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=


ANSWER

$$7$$


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
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