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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)}

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

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

A function f is a relation from a non-empty set A to a non-empty set B such that the domain of f is A and no two distinct ordered pairs in f have the same first element. |

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

Since 2, 5, 8, 11, 14, and 17 are the elements of the domain of the given relation having their unique images, this relation is a function.

Here,domain = {2, 5, 8, 11, 14, 17} and range = {1}

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

Since 2, 4, 6, 8, 10, 12, and 14 are the elements of the domain of the given relation having their unique images, this relation is a function.

Here, domain = {2, 4, 6, 8, 10, 12, 14} and range = {1, 2, 3, 4, 5, 6, 7}

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

Since the same first element i.e., 1 corresponds to two different images i.e., 3 and 5, So this relation is not a function.

Find the domain and range of the following real function:

(i) f(x) = –|x|

(ii) f(x) = √9 − x2

(i) f(x) = –|x|, x ∈ R

This can be written as

f(x) = −|x| =

{ −x, if x ≥ 0

x, if x < 0

Here we see that f(x) is defined for x ∈ R. So, the domain of f is R.

From the Function, we can see that, it returns negative real number for all positive and negative real number

Therefore, the range of f is (−∞, 0].

(ii) f(x) = √(9 – x

It means 9 – x

=> 3 ≤ x ≤ 3

So, the domain of f(x) is {x: –3 ≤ x ≤ 3} or [–3, 3].

For any value of x such that –3 ≤ x ≤ 3, the value of f(x) will lie between

0 and 3. ∴The range of f(x) is {x: 0 ≤ x ≤ 3} or [0, 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)

The given function is f(x) = 2x – 5.

Therefore,

(i) f (0) = 2 × 0 – 5 = 0 – 5 = –5

(ii) f (7) = 2 × 7 – 5 = 14 – 5 = 9

(iii) f (–3) = 2 × (–3) – 5 = – 6 – 5 = –11

The function ‘t’ which maps temperature in degree Celsius into temperature

in degree Fahrenheit is defined by

t(C) = 9C/5+ 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) = 9C/5+ 32.

Therefore,

(i) t (0) = 9(0)/5+ 32 =32

(ii) t (28) = 9(28)/5+ 32 =412/5

(iii) t (-10) = 9(-10)/5+ 32 =14

(iv) It is given that t(C) = 212

9C/5+ 32=212

9C/5=180

C=100

Hence, the value of t, when t(C) = 212, is 100.

Find the range of each of the following functions.

(i) f(x) = 2 – 3x,

(ii) f(x) = x

(iii) f(x) = x, x is a real number

(i) f(x) = 2 – 3x,

Let x > 0

⇒ 3x > 0

⇒ 2 –3x < 2

⇒ f(x) < 2

So, Range of f = (-∞ , 2)

(ii) f(x) = x

For any real number, we know that

x

⇒ x

⇒ f(x) ≥ 2

So, Range of f = [2, ∞ )

(iii) f(x) = x, x is a real number

Range of f is the set of all real numbers. So, Range of f = R

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- Identity Function
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- Linear Function
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- Algebra of Real Function

Class 11 Maths Class 11 Physics