Q. 375.0( 1 Vote )

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 :


Answer :

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


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