Question

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

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

Answer 1

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

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

first solve $1+x^2+x^4$

$=x^4+x^2++1+x^2-x^2$

$=x^4+(1{)}^2+2x^2-x^2$

$=(x^2+1{)}^2-x^2$

$=(x^2+1+x)(x^2+1-x)$

Put it in the given question

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

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

$=2x-1$

Compare it with $ax+b$

Then $a=2,b=-1$

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