If ${{ax - 1} \over {(1 - x + {x^2})\,(2 + x)}} = {x \over {1 - x + {x^2}}} - {1 \over {2 + x}}$, then $a = $
$2$
$3$
$4$
$5$
(b) $ax – 1 = x(2 + x) – (1 – x + {x^2}) = 3x – 1$
$\therefore a = 3$.
If ${9 \over {(x – 1)\,{{(x + 2)}^2}}} = {A \over {x – 1}} + {B \over {x + 2}} + {C \over {{{(x + 2)}^2}}}$ then $A – B – C = $
If ${{2x} \over {{x^3} – 1}} = {A \over {x – 1}} + {{Bx + C} \over {{x^2} + x + 1}}$, then
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
The partial fractions of ${{3x – 1} \over {(1 – x + {x^2})\,(2 + x)}}$ are
If ${{{{(x + 1)}^2}} \over {{x^3} + x}} = {A \over x} + {{Bx + C} \over {{x^2} + 1}}$, then ${\sin ^{ – 1}}\left( {{A \over C}} \right) = $
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