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Question Number 111535 by Aina Samuel Temidayo last updated on 04/Sep/20

If f(x)=ax^2 −c satisfy −4≤f(1)≤−1 and −1≤f(2)≤5, then A. 7≤f(3)≤26 B. −1≤f(3)≤20 C. −4≤f(3)≤15 D. ((−28)/3)≤f(3)≤((35)/3) E. (8/3)≤f(3)≤((13)/3)

Iff(x)=ax2−csatisfy−4⩽f(1)⩽−1and−1⩽f(2)⩽5,thenA.7⩽f(3)⩽26B.−1⩽f(3)⩽20C.−4⩽f(3)⩽15D.−283⩽f(3)⩽353E.83⩽f(3)⩽133

Answered by 1549442205PVT last updated on 04/Sep/20

−4≤f(1)≤−1⇔−4≤a−c≤−1(1) −1≤f(2)≤5⇔−1≤4a−c≤5.We have { ((−1≤4a−c≤−1)),((−1≥a−c≥−4)) :} Substracting two inequaliries we get 0≤3a≤3⇔0≤a≤1(2). (1)⇔1≤c−a≤4 (3).Adding up (2) (3) we get 1≤c≤5(5) f(3)=9a−c≤9−1=8 f(3)=9a−c≥0−5=−4 Hence −4≤f(3)≤8⇒Choose C

−4⩽f(1)⩽−1⇔−4⩽a−c⩽−1(1)−1⩽f(2)⩽5⇔−1⩽4a−c⩽5.Wehave{−1⩽4a−c⩽−1−1⩾a−c⩾−4Substractingtwoinequaliriesweget0⩽3a⩽3⇔0⩽a⩽1(2).(1)⇔1⩽c−a⩽4(3).Addingup(2)(3)weget1⩽c⩽5(5)f(3)=9a−c⩽9−1=8f(3)=9a−c⩾0−5=−4Hence−4⩽f(3)⩽8⇒ChooseC

Commented by Aina Samuel Temidayo last updated on 04/Sep/20

Thanks.

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