Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 158988 by cortano last updated on 11/Nov/21

Answered by Rasheed.Sindhi last updated on 11/Nov/21

The graph of f(x) intersects y=2x−1  at x=1,2,3  ∴ The points of intersections:        (1,2(1)−1) , (2,2(2)−1) &(3,2(3)−1)     Or (1,1),(2,3) & (3,5)       also the points on f(x):  (1,1):  f(1)=(1)^4 +a(1)^3 +b(1)^2 +c(1)+d=1           ⇒a+b+c+d=0............A  (2,3):  f(2)=(2)^4 +a(2)^3 +b(2)^2 +c(2)+d=3           ⇒8a+4b+2c+d=−13.....B  (3,5):  f(3)=(3)^4 +a(3)^3 +b(3)^2 +c(3)+d=5      ⇒27a+9b+3c+d=−76........C  f(0)=d  f(4)=4^4 +a(4)^3 +b(4)^2 +c(4)+d=e(say)           =256+64a+16b+4c+d  f(0)+f(4)=256+64a+16b+4c+2d           =256+64a+16b+4c+2d           =256+64a+16b+2(−13−8a−4b)            =230+48a+8b   f(0)+f(4)=230+8(6a+b)  B−A:7a+3b+c=−13...........D  C−B: 19a+5b+c=−63........E  E−D:12a+2b=−50⇒6a+b=−25   f(0)+f(4)=230+8(6a+b)            =230+8(−25)=30

$$\mathcal{T}{he}\:{graph}\:{of}\:{f}\left({x}\right)\:{intersects}\:{y}=\mathrm{2}{x}−\mathrm{1} \\ $$$${at}\:{x}=\mathrm{1},\mathrm{2},\mathrm{3} \\ $$$$\therefore\:\mathcal{T}{he}\:{points}\:{of}\:{intersections}: \\ $$$$\:\:\:\:\:\:\left(\mathrm{1},\mathrm{2}\left(\mathrm{1}\right)−\mathrm{1}\right)\:,\:\left(\mathrm{2},\mathrm{2}\left(\mathrm{2}\right)−\mathrm{1}\right)\:\&\left(\mathrm{3},\mathrm{2}\left(\mathrm{3}\right)−\mathrm{1}\right) \\ $$$$\:\:\:{Or}\:\left(\mathrm{1},\mathrm{1}\right),\left(\mathrm{2},\mathrm{3}\right)\:\&\:\left(\mathrm{3},\mathrm{5}\right)\: \\ $$$$\:\:\:\:{also}\:{the}\:{points}\:{on}\:{f}\left({x}\right): \\ $$$$\left(\mathrm{1},\mathrm{1}\right): \\ $$$${f}\left(\mathrm{1}\right)=\left(\mathrm{1}\right)^{\mathrm{4}} +{a}\left(\mathrm{1}\right)^{\mathrm{3}} +{b}\left(\mathrm{1}\right)^{\mathrm{2}} +{c}\left(\mathrm{1}\right)+{d}=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\Rightarrow{a}+{b}+{c}+{d}=\mathrm{0}............{A} \\ $$$$\left(\mathrm{2},\mathrm{3}\right): \\ $$$${f}\left(\mathrm{2}\right)=\left(\mathrm{2}\right)^{\mathrm{4}} +{a}\left(\mathrm{2}\right)^{\mathrm{3}} +{b}\left(\mathrm{2}\right)^{\mathrm{2}} +{c}\left(\mathrm{2}\right)+{d}=\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:\Rightarrow\mathrm{8}{a}+\mathrm{4}{b}+\mathrm{2}{c}+{d}=−\mathrm{13}.....{B} \\ $$$$\left(\mathrm{3},\mathrm{5}\right): \\ $$$${f}\left(\mathrm{3}\right)=\left(\mathrm{3}\right)^{\mathrm{4}} +{a}\left(\mathrm{3}\right)^{\mathrm{3}} +{b}\left(\mathrm{3}\right)^{\mathrm{2}} +{c}\left(\mathrm{3}\right)+{d}=\mathrm{5} \\ $$$$\:\:\:\:\Rightarrow\mathrm{27}{a}+\mathrm{9}{b}+\mathrm{3}{c}+{d}=−\mathrm{76}........{C} \\ $$$${f}\left(\mathrm{0}\right)={d} \\ $$$${f}\left(\mathrm{4}\right)=\mathrm{4}^{\mathrm{4}} +{a}\left(\mathrm{4}\right)^{\mathrm{3}} +{b}\left(\mathrm{4}\right)^{\mathrm{2}} +{c}\left(\mathrm{4}\right)+{d}={e}\left({say}\right) \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{256}+\mathrm{64}{a}+\mathrm{16}{b}+\mathrm{4}{c}+{d} \\ $$$${f}\left(\mathrm{0}\right)+{f}\left(\mathrm{4}\right)=\mathrm{256}+\mathrm{64}{a}+\mathrm{16}{b}+\mathrm{4}{c}+\mathrm{2}{d} \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{256}+\mathrm{64}{a}+\mathrm{16}{b}+\mathrm{4}{c}+\mathrm{2}{d} \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{256}+\mathrm{64}{a}+\mathrm{16}{b}+\mathrm{2}\left(−\mathrm{13}−\mathrm{8}{a}−\mathrm{4}{b}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{230}+\mathrm{48}{a}+\mathrm{8}{b} \\ $$$$\:{f}\left(\mathrm{0}\right)+{f}\left(\mathrm{4}\right)=\mathrm{230}+\mathrm{8}\left(\mathrm{6}{a}+{b}\right) \\ $$$${B}−{A}:\mathrm{7}{a}+\mathrm{3}{b}+{c}=−\mathrm{13}...........{D} \\ $$$${C}−{B}:\:\mathrm{19}{a}+\mathrm{5}{b}+{c}=−\mathrm{63}........{E} \\ $$$${E}−{D}:\mathrm{12}{a}+\mathrm{2}{b}=−\mathrm{50}\Rightarrow\mathrm{6}{a}+{b}=−\mathrm{25} \\ $$$$\:{f}\left(\mathrm{0}\right)+{f}\left(\mathrm{4}\right)=\mathrm{230}+\mathrm{8}\left(\mathrm{6}{a}+{b}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{230}+\mathrm{8}\left(−\mathrm{25}\right)=\mathrm{30} \\ $$

Commented by cortano last updated on 13/Nov/21

thank you sir

$${thank}\:{you}\:{sir} \\ $$

Answered by mr W last updated on 11/Nov/21

say g(x)=f(x)−(2x−1)  g(1)=g(2)=g(3)=0  ⇒g(x)=(x−1)(x−2)(x−3)(x+p)  ⇒f(x)−(2x−1)=(x−1)(x−2)(x−3)(x+p)  ⇒f(x)=(x−1)(x−2)(x−3)(x+p)+2x−1  f(0)=(0−1)(0−2)(0−3)(0+p)+2×0−1  ⇒f(0)=−6p−1  f(4)=(4−1)(4−2)(4−3)(4+p)+2×4−1  ⇒f(4)=31+6p  f(0)+f(4)=−6p−1+31+6p=30

$${say}\:{g}\left({x}\right)={f}\left({x}\right)−\left(\mathrm{2}{x}−\mathrm{1}\right) \\ $$$${g}\left(\mathrm{1}\right)={g}\left(\mathrm{2}\right)={g}\left(\mathrm{3}\right)=\mathrm{0} \\ $$$$\Rightarrow{g}\left({x}\right)=\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)\left({x}−\mathrm{3}\right)\left({x}+{p}\right) \\ $$$$\Rightarrow{f}\left({x}\right)−\left(\mathrm{2}{x}−\mathrm{1}\right)=\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)\left({x}−\mathrm{3}\right)\left({x}+{p}\right) \\ $$$$\Rightarrow{f}\left({x}\right)=\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)\left({x}−\mathrm{3}\right)\left({x}+{p}\right)+\mathrm{2}{x}−\mathrm{1} \\ $$$${f}\left(\mathrm{0}\right)=\left(\mathrm{0}−\mathrm{1}\right)\left(\mathrm{0}−\mathrm{2}\right)\left(\mathrm{0}−\mathrm{3}\right)\left(\mathrm{0}+{p}\right)+\mathrm{2}×\mathrm{0}−\mathrm{1} \\ $$$$\Rightarrow{f}\left(\mathrm{0}\right)=−\mathrm{6}{p}−\mathrm{1} \\ $$$${f}\left(\mathrm{4}\right)=\left(\mathrm{4}−\mathrm{1}\right)\left(\mathrm{4}−\mathrm{2}\right)\left(\mathrm{4}−\mathrm{3}\right)\left(\mathrm{4}+{p}\right)+\mathrm{2}×\mathrm{4}−\mathrm{1} \\ $$$$\Rightarrow{f}\left(\mathrm{4}\right)=\mathrm{31}+\mathrm{6}{p} \\ $$$${f}\left(\mathrm{0}\right)+{f}\left(\mathrm{4}\right)=−\mathrm{6}{p}−\mathrm{1}+\mathrm{31}+\mathrm{6}{p}=\mathrm{30} \\ $$

Commented by Rasheed.Sindhi last updated on 11/Nov/21

A Novel Method!  Wonderful Sir!

$${A}\:{Novel}\:{Method}! \\ $$$${Wonderful}\:\mathcal{S}{ir}! \\ $$

Commented by mr W last updated on 12/Nov/21

thanks sir! we see again:  many roads lead to Rome!

$${thanks}\:{sir}!\:{we}\:{see}\:{again}: \\ $$$${many}\:{roads}\:{lead}\:{to}\:{Rome}! \\ $$

Commented by Rasheed.Sindhi last updated on 12/Nov/21

Of course sir!  I′m always interested in ′many  roads′  that is variety of solutions.  But after all some roads are better  than others!

$${Of}\:{course}\:{sir}! \\ $$$${I}'{m}\:{always}\:{interested}\:{in}\:'{many} \\ $$$${roads}'\:\:{that}\:{is}\:{variety}\:{of}\:{solutions}. \\ $$$${But}\:{after}\:{all}\:{some}\:{roads}\:{are}\:{better} \\ $$$${than}\:{others}! \\ $$

Commented by cortano last updated on 13/Nov/21

nice

$${nice} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com