{{(x - a)(x - b)} over {(x - c)(x - d)}} = {A over {x - c}} - {B over {(x - d)}} + C , then C =

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Basic of Logarithms

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${{(x - a)(x - b)} \over {(x - c)(x - d)}} = {A \over {x - c}} - {B \over {(x - d)}} + C$, then $C =$

A

$5$

B

$4$

C

$3$

D

$1$

Solution

(d) $A(x – d) – B(x – c) + C(x – c)\,(x – d) = (x – a)$$(x – b)$

Equating coefficient of ${x^2},\,C = 1$.

Standard 11

Mathematics

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