Question

Given a real value x such that $0\le x\le 1$. Find the minimum value of $\frac{x^2}{x+1}+\frac{(1-x{)}^2}{2-x}$.

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $\frac{1}{2}$
• D. None of these

Answer 1

The correct option is A $\frac{1}{3}$

Given equation is $\frac{x^2}{x+1}+\frac{(1-x{)}^2}{2-x}$

On solving given equation we get

$=$$\frac{x^2-x+1}{-x^2+x+2}$

$=\frac{x^2-x-2+3}{-x^2+x+2}$

$=-1-\frac{3}{x^2-x-2}$

min value of $x^2-x-2$ is $\frac{-D^2}{4a}$ which is equal to $\frac{-9}{4}$

Therefore put this value in place of $x^2-x-2$, then we get final value is $\frac{1}{3}$