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Question Number 208199 by efronzo1 last updated on 07/Jun/24
Answered by Frix last updated on 07/Jun/24
y=(√(18+3x−x^2 )) is a semi circle with r=(9/2) (√(x+3))+(√(6−x)) has the maximum (((3/2)),((3(√2))) ) We have 2 solutions for 0≤m<((3(√2))/4) and exactly one solution at m=((3(√2))/4)
$${y}=\sqrt{\mathrm{18}+\mathrm{3}{x}−{x}^{\mathrm{2}} }\:\mathrm{is}\:\mathrm{a}\:\mathrm{semi}\:\mathrm{circle}\:\mathrm{with}\:{r}=\frac{\mathrm{9}}{\mathrm{2}} \\ $$$$\sqrt{{x}+\mathrm{3}}+\sqrt{\mathrm{6}−{x}}\:\mathrm{has}\:\mathrm{the}\:\mathrm{maximum}\:\begin{pmatrix}{\frac{\mathrm{3}}{\mathrm{2}}}\\{\mathrm{3}\sqrt{\mathrm{2}}}\end{pmatrix} \\ $$$$\mathrm{We}\:\mathrm{have}\:\mathrm{2}\:\mathrm{solutions}\:\mathrm{for}\:\mathrm{0}\leqslant{m}<\frac{\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{4}}\:\mathrm{and} \\ $$$$\mathrm{exactly}\:\mathrm{one}\:\mathrm{solution}\:\mathrm{at}\:{m}=\frac{\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$
Commented by efronzo1 last updated on 07/Jun/24
how ≅
$$\mathrm{how}\:\:\cancel{\underbrace{\cong}} \\ $$
Commented by Frix last updated on 07/Jun/24
Draw it x∈[−3, 6] f(x)=3≤(√(x+3))+(√(6−x))≤3(√2) g(x)=0≤(√((x+3)(6−x)))≤(9/2)
$$\mathrm{Draw}\:\mathrm{it}\:{x}\in\left[−\mathrm{3},\:\mathrm{6}\right] \\ $$$${f}\left({x}\right)=\mathrm{3}\leqslant\sqrt{{x}+\mathrm{3}}+\sqrt{\mathrm{6}−{x}}\leqslant\mathrm{3}\sqrt{\mathrm{2}} \\ $$$${g}\left({x}\right)=\mathrm{0}\leqslant\sqrt{\left({x}+\mathrm{3}\right)\left(\mathrm{6}−{x}\right)}\leqslant\frac{\mathrm{9}}{\mathrm{2}} \\ $$
Answered by mr W last updated on 07/Jun/24
m((√(3+x))+(√(6−x)))=(√((3+x)(6−x))) let u=x−(3/2) m((√((9/2)+u))+(√((9/2)−u)))=(√(((9/2)+u)((9/2)−u))) both LHS and RHS are even functions with respect to u. due to symmetry there are (is) 1) none solution or 2) one solution at u=0 or 3) two solutions at u=±a (a≠0) or 4) three solutions at u=0 ∧ u=±a (a≠0) or 5) four solutions at u=±a ∧ u=±b (a≠b≠0) or etc. since there is only one solution, it must be at u=0, i.e. m((√((9/2)+0))+(√((9/2)−0)))=(√(((9/2)+0)((9/2)−0))) 2m(√(9/2))=(9/2) ⇒m=(1/2)(√(9/2))=((3(√2))/4) ✓
$${m}\left(\sqrt{\mathrm{3}+{x}}+\sqrt{\mathrm{6}−{x}}\right)=\sqrt{\left(\mathrm{3}+{x}\right)\left(\mathrm{6}−{x}\right)} \\ $$$${let}\:{u}={x}−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${m}\left(\sqrt{\frac{\mathrm{9}}{\mathrm{2}}+{u}}+\sqrt{\frac{\mathrm{9}}{\mathrm{2}}−{u}}\right)=\sqrt{\left(\frac{\mathrm{9}}{\mathrm{2}}+{u}\right)\left(\frac{\mathrm{9}}{\mathrm{2}}−{u}\right)} \\ $$$${both}\:{LHS}\:{and}\:{RHS}\:{are}\:{even}\:{functions} \\ $$$${with}\:{respect}\:{to}\:{u}. \\ $$$${due}\:{to}\:{symmetry}\:{there}\:{are}\:\left({is}\right) \\ $$$$\left.\mathrm{1}\right)\:{none}\:{solution}\:{or} \\ $$$$\left.\mathrm{2}\right)\:{one}\:{solution}\:{at}\:{u}=\mathrm{0}\:{or} \\ $$$$\left.\mathrm{3}\right)\:{two}\:{solutions}\:{at}\:{u}=\pm{a}\:\left({a}\neq\mathrm{0}\right)\:{or} \\ $$$$\left.\mathrm{4}\right)\:{three}\:{solutions}\:{at}\:{u}=\mathrm{0}\:\wedge\:{u}=\pm{a}\:\left({a}\neq\mathrm{0}\right)\:{or} \\ $$$$\left.\mathrm{5}\right)\:{four}\:{solutions}\:{at}\:{u}=\pm{a}\:\wedge\:{u}=\pm{b}\:\left({a}\neq{b}\neq\mathrm{0}\right)\:{or} \\ $$$${etc}. \\ $$$${since}\:{there}\:{is}\:{only}\:{one}\:{solution}, \\ $$$${it}\:{must}\:{be}\:{at}\:{u}=\mathrm{0},\:{i}.{e}. \\ $$$${m}\left(\sqrt{\frac{\mathrm{9}}{\mathrm{2}}+\mathrm{0}}+\sqrt{\frac{\mathrm{9}}{\mathrm{2}}−\mathrm{0}}\right)=\sqrt{\left(\frac{\mathrm{9}}{\mathrm{2}}+\mathrm{0}\right)\left(\frac{\mathrm{9}}{\mathrm{2}}−\mathrm{0}\right)} \\ $$$$\mathrm{2}{m}\sqrt{\frac{\mathrm{9}}{\mathrm{2}}}=\frac{\mathrm{9}}{\mathrm{2}} \\ $$$$\Rightarrow{m}=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{9}}{\mathrm{2}}}=\frac{\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{4}}\:\checkmark \\ $$
Commented by mr W last updated on 07/Jun/24
possible cases for the intersection from two even functions: f(x)=f(−x) g(x)=g(−x)
$${possible}\:{cases}\:{for}\:{the}\:{intersection} \\ $$$${from}\:{two}\:{even}\:{functions}: \\ $$$${f}\left({x}\right)={f}\left(−{x}\right) \\ $$$${g}\left({x}\right)={g}\left(−{x}\right) \\ $$
Commented by mr W last updated on 07/Jun/24
Commented by Frix last updated on 08/Jun/24
But in this special case obviously 1. 3≤(√(x+3))+(√(6−x))≤3(√2) 2. 0≤(√((x+3)(6−x)))≤(9/2) ⇒ 3. m≥0 4. We never get more than 2 real solutions These solutions are the solutions of x^2 −3x+(m^2 +6)(2m^2 −3)+2m^3 (√(m^2 +9))=0 x∈R ⇔ 0≤m≤((3(√2))/4)
$$\mathrm{But}\:\mathrm{in}\:\mathrm{this}\:\mathrm{special}\:\mathrm{case}\:\mathrm{obviously} \\ $$$$\mathrm{1}.\:\mathrm{3}\leqslant\sqrt{{x}+\mathrm{3}}+\sqrt{\mathrm{6}−{x}}\leqslant\mathrm{3}\sqrt{\mathrm{2}} \\ $$$$\mathrm{2}.\:\mathrm{0}\leqslant\sqrt{\left({x}+\mathrm{3}\right)\left(\mathrm{6}−{x}\right)}\leqslant\frac{\mathrm{9}}{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{3}.\:{m}\geqslant\mathrm{0} \\ $$$$\mathrm{4}.\:\mathrm{We}\:\mathrm{never}\:\mathrm{get}\:\mathrm{more}\:\mathrm{than}\:\mathrm{2}\:\mathrm{real}\:\mathrm{solutions} \\ $$$$ \\ $$$$\mathrm{These}\:\mathrm{solutions}\:\mathrm{are}\:\mathrm{the}\:\mathrm{solutions}\:\mathrm{of} \\ $$$${x}^{\mathrm{2}} −\mathrm{3}{x}+\left({m}^{\mathrm{2}} +\mathrm{6}\right)\left(\mathrm{2}{m}^{\mathrm{2}} −\mathrm{3}\right)+\mathrm{2}{m}^{\mathrm{3}} \sqrt{{m}^{\mathrm{2}} +\mathrm{9}}=\mathrm{0} \\ $$$${x}\in\mathbb{R}\:\Leftrightarrow\:\mathrm{0}\leqslant{m}\leqslant\frac{\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$
Commented by mr W last updated on 07/Jun/24
i was talking about general case. i didn′t mean that every equation may have no solution or one solutions or three solutions etc. if f(x)=g(x) has only one solution and f(x) and g(x) are even, then f(0)=g(0). this is the easiest way to solve this question, i think.
$${i}\:{was}\:{talking}\:{about}\:{general}\:{case}.\:{i} \\ $$$${didn}'{t}\:{mean}\:{that}\:{every}\:{equation}\:{may} \\ $$$${have}\:{no}\:{solution}\:{or}\:{one}\:{solutions} \\ $$$${or}\:{three}\:{solutions}\:{etc}. \\ $$$${if}\:{f}\left({x}\right)={g}\left({x}\right)\:{has}\:{only}\:{one}\:{solution} \\ $$$${and}\:{f}\left({x}\right)\:{and}\:{g}\left({x}\right)\:{are}\:{even},\:{then} \\ $$$${f}\left(\mathrm{0}\right)={g}\left(\mathrm{0}\right).\:{this}\:{is}\:{the}\:{easiest}\:{way} \\ $$$${to}\:{solve}\:{this}\:{question},\:{i}\:{think}. \\ $$
Commented by Frix last updated on 08/Jun/24
Yes. I thought “a solution” ≠ “exactly 1 solution”
$$\mathrm{Yes}. \\ $$$$\mathrm{I}\:\mathrm{thought}\:“\mathrm{a}\:\mathrm{solution}''\:\neq\:“\mathrm{exactly}\:\mathrm{1}\:\mathrm{solution}'' \\ $$
Commented by mr W last updated on 08/Jun/24
you are basically right. but i think many questioners here are non−native english speakers. they say “a solution” and mean “one solution”.
$${you}\:{are}\:{basically}\:{right}.\:{but}\:{i}\:{think} \\ $$$${many}\:{questioners}\:{here}\:{are}\: \\ $$$${non}−{native}\:{english}\:{speakers}.\:{they} \\ $$$${say}\:“{a}\:{solution}''\:{and}\:{mean}\: \\ $$$$“{one}\:{solution}''. \\ $$

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