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Basic of Logarithms
easy
જો ${{{{(x - 1)}^2}} \over {{x^3} + x}} = {A \over x} + {{Bx + C} \over {{x^2} + 1}}$, તો
A
$A = 1,\,B = 0,\,C = 2$
B
$A = 1,\,B = 0,\,C = - 2$
C
$A = - 1,\,B = 0,\,C = - 2$
D
એકપણ નહીં
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 ${x^2}$,
$A + B = 1 \Rightarrow A = 1 – B = 1 – 0 = 1$;
$\therefore A = 1,\,B = 0,\,C = – 2$.
Standard 11
Mathematics
