Q104 / 160 · AP EAMCET 2022 (07 Jul Shift 2) · Mathematics

$\begin{aligned} & \text { If } \frac{d}{d x}\left(\frac{x^2+1}{\left(x^2+5\right)\left(x^2+9\right)}\right) \\ & =\frac{2 x\left(x^2+1\right)}{\left(x^2+5\right)\left(x^2+9\right)}\left[\frac{1}{f(x)}-\frac{1}{g(x)}-\frac{1}{h(x)}\right] \\ & \text { then } 2 h(x)-f(x)-g(x)=\end{aligned}$

  • Correct answer
    A
    $12$
  • B
    $16$
  • C
    $18$
  • D
    $20$

Solution

$\begin{aligned} & y=\frac{x^2+1}{\left(x^2+5\right)\left(x^2+9\right)} \\ \Rightarrow & \log y=\log \left(x^2+1\right)-\log \left(x^2+5\right)-\log \left(x^2+9\right) \\ \Rightarrow & \frac{1}{y} \cdot \frac{d y}{d x}=2 x\left[\frac{1}{x^2+1}-\frac{1}{x^2+5}-\frac{1}{x^2+9}\right] \\ \Rightarrow & \frac{d y}{d x}=\frac{2 x\left(x^2+1\right)}{\left(x^2+5\right)\left(x^2+9\right)}\left[\frac{1}{x^2+1}-\frac{1}{x^2+5}-\frac{1}{x^2+9}\right] \\ \Rightarrow & f(x)=x^2+1, g(x)=x^2+5, h(x)=x^2+9 \\ \therefore & 2 h(x)-g(x)-f(x)=12\end{aligned}$

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

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