QJEEM19Macb90: JEE Main 2019 Mathematics - Class 12 JEEM19Macb90 Solution: A value of such that is :

Question

Verified Solution • 2025–26 Session moderate

A value of

\alpha

such that

\int\limits_\alpha ^{\alpha + 1} {{{dx} \over {\left( {x + \alpha } \right)\left( {x + \alpha + 1} \right)}}} = {\log _e}\left( {{9 \over 8}} \right)

is :

Accepted Answer

I = \int\limits_\alpha ^{\alpha + 1} {{{dx} \over {(x + \alpha )(x + \alpha + 1)}}}

Let x + $\alpha$ = t and dx = dt

I = \int\limits_{2\alpha }^{2\alpha + 1} {{{dt} \over {t(t + 1)}}} = \int\limits_{2\alpha }^{2\alpha + 1} {\left( {{1 \over t} - {1 \over {t + 1}}} \right)dx}

\Rightarrow \left( {\ln t - \ln (t + 1)} \right)|_{2\alpha }^{2\alpha + 1} = \ln \left( {{t \over {t + 1}}} \right)|_{2\alpha }^{2\alpha + 1}

\Rightarrow \ln \left( {{{2\alpha + 1} \over {2\alpha + 2}}} \right) - \ln \left( {{{2\alpha } \over {2\alpha + 1}}} \right)

\Rightarrow \ln \left( {{{2\alpha + 1} \over {2\alpha + 2}} \times {{2\alpha + 1} \over {2\alpha }}} \right) = \ln \left( {{9 \over 8}} \right)

\Rightarrow {{{{\left( {2\alpha + 1} \right)}^2}} \over {\left( {2\alpha + 2} \right)}} = {{9\alpha } \over 4}

4(4{\alpha ^2} + 1 + 4\alpha ) = 18{\alpha ^2} + 18\alpha

\Rightarrow 2{\alpha ^2} - 2\alpha - 4 = 0

\Rightarrow {\alpha ^2} + \alpha - 2 = 0

\Rightarrow (\alpha + 2)(\alpha - 1) = 0

\alpha

= 1 or

\alpha

= -2

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