Question

If $y=\frac{1+x^2+x^4}{1+x+x^2}$ and $\frac{dy}{dx}=ax+b$, then the values of $a$ and $b$ are,

• A. $a=2,b=1$
• B. $a=-2,b=1$
• C. $a=2,b=-1$
• D. $a=-2,b=-1$

Answer 1

Answer: (C)

$a=2,b=-1$

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

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

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

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

$\left(\frac{dy}{dx}\right)=\frac{(2x^5+4x^4+6x^3+4x^2+2x)-(x^4+2x^3+3x^2+2x+1)}{(x^4+2x^3+3x^2+2x+1)}$

$\left(\frac{dy}{dx}\right)=\frac{2x[x^4+2x^3+3x^2+2x+1]}{(x^4+2x^3+3x^2+2x+1)}-1$

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

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