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

Answer 1

In order to solve this problem we need to know about quite a few topics, as the solution of the problem deals with miscellaneous concepts in mathematics. We need to be familiar with differentiation and trigonometry and also with finding the minimum and maximum values of a given function. Here, given an expression, we are asked to find the difference of the square of maximum and minimum values of the given expression.

Complete step-by-step solution:
In order to find the maximum and minimum value we differentiate the given expression which is :
$\Rightarrow u = \sqrt {{a^2}{{\cos }^2}\theta + {b^2}{{\sin }^2}\theta } + \sqrt {{a^2}{{\sin }^2}\theta + {b^2}{{\cos }^2}\theta }$
Differentiating the expression w.r.t $\theta$, as given below:
$\Rightarrow \dfrac{{du}}{{d\theta }} = \dfrac{{(2{a^2}\cos \theta ( - \sin \theta ) + 2{b^2}\sin \theta \cos \theta )}}{{2\sqrt {{a^2}{{\cos }^2}\theta + {b^2}{{\sin }^2}\theta } }} + \dfrac{{(2{a^2}\sin \theta \cos \theta + 2{b^2}\cos \theta ( - \sin \theta ))}}{{2\sqrt {{a^2}{{\sin }^2}\theta + {b^2}{{\cos }^2}\theta } }}$
As the differentiation of $\sin \theta$ is $\cos \theta$, whereas the derivative of $\cos \theta$is $\sin \theta$, as given below:
$\Rightarrow \dfrac{{du}}{{d\theta }} = \dfrac{{ - 2{a^2}\sin \theta \cos \theta + 2{b^2}\sin \theta \cos \theta }}{{2\sqrt {{a^2}{{\cos }^2}\theta + {b^2}{{\sin }^2}\theta } }} + \dfrac{{2{a^2}\sin \theta \cos \theta - 2{b^2}\sin \theta \cos \theta }}{{2\sqrt {{a^2}{{\sin }^2}\theta + {b^2}{{\cos }^2}\theta } }}$
$\Rightarrow \dfrac{{du}}{{d\theta }} = \dfrac{{2\sin \theta \cos \theta ({b^2} - {a^2})}}{{2\sqrt {{a^2}{{\cos }^2}\theta + {b^2}{{\sin }^2}\theta } }} + \dfrac{{2\sin \theta \cos \theta ({a^2} - {b^2})}}{{2\sqrt {{a^2}{{\sin }^2}\theta + {b^2}{{\cos }^2}\theta } }}$
Dividing the whole expression by 2, as given below:
$\Rightarrow \dfrac{{du}}{{d\theta }} = \dfrac{{\sin \theta \cos \theta ({b^2} - {a^2})}}{{\sqrt {{a^2}{{\cos }^2}\theta + {b^2}{{\sin }^2}\theta } }} + \dfrac{{\sin \theta \cos \theta ({a^2} - {b^2})}}{{\sqrt {{a^2}{{\sin }^2}\theta + {b^2}{{\cos }^2}\theta } }}$
Now the differentiated expression is equated to zero, in order to find the maximum value of $\theta$, as given,
$\Rightarrow \dfrac{{du}}{{d\theta }} = 0$
$\Rightarrow \dfrac{{\sin \theta \cos \theta ({b^2} - {a^2})}}{{\sqrt {{a^2}{{\cos }^2}\theta + {b^2}{{\sin }^2}\theta } }} + \dfrac{{\sin \theta \cos \theta ({a^2} - {b^2})}}{{\sqrt {{a^2}{{\sin }^2}\theta + {b^2}{{\cos }^2}\theta } }} = 0$
$\Rightarrow \dfrac{{\sin \theta \cos \theta ({b^2} - {a^2})}}{{\sqrt {{a^2}{{\cos }^2}\theta + {b^2}{{\sin }^2}\theta } }} = \dfrac{{\sin \theta \cos \theta ({b^2} - {a^2})}}{{\sqrt {{a^2}{{\sin }^2}\theta + {b^2}{{\cos }^2}\theta } }}$
Here $\sin \theta \cos \theta ({b^2} - {a^2})$ gets cancelled on both sides, as given below:
$\Rightarrow \dfrac{1}{{\sqrt {{a^2}{{\cos }^2}\theta + {b^2}{{\sin }^2}\theta } }} = \dfrac{1}{{\sqrt {{a^2}{{\sin }^2}\theta + {b^2}{{\cos }^2}\theta } }}$
By equating the denominators, as given below:
$\Rightarrow \sqrt {{a^2}{{\cos }^2}\theta + {b^2}{{\sin }^2}\theta } = \sqrt {{a^2}{{\sin }^2}\theta + {b^2}{{\cos }^2}\theta }$
Squaring on both sides of the equation as given below:
$\Rightarrow {a^2}{\cos ^2}\theta + {b^2}{\sin ^2}\theta = {a^2}{\sin ^2}\theta + {b^2}{\cos ^2}\theta$
Rearranging or grouping the like terms together, as given below:
$\Rightarrow {a^2}{\cos ^2}\theta - {b^2}{\cos ^2}\theta = {a^2}{\sin ^2}\theta - {b^2}{\sin ^2}\theta$
$\Rightarrow ({a^2} - {b^2}){\cos ^2}\theta = ({a^2} - {b^2}){\sin ^2}\theta$
Here $({a^2} - {b^2})$ term on both sides of the equation, as given below:
$\Rightarrow {\cos ^2}\theta = {\sin ^2}\theta$
$\Rightarrow \dfrac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }} = 1$
$\Rightarrow {\tan ^2}\theta = 1$
$\therefore \tan \theta = \pm 1$
$\Rightarrow \theta = {45^ \circ }$, $\theta = {135^ \circ }$
Substituting either of the above values in the $u = \sqrt {{a^2}{{\cos }^2}\theta + {b^2}{{\sin }^2}\theta } + \sqrt {{a^2}{{\sin }^2}\theta + {b^2}{{\cos }^2}\theta }$, gives the same maximum value of $u$, which is ${u_{\max }}$, substitution as given below:
We know that $\cos {45^ \circ } = \dfrac{1}{{\sqrt 2 }}$ and $\sin {45^ \circ } = \dfrac{1}{{\sqrt 2 }}$, substituting these values in the expression below:
$\Rightarrow {u_{\max }} = \sqrt {{a^2}{{\left( {\dfrac{1}{{\sqrt 2 }}} \right)}^2} + {b^2}{{\left( {\dfrac{1}{{\sqrt 2 }}} \right)}^2}} + \sqrt {{a^2}{{\left( {\dfrac{1}{{\sqrt 2 }}} \right)}^2} + {b^2}{{\left( {\dfrac{1}{{\sqrt 2 }}} \right)}^2}}$
$\Rightarrow {u_{\max }} = \sqrt {{a^2}\left( {\dfrac{1}{2}} \right) + {b^2}\left( {\dfrac{1}{2}} \right)} + \sqrt {{a^2}\left( {\dfrac{1}{2}} \right) + {b^2}\left( {\dfrac{1}{2}} \right)}$
$\Rightarrow {u_{\max }} = \sqrt {\dfrac{{{a^2} + {b^2}}}{2}} + \sqrt {\dfrac{{{a^2} + {b^2}}}{2}}$
Adding the two similar square roots, which gives as given below:
$\Rightarrow {u_{\max }} = 2\sqrt {\dfrac{{{a^2} + {b^2}}}{2}}$
$\Rightarrow {u_{\max }} = \sqrt {2({a^2} + {b^2})}$
$\therefore {u_{\max }} = \sqrt {2({a^2} + {b^2})}$
Now the expression $u = \sqrt {{a^2}{{\cos }^2}\theta + {b^2}{{\sin }^2}\theta } + \sqrt {{a^2}{{\sin }^2}\theta + {b^2}{{\cos }^2}\theta }$ is minimum at $\theta = {0^ \circ }$or ${90^ \circ }$, when either of these values of $\theta$ are substituted the minimum value of $u$is obtained which is ${u_{\min }}$:
We know that $\cos {0^ \circ } = 1$ and $\sin {0^ \circ } = 0$, substituting these values, as given below:
$\Rightarrow {u_{min}} = \sqrt {{a^2}{{(1)}^2} + {b^2}(0)} + \sqrt {{a^2}(0) + {b^2}{{(1)}^2}}$
$\Rightarrow {u_{min}} = \sqrt {{a^2}} + \sqrt {{b^2}}$
$\Rightarrow {u_{min}} = a + b$
$\therefore {u_{min}} = a + b$
Now calculating the difference between $u_{\max }^2$ and $u_{\min }^2$, as given below:
$\Rightarrow u_{\max }^2 = {\left( {\sqrt {2({a^2} + {b^2})} } \right)^2}$
$\Rightarrow u_{\max }^2 = 2({a^2} + {b^2})$
$\Rightarrow u_{\min }^2 = {(a + b)^2}$
$\therefore u_{\max }^2 - u_{\min }^2 = 2({a^2} + {b^2}) - {(a + b)^2}$
On expanding the expression, as given below:
$\Rightarrow u_{\max }^2 - u_{\min }^2 = 2{a^2} + 2{b^2} - ({a^2} + {b^2} + 2ab)$
$\Rightarrow u_{\max }^2 - u_{\min }^2 = {a^2} + {b^2} - 2ab$
$\therefore u_{\max }^2 - u_{\min }^2 = {(a - b)^2}$
The difference between the maximum and minimum values of ${u^2}$ is $u_{\max }^2 - u_{\min }^2 = {(a - b)^2}$.
Option D is the correct answer.

Note: While solving this problem, please note that while calculating the maximum value of u, in that theta had two solutions 45 and 135 degrees, but as there is a square present for the trigonometric functions in the u expression, so the final value of the u expression will be the same. Similarly while finding the minimum value of u, theta had two solutions either 0 or 90 degrees, but as there is a square present for the trigonometric functions in the u expression, so the final value of the u expression will be the same.