Question

If $f\left(x\right)=\left(\frac{x^{12}+x^6+1}{x^6+x^3+1}\right),$ then the derivative of $f\left(x\right)$ w.r.t. $x$ at $x=1$ is

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is D $3$

Given,$f\left(x\right)=\left(\frac{x^{12}+x^6+1}{x^6+x^3+1}\right)$
$x^{12}+x^6+1=(x^6+x^3+1)(x^6-x^3+1)$
$(\because a^4+a^2{b}^2+b^4=(a^2+ab+b^2\left)\right(a^2-ab+b^2\left)\right)$
$\Rightarrow f\left(x\right)=x^6-x^3+1f^{\prime }\left(x\right)=6x^5-3x^2\Rightarrow f^{\prime }\left(1\right)=6-3=3$