Question

The rod of fixed length $k$ slides along the coordinate axes. If it meets the axes at $A(a,0)$ and $B(0,b)$, then the minimum value of ${(a+\frac{1}{a})}^2+{(b+\frac{1}{b})}^2$ is

• A. $0$
• B. $8$
• C. $k^2-4+\frac{4}{k^2}$
• D. $k^2+4+\frac{4}{k^2}$

Answer 1

The correct option is B $8$

since, $A.M.\ge G.M.⟹\frac{a+b}{2}\ge \sqrt{ab}$

let $a=1$ and $b=a^2$

Therefore, $\frac{1+a^2}{2}\ge \sqrt{a^2}$

$⟹\frac{1+a^2}{2}\ge \sqrt{a^2}$

$⟹\frac{1+a^2}{2}\ge a$

$⟹\frac{1+a^2}{a}\ge 2$

$⟹\frac{1}{a}+\frac{a^2}{a}\ge 2$

$⟹a+\frac{1}{a}\ge 2$

squaring on both sides

$⟹(a+\frac{1}{a}{)}^2\ge 4$ ------- (1)

similarly, $(b+\frac{1}{b}{)}^2\ge 4$ --------(2)

adding (1) and (2) we get

$(a+\frac{1}{a}{)}^2+(b+\frac{1}{b}{)}^2\ge 4+4$

$(a+\frac{1}{a}{)}^2+(b+\frac{1}{b}{)}^2\ge 8$

Therefore the minimum value is $8$