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Question Number 107018 by bemath last updated on 08/Aug/20

Given f(x)=((10)/(2−sin 2x)). Find maximum value f(x).

$$\mathcal{G}{iven}\:{f}\left({x}\right)=\frac{\mathrm{10}}{\mathrm{2}−\mathrm{sin}\:\mathrm{2}{x}}.\:{Find}\:{maximum}\:{value}\: \\ $$$${f}\left({x}\right). \\ $$

Commented by PRITHWISH SEN 2 last updated on 08/Aug/20

for f(x) to be max. 2−sin 2x must be min. i.e sin 2x must be max.i.e 1 ∴ max. f(x)=((10)/(2−1)) = 10 and for min. sin 2x must be min. i.e −1 ∴ min.f(x)=((10)/(2−(−1)))= 3.33..

$$\mathrm{for}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{to}\:\mathrm{be}\:\mathrm{max}.\:\mathrm{2}−\mathrm{sin}\:\mathrm{2x}\:\mathrm{must}\:\mathrm{be}\:\mathrm{min}.\: \\ $$$$\mathrm{i}.\mathrm{e}\:\mathrm{sin}\:\mathrm{2x}\:\mathrm{must}\:\mathrm{be}\:\mathrm{max}.\mathrm{i}.\mathrm{e}\:\mathrm{1} \\ $$$$\therefore\:\mathrm{max}.\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{10}}{\mathrm{2}−\mathrm{1}}\:=\:\mathrm{10} \\ $$$$\mathrm{and}\:\mathrm{for}\:\mathrm{min}.\:\mathrm{sin}\:\mathrm{2x}\:\mathrm{must}\:\mathrm{be}\:\mathrm{min}.\:\mathrm{i}.\mathrm{e}\:−\mathrm{1} \\ $$$$\therefore\:\mathrm{min}.\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{10}}{\mathrm{2}−\left(−\mathrm{1}\right)}=\:\mathrm{3}.\mathrm{33}.. \\ $$

Commented by kaivan.ahmadi last updated on 08/Aug/20

f is maximum if 2−sin2x is minimum.so since −1≤sin2x≤1⇒−1≤−sin2x≤1 ⇒1≤2−sin2x≤3⇒ min(2−sin2x)=1⇒ max(f)=((10)/1)=10

$${f}\:{is}\:{maximum}\:{if}\:\mathrm{2}−{sin}\mathrm{2}{x}\:{is}\:{minimum}.{so} \\ $$$${since}\:−\mathrm{1}\leqslant{sin}\mathrm{2}{x}\leqslant\mathrm{1}\Rightarrow−\mathrm{1}\leqslant−{sin}\mathrm{2}{x}\leqslant\mathrm{1} \\ $$$$\Rightarrow\mathrm{1}\leqslant\mathrm{2}−{sin}\mathrm{2}{x}\leqslant\mathrm{3}\Rightarrow\:{min}\left(\mathrm{2}−{sin}\mathrm{2}{x}\right)=\mathrm{1}\Rightarrow \\ $$$${max}\left({f}\right)=\frac{\mathrm{10}}{\mathrm{1}}=\mathrm{10} \\ $$

Answered by bemath last updated on 08/Aug/20

we claim −1≤−sin 2x≤1 −1+2≤2−sin 2x≤2+1 1 ≤2−sin 2x≤3 ⇒(1/3)≤(1/(2−sin 2x))≤1 ((10)/3)≤((10)/(2−sin 2x))≤10 ⇒((10)/3)≤f(x)≤10 → { ((max=10)),((min=((10)/3))) :}

$$\:{we}\:{claim}\:−\mathrm{1}\leqslant−\mathrm{sin}\:\mathrm{2}{x}\leqslant\mathrm{1} \\ $$$$−\mathrm{1}+\mathrm{2}\leqslant\mathrm{2}−\mathrm{sin}\:\mathrm{2}{x}\leqslant\mathrm{2}+\mathrm{1} \\ $$$$\mathrm{1}\:\leqslant\mathrm{2}−\mathrm{sin}\:\mathrm{2}{x}\leqslant\mathrm{3}\:\Rightarrow\frac{\mathrm{1}}{\mathrm{3}}\leqslant\frac{\mathrm{1}}{\mathrm{2}−\mathrm{sin}\:\mathrm{2}{x}}\leqslant\mathrm{1} \\ $$$$\frac{\mathrm{10}}{\mathrm{3}}\leqslant\frac{\mathrm{10}}{\mathrm{2}−\mathrm{sin}\:\mathrm{2}{x}}\leqslant\mathrm{10}\:\Rightarrow\frac{\mathrm{10}}{\mathrm{3}}\leqslant{f}\left({x}\right)\leqslant\mathrm{10} \\ $$$$\rightarrow\begin{cases}{{max}=\mathrm{10}}\\{{min}=\frac{\mathrm{10}}{\mathrm{3}}}\end{cases} \\ $$

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