If {{3x + 4} over {{{(x + 1)}^2}(x - 1)}} = {A over {(x - 1)}} + {B over {(x + 1)}} + {C over {{{(x

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Basic of Logarithms

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If ${{3x + 4} \over {{{(x + 1)}^2}(x - 1)}} = {A \over {(x - 1)}} + {B \over {(x + 1)}} + {C \over {{{(x + 1)}^2}}}$, then $A = $

A

$-0.5$

B

$3.75$

C

$1.75$

D

$-0.25$

Solution

(c) We have, ${{3x + 4} \over {{{(x + 1)}^2}(x – 1)}} = {A \over {(x – 1)}} + {B \over {(x + 1)}} + {C \over {{{(x + 1)}^2}}}$

==> $3x + 4 = A{(x + 1)^2}$+$B(x + 1)\,(x – 1) + C(x – 1)$

Putting $x = 1$, we get $7 = A{(2)^2}$ ==> $A = {7 \over 4}$.

Standard 11

Mathematics

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