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Basic of Logarithms

easy

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

A

$A = 1,\,B = 0,\,C = 2$

B

$A = 1,\,B = 0,\,C = - 2$

C

$A = - 1,\,B = 0,\,C = - 2$

D

None of these

Solution

(b) $A({x^2} + 1) + x(Bx + C) = {(x – 1)^2}$

For $x = i,\, – B + Ci = – 2i$ $ \Rightarrow $ $B = 0,\,C = – 2$

Equating coefficient of ${x^2}$,

$A + B = 1 \Rightarrow A = 1 – B = 1 – 0 = 1$;

$\therefore A = 1,\,B = 0,\,C = – 2$.

Std 11

Mathematics

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