Question

The minimum value of $x^2+y^2$, when $x+y=a$, where $a$ is a positive odd integer and $x,y$ are non negative integers is

• A. $\frac{a^2}{2}$
• B. $\frac{a^2+1}{2}$
• C. $\frac{a^2-1}{2}$
• D. $\frac{a^2}{2}-1$

Answer 1

The correct option is B $\frac{a^2+1}{2}$

We know that,
$(x-y{)}^2\ge 0\Rightarrow x^2+y^2\ge 2xy\Rightarrow 2(x^2+y^2)\ge x^2+y^2+2xy\Rightarrow x^2+y^2\ge \frac{(x+y{)}^2}{2}\Rightarrow x^2+y^2\ge \frac{a^2}{2}$
As $x,y$ are integer but $\frac{a^2}{2}$ is non integer so, the next integer value will be $\frac{a^2+1}{2}$
$\therefore$ Minimum value of $x^2+y^2$ is $\frac{a^2+1}{2}$