Question

For all real values of x, the minimum value of $\frac{1-x+x^2}{1+x+x^2}$ is

a) zero b)1 c) 3 d) $\frac{1}{3}$

Answer 1

Let $y=\frac{1-x+x^2}{1+x+x^2}$
On differentiating w.r.t. x, we get
$\frac{dy}{dx}=\frac{(1+x+x^2)(-1+2x)-(1-x+x^2)(1+2x)}{(1+x+x^2{)}^2}=\frac{-1+2x-x+2x^2-x^2+2x^3-1-2x+x+2x^2-x^2-2x^3}{(1+x+x^2{)}^2}=\frac{-2+2x^2-2x^2+2x^2}{(1+x+x^2{)}^2}=\frac{2x^2-2}{(1+x+x^2)}=\frac{2(x^2-1)}{(1+x+x^2{)}^2}$
Put $\frac{dy}{dx}=0\Rightarrow x^2=1\Rightarrow x=\pm 1$
Now, $\frac{d^{2}y}{d{x}^2}=\frac{2(1+x+x^2{)}^2(2x)-(x^2-1\left)\right(2\left)\right(1+x+x^2\left)\right(1+2x)}{(1+x+x^2{)}^4}$
$=\frac{4(1+x+x^2)(1+x+x^2)x-(x^2-1)(1+2x)}{(1+x+x^2{)}^4}=\frac{4[x+x^2+x^3-x^2-2x^3+1+2x]}{(1+x+x^2)}=\frac{4(1+3x-x^3)}{(1+x+x^2{)}^3}$
At x=1, ${\left(\frac{d^{2}y}{d{x}^2}\right)}_{x=1}=\frac{4(1+3(1)-1^3)}{(1+1+1^2{)}^3}=\frac{4\left(3\right)}{3^3}=\frac{4}{9}>0$
At x=-1, ${\left(\frac{d^{2}y}{d{x}^2}\right)}_{x=1}=\frac{4(1+3(-1)-(-1){)}^3}{[1+(-1)+(-1{)}^2{]}^3}=\frac{4(1-3+1)}{(1-1+1{)}^3}=4(-1)=-4<0$
$\therefore$ By second derivative test, f is minimum at x=1 and the minimum value is given by $y=\frac{1-1+1}{1+1+1}=\frac{1}{3}$
Hence, (d) is the correct option.