if for nonzero x, `af(x) + bf(1/x) =1/x-5`, where `a!=b` then f(2) =A. `(3(2b+3a))/(2(a^(2)-b^(2))`B. `(3(2b-3a))/(2(a^(2)-b^(2))`C. `(3(3a-2b))/(2(a^(2)-b^(2))`D. `(6)/(a+b)`

if for nonzero x, `af(x) + bf(1/x) =1/x-5`, where `a!=b` then f(2) =A.

1.	if for nonzero x, `af(x) + bf(1/x) =1/x-5`, where `a!=b` then f(2) =A. `(3(2b+3a))/(2(a^(2)-b^(2))`B. `(3(2b-3a))/(2(a^(2)-b^(2))`C. `(3(3a-2b))/(2(a^(2)-b^(2))`D. `(6)/(a+b)`
Answer» Correct Answer - B We have, `a f(x)+bf((1)/(x))=(1)/(x)-5" "`…(i) On replacing x by `(1)/(x)`, we get `af((1)/(x))+bf(x)=x-5` `implies bf(x)+af((1)/(x))=x-5" "`……(ii) Multiplying (i) by a (ii) by b and then subtracting, we get `(a^(2)-b^(2))f(x)=((a)/(x)-bx)-5(a-b)` `implies f(x)=(1)/(a^(2)-b^(2))((1)/(x)-bx)-(5)/(a+b)` `implies f(2)=(3(2b-3a))/(2(a^(2)-b^(2))`

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