Question

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

• A. $\left(-1,2\right)$
• B. $\left(-2,1\right)$
• C. $\left(2,-1\right)$
• D. $\left(1,2\right)$

Answer 1

The correct option is C $\left(2,-1\right)$

$\frac{d}{dx}\left\{\frac{1+x^2+x^4}{1+x+x^2}\right\}=ax+b$

$\frac{d}{dx}\left\{\frac{1+2x^2+x^4-x^2}{1+x+x^2}\right\}=ax+b$

$\Rightarrow \frac{d}{dx}\left\{\frac{{(1+x^2)}^2-x^2}{1+x+x^2}\right\}=ax+b$

$\Rightarrow \frac{d}{dx}\left\{\frac{(1+x^2+x)(1+x^2-x)}{1+x+x^2}\right\}=ax+b$

$\Rightarrow \frac{d}{dx}(1+x^2-x)=ax+b$

$\Rightarrow 2x-1=ax+b$

$\Rightarrow a=2,b=-1$

$\therefore (a,b)=(2,-1)$