Discussion Forum

1)

If  $\frac{2x+7}{(x^{2}+4)(x^{2}+9)(x^{2}+16)}= \frac{Ax+1}{x^{2}+4}+\frac{Bx+m}{x^{2}+9}+\frac{Cx+n}{x^{2}+16}$  , then 

$\frac{1}{A}+\frac{1}{B}+\frac{1}{C}$=........

Answer:

Option D

Explanation:

We have,

 $\frac{2x+7}{(x^{2}+4)(x^{2}+9)(x^{2}+16)}= \frac{Ax+1}{x^{2}+4}+\frac{Bx+m}{x^{2}+9}+\frac{Cx+n}{x^{2}+16}$ 

$2x+7= (Ax+l)(x^{2}+9)(x^{2}+16) +(Bx+m)(x^{2}+4)$

$(x^{2}+16)+(Cx+n)(x^{2}+4)(x^{2}+9)$

put x=2i

      4i+7=120Ai+60l  $\Rightarrow \frac{1}{A} =\frac{60}{2}$

 put x=3i

     6(i) +7 =-105 Bi-35m $\Rightarrow \frac{1}{B}=\frac{-35}{2}$

 pit x=4i

 8i+7=336 Ci+84n

$\Rightarrow$     $\frac{1}{C} = \frac{84}{2}$

 $\therefore$     $\frac{1}{A}+\frac{1}{B}+\frac{1}{C}$= $\frac{60-35+84}{2}= \frac{109}{2}$