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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)

$$\mathrm{If}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{ax}^{\mathrm{2}} −\mathrm{c}\:\mathrm{satisfy}\:−\mathrm{4}\leqslant\mathrm{f}\left(\mathrm{1}\right)\leqslant−\mathrm{1} \\ $$$$\mathrm{and}\:−\mathrm{1}\leqslant\mathrm{f}\left(\mathrm{2}\right)\leqslant\mathrm{5},\:\mathrm{then} \\ $$$$ \\ $$$$\mathrm{A}.\:\mathrm{7}\leqslant\mathrm{f}\left(\mathrm{3}\right)\leqslant\mathrm{26}\:\mathrm{B}.\:−\mathrm{1}\leqslant\mathrm{f}\left(\mathrm{3}\right)\leqslant\mathrm{20}\:\mathrm{C}. \\ $$$$−\mathrm{4}\leqslant\mathrm{f}\left(\mathrm{3}\right)\leqslant\mathrm{15}\:\mathrm{D}.\:\frac{−\mathrm{28}}{\mathrm{3}}\leqslant\mathrm{f}\left(\mathrm{3}\right)\leqslant\frac{\mathrm{35}}{\mathrm{3}}\:\mathrm{E}. \\ $$$$\frac{\mathrm{8}}{\mathrm{3}}\leqslant\mathrm{f}\left(\mathrm{3}\right)\leqslant\frac{\mathrm{13}}{\mathrm{3}} \\ $$

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

$$−\mathrm{4}\leqslant\mathrm{f}\left(\mathrm{1}\right)\leqslant−\mathrm{1}\Leftrightarrow−\mathrm{4}\leqslant\mathrm{a}−\mathrm{c}\leqslant−\mathrm{1}\left(\mathrm{1}\right) \\ $$$$−\mathrm{1}\leqslant\mathrm{f}\left(\mathrm{2}\right)\leqslant\mathrm{5}\Leftrightarrow−\mathrm{1}\leqslant\mathrm{4a}−\mathrm{c}\leqslant\mathrm{5}.\mathrm{We}\:\mathrm{have} \\ $$$$\begin{cases}{−\mathrm{1}\leqslant\mathrm{4a}−\mathrm{c}\leqslant−\mathrm{1}}\\{−\mathrm{1}\geqslant\mathrm{a}−\mathrm{c}\geqslant−\mathrm{4}}\end{cases} \\ $$$$\mathrm{Substracting}\:\mathrm{two}\:\mathrm{inequaliries}\:\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{0}\leqslant\mathrm{3a}\leqslant\mathrm{3}\Leftrightarrow\mathrm{0}\leqslant\mathrm{a}\leqslant\mathrm{1}\left(\mathrm{2}\right). \\ $$$$\left(\mathrm{1}\right)\Leftrightarrow\mathrm{1}\leqslant\mathrm{c}−\mathrm{a}\leqslant\mathrm{4}\:\left(\mathrm{3}\right).\mathrm{Adding}\:\mathrm{up}\:\left(\mathrm{2}\right)\:\left(\mathrm{3}\right) \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{1}\leqslant\mathrm{c}\leqslant\mathrm{5}\left(\mathrm{5}\right) \\ $$$$\mathrm{f}\left(\mathrm{3}\right)=\mathrm{9a}−\mathrm{c}\leqslant\mathrm{9}−\mathrm{1}=\mathrm{8} \\ $$$$\mathrm{f}\left(\mathrm{3}\right)=\mathrm{9a}−\mathrm{c}\geqslant\mathrm{0}−\mathrm{5}=−\mathrm{4} \\ $$$$\mathrm{Hence}\:−\mathrm{4}\leqslant\mathrm{f}\left(\mathrm{3}\right)\leqslant\mathrm{8}\Rightarrow\boldsymbol{\mathrm{Choose}}\:\boldsymbol{\mathrm{C}} \\ $$

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

Thanks.

$$\mathrm{Thanks}. \\ $$

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