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Question Number 182721 by mnjuly1970 last updated on 13/Dec/22

Answered by Acem last updated on 13/Dec/22

It period: T= (1/a) f(x)=1 ⇒ b=1 lim_(x→((15^( −) )/2)) f(x)= −∞ ⇒ cos πax= cos (π^( +) /2) ⇒ a<0 π a x= π/2 ± 2πk when x= ((15)/2) : a= (1/(15)) ± (4/(15)) k^(am at problem of k) this term should be negative cause a< 0 so for k=1 and with minus mark_( But why not 2, 3, ..?) a= (1/(15)) −(4/(15)) ⇒ a= ((−1)/5) f(x)= tan ((−π)/5) x +1 _(continue...) f(x)= 0 ⇒ tan (π/(15)) x = −1 = tan (((3 π)/4) +πk) According to the diagram k=0 , k=1 x_A = ((45)/4) , x_B = ((105)/4) ⇒ Σ= ((75)/2) ≠ {a, b, c, d}

$$\:{It}\:{period}:\:{T}=\:\frac{\mathrm{1}}{{a}} \\ $$$$\:{f}\left({x}\right)=\mathrm{1}\:\Rightarrow\:{b}=\mathrm{1} \\ $$$$\underset{{x}\rightarrow\frac{\mathrm{15}^{\:−} }{\mathrm{2}}} {\mathrm{lim}}\:{f}\left({x}\right)=\:−\infty\:\Rightarrow\:\mathrm{cos}\:\pi{ax}=\:\mathrm{cos}\:\frac{\pi^{\:+} }{\mathrm{2}}\:\Rightarrow\:{a}<\mathrm{0} \\ $$$$\:\pi\:{a}\:{x}=\:\pi/\mathrm{2}\:\pm\:\mathrm{2}\pi{k} \\ $$$$\:{when}\:{x}=\:\frac{\mathrm{15}}{\mathrm{2}}\::\:{a}=\:\frac{\mathrm{1}}{\mathrm{15}}\:\underbrace{\pm\:\frac{\mathrm{4}}{\mathrm{15}}\:{k}}\:^{{am}\:{at}\:{problem}\:{of}\:{k}} \\ $$$$\:{this}\:{term}\:{should}\:{be}\:{negative}\:{cause}\:{a}<\:\mathrm{0} \\ $$$$\:{so}\:{for}\:{k}=\mathrm{1}\:{and}\:{with}\:{minus}\:{mark}_{\:{But}\:{why}\:{not}\:\mathrm{2},\:\mathrm{3},\:..?} \\ $$$$\:{a}=\:\frac{\mathrm{1}}{\mathrm{15}}\:−\frac{\mathrm{4}}{\mathrm{15}}\:\Rightarrow\:{a}=\:\frac{−\mathrm{1}}{\mathrm{5}} \\ $$$$\:{f}\left({x}\right)=\:\mathrm{tan}\:\frac{−\pi}{\mathrm{5}}\:{x}\:+\mathrm{1}\:\:_{{continue}...} \\ $$$$\: \\ $$$$\cancel{\:{f}\left({x}\right)=\:\mathrm{0}\:\Rightarrow\:\mathrm{tan}\:\frac{\pi}{\mathrm{15}}\:{x}\:=\:−\mathrm{1}\:=\:\mathrm{tan}\:\left(\frac{\mathrm{3}\:\pi}{\mathrm{4}}\:+\pi{k}\right)} \\ $$$$\:{According}\:{to}\:{the}\:{diagram}\:{k}=\mathrm{0}\:,\:{k}=\mathrm{1} \\ $$$$\cancel{\:{x}_{{A}} =\:\frac{\mathrm{45}}{\mathrm{4}}\:,\:{x}_{{B}} =\:\frac{\mathrm{105}}{\mathrm{4}}\:\Rightarrow\:\Sigma=\:\frac{\mathrm{75}}{\mathrm{2}}\:\neq\:\left\{{a},\:{b},\:{c},\:{d}\right\}} \\ $$

Answered by mahdipoor last updated on 13/Dec/22

f(0)=b=1 ⇒ x^− →((15)/2) ⇒ f→−∞ ⇒ a<0 lim f(x)=±∞ ⇔ πax→(π/2)+πk k∈Z ⇒x→(1/(2a))+(k/a) ⇒ x→((15)/2)=(1/(2a))+((−2)/a)=((−3)/(2a)) ⇒a=−(1/5) f(x)=tan(−(π/5)x)+1=0 ⇒ x=(5/4),5+(5/4),... x_A +x_B =((5/4))+((5/4)+5)=7.5

$${f}\left(\mathrm{0}\right)={b}=\mathrm{1}\:\Rightarrow\: \\ $$$${x}^{−} \rightarrow\frac{\mathrm{15}}{\mathrm{2}}\:\:\Rightarrow\:{f}\rightarrow−\infty\:\Rightarrow\:\:{a}<\mathrm{0} \\ $$$${lim}\:{f}\left({x}\right)=\pm\infty\:\Leftrightarrow\:\pi{ax}\rightarrow\frac{\pi}{\mathrm{2}}+\pi{k}\:\:\:\:{k}\in{Z} \\ $$$$\Rightarrow{x}\rightarrow\frac{\mathrm{1}}{\mathrm{2}{a}}+\frac{{k}}{{a}}\:\:\:\:\Rightarrow\:\:\:\:{x}\rightarrow\frac{\mathrm{15}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}{a}}+\frac{−\mathrm{2}}{{a}}=\frac{−\mathrm{3}}{\mathrm{2}{a}} \\ $$$$\Rightarrow{a}=−\frac{\mathrm{1}}{\mathrm{5}} \\ $$$${f}\left({x}\right)={tan}\left(−\frac{\pi}{\mathrm{5}}{x}\right)+\mathrm{1}=\mathrm{0}\:\Rightarrow\:\:\:{x}=\frac{\mathrm{5}}{\mathrm{4}},\mathrm{5}+\frac{\mathrm{5}}{\mathrm{4}},... \\ $$$${x}_{{A}} +{x}_{{B}} =\left(\frac{\mathrm{5}}{\mathrm{4}}\right)+\left(\frac{\mathrm{5}}{\mathrm{4}}+\mathrm{5}\right)=\mathrm{7}.\mathrm{5} \\ $$

Commented by Acem last updated on 13/Dec/22

Hello Sir! why k= −2 when x→ ((15)/2)

$${Hello}\:{Sir}!\:{why}\:{k}=\:−\mathrm{2}\:{when}\:{x}\rightarrow\:\frac{\mathrm{15}}{\mathrm{2}} \\ $$

Commented by mahdipoor last updated on 13/Dec/22

Hi , when a<0 and x>0 ⇒ πax<0 in first vertical line : πax=−(π/2) and in secend line : πax=((−π)/2)−π=((−3π)/2) ... x=((15)/2) is secend line so πax=(π/2)+kπ=((−3π)/2) ⇒ k=−2

$${Hi}\:,\:{when}\:{a}<\mathrm{0}\:{and}\:{x}>\mathrm{0}\:\Rightarrow\:\pi{ax}<\mathrm{0} \\ $$$${in}\:{first}\:{vertical}\:{line}\::\:\pi{ax}=−\frac{\pi}{\mathrm{2}} \\ $$$${and}\:{in}\:{secend}\:{line}\::\:\pi{ax}=\frac{−\pi}{\mathrm{2}}−\pi=\frac{−\mathrm{3}\pi}{\mathrm{2}} \\ $$$$... \\ $$$${x}=\frac{\mathrm{15}}{\mathrm{2}}\:{is}\:{secend}\:{line}\:{so} \\ $$$$\pi{ax}=\frac{\pi}{\mathrm{2}}+{k}\pi=\frac{−\mathrm{3}\pi}{\mathrm{2}}\:\Rightarrow\:{k}=−\mathrm{2} \\ $$

Commented by Acem last updated on 13/Dec/22

Hi Sir! thank you very much In fact there are still tow problems: a• how do i know that it′s the first assymptot b• you said that πax= ((−π)/2) that mean k=−1 how do i know that k here = −1 Thought that we may can say that the 1st assymptot is when k=0 not −1

$${Hi}\:{Sir}!\:{thank}\:{you}\:{very}\:{much} \\ $$$$\:{In}\:{fact}\:{there}\:{are}\:{still}\:{tow}\:{problems}: \\ $$$$\:\:\:\:\:{a}\bullet\:{how}\:{do}\:{i}\:{know}\:{that}\:{it}'{s}\:{the}\:{first}\:{assymptot} \\ $$$$\:\:\:\:\:{b}\bullet\:{you}\:{said}\:{that}\:\pi{ax}=\:\frac{−\pi}{\mathrm{2}}\:{that}\:{mean}\:{k}=−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:{how}\:{do}\:{i}\:{know}\:{that}\:{k}\:{here}\:=\:−\mathrm{1} \\ $$$$ \\ $$$$\:\:{Thought}\:{that}\:{we}\:{may}\:{can}\:{say}\:{that}\:{the}\:\mathrm{1}{st} \\ $$$$\:{assymptot}\:{is}\:{when}\:{k}=\mathrm{0}\:{not}\:−\mathrm{1} \\ $$$$\: \\ $$

Commented by mahdipoor last updated on 14/Dec/22

[k] is not important , [πax_2 =πa((15)/2)=((−3π)/2)] is important, im say [πax=(π/2)+kπ] so k=−2 when [πax_2 =((−3π)/2)=(π/2)+(−2)π]

$$\left[{k}\right]\:{is}\:{not}\:{important}\:,\:\left[\pi{ax}_{\mathrm{2}} =\pi{a}\frac{\mathrm{15}}{\mathrm{2}}=\frac{−\mathrm{3}\pi}{\mathrm{2}}\right]\: \\ $$$${is}\:{important},\:{im}\:{say}\:\left[\pi{ax}=\frac{\pi}{\mathrm{2}}+{k}\pi\right]\:{so}\: \\ $$$${k}=−\mathrm{2}\:{when}\:\left[\pi{ax}_{\mathrm{2}} =\frac{−\mathrm{3}\pi}{\mathrm{2}}=\frac{\pi}{\mathrm{2}}+\left(−\mathrm{2}\right)\pi\right]\: \\ $$

Commented by mnjuly1970 last updated on 14/Dec/22

mamnoon jenabe mahdipoor ali bood

$${mamnoon}\:{jenabe}\:{mahdipoor}\: \\ $$$$\:\:{ali}\:{bood} \\ $$

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