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

1) lim_(x→0) ((e^(ax) −bx−1)/x^2 )=2 . find a,b ? 2) 6 balls are placed randomly into 6 cells. Then the probability that exactly one cell remains empty is ?

$$\left.\mathrm{1}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{e}^{{ax}} −{bx}−\mathrm{1}}{{x}^{\mathrm{2}} }=\mathrm{2}\:. \\ $$$${find}\:{a},{b}\:? \\ $$$$ \\ $$$$\left.\mathrm{2}\right)\:\mathrm{6}\:{balls}\:{are}\:{placed}\:{randomly}\:{into} \\ $$$$\mathrm{6}\:{cells}.\:{Then}\:{the}\:{probability}\:{that}\:{exactly} \\ $$$${one}\:{cell}\:{remains}\:{empty}\:{is}\:? \\ $$

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

⇒lim_(x→0) ((e^(ax) −bx−1−2x^2 )/x^2 ) =0 ⇒f(0)=f^′ (x) =f^((2)) (0) =0 we have f(0) =0(verified) f^′ (x)=a e^(ax) −b−4x f^′ (0) =0 ⇒a−b =0 ⇒a=b f^((2)) (x) =a^2 e^(ax) −4 and f^((2)) (0) =0 ⇒a^2 −4 =0 ⇒a =+^− 2

$$\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} \:\:\frac{{e}^{{ax}} −{bx}−\mathrm{1}−\mathrm{2}{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} }\:=\mathrm{0}\:\Rightarrow{f}\left(\mathrm{0}\right)={f}^{'} \left({x}\right)\:={f}^{\left(\mathrm{2}\right)} \left(\mathrm{0}\right)\:=\mathrm{0} \\ $$$${we}\:{have}\:{f}\left(\mathrm{0}\right)\:=\mathrm{0}\left({verified}\right)\:\:{f}^{'} \left({x}\right)={a}\:{e}^{{ax}} −{b}−\mathrm{4}{x}\: \\ $$$${f}^{'} \left(\mathrm{0}\right)\:=\mathrm{0}\:\Rightarrow{a}−{b}\:=\mathrm{0}\:\Rightarrow{a}={b} \\ $$$${f}^{\left(\mathrm{2}\right)} \left({x}\right)\:={a}^{\mathrm{2}} {e}^{{ax}} −\mathrm{4}\:\:\:{and}\:{f}^{\left(\mathrm{2}\right)} \left(\mathrm{0}\right)\:=\mathrm{0}\:\Rightarrow{a}^{\mathrm{2}} −\mathrm{4}\:=\mathrm{0}\:\Rightarrow{a}\:=\overset{−} {+}\mathrm{2} \\ $$

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

⇒the couple (a,b) is(2,2) or (−2,−2) .

$$\Rightarrow{the}\:{couple}\:\left({a},{b}\right)\:{is}\left(\mathrm{2},\mathrm{2}\right)\:{or}\:\left(−\mathrm{2},−\mathrm{2}\right)\:. \\ $$

Answered by tanmay last updated on 22/Apr/19

1)lim_(x→0) (((1+ax+((a^2 x^2 )/(2!))+((a^3 x^3 )/(3!))+..)−bx−1)/x^2 ) =lim_(x→0) ((x(b−a)+((a^2 x^2 )/2)+terms containing x^r )/x^2 ) when r>2 foe existence of limit b−a=0 (a^2 /2)=2 a=±2 slso b±2

$$\left.\mathrm{1}\right)\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}+{ax}+\frac{{a}^{\mathrm{2}} {x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{a}^{\mathrm{3}} {x}^{\mathrm{3}} }{\mathrm{3}!}+..\right)−{bx}−\mathrm{1}}{{x}^{\mathrm{2}} } \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}\left({b}−{a}\right)+\frac{{a}^{\mathrm{2}} {x}^{\mathrm{2}} }{\mathrm{2}}+{terms}\:{containing}\:{x}^{{r}} }{{x}^{\mathrm{2}} } \\ $$$${when}\:{r}>\mathrm{2} \\ $$$${foe}\:{existence}\:{of}\:{limit}\:\:{b}−{a}=\mathrm{0} \\ $$$$\frac{{a}^{\mathrm{2}} }{\mathrm{2}}=\mathrm{2}\:\:\:{a}=\pm\mathrm{2}\:\:{slso}\:{b}\pm\mathrm{2} \\ $$

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

thanks sir!

$${thanks}\:{sir}! \\ $$

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

(2) to place 6 balls into 6 cells C_5 ^(11) =462 ways to place 6 balls into 5 cells 5 ways ⇒p=((6×5)/(462))=(5/(77))

$$\left(\mathrm{2}\right) \\ $$$${to}\:{place}\:\mathrm{6}\:{balls}\:{into}\:\mathrm{6}\:{cells} \\ $$$${C}_{\mathrm{5}} ^{\mathrm{11}} =\mathrm{462}\:{ways} \\ $$$$ \\ $$$${to}\:{place}\:\mathrm{6}\:{balls}\:{into}\:\mathrm{5}\:{cells} \\ $$$$\mathrm{5}\:{ways} \\ $$$$ \\ $$$$\Rightarrow{p}=\frac{\mathrm{6}×\mathrm{5}}{\mathrm{462}}=\frac{\mathrm{5}}{\mathrm{77}} \\ $$

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

yes , it should be although i′ve typed the whole Q.

$${yes}\:,\:{it}\:{should}\:{be}\:{although}\:{i}'{ve}\:{typed} \\ $$$${the}\:{whole}\:{Q}. \\ $$

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

thanks

$${thanks} \\ $$

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

is the answer correct?

$${is}\:{the}\:{answer}\:{correct}? \\ $$

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

Sir ans given is ((25)/(108)).

$${Sir}\:{ans}\:{given}\:{is}\:\frac{\mathrm{25}}{\mathrm{108}}. \\ $$

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

assume the 6 balls are identical and the 6 cells are different. right?

$${assume}\:{the}\:\mathrm{6}\:{balls}\:{are}\:{identical}\:{and} \\ $$$${the}\:\mathrm{6}\:{cells}\:{are}\:{different}.\:{right}? \\ $$

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

rahul sir: i checked my solution and found no error. to put 6 balls into 6 boxes (empty boxes are allowed), there are C_(6−1) ^(6+6−1) =C_5 ^(11) =462 ways. when exactly one box (say box A) remains empty, there are 5 ways to put the 6 balls into the other 5 boxes. since each of the 6 boxes can be the empty one, there are totally 6×5=30 ways. the probability is then ((30)/(462))=(5/(77)).

$${rahul}\:{sir}:\:{i}\:{checked}\:{my}\:{solution}\:{and} \\ $$$${found}\:{no}\:{error}. \\ $$$${to}\:{put}\:\mathrm{6}\:{balls}\:{into}\:\mathrm{6}\:{boxes}\:\left({empty}\:{boxes}\right. \\ $$$$\left.{are}\:{allowed}\right),\:{there}\:{are}\:{C}_{\mathrm{6}−\mathrm{1}} ^{\mathrm{6}+\mathrm{6}−\mathrm{1}} ={C}_{\mathrm{5}} ^{\mathrm{11}} =\mathrm{462} \\ $$$${ways}.\:{when}\:{exactly}\:{one}\:{box}\:\left({say}\:{box}\:{A}\right) \\ $$$${remains}\:{empty},\:{there}\:{are}\:\mathrm{5}\:{ways}\:{to} \\ $$$${put}\:{the}\:\mathrm{6}\:{balls}\:{into}\:{the}\:{other}\:\mathrm{5}\:{boxes}. \\ $$$${since}\:{each}\:{of}\:{the}\:\mathrm{6}\:{boxes}\:{can}\:{be}\:{the} \\ $$$${empty}\:{one},\:{there}\:{are}\:{totally}\:\mathrm{6}×\mathrm{5}=\mathrm{30} \\ $$$${ways}.\:{the}\:{probability}\:{is}\:{then} \\ $$$$\frac{\mathrm{30}}{\mathrm{462}}=\frac{\mathrm{5}}{\mathrm{77}}. \\ $$

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

Thank U so much sir!

$${Thank}\:{U}\:{so}\:{much}\:{sir}! \\ $$

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

(1) lim_(x→0) ((e^(ax) −bx−1)/x^2 ) (=(0/0)) =lim_(x→0) ((ae^(ax) −b)/(2x)) (⇒b=a, otherwise →∞) =lim_(x→0) ((a^2 e^(ax) )/2) =(a^2 /2)=2 ⇒a=±2=b

$$\left(\mathrm{1}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{e}^{{ax}} −{bx}−\mathrm{1}}{{x}^{\mathrm{2}} }\:\:\:\:\left(=\frac{\mathrm{0}}{\mathrm{0}}\right) \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{ae}^{{ax}} −{b}}{\mathrm{2}{x}}\:\:\:\left(\Rightarrow{b}={a},\:{otherwise}\:\rightarrow\infty\right) \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{a}^{\mathrm{2}} {e}^{{ax}} }{\mathrm{2}} \\ $$$$=\frac{{a}^{\mathrm{2}} }{\mathrm{2}}=\mathrm{2} \\ $$$$\Rightarrow{a}=\pm\mathrm{2}={b} \\ $$

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

thanks sir.

$${thanks}\:{sir}. \\ $$

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