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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Trignometry
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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

$\begin{array}{1 1}(a)\;(a-b)^2&(b)\;2\sqrt{a^2+b^2}\\(c)(a+b)^2&(d)\;2(a^2+b^2)\end{array}$

Can you answer this question?
 
 

1 Answer

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Step 1:
$u^2=a^2+b^2+2\sqrt{(a^4+b^4)\cos^2\theta\sin^2\theta+a^2b^2(\cos^4\theta+\sin^4\theta)}$------(1)
Now $(a^4+b^4)\cos^2\theta\sin^2\theta+a^2b^2(\cos^4\theta+\sin^4\theta)$
$\Rightarrow (a^4+b^4)\cos^2\theta\sin^2\theta+a^2b^2(1-2\cos^2\theta\sin^2\theta)$
$\Rightarrow (a^4+b^4-2a^2b^2)\cos^2\theta\sin^2\theta+a^2b^2$
$\Rightarrow (a^2-b^2)^2.\large\frac{\sin^22\theta}{4}$$+a^2b^2$-----(2)
$\therefore o\leq \sin^22\theta\leq 1$
$\Rightarrow 0\leq (a^2-b^2)^2\large\frac{\sin^22\theta}{4}$$\leq \large\frac{(a^2-b^2)^2}{4}$
$a^2b^2\leq (a^2-b^2)^2\large\frac{\sin^22\theta}{4}$$+a^2b^2$
$\leq (a^2-b^2).\large\frac{1}{4}$$+a^2b^2$----(3)
Step 2:
From (1) ,(2) &(3)
Minimum value of $u^2=a^2+b^2+2\sqrt{a^2b^2}=(a+b)^2$
Maximum value of $u^2$
$\Rightarrow a^2+b^2+2\sqrt{(a^2-b^2)^2.\large\frac{1}{4}\normalsize +a^2b^2}$
$\Rightarrow a^2+b^2+\large\frac{2}{2}$$\sqrt{(a^2+b^2)^2}$
$\Rightarrow 2(a^2+b^2)$
Step 3:
Maximum value -Minimum value =$2(a^2+b^2)-(a+b)^2$
$\Rightarrow (a-b)^2$
Hence (a) is the correct answer.
answered Oct 17, 2013 by sreemathi.v
 

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