જો ${{ax + b} \over {{{(3x + 4)}^2}}} = {1 \over {3x + 4}} - {3 \over {{{(3x + 4)}^2}}}$ તો
$a = 2$
$b = 1$
$a = 3$
$(b)$ અને $(c)$ બંને
(d) $ax + b = (3x + 4) – 3$ $ \Rightarrow $ $a = 3,\,b = 4 – 3 = 1$.
જો ${{{x^2}} \over {({x^2} + {a^2})\,({x^2} + {b^2})}} = k\left( {{{{a^2}} \over {{x^2} + {a^2}}} – {{{b^2}} \over {{x^2} + {b^2}}}} \right)$ તો $k =$
${{x + 1} \over {(x – 1)\,(x – 2)\,(x – 3)}} = $
જો ${{{{\sin }^2}x + 1} \over {2{{\sin }^2}x – 5\sin x + 3}}$=${A \over {(2\sin x – 3)}} + {B \over {(\sin x – 1)}} + C$, તો
જો ${9 \over {(x – 1)\,{{(x + 2)}^2}}} = {A \over {x – 1}} + {B \over {x + 2}} + {C \over {{{(x + 2)}^2}}}$ તો $A – B – C = $
${{{x^2} – 5} \over {{x^2} – 3x + 2}}$ નું આંશિક અપૂર્ણાક મેળવો.
