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RE: about IITJEE

This letter and its reply may be of general interest:
___________________________________________________________________________

The cut-off for screening has been around 40-45% which may change depending upon the toughness of the paper. Detailed statistics can be found at http://www.123iitjee.com/iitjee_stats.htm .

The preparation for mains should automatically take care of preparation for screening. However, here are the recommendations for the objective questions since you asked:

Physics-H.C. Verma and MCQ (Bharti Bhavan)
Chemistry-Arihant
Mathematics-Arihant

Your job is to prepare and not to worry about the result. Fear will find no place if you think that way.
From: hemanth reddy
Sent: Monday, February 07, 2005 2:40 PM
To: 123iitjee
Subject: about IITJEE

Sir,My name is hemanth.i am preparing for JEE-2005 IN GENERAL CATAGEREY. which boks are prepared for screening.i had no idea about SCREENING CUTOFF.Nobody help me,so i am very feared about screening. what is the screening cutoff for JEE-2004,JEE-2003,JEE-2002.Atleast give roundoff figures.please help.......

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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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