Question

If $a,b$ are positive real numbers such that $a-b=2$, then find the smallest value of the constant $L$ for which $\sqrt{x^2+ax}-\sqrt{x^2+bx}<1$ for all $x>0$.

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is B $1$

Consider the given expression,

$\sqrt{x^2+ax}-\sqrt{x^2+bx}$

Multiply and divide by $\sqrt{x^2+ax}+\sqrt{x^2+bx}$

$=\frac{(\sqrt{x^2+ax}-\sqrt{x^2+bx})(\sqrt{x^2+ax}+\sqrt{x^2+bx})}{(\sqrt{x^2+ax}+\sqrt{x^2+bx})}$

$=\frac{(\sqrt{x^2+ax}{)}^2-(\sqrt{x^2+bx}{)}^2}{(\sqrt{x^2+ax}+\sqrt{x^2+bx})}$

$=\frac{x^2+ax-x^2-bx}{(\sqrt{x^2+ax}+\sqrt{x^2+bx})}$

$=\frac{x(a-b)}{(\sqrt{x^2+ax}+\sqrt{x^2+bx})}$

Given $a-b=2$. Divide Numerator and Denominator by $x$. We get,

$=\frac{\frac{2x}{x}}{\frac{(\sqrt{x^2+ax}+\sqrt{x^2+bx})}{x}}$

$=\frac{2}{\sqrt{\frac{x^2+ax}{x^2}}+\sqrt{\frac{x^2+bx}{x^2}}}$

$=\frac{2}{\sqrt{1+\frac{a}{x}}+\sqrt{1+\frac{b}{x}}}$ --------- (1)

Given, (1) $<L$ for all $x>0$

For (1) to be maximum, its denominator should be minimum. Which means $\frac{a}{x}and\frac{b}{x}$ should be minimum.

So, lets take ${lim}_{x->\infty }$

$={lim}_{x->\infty }\frac{2}{\sqrt{1+\frac{a}{x}}+\sqrt{1+\frac{b}{x}}}$

Consider, $\frac{a}{x}=\frac{a}{\infty }$ and $\frac{b}{x}=\frac{b}{\infty }$

These two are slightly greater than $0$, lets denote $0^{+}$

$\therefore \sqrt{1+\frac{a}{x}}$ becomes $\sqrt{1+0^{+}}$

and $\sqrt{1+\frac{b}{x}}$ becomes $\sqrt{1+0^{+}}$

$=\frac{2}{\sqrt{1+0^{+}}+\sqrt{1+0^{+}}}$

On simplifying, we get the value slightly less than $1$

Hence, the smallest value of $L$ is $1$