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AIEEE 2005 Preparation


The letter from a student along with its reply is being put up here as it is
of general interest to all the students.

> Dear Zuhaib,
>
> Right now only the talks are going on. If any declaration comes
officially,
> it will appear at 123iitjee.com .
>
> Right now, the preparation should be targeted towards subjective which
gives
> in depth understanding required at this stage.
>
> ----- Original Message -----
> Sent: Sunday, August 01, 2004 2:10 PM
>
>
> > name = Zuhaib Ahmad
> > comments = Respected Sir,
> > With due respect i want to bring in ur notice that
recently
> i came to know that from 2005 the examination of AIEEE will be divided
into
> two phases i.e screening and mains.It will be ur most kindness if u will
> inform me whether this news is true or not so that i could start my
> preparation likewise
> > thank you
> >
>

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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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