Question

Find the intervals of monotonicity of $f\left(x\right)=\frac{1-x+x^2}{1+x+x^2}$

• A. Increasing in $(-\infty ,-1)\cup (1,\infty )$
• B. Decreasing in $(-1,1)$
• C. Decreasing in $(-2,2)$
• D. Increasing in $(-\infty ,-2)\cup (2,\infty )$

Answer 1

Answer: (A), (B)

Increasing in $(-\infty ,-1)\cup (1,\infty )$
Decreasing in $(-1,1)$

Let $y=\frac{1+x^2+x-2x}{1+x^2+x}=1-\frac{2.x}{1+x+x^2}$
$\therefore \frac{dy}{dx}=-2\frac{[1+x+x^2].1-x(1+2x)}{(1+x+x^2)}$
$\Rightarrow \frac{dy}{dx}=\frac{2(x+1)(x-1)}{(1+x+x^2)}$
$\frac{dy}{dx}=+ive$ for $x<-1$ or$x>1$ and =-ive $for$ $-1<x<1$
Thus $f$ is Increasing in $(-\infty ,-1)(1,\infty )$ and decreasing in $(-1,1)$