**Q.1: Which of the given relations are functions? Answer with reason. If it is a function, find its range and domain.**

**(i) {(12, 11), (15, 11), (18, 11), (1, 11), (4, 11), (7, 11)}**

**(ii) {(12, 11), (14, 12), (16, 13), (18, 14), (0, 15), (2, 16), (4, 17)}**

**(iii) {(11, 13), (11, 15), (12, 15)}**

**Sol:**

**(i). {(12, 11), (15, 11), (18, 11), (1, 11), (4, 11), (7, 11)}**

Since, 12, 15, 18, 1, 4 and 7 are the elements of the domain of the given relation which have unique images.

Therefore, the given relation **is a function**.

**Domain = {12, 15, 18, 1, 4, 7}**

**Range = {11}**

** **

**(ii). {(12, 11), (14, 12), (16, 13), (18, 14), (0, 15), (2, 16), (4, 17)}**

Since, 12, 14, 16, 18, 0, 2 and 4 are the elements of the domain of the given relation which have unique images.

Therefore, the given relation **is a function**.

**Domain = {12, 14, 16, 18, 0, 2, 4}**

**Range = {11, 12, 13, 14, 15, 16, 17}**

**(iii). {(11, 13), (11, 15), (12, 15)}**

Since, the same first element that is 11 corresponds to two images that is 13 and 15.

**Therefore, the given relation is not a function.**

**Q.2: Find the range and domain of the given real function:**

**(i) f(y) = -|y|**

**(ii) f(y) = 9–y2−−−−√**

**Sol:**

**(i)** f(y) = -|y|, y

**As we know:**

**|y| =**

Therefore,

**f(y) = -|y| =**

Since, f(y) is defined for y **domain of ‘f’ is R.**

The range of **f(y) = -|y|**is all the real number except positive real number.

**Therefore, range of ‘f’ is (−∞,0].**

**(ii)** f(y) =

Therefore, **Domain of f(y) =** {y: -3

For any **value** **of** **‘y’** such that **0 and 3.**

**Therefore, the Range of f(y) = {y: 0 ≤ y ≤ 3} or [0, 3]**

**Q.3: A function f is f(y) = 3y – 6. Find the values of the following:**

**(i) f(1)**

**(ii) f(8)**

**(iii) f(-2)**

**Sol:**

**The function of ‘f’ is:**

**f(y) = 3y – 6**

**(i) f(1)** = (3 × 1) – 6

= 3 – 6** = -3**

**(ii) f(8)** = (3 × 8) – 6

= 24 – 6**= 18**

**(iii) f(-2)** = (3 × -2) – 6

= -6 – 6**= -12**

**Q.4: The function ‘f’ which shows temperature in degree Celsius into temperature in degree Fahrenheit is expressed as: 9C5+32.**

**Find for the following values:**

**(i) f(0)**

**(ii) f(28)**

**(iii) f(-10)**

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

**Sol:**

F =

**(i) f(0):**

=

= 0 + 32**= 32**

**(ii) f(28):**

=

=

=

**(iii) f(-10):**

=

=

= -18 + 32**= 14**

**(iv) The value of C, when f(C) = 212:**

**i.e. C=100**

**Therefore, The value of f, when f(C) = 212 is 100.**

**Q.5: Calculate range of the given functions:**

**(i) f(y) = 2 – 3y, y ∈ R, y > 0.**

**(ii) f(y) = y2+2, is a real no.**

**(iii) f(y) = y, y is a real no.**

**Sol:**

**(i) **

**We can write the value of f(y) for different real numbers x > 0 in tabular form as:**

y | 0.01 | 0.1 | 0.9 | 1 | 2 | 2.5 | 4 | 5 | … |

f(y) | 1.97 | 1.7 | –
0.7 |
–
1 |
–
4 |
–
5.5 |
–
10 |
–
13 |
… |

Thus, we can clearly observe that **the range of ‘f’ forms the set for all real numbers which are less than 2. **

i.e. Range of f =

**Alternative:**

Let, y > 0

3y > 0

2 – 3y < 2

i.e f(y) < 2

**Therefore, Range of f = (−∞,2)**

**(ii) **

**We can write the value of f(y) for different real numbers x, in tabular form as: **

y | 0 | . . . | |||||

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

Thus, we can clearly observe **that the range of f forms the set for all real numbers which are less than 2. **

**i.e. Range of f = (−∞,2)**

**Alternative:**

Let ‘y’ be any **real** **number**. Then,

**Therefore, Range of f = [2,∞)**

**(iii) f(y) = y, where y is a real number**

Here, we can see that the range of f is the set of all the real numbers.

**Therefore, Range of f = R**