Question Number 178059 by DAVONG last updated on 12/Oct/22
Commented by Rasheed.Sindhi last updated on 12/Oct/22
$$\left(\mathrm{B}\right)\:−\mathrm{17} \\ $$
Answered by Rasheed.Sindhi last updated on 13/Oct/22
![∣x+k∣+∣x−3∣=8 Let ∣x−3∣=m⇒0≤m≤8 [If ∣x−3∣=m were greater than 8, ∣x+k∣ were negative.] ⇒x=3±m⇒∣x+k∣=8−m ⇒∣3±m+k∣=8−m ⇒3±m+k=±(8−m)∨3±m+k=∓(8−m) k=±8∓m−3∓m { ((k=±8∓m−3∓m⇒k=5−2m,−11+2m)),(( ∨)),((k=∓8±m−3∓m⇒k=−11,5)) :} a=min(k)=min(−11+2m) =−11+2(0)=−11 [or a=min(k)=min(5−2m)=5−2(8)=−11] b=max(k)=max(5−2m)=5−2(0)=5 [or b=max(k)=max(−11+2m)=−11+2(8)=5] a=min(k)=−11 b=max(k)=5 or −11≤k≤5 2a+b=2(−11)+5=−17](https://www.tinkutara.com/question/Q178157.png)
$$\mid{x}+{k}\mid+\mid{x}−\mathrm{3}\mid=\mathrm{8} \\ $$$${Let}\:\mid{x}−\mathrm{3}\mid={m}\Rightarrow\mathrm{0}\leqslant{m}\leqslant\mathrm{8} \\ $$$$\left[{If}\:\mid{x}−\mathrm{3}\mid={m}\:{were}\:{greater}\:{than}\:\mathrm{8},\:\mid{x}+{k}\mid\:{were}\:{negative}.\right] \\ $$$$\Rightarrow{x}=\mathrm{3}\pm{m}\Rightarrow\mid{x}+{k}\mid=\mathrm{8}−{m} \\ $$$$\Rightarrow\mid\mathrm{3}\pm{m}+{k}\mid=\mathrm{8}−{m} \\ $$$$\Rightarrow\mathrm{3}\pm{m}+{k}=\pm\left(\mathrm{8}−{m}\right)\vee\mathrm{3}\pm{m}+{k}=\mp\left(\mathrm{8}−{m}\right) \\ $$$$\:\:{k}=\pm\mathrm{8}\mp{m}−\mathrm{3}\mp{m} \\ $$$$\begin{cases}{{k}=\pm\mathrm{8}\mp{m}−\mathrm{3}\mp{m}\Rightarrow{k}=\mathrm{5}−\mathrm{2}{m},−\mathrm{11}+\mathrm{2}{m}}\\{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\vee}\\{{k}=\mp\mathrm{8}\pm{m}−\mathrm{3}\mp{m}\Rightarrow{k}=−\mathrm{11},\mathrm{5}}\end{cases} \\ $$$${a}={min}\left({k}\right)={min}\left(−\mathrm{11}+\mathrm{2}{m}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=−\mathrm{11}+\mathrm{2}\left(\mathrm{0}\right)=−\mathrm{11} \\ $$$$\left[{or}\:{a}={min}\left({k}\right)={min}\left(\mathrm{5}−\mathrm{2}{m}\right)=\mathrm{5}−\mathrm{2}\left(\mathrm{8}\right)=−\mathrm{11}\right] \\ $$$${b}={max}\left({k}\right)={max}\left(\mathrm{5}−\mathrm{2}{m}\right)=\mathrm{5}−\mathrm{2}\left(\mathrm{0}\right)=\mathrm{5} \\ $$$$\left[{or}\:{b}={max}\left({k}\right)={max}\left(−\mathrm{11}+\mathrm{2}{m}\right)=−\mathrm{11}+\mathrm{2}\left(\mathrm{8}\right)=\mathrm{5}\right] \\ $$$$ \\ $$$${a}={min}\left({k}\right)=−\mathrm{11} \\ $$$${b}={max}\left({k}\right)=\mathrm{5} \\ $$$${or}\:\:\:\:\:−\mathrm{11}\leqslant{k}\leqslant\mathrm{5} \\ $$$$\mathrm{2}{a}+{b}=\mathrm{2}\left(−\mathrm{11}\right)+\mathrm{5}=−\mathrm{17} \\ $$