જો {9 over {(x - 1),{{(x + 2)}^2}}} = {A over {x - 1}} + {B over {x + 2}} + {C over {{{(x + 2)}^

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Basic of Logarithms

easy

જો ${9 \over {(x - 1)\,{{(x + 2)}^2}}} = {A \over {x - 1}} + {B \over {x + 2}} + {C \over {{{(x + 2)}^2}}}$ તો $A - B - C = $

A

$3$

B

$-1$

C

$5$

D

એકપણ નહીં

Solution

(c) $9 = A{(x + 2)^2} + B(x – 1)\,(x + 2)\, + C\,(x – 1)$

For $x = 1,\,\,9 = 9A \Rightarrow A = 1$

For $x = – 2,\,9 = – 3C \Rightarrow C = – 3$

Equating coefficient of ${x^2},\,0 = A + B \Rightarrow B = – A = – 1$

$\therefore A – B – C = 1 – ( – 1) – ( – 3) = 1 + 1 + 3 = 5$.

Standard 11

Mathematics

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