# Find the range of each of the following functions

(i) f (x) = 2 – 3x, x ∈ R, x > 0

(ii) f (x) = x^{2} + 2, x is a real number

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

**Solution:**

**(i) **f (x) = 2 - 3x, x ∈ R, x > 0

The values of f (x) for various values of real numbers x > 0 can be written in the tabular form as:

x | 0.01 | 0.1 | 0.9 | 1 | 2 | 2.5 | 4 | 5 | .... |

f ( x) | 1.97 | 1.7 | - 0.7 | - 1 | - 4 | - 5.5 | - 10 | - 13 | .... |

Thus, it can be clearly observed that the range of f is the set of all real numbers less than 2.

that is range of f = (- ∞, 2)

Alternative Method:

Let x > 0

⇒ 3x > 0

⇒ 2 - 3x < 2

⇒ f ( x) < 2

Therefore, Range of f = (- ∞, 2)

**(ii) **f (x)= x^{2} + 2, x is a real number.

The values of f(x) for various values of real numbers x can be written in the tabular form as:

x | 0 | ± 0.3 | ± 0.8 | ± 1 | ± 2 | ± 3 | .... |

f ( x) | 2 | 2.09 | 2.64 | 3 | 6 | 11 | .... |

Thus, it can be clearly observed that the range of f is the set of all real numbers greater than 2.

that is range of f = [2, ∞]

Alternative Method:

Let x be any real number i.e., x^{2} ≥ 0

Accordingly,

⇒ x^{2} + 2 ≥ 0 + 2

⇒ x^{2} + 2 ≥ 2

⇒ x^{2} ≥ 0.

⇒ f (x) ≥ 2

Therefore, Range of f = [2, ∞)

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

It is clear that the range of f is the set of all real numbers. Therefore, the Range of f = R.

NCERT Solutions Class 11 Maths Chapter 2 Exercise 2.3 Question 5

## Find the range of each of the following functions.(i) f (x) = 2 - 3x, x ∈ R, x > 0(ii) f (x)= x^{2} + 2, x is a real number.(iii) f (x) = x, x is a real number

**Summary:**

A few functions are given. We have found that the range of f is the set of all real numbers. Therefore, the Range of f (x) = x is R. Range of f (x)= x^{2} + 2 is [2, ∞). Range of f (x) = 2 – 3x is (- ∞, 2)

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