Q103 / 160 · AP EAMCET 2024 (21 May Shift 2) · Mathematics

If $\frac{d}{d x}\left(\frac{1+x^2+x^4}{1+x+x^2}\right)=a x+b$ then $(a, b)=$

  • A
    $(-1,2)$
  • B
    $(-2,1)$
  • Correct answer
    C
    $(2,-1)$
  • D
    $(1,2)$

Solution

$\begin{aligned} & \text { } \frac{1+x^2+x^4}{1+x+x^2}=\frac{\left(1+x+x^2\right)\left(1-x+x^2\right)}{1+x+x^2}=x^2-x+1 \\ & \frac{d}{d x}\left(\frac{1+x^2+x^4}{1+x+x^2}\right)=2 x-1 \Rightarrow a=2, b=-1\end{aligned}$

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

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