NCERT Solution (Miscellaneous) - Relations and Functions

# NCERT Solutions Class 11 Maths Chapter 2 - Relations and Functions

Ques 1: The relation f is defined by  The relation g is defined by  Show that f is a function and g is not a function.
Ans:  The relation f is defined as

It is observed that for

0 ≤ x < 3, f(x) = x2

3 < x ≤ 10, f(x) = 3x

Also, at x = 3, f(x) = 32 = 9 or f(x) = 3 × 3 = 9

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

Therefore, for 0 ≤ x ≤ 10, the images of f(x) are unique.

Thus, the given relation is a function.

The relation g is defined as

It can be observed that for x = 2, g(x) = 22 = 4 and g(x) = 3 × 2 = 6

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

Ques 2: If f(x) = x2, find .
Ans:

Ques 3: Find the domain of the function
Ans:  The given function is .

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 R – {2, 6}.

Ques 4: Find the domain and the range of the real function f defined by .
Ans:   The given real function is .
It can be seen that  is defined for (x – 1) ≥ 0.
i.e.,  is defined for x ≥ 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,).
As x ≥ 1 ⇒ (x – 1) ≥ 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,).

Ques 5: Find the domain and the range of the real function f defined by f (x) = |x – 1|.
Ans:  The given real function is f (x) = |x – 1|.
It is clear that |x – 1| is defined for all real numbers.
∴Domain of f = R
Also, for xR, |x – 1| assumes all real numbers.
Hence, the range of f is the set of all non-negative real numbers.

Ques 6: Let  be a function from R into R. Determine the range of f.
Ans:

The range of f is 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)

Ques 7: Let f, g: R → R be defined, respectively by f(x) = x  +1, g(x) = 2x – 3. Find f +  g, fg and f/g.
Ans:   f, g: RR is defined as f(x) = x + 1, g(x) = 2x – 3

(f  + g) (x) = f(x) + g(x) = (x  +1) (2x – 3) = 3x – 2

∴(f+ g) (x) = 3x – 2

(f – g) (x) = f(x) – g(x) = (x + 1) – (2x – 3) = x + 1 – 2x + 3 = – x + 4

∴ (f – g) (x) = –x + 4

Ques 8: Let f = {(1, 1), (2, 3), (0, –1), (–1, –3)} be a function from Z to Z defined by f(x) = ax +  b, for some integers a, b. Determine a, b.
Ans:  f = {(1, 1), (2, 3), (0, –1), (–1, –3)}

f(x) = ax +  b

(1, 1) ∈ f

f(1) = 1

a × 1 + b = 1

a  + b = 1

(0, –1) ∈ f

f(0) = –1

a × 0 + b = –1

b = –1

On substituting b = –1 in a  + b = 1, we obtain a  (–1) = 1 ⇒ a = 1 + 1 = 2.

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

Ques 9: Let R be a relation from N to N defined by R = {(a, b): a, b ∈ N and a = b2}. Are the following true?
(i) (a, a) ∈ R, for all a ∈ N
(ii) (a, b) ∈ R, implies (b, a) ∈ R
(iii) (a, b) ∈ R, (b, c) ∈ R implies (a, c) ∈ R.
Ans:  R = {(a, b): a, bN and a = b2}

(i) It can be seen that 2 ∈ N;however, 2 ≠ 22 = 4.

Therefore, the statement “(a, a) ∈ R, for all aN” is not true.

(ii) It can be seen that (9, 3) ∈ N because 9, 3 ∈ N and 9 = 32.

Now, 3 ≠ 92 = 81; therefore, (3, 9) ∉ N

Therefore, the statement “(a, b) ∈ R, implies (b, a) ∈ R” is not true.

(iii) It can be seen that (9, 3) ∈ R, (16, 4) ∈ R because 9, 3, 16, 4 ∈ N and 9 = 32 and 16 = 42.

Now, 9 ≠ 42 = 16; therefore, (9, 4) ∉ N

Therefore, the statement “(a, b) ∈ R, (b, c) ∈ R implies (a, c) ∈ R” is not true.

Ques 10: Let A = {1, 2, 3, 4}, B = {1, 5, 9, 11, 15, 16} and f = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}. Are the following true?
(i) f is a relation from A to B
(ii) f is a function from A to B.
Ans:  A = {1, 2, 3, 4} and B = {1, 5, 9, 11, 15, 16}

∴A × 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 × B.

It is observed that f is a subset of A × 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.

Ques 11: Let f be the subset of Z × Z defined by f = {(ab, a  + b): a, b ∈ Z}. Is f a function from Z to Z: justify your answer.
Ans:  The relation f is defined as f = {(ab, a +  b): a, b ∈  Z}
We know that a relation f 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 2, 6, –2, –6 ∈ Z, (2 × 6, 2 + 6), (–2 × –6, –2 + (–6)) ∈ f
i.e., (12, 8), (12, –8) ∈ 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.

Ques 12: Let A = {9, 10, 11, 12, 13} and let f: A → N be defined by f(n) = the highest prime factor of n. Find the range of f.
Ans:       A = {9, 10, 11, 12, 13}
f: A → N is 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
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 f is the set of all f(n), where n ∈ A.
∴Range of f = {3, 5, 11, 13}

The document NCERT Solutions Class 11 Maths Chapter 2 - Relations and Functions is a part of the JEE Course Mathematics (Maths) for JEE Main & Advanced.
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## Mathematics (Maths) for JEE Main & Advanced

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## FAQs on NCERT Solutions Class 11 Maths Chapter 2 - Relations and Functions

 1. What are relations and functions in mathematics?
Ans. Relations and functions are fundamental concepts in mathematics. A relation is a set of ordered pairs, where each pair consists of two elements from different sets. On the other hand, a function is a special type of relation where each element from the domain set is related to exactly one element in the range set.
 2. How do you determine if a relation is a function?
Ans. To determine if a relation is a function, we use the vertical line test. This test states that if a vertical line intersects the graph of the relation at more than one point, then the relation is not a function. If every vertical line intersects the graph at most once, then the relation is a function.
 3. What is the difference between a one-to-one function and an onto function?
Ans. A one-to-one function is a function in which each element in the domain set corresponds to a unique element in the range set. In other words, no two elements in the domain can have the same image in the range. On the other hand, an onto function is a function in which every element in the range set has at least one pre-image in the domain set.
 4. How do you find the domain and range of a function?
Ans. To find the domain of a function, we identify all possible values of the independent variable (usually denoted as x) for which the function is defined. The range of a function, on the other hand, refers to all possible values of the dependent variable (usually denoted as y) that the function can take.
 5. Can a relation be both reflexive and symmetric but not transitive?
Ans. No, a relation cannot be both reflexive and symmetric but not transitive. If a relation is reflexive, it means that every element of the set is related to itself. If a relation is symmetric, it means that whenever an element a is related to an element b, then b is also related to a. Transitivity, on the other hand, states that if a is related to b and b is related to c, then a must also be related to c. Therefore, if a relation is reflexive and symmetric, it must also be transitive.

## Mathematics (Maths) for JEE Main & Advanced

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