Question

Let $a , b$ are positive real numbers such that $a - b =10$, then the smallest value of the constant $K$ for which $\sqrt{\left( x ^2+a x\right)}-\sqrt{\left( x ^2+b x\right)}< K$ for all $x >0$, is

• A. 2
• B. 3
• C. 4
• D. 5

Answer 1

Answer: (D)

$f ( x )=\sqrt{ x ^2+ ax }-\sqrt{ x ^2+ bx }$
$\therefore f ( x )=\frac{( a - b )}{\sqrt{1+( a / x )}+\sqrt{1+( b / x )}} $ (on rationalising)
$\therefore f ( x )_{\max }=\frac{ a - b }{2}$
Hence, $k=5$