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Question Number 163048 by mnjuly1970 last updated on 03/Jan/22

	Answered by mahdipoor last updated on 03/Jan/22
	$sin^2 x=m cos^2 x=n ⇒m^′ =−n^′ =2cosx.sinx f ′=(nm^3 +mn^3 ) ′=m′(3nm^2 +n^3 )+n′(3mn^2 +m^3 )= 2cosx.sinx(n^3 −3mn^2 +3m^2 n−m^3 )= sin2x(n−m)^3 = sin2x(cos^2 x−sin^2 x)^3 = sin2x.cos^3 2x=0 ⇒x=((kπ)/4)+(π/2) { ((sin(((kπ)/4)+(π/2))=cos(((kπ)/4)))),((cos(((kπ)/4)+(π/2))=−sin(((kπ)/2)))) :} f(((kπ)/4)+(π/2))=(1/4)sin^2 (((kπ)/2))(sin^4 (((kπ)/4))+cos^4 (((kπ)/4))) = { ((2∣k f=0)),((2∤k f=+(1/8))) :} R_f =[0,0.125]$
	${s i n}^{2} x = m {c o s}^{2} x = n$ $\Rightarrow m^{^{'}} = - n^{^{'}} = 2 c o s x . s i n x$ $f^{'} = ({n m}^{3} + {m n}^{3})^{'} = m^{'} (3 {n m}^{2} + n^{3}) + n^{'} (3 {m n}^{2} + m^{3}) =$ $2 c o s x . s i n x (n^{3} - 3 {m n}^{2} + 3 m^{2} n - m^{3}) =$ $s i n 2 x {(n - m)}^{3} =$ $s i n 2 x {({c o s}^{2} x - {s i n}^{2} x)}^{3} =$ $s i n 2 x . {c o s}^{3} 2 x = 0$ $\Rightarrow x = \frac{k π}{4} + \frac{π}{2}$ ${\begin{cases} s i n (\frac{k π}{4} + \frac{π}{2}) = c o s (\frac{k π}{4}) \\ c o s (\frac{k π}{4} + \frac{π}{2}) = - s i n (\frac{k π}{2}) \end{cases}$ $f (\frac{k π}{4} + \frac{π}{2}) = \frac{1}{4} {s i n}^{2} (\frac{k π}{2}) ({s i n}^{4} (\frac{k π}{4}) + {c o s}^{4} (\frac{k π}{4}))$ $= {\begin{cases} 2 ∣ k f = 0 \\ 2 ∤ k f = + \frac{1}{8} \end{cases}$ $R_{f} = [0, 0.125]$
	Commented by mnjuly1970 last updated on 03/Jan/22

	$m e r c e y s i r m a h d i p o o r$
	Answered by tounghoungko last updated on 03/Jan/22
	$f(x)= sin^2 x cos^2 x (sin^4 x+cos^4 x) f(x)=(1/4)sin^2 2x(1−2sin^2 x cos^2 x) f(x)=(1/4)sin^2 2x (1−(1/2)sin^2 2x) f(x)= (1/4)sin^2 2x−(1/8)sin^4 2x f(x)=−(1/8)(sin^4 2x−2sin^2 2x+1)+(1/8) f(x)=−(1/8)(sin^2 2x−1)^2 +(1/8) where 0≤sin^2 2x ≤1 when { ((sin^2 2x =0 ⇒f = 0)),((sin^2 2x=1⇒f= (1/8))) :} Range f ⇒ 0 ≤f ≤ (1/8)$
	$f (x) = \sin^{2} x \cos^{2} x (\sin^{4} x + \cos^{4} x)$ $f (x) = \frac{1}{4} \sin^{2} 2 x (1 - 2 \sin^{2} x \cos^{2} x)$ $f (x) = \frac{1}{4} \sin^{2} 2 x (1 - \frac{1}{2} \sin^{2} 2 x)$ $f (x) = \frac{1}{4} \sin^{2} 2 x - \frac{1}{8} \sin^{4} 2 x$ $f (x) = - \frac{1}{8} (\sin^{4} 2 x - 2 \sin^{2} 2 x + 1) + \frac{1}{8}$ $f (x) = - \frac{1}{8} {(\sin^{2} 2 x - 1)}^{2} + \frac{1}{8}$ $w h e r e 0 ⩽ \sin^{2} 2 x ⩽ 1$ $w h e n {\begin{cases} \sin^{2} 2 x = 0 \Rightarrow f = 0 \\ \sin^{2} 2 x = 1 \Rightarrow f = \frac{1}{8} \end{cases}$ $R a n g e f \Rightarrow 0 ⩽ f ⩽ \frac{1}{8}$
	Commented by mnjuly1970 last updated on 03/Jan/22

	$v e r y n i c e s o l u t i o n s i r$
	Answered by abdullahhhhh last updated on 03/Jan/22

	Commented by mnjuly1970 last updated on 03/Jan/22

	$t h a n k s a l o t m a s t e r$
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