If _2^e [ 1 x - 1 ( x) ^2 ] ,dx = + 2 , then - Vidyadip

MCQ

If $\int_2^e {\left[ {\frac{1}{{\log x}} - \frac{1}{{{{(\log x)}^2}}}} \right]} \,dx = \alpha + \frac{\beta }{{\log 2}},$ then

✓
$\alpha = e,\,\,\,\beta = - 2$
B
$\alpha = e,\,\,\,\beta = 2$
C
$\alpha = - e,\,\,\,\beta = 2$
D
$\alpha = - e,\,\,\,\beta = - 2$

✓

Answer

Correct option: A.

$\alpha = e,\,\,\,\beta = - 2$

a
(a) $\int_2^e {\left[ {\frac{1}{{\log x}} - \frac{1}{{{{(\log x)}^2}}}} \right]} \,dx = \alpha + \frac{\beta }{{\log 2}}$
$L.H.S.$ $ = \int_2^e {\left[ {\frac{1}{{\log x}} - \frac{1}{{{{(\log x)}^2}}}} \right]} \,dx = \int_2^e {\frac{1}{{\log x}}dx - \int_2^e {\frac{1}{{{{(\log x)}^2}}}} dx} $
$ = \left[ {\left( {\frac{x}{{\log x}}} \right)_2^e - \int_2^e {\left\{ { - \frac{1}{{x{{(\log x)}^2}}}} \right\}x\,\,dx} } \right]\, - \int_2^e {\frac{1}{{{{(\log x)}^2}}}dx} $
= $\left| {\frac{x}{{\log x}}} \right|_2^e = e - \frac{2}{{\log 2}}$
Comparing it with the given value, we get $\alpha = e$, $\beta = - 2$.

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