Question

ex 2 si 3
dau coroana.............​

Answer 1

Răspuns:

2. f:R -> R, f(x)= (m²-2)x-3 - str. descrescatoare <=> m²-2<0

m²-2=0 <=> (m+2)(m-2)=0 => m=±2

m:       -∞                       -2       2               +∞  

m²-2:  -∞       +++++        0    -   0    ++++

m²-2<0 => m∈ (-2,2)

3. arctg (x/3) + arctg( 1/√3) = π/3

<=> tg ( arctg(x/3) + arctg(√3/3) ) = tg (π/3)

<=> tg(arctg (x/3)) + tg( arctg(√3/3)) / 1- tg(arctg (x/3)) * tg( arctg(√3/3)) = √3

<=> x/3 + √3/3 / 1- (x/3)*(√3/3) = √3

<=> x+√3/ 3*(1- x√3/9)=√3

<=> x+√3 / 3* (9-x√3)/3 = √3

<=> x+√3 / 9-x√3 =√3

<=> x+√3 = √3* (9-x√3)

<=> x+√3= 9√3 - 3x

<=> 4x= 8√3

<=> x= 2√3

Answer 2

3)

$\it arctg\dfrac{x}{3}+arctg\dfrac{1}{\sqrt3} =\dfrac{\pi}{3} \Rightarrow arctg\dfrac{x}{3}+\dfrac{\pi}{6}=\dfrac{\pi}{3} \Rightarrow arctg \dfrac{x}{3}=\dfrac{\pi}{6} \Rightarrow\\ \\ \\ \Rightarrow \dfrac{x}{3}=\dfrac{\sqrt3}{3} \Rightarrow x=\sqrt3$

2)

$\it m^2-2<0 \Rightarrow m^2<2 \Rightarrow \sqrt{m^2}<\sqrt2 \Rightarrow |m|<\sqrt2 \Rightarrow -\sqrt2<m<\sqrt2 \Rightarrow \\ \\ \Rightarrow m\in(-\sqrt2,\ \sqrt2)$