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Question-197848

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Question Number 197848 by cortano12 last updated on 01/Oct/23
Answered by mr W last updated on 01/Oct/23
f(x)=(a^2 /(cos^2 x))+(b^2 /(sin^2 x))≥(((a+b)^2 )/(cos^2 x+sin^2 x))=(a+b)^2 ⇒f(x)_(min) =(a+b)^2
$${f}\left({x}\right)=\frac{{a}^{\mathrm{2}} }{\mathrm{cos}^{\mathrm{2}} \:{x}}+\frac{{b}^{\mathrm{2}} }{\mathrm{sin}^{\mathrm{2}} \:{x}}\geqslant\frac{\left({a}+{b}\right)^{\mathrm{2}} }{\mathrm{cos}^{\mathrm{2}} \:{x}+\mathrm{sin}^{\mathrm{2}} \:{x}}=\left({a}+{b}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{f}\left({x}\right)_{{min}} =\left({a}+{b}\right)^{\mathrm{2}} \\ $$
Commented by cortano12 last updated on 01/Oct/23
Titus Lemma sir ?
$$\:\mathrm{Titus}\:\mathrm{Lemma}\:\mathrm{sir}\:? \\ $$
Commented by mr W last updated on 01/Oct/23
Answered by MM42 last updated on 01/Oct/23
f(x)=a^2 +b^2 +b^2 tan^2 x+a^2 cot^2 x f^′ (x)=2b^2 tanx(1+tan^2 x)−2a^2 cotx(1+cot^2 x) f′(x)=0⇒tan^2 x=(a/b) ⇒minf=a^2 +b^2 +b^2 ×(a/b)+a^2 ×(b/a) =(a+b)^2 ✓
$${f}\left({x}\right)={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{b}^{\mathrm{2}} {tan}^{\mathrm{2}} {x}+{a}^{\mathrm{2}} {cot}^{\mathrm{2}} {x} \\ $$$${f}^{'} \left({x}\right)=\mathrm{2}{b}^{\mathrm{2}} {tanx}\left(\mathrm{1}+{tan}^{\mathrm{2}} {x}\right)−\mathrm{2}{a}^{\mathrm{2}} {cotx}\left(\mathrm{1}+{cot}^{\mathrm{2}} {x}\right) \\ $$$${f}'\left({x}\right)=\mathrm{0}\Rightarrow{tan}^{\mathrm{2}} {x}=\frac{{a}}{{b}} \\ $$$$\Rightarrow{minf}={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{b}^{\mathrm{2}} ×\frac{{a}}{{b}}+{a}^{\mathrm{2}} ×\frac{{b}}{{a}} \\ $$$$=\left({a}+{b}\right)^{\mathrm{2}} \:\:\checkmark \\ $$
Answered by dimentri last updated on 01/Oct/23
f(x)= ((a^2 sin^2 x)/(cos^2 x)) +((b^2 cos^2 x)/(sin^2 x)) +a^2 +b^2 f(x)= a^2 tan^2 x + b^2 cot^2 x + a^2 +b^2 f(x) ≥ 2(√(a^2 tan^2 x. b^2 cot^2 x)) +a^2 +b^2 f(x) ≥ 2ab+a^2 +b^2 f(x) ≥ (a+b)^2 min f(x) = (a+b)^2
$${f}\left({x}\right)=\:\frac{{a}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} {x}}{\mathrm{cos}\:^{\mathrm{2}} {x}}\:+\frac{{b}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} {x}}{\mathrm{sin}\:^{\mathrm{2}} {x}}\:+{a}^{\mathrm{2}} +{b}^{\mathrm{2}} \\ $$$$\:{f}\left({x}\right)=\:{a}^{\mathrm{2}} \:\mathrm{tan}\:^{\mathrm{2}} {x}\:+\:{b}^{\mathrm{2}} \:\mathrm{cot}\:^{\mathrm{2}} {x}\:+\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} \: \\ $$$$\:{f}\left({x}\right)\:\geqslant\:\mathrm{2}\sqrt{{a}^{\mathrm{2}} \mathrm{tan}\:^{\mathrm{2}} {x}.\:{b}^{\mathrm{2}} \mathrm{cot}\:^{\mathrm{2}} {x}}\:+{a}^{\mathrm{2}} +{b}^{\mathrm{2}} \\ $$$$\:{f}\left({x}\right)\:\geqslant\:\mathrm{2}{ab}+{a}^{\mathrm{2}} +{b}^{\mathrm{2}} \\ $$$$\:{f}\left({x}\right)\:\geqslant\:\left({a}+{b}\right)^{\mathrm{2}} \\ $$$$\:\:{min}\:{f}\left({x}\right)\:=\:\left({a}+{b}\right)^{\mathrm{2}} \\ $$
Answered by mnjuly1970 last updated on 01/Oct/23
f(x)= a^2 + b^( 2) + a^2 tan^2 (x)+b^( 2) cot^2 (x) −−−−− a^2 tan^2 (x) +b^2 cot^2 (x)≥ 2(√(a^2 b^2 )) −− f (x) ≥^(a,b>0) a^( 2 ) +b^( 2) + 2ab = (a+b )^( 2) min_( f) = (a+ b )^( 2)
$$\:\:\:\:\:{f}\left({x}\right)=\:{a}^{\mathrm{2}} +\:{b}^{\:\mathrm{2}} \:+\:{a}^{\mathrm{2}} {tan}^{\mathrm{2}} \left({x}\right)+{b}^{\:\mathrm{2}} {cot}^{\mathrm{2}} \left({x}\right) \\ $$$$\:\:\:−−−−− \\ $$$$\:\:\:\:\:\:{a}^{\mathrm{2}} {tan}^{\mathrm{2}} \left({x}\right)\:+{b}^{\mathrm{2}} {cot}^{\mathrm{2}} \left({x}\right)\geqslant\:\mathrm{2}\sqrt{{a}^{\mathrm{2}} {b}^{\mathrm{2}} } \\ $$$$−− \\ $$$$\:\:\:\:\:\:{f}\:\left({x}\right)\:\overset{{a},{b}>\mathrm{0}} {\geqslant}\:{a}^{\:\mathrm{2}\:} \:+{b}^{\:\mathrm{2}} \:+\:\mathrm{2}{ab}\:=\:\left({a}+{b}\:\right)^{\:\mathrm{2}} \\ $$$$\:\:\:\:\:{min}_{\:{f}} \:\:=\:\left({a}+\:{b}\:\right)^{\:\mathrm{2}} \\ $$

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