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Evaluate,

$\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{{{(\cos x + \cos 3x + \sin 2x)}^2}}}{{{{(\sin x - \sin 3x + 2\cos 2x)}^3}}}$

The given limit,

$ = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{{{(2\cos 2x\cos x + 2\sin x\cos x)}^2}}}{{{{( - 2\cos 2x\sin x + 2\cos 2x)}^3}}}$

$ = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{{{\cos }^2}x{{(\cos 2x + \sin x)}^2}}}{{2{{\cos }^3}2x{{( - \sin x + 1)}^3}}}$

$\mathop { = \lim }\limits_{x \to \frac{\pi }{2}} \frac{{(1 + \sin x)(1 - \sin x){{(\cos 2x + \sin x)}^2}}}{{2{{\cos }^3}2x{{( - \sin x + 1)}^3}}}$

$ = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{(1 + \sin x){{(\cos 2x + \sin x)}^2}}}{{2{{\cos }^3}2x{{(1 - \sin x)}^2}}}$

$ = {\left. {\frac{{(1 + \sin x)}}{{2{{\cos }^3}2x}}} \right|_{x = \frac{\pi }{2}}}\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{{{(\cos 2x + \sin x)}^2}}}{{{{(1 - \sin x)}^2}}}$

$ = ( - 1){\left[ {\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{(\cos 2x + \sin x)}}{{(1 - \sin x)}}} \right]^2}$

$ =  - {\left[ {\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{ - 2\sin 2x + \cos x}}{{ - \cos x}}} \right]^2}$

[Using L.H. Rule as the limit has $\frac {0}{0}$ form]

$ =  - {\left[ {\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{ - 4\sin x\cos x + \cos x}}{{ - \cos x}}} \right]^2}$

$ =  - {\left[ {\mathop {\lim }\limits_{x \to \frac{\pi }{2}} (4\sin x - 1)} \right]^2} =  - 9$

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${\log _{\sqrt 5 }}\left[ {3 + \cos \left( {\frac{{3\pi }}{4} + x} \right) + \cos \left( {\frac{\pi }{4} + x} \right) + \cos \left( {\frac{\pi }{4} - x} \right) - \cos \left( {\frac{{3\pi }}{4} - x} \right)} \right]$

The range of the function $f(x) = {\log _{\sqrt 5 }}\left[ {3 + \cos \left( {\frac{{3\pi }}{4} + x} \right) + \cos \left( {\frac{\pi }{4} + x} \right) + \cos \left( {\frac{\pi }{4} - x} \right) - \cos \left( {\frac{{3\pi }}{4} - x} \right)} \right]$ is: (A) $[ - 2,2]$ (B) $\left[ {\frac{1}{{\sqrt 5 }},\sqrt 5 } \right]$ (C) $(0,\sqrt 5 )$ (D) $[ 0,2]$ Solution We have, $f(x) = {\log _{\sqrt 5 }}\left( {3 - 2\sin \frac{{3\pi }}{4}\sin x + 2\cos \frac{\pi }{4}\cos x} \right)$ $ \Rightarrow f(x) = {\log _{\sqrt 5 }}\left[ {3 + \sqrt 2 (\cos x - \sin x)} \right]$ Now, $ - \sqrt 2  \le \cos x - \sin x \le \sqrt 2 $ $\therefore - 2 \le \sqrt 2 (\cos x - \sin x) \le 2$ $\therefore 1 \le 3 + \sqrt 2 (\cos x - \sin x) \le 5$ $\therefore{\log _{\sqrt 5 }}1 \le {\log _{\sqrt 5 }}[3 + \sqrt 2 (\cos x - \sin x)] \le {\log _{\sqrt 5 }}5$ $ \Rightarrow 0 \le f(x) \le 2$ Answer: (D)
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