**Exercise 2.3 Relations and Functions**

**Question and Answers**

**Class 11 – Maths**

Class | Class 11 |

Subject | Mathematics |

Chapter Name | Relations and Functions |

Chapter No. | Chapter 2 |

Exercise | Exercise 2.3 |

Category | Class 11 Maths NCERT Solutions |

**Question 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)}**

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

As 2, 5, 8, 11, 14, and 17 are the elements of the domain of the given relation having their unique images, this relation can be called 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)}**

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

As 2, 4, 6, 8, 10, 12, and 14 are the elements of the domain of the given relation having their unique images, this relation can be called 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)}.**

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

It’s seen that the same first element, i.e., 1, corresponds to two different images, i.e., 3 and 5; this relation cannot be called a function.

**Question 2 Find the domain and range of the following real functions: **

**(i) f(x) = – x**

**(i)** We have the given function as, f (x) = − |x|

It is also know that, | x | = x , if x ≥ 0

−x ,if x < 0

Thus, f(x) = − | x | = − x ,if x ≥ 0

x ,if x < 0

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

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

Therefore, the range of *f* is given by (–∞, 0].

**(ii) f(x) = √9 − x ^{2} **

**(ii)** As √(9 – x^{2}) is defined for all real numbers that are greater than or equal to –3 and less than or equal to 3, for 9 – x^{2} ≥ 0.

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

For any value of *x* in the range [–3, 3], the value of *f*(*x*) will lie between 0 and 3.

Therefore, the range of *f*(*x*) is {*x*: 0 ≤ *x* ≤ 3} or [0, 3].

**Question 3 A function f is defined by f(x) = 2x –5. Write down the values of **

**(i) f (0)**

**(i)** Function, *f*(*x*) = 2*x* – 5

*f *(0) = 2 × 0 – 5 = 0 – 5 = –5

** (ii) f (7)**

**(ii)** Function, *f*(*x*) = 2*x* – 5

*f*(7) = 2 × 7 – 5 = 14 – 5 = 9

**(iii) f (–3)**

**(iii)** Function, *f*(*x*) = 2*x* – 5

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

**Question 4 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)**

**(i)** We have our given function as, t(C)= (9C/5) +32

To find our needed values of the function we just have to put the values in the given function and simplify it.

So, we get now,

t (0) =[ (9 × o ) /5 ] + 32

t (0)= 32

**(ii) t(28)**

**(ii)** We have our given function as, t(C)= (9C/5) +32

To find our needed values of the function we just have to put the values in the given function and simplify it.

So, we get now,

t (28) =[ (9 × 28 ) /5 ] + 32

t (28 ) =( 252 + 160) /5

t (28 ) =(412 /5 ]

t (28 ) = 82.4

**(iii) t(–10)**

**(iii)** We have our given function as, t(C)= (9C/5) +32

To find our needed values of the function we just have to put the values in the given function and simplify it.

So, we get now,

t ( -10 ) = [ (9 × (-10) ) /5 ] + 32

t (-10) =−18+32

t (-10) = 14

**(iv) The value of C, when t(C) = 212.**

**(iv)** We have our given function as, t(C)= (9C/5) +32

we are given that, t(C)=212

212 = (9C/5) +32

9C/5 = 212−32

9C/5 = 180

C = 900 / 9 = 100

Thus , it can be said that, for t(C)=212 the value of t is 100

**Question 5 Find the range of each of the following functions. **

**(i) f (x) = 2 – 3x, x ∈ R, x > 0.**

**(i)** f(x) = 2 – 3*x*, *x* ∈ R, *x* > 0

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 |
… |

We can now see, it can be seen that the elements of the range is less than 2.

So, the range will be, f = (−∞ ,2 )

x > 0, So , 3x > 0

-3x < 0 [Multiplying by -1 on both sides, the inequality sign changes]

2 – 3x < 2

Therefore, the value of 2 – 3x is less than 2.

Hence, Range = (–∞, 2)

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

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

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

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

So, we see that the range of the function f is the set of all numbers which are greater than or equal to 2.

Thus we can conclude that the range of the function is, [2,∞]

*x*^{2} ≥ 0

So,

*x*^{2} + 2 ≥ 2 [Adding 2 on both sides]

Therefore, the value of *x*^{2} + 2 is always greater or equal to 2, for x is a real number.

Hence, Range = [2, ∞)

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

* (iii) f*(

*x*) =

*x, x*is a real number

Clearly, the range of *f* is the set of all real numbers.

Thus,

Range of *f* = R

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