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Question Number 161396 by abdullah_ff last updated on 17/Dec/21

If y = cos^2 x + sin^4 x for all values of x,  then prove that   (3/4) ≤ y ≤ 1

$$\mathrm{If}\:{y}\:=\:{cos}^{\mathrm{2}} {x}\:+\:{sin}^{\mathrm{4}} {x}\:\mathrm{for}\:\mathrm{all}\:\mathrm{values}\:\mathrm{of}\:\mathrm{x}, \\ $$$$\mathrm{then}\:\mathrm{prove}\:\mathrm{that}\: \\ $$$$\frac{\mathrm{3}}{\mathrm{4}}\:\leq\:{y}\:\leq\:\mathrm{1} \\ $$

Commented by cortano last updated on 17/Dec/21

 y=cos^2 x+(1−cos^2 x)^2    y=cos^4 x−cos^2 x+1   y=(cos^2 x−(1/2))^2 +(3/4)   ⇒0≤cos^2 x≤1    { ((min = (3/4) when cos^2 x=(1/4))),((max =1 when cos^2 x=1)) :}

$$\:{y}=\mathrm{cos}\:^{\mathrm{2}} {x}+\left(\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} {x}\right)^{\mathrm{2}} \\ $$$$\:{y}=\mathrm{cos}\:^{\mathrm{4}} {x}−\mathrm{cos}\:^{\mathrm{2}} {x}+\mathrm{1} \\ $$$$\:{y}=\left(\mathrm{cos}\:^{\mathrm{2}} {x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\:\Rightarrow\mathrm{0}\leqslant\mathrm{cos}\:^{\mathrm{2}} {x}\leqslant\mathrm{1} \\ $$$$\:\begin{cases}{{min}\:=\:\frac{\mathrm{3}}{\mathrm{4}}\:{when}\:\mathrm{cos}\:^{\mathrm{2}} {x}=\frac{\mathrm{1}}{\mathrm{4}}}\\{{max}\:=\mathrm{1}\:{when}\:\mathrm{cos}\:^{\mathrm{2}} {x}=\mathrm{1}}\end{cases} \\ $$

Commented by abdullah_ff last updated on 17/Dec/21

thank you my dear sir..

$${thank}\:{you}\:{my}\:{dear}\:{sir}.. \\ $$

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