First slide

Extreme values and Periodicity of Trigonometric functions

Question

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 between the maximum and minimum of $u^{2}$ is given by

Moderate

A

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

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

${(a - b)}^{2}$
D

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

Solution

$u^{2} = a^{2} + b^{2} + 2 \sqrt{(a^{2} \cos^{2} θ + b^{2} \sin^{2} θ) (a^{2} \sin^{2} θ + b^{2} \cos^{2} θ)} = a^{2} + b^{2} +$
$2 \sqrt{(a^{4} + b^{4}) \sin^{2} θ \cos^{2} θ + a^{2} b^{2} (\cos^{4} θ + \sin^{4} θ)} = a^{2} + b^{2} +$
$2 \sqrt{(a^{4} + b^{4}) \sin^{2} θ \cos^{2} θ + a^{2} b^{2} (1 - 2 s i n^{2} θ \cos^{2} θ)} = a^{2} + b^{2} +$ $2 \sqrt{a^{2} b^{2} + {(a^{2} - b^{2})}^{2} \sin^{2} θ \cos^{2} θ}$
$= a^{2} + b^{2} + 2 \sqrt{a^{2} b^{2} + \frac{{(a^{2} - b^{2})}^{2}}{4} \sin^{2} 2 θ}$
$∴$ difference =(maximum of $u^{2}$ )
-(minimum of $u^{2}$ )
$= 2 \sqrt{a^{2} b^{2} + \frac{(a^{2} - b^{2})}{4}} - 2 a b$
$= \sqrt{{(a^{2} + b^{2})}^{2}} - 2 a b = {(a - b)}^{2}$

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