Given-f-x-10-2-sin-2x-Find-maximum-value-f-x-

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

G i v e n f (x) = \frac{10}{2 - \sin 2 x} . F i n d m a x i m u m v a l u e

f (x) .

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

for f (x) to be \max . 2 - \sin 2 x must be \min .

i . e \sin 2 x must be \max . i . e 1

∴ \max . f (x) = \frac{10}{2 - 1} = 10

and for \min . \sin 2 x must be \min . i . e - 1

∴ \min . f (x) = \frac{10}{2 - (- 1)} = 3.33 . .

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

f i s m a x i m u m i f 2 - s i n 2 x i s m i n i m u m . s o

s i n c e - 1 ⩽ s i n 2 x ⩽ 1 \Rightarrow - 1 ⩽ - s i n 2 x ⩽ 1

\Rightarrow 1 ⩽ 2 - s i n 2 x ⩽ 3 \Rightarrow m i n (2 - s i n 2 x) = 1 \Rightarrow

m a x (f) = \frac{10}{1} = 10

Answered by bemath last updated on 08/Aug/20

w e c l a i m - 1 ⩽ - \sin 2 x ⩽ 1

- 1 + 2 ⩽ 2 - \sin 2 x ⩽ 2 + 1

1 ⩽ 2 - \sin 2 x ⩽ 3 \Rightarrow \frac{1}{3} ⩽ \frac{1}{2 - \sin 2 x} ⩽ 1

\frac{10}{3} ⩽ \frac{10}{2 - \sin 2 x} ⩽ 10 \Rightarrow \frac{10}{3} ⩽ f (x) ⩽ 10

\to {\begin{cases} m a x = 10 \\ m i n = \frac{10}{3} \end{cases}

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