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Basic of Logarithms
easy
If ${{{{\sin }^2}x + 1} \over {2{{\sin }^2}x - 5\sin x + 3}}$=${A \over {(2\sin x - 3)}} + {B \over {(\sin x - 1)}} + C$, then
A
$A = {{13} \over 2}$
B
$A + B + C = 5$
C
$C = 1$
D
$(a)$ and $(b)$ both
Solution
(d) ${\sin ^2}x + 1 = A(\sin x – 1) + B(2\sin x – 3)$
$ + C(\sin x – 1)\,(2\sin x – 3)$
$ \Rightarrow $ $1 = 2C \Rightarrow C = {1 \over 2}$
$0 = A + 2B – 5C,\,1 = – A – 3B + 3C$.
$\therefore A = {{13} \over 2}$, $B = – 2$, $C = {1 \over 2}$, $A + B + C = 5$.
Std 11
Mathematics
