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Question Number 130670 by EDWIN88 last updated on 27/Jan/21 | ||
$$\:{Given}\:{f}\left({x}\right)=\sqrt{{ax}+{b}}\:;\:{g}\left({x}\right)={px}+{q} \\ $$ $${and}\:{f}\left({x}\right)<{g}\left({x}\right)\:{has}\:{solution} \\ $$ $$\mathrm{1}<{x}<\mathrm{5}.\:{Give}\:{one}\:{example}\:{f}\left({x}\right) \\ $$ $${and}\:{g}\left({x}\right)\:{satisfy}\:{the}\:{condition} \\ $$ | ||
Commented byMJS_new last updated on 28/Jan/21 | ||
$$\mathrm{do}\:\mathrm{you}\:\mathrm{have}\:\mathrm{an}\:\mathrm{example}?\:\mathrm{to}\:\mathrm{me}\:\mathrm{it}\:\mathrm{seems} \\ $$ $$\mathrm{impossible}\:\mathrm{with}\:{a},\:{b},\:{p},\:{q},\:{x}\:\in\mathbb{R} \\ $$ | ||
Commented byEDWIN88 last updated on 28/Jan/21 | ||
$${yes}\:{sir}.\:{i}\:{think}\:{it}'{s}\:{possible} \\ $$ $${if}\:{f}\left({x}\right)>{g}\left({x}\right) \\ $$ | ||
Answered by MJS_new last updated on 28/Jan/21 | ||
$$\mathrm{with}\:{f}\left({x}\right)>{g}\left({x}\right)\:\mathrm{we}\:\mathrm{start}\:\mathrm{at}\:\mathrm{the}\:\mathrm{borders} \\ $$ $${x}=\mathrm{1}\:\Rightarrow\:\sqrt{{a}+{b}}={p}+{q} \\ $$ $${x}=\mathrm{5}\:\Rightarrow\:\sqrt{\mathrm{5}{a}+{b}}=\mathrm{5}{p}+{q} \\ $$ $$\Rightarrow \\ $$ $${p}=\frac{\sqrt{\mathrm{5}{a}+{b}}−\sqrt{{a}+{b}}}{\mathrm{4}}\wedge{q}=\frac{\mathrm{5}\sqrt{{a}+{b}}−\sqrt{\mathrm{5}{a}+{b}}}{\mathrm{4}} \\ $$ $$\Rightarrow \\ $$ $$\sqrt{{ax}+{b}}>\frac{\sqrt{\mathrm{5}{a}+{b}}−\sqrt{{a}+{b}}}{\mathrm{4}}{x}+\frac{\mathrm{5}\sqrt{{a}+{b}}−\sqrt{\mathrm{5}{a}+{b}}}{\mathrm{4}};\:\mathrm{1}<{x}<\mathrm{5} \\ $$ $$\mathrm{test}\:\mathrm{this}\:\mathrm{for}\:{x}=\mathrm{3} \\ $$ $$\sqrt{\mathrm{3}{a}+{b}}>\frac{\sqrt{\mathrm{5}{a}+{b}}+\sqrt{{a}+{b}}}{\mathrm{2}} \\ $$ $$\mathrm{2}\sqrt{\mathrm{3}{a}+{b}}>\sqrt{\mathrm{5}{a}+{b}}+\sqrt{{a}+{b}} \\ $$ $${t}^{\mathrm{2}} =\mathrm{3}{a}+{b}\wedge{t}>\mathrm{0}\:\Rightarrow\:{b}={t}^{\mathrm{2}} −\mathrm{3}{a} \\ $$ $$\mathrm{2}{t}>\sqrt{{t}^{\mathrm{2}} +\mathrm{2}{a}}+\sqrt{{t}^{\mathrm{2}} −\mathrm{2}{a}} \\ $$ $$\mathrm{4}{t}^{\mathrm{2}} >\mathrm{2}{t}^{\mathrm{2}} +\mathrm{2}\sqrt{{t}^{\mathrm{2}} +\mathrm{2}{a}}\sqrt{{t}^{\mathrm{2}} −\mathrm{2}{a}} \\ $$ $${t}^{\mathrm{2}} >\sqrt{{t}^{\mathrm{2}} +\mathrm{2}{a}}\sqrt{{t}^{\mathrm{2}} −\mathrm{2}{a}} \\ $$ $${t}^{\mathrm{4}} >{t}^{\mathrm{4}} −\mathrm{4}{a}^{\mathrm{4}} \\ $$ $$\mathrm{true}\:\mathrm{for}\:{a}\neq\mathrm{0} \\ $$ $$\Rightarrow \\ $$ $$\mathrm{we}\:\mathrm{can}\:\mathrm{choose}\:{a},\:{b}\:\mathrm{with}\:{ax}+{b}\geqslant\mathrm{0}\wedge\mathrm{1}<{x}<\mathrm{5}\:\Leftrightarrow \\ $$ $$\Leftrightarrow\:{b}\geqslant−{ax}\wedge\mathrm{1}<{x}<\mathrm{5}\:\Rightarrow\:\begin{cases}{{b}>−\mathrm{5}{a};\:{a}<\mathrm{0}}\\{{b}>−{a};\:{a}>\mathrm{0}}\end{cases} \\ $$ $$\mathrm{examples} \\ $$ $$\left(\mathrm{1}\right)\:{a}=−\mathrm{3}\wedge{b}=\mathrm{16}\:\Rightarrow\:{p}=\frac{\mathrm{1}−\sqrt{\mathrm{13}}}{\mathrm{4}}\wedge{q}=\frac{−\mathrm{1}+\mathrm{5}\sqrt{\mathrm{13}}}{\mathrm{4}} \\ $$ $$\sqrt{−\mathrm{3}{x}+\mathrm{16}}>\frac{\mathrm{1}−\sqrt{\mathrm{13}}}{\mathrm{4}}{x}−\frac{\mathrm{1}−\mathrm{5}\sqrt{\mathrm{13}}}{\mathrm{4}};\:\mathrm{1}<{x}<\mathrm{5} \\ $$ $$\left(\mathrm{2}\right)\:{a}=\mathrm{5}\wedge{b}=−\mathrm{1}\:\Rightarrow\:{p}=\frac{−\mathrm{1}+\sqrt{\mathrm{6}}}{\mathrm{2}}\wedge{q}=\frac{\mathrm{5}−\sqrt{\mathrm{6}}}{\mathrm{2}} \\ $$ $$\sqrt{\mathrm{5}{x}−\mathrm{1}}>−\frac{\mathrm{1}−\sqrt{\mathrm{6}}}{\mathrm{2}}{x}+\frac{\mathrm{5}−\sqrt{\mathrm{6}}}{\mathrm{2}};\:\mathrm{1}<{x}<\mathrm{5} \\ $$ $$\mathrm{plot}\:\mathrm{them}\:\mathrm{to}\:\mathrm{see}\:\mathrm{it}'\mathrm{s}\:\mathrm{true} \\ $$ | ||
Commented byliberty last updated on 28/Jan/21 | ||
Commented byliberty last updated on 28/Jan/21 | ||
$$\mathrm{correct}\:\mathrm{sir} \\ $$ | ||
Commented byEDWIN88 last updated on 28/Jan/21 | ||
$${i}\:{got}\:\sqrt{−\mathrm{4}{x}+\mathrm{20}}\:>\:−{x}+\mathrm{5}\:{sir} \\ $$ | ||
Commented byEDWIN88 last updated on 28/Jan/21 | ||