If u = a^2 ^2 + b^2 ^2 + a^2 ^2 + b^2 ^2 , then difference betwee… - Vidyadip

MCQ

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

✓
${(a - b)^2}$
B
$2\sqrt {{a^2} + {b^2}} $
C
${(a + b)^2}$
D
$2({a^2} + {b^2})$

✓

Answer

Correct option: A.

${(a - b)^2}$

a
(a) $2u = \sqrt {{a^2}{{\cos }^2}\theta + {b^2}{{\sin }^2}\theta } + \sqrt {{a^2}{{\sin }^2}\theta + {b^2}{{\cos }^2}\theta } $
${u^2} = {a^2} + {b^2} + 2\sqrt {({a^2}{{\cos }^2}\theta + {b^2}{{\sin }^2}\theta )({a^2}{{\sin }^2}\theta + {b^2}{{\cos }^2}\theta )} $
$ = {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} $
where $t = {a^2}{\cos ^2}\theta + {b^2}{\sin ^2}\theta ,\,({a^2} > {b^2})$
${t_{\max }} = {a^2}$ and ${t_{\min }} = {b^2}.$
Let $y = - {t^2} + ({a^2} + {b^2})t$
Now $\frac{{dy}}{{dt}} = 0 \Rightarrow - 2t + ({a^2} + {b^2}) = 0 \Rightarrow t = \frac{{{a^2} + {b^2}}}{2}$
Sign scheme for $\frac{{dy}}{{dt}}$
$\therefore$ ${({u_{\max }})^2} - {({u_{\min }})^2} = 2({a^2} + {b^2}) - {(a + b)^2} = {(a - b)^2}$

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