Question Number 162728 by mkam last updated on 31/Dec/21
Commented by mkam last updated on 31/Dec/21
$$???\: \\ $$
Commented by tounghoungko last updated on 01/Jan/22
![f(x)=(a^2 /u)+(b^2 /(1−u)) ; u=sin^2 x ; 0<u<1 f ′(x)=[−(a^2 /u^2 )+(b^2 /((1−u)^2 )) ].sin 2x =0 ⇔ (bu)^2 −(a−au)^2 = 0 ⇔ ((b−a)u+a)((b+a)u−a)=0 { ((u=−(a/(b−a))=(a/(a−b)))),((u=(a/(a+b)))) :} f(x)_(min) = (a^2 /(((a/(a+b))))) + (b^2 /((1−(a/(a+b))))) = a(a+b)+b(a+b)= (a+b)^2 =∣a+b∣^2](Q162751.png)
$$\:{f}\left({x}\right)=\frac{{a}^{\mathrm{2}} }{{u}}+\frac{{b}^{\mathrm{2}} }{\mathrm{1}−{u}}\:;\:{u}=\mathrm{sin}\:^{\mathrm{2}} {x}\:;\:\mathrm{0}<{u}<\mathrm{1} \\ $$$$\:{f}\:'\left({x}\right)=\left[−\frac{{a}^{\mathrm{2}} }{{u}^{\mathrm{2}} }+\frac{{b}^{\mathrm{2}} }{\left(\mathrm{1}−{u}\right)^{\mathrm{2}} }\:\right].\mathrm{sin}\:\mathrm{2}{x}\:=\mathrm{0}\: \\ $$$$\:\Leftrightarrow\:\left({bu}\right)^{\mathrm{2}} −\left({a}−{au}\right)^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\:\Leftrightarrow\:\left(\left({b}−{a}\right){u}+{a}\right)\left(\left({b}+{a}\right){u}−{a}\right)=\mathrm{0} \\ $$$$\:\begin{cases}{{u}=−\frac{{a}}{{b}−{a}}=\frac{{a}}{{a}−{b}}}\\{{u}=\frac{{a}}{{a}+{b}}}\end{cases}\: \\ $$$$\:{f}\left({x}\right)_{{min}} \:=\:\frac{{a}^{\mathrm{2}} }{\left(\frac{{a}}{{a}+{b}}\right)}\:+\:\frac{{b}^{\mathrm{2}} }{\left(\mathrm{1}−\frac{{a}}{{a}+{b}}\right)} \\ $$$$\:=\:{a}\left({a}+{b}\right)+{b}\left({a}+{b}\right)=\:\left({a}+{b}\right)^{\mathrm{2}} \:=\mid{a}+{b}\mid^{\mathrm{2}} \: \\ $$
Answered by mnjuly1970 last updated on 31/Dec/21
$$\:\:{f}\left({x}\right)\:=\:{a}^{\:\mathrm{2}} +{b}^{\:\mathrm{2}} +\:{a}^{\:\mathrm{2}} {cot}^{\:\mathrm{2}} \left({x}\right)+{b}^{\:\mathrm{2}} {tan}^{\:\mathrm{2}} \left({x}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\geqslant\:{a}^{\:\mathrm{2}} +\:{b}^{\:\mathrm{2}} +\:\mathrm{2}\:\sqrt{\:{a}^{\:\mathrm{2}} .{b}^{\:\mathrm{2}} }\:=\:\left(\mid{a}\mid+\mid{b}\mid\right)^{\:\mathrm{2}} \\ $$$$\:\:\:\:{f}_{\:{min}} \:=\:\left(\:\mid{a}\mid\:+\:\mid\:{b}\mid\right)^{\:\mathrm{2}} \\ $$
Answered by safmoh22 last updated on 31/Dec/21
![665n[x((66x)/([22]))b]](Q162733.png)
$$\mathrm{665}{n}\left[{x}\frac{\mathrm{66}{x}}{\left[\mathrm{22}\right]}{b}\right] \\ $$