Question Number 161406 by mnjuly1970 last updated on 17/Dec/21
![f (x ) = cos^( 2) ( x ) + sin^( 4) ( x ) R_( f) = ? −−−solution−−− y = cos^( 2) (x ) + sin^( 2) (x) .( 1−cos^( 2) (x)) = 1 − (1/4) sin^( 2) ( 2x) we know that : 0≤ sin^( 2) ( αx) ≤1 therefore −1≤− sin^( 2) (2x) ≤ 0 1−(1/4) ≤ 1− (1/4) sin^( 2) (2x) ≤1 R_( f) = [ (3/4) , 1 ] ◂ ★ ▶](https://www.tinkutara.com/question/Q161406.png)
$$ \\ $$$$\:\:\:\:\:\:\:\:\:{f}\:\left({x}\:\right)\:=\:{cos}^{\:\mathrm{2}} \left(\:{x}\:\right)\:+\:{sin}^{\:\mathrm{4}} \left(\:{x}\:\right) \\ $$$$\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{R}_{\:{f}} \:=\:? \\ $$$$\:\:\:\:−−−{solution}−−− \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{y}\:=\:{cos}^{\:\mathrm{2}} \left({x}\:\right)\:+\:{sin}^{\:\mathrm{2}} \left({x}\right)\:.\left(\:\mathrm{1}−{cos}^{\:\mathrm{2}} \left({x}\right)\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{1}\:−\:\frac{\mathrm{1}}{\mathrm{4}}\:{sin}^{\:\mathrm{2}} \left(\:\mathrm{2}{x}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:{we}\:{know}\:\:{that}\:: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\leqslant\:{sin}^{\:\mathrm{2}} \left(\:\alpha{x}\right)\:\leqslant\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{therefore} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{1}\leqslant−\:{sin}^{\:\mathrm{2}} \left(\mathrm{2}{x}\right)\:\leqslant\:\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\:\leqslant\:\mathrm{1}−\:\frac{\mathrm{1}}{\mathrm{4}}\:{sin}^{\:\mathrm{2}} \left(\mathrm{2}{x}\right)\:\leqslant\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{R}_{\:{f}} \:=\:\left[\:\frac{\mathrm{3}}{\mathrm{4}}\:\:,\:\mathrm{1}\:\right]\:\:\:\:\:\:\:\:\:\:\blacktriangleleft\:\bigstar\:\blacktriangleright \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\: \\ $$$$ \\ $$