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Question Number 95662 by john santu last updated on 26/May/20

f(x+p) + f(x−p) = 6x−4 f(20) = 29p ((f(p))/(2p)) = ?

$$\mathrm{f}\left(\mathrm{x}+\mathrm{p}\right)\:+\:\mathrm{f}\left(\mathrm{x}−\mathrm{p}\right)\:=\:\mathrm{6x}−\mathrm{4} \\ $$$$\mathrm{f}\left(\mathrm{20}\right)\:=\:\mathrm{29p} \\ $$$$\frac{\mathrm{f}\left(\mathrm{p}\right)}{\mathrm{2p}}\:=\:? \\ $$

Commented by i jagooll last updated on 26/May/20

replace x+p = x (1) f(x)+f(x−2p)=6x−6p−4 replace x−p=x (2)f(x+2p)+f(x)=6x+6p−4 ⇒(1)−(2) f(x−2p)−f(x+2p) = −12p let x−2p = 20 ⇒x=20+2p f(20)−f(20+4p) = −12p 29p+12p = f(20+4p) f(20+4p) = 41p let 20+4p = t ⇒p = ((t−20)/4) f(t) = 41(((t−20)/4)) = ((41t−820)/4) ((f(t))/(2t)) = ((41t−820)/(8t))

$$\mathrm{replace}\:\mathrm{x}+\mathrm{p}\:=\:\mathrm{x}\: \\ $$$$\left(\mathrm{1}\right)\:\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(\mathrm{x}−\mathrm{2p}\right)=\mathrm{6x}−\mathrm{6p}−\mathrm{4} \\ $$$$\mathrm{replace}\:\mathrm{x}−\mathrm{p}=\mathrm{x} \\ $$$$\left(\mathrm{2}\right)\mathrm{f}\left(\mathrm{x}+\mathrm{2p}\right)+\mathrm{f}\left(\mathrm{x}\right)=\mathrm{6x}+\mathrm{6p}−\mathrm{4} \\ $$$$\Rightarrow\left(\mathrm{1}\right)−\left(\mathrm{2}\right)\: \\ $$$$\mathrm{f}\left(\mathrm{x}−\mathrm{2p}\right)−\mathrm{f}\left(\mathrm{x}+\mathrm{2p}\right)\:=\:−\mathrm{12p} \\ $$$$\mathrm{let}\:\mathrm{x}−\mathrm{2p}\:=\:\mathrm{20}\:\Rightarrow\mathrm{x}=\mathrm{20}+\mathrm{2p} \\ $$$$\mathrm{f}\left(\mathrm{20}\right)−\mathrm{f}\left(\mathrm{20}+\mathrm{4p}\right)\:=\:−\mathrm{12p} \\ $$$$\mathrm{29p}+\mathrm{12p}\:=\:\mathrm{f}\left(\mathrm{20}+\mathrm{4p}\right) \\ $$$$\mathrm{f}\left(\mathrm{20}+\mathrm{4p}\right)\:=\:\mathrm{41p} \\ $$$$\mathrm{let}\:\mathrm{20}+\mathrm{4p}\:=\:\mathrm{t}\:\Rightarrow\mathrm{p}\:=\:\frac{\mathrm{t}−\mathrm{20}}{\mathrm{4}} \\ $$$$\mathrm{f}\left(\mathrm{t}\right)\:=\:\mathrm{41}\left(\frac{\mathrm{t}−\mathrm{20}}{\mathrm{4}}\right)\:=\:\frac{\mathrm{41t}−\mathrm{820}}{\mathrm{4}} \\ $$$$\frac{\mathrm{f}\left(\mathrm{t}\right)}{\mathrm{2t}}\:=\:\frac{\mathrm{41t}−\mathrm{820}}{\mathrm{8t}} \\ $$$$\:\:\:\:\:\: \\ $$

Commented by PRITHWISH SEN 2 last updated on 26/May/20

put p=0 2f(x)=6x−4 ⇒f(x)=3x−2⇒f(20)=58 ⇒p=2 ∴((f(p))/(2p)) = ((3×2−2)/(2×2)) =1 please check

$$\mathrm{put}\:\mathrm{p}=\mathrm{0} \\ $$$$\mathrm{2f}\left(\mathrm{x}\right)=\mathrm{6x}−\mathrm{4}\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)=\mathrm{3x}−\mathrm{2}\Rightarrow\mathrm{f}\left(\mathrm{20}\right)=\mathrm{58} \\ $$$$\Rightarrow\mathrm{p}=\mathrm{2} \\ $$$$\therefore\frac{\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{p}}\right)}{\mathrm{2}\boldsymbol{\mathrm{p}}}\:=\:\frac{\mathrm{3}×\mathrm{2}−\mathrm{2}}{\mathrm{2}×\mathrm{2}}\:=\mathrm{1}\:\:\boldsymbol{\mathrm{please}}\:\boldsymbol{\mathrm{check}} \\ $$

Commented by i jagooll last updated on 27/May/20

how can p = 0 and p = 2 ? it meant 0 = 2 ?

$$\mathrm{how}\:\mathrm{can}\:\mathrm{p}\:=\:\mathrm{0}\:\mathrm{and}\:\mathrm{p}\:=\:\mathrm{2}\:? \\ $$$$\mathrm{it}\:\mathrm{meant}\:\mathrm{0}\:=\:\mathrm{2}\:? \\ $$

Commented by PRITHWISH SEN 2 last updated on 27/May/20

for the 1st equation p is the point which can take any value. But for the second equation it is a constant. now fixed the value of p at 2 then you get for f(x)=3x−2 then f(x+2)+f(x−2)=3(x+2)−2+3(x−2)−2 = 6x−4 and f(20)= 3×20−2=58=29p where p=2

$$\mathrm{for}\:\mathrm{the}\:\mathrm{1st}\:\mathrm{equation}\:\mathrm{p}\:\mathrm{is}\:\mathrm{the}\:\mathrm{point}\:\mathrm{which}\:\mathrm{can}\: \\ $$$$\mathrm{take}\:\mathrm{any}\:\mathrm{value}. \\ $$$$\mathrm{But}\:\mathrm{for}\:\mathrm{the}\:\mathrm{second}\:\mathrm{equation}\:\mathrm{it}\:\mathrm{is}\:\mathrm{a}\:\mathrm{constant}.\: \\ $$$$\mathrm{now}\:\mathrm{fixed}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{p}\:\mathrm{at}\:\mathrm{2}\:\mathrm{then}\:\mathrm{you}\:\mathrm{get} \\ $$$$\mathrm{for}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{3x}−\mathrm{2} \\ $$$$\mathrm{then}\:\mathrm{f}\left(\mathrm{x}+\mathrm{2}\right)+\mathrm{f}\left(\mathrm{x}−\mathrm{2}\right)=\mathrm{3}\left(\mathrm{x}+\mathrm{2}\right)−\mathrm{2}+\mathrm{3}\left(\mathrm{x}−\mathrm{2}\right)−\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{6x}−\mathrm{4} \\ $$$$\mathrm{and}\:\mathrm{f}\left(\mathrm{20}\right)=\:\mathrm{3}×\mathrm{20}−\mathrm{2}=\mathrm{58}=\mathrm{29p}\:\:\mathrm{where}\:\mathrm{p}=\mathrm{2} \\ $$

Commented by i jagooll last updated on 27/May/20

alright sir. thank you

$$\mathrm{alright}\:\mathrm{sir}.\:\mathrm{thank}\:\mathrm{you} \\ $$

Answered by john santu last updated on 28/May/20

let f(x) = ax+b f(x+p) = ax+ap+b (i) f(x−p)= ax−ap+b (ii) (i)+(ii) ⇒2ax+2b = 6x−4 a = 3 , b = −2 f(x) = 3x−2 f(20) = 58 = 29p ; p = 2 ((f(p))/(2p)) = ((f(2))/4) = ((6−2)/4) = 1

$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{ax}+\mathrm{b}\: \\ $$$$\mathrm{f}\left(\mathrm{x}+\mathrm{p}\right)\:=\:\mathrm{ax}+\mathrm{ap}+\mathrm{b}\:\:\left(\mathrm{i}\right) \\ $$$$\mathrm{f}\left(\mathrm{x}−\mathrm{p}\right)=\:\mathrm{ax}−\mathrm{ap}+\mathrm{b}\:\:\left(\mathrm{ii}\right) \\ $$$$\left(\mathrm{i}\right)+\left(\mathrm{ii}\right)\:\Rightarrow\mathrm{2ax}+\mathrm{2b}\:=\:\mathrm{6x}−\mathrm{4}\: \\ $$$$\mathrm{a}\:=\:\mathrm{3}\:,\:\mathrm{b}\:=\:−\mathrm{2}\: \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{3x}−\mathrm{2}\: \\ $$$$\mathrm{f}\left(\mathrm{20}\right)\:=\:\mathrm{58}\:=\:\mathrm{29p}\:;\:\mathrm{p}\:=\:\mathrm{2} \\ $$$$\frac{\mathrm{f}\left(\mathrm{p}\right)}{\mathrm{2p}}\:=\:\frac{\mathrm{f}\left(\mathrm{2}\right)}{\mathrm{4}}\:=\:\frac{\mathrm{6}−\mathrm{2}}{\mathrm{4}}\:=\:\mathrm{1}\: \\ $$

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