Mathematics

If $\displaystyle \frac{(e^x+2)}{(e^x-1)(2e^x-3)} = \frac{-3}{e^x-1}+\frac{B}{2e^x-3}$ then B=

$7$

SOLUTION
Given, $\displaystyle \frac{(e^x+2)}{(e^x-1)(2e^x-3)} = \frac{3}{e^x-1}+\frac{B}{2e^x-3}$         ....(1)
Consider, $\displaystyle \frac { (e^{ x }+2) }{ (e^{ x }-1)(2e^{ x }-3) }$
Let $e^{x}=t$ and resolving into partial fractions
$\displaystyle \frac { (t+2) }{ (t-1)(2t-3) } =\frac { A }{ t-1 } +\frac { B }{ 2t-3 }$     ....(2)
$t+2=A(2t-3)+B(t-1)$
$\Rightarrow 2A+B=1, -3A-B=2$
$\Rightarrow A=-3, B=7$
Put this value in (2)
$\displaystyle \frac { (t+2) }{ (t-1)(2t-3) } =\frac { -3 }{ t-1 } +\frac { 7 }{ 2t-3 }$
$\displaystyle \frac { (e^x+2) }{ (e^x-1)(2e^x-3) } =\frac { -3 }{ e^x-1 } +\frac { 7 }{ 2e^x-3 }$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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