Definite Integration Ques 1
- A value of $\alpha$ such that $\int_\alpha^{\alpha+1} \frac{d x}{(x+\alpha)(x+\alpha+1)}=\log _e\left(\frac{9}{8}\right)$ is
(2019 Main, 12 April II)
(a) $-2$
(b) $\frac{1}{2}$
(c) $-\frac{1}{2}$
(d) $2$
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Answer:
Correct Answer: 1.(a)
Solution: (a) Let $I=\int_\alpha^{\alpha+1} \frac{d x}{(x+\alpha)(x+\alpha+1)}$
$= \quad \int_\alpha^{\alpha+1} \frac{(x+\alpha+1)-(x+\alpha)}{(x+\alpha)(x+\alpha+1)} d x$
$= \quad \int_\alpha^{\alpha+1}\left(\frac{1}{x+\alpha}-\frac{1}{x+\alpha+1}\right) d x$
$= \quad \left[\log _e(x+\alpha)-\log _e(x+\alpha+1) _\alpha^{\alpha+1}\right]$
$= \quad \left[\log _e\left(\frac{x+\alpha}{x+\alpha+1}\right)\right] _\alpha^{\alpha+1} $
$= \quad \log _e \frac{2 \alpha+1}{2 \alpha+2}-\log _e \frac{2 \alpha}{2 \alpha+1}$
$= \quad \log _e\left(\frac{2 \alpha+1}{2 \alpha+2} \times \frac{2 \alpha+1}{2 \alpha}\right)=\log _e\left(\frac{9}{8}\right) \quad $ (given)
$\Rightarrow \quad \frac{(2 \alpha+1)^2}{4 \alpha(\alpha+1)}=\frac{9}{8} \Rightarrow \quad 8\left[4 \alpha^2+4 \alpha+1\right]=36\left(\alpha^2+\alpha\right)$
$\Rightarrow \quad 8 \alpha^2+8 \alpha+2=9 \alpha^2+9 \alpha$
$\Rightarrow \quad \alpha^2+\alpha-2=0$
$\Rightarrow \quad(\alpha+2)(\alpha-1)=0$
$\Rightarrow \quad \alpha=1,-2$
From the options we get $\alpha=-2$
BETA
