Definite Integration Question 492
Question: If for non-zero $ x, $
$ af(x)+bf( \frac{1}{x} )=\frac{1}{x}-5, $ where $ a\ne b, $ then $ \int_1^{2}{f(x)dx=} $
[IIT 1996]
Options:
A) $ \frac{1}{(a^{2}+b^{2})}[ a\log 2-5a+\frac{7}{2}b ] $
B) $ \frac{1}{(a^{2}-b^{2})}[ a\log 2-5a+\frac{7}{2}b ] $
C) $ \frac{1}{(a^{2}-b^{2})}[ a\log 2-5a-\frac{7}{2}b ] $
D) $ \frac{1}{(a^{2}+b^{2})}[ a\log 2-5a-\frac{7}{2}b ] $
Show Answer
Answer:
Correct Answer: B
Solution:
$ af(x)+bf( \frac{1}{x} )=\frac{1}{x}-5 $ (for each $ x\ne 0 $ ) …..(i) Replacing x by $ \frac{1}{x} $ in (i), we get $ af( \frac{1}{x} )+bf(x)=x-5 $ ……(ii) Eliminating $ f( \frac{1}{x} ) $ from (i) and (ii), we get $ (a^{2}-b^{2})f(x)=\frac{a}{x}-bx-5a+5b $
therefore $ (a^{2}-b^{2})\int_1^{2}{f(x)}dx=[ ( a\log |x|-\frac{b}{2}x^{2}-5(a-b)x ) ]_1^{2} $
$ =a\log 2-2b-10(a-b)-a\log 1+\frac{b}{2}+5(a-b) $
$ =a\log 2-5a+\frac{7}{2}b $
therefore $ \int_1^{2}{f(x)dx=\frac{1}{a^{2}-b^{2}}[ a\log 2-5a+\frac{7}{2}b ]} $ .
BETA
