Question

The value of $\alpha$ such that :
$\sf \: \displaystyle\int\limits^{ \alpha + 1}_{ \alpha } \dfrac{dx}{(x + \alpha )(x + \alpha + 1)} = log_{e}( \dfrac{9}{8} )$
1. 1/2
2. 2
3. -1/2
4. -2​

Answer 1

Answer:-

$\displaystyle \small\int \limits^{ \alpha + 1}_{ \alpha } \sf\frac{dx}{(x + \alpha )(x + \alpha + 1)}$

$\displaystyle = \int \limits^{ \alpha + 1}_{ \alpha } \bigg[ \sf \frac{1}{x + \alpha } - \frac{1}{x + \alpha + 1} \bigg] \sf dx$

$\displaystyle = \ln \bigg( \sf\frac{x + \alpha }{x + \alpha + 1} \bigg) \bigg|^{ \small\alpha + 1}_{ \small \alpha }$

$\displaystyle = \ln \bigg ( \sf \frac{2 \alpha + 1}{2 \alpha + 2} \times \frac{2 \alpha + 1}{2 \alpha } \bigg)$

$\displaystyle = \ln \bigg( \sf\frac{9}{8} \bigg)$

$\implies \displaystyle \frac{ \sf{(2 \alpha + 1)}^{2} }{4 \alpha ( \alpha + 1)} = \frac{9}{8}$

$\implies \displaystyle \frac{ {(2 \alpha + 1)}^{2} }{ \alpha ( \alpha + 1)} = \frac{9}{2}$

$\implies \displaystyle \sf\frac{4 { \alpha }^{2} + 4\alpha + 1 }{ \alpha^{2} + \alpha } = \frac{9}{2}$

$\implies \displaystyle ( \sf8 { \alpha }^{2} + 8\alpha + 2 ) = 9 { \alpha }^{2} + 9 \alpha$

$\implies \displaystyle { \sf \alpha }^{2} + \alpha - 2 = 0$

On solving the equation, we get,

$\sf \alpha = 1 \: or - 2$

So, Option 4 is the answer for the problem.

Answer 2

Answer:

Answer:-

\displaystyle \small\int \limits^{ \alpha + 1}_{ \alpha } \sf\frac{dx}{(x + \alpha )(x + \alpha + 1)}

α

∫

α+1

(x+α)(x+α+1)

dx

\displaystyle = \int \limits^{ \alpha + 1}_{ \alpha } \bigg[ \sf \frac{1}{x + \alpha } - \frac{1}{x + \alpha + 1} \bigg] \sf dx=

α

∫

α+1

[

x+α

1

−

x+α+1

1

]dx

\displaystyle = \ln \bigg( \sf\frac{x + \alpha }{x + \alpha + 1} \bigg) \bigg|^{ \small\alpha + 1}_{ \small \alpha }=ln(

x+α+1

x+α

)

∣

∣

∣

∣

∣

α

α+1

\displaystyle = \ln \bigg ( \sf \frac{2 \alpha + 1}{2 \alpha + 2} \times \frac{2 \alpha + 1}{2 \alpha } \bigg)=ln(

2α+2

2α+1

×

2α

2α+1

)

\displaystyle = \ln \bigg( \sf\frac{9}{8} \bigg)=ln(

8

9

)

\implies \displaystyle \frac{ \sf{(2 \alpha + 1)}^{2} }{4 \alpha ( \alpha + 1)} = \frac{9}{8}⟹

4α(α+1)

(2α+1)

2

=

8

9

\implies \displaystyle \frac{ {(2 \alpha + 1)}^{2} }{ \alpha ( \alpha + 1)} = \frac{9}{2}⟹

α(α+1)

(2α+1)

2

=

2

9

\implies \displaystyle \sf\frac{4 { \alpha }^{2} + 4\alpha + 1 }{ \alpha^{2} + \alpha } = \frac{9}{2}⟹

α

2

+α

4α

2

+4α+1

=

2

9

\implies \displaystyle ( \sf8 { \alpha }^{2} + 8\alpha + 2 ) = 9 { \alpha }^{2} + 9 \alpha⟹(8α

2

+8α+2)=9α

2

+9α

\implies \displaystyle { \sf \alpha }^{2} + \alpha - 2 = 0⟹α

2

+α−2=0

On solving the equation, we get,

\sf \alpha = 1 \: or - 2α=1or−2

So, Option 4 is the answer for the problem.