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 178059 by DAVONG last updated on 12/Oct/22

Commented by Rasheed.Sindhi last updated on 12/Oct/22

(B) −17

(B)−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

∣x+k∣+∣x−3∣=8Let∣x−3∣=m⇒0⩽m⩽8[If∣x−3∣=mweregreaterthan8,∣x+k∣werenegative.]⇒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=−11,5a=min(k)=min(−11+2m)=−11+2(0)=−11[ora=min(k)=min(5−2m)=5−2(8)=−11]b=max(k)=max(5−2m)=5−2(0)=5[orb=max(k)=max(−11+2m)=−11+2(8)=5]a=min(k)=−11b=max(k)=5or−11⩽k⩽52a+b=2(−11)+5=−17

Terms of Service

Privacy Policy

Contact: info@tinkutara.com