Question

Let f = {(2, 4), (5, 6), (8, – 1), (10, – 3)}
g = {(2, 5), (7, 1), (8, 4), (10, 13), (11, 5)}
be two real functions. Then Match the following :

Column A	Column B
f – g	`{(2, 4/5), (8, (-1)/4), (10, (-3)/13)}`
f + g	{(2, 20), (8, −4), (10, −39)}
f . g	{(2, −1), (8, −5), (10, −16)}
`f/g`	{(2, 9), (8, 3), (10, 10)}

Answer 1

Column A	Column B
f – g	{(2, −1), (8, −5), (10, −16)}
f + g	{(2, 9), (8, 3), (10, 10)}
f . g	{(2, 20), (8, −4), (10, −39)}
`f/g`	`{(2, 4/5), (8, (-1)/4), (10, (-3)/13)}`

Explanation:

Given that: f = {(2, 4), (5, 6), (8, – 1), (10, – 3)}

And g = {(2, 5), (7, 1), (8, 4), (10, 13), (11, 5)}

f – g, f + g, f.g, `"f"/"g"` are defined in the domain

(domain of f ∩ domain of g)

⇒ {2, 5, 8, 10} ∩ {2, 7, 8, 10, 11}

⇒ {2, 8, 10}

(i) (f – g)² = f(2) – g(2)

= 4 – 5

= – 1

(f – g)⁸ = f(8) – g(8)

= – 1 – 4

= – 5

(f – g)¹⁰ = f(10) – g(10)

= – 3 – 13

= – 16

∴ (f – g) = {(2, – 1), (8, – 5), (10, – 16)}

(ii) (f + g)² = f(2) + g(2)

= 4 + 5

= 9

(f + g)⁸ = f(8) + g(8)

= – 1 + 4

= 3

(f + g)10 = f(10) + g(10)

= – 3 + 13

= 10

∴ (f + g) = {(2, 9), (8, 3), (10, 10)}

(iii) (f . g)² = f(2) . g(2)

= 4 . 5

= 20

(f . g)⁸ = f(8) . g(8)

= (– 1) . (4)

= – 4

(f . g)² = f(10) . g(10)

= – 3 . 13

= – 39

∴ (f . g) = {(2, 20), (8, – 4), (10, – 39)}

(iv) `(f/g)(2) = (f(2))/(g(2)) = 4/5`

`(f/g)(8) = (f(8))/(g(8)) = (-1)/4`

`(f/g)(10) = (f(10))/(g(10)) = (-3)/3`

∴ `(f/g) = {(2, 4/5), (8, (-1)/4), (10, (-3)/13)}`