**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)}**

Solution:

(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.

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

**(i) f(x) = β|x|**

**(ii) π(π₯) = β(9 β π₯ ^{2})**

Solution:

(i) f(x) = β|x|, x β R

We know that |π₯| = { π₯, ππ π₯ β₯ 0

{βπ₯, ππ π₯ < 0

β΄ π(π₯) = β|π₯| = { βπ₯, ππ π₯ β₯ 0

{ Β π₯, ππ π₯ < 0

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

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

β΄ The range of f is (ββ, 0]. (ii) π(π₯) = β(9 β π₯^{2})

Since β(9 β π₯^{2}) is defined for all real numbers that are greater than or equal to β3 and less than or equal to 3, the domain of f(x) is {x : β3 β€ x β€ 3} or [β3, 3].

For any value of x such that β3 β€ x β€ 3, the value of f(x) will lie between 0 and 3. β΄The range of f(x) is {x: 0 β€ x β€ 3} or [0, 3].

**3: A function f is defined by f(x) = 2x β 5. Write down the values of (i) f(0), (ii) f(7), (iii) f(β3)**

Solution:

The given function is f(x) = 2x β 5.

Therefore,

(i) f(0) = 2 Γ 0 β 5

= 0 β 5

= β5

(ii) f(7) = 2 Γ 7 β 5

= 14 β 5

= 9

(iii) f(β3) = 2 Γ (β3) β 5

= β 6 β 5

= β11

**The function βtβ which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by**

**π(πΆ) = ^{9}^{πΆ}/_{5} + 32. Find**

**(i) t (0)**

**(ii) t (28)**

**(iii) t (β10)**

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

Solution:

The given function is π(πΆ) = 9πΆ 5 + 32. Therefore,

(i) t(0) = ^{9×0}/_{5} + 32 = 0 + 32 = 32

(ii) t(28) = ^{9×28}/_{5} + 32 = ^{252+160}/_{5} = ^{412}/_{5}

(iii) t(-10) = ^{9x(-10)}/_{5} + 32 = 9(-2)+32 = -18+32 = 14

(iv) it is given that t(c) = 212

212 = ^{9c}/_{5} + 32

^{9c}/_{5} = 212 β 32

^{9c}/_{5} = 180

9C = 180 x 5

C = ^{180×5}/_{9} = 100

Thus the value of t, when tΒ© = 212, is 100

**Find the range of each of the following functions**

**(i) f(x) = 2 β 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 β 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 | β¦ |

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

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

i.e., range of f = (ββ , 2)