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STD 12 - 7.2 definite integral — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 7.2 definite integral1 Mark
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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