MCQ
A value of $\alpha $ such that $\int\limits_\alpha ^{\alpha + 1} {\frac{{dx}}{{\left( {x + \alpha } \right)\left( {x + \alpha + 1} \right)}} = {{\log }_e}\left( {\frac{9}{8}} \right)} $ is
- A$-\frac{1}{2}$
- ✓$-2$
- C$\frac{1}{2}$
- D$2$
$\Rightarrow \int_{\alpha}^{\alpha+1} \frac{(x+\alpha+1)-(x+\alpha)}{(x+\alpha)(x+\alpha+1)} d x=\log _{e}\left(\frac{9}{8}\right)$
$\Rightarrow \int_{a}^{a+1} \frac{d x}{x+\alpha}-\int_{a}^{a+1} \frac{d x}{x+\alpha+1}=\log _{e}\left(\frac{9}{8}\right)$
$\left.\Rightarrow \log _{e}\left(\frac{x+\alpha}{x+\alpha+1}\right)\right|_{\alpha} ^{\alpha+1}=\log _{e}\left(\frac{9}{8}\right)$
$\Rightarrow \log _{e}\left(\frac{2 \alpha+1}{2 \alpha+2}\right)-\log \left(\frac{2 \alpha}{2 \alpha+1}\right)=\log _{e}\left(\frac{9}{8}\right.$
$\Rightarrow \log \left[\left(\frac{2 \alpha+1}{2 \alpha+2}\right)\left(\frac{2 \alpha+1}{2 \alpha}\right)\right]=\log _{e} \frac{9}{8}$
$\Rightarrow \frac{(2 \alpha+1)^{2}}{4 \alpha(\alpha+1)}=\frac{9}{8}$
$\Rightarrow 8\left[4 \alpha^{2}+4 \alpha+1\right]=9\left[4 \alpha^{2}+4 \alpha\right]$
$\Rightarrow 32 \alpha^{2}+32 \alpha+8=36 \alpha^{2}+36 \alpha$
$\Rightarrow 4 \alpha^{2}+4 \alpha-8=0$
$\Rightarrow \alpha^{2}+\alpha-2=0$
$=(\alpha+2)(\alpha-1)=0$
$\Rightarrow \alpha=1,-2$
Hence the correct answer is option $(B).$
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