h-x-sin-4-x-cos-4-x-2msinxcosx-Find-all-the-values-of-the-parameter-m-for-the-funtion-denined-on-R-

Question Number 41911 by lucha116 last updated on 15/Aug/18

h (x) = \sqrt{{s i n}^{4} x + {c o s}^{4} x - 2 m s i n x c o s x}

F i n d a l l t h e v a l u e s o f t h e p a r a m e t e r m f o r t h e f u n t i o n d e n i n e d o n R

Commented by MJS last updated on 15/Aug/18

draw the function for ∣ m ∣> \frac{1}{2}, you will see

that {it}^{'} s not defined \forall x \in R

m = 1 \Rightarrow \sin^{4} \frac{π}{4} + \cos^{4} \frac{π}{4} - 2 \sin \frac{π}{4} \cos \frac{π}{4} =

= \frac{1}{4} + \frac{1}{4} - 2 \times \frac{\sqrt{2}}{2} \times \frac{\sqrt{2}}{2} = - \frac{1}{2}

as I hopefully showed the minimum of

\sin^{4} x + \cos^{4} x - 2 m \sin x \cos x must be ⩾ 0

and {it}^{'} s exactly zero for m = \pm \frac{1}{2} . for ∣ m ∣> \frac{1}{2}

the minimum is below zero

Commented by math khazana by abdo last updated on 15/Aug/18

w e h a v e h (x) = \sqrt{{({c o s}^{2} x + {s i n}^{2} x)}^{2} - 2 {c o s}^{2} {x s i n}^{2} x - 2 m s i n x c o s x}

= \sqrt{1 - \frac{1}{2} {s i n}^{2} (2 x) - m s i n (2 x)}

= \frac{1}{\sqrt{2}} \sqrt{2 - {s i n}^{2} (2 x) - 2 m s i n (2 x)} d u e t o D_{f} w e m i s t

h a v e - {s i n}^{2} (2 x) - 2 m s i n (2 x) + 2 ⩾ 0 \Rightarrow

{s i n}^{2} (2 x) + 2 m s i n (2 x) - 2 ⩽ 0 l e t s i n (2 x) = t \Rightarrow

t^{2} + 2 m t - 2 ⩽ 0 a n d ∣ t ∣⩽ 1 \Rightarrow

Δ^{^{'}} = m^{2} + 2 > 0 \Rightarrow t_{1} = - m + \sqrt{m^{2} + 2} a n d

t_{2} = - m - \sqrt{m^{2} + 2}

∣ t_{1} ∣ ⩽ 1 \Rightarrow - 1 ⩽ - m + \sqrt{m^{2} + 2} ⩽ 1 \Rightarrow

m - 1 ⩽ \sqrt{m^{2} + 2} ⩽ m + 1 i f m > 1 \Rightarrow

m^{2} - 2 m + 1 ⩽ m^{2} + 2 ⩽ m^{2} + 2 m + 2 \Rightarrow

- 2 m + 1 ⩽ 2 ⩽ 2 m + 2 \Rightarrow - m + \frac{1}{2} ⩽ 1 ⩽ m + 1 \Rightarrow

- m ⩽ \frac{1}{2} a n d m ⩾ 0 \Rightarrow m ⩾ 0 \Rightarrow m ⩾ 1

i f - 1 < m < 1 \Rightarrow - (1 - m) ⩽ \sqrt{m^{2} + 2} ⩽ m + 1 \Rightarrow

m^{2} - 2 m + 1 ⩽ m^{2} + 2 ⩽ m^{2} + 2 m + 1 \Rightarrow

- m ⩽ \frac{1}{2} a n d m ⩾ 0 \Rightarrow m ⩾ 0 \Rightarrow 0 ⩽ m ⩽ 1

∣ t_{2} ∣⩽ 1 \Rightarrow ∣ - m - \sqrt{m^{2} + 2} ∣⩽ 1 \Rightarrow

- 1 ⩽ m + \sqrt{m^{2} + 2} ⩽ 1 \Rightarrow - m - 1 ⩽ \sqrt{m^{2} + 2} ⩽ 1 - m s o

m ⩽ 1 i f - m - 1 > 0 \Rightarrow m ⩽ - 1 \Rightarrow

m^{2} + 2 m + 1 ⩽ m^{2} + 2 ⩽ m^{2} - 2 m + 1 \Rightarrow

2 m + 1 ⩽ 2 ⩽ - 2 m + 1 \Rightarrow 2 m ⩽ 1 ⩽ - 2 m \Rightarrow

m ⩽ \frac{1}{2} a n d \frac{1}{2} ⩽ - m \Rightarrow m ⩽ \frac{1}{2} a n d m ⩽ - \frac{1}{2} \Rightarrow

m ⩽ - \frac{1}{2} \Rightarrow m ⩽ - 1 d u e t o i n i t i a l c o n d i t i o n

i f - m - 1 ⩽ 0 \Rightarrow - 1 ⩽ m ⩽ 1 s o

∣ t_{2} ∣ ⩽ 1 \Rightarrow - 1 ⩽ m ⩽ 1 f i n a l l y

∣ t_{1} ∣⩽ 1 a n d ∣ t_{2} ∣⩽ 1 \Rightarrow 0 ⩽ m ⩽ 1 .

Commented by math khazana by abdo last updated on 15/Aug/18

a n o t h e r w a y b u t e a s y w e h a v e ∣ t_{1} ∣⩽ 1 a n d

∣ t_{2} ∣⩽ 1 \Rightarrow - 1 ⩽ - m + \sqrt{m^{2} + 2} ⩽ 1 a n d

- 1 ⩽ - m - \sqrt{m^{2} + 2} ⩽ 1 \Rightarrow

- 2 ⩽ - 2 m ⩽ 2 \Rightarrow - 1 ⩽ - m ⩽ 1 \Rightarrow - 1 ⩽ m ⩽ 1

Commented by math khazana by abdo last updated on 15/Aug/18

- 1 ⩽ m ⩽ 1

Commented by maxmathsup by imad last updated on 15/Aug/18

f o r m = 1 h (x) = \sqrt{{s i n}^{4} x + {c o s}^{4} x - 2 s i n x c o s x}

= \sqrt{{({s i n}^{2} x + {c o s}^{2} x)}^{2} - 2 {s i n}^{2} {x c o s}^{2} x - s i n (2 x)}

= \sqrt{1 - \frac{1}{2} {s i n}^{2} (2 x) - s i n (2 x)} = \frac{1}{\sqrt{2}} \sqrt{2 - {s i n}^{2} (2 x) - 2 s i n (2 x)}

= \frac{1}{\sqrt{2}} \sqrt{- {s i n}^{2} (2 x) - 2 s i n (2 x) + 2} l e t t = s i n (2 x)

Δ^{^{'}} = 1 + 2 = 3 \Rightarrow t_{1} = \frac{1 + \sqrt{3}}{- 1} a n d t_{2} = \frac{1 - \sqrt{3}}{- 1} \Rightarrow t_{1} = - \sqrt{3} - 1 a n d

t_{2} = - 1 - \sqrt{3} s o ∣ t_{1} ∣< 1 b u t ∣ t_{2} ∣ > 1 a n y w a y w e c a n t a k e - \frac{1}{2} ⩽ m ⩽ \frac{1}{2}

t h a n k y o u s i r f o r t h i s r e m a r k \dots

Answered by MJS last updated on 15/Aug/18

\sin^{4} x + \cos^{4} x - 2 m \sin x \cos x ⩾ 0

\dots after some trigonometric transformations :

\frac{1}{4} \cos 4 x - m \sin 2 x + \frac{3}{4} ⩾ 0

\frac{d}{d x} [\frac{1}{4} \cos 4 x - m \sin 2 x + \frac{3}{4}] = 0

- \sin 4 x - 2 m \cos 2 x = 0

x = \arctan t

\frac{2 {m t}^{4} + 4 t^{3} - 4 t^{2} - 2 m}{{(t^{2} + 1)}^{2}} = 0

t^{4} + \frac{2 t^{3}}{m} - \frac{2 t}{m} - 1 = 0

t = \pm 1 \lor t = - \frac{1}{m} \pm \frac{\sqrt{1 - m^{2}}}{m} \Rightarrow - 1 ⩽ m ⩽ 1

\Rightarrow x = \pm \frac{π}{4} \lor x = \arctan (- \frac{1}{m} \pm \frac{\sqrt{1 - m^{2}}}{m})

\frac{1}{4} \cos 4 x - m \sin 2 x + \frac{3}{4} ⩾ 0

x_{1} = - \frac{π}{4} \Rightarrow m + \frac{1}{2} ⩾ 0 \Rightarrow m ⩾ - \frac{1}{2}

x_{2} = \frac{π}{4} \Rightarrow - m + \frac{1}{2} ⩾ 0 \Rightarrow m ⩽ \frac{1}{2}

x_{3, 4} = \arctan (- \frac{1}{m} \pm \frac{\sqrt{1 - m^{2}}}{m}) \Rightarrow \frac{m^{2}}{2} + 1 ⩾ 0 \Rightarrow m \in R

answer is - \frac{1}{2} ⩽ m ⩽ \frac{1}{2}

Commented by lucha116 last updated on 15/Aug/18

i t i s t r u e b u t i h a v e s t u d i e d g r a d e 11 a n d i h a v e n o t l e a r n t \frac{d}{d x} . C a n y o u s o l v e b y o t h e r w a y ?

Commented by maxmathsup by imad last updated on 15/Aug/18

t h i s q u e s t i o n i s m o r e d i f f i c u l t f o r g r a d e 11 a n d g e n e r a l l y t h e c o n s e p t o f

t r i g o n o m e t r y i s m o r e d i f f i c u l t \dots

Commented by lucha116 last updated on 15/Aug/18

y e s . i k n o w . t h k s \hat{} 0 \hat{}

Answered by MJS last updated on 15/Aug/18

hopefully an easier explanation

\sin^{2} x = \frac{1}{2} (1 - \cos 2 x)

\cos^{2} x = \frac{1}{2} (1 + \cos 2 x)

\sin^{4} x = {(\frac{1}{2} (1 - \cos 2 x))}^{2} =

= \frac{1}{4} (1 - 2 \cos 2 x + \cos^{2} 2 x) =

= \frac{1}{4} (1 - 2 \cos 2 x + \frac{1}{2} (1 + \cos 4 x)) =

= \frac{3}{8} + \frac{1}{8} \cos 4 x - \frac{1}{2} \cos 2 x

similar \cos^{4} x = \frac{3}{8} + \frac{1}{8} \cos 4 x + \frac{1}{2} \cos 2 x

\Rightarrow \sin^{4} x + \cos^{4} x = \frac{3}{4} + \frac{1}{4} \cos 4 x

this has a period of \frac{π}{2} with \min = (\begin{matrix} \frac{π}{4} + \frac{π}{2} z \\ \frac{1}{2} \end{matrix}); z \in Z

- 2 m \sin x \cos x = - m \sin 2 x

this has a period of π with \min = (\begin{matrix} \frac{π}{4} + π z \\ - m \end{matrix}); z \in Z

the sum of both gives (\begin{matrix} \frac{π}{4} \\ \frac{1}{2} - m \end{matrix}) within the

first period . we know that \frac{1}{2} - m ⩾ 0 \Rightarrow m ⩽ \frac{1}{2}

if m < 0 the \min of - m \sin 2 x changes to the

2^{nd} half of its period because the curve now

is upside down : \min = (\begin{matrix} \frac{3 π}{4} + π z \\ m \end{matrix})

so the sum now is (\begin{matrix} \frac{3 π}{4} \\ \frac{1}{2} + m \end{matrix}) \Rightarrow m ⩾ - \frac{1}{2}

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