Question

Which of the following is/are true for $f\left(x\right)=\frac{1-x+x^2}{1+x+x^2}(x\in R)$

• A. Minimum value of $f\left(x\right)$ is $1$
• B. Maximum value of $f\left(x\right)$ is $3$
• C. $x=1$ is point of minima
• D. $x=-1$ is point of maxima

Answer 1

Answer: (B), (C), (D)

$x=-1$ is point of maxima

$f\left(x\right)=\frac{1-x+x^2}{1+x+x^2}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{(1+x+x^2)(-1+2x)-(1-x+x^2)(1+2x)}{(1+x+x^2{)}^2}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{2(x+1)(x-1)}{(1+x+x^2{)}^2}$
$f^{\prime }\left(x\right)=0$
$\Rightarrow x=-1,x=1$
$f^{\prime }\left(x\right)$ changes from positive to negative at $x=-1$
$\Rightarrow x=-1$ is point of maxima
$f^{\prime }\left(x\right)$ changes from negative to positive at $x=1$
$\Rightarrow x=1$ is point of minima
$f_{max}=f(-1)=3$
$f_{min}=f\left(1\right)=\frac{1}{3}$