If u = a^2^2+b^2^2 + a^2^2+b^2^2, then the difference between the maximum and minimum valu…

Question

If $u = \sqrt{a^2\cos^2	heta+b^2\sin^2	heta} + \sqrt{a^2\sin^2	heta+b^2\cos^2	heta}$, then the difference between the maximum and minimum values of $u^2$ is given by:

Accepted Answer

Given $$u = \sqrt {{a^2}{{\cos }^2}\theta  + {b^2}{{\sin }^2}\theta }  $$ $+ \sqrt {{a^2}{{\sin }^2}\theta  + {b^2}{{\cos }^2}\theta } $$\therefore$${u^2} = {a^2}{\cos ^2}\theta  + {b^2}{\sin ^2}\theta  + {a^2}{\sin ^2}\theta  + {b^2}{\cos ^2}\theta $$ + 2\sqrt {\left( {{a^2}{{\cos }^2}\theta  + {b^2}{{\sin }^2}\theta } \right)}  \times \sqrt {\left( {{a^2}{{\sin }^2}\theta  + {b^2}{{\cos }^2}\theta } \right)} $$ \Rightarrow {u^2} = $${a^2} + {b^2}$$ +$ $2\sqrt {\left( {{a^4} + {b^4}} \right){{\cos }^2}\theta {{\sin }^2}\theta  + {a^2}{b^2}\left( {{{\cos }^4}\theta  + {{\sin }^4}\theta } \right)} $$ \Rightarrow {u^2} = $${a^2} + {b^2}$$ +$ $2\sqrt {\left( {{a^4} + {b^4}} \right){{\cos }^2}\theta {{\sin }^2}\theta  + {a^2}{b^2}\left( {1 - 2{{\cos }^2}\theta {{\sin }^2}\theta } \right)} $$ \Rightarrow {u^2} = $${a^2} + {b^2}$$ +$ $2\sqrt {\left( {{a^4} + {b^4} - 2{a^2}{b^2}} \right){{\cos }^2}\theta {{\sin }^2}\theta  + {a^2}{b^2}} $$ \Rightarrow {u^2} = $${a^2} + {b^2}$$ +$ $2\sqrt {{{\left( {{a^2} - {b^2}} \right)}^2}{{{{\left( {2\cos \theta \sin \theta } \right)}^2}} \over 4} + {a^2}{b^2}} $$ \Rightarrow {u^2} = $${a^2} + {b^2}$$ +$ $2\sqrt {{{\left( {{a^2} - {b^2}} \right)}^2}{{{{\sin }^2}2\theta } \over 4} + {a^2}{b^2}} $We know$0 \le {\sin ^2}2\theta  \le 1$$\therefore$$0 \le {\left( {{a^2} - {b^2}} \right)^2}{{{{\sin }^2}2\theta } \over 4} \le {{{{\left( {{a^2} - {b^2}} \right)}^2}} \over 4}$$ \Rightarrow $${a^2}{b^2} \le $${\left( {{a^2} - {b^2}} \right)^2}{{{{\sin }^2}2\theta } \over 4} + {a^2}{b^2}$$ $ \le {{{{\left( {{a\^2} - {b\^2}} \right)}\^2}} \over 4} + {a\^2}{b\^2}$$

$$\therefore$$ Min value of ${u^2} = {a^2} + {b^2}$ $$ + 2\sqrt {{a^2}{b^2}} $$ = $${\left( {a + b} \right)^2}$$

and Max value of ${u^2} = {a^2} + {b^2}$ $$ + 2\sqrt {{{{{\left( {{a^2} - {b^2}} \right)}^2}} \over 4} + {a^2}{b^2}} $$

$= {a^2} + {b^2}$ $$ + 2\sqrt {{{{{\left( {{a^2} - {b^2}} \right)}^2} + 4{a^2}{b^2}} \over 4}} $$

$= {a^2} + {b^2}$ $$ + 2\sqrt {{{{{\left( {{a^2} + {b^2}} \right)}^2}} \over 4}} $$

$= {a^2} + {b^2}$ $$+\, {a^2} + {b^2}$$

$${ = 2\left( {{a^2} + {b^2}} \right)}$$

Max of ${u^2}$ - Min of ${u^2}$ = $${2\left( {{a^2} + {b^2}} \right)}$$ - $${\left( {a + b} \right)^2}$$

= $${2\left( {{a^2} + {b^2}} \right)}$$ - $${\left( {{a^2} + {b^2} + 2ab} \right)}$$

= $$\sqrt {{{ = 2\left( {{a^2} + {b^2} - 2ab} \right)} \over 4}} $$

= $${\left( {a - b} \right)^2}$$

Solutions · 1