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Question Number 153257 by naka3546 last updated on 06/Sep/21

Find  set  of  k  value  so  that         ∣x∣ + ∣x−1∣ + ∣x−4∣ = k  a. has  one  solution  b. has  two  solutions  c. has  many  solutions  d. has  no  solution

$${Find}\:\:{set}\:\:{of}\:\:{k}\:\:{value}\:\:{so}\:\:{that} \\ $$$$\:\:\:\:\:\:\:\mid{x}\mid\:+\:\mid{x}−\mathrm{1}\mid\:+\:\mid{x}−\mathrm{4}\mid\:=\:{k} \\ $$$${a}.\:{has}\:\:{one}\:\:{solution} \\ $$$${b}.\:{has}\:\:{two}\:\:{solutions} \\ $$$${c}.\:{has}\:\:{many}\:\:{solutions} \\ $$$${d}.\:{has}\:\:{no}\:\:{solution} \\ $$

Answered by ajfour last updated on 06/Sep/21

let  x−2=t , then  ∣x∣+∣x−1∣+∣x−3∣+∣x−4∣     =k+∣x−3∣  ⇒  ∣t+2∣+∣t+1∣+∣t−1∣+∣t−2∣     =k+∣t−1∣  now we better use graph:     blue graph = red graph  so for no solution k<4  one solution    k=4  for three solutions,(two equal)        k=6  two solutions   k>4  For  more than two (distinct)           no real value of k.

$${let}\:\:{x}−\mathrm{2}={t}\:,\:{then} \\ $$$$\mid{x}\mid+\mid{x}−\mathrm{1}\mid+\mid{x}−\mathrm{3}\mid+\mid{x}−\mathrm{4}\mid \\ $$$$\:\:\:={k}+\mid{x}−\mathrm{3}\mid \\ $$$$\Rightarrow\:\:\mid{t}+\mathrm{2}\mid+\mid{t}+\mathrm{1}\mid+\mid{t}−\mathrm{1}\mid+\mid{t}−\mathrm{2}\mid \\ $$$$\:\:\:={k}+\mid{t}−\mathrm{1}\mid \\ $$$${now}\:{we}\:{better}\:{use}\:{graph}: \\ $$$$\:\:\:{blue}\:{graph}\:=\:{red}\:{graph} \\ $$$${so}\:{for}\:{no}\:{solution}\:{k}<\mathrm{4} \\ $$$${one}\:{solution}\:\:\:\:{k}=\mathrm{4} \\ $$$${for}\:{three}\:{solutions},\left({two}\:{equal}\right) \\ $$$$\:\:\:\:\:\:{k}=\mathrm{6} \\ $$$${two}\:{solutions}\:\:\:{k}>\mathrm{4} \\ $$$${For}\:\:{more}\:{than}\:{two}\:\left({distinct}\right) \\ $$$$\:\:\:\:\:\:\:\:\:{no}\:{real}\:{value}\:{of}\:{k}. \\ $$$$ \\ $$

Commented by ajfour last updated on 06/Sep/21

Answered by MJS_new last updated on 06/Sep/21

f(x)=∣x∣+∣x−1∣+∣x−4∣  f(x)= { ((−3x+5; x≤0 ∧ 5≤f(x)<+∞)),((−x+5; 0<x≤1 ∧ 4≤f(x)<5 )),((x+3; 1<x≤4 ∧ 4<f(x)≤7)),((3x−5; 4<x ∧ 7<f(x)<+∞)) :}  ⇒  k<4 no solution  k=4 one solution  k>4 two solutions

$${f}\left({x}\right)=\mid{x}\mid+\mid{x}−\mathrm{1}\mid+\mid{x}−\mathrm{4}\mid \\ $$$${f}\left({x}\right)=\begin{cases}{−\mathrm{3}{x}+\mathrm{5};\:{x}\leqslant\mathrm{0}\:\wedge\:\mathrm{5}\leqslant{f}\left({x}\right)<+\infty}\\{−{x}+\mathrm{5};\:\mathrm{0}<{x}\leqslant\mathrm{1}\:\wedge\:\mathrm{4}\leqslant{f}\left({x}\right)<\mathrm{5}\:}\\{{x}+\mathrm{3};\:\mathrm{1}<{x}\leqslant\mathrm{4}\:\wedge\:\mathrm{4}<{f}\left({x}\right)\leqslant\mathrm{7}}\\{\mathrm{3}{x}−\mathrm{5};\:\mathrm{4}<{x}\:\wedge\:\mathrm{7}<{f}\left({x}\right)<+\infty}\end{cases} \\ $$$$\Rightarrow \\ $$$${k}<\mathrm{4}\:\mathrm{no}\:\mathrm{solution} \\ $$$${k}=\mathrm{4}\:\mathrm{one}\:\mathrm{solution} \\ $$$${k}>\mathrm{4}\:\mathrm{two}\:\mathrm{solutions} \\ $$

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