Question

Maximum value of $f\left(x\right)=\frac{x^2-x+1}{x^2+x+1}$ is

• A. $\frac{1}{3}$
• B. $3$
• C. $\frac{3}{7}$
• D. $\frac{7}{3}$

Answer 1

The correct option is B $3$

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

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

$\Rightarrow f^{\prime }\left(x\right)=-\left[\frac{2(x^2+x+1)-2x(2x+1)}{{(x^2+x+1)}^2}\right]$

$\Rightarrow f^{\prime }\left(x\right)=0$ ( for local maxima )

$\Rightarrow (2x^2+2x+2)-(4x^2+2x)=0$

$2x^2+2x+2-4x^2-2x=0$

$-2x^2+2=0$

$x^2=1$

$\Rightarrow x=\pm 1$

When $x=1,f\left(x\right)=\frac{1}{3}$

$\Rightarrow x=-1,f\left(x\right)=3$ ( maxima )

Hence, the answer is $3.$