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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Trignometry
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The maximum value of the expression $\large\frac{1}{\sin^2\theta+3\sin\theta\cos\theta+5\cos^2\theta}$ is

$(a)\;\large\frac{1}{2}$$\qquad(b)\;2\qquad(c)\;0\qquad(d)\;3$

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1 Answer

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Let $f(\theta)=\large\frac{1}{\sin^2\theta+3\sin\theta\cos\theta+5\cos^2\theta}$
Let $f(\phi)=\sin^2\theta+3\sin\theta\cos\theta+5\cos^2\theta$
$\qquad\quad\;\;=\large\frac{1-\cos 2\theta}{2}+$$5\big(\large\frac{1+\cos 2\theta}{2}\big)+\large\frac{3}{2}$$\sin 2\theta$
$\qquad\qquad\;\;=3+2\cos 2\theta+\large\frac{3}{2}$$\sin 2\theta$
$\therefore f(\phi)=3-\sqrt{4+\large\frac{9}{4}}$
$\qquad\quad=3-\sqrt{\large\frac{16+9}{4}}$
$\qquad\quad=3-\sqrt{\large\frac{25}{4}}$
$\qquad\quad=3-\large\frac{5}{2}$
$\qquad\quad=\large\frac{1}{2}$
$f(\theta)=\large\frac{1}{f(\phi)}$
$\qquad=2$
Hence (b) is the correct answer.
answered Oct 15, 2013 by sreemathi.v
 

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