The document NCERT Solution(Ex - 2.3) - Relations and Functions JEE Notes | EduRev is a part of the JEE Course Mathematics (Maths) Class 11.

All you need of JEE at this link: JEE

**Ex - 2.3**

**Ques 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)}**

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

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

**Ans: **(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, this relation is not a function.

**Ques 2: ****Find the domain and range of the following real function:**

**(i) f(x) = –|x| (ii)**

**Ans: **(i) *f*(*x*) = –|*x*|, *x* ∈ R

We know that |*x*| =

Since *f*(*x*) is defined for *x* ∈ **R**, the domain of *f* is **R.**

It can be observed that the range of *f*(*x*) = –|*x*| is all real numbers except positive real numbers.

∴The range of *f* is (–, 0].

(ii)

Since is defined for all real numbers that are greater than or equal to –3 and less than or equal to 3, 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].

**Ques 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)**

**Ans: **The given function is *f*(*x*) = 2*x* – 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

**Ques 4:****The function ‘ t’ which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by **

**Find (i) t (0) (ii) t (28) (iii) t (–10) (iv) The value of C, when t(C) = 212**

**Ans: **The given function is .

Therefore,

(i)

(ii)

(iii)

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

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

**Ques 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**

**Ans: **(i) Let *x* > 0

⇒ 3*x* > 0

⇒ 2 –3*x* < 2

⇒ *f*(*x*) < 2

∴Range of *f* = (–, 2)

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

We know x^{2}≥ 0 ⇒ x^{2} + 2 ≥ 0 + 2

⇒ x^{2} + 2 > 2 ∴ f (x) ≥ 2

∴ The range of f (x) is [2, ∞).

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

Let y =f (x) = x ⇒ y = x

∴ Range of f (x) = Domain of f (x)

∴ Range of f (x) is R.

Offer running on EduRev: __Apply code STAYHOME200__ to get INR 200 off on our premium plan EduRev Infinity!

158 videos|186 docs|161 tests