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Question Number 98806 by bemath last updated on 16/Jun/20
for a is integer number such that ∣∣x−1∣ −2∣ ≤ a exactly has 2013 solution
foraisintegernumbersuchthat∣∣x−1∣−2∣⩽aexactlyhas2013solution
Commented by Rasheed.Sindhi last updated on 16/Jun/20
Assuming x an integer: ∣∣x−1∣ −2∣ ≤ a ⇒ { ((∣x−1−2∣≤a⇒∣x−3∣≤a...(i))),((∣−x+1−2∣≤a⇒∣−x−1∣≤a..(ii))) :} (i)⇒ { ((x−3≤a⇒x≤a+3)),((−x+3≤a⇒x≥−a+3)) :} ..(iii) (ii)⇒ { ((−x−1≤a⇒x≥−a−1)),((x+1≤a⇒x≤a−1)) :}..(iv) (iii)⇒−a+3≤x≤a+3......(v) (iv)⇒−a−1≤x≤a−1......(vi) (v)&(vi)⇒−a−1≤x≤a+3 −1004−1≤x≤1004+3 −1005≤x≤1007 x=−1005,−1004,...,−2,−1,0,1,2...,1006,1007 a=1004
Assumingxaninteger:∣∣x−1∣−2∣⩽a⇒{∣x−1−2∣⩽a⇒∣x−3∣⩽a…(i)∣−x+1−2∣⩽a⇒∣−x−1∣⩽a..(ii)(i)⇒{x−3⩽a⇒x⩽a+3−x+3⩽a⇒x⩾−a+3..(iii)(ii)⇒{−x−1⩽a⇒x⩾−a−1x+1⩽a⇒x⩽a−1..(iv)(iii)⇒−a+3⩽x⩽a+3……(v)(iv)⇒−a−1⩽x⩽a−1……(vi)(v)&(vi)⇒−a−1⩽x⩽a+3−1004−1⩽x⩽1004+3−1005⩽x⩽1007x=−1005,−1004,…,−2,−1,0,1,2…,1006,1007a=1004

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