Let $y=\frac{x^8+x^4+1}{x^4+x^2+1}$. If $\frac{d y}{d x}=a x^3+b x$. Then,
Answer & explanation
Correct answer: option 2
We have,
$y =\frac{x^8+x^4+1}{x^4+x^2+1}$
$\Rightarrow y =\frac{\left(x^4+1\right)^2-x^4}{x^4+x^2+1} = \frac{\left(x^4+1+x^2\right) \left(x^4+1-x^2\right)} {x^4+x^2+1}$
$\Rightarrow y =x^4-x^2+1$
$\Rightarrow \frac{d y}{d x}=4 x^3-2 x $
$\Rightarrow a x^3+b x=4 x^3-2 x$ for all x
$\Rightarrow a=4, b=-2$