Q59 / 160 · AP EAMCET 2022 (07 Jul Shift 1) · Mathematics

If $\frac{4 x^3+16 x+7}{\left(x^2+4\right)^2}=\frac{A x+B}{x^2+4}+\frac{C x+D}{\left(x^2+4\right)^2}$, then the number of non-zero values in $A, B, C, D$ is

  • A
    $1$
  • Correct answer
    B
    $2$
  • C
    $3$
  • D
    $4$

Solution

$\frac{4 x^3+16 x+7}{\left(x^2+4\right)^2}=\frac{A x+B}{x^2+4}+\frac{C x+D}{\left(x^2+4\right)^2}$ $=\frac{(A x+B)\left(x^2+4\right)+(C x+D)}{\left(x^2+4\right)^2}$ $\begin{aligned} & \Rightarrow 4 x^3+16 x+7=A x^3+B x^2+(4 A+C) x+(4 B+D) \\ & \Rightarrow \quad A=4, B=0 \Rightarrow 4 A+C=16 \\ & \Rightarrow \quad C=0 \text { and } 4 B+D=7 \Rightarrow D=7\end{aligned}$ $\therefore$ The number of non-zero values in $A, B, C$ and $D$ $=2$

Hence, the correct answer is option: 'B'.

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