RD Sharma Class 11 Solutions Chapter - Functions

 Page 1


3. Functions
Exercise 3.1
1. Question
Define a function as a set of ordered pairs.
Answer
A function from is defined by a set of ordered pairs such that any two ordered pairs should not have the
same first component and the different second component.
This means that each element of a set, say X is assigned exactly to one element of another set, say Y.
The set X containing the first components of a function is called the domain of the function.
The set Y containing the second components of a function is called the range of the function.
For example, f = {(a, 1), (b, 2), (c, 3)} is a function.
Domain of f = {a, b, c}
Range of f = {1, 2, 3}
2. Question
Define a function as a correspondence between two sets.
Answer
A function from a set X to a set Y is defined as a correspondence between sets X and Y such that for each
element of X, there is only one corresponding element in Y.
The set X is called the domain of the function.
The set Y is called the range of the function.
For example, X = {a, b, c}, Y = {1, 2, 3, 4, 5} and f be a correspondence which assigns the position of a
letter in the set of alphabets.
Therefore, f(a) = 1, f(b) = 2 and f(c) = 3.
As there is only one element of Y for each element of X, f is a function with domain X and range Y.
3. Question
What is the fundamental difference between a relation and a function? Is every relation a function?
Answer
Let f be a function and R be a relation defined from set X to set Y.
The domain of the relation R might be a subset of the set X, but the domain of the function f must be equal to
X. This is because each element of the domain of a function must have an element associated with it,
whereas this is not necessary for a relation.
In relation, one element of X might be associated with one or more elements of Y, while it must be
associated with only one element of Y in a function.
Thus, not every relation is a function. However, every function is necessarily a relation.
4. Question
Let A = {–2, –1, 0, 1, 2} and f : A ? Z be a function defined by f(x) = x
2
 – 2x – 3. Find:
i. range of f i.e. f(A)
ii. pre-images of 6, –3 and 5
Answer
Given A = {–2, –1, 0, 1, 2}
Page 2


3. Functions
Exercise 3.1
1. Question
Define a function as a set of ordered pairs.
Answer
A function from is defined by a set of ordered pairs such that any two ordered pairs should not have the
same first component and the different second component.
This means that each element of a set, say X is assigned exactly to one element of another set, say Y.
The set X containing the first components of a function is called the domain of the function.
The set Y containing the second components of a function is called the range of the function.
For example, f = {(a, 1), (b, 2), (c, 3)} is a function.
Domain of f = {a, b, c}
Range of f = {1, 2, 3}
2. Question
Define a function as a correspondence between two sets.
Answer
A function from a set X to a set Y is defined as a correspondence between sets X and Y such that for each
element of X, there is only one corresponding element in Y.
The set X is called the domain of the function.
The set Y is called the range of the function.
For example, X = {a, b, c}, Y = {1, 2, 3, 4, 5} and f be a correspondence which assigns the position of a
letter in the set of alphabets.
Therefore, f(a) = 1, f(b) = 2 and f(c) = 3.
As there is only one element of Y for each element of X, f is a function with domain X and range Y.
3. Question
What is the fundamental difference between a relation and a function? Is every relation a function?
Answer
Let f be a function and R be a relation defined from set X to set Y.
The domain of the relation R might be a subset of the set X, but the domain of the function f must be equal to
X. This is because each element of the domain of a function must have an element associated with it,
whereas this is not necessary for a relation.
In relation, one element of X might be associated with one or more elements of Y, while it must be
associated with only one element of Y in a function.
Thus, not every relation is a function. However, every function is necessarily a relation.
4. Question
Let A = {–2, –1, 0, 1, 2} and f : A ? Z be a function defined by f(x) = x
2
 – 2x – 3. Find:
i. range of f i.e. f(A)
ii. pre-images of 6, –3 and 5
Answer
Given A = {–2, –1, 0, 1, 2}
f : A ? Z such that f(x) = x
2
 – 2x – 3
i. range of f i.e. f(A)
A is the domain of the function f. Hence, range is the set of elements f(x) for all x ? A.
Substituting x = –2 in f(x), we get
f(–2) = (–2)
2
 – 2(–2) – 3
? f(–2) = 4 + 4 – 3
? f(–2) = 5
Substituting x = –1 in f(x), we get
f(–1) = (–1)
2
 – 2(–1) – 3
? f(–1) = 1 + 2 – 3
? f(–1) = 0
Substituting x = 0 in f(x), we get
f(0) = (0)
2
 – 2(0) – 3
? f(0) = 0 – 0 – 3
? f(0) = –3
Substituting x = 1 in f(x), we get
f(1) = 1
2
 – 2(1) – 3
? f(1) = 1 – 2 – 3
? f(1) = –4
Substituting x = 2 in f(x), we get
f(2) = 2
2
 – 2(2) – 3
? f(2) = 4 – 4 – 3
? f(2) = –3
Thus, the range of f is {5, 0, –3, –4}.
ii. pre-images of 6, –3 and 5
Let x be the pre-image of 6 ? f(x) = 6
? x
2
 – 2x – 3 = 6
? x
2
 – 2x – 9 = 0
Page 3


3. Functions
Exercise 3.1
1. Question
Define a function as a set of ordered pairs.
Answer
A function from is defined by a set of ordered pairs such that any two ordered pairs should not have the
same first component and the different second component.
This means that each element of a set, say X is assigned exactly to one element of another set, say Y.
The set X containing the first components of a function is called the domain of the function.
The set Y containing the second components of a function is called the range of the function.
For example, f = {(a, 1), (b, 2), (c, 3)} is a function.
Domain of f = {a, b, c}
Range of f = {1, 2, 3}
2. Question
Define a function as a correspondence between two sets.
Answer
A function from a set X to a set Y is defined as a correspondence between sets X and Y such that for each
element of X, there is only one corresponding element in Y.
The set X is called the domain of the function.
The set Y is called the range of the function.
For example, X = {a, b, c}, Y = {1, 2, 3, 4, 5} and f be a correspondence which assigns the position of a
letter in the set of alphabets.
Therefore, f(a) = 1, f(b) = 2 and f(c) = 3.
As there is only one element of Y for each element of X, f is a function with domain X and range Y.
3. Question
What is the fundamental difference between a relation and a function? Is every relation a function?
Answer
Let f be a function and R be a relation defined from set X to set Y.
The domain of the relation R might be a subset of the set X, but the domain of the function f must be equal to
X. This is because each element of the domain of a function must have an element associated with it,
whereas this is not necessary for a relation.
In relation, one element of X might be associated with one or more elements of Y, while it must be
associated with only one element of Y in a function.
Thus, not every relation is a function. However, every function is necessarily a relation.
4. Question
Let A = {–2, –1, 0, 1, 2} and f : A ? Z be a function defined by f(x) = x
2
 – 2x – 3. Find:
i. range of f i.e. f(A)
ii. pre-images of 6, –3 and 5
Answer
Given A = {–2, –1, 0, 1, 2}
f : A ? Z such that f(x) = x
2
 – 2x – 3
i. range of f i.e. f(A)
A is the domain of the function f. Hence, range is the set of elements f(x) for all x ? A.
Substituting x = –2 in f(x), we get
f(–2) = (–2)
2
 – 2(–2) – 3
? f(–2) = 4 + 4 – 3
? f(–2) = 5
Substituting x = –1 in f(x), we get
f(–1) = (–1)
2
 – 2(–1) – 3
? f(–1) = 1 + 2 – 3
? f(–1) = 0
Substituting x = 0 in f(x), we get
f(0) = (0)
2
 – 2(0) – 3
? f(0) = 0 – 0 – 3
? f(0) = –3
Substituting x = 1 in f(x), we get
f(1) = 1
2
 – 2(1) – 3
? f(1) = 1 – 2 – 3
? f(1) = –4
Substituting x = 2 in f(x), we get
f(2) = 2
2
 – 2(2) – 3
? f(2) = 4 – 4 – 3
? f(2) = –3
Thus, the range of f is {5, 0, –3, –4}.
ii. pre-images of 6, –3 and 5
Let x be the pre-image of 6 ? f(x) = 6
? x
2
 – 2x – 3 = 6
? x
2
 – 2x – 9 = 0
However, 
Thus, there exists no pre-image of 6.
Now, let x be the pre-image of –3 ? f(x) = –3
? x
2
 – 2x – 3 = –3
? x
2
 – 2x = 0
? x(x – 2) = 0
? x = 0 or 2
Clearly, both 0 and 2 are elements of A.
Thus, 0 and 2 are the pre-images of –3.
Now, let x be the pre-image of 5 ? f(x) = 5
? x
2
 – 2x – 3 = 5
? x
2
 – 2x – 8= 0
? x
2
 – 4x + 2x – 8= 0
? x(x – 4) + 2(x – 4) = 0
? (x + 2)(x – 4) = 0
? x = –2 or 4
However, 4 ? A but –2 ? A
Thus, –2 is the pre-images of 5.
5. Question
If a function f: R ? R be defined by
Find: f(1), f(–1), f(0), f(2).
Answer
Given 
We need to find f(1), f(–1), f(0) and f(2).
When x > 0, f(x) = 4x + 1
Substituting x = 1 in the above equation, we get
f(1) = 4(1) + 1
? f(1) = 4 + 1
? f(1) = 5
When x < 0, f(x) = 3x – 2
Substituting x = –1 in the above equation, we get
f(–1) = 3(–1) – 2
Page 4


3. Functions
Exercise 3.1
1. Question
Define a function as a set of ordered pairs.
Answer
A function from is defined by a set of ordered pairs such that any two ordered pairs should not have the
same first component and the different second component.
This means that each element of a set, say X is assigned exactly to one element of another set, say Y.
The set X containing the first components of a function is called the domain of the function.
The set Y containing the second components of a function is called the range of the function.
For example, f = {(a, 1), (b, 2), (c, 3)} is a function.
Domain of f = {a, b, c}
Range of f = {1, 2, 3}
2. Question
Define a function as a correspondence between two sets.
Answer
A function from a set X to a set Y is defined as a correspondence between sets X and Y such that for each
element of X, there is only one corresponding element in Y.
The set X is called the domain of the function.
The set Y is called the range of the function.
For example, X = {a, b, c}, Y = {1, 2, 3, 4, 5} and f be a correspondence which assigns the position of a
letter in the set of alphabets.
Therefore, f(a) = 1, f(b) = 2 and f(c) = 3.
As there is only one element of Y for each element of X, f is a function with domain X and range Y.
3. Question
What is the fundamental difference between a relation and a function? Is every relation a function?
Answer
Let f be a function and R be a relation defined from set X to set Y.
The domain of the relation R might be a subset of the set X, but the domain of the function f must be equal to
X. This is because each element of the domain of a function must have an element associated with it,
whereas this is not necessary for a relation.
In relation, one element of X might be associated with one or more elements of Y, while it must be
associated with only one element of Y in a function.
Thus, not every relation is a function. However, every function is necessarily a relation.
4. Question
Let A = {–2, –1, 0, 1, 2} and f : A ? Z be a function defined by f(x) = x
2
 – 2x – 3. Find:
i. range of f i.e. f(A)
ii. pre-images of 6, –3 and 5
Answer
Given A = {–2, –1, 0, 1, 2}
f : A ? Z such that f(x) = x
2
 – 2x – 3
i. range of f i.e. f(A)
A is the domain of the function f. Hence, range is the set of elements f(x) for all x ? A.
Substituting x = –2 in f(x), we get
f(–2) = (–2)
2
 – 2(–2) – 3
? f(–2) = 4 + 4 – 3
? f(–2) = 5
Substituting x = –1 in f(x), we get
f(–1) = (–1)
2
 – 2(–1) – 3
? f(–1) = 1 + 2 – 3
? f(–1) = 0
Substituting x = 0 in f(x), we get
f(0) = (0)
2
 – 2(0) – 3
? f(0) = 0 – 0 – 3
? f(0) = –3
Substituting x = 1 in f(x), we get
f(1) = 1
2
 – 2(1) – 3
? f(1) = 1 – 2 – 3
? f(1) = –4
Substituting x = 2 in f(x), we get
f(2) = 2
2
 – 2(2) – 3
? f(2) = 4 – 4 – 3
? f(2) = –3
Thus, the range of f is {5, 0, –3, –4}.
ii. pre-images of 6, –3 and 5
Let x be the pre-image of 6 ? f(x) = 6
? x
2
 – 2x – 3 = 6
? x
2
 – 2x – 9 = 0
However, 
Thus, there exists no pre-image of 6.
Now, let x be the pre-image of –3 ? f(x) = –3
? x
2
 – 2x – 3 = –3
? x
2
 – 2x = 0
? x(x – 2) = 0
? x = 0 or 2
Clearly, both 0 and 2 are elements of A.
Thus, 0 and 2 are the pre-images of –3.
Now, let x be the pre-image of 5 ? f(x) = 5
? x
2
 – 2x – 3 = 5
? x
2
 – 2x – 8= 0
? x
2
 – 4x + 2x – 8= 0
? x(x – 4) + 2(x – 4) = 0
? (x + 2)(x – 4) = 0
? x = –2 or 4
However, 4 ? A but –2 ? A
Thus, –2 is the pre-images of 5.
5. Question
If a function f: R ? R be defined by
Find: f(1), f(–1), f(0), f(2).
Answer
Given 
We need to find f(1), f(–1), f(0) and f(2).
When x > 0, f(x) = 4x + 1
Substituting x = 1 in the above equation, we get
f(1) = 4(1) + 1
? f(1) = 4 + 1
? f(1) = 5
When x < 0, f(x) = 3x – 2
Substituting x = –1 in the above equation, we get
f(–1) = 3(–1) – 2
? f(–1) = –3 – 2
? f(–1) = –5
When x = 0, f(x) = 1
? f(0) = 1
When x > 0, f(x) = 4x + 1
Substituting x = 2 in the above equation, we get
f(2) = 4(2) + 1
? f(2) = 8 + 1
? f(2) = 9
Thus, f(1) = 5, f(–1) = –5, f(0) = 1 and f(2) = 9.
6. Question
A function f : R ? R is defined by f(x) = x
2
. Determine
i. range of f
ii. {x: f(x) = 4}
iii. {y: f(y) = –1}
Answer
Given f : R ? R and f(x) = x
2
.
i. range of f
Domain of f = R (set of real numbers)
We know that the square of a real number is always positive or equal to zero.
Hence, the range of f is the set of all non-negative real numbers.
Thus, range of f = R
+
? {0}
ii. {x: f(x) = 4}
Given f(x) = 4
? x
2
 = 4
? x
2
 – 4 = 0
? (x – 2)(x + 2) = 0
? x = ±2
Thus, {x: f(x) = 4} = {–2, 2}
iii. {y: f(y) = –1}
Given f(y) = –1
? y
2
 = –1
However, the domain of f is R, and for every real number y, the value of y
2
 is non-negative.
Hence, there exists no real y for which y
2
 = –1.
Thus, {y: f(y) = –1} = Ø
7. Question
Page 5


3. Functions
Exercise 3.1
1. Question
Define a function as a set of ordered pairs.
Answer
A function from is defined by a set of ordered pairs such that any two ordered pairs should not have the
same first component and the different second component.
This means that each element of a set, say X is assigned exactly to one element of another set, say Y.
The set X containing the first components of a function is called the domain of the function.
The set Y containing the second components of a function is called the range of the function.
For example, f = {(a, 1), (b, 2), (c, 3)} is a function.
Domain of f = {a, b, c}
Range of f = {1, 2, 3}
2. Question
Define a function as a correspondence between two sets.
Answer
A function from a set X to a set Y is defined as a correspondence between sets X and Y such that for each
element of X, there is only one corresponding element in Y.
The set X is called the domain of the function.
The set Y is called the range of the function.
For example, X = {a, b, c}, Y = {1, 2, 3, 4, 5} and f be a correspondence which assigns the position of a
letter in the set of alphabets.
Therefore, f(a) = 1, f(b) = 2 and f(c) = 3.
As there is only one element of Y for each element of X, f is a function with domain X and range Y.
3. Question
What is the fundamental difference between a relation and a function? Is every relation a function?
Answer
Let f be a function and R be a relation defined from set X to set Y.
The domain of the relation R might be a subset of the set X, but the domain of the function f must be equal to
X. This is because each element of the domain of a function must have an element associated with it,
whereas this is not necessary for a relation.
In relation, one element of X might be associated with one or more elements of Y, while it must be
associated with only one element of Y in a function.
Thus, not every relation is a function. However, every function is necessarily a relation.
4. Question
Let A = {–2, –1, 0, 1, 2} and f : A ? Z be a function defined by f(x) = x
2
 – 2x – 3. Find:
i. range of f i.e. f(A)
ii. pre-images of 6, –3 and 5
Answer
Given A = {–2, –1, 0, 1, 2}
f : A ? Z such that f(x) = x
2
 – 2x – 3
i. range of f i.e. f(A)
A is the domain of the function f. Hence, range is the set of elements f(x) for all x ? A.
Substituting x = –2 in f(x), we get
f(–2) = (–2)
2
 – 2(–2) – 3
? f(–2) = 4 + 4 – 3
? f(–2) = 5
Substituting x = –1 in f(x), we get
f(–1) = (–1)
2
 – 2(–1) – 3
? f(–1) = 1 + 2 – 3
? f(–1) = 0
Substituting x = 0 in f(x), we get
f(0) = (0)
2
 – 2(0) – 3
? f(0) = 0 – 0 – 3
? f(0) = –3
Substituting x = 1 in f(x), we get
f(1) = 1
2
 – 2(1) – 3
? f(1) = 1 – 2 – 3
? f(1) = –4
Substituting x = 2 in f(x), we get
f(2) = 2
2
 – 2(2) – 3
? f(2) = 4 – 4 – 3
? f(2) = –3
Thus, the range of f is {5, 0, –3, –4}.
ii. pre-images of 6, –3 and 5
Let x be the pre-image of 6 ? f(x) = 6
? x
2
 – 2x – 3 = 6
? x
2
 – 2x – 9 = 0
However, 
Thus, there exists no pre-image of 6.
Now, let x be the pre-image of –3 ? f(x) = –3
? x
2
 – 2x – 3 = –3
? x
2
 – 2x = 0
? x(x – 2) = 0
? x = 0 or 2
Clearly, both 0 and 2 are elements of A.
Thus, 0 and 2 are the pre-images of –3.
Now, let x be the pre-image of 5 ? f(x) = 5
? x
2
 – 2x – 3 = 5
? x
2
 – 2x – 8= 0
? x
2
 – 4x + 2x – 8= 0
? x(x – 4) + 2(x – 4) = 0
? (x + 2)(x – 4) = 0
? x = –2 or 4
However, 4 ? A but –2 ? A
Thus, –2 is the pre-images of 5.
5. Question
If a function f: R ? R be defined by
Find: f(1), f(–1), f(0), f(2).
Answer
Given 
We need to find f(1), f(–1), f(0) and f(2).
When x > 0, f(x) = 4x + 1
Substituting x = 1 in the above equation, we get
f(1) = 4(1) + 1
? f(1) = 4 + 1
? f(1) = 5
When x < 0, f(x) = 3x – 2
Substituting x = –1 in the above equation, we get
f(–1) = 3(–1) – 2
? f(–1) = –3 – 2
? f(–1) = –5
When x = 0, f(x) = 1
? f(0) = 1
When x > 0, f(x) = 4x + 1
Substituting x = 2 in the above equation, we get
f(2) = 4(2) + 1
? f(2) = 8 + 1
? f(2) = 9
Thus, f(1) = 5, f(–1) = –5, f(0) = 1 and f(2) = 9.
6. Question
A function f : R ? R is defined by f(x) = x
2
. Determine
i. range of f
ii. {x: f(x) = 4}
iii. {y: f(y) = –1}
Answer
Given f : R ? R and f(x) = x
2
.
i. range of f
Domain of f = R (set of real numbers)
We know that the square of a real number is always positive or equal to zero.
Hence, the range of f is the set of all non-negative real numbers.
Thus, range of f = R
+
? {0}
ii. {x: f(x) = 4}
Given f(x) = 4
? x
2
 = 4
? x
2
 – 4 = 0
? (x – 2)(x + 2) = 0
? x = ±2
Thus, {x: f(x) = 4} = {–2, 2}
iii. {y: f(y) = –1}
Given f(y) = –1
? y
2
 = –1
However, the domain of f is R, and for every real number y, the value of y
2
 is non-negative.
Hence, there exists no real y for which y
2
 = –1.
Thus, {y: f(y) = –1} = Ø
7. Question
Let f: R
+
? R, where R
+
 is the set of all positive real numbers, be such that f(x) = log
e
x. Determine
i. the image set of the domain of f
ii. {x: f(x) = –2}
iii. whether f(xy) = f(x) + f(y) holds.
Answer
Given f : R
+
? R and f(x) = log
e
x.
i. the image set of the domain of f
Domain of f = R
+
 (set of positive real numbers)
We know the value of logarithm to the base e (natural logarithm) can take all possible real values.
Hence, the image set of f is the set of real numbers.
Thus, the image set of f = R
ii. {x: f(x) = –2}
Given f(x) = –2
? log
e
x = –2
? x = e
-2
 [? log
b
a = c ? a = b
c
]
Thus, {x: f(x) = –2} = {e
–2
}
iii. whether f(xy) = f(x) + f(y) holds.
We have f(x) = log
e
x ? f(y) = log
e
y
Now, let us consider f(xy).
f(xy) = log
e
(xy)
? f(xy) = log
e
(x × y) [? log
b
(a×c) = log
b
a + log
b
c]
? f(xy) = log
e
x + log
e
y
? f(xy) = f(x) + f(y)
Hence, the equation f(xy) = f(x) + f(y) holds.
8. Question
Write the following relations as sets of ordered pairs and find which of them are functions:
i. {(x, y): y = 3x, x ? {1, 2, 3}, y ? {3, 6, 9, 12}}
ii. {(x, y): y > x + 1, x = 1, 2 and y = 2, 4, 6}
iii. {(x, y): x + y = 3, x, y ? {0, 1, 2, 3}}
Answer
i. {(x, y): y = 3x, x ? {1, 2, 3}, y ? {3, 6, 9, 12}}
When x = 1, we have y = 3(1) = 3
When x = 2, we have y = 3(2) = 6
When x = 3, we have y = 3(3) = 9
Thus, R = {(1, 3), (2, 6), (3, 9)}
Every element of set x has an ordered pair in the relation and no two ordered pairs have the same first
component but different second components.