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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Trignometry
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Let $f(x)=\cos x(\sin x+\sqrt{\sin^2 x+\sin^2\theta})$ where $\theta$ is a given constant then maximum value of $f(x)$ is

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

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1 Answer

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$f(x)=\cos x(\sin x+\sqrt{\sin^2x+\sin^2\theta})$
$\Rightarrow (f(x).\sec x-\sin x)^2$
$\Rightarrow \sin^2x+\sin^2\theta$
$f^2(x).\sec^2x=2f(x).\tan x=\sin^2\theta$
$f^2(x).\tan^2x-2f(x).\tan x+f^2(x)-\sin^2\theta=0$
$\Rightarrow 4f^2(x)\geq 4f^2(x)(f^2(x)-\sin^2\theta)$
$\Rightarrow f^2(x)\leq 1+\sin^2\theta$
$\Rightarrow \mid f(x)\mid\leq \sqrt{1+\sin^2\theta}$
Hence (b) is the correct option.
answered Oct 15, 2013 by sreemathi.v
 

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