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

Answer: (C)

$(2,-1)$

Given,

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

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

$=\frac{(4x^3+2x)(1+x+x^2)-(2x+1)(1+x^2+x^4)}{{(1+x+x^2)}^2}$

$=\frac{2x^5+3x^4+4x^3+x^2-1}{{(1+x+x^2)}^2}$

$=\frac{2x^5+3x^4+4x^3+x^2-1}{x^4+2x^3+3x^2+2x+1}$

$=2x+\frac{-x^4-2x^3-3x^2-2x-1}{x^4+2x^3+3x^2+2x+1}$

$=2x-1$

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

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