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If for nonzero `x ,af(x)+bf(1/x)=1/x-5,`where `a!=b`then `int_1^2f(x)dx=1____` |
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Answer» Correct Answer - `(1)/(a^(2)-b^(2))[a log 2-5a+(7)/(2)b]` Given , of ` (x)+bf (1//x)=(1)/(x)-5` . . . (i) Replacing x by `1//x` in Eq . (i) , we get af `(1//x) +bf (x)=x-5` .. . (ii) On multiplying Eq . (i) by a and Eqs . (ii) by b , we get `a^(2)f(x)+abf(1//x)=a((1)/(x)-5)`. . . (iii) `abf(1//x)+b^(2)f(x)=b(x-5)` ... (iv) On subtracting Eq . (iv) from Eq . (iii) , we get `(a^(2)-b^(2)) f(x)=(a)/(x)-bx-5a+5b` `rArr f(x)=(1)/((a^(2)-b^(2)))((a)/(x)-bx-5a+5b)` `rArr int_(1)^(2)f(x)dx=(1)/((a^(2)-b^(2)))int_(1)^(2)((a)/(x)-bx-5a+5b)dx` `=(1)/((a^(2)-b^(2)))[alog|x|-(b)/(2)x^(2)-5(a-b)x]_(1)^(2)` `=(1)/((a^(2)-b^(2)))[alog2-2b-10(a-b)-alog1+(b)/(2)+5(a-b)]` `=(1)/((a^(2)-b^(2)))[alog 2-5a+(7)/(2)b]` |
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