Fill in the blanks:

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 :

The correct answer is (a)-(iii), (b)-(iv), (c)-(ii), (d) – (i)

Explanation:

given two functions are

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

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

So the domain of f = {2, 5, 8, 10}

And the domain of g = {2, 7, 8, 10, 11}

(i) f-g

Domain of (f - g) = (Domain of f) ∩ (Domain of g) = {2, 8, 10}

So (f-g) = (f-g)(x) = f(x)-g(x)

When x = 2, (f-g) = (f-g)(2) = f(2)-g(2) = 4-5 = -1⇒ (2, -1)

When x = 8, (f-g) = (f-g)(8) = f(8)-g(8) = -1-4 = -5⇒ (8, -5)

When x = 10, (f-g) = (f-g)(10) = f(10)-g(10) = -3-13 = -16⇒ (10, -16)

So f-g = {(2, -1), (8,-5), (10,-16)}

(ii) f + g

Domain of (f + g) = (Domain of f) ∩ (Domain of g) = {2, 8, 10}

So (f + g) = (f + g)(x) = f(x) + g(x)

When x = 2, (f + g) = (f + g)(2) = f(2) + g(2) = 4 + 5 = 9⇒ (2, 9)

When x = 8, (f + g) = (f + g)(8) = f(8) + g(8) = -1 + 4 = 3⇒ (8, 3)

When x = 10, (f + g) = (f + g)(10) = f(10) + g(10) = -3 + 13 = 10⇒ (10, 10)

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

(iii) f.g

Domain of (f . g) = (Domain of f) ∩ (Domain of g) = {2, 8, 10}

So (f.g) = (f.g)(x) = f(x) g(x)

When x = 2, (f.g) = (f.g)(2) = f(2) g(2) = 4× 5 = 20⇒ (2, 20)

When x = 8, (f.g) = (f.g)(8) = f(8) g(8) = -1× 4 = -4⇒ (8, -4)

When x = 10, (f.g) = (f.g)(10) = f(10) g(10) = -3×13 = -39⇒ (10, -39)

So f.g = {(2, 20), (8,-4), (10,-39)}

(iv)

Domain of = (Domain of f) ∩ (Domain of g) = {2, 8, 10}

So

When x = 2,

When x = 8,

When x = 10,

So