Let-a-b-are-positive-real-numbers-such-that-a-b-10-then-the-smallest-value-of-the-constant-k-for-which-x-2-ax-x-2-bx-lt-k-for-all-x-gt-0-is-

Question Number 33940 by rahul 19 last updated on 28/Apr/18

L e t a, b a r e p o s i t i v e r e a l n u m b e r s s u c h

t h a t a - b = 10, t h e n t h e s m a l l e s t v a l u e

o f t h e c o n s t a n t k f o r w h i c h

\sqrt{(x^{2} + a x)} - \sqrt{(x^{2} + b x)} < k f o r a l l x > 0

i s ?

Answered by MJS last updated on 28/Apr/18

a = r + 5

b = r - 5

\sqrt{x^{2} + r x + 5 x} - \sqrt{x^{2} + r x - 5 x} < k

we have to show that

\lim_{x \to \infty} \sqrt{x^{2} + r x + 5 x} - \sqrt{x^{2} + r x - 5 x} = 5

sorry I have no time right now \dots

\sqrt{x^{2} + r x + 5 x} - \sqrt{x^{2} + r x - 5 x} = k (I .)

[x = y \Rightarrow \frac{1}{x} = \frac{1}{y}]

\frac{1}{\sqrt{x^{2} + r x + 5 x} - \sqrt{x^{2} + r x - 5 x}} = \frac{1}{k}

[\frac{1}{u - v} = \frac{u + v}{(u - v) (u + v)} = \frac{u + v}{u^{2} - v^{2}}]

\frac{\sqrt{x^{2} + r x + 5 x} + \sqrt{x^{2} + r x - 5 x}}{x^{2} + r x + 5 x - (x^{2} + r x - 5 x)} = \frac{1}{k}

\frac{\sqrt{x^{2} + r x + 5 x} + \sqrt{x^{2} + r x - 5 x}}{10 x} = \frac{1}{k}

\sqrt{x^{2} + r x + 5 x} + \sqrt{x^{2} + r x - 5 x} = \frac{10 x}{k} (I I .)

2 \sqrt{x^{2} + r x + 5 x} = k + \frac{10 x}{k} (I . + I I .)

4 (x^{2} + r x + 5 x) = k^{2} + 20 x + \frac{100 x^{2}}{k^{2}}

(4 - \frac{100}{k^{2}}) x^{2} + 4 r x - k^{2} = 0

x^{2} + \frac{k^{2} r}{k^{2} - 25} x - \frac{k^{4}}{4 (k^{2} - 25)} = 0

x = - \frac{k^{2} r}{2 (k^{2} - 25)} \pm \frac{k^{2}}{2 (k^{2} - 25)} \sqrt{k^{2} + r^{2} - 25} =

= \frac{r \pm \sqrt{k^{2} + r^{2} - 25}}{2 (k^{2} - 25)} k^{2}

\frac{1}{x} = \frac{2 (k^{2} - 25)}{r \pm \sqrt{k^{2} + r^{2} - 25}} \times \frac{1}{k^{2}}

x \to \infty \Rightarrow k^{2} - 25 = 0 \Rightarrow k = \pm 5

\sqrt{x^{2} + r x + 5 x} - \sqrt{x^{2} + r x - 5 x} > 0 \Rightarrow k = 5

Commented by rahul 19 last updated on 28/Apr/18

N o problem sir .

btw, h o w y o u c a m e t o k n o w i t s h o u l d

b e e q u a l t o 5 ?

Commented by MJS last updated on 28/Apr/18

this is what seems obvious to me :

\sqrt{x (x + c)} = \sqrt{{(x + \frac{c}{2})}^{2} - \frac{c^{2}}{4}} \underset{x \to \infty}{\approx} x + \frac{c}{2}

\sqrt{x (x + a)} - \sqrt{x (x + b)} \underset{x \to \infty}{\approx} x + \frac{a}{2} - (x + \frac{b}{2}) = \frac{a - b}{2} = \frac{a - (a - 10)}{2} = 5

not sure if this counts as proof \dots

Commented by rahul 19 last updated on 28/Apr/18

T h a n k you sir .

Answered by tanmay.chaudhury50@gmail.com last updated on 28/Apr/18

\sqrt{(x^{2} + a x)} - \sqrt{(x^{2} + b x)} < k

x (a - b) / \sqrt{(x_{}^{2} + a x)} + \sqrt{(x^{2} + b x)} < k

(a - b) / \sqrt{(1 + a / x)} + \sqrt{(1 + b / x)} < k

10 / \sqrt{(1 + a / x)} + \sqrt{(1 + b / x)} < k

f o r s m a l l e s t v a l u e o f k t h e d e n o m i n a t o r s h o u l b e

b e m a x i m u m v a l u e

m a x i m u m v a l u e \sqrt{(1 + a / x)} + \sqrt{(1 + b / x)}

p l s c h e c k t h e q u e s t i o n

Commented by rahul 19 last updated on 28/Apr/18

W o w! n i c e s o l u t i o n s i r!

Answered by tanmay.chaudhury50@gmail.com last updated on 28/Apr/18

f x \to \infty t h e v a l u e o f \sqrt{(1 + a / x)} + \sqrt{(1 + b / x)}

i s 2 s o t h e v a l u e o f k i s 10 / 2 = 5

Commented by rahul 19 last updated on 29/Apr/18

d o u b t :

w e n e e d t o f i n d m a x . v a l u e o f

\sqrt{1 + \frac{a}{x}} + \sqrt{1 + \frac{b}{x}} \to m i n . v a l u e o f x

s o w h y x \to \infty ? ? ? i t s h o u l d b e x \to 0!

Commented by tanmay.chaudhury50@gmail.com last updated on 29/Apr/18

c h e c k t h e q u e s t i o n \dots . w h e n x \to \infty t h e e x p r e s s i o n

h a s m i n v a l u e t h a t i s 2

t h e v a l u e o f k i s 10 / 2 = 5

w h e n x \to 0 t h e e x p r e s s i o n b e c o m e \infty, t h e n

v a l u e o f k i s 10 / \infty = 0

m i n v a l u e o f k = 0 a n d m a x v a l u e o f k = 5

t h e q u e s t i o n p o s t e d b y y o u h a s s o m e e r r o r