Let 5 f ( x ) + 4 f ( 1 x ) = 1 x + 3 , ? x 0 . Then 18 1 2 f ( x ) ? d x is equal to [2023]

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Solution

Q.
$Let 5 f (x) + 4 f (\frac{1}{x}) = \frac{1}{x} + 3, x > 0 .$ $Then 18 \int_{1}^{2} f (x) d x is equal to$ [2023]

1 $5 \log_{e} 2 + 3$

2 $10 \log_{e} 2 + 6$

3 $5 \log_{e} 2 - 3$

4 $10 \log_{e} 2 - 6$

Ans.
(4)

$We have, 5 f (x) + 4 f (\frac{1}{x}) = \frac{1}{x} + 3 ...(i)$

$Replace x by \frac{1}{x}, we get 5 f (\frac{1}{x}) + 4 f (x) = x + 3 ...(ii)$

$Solving (i) and (ii), we get 25 f (x) - 16 f (x) = \frac{5}{x} + 15 - 4 x - 12$

$\Rightarrow 9 f (x) = \frac{5}{x} - 4 x + 3$ $\Rightarrow f (x) = \frac{1}{9} {\frac{5}{x} - 4 x + 3}$

$∴$ $18 \int_{1}^{2} f (x) d x = 18 \times \frac{1}{9} \int_{1}^{2} (\frac{5}{x} - 4 x + 3) d x$

$= 2 {[5 \log x - \frac{4 x^{2}}{2} + 3 x]}_{1}^{2}$

$= 2 [5 \log 2 - 6 + 3]$ $= 10 \log 2 - 6$

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