Question
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
A.
$${\left( {a - b} \right)^2}$$
B.
$$2\sqrt {{a^2} + {b^2}} $$
C.
$${\left( {a + b} \right)^2}$$
D.
$$2\left( {{a^2} + {b^2}} \right)$$
Answer :
$${\left( {a - b} \right)^2}$$
Solution :
$$\eqalign{
& {u^2} = {a^2} + {b^2} + 2\sqrt {\left( {{a^4} + {b^4}} \right){{\cos }^2}\theta {{\sin }^2}\theta + {a^2}{b^2}\left( {{{\cos }^4}\theta + {{\sin }^4}\theta } \right)} \,\,\,......\left( 1 \right) \cr
& {\text{Now}}\,\left( {{a^4} + {b^4}} \right){\cos ^2}\theta {\sin ^2}\theta + {a^2}{b^2}\left( {{{\cos }^4}\theta + {{\sin }^4}\theta } \right) \cr
& = \left( {{a^4} + {b^4}} \right){\cos ^2}\theta {\sin ^2}\theta + {a^2}{b^2}\left( {1 - 2{{\cos }^2}\theta {{\sin }^2}\theta } \right) \cr
& = \left( {{a^4} + {b^4} - 2{a^2}{b^2}} \right){\cos ^2}\theta {\sin ^2}\theta + {a^2}{b^2} \cr
& = {\left( {{a^2} - {b^2}} \right)^2}.\frac{{{{\sin }^2}2\theta }}{4} + {a^2}{b^2}\,\,\,.....\left( 2 \right) \cr
& \because \,\,0 \leqslant {\sin ^2}2\theta \leqslant 1 \cr
& \Rightarrow \,\,0 \leqslant {\left( {{a^2} - {b^2}} \right)^2}\frac{{{{\sin }^2}2\theta }}{4} \leqslant \frac{{{{\left( {{a^2} - {b^2}} \right)}^2}}}{4} \cr
& \Rightarrow \,\,{a^2}{b^2} \leqslant {\left( {{a^2} - {b^2}} \right)^2}\frac{{{{\sin }^2}2\theta }}{4} + {a^2}{b^2} \leqslant {\left( {{a^2} - {b^2}} \right)^2}.\frac{1}{4} + {a^2}{b^2}\,\,\,\,\,.....\left( 3 \right) \cr} $$
∴ from (1), (2) and (3)
Minimum value of $${u^2} = {a^2} + {b^2} + 2\sqrt {{a^2}{b^2}} = {\left( {a + b} \right)^2}$$
Maximum value of $${u^2}$$
$$\eqalign{
& = {a^2} + {b^2} + 2\sqrt {{{\left( {{a^2} - {b^2}} \right)}^2}.\frac{1}{4} + {a^2}{b^2}} \cr
& = {a^2} + {b^2} + \frac{2}{2}\sqrt {{{\left( {{a^2} + {b^2}} \right)}^2}} \cr
& = 2\left( {{a^2} + {b^2}} \right) \cr} $$
∴ Max value - Min value
$$\eqalign{
& = 2\left( {{a^2} + {b^2}} \right) - \left( {a + {b^2}} \right) \cr
& = {\left( {a - b} \right)^2} \cr} $$