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Question Number 58405 by rahul 19 last updated on 22/Apr/19

1) \lim_{x \to 0} \frac{e^{a x} - b x - 1}{x^{2}} = 2 .

f i n d a, b ?

2) 6 b a l l s a r e p l a c e d r a n d o m l y i n t o

6 c e l l s . T h e n t h e p r o b a b i l i t y t h a t e x a c t l y

o n e c e l l r e m a i n s e m p t y i s ?

Commented by maxmathsup by imad last updated on 24/Apr/19

\Rightarrow {l i m}_{x \to 0} \frac{e^{a x} - b x - 1 - 2 x^{2}}{x^{2}} = 0 \Rightarrow f (0) = f^{^{'}} (x) = f^{(2)} (0) = 0

w e h a v e f (0) = 0 (v e r i f i e d) f^{^{'}} (x) = a e^{a x} - b - 4 x

f^{^{'}} (0) = 0 \Rightarrow a - b = 0 \Rightarrow a = b

f^{(2)} (x) = a^{2} e^{a x} - 4 a n d f^{(2)} (0) = 0 \Rightarrow a^{2} - 4 = 0 \Rightarrow a = \bar{+} 2

Commented by maxmathsup by imad last updated on 24/Apr/19

\Rightarrow t h e c o u p l e (a, b) i s (2, 2) o r (- 2, - 2) .

Answered by tanmay last updated on 22/Apr/19

1) \lim_{x \to 0} \frac{(1 + a x + \frac{a^{2} x^{2}}{2!} + \frac{a^{3} x^{3}}{3!} + . .) - b x - 1}{x^{2}}

= \lim_{x \to 0} \frac{x (b - a) + \frac{a^{2} x^{2}}{2} + t e r m s c o n t a i n i n g x^{r}}{x^{2}}

w h e n r > 2

f o e e x i s t e n c e o f l i m i t b - a = 0

\frac{a^{2}}{2} = 2 a = \pm 2 s l s o b \pm 2

Commented by rahul 19 last updated on 23/Apr/19

t h a n k s s i r!

Answered by mr W last updated on 22/Apr/19

(2)

t o p l a c e 6 b a l l s i n t o 6 c e l l s

C_{5}^{11} = 462 w a y s

t o p l a c e 6 b a l l s i n t o 5 c e l l s

5 w a y s

\Rightarrow p = \frac{6 \times 5}{462} = \frac{5}{77}

Commented by rahul 19 last updated on 23/Apr/19

y e s, i t s h o u l d b e a l t h o u g h i^{'} v e t y p e d

t h e w h o l e Q .

Commented by peter frank last updated on 22/Apr/19

t h a n k s

Commented by mr W last updated on 23/Apr/19

i s t h e a n s w e r c o r r e c t ?

Commented by rahul 19 last updated on 23/Apr/19

S i r a n s g i v e n i s \frac{25}{108} .

Commented by mr W last updated on 23/Apr/19

a s s u m e t h e 6 b a l l s a r e i d e n t i c a l a n d

t h e 6 c e l l s a r e d i f f e r e n t . r i g h t ?

Commented by mr W last updated on 23/Apr/19

r a h u l s i r : i c h e c k e d m y s o l u t i o n a n d

f o u n d n o e r r o r .

t o p u t 6 b a l l s i n t o 6 b o x e s (e m p t y b o x e s

a r e a l l o w e d), t h e r e a r e C_{6 - 1}^{6 + 6 - 1} = C_{5}^{11} = 462

w a y s . w h e n e x a c t l y o n e b o x (s a y b o x A)

r e m a i n s e m p t y, t h e r e a r e 5 w a y s t o

p u t t h e 6 b a l l s i n t o t h e o t h e r 5 b o x e s .

s i n c e e a c h o f t h e 6 b o x e s c a n b e t h e

e m p t y o n e, t h e r e a r e t o t a l l y 6 \times 5 = 30

w a y s . t h e p r o b a b i l i t y i s t h e n

\frac{30}{462} = \frac{5}{77} .

Commented by rahul 19 last updated on 23/Apr/19

T h a n k U s o m u c h s i r!

Answered by mr W last updated on 23/Apr/19

(1)

\lim_{x \to 0} \frac{e^{a x} - b x - 1}{x^{2}} (= \frac{0}{0})

= \lim_{x \to 0} \frac{{a e}^{a x} - b}{2 x} (\Rightarrow b = a, o t h e r w i s e \to \infty)

= \lim_{x \to 0} \frac{a^{2} e^{a x}}{2}

= \frac{a^{2}}{2} = 2

\Rightarrow a = \pm 2 = b

Commented by rahul 19 last updated on 23/Apr/19

t h a n k s s i r .