Question

A rod of fixed length k has its ends sliding along the co-ordinate axes in $1^{st}$ quadrant such that rod meets the x-axis at $A(a,0)$ and y-axis at $B(0,b)$ then the minimum value of ${(a+\frac{1}{b})}^2+{(b+\frac{1}{a})}^2$ is equal to

• A. $k^2+2+\frac{4}{k^2}$
• B. ${(k+\frac{2}{k})}^2$
• C. Minimum value of ${(a+\frac{1}{a})}^2+{(b+\frac{1}{b})}^2$
• D. 8

Answer 1

Answer: (B), (C)

The correct options are
B ${(k+\frac{2}{k})}^2$
C Minimum value of ${(a+\frac{1}{a})}^2+{(b+\frac{1}{b})}^2$

Clearly $a^2+b^2=k^2$
Again ${(a+\frac{1}{b})}^2+{(b+\frac{1}{a})}^2=a^2+b^2+\frac{1}{a^2}+\frac{1}{b^2}+2(\frac{a}{b}+\frac{b}{a})$
Using $AM\ge HM$

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