2, 3} be defined by R = {(1, 2)}, then R is

(A) reflexive (B) transitive

(C) symmetric (D) none of these

Answer:

(D) none of these

If, R be defined on the set {1, 2, 3} by R = {(1, 2)}

Then, we can say that R is not reflexive, transitive, and symmetric.

Question:32

Let us define a relation R in R as aRb if a \geq b. Then R is

(a) an equivalence relation (b) reflexive, transitive but not symmetric (c) symmetric, transitive but not reflexive (d) neither transitive nor reflexive but symmetric.

Answer:

(b).
The defined relation R in R as aRb if a \geq b.\\
Similarly, aRa implies a \geq a which is true.
Therefore, it is reflexive.
Let aRb \rightarrow a \geq b, but, b \ngtr a.\\
Therefore, we can’t write it as Rba
Hence, R is not symmetric.
Now, a \geq b, b \geq c. So, a \geq c, and this is true.
Therefore, R is transitive.

Question:33

Let A = {1, 2, 3} and consider the relation R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}, then R is

(a) reflexive but not symmetric (b) reflexive but not transitive (c) symmetric and transitive (d) neither symmetric nor transitive.

Answer:

(a)

Given, R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}

Therefore, it can be written as: 1R1, 2R2 and 3R3.

Therefore, R is reflexive.

Here, 1R2 is not the same as 2R1 and 2R3 is not the same as 3R2.

Therefore, R is not symmetric.

Again, 1R1 and 1R2 \rightarrow 1R3\\

Therefore, R is transitive.

Question:34

The identity element for the binary operation * defined on Q ~ {0} as a * b = 2 ab a, b Q ~ {0} is

(a) 1 (b) 0 (c) 2 (d) None of these

Answer:

(c)
Here: a \ast b = ab/2 a, b Q \sim \{ 0 \} \\
Assume ‘e’ as the identity element.
Therefore, a \ast e = ae/2 = a \rightarrow e = 2 \\

Question:35

If the set A contains 5 elements and set B contains 6 elements, then the number of one-one and onto mapping from A to B is

(a) 720 (b) 120 (c) 0 (d) None of these

Answer:

Let, the number of elements in A and B set are m and n respectively. Therefore, one-one and onto mapping

from A to B is n! when m = n

and, 0 if m ≠ n

It is given that, m = 5 and n = 6.

As,5 ≠ 6 So, from A to B mapping = 0

Question:36

Let A = \{ 1, 2, 3, ..., n \} and B = \{ a, b \} . Then the number of surjections from A to B is
(a) \textsuperscript{n}P\textsubscript{2}
(b) 2\textsuperscript{n} - 2
(c)2\textsuperscript{n} - 1
(d) None of these

Answer:

(d)
It is given that, A = \{ 1, 2, 3, ..., n \} and B = \{ a, b \} \\
Say, the number of elements in set A and B are m and n respectively.
Therefore, \textsuperscript{n}C\textsubscript{m} \times m! is the number of surjections from A to B, for n \ngeq m \\
Given, m=2.
Therefore, the number of surjections from A to B
= nC_2 \times 2! = n!/[2! \times (n-2)!] = n(n - 1) = n\textsuperscript{2} - n \\

Question:37

Let f : R \rightarrow R be defined by f(x) = 1/x \forall x \in R then f is
(a) one-one (b) onto (c) bijective (d) f is not defined

Answer:

(d)
Here, f(x) = 1/x
Say, x = 0. Then f(x) = 1/0 = undefined
Hence, f(x) is not defined.

Question:38

Let f : R \rightarrow R be defined by f(x) = 3x\textsuperscript{2} - 5 and g : R \rightarrow R by g(x) = x/(x\textsuperscript{2}+1), then gof is
\\(a) (3x\textsuperscript{2}-5)/(9x\textsuperscript{4}-30x\textsuperscript{2}+ 26)\\ (b) (3x\textsuperscript{2}-5)/(9x\textsuperscript{4}-6x\textsuperscript{2}+ 26) \\(c) 3x\textsuperscript{2}/(x\textsuperscript{4}+2x\textsuperscript{2}-4) \\(d) (3x\textsuperscript{2})/(9x\textsuperscript{4}-30\textsuperscript{2}+2)\\

Answer:

(a)
Given, f(x) = 3x\textsuperscript{2} - 5 \ and \ g(x) = x/(x\textsuperscript{2}+1)\\
Therefore,
\\gof = gof (x) = g(3x\textsuperscript{2}-5) = (3x\textsuperscript{2}-5)/[(3x\textsuperscript{2}-5)\textsuperscript{2}+1]\\ = (3x\textsuperscript{2}-5)/(9x\textsuperscript{4}-30x\textsuperscript{2}+ 26)\\

Question:39

Which of the following functions from Z to Z are bijections?
\\(a) f(x) = x\textsuperscript{3 }\\(b) f(x) = x + 2 \\(c) f(x) = 2x + 1 \\(d) f(x) = x\textsuperscript{2} + 1 \\

Answer:

(b)
Given that f : Z \rightarrow Z \\
Say, x\textsubscript{1}, x\textsubscript{2} \in f(x) \rightarrow f(x\textsubscript{1}) = x\textsubscript{1}+ 2; f(x\textsubscript{2}) = x\textsubscript{2}+ 2;\\
For, f(x\textsubscript{1}) = f(x\textsubscript{2})\ \rightarrow x\textsubscript{1}+ 2 = x\textsubscript{2}+\ 2 \\
or, x\textsubscript{1 }= x\textsubscript{2}\\
Therefore, the function f(x) is one-one.
Say, y = x + 2
or, x = y - 2 \forall y \in Z \\
Hence, f(x) is onto.
Therefore, the function f(x) is bijective.

Question:40

Let f : R \rightarrow R be the functions defined by f(x) = x\textsuperscript{3} + 5. Then f\textsuperscript{-1}(x) is
\\(a) (x + 5)\textsuperscript{1/3}\\ (b) (x - 5)\textsuperscript{1/3}\\ (c) (5 - x)\textsuperscript{1/3}\\ (d) 5 - x \\

Answer:

(b)Here,
\\f(x) = x\textsuperscript{3} + 5\\ Say, y = x\textsuperscript{3} + 5\\ or, x\textsuperscript{3} = y - 5 \\ or, x = (y - 5)\textsuperscript{1/3}\ f\textsuperscript{-1}1(x) = (x - 5)\textsuperscript{1/3} \\

Question:41

Let f : A \rightarrow B and g : B \rightarrow C be the bijective functions. Then(gof)\textsuperscript{-1} is
\\(a) f\textsuperscript{-1 }og\textsuperscript{-1} \\(b) fog \\(c) g\textsuperscript{-1}of\textsuperscript{-1}\\ (d) gof \\
Answer:

(a)
Given that, f : A \rightarrow B \text{ and } g : B \rightarrow C \\
Then, (gof)\textsuperscript{-1} = f\textsuperscript{-1}og\textsuperscript{-1}\\

Question:42

Let f: \{ R-3/5 \} \rightarrow R be defined by f(x) = (3x + 2)/(5x - 3) then
\\(a) f(x) = f(x); \\(b) f\textsuperscript{-1}(x) = - f(x) \\(c) (fof)x = - x \\(d) f\textsuperscript{-1}(x) = f(x)/19\\

Answer:

(a)
Here,f(x) = (3x + 2)/(5x - 3) \forall x \neq 3/5\\
\\So, y = (3x + 2)/(5x - 3)\\ or, y (5x-3) = (3x + 2)\\ or, 5xy-3y = 3x + 2\\ or, x = (3y+2)/(5y-3)\\ Therefore, f\textsuperscript{-1}(x) = (3x+2)/(5x-3)\\ Hence, f\textsuperscript{-1}(x) = f(x)\\

Question:43

Let f : [0, 1] \rightarrow [0, 1] be defined by

f(x) = { x, if is rational

{ 1-x , if is irrational
Then (fof)x is

(a) constant

(b) 1 + x

(c) x

(d) None of these

Answer:

(c)
Here, f : [0, 1] \rightarrow [0, 1]\\
\\f = f\textsuperscript{-1} \\ Therefore, (fof)x = x\\

Question:44

Let f : [2, \infty ) and R be the function defined by f(x) = x\textsuperscript{2} - 4x + 5, then the range of f is
\\(a) R\\ (b) [1, \infty ) \\(c) [4, \infty ) \\(d) [5, \infty ) \\
Answer:

(b)
\\Here, f(x) = x\textsuperscript{2} - 4x + 5 \\ Assume, y = x\textsuperscript{2}\ - 4x + 5 \\ or, x\textsuperscript{2}\ - 4x + 5 - y = 0 \\ or, x = [-4 \pm \sqrt \{ (-4)\textsuperscript{2}- 4(-4)(5-y) \} ]/2x = 2 \pm \sqrt (y-1)\\ \text{When x is real}, y-1 \geq 0\\ Therefore, y \geq 1\\ \text{Hence the range is}[1, \infty )\\

Question:45

Let f : N \rightarrow R be the function defined by f(x) = (2x-1)/2 and g : Q \rightarrow R be another function defined by g(x) = x + 2 then, gof(3/2) is
\\(a) 1 \\(b) - 1 \\(c) 7/2 \\(d) None of these \\

Answer:

(d)
Given that, f(x) = (2x-1)/2 \: \: and\: \: g(x) = x + 2 \\
Therefore, gof(x) = g[(f(x)] = f(x) + 2 = (2x-1)/2 = (2x+3)/2\\
Hence, gof(3/2) = 3 \\

Question:46

Let f : R \rightarrow R be defined by

\\f(x) = \{2x, when x> 3 \\ \ \ \ \: \: \: \: \: \: \: \: \ \ \ \ \ \{ x\textsuperscript{2}, when 1< x \leq 3\\ \ \ \ \ \ \ \ \ \ \ \ \{ 3x, when x \leq 1\\ then f(- 1) + f(2) + f(4) is\\

\\ (a) 9 \\(b) 14 \\(c) 5 \\(d) None of these \\

Answer:

(a)
Here, f(- 1) + f(2) + f(4) = 3(- 1) + (2)\textsuperscript{2} + 2(4) = - 3 + 4 + 8 = 9 \\

Question:47

If f : R \rightarrow R be given by f(x) = tan x, then f-1 is
\\(a) \frac{\pi}{4} \\ (b) (n \pi + \pi /4) where\ n \in Z \\(c) \ does \ not \ exist\\ (d) None\ of \ these \\
Answer:

(a)
\\Here, f(x) = tan x \\ Assume, f(x) = y = tan x\\ Then,\ \ x = tan\textsuperscript{-1} (y)\\ or, f\textsuperscript{-1}(x) = tan\textsuperscript{-1}\ (x) \\ or, f\textsuperscript{-1}(1) = tan\textsuperscript{-1} (1)\\ or,\ f\textsuperscript{-1}(1) = tan\textsuperscript{-1} tan ( \pi /4)=\frac{\pi}{4}\\

Question:48

Let the relation R be defined in N by aRb if 2a + 3b = 30. Then R = ............

Answer:

Here, aRb : 2a + 3b = 30

Or, 3b = 30 – 2a

Or, b = (30 - 2a)/ 3

for a = 3, b = 8

a = 6, b = 6

a = 9, b = 4

a = 12, b = 2

Therefore, R = {(3, 8), (6, 6), (9, 4), (12, 2)}

Question:49

Let the relation R be defined on the set A = \{ 1, 2, 3, 4, 5 \} by R = \{ (a, b) : \vert a\textsuperscript{2} - b\textsuperscript{2} \vert < 8 \}. Then R is given by ............

Answer:

Here
\\A = \{ 1, 2, 3, 4, 5 \} and R = \{ (a, b) : \vert a\textsuperscript{2} - b\textsuperscript{2} \vert < 8 \} \\ \text{Therefore, we can say}, R = \{ (1, 1), (1, 2), (2, 1), (2, 2), (2, 3), (3, 2), (4, 3) (3, 4), (4, 4), (5, 5) \} \\

Question:50

Let f = {(1, 2), (3, 5), (4, 1)} and g = {(2, 3), (5, 1), (1, 3)}. Then gof = ................ and fog = ...................

Answer:

\\Here, f = \{ (1, 2), (3, 5), (4, 1) \} and g = \{ (2, 3), (5, 1), (1, 3) \} \\ Then, gof(1) = g[f(1)] = g(2) = 3\\ Again, gof(3) = g[f(3)] = g(5) = 1 \\ Again, gof(4) = g[f(4)] = g(1) = 3\\ Therefore, gof = \{ (1, 3), (3, 1), (4, 3) \} \\ Again, fog(2) = f[g(2)] = f(3) = 5\\ Again, fog(5) = f[g(5)] = f(1) = 2\\ Again, fog(1) = f[g(1)] = f(3) = 5\\ Hence, fog = \{ (2, 5), (5, 2), (1, 5) \} \\

Question:51

Let f : R \rightarrow R be defined by f(x) = x/ \sqrt (1+x\textsuperscript{2}) then (fofof)(x) = ................. \\

Answer:

\begin{aligned} (\text { fofof })(x) &=f[f(x))] \\ &=f\left[\begin{array}{l} f\left(\frac{x}{\sqrt{1+x^{2}}}\right) \end{array}\right] \\ &=f\left(\frac{\frac{x}{\sqrt{1+x^{2}}}}{\sqrt{1+\frac{x^{2}}{1+x^{2}}}}\right) \\ &=f\left(\frac{x}{\sqrt{1+2 x^{2}}}\right) \end{aligned}
\\\text { fofof }(x) =\frac{\frac{x}{\sqrt{1+2 x^{2}}}}{\sqrt{1+\frac{x^{2}}{1+2 x^{2}}}} \\ =\frac{x}{\sqrt{1+3 x^{2}}}

Question:52

If, f(x) = [4 - (x - 7)\textsuperscript{3}], then f\textsuperscript{-1}(x) = .............

Answer:

\\Here, f(x) = [4 - (x - 7)\textsuperscript{3}]\\ Say, y = [4 - (x - 7)\textsuperscript{3}]\\ Or, (x - 7)\textsuperscript{3} = (4 - y)\\ Or, (x-7) = (4-y)\textsuperscript{1/3}\\ Or, x = 7 + (4-y)\textsuperscript{1/3}\\ Therefore, f\textsuperscript{-1}(x) = 7 + (4 - x)\textsuperscript{1/3}\\

Question:53

Let R = {(3,1), (1,3), (3,3)} be a relation defined on the set A = {1, 2, 3}. Then R is symmetric, transitive and not reflexive.

Answer:

False Here in the question, R = {(3,1), (1,3), (3,3)} which is defined on the set A = {1, 2, 3}

As (1 ,1) R, R is not a reflexive one.

As (3, 1) R and (1, 3) belongs to R, then R is symmetric.

Again, (1, 3) R, (3, 1) R. However (1, 1) does not belong to R. Then R is not transitive.

Question:54

Let f: R→R be the function defined by f(x) = sin(3x+2), x R. Then f is invertible.

Answer:

False
Here, f(x) = sin(3x+2), x R. This function is not a one-one onto function for all x \in R.\\
Again, f(x) = sin(3x+2) = 0\\
or, 3x + 2 = n\pi, n \in \ Z.
Hence, the function is not invertible.

Question:55

Every relation which is symmetric and transitive is also reflexive.

Answer:

False

Assume, a relation R : R= {(1,2), (2,1), (2,2)} on the set A = {1,2}

Therefore, it is clear that (1,1) ∉ R. Hence, it is not reflexive.

Question:56

An integer m is said to be related to another integer n, if m is an integral multiple of n. This relation Z is reflexive, symmetric, and transitive.

Answer:

False.

Here, the given relation in the question is reflexive and transitive. However, it is not symmetric

Question:57

Let A = \{ 0,1 \} and N be the set of natural numbers. Then the mapping f: N \rightarrow A defined by f(2n - 1) = 0, f(2n) = 1, n \in N, \\is onto.

Answer:

True
Here, A = \{ 0,1 \} , and\: \: f(2n - 1) = 0, f(2n) = 1, \forall A n \in N\\
Therefore, the range of f is \{ 0, 1 \} \\
Hence, f: N \rightarrow A mapping is onto.

Question:58

The relation R on the set A = {1, 2, 3} defined as R = {(1,1), (1,2), (2,1), (3,3)}

Answer:

False

As, R = \{ (1,1), (1,2), (2,1), (3,3) \} \\

Therefore, (2,2) \notin R\\

Hence, R is not reflexive.

Question:59

The composition function is cumulative.
Answer:

\\ False \\ Let, f(x) = x+1, g(x) = x \textsuperscript{2} \\ Therefore, fog(x) = f(g(x))\\ = f(x\textsuperscript{2})\\ \ \ \ \ \ = x\textsuperscript{2 }+ 1\\ Again, gof(x) = g(f(x))\\ =g(x+1)\\ =(x+1)\textsuperscript{2}\\ = (x\textsuperscript{2} + 2x + 1)\\ Hence, fog(x) \neq gof(x) \\

Question:60

The composition of function is associative.

Answer:

True.

fo(goh)(x) = (fog)oh

Question:61

Every function is invertible.

Answer:

False.

Only bijective functions are invertible.

Question:62

A binary operation on a set always has the identity element.

Answer:

False

‘+’ is a binary operation on the set N but it has no identity element.

Subtopics Covered in NCERT Exemplar Solutions for Class 12 Maths Chapter 1

This chapter is divided into various sub-topics that are related to relations and functions and its operations. They are mentioned below:

  • Introduction
  • Types of relations
  • Types of functions
  • Composition of functions & invertible functions
  • Binary operations

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What student will learn in NCERT Exemplar Class 12 Maths Solutions Chapter 1?

  • NCERT Exemplar solutions for Class 12 Maths chapter 1, students will learn about the basic concept of functions and relations. They will also get a better view of domains, co-domains, range etc. They will solve questions that will help in understanding these relations in a much more detailed way with help on lemmas and theorems.
  • NCERT Exemplar Class 12 Maths solutions chapter 1 will throw light on how to operate these functions by adding, multiplying, subtracting etc. Students will also get a more detailed view of various function types, injectivity and surjectivity. One can understand what all elements and factors comprise a function. Also, it throws light on a topic called invertible functions.
  • Topics about binary operations of functions using two operands will be covered in Class 12 Maths NCERT Exemplar solutions chapter 1 as well. It will explain in detail how commutativity, symmetry, associativity is used, in binary operations.
  • We have a team of expert teachers and guides who have solved these questions and have provided the same for the students to understand.
  • Solving NCERT questions is important, to understand the topic from the very beginning. We have solved the questions in a very simple way so that it is easier for the student to understand and grasp the topic and idea with much ease.
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NCERT Exemplar Class 12 Maths Solutions

Important Topics To Cover From NCERT Exemplar Class 12 Maths Solutions Chapter 1

· This is the very start of high school algebra, which holds the very root of higher education mathematics. In NCERT Exemplar Class 12 Maths chapter 1 solutions, students will learn about relations and functions, which is a type of relation. The relation is nothing but the relationship between two groups or the relationship between input and output.

· The function is a relation, but is used for obtaining the output from a set input. Learning the relations and functions at the very start of 12 Class will help in setting one's roots in calculus mathematics for cracking entrance exams and higher education.

· Several things are covered in Class 12 Maths NCERT Exemplar solutions chapter 1 which goes into the types of functions and relations. One will learn about various relations that will be used, like universal relations, reflective relations, empty relation, symmetric relation, etc.

NCERT Exemplar Class 12 Solutions

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Frequently Asked Question (FAQs)

1. Are these solutions helpful for board examinations?

Yes, NCERT exemplar solutions for Class 12 Maths chapter 1  are designed well to prepare you for board examinations.

2. What are the important topics of this chapter?

The important topics of this NCERT Exemplar Class 12 Maths solutions chapter 1 include Types of Relations, Types of Functions, Binary Operations and Composition of Functions and Invertible Function.

3. How to remember the formulae well?

The best ways to remember formulae is to keep revising them and solving related problems. Other methods include reading them on a daily basis. You can also prepare charts to stick near your study table or bed to go through in your leisure time.

4. How many questions are there in this chapter?

There are a total 70 questions in NCERT Exemplar Class 12 Maths solutions chapter 1 based on different concepts.

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?
i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Read Complete Answer
Ashvadha 12 July,2024
Kya mujhe class 12 ka important question milega jo exam mai per year puche jate hai

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

Read Complete Answer
kumardurgesh1802 10 July,2024
I have completed my 12th with arts cbse. I want to repeat my 12th with pcmb. How to become a private candidate and what are deadline to be able to appear in 2025 exam?

Hello student,

If you are planning to appear again for class 12th board exam with PCMB as a private candidate here is the right information you need:

  • No school admission needed! Register directly with CBSE. (But if you want to attend the school then you can take admission in any private school of your choice but it will be waste of money)
  • You have to appear for the 2025 12th board exams.
  • Registration for class 12th board exam starts around September 2024 (check CBSE website for exact dates).
  • Aim to register before late October to avoid extra fees.
  • Schools might not offer classes for private students, so focus on self-study or coaching.

Remember , these are tentative dates based on last year. Keep an eye on the CBSE website ( https://www.cbse.gov.in/ ) for the accurate and official announcement.

I hope this answer helps you. If you have more queries then feel free to share your questions with us, we will be happy to help you.

Good luck with your studies!

Read Complete Answer
kumardurgesh1802 10 July,2024
I passed my 12th cbse boards in 2023 , and first time appear for jee mains and advance in 2024 so , am i eligible for 2025 jee mains and advance ?

Hello Aspirant , Hope your doing great . As per your query , your eligible for JEE mains in the year of 2025 , Every candidate can appear for the JEE Main exam 6 times over three consecutive years . The JEE Main exam is held two times every year, in January and April.

Read Complete Answer
satyamercedes0626 23 July,2024
I've passed my class 12th board exam from cbse in PCB in 2024. Now I'll give mathematics exam from cbse as a additional subject in 2025. However, they'll provide separate marksheet for maths. Will i be eligible for jee?? And how I'll fill the form??

Hi there,

Hope you are doing fine

Yes you are certainly eligible for giving the jee exam in the year 2025. You must pass the maths exam with at least 75% criteria as required by jee and provide the marksheet and the passing certificate while registering for the exam.


Pursuing maths as an additional subject while taking biology as your main subject does not offer any hindrance in you appearing for the jee examination. It is indeed an privilege to pursue both maths and biology as the subjects and prepare for the same.

There will be no issue in filling the form while registering for the exam as it will only require your basic details and marksheet which you can provide by attaching the marksheet of maths also. Also, a detailed roadmap is also available on the official websites on how to fill the registration form. So you can fill the form easily.


Hope this resolves your query.

Read Complete Answer
Abhishek 12 July,2024
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Popular CBSE Class 12th Questions

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

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1.00\; J

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2.45×10−3 kg

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 6.45×10−3 kg
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In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

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0.02

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remain unchanged

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A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

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less than 3

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more than 3 but less than 6
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more than 6 but less than 9

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more than 9
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