If $u = \sqrt{a^{2} \cos^{2} θ + b^{2} \sin^{2} θ} + \sqrt{a^{2} \sin^{2} θ + b^{2} \cos^{2} θ}$ then the difference the maximum and minimum values of u² is given by

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$2 (a^{2} + b^{2})$

$2 \sqrt{a^{2} + b^{2}}$

$(a + b)^{2}$

$(a - b)^{2}$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$u^{2} = a^{2} + b^{2} + 2 \sqrt{a^{2} \cos^{2} θ + b^{2} \sin^{2} θ} \times \sqrt{a^{2} \sin^{2} θ + b^{2} \cos^{2} θ}$
$\begin{array}{l} = (a^{2} + b^{2}) + 2 \sqrt{a^{2} b^{2} + \frac{{(a^{2} - b^{2})}^{2}}{4} \sin^{2} 2 θ} \\ Max. u^{2} = (a^{2} + b^{2}) = 2 \sqrt{a^{2} b^{2} + \frac{{(a^{2} - b^{2})}^{2}}{4}} \\ \Rightarrow Min. u^{2} = (a^{2} + b^{2}) + 2 a b \\ \Rightarrow Difference = 2 \sqrt{a^{2} b^{2} + \frac{{(a^{2} - b^{2})}^{2}}{4}} - 2 a b \\ = \sqrt{4 a^{2} b^{2} + a^{4} + b^{4} - 2 a^{2} b^{2}} - 2 a b \\ = \sqrt{{(a^{2} + b^{2})}^{2}} - 2 a b = a^{2} + b^{2} - 2 a b = (a - b)^{2} \end{array}$