JEE MAIN - Mathematics (2004 - No. 29)
If $$u = \sqrt {{a^2}{{\cos }^2}\theta + {b^2}{{\sin }^2}\theta } + \sqrt {{a^2}{{\sin }^2}\theta + {b^2}{{\cos }^2}\theta } $$
then the difference between the maximum and minimum values of $${u^2}$$ is given by :
then the difference between the maximum and minimum values of $${u^2}$$ is given by :
$${\left( {a - b} \right)^2}$$
$$2\sqrt {{a^2} + {b^2}} $$
$${\left( {a + b} \right)^2}$$
$$2\left( {{a^2} + {b^2}} \right)$$
Explanation
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 $${u^2}$$ = $${2\left( {{a^2} + {b^2}} \right)}$$ - $${\left( {a + b} \right)^2}$$
= $${2\left( {{a^2} + {b^2}} \right)}$$ - $${\left( {{a^2} + {b^2} + 2ab} \right)}$$
= $$\sqrt {{{ = 2\left( {{a^2} + {b^2} - 2ab} \right)} \over 4}} $$
= $${\left( {a - b} \right)^2}$$
$$\therefore$$ $${u^2} = {a^2}{\cos ^2}\theta + {b^2}{\sin ^2}\theta + {a^2}{\sin ^2}\theta + {b^2}{\cos ^2}\theta $$
$$ + 2\sqrt {\left( {{a^2}{{\cos }^2}\theta + {b^2}{{\sin }^2}\theta } \right)} \times \sqrt {\left( {{a^2}{{\sin }^2}\theta + {b^2}{{\cos }^2}\theta } \right)} $$
$$ \Rightarrow {u^2} = $$ $${a^2} + {b^2}$$$$ + $$$$2\sqrt {\left( {{a^4} + {b^4}} \right){{\cos }^2}\theta {{\sin }^2}\theta + {a^2}{b^2}\left( {{{\cos }^4}\theta + {{\sin }^4}\theta } \right)} $$
$$ \Rightarrow {u^2} = $$ $${a^2} + {b^2}$$$$ + $$$$2\sqrt {\left( {{a^4} + {b^4}} \right){{\cos }^2}\theta {{\sin }^2}\theta + {a^2}{b^2}\left( {1 - 2{{\cos }^2}\theta {{\sin }^2}\theta } \right)} $$
$$ \Rightarrow {u^2} = $$ $${a^2} + {b^2}$$$$ + $$$$2\sqrt {\left( {{a^4} + {b^4} - 2{a^2}{b^2}} \right){{\cos }^2}\theta {{\sin }^2}\theta + {a^2}{b^2}} $$
$$ \Rightarrow {u^2} = $$ $${a^2} + {b^2}$$$$ + $$$$2\sqrt {{{\left( {{a^2} - {b^2}} \right)}^2}{{{{\left( {2\cos \theta \sin \theta } \right)}^2}} \over 4} + {a^2}{b^2}} $$
$$ \Rightarrow {u^2} = $$ $${a^2} + {b^2}$$$$ + $$$$2\sqrt {{{\left( {{a^2} - {b^2}} \right)}^2}{{{{\sin }^2}2\theta } \over 4} + {a^2}{b^2}} $$
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 $${u^2}$$ = $${2\left( {{a^2} + {b^2}} \right)}$$ - $${\left( {a + b} \right)^2}$$
= $${2\left( {{a^2} + {b^2}} \right)}$$ - $${\left( {{a^2} + {b^2} + 2ab} \right)}$$
= $$\sqrt {{{ = 2\left( {{a^2} + {b^2} - 2ab} \right)} \over 4}} $$
= $${\left( {a - b} \right)^2}$$
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