{{x + 1} over {(x - 1),(x - 2),(x - 3)}} =

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Basic of Logarithms

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${{x + 1} \over {(x - 1)\,(x - 2)\,(x - 3)}} = $

A

${1 \over {x - 1}} + {3 \over {x - 2}} + {1 \over {x - 3}}$

B

$ - {3 \over {x - 1}} + {1 \over {x - 2}} + {2 \over {x - 3}}$

C

${1 \over {x - 1}} - {3 \over {x - 2}} + {2 \over {x - 3}}$

D

None of these

(IIT-1996)

Solution

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

$ \Rightarrow $$x + 1 = A\,(x – 2)\,(x – 3)\, + B(x – 1)\,(x – 3) + C(x – 1)\,(x – 2)$

Putting $x = 1,\,A = 1$ ; $x=2$ gives $B = – 3$,

For $x = 3,\,C = 2$

$\therefore$ Given expression = ${1 \over {x – 1}} – {3 \over {x – 2}} + {2 \over {x – 3}}$.

Standard 11

Mathematics

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