If 2f(x) - 3f ( 1 x ) = x, then _1^2 f(x) ;dx is equal to - Vidyadip

MCQ

If $2f(x) - 3f\left( {\frac{1}{x}} \right) = x$, then $\int_1^2 {f(x)} \;dx$ is equal to

A
$\frac{3}{5}\ln 2$
✓
$\frac{{ - 3}}{5}(1 + \ln 2)$
C
$\frac{{ - 3}}{5}\ln 2$
D
None of these

✓

Answer

Correct option: B.

$\frac{{ - 3}}{5}(1 + \ln 2)$

b
(b) $2f(x) - 3f\left( {\frac{1}{x}} \right) = x$....$(i)$

Replacing $x$ by $\left( {\frac{1}{x}} \right)$ in $(i),$ we get

$2f{\rm{ }}\left( {\frac{1}{x}} \right) - 3f(x) = \frac{1}{x}$......$(ii)$

Eliminating $f{\rm{ }}\left( {\frac{1}{x}} \right)$ from $(i)$ and $(ii),$ we get

$ - 5\;f(x) = 2x + \frac{3}{x} = \frac{{2{x^2} + 3}}{3}$

==> $f(x) = - \,\left( {\frac{{2{x^2} + 3}}{{5x}}} \right)$

$\int_1^2 {f(x)dx = } $$ - \int_1^2 {\left( {\frac{{2{x^2} + 3}}{{5x}}} \right)} \;dx = - \frac{1}{5}[{x^2} + 3{\log _e}x]_1^2$

$ = - \frac{3}{5}[1 + {\log _e}2] = - \frac{3}{5}\left[ {1 + \ln 2} \right]$.

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