Menu Close

Question-132935

Tinku Tara 0 Comments
Pin




Question Number 132935 by liberty last updated on 17/Feb/21
Answered by EDWIN88 last updated on 17/Feb/21
let x=2 ⇒ f(2).f((1/2))=f(2)+f((1/2)) ⇒ 65.f((1/2)) = 65+f((1/2)) ⇒ f((1/2)) = ((65)/(64)) ⇒ f((1/2))= 1+(1/(64)) ⇒f((1/2)) = 1+(1/2^6 ) ⇒f(x)=1+x^6 then f(−3)=1+(−3)^6 = 730
$$\:\mathrm{let}\:\mathrm{x}=\mathrm{2}\:\Rightarrow\:\mathrm{f}\left(\mathrm{2}\right).\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{f}\left(\mathrm{2}\right)+\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\Rightarrow\:\mathrm{65}.\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:=\:\mathrm{65}+\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\Rightarrow\:\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:=\:\frac{\mathrm{65}}{\mathrm{64}}\:\Rightarrow\:\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{64}} \\ $$$$\Rightarrow\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:=\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{6}} }\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)=\mathrm{1}+\mathrm{x}^{\mathrm{6}} \\ $$$$\mathrm{then}\:\mathrm{f}\left(−\mathrm{3}\right)=\mathrm{1}+\left(−\mathrm{3}\right)^{\mathrm{6}} \:=\:\mathrm{730} \\ $$
Answered by Ar Brandon last updated on 17/Feb/21
f(2)=65 therefore f(x) is a non zero polynomial Let f(x)=a_0 +a_1 x+a_2 x^2 +∙∙∙+a_n x^n , a_n ≠0 ⇒(Σ_(r=0) ^n a_r x^r )(Σ_(r=0) ^n (a_r /x^r ))=Σ_(r=0) ^n a_r x^r +Σ_(r=0) ^n (a_r /x^r ) ⇒(Σ_(r=0) ^n a_r x^r )(Σ_(r=0) ^n a_r x^(n−r) )=Σ_(r=0) ^n a_r x^(n+r) +Σ_(r=0) ^n a_r x^(n−r) ⇒(a_0 +a_1 x+a_2 x^2 +∙∙∙+a_n x^n )(a_0 x^n +a_1 x^(n−1) +a_2 x^(n−2) +∙∙∙a_n ) =(a_0 x^n +a_1 x^(n+1) +a_2 x^(n+2) +∙∙∙+a_n x^(2n) )+(a_0 x^n +a_1 x^(n−1) +a_2 x^(n−2) +∙∙∙+a_n ) Comparing coefs we get: a_0 a_n =a_n , a_n ≠0 ⇒a_0 =1 a_0 a_(n−1) +a_1 a_n =a_(n−1) ⇒a_1 =0 a_0 a_(n−2) +a_1 a_(n−1) +a_2 a_n =a_(n−2) ⇒a_2 =0 Continuing this process we get a_1 =a_2 =a_3 =∙∙∙=a_(n−1) =0 For coefs of x^n we have; a_0 ^2 +a_1 ^2 +a_2 ^2 +∙∙∙+a_(n−1) ^2 +a_n ^2 =2a_0 ⇒a_0 ^2 +a_n ^2 =2a_0 ⇒ a_n ^2 =1 ⇒a_n =±1 ⇒f(x)=1±x^n , f(2)=65 ⇒ f(x)=1+x^n ⇒1+2^n =65 ⇒n=6 ⇒f(x)=1+x^6 ⇒f(−3)=1+3^6 =730
$$\mathrm{f}\left(\mathrm{2}\right)=\mathrm{65}\:\mathrm{therefore}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{non}\:\mathrm{zero}\:\mathrm{polynomial} \\ $$$$\mathrm{Let}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{a}_{\mathrm{0}} +\mathrm{a}_{\mathrm{1}} \mathrm{x}+\mathrm{a}_{\mathrm{2}} \mathrm{x}^{\mathrm{2}} +\centerdot\centerdot\centerdot+\mathrm{a}_{\mathrm{n}} \mathrm{x}^{\mathrm{n}} ,\:\mathrm{a}_{\mathrm{n}} \neq\mathrm{0} \\ $$$$\Rightarrow\left(\underset{\mathrm{r}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\mathrm{a}_{\mathrm{r}} \mathrm{x}^{\mathrm{r}} \right)\left(\underset{\mathrm{r}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{a}_{\mathrm{r}} }{\mathrm{x}^{\mathrm{r}} }\right)=\underset{\mathrm{r}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\mathrm{a}_{\mathrm{r}} \mathrm{x}^{\mathrm{r}} +\underset{\mathrm{r}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{a}_{\mathrm{r}} }{\mathrm{x}^{\mathrm{r}} } \\ $$$$\Rightarrow\left(\underset{\mathrm{r}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\mathrm{a}_{\mathrm{r}} \mathrm{x}^{\mathrm{r}} \right)\left(\underset{\mathrm{r}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\mathrm{a}_{\mathrm{r}} \mathrm{x}^{\mathrm{n}−\mathrm{r}} \right)=\underset{\mathrm{r}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\mathrm{a}_{\mathrm{r}} \mathrm{x}^{\mathrm{n}+\mathrm{r}} +\underset{\mathrm{r}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\mathrm{a}_{\mathrm{r}} \mathrm{x}^{\mathrm{n}−\mathrm{r}} \\ $$$$\Rightarrow\left(\mathrm{a}_{\mathrm{0}} +\mathrm{a}_{\mathrm{1}} \mathrm{x}+\mathrm{a}_{\mathrm{2}} \mathrm{x}^{\mathrm{2}} +\centerdot\centerdot\centerdot+\mathrm{a}_{\mathrm{n}} \mathrm{x}^{\mathrm{n}} \right)\left(\mathrm{a}_{\mathrm{0}} \mathrm{x}^{\mathrm{n}} +\mathrm{a}_{\mathrm{1}} \mathrm{x}^{\mathrm{n}−\mathrm{1}} +\mathrm{a}_{\mathrm{2}} \mathrm{x}^{\mathrm{n}−\mathrm{2}} +\centerdot\centerdot\centerdot\mathrm{a}_{\mathrm{n}} \right) \\ $$$$\:\:\:\:\:=\left(\mathrm{a}_{\mathrm{0}} \mathrm{x}^{\mathrm{n}} +\mathrm{a}_{\mathrm{1}} \mathrm{x}^{\mathrm{n}+\mathrm{1}} +\mathrm{a}_{\mathrm{2}} \mathrm{x}^{\mathrm{n}+\mathrm{2}} +\centerdot\centerdot\centerdot+\mathrm{a}_{\mathrm{n}} \mathrm{x}^{\mathrm{2n}} \right)+\left(\mathrm{a}_{\mathrm{0}} \mathrm{x}^{\mathrm{n}} +\mathrm{a}_{\mathrm{1}} \mathrm{x}^{\mathrm{n}−\mathrm{1}} +\mathrm{a}_{\mathrm{2}} \mathrm{x}^{\mathrm{n}−\mathrm{2}} +\centerdot\centerdot\centerdot+\mathrm{a}_{\mathrm{n}} \right) \\ $$$$\mathrm{Comparing}\:\mathrm{coefs}\:\mathrm{we}\:\mathrm{get}: \\ $$$$\mathrm{a}_{\mathrm{0}} \mathrm{a}_{\mathrm{n}} =\mathrm{a}_{\mathrm{n}} ,\:\mathrm{a}_{\mathrm{n}} \neq\mathrm{0}\:\Rightarrow\mathrm{a}_{\mathrm{0}} =\mathrm{1} \\ $$$$\mathrm{a}_{\mathrm{0}} \mathrm{a}_{\mathrm{n}−\mathrm{1}} +\mathrm{a}_{\mathrm{1}} \mathrm{a}_{\mathrm{n}} =\mathrm{a}_{\mathrm{n}−\mathrm{1}} \:\Rightarrow\mathrm{a}_{\mathrm{1}} =\mathrm{0} \\ $$$$\mathrm{a}_{\mathrm{0}} \mathrm{a}_{\mathrm{n}−\mathrm{2}} +\mathrm{a}_{\mathrm{1}} \mathrm{a}_{\mathrm{n}−\mathrm{1}} +\mathrm{a}_{\mathrm{2}} \mathrm{a}_{\mathrm{n}} =\mathrm{a}_{\mathrm{n}−\mathrm{2}} \:\Rightarrow\mathrm{a}_{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{Continuing}\:\mathrm{this}\:\mathrm{process}\:\mathrm{we}\:\mathrm{get}\:\mathrm{a}_{\mathrm{1}} =\mathrm{a}_{\mathrm{2}} =\mathrm{a}_{\mathrm{3}} =\centerdot\centerdot\centerdot=\mathrm{a}_{\mathrm{n}−\mathrm{1}} =\mathrm{0} \\ $$$$\mathrm{For}\:\mathrm{coefs}\:\mathrm{of}\:\mathrm{x}^{\mathrm{n}} \:\mathrm{we}\:\mathrm{have}; \\ $$$$\mathrm{a}_{\mathrm{0}} ^{\mathrm{2}} +\mathrm{a}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{a}_{\mathrm{2}} ^{\mathrm{2}} +\centerdot\centerdot\centerdot+\mathrm{a}_{\mathrm{n}−\mathrm{1}} ^{\mathrm{2}} +\mathrm{a}_{\mathrm{n}} ^{\mathrm{2}} =\mathrm{2a}_{\mathrm{0}} \\ $$$$\Rightarrow\mathrm{a}_{\mathrm{0}} ^{\mathrm{2}} +\mathrm{a}_{\mathrm{n}} ^{\mathrm{2}} =\mathrm{2a}_{\mathrm{0}} \:\Rightarrow\:\mathrm{a}_{\mathrm{n}} ^{\mathrm{2}} =\mathrm{1}\:\Rightarrow\mathrm{a}_{\mathrm{n}} =\pm\mathrm{1} \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{x}\right)=\mathrm{1}\pm\mathrm{x}^{\mathrm{n}} ,\:\mathrm{f}\left(\mathrm{2}\right)=\mathrm{65}\:\Rightarrow\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{1}+\mathrm{x}^{\mathrm{n}} \\ $$$$\Rightarrow\mathrm{1}+\mathrm{2}^{\mathrm{n}} =\mathrm{65}\:\Rightarrow\mathrm{n}=\mathrm{6}\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)=\mathrm{1}+\mathrm{x}^{\mathrm{6}} \\ $$$$\Rightarrow\mathrm{f}\left(−\mathrm{3}\right)=\mathrm{1}+\mathrm{3}^{\mathrm{6}} =\mathrm{730} \\ $$
Commented by Ar Brandon last updated on 17/Feb/21
What level ?
Commented by Dwaipayan Shikari last updated on 17/Feb/21
When you have added?
$${When}\:{you}\:{have}\:{added}? \\ $$
Commented by Dwaipayan Shikari last updated on 17/Feb/21
A year in your life?
$${A}\:{year}\:{in}\:{your}\:{life}? \\ $$
Commented by Ar Brandon last updated on 17/Feb/21
Haha ! Let's get inbox, bro. You'll tell me what you're talking about
Commented by Dwaipayan Shikari last updated on 17/Feb/21
What is your birthday? I know it has gone
$${What}\:{is}\:{your}\:{birthday}?\:{I}\:{know}\:{it}\:{has}\:{gone} \\ $$
Commented by Ar Brandon last updated on 18/Feb/21
13/01
Commented by Ar Brandon last updated on 18/Feb/21
When is that ? ��
Commented by Dwaipayan Shikari last updated on 20/Feb/21
(((5^2 −1)/4)+(1/8)(((√(5^2 −1))/(2.1!)))^2 +(1/(12))(((1.3(√(5^2 −1)))/(2^2 .2!)))^2 +(1/(16))(((1.3.5(√(5^2 −1)))/(2^3 .3!)))^2 +....)(∫_(−∞) ^∞ ((sinx)/x)dx)^2
$$\left(\frac{\mathrm{5}^{\mathrm{2}} −\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{8}}\left(\frac{\sqrt{\mathrm{5}^{\mathrm{2}} −\mathrm{1}}}{\mathrm{2}.\mathrm{1}!}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{12}}\left(\frac{\mathrm{1}.\mathrm{3}\sqrt{\mathrm{5}^{\mathrm{2}} −\mathrm{1}}}{\mathrm{2}^{\mathrm{2}} .\mathrm{2}!}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{16}}\left(\frac{\mathrm{1}.\mathrm{3}.\mathrm{5}\sqrt{\mathrm{5}^{\mathrm{2}} −\mathrm{1}}}{\mathrm{2}^{\mathrm{3}} .\mathrm{3}!}\right)^{\mathrm{2}} +….\right)\left(\int_{−\infty} ^{\infty} \frac{{sinx}}{{x}}{dx}\right)^{\mathrm{2}} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *