Question

Given that $\frac{d}{dx}\left(\frac{1+x^2+x^4}{1+x+x^2}\right)=Ax+B$.

What is the value of $B$?

• A. $-1$
• B. $1$
• C. $2$
• D. $4$

Answer 1

The correct option is A $-1$

Given $\frac{d}{dx}\left(\frac{1+x^2+x^4}{1+x+x^2}\right)=Ax+B$ ..... $\left(i\right)$

Now, $\frac{d}{dx}\left(\frac{x^4+x^2+1}{x^2+x+1}\right)=\frac{d}{dx}\left(\frac{{(x}^2+x+1){(x}^2-x+1)}{x^2+x+1}\right)$

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

$=2x-1$

$\Rightarrow Ax+B=2x-1$ ..... From $\left(i\right)$

Hence, $B=-1$.