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If-M-and-m-are-respectively-the-largest-and-the-smallest-integers-that-satisfying-the-inequality-6n-2-5n-99-find-the-value-of-M-m-

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Question Number 119835 by ZiYangLee last updated on 27/Oct/20
If M and m are respectively the largest and the smallest integers that satisfying the inequality 6n^2 −5n≤99, find the value of M−m.
IfMandmarerespectivelythelargestandthesmallestintegersthatsatisfyingtheinequality6n2−5n⩽99,findthevalueofM−m.
Commented by talminator2856791 last updated on 27/Oct/20
is the question correctly stated?
isthequestioncorrectlystated?
Answered by TANMAY PANACEA last updated on 27/Oct/20
6n^2 −5n−99≤0 6n^2 −27n+22n−99≤0 3n(2n−9)+11(2n−9)≤0 (2n−9)(3n+11)≤0 critical value of n=(9/2) and ((−11)/3) f(n)=6n^2 −5n−99 when n>(9/2) f(n)>0 when n<((−11)/3) f(n)>0 when (9/2)>n>((−11)/3) f(n)<0 at n=((−11)/3) f(n)=0 at n=(9/2) f(n)=0 M=4 m=−3 M−m 4−(−3)=7
6n2−5n−99⩽06n2−27n+22n−99⩽03n(2n−9)+11(2n−9)⩽0(2n−9)(3n+11)⩽0criticalvalueofn=92and−113f(n)=6n2−5n−99whenn>92f(n)>0whenn<−113f(n)>0when92>n>−113f(n)<0atn=−113f(n)=0atn=92f(n)=0M=4m=−3M−m4−(−3)=7
Commented by ZiYangLee last updated on 27/Oct/20
Correct!
Correct!
Commented by TANMAY PANACEA last updated on 27/Oct/20
thank you sir
thankyousir
Answered by Olaf last updated on 27/Oct/20
6n^2 −5n−99 = 0 Δ = 25−4×6×(−99) = 2401 = 49^2 n_1 = ((5−49)/(12)) = −((11)/3) n_2 = ((5+49)/(12)) = (9/2) ⇒ 6n^2 −5n ≤ 99 if n∈[−((11)/3);+(9/2)] M = n_(max) = 4 m = n_(min) = −3 (if m∈Z) M−m = 4−(−3) = 7 (or 4 if m∈N)
6n2−5n−99=0Δ=25−4×6×(−99)=2401=492n1=5−4912=−113n2=5+4912=92⇒6n2−5n⩽99ifn∈[−113;+92]M=nmax=4m=nmin=−3(ifm∈Z)M−m=4−(−3)=7(or4ifm∈N)

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