If u = sqrt {{a^2}{{cos }^2}theta + {b^2}{{si | vedclass

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(A) Given $u = \sqrt {{a^2}{{\cos }^2}	heta + {b^2}{{\sin }^2}	heta } + \sqrt {{a^2}{{\sin }^2}	heta + {b^2}{{\cos }^2}	heta } $.Squaring both sid

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${(a - b)^2}$

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(A) Given $u = \sqrt {{a^2}{{\cos }^2}	heta + {b^2}{{\sin }^2}	heta } + \sqrt {{a^2}{{\sin }^2}	heta + {b^2}{{\cos }^2}	heta } $.Squaring both sides,we get:${u^2} = ({a^2}{{\cos }^2}	heta + {b^2}{{\sin }^2}	heta ) + ({a^2}{{\sin }^2}	heta + {b^2}{{\cos }^2}	heta ) + 2\sqrt {({a^2}{{\cos }^2}	heta + {b^2}{{\sin }^2}	heta )({a^2}{{\sin }^2}	heta + {b^2}{{\cos }^2}	heta )} $${u^2} = {a^2} + {b^2} + 2\sqrt {({a^2}{{\cos }^2}	heta + {b^2}{{\sin }^2}	heta )({a^2}{{\sin }^2}	heta + {b^2}{{\cos }^2}	heta )} $Let $t = {a^2}{{\cos }^2}	heta + {b^2}{{\sin }^2}	heta $. Then ${a^2}{{\sin }^2}	heta + {b^2}{{\cos }^2}	heta = {a^2} + {b^2} - t$.So,${u^2} = {a^2} + {b^2} + 2\sqrt {t({a^2} + {b^2} - t)} = {a^2} + {b^2} + 2\sqrt { - {t^2} + ({a^2} + {b^2})t} $.Let $f(t) = - {t^2} + ({a^2} + {b^2})t$. The maximum value of $f(t)$ occurs at $t = \frac{{{a^2} + {b^2}}}{2}$,which is $f\left( \frac{{{a^2} + {b^2}}}{2} ight) = \frac{{{({a^2} + {b^2})^2}}}{4}$.Thus,${({u^2})_{\max }} = {a^2} + {b^2} + 2\sqrt {\frac{{{({a^2} + {b^2})^2}}}{4}} = {a^2} + {b^2} + ({a^2} + {b^2}) = 2({a^2} + {b^2})$.The minimum value of $f(t)$ occurs at the boundaries of $t$,i.e.,$t = {a^2}$ or $t = {b^2}$.At $t = {a^2}$,$f({a^2}) = - {a^4} + ({a^2} + {b^2}){a^2} = {a^2}{b^2}$.Thus,${({u^2})_{\min }} = {a^2} + {b^2} + 2\sqrt {{a^2}{b^2}} = {a^2} + {b^2} + 2ab = {(a + b)^2}$.The difference is ${({u^2})_{\max }} - {({u^2})_{\min }} = 2{a^2} + 2{b^2} - ({a^2} + {b^2} + 2ab) = {a^2} + {b^2} - 2ab = {(a - b)^2}$.

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

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