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Basic of Logarithms
easy
${1 \over {x({x^2} + 1)}} = {A \over x} + {{Bx + C} \over {({x^2} + 1)}}$, then $(A,\,B,\,C) = $
A
$(1, \,-1,\, 0)$
B
$( - 1,\,0,\, - 1)$
C
$(0,\,1,\,1)$
D
None of these
(IIT-1995)
Solution
(a) $A({x^2} + 1) + (Bx + C)x = 1$
For $x = 0,\,A = 1$ and for $x = i$, $ – B + Ci = 1$
$ \Rightarrow B = – 1,\,C = 0$ $ \Rightarrow $ $(A,B,C) = (1,\, – 1,\,0)$.
Std 11
Mathematics
