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I wish I could get inside info about Princeton or Wisconsin University .But that guy Ken has gone offline .
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can anyone give me a link to last years WBJEE paper ...the paper of 2007 ?????????????????????
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can anyone give me a link to last years WBJEE paper ...the paper of 2007 ?????????????????????
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can anyone give me a link to last years WBJEE paper ...the paper of 2007 ?????????????????????
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can anyone give me a link to last years WBJEE paper ...the paper of 2007 ?????????????????????
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dheer_07 -> the solution has many steps , ..the main point is integration of log | sec theta | d ( theta ) .. Integrate this one by breaking sec into cos ..u will get minus of integration 1/2 log cos theta d ( theta) within the limits 0 to pie / 4 .
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No the answer is ( pie )square / 4 + pie ln 2 -( pie/2) ln 2 ..you must have missed some 2 somewhere .And " sandeepramesh" yes, u are right ..this question is abt a type of IIT JEE subjective pattern ....it is not right to put up this question in board level exam .
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yeah , but he is the only guy here in Kolkata who has solved and got this answer .I called him when I reached home ...it was around 3 pm , he was probably having some ice cream somewhere .., there only he took the back of the question paper ans started solving ......but right now he called me and said that the question is not at all difficult , but it is a perfect 6 marks question .
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yes it is 100 percent right ..but heay "sandeepramesh " ...check my previous posts ...I dint mean that the answer can come in 2 steps ..I means the expression
2 theta2 sec2theta d (theta)
can be brought in 2 steps only by putting x = tan theta . My friend came to this expression by breaking 2 tan inverses and then bringing one of them to sin and the other one to tan .What the hell was he thinking ...anyways , that expression is the key to answer of this question . As I said before , cbse is going to give marks ( at least 4 ) to the students who have got that expression .
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Use of Catalian constant is not in CBSE syllabus and by the way , that answer is a bit wrong ...check the third step again please . Yeah , as I was before ..I told my friend soumyajit that what he got after so many substitutions can be brought in just 2 step by putting x = tan theta . so here is the after math .........->
The answer is ( pie )2 / 4 + pie ln 2 -( pie/2) ln 2
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yes "eistien " you are right , our sir said that 6 marks will be given to all the students who have done with that technique and reached till
= 2 theta2 sec2theta d (theta)
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ya but then you eventually reach the same question given ....besides that its a 6 marks ...u won't get 6 marks just in 3 steps .
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No friends ..this question can be solved . My friend Soumyajit has solved this question , he dint get this set ..but he tried the question when I asked him to do so when I called him after reaching home . Check this out- >
Let I = [0] [1] 2( tan -1 x) 2 dx
=[0][1] (2 tan-1 x ) . (tan-1 x)
= [0][1] sin-1 (2x / (1 + x2)) . tan-1 x dx
Let us assume x = tan theta (.................so theta = tan-1 x )
so dx = sec2 theta d(theta)
when x = 0 , theta =0
when x = 1 , theta = pie / 4
Again, I = [0][pie/4] sin-1 ( 2 tan theta / 1 + tan2 theta ). tan-1 (tan theta ) sec2 theta d (theta)
= [0][pie/4] sin-1 ( sin-1 (sin 2 theta ) . tan-1(tan theta ) sec2 theta d ( theta )
= 2 theta .theta sec 2 theta d (theta)
= 2 theta 2 sec2theta d (theta)
Now , solve this normal integration as integration by parts and then put the limits to get the answer as
I = (pie)2/ 8 + 8 - 2 pie
The answer may not be fully correct because this is huge integration by parts ..would take at least 2 pages for each correct step ...however he said ..this is the most reliable answer if we use the series of tan inverse or log series ...which would make us decide on the basis of limits given till which boundary we should assume the limit . But cbse has not done a gud job by giving this question ...this is an extremely tough HOT ( not of any type given in ncert )
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Vriti Edustore
