Exercise 2.3 NCERT Class 11
Q.1 Q.2 Q.3 Q.4 Q.5

Exercise 2.3 NCERT Class 11 from chapter Relations and Functions contains 5 questions.
In this exercise, the questions are based on the topic functions, domain and range.

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Exercise 1.1 Exercise 1.2 Exercise 1.3 Exercise 1.4 Exercise 1.5 Miscellaneous Exercise on Chapter 1 Exercise 2.1 Exercise 2.2 Exercise 2.3 Miscellaneous Exercise on Chapter 2 Exercise 3.1 Exercise 3.2 Exercise 3.3 Miscellaneous Exercise on Chapter 3 Exercise 4.1 Miscellaneous Exercise on Chapter 4 Exercise 5.1 Miscellaneous Exercise on Chapter 5 Exercise 6.1 Exercise 6.2 Exercise 6.3 Exercise 6.4 Miscellaneous Exercise on Chapter 6 Exercise 7.1 Miscellaneous Exercise on Chapter 7 Exercise 8.1 Exercise 8.2 Miscellaneous Exercise on Chapter 8 Exercise 9.1 Exercise 9.2 Exercise 9.3 Miscellaneous Exercise on Chapter 9 Exercise 10.1 Exercise 10.2 Exercise 10.3 Exercise 10.4 Miscellaneous Exercise on Chapter 10 Exercise 11.1 Exercise 11.2 Miscellaneous Exercise on Chapter 11 Exercise 12.1 Exercise 12.2 Miscellaneous Exercise on Chapter 12 Exercise 13.1 Exercise 13.2 Miscellaneous Exercise on Chapter 13 Exercise 14.1 Exercise 14.2 Miscellaneous Exercise on Chapter 14

Exercise 2.3 NCERT Class 11 Mathematics Solutions

The first question of Exercise 2.3 NCERT Class 11 is related to functions, domain, codomain and range.

1. Which of the following relations are functions? Give reasons. If it is a function, determine its domain and range.

(i) {(2,1), (5,1), (8,1), (11,1), (14,1), (17,1)}

The given relation {(2,1), (5,1), (8,1), (11,1), (14,1), (17,1)} is a function because the first elements of every ordered pair is unique i.e. the same first element of the ordered pair has unique image.
Domain = {2, 5, 8, 11, 14, 17}
Range = {1}

(ii) {(2,1), (4,2), (6,3), (8,4), (10,5), (12, 6), (14,7)}

{(2,1), (4,2), (6,3), (8,4), (10,5), (12,6), (14,7)}
This relation is also a function because the first element of every ordered pair is unique.
Domain = {2, 4, 6, 8, 10, 12, 14}
Range = {1, 2, 3, 4, 5, 6, 7}

(iii) {(1,3), (1,5), (2,5)}

{(1, 3), (1, 5), (2, 5)}
This relation is not a function, as the first element '1' of the ordered pairs (1, 3) and (1, 5) has two different images, '3' and '5'.

The second question of Exercise 2.3 NCERT Class 11 is related to real functions, domain, and range.

2. Find the domain and range of the following real functions:

(i) f(x)=−|x|

The given function is
f(x)=−|x|
We know that domain is the set of values of x for which f(x) is defined.
∵ f(x) is defined for all values of x ∈ R.
∴ Domain of f(x) = R.

∵ Range is the set of values of f(x) for all values of x belonging to its domain R.
Here, x ∈ R ... (∵ Domain of f(x)).
⇒ |x| ∈ [0, ∞) ... (∵ |x| is never negative.)
⇒ −|x| ∈ (−∞, 0]
⇒ f(x) ∈ (−∞, 0] ... (∵ f(x)=−|x|)
⇒ Range = (−∞, 0]

(ii) f(x)=9−x2

The given function is
f(x)=9−x2
∵ Domain is the set of values of x for which f(x) is defined.
So, f(x)=9−x2 is defined if
9−x2≥0 ... (∵ A number inside a square root)
⇒x2−9≤0
⇒x2≤9
⇒|x|≤3
⇒x≤3or−x≤3
⇒x≤3orx≥−3
⇒x∈[−3,3]
Thus, f(x) is defined for all x∈[−3,3]
∴ Domain of f(x)= [−3,3].

∵ Range of a function is the set of values of f(x) for all x belonging to its domain.
So, x∈[−3,3]
⇒x2∈ [0,9]
⇒−x2∈[−9,0]
⇒9−x2∈[0,9]
⇒9−x2∈[0,3] or 9−x2∈[0,−3] (Not possible).
⇒f(x)∈[0,3]
Hence, Range of f(x) = [0, 3]

The third question of Exercise 2.3 NCERT Class 11 asks to find the value of a function at a point.

3. A function f is defined by f(x)=2x−5. Write down the values of
(i) f(0)
(ii) f(7)
(iii) f(−3)

We have,
f(x)=2x−5.

(i) f(0)=2×0−5=−5

(ii) f(7)=2×7−5=9

(iii) f(−3)=2×(−3)−5=−11

The fourth question of Exercise 2.3 NCERT Class 11 asks to find the value of a function at a point.

4. The function 't' which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by
t(C)=9C5+32.
Find
(i) t(0)
(ii) t(28)
(iii) t(−10)
(iv) The value of C, when t(C)=212.

The given function is
t(C)=9C5+32.

(i) t(0)=9×05+32=32.

(ii) t(28)=9×285+32=50.4+32=82.4.

(iii) t(−10)=9×(−10)5+32=−18+32=14.

(iv) ∵t(C)=212
⇒9C5+32=212
⇒9C5=212−32
⇒9C5=180
⇒9C=180×5=900
⇒C=9009=100
Thus, C = 100°.

The fifth question of Exercise 2.3 NCERT Class 11 asks to find the range of functions.

5. Find the range of the following functions.

(i) f(x)=2−3x,x∈R,x>0

∵ Range is the set of values of f(x) for x∈R and x>0.
Here, the domain of the given function is R ∩ (0, ∞) = (0, ∞).
So, x∈(0, ∞)
⇒x>0
⇒3x>0
⇒−3x<0
⇒2−3x<2
⇒f(x)<2
⇒f(x)∈[−∞,2]
Hence, range of f(x)=[−∞,2].

(ii) f(x)=x2+2,xis a real number.

∵ The domain of the function is R.
∴ x∈R
⇒x∈(−∞,∞)
⇒x2∈[0,∞]
⇒(x2+2)∈[2,∞]
⇒f(x)∈[2,∞]
∴ Range of f(x)=[2,∞].

(iii) f(x)=x,xis a real number.

∵ x∈ R
⇒ f(x)∈R
⇒ Range of f(x)=R.

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