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Q & A (IIT-JEE 2006 Pattern)


Here in Chennai, there is a strong belief that the IIT JEE 2006 paper is going to have more than 90%, (or even the entire paper) with more than one correct answer. If it is so, then it would be terribly low scoring. If negative matks go along with it,then it would be even more so. What is your opinion?

Two things are mentioned clearly in JEE site, (i) test of comprehension & analytical abilities (ii) negative marking. More than one answer correct type of questions test more of, "real stuff" and are good to test these abilities. That could be one of the reasons giving rise to belief that they may form a significant percentage of the test. On the other hand if we look at negative marking aspect which is going to be there, negative marking is meant to discourage wild guessing. More than one answer correct questions also serve the same purpose. Threfore, if more than one correct questions are asked then it makes no sense to keep negative marking for them. (Note: There is another way to keep negative marking in more than one answer type of questions that has not exercised earlier but makes more sense if question setter wants to keep it in these types of questions as well. In this system, marks are awarded for every correct option selected and marks are deducted for every wrong option selected. Let us consider two examples assuming that a question carries 4 marks. Let us say that a question has (a), (c) & (d) options correct and a student chooses (c) & (d). In this case, 4 marks are divided in three answers and he may be given 4/3 marks for every right answer marked by him and hence he gets 8/3. Since he does not choose a wrong answer he does not get negative marks. In another question, let us say that the correct options are (a) & (b) and the student chooses (a) & (c). In this case 4 marks are divided into two answers and he gets 2 marks for choosing (a). Also he gets -1 (say 1/4'th of the total marks) as negative marks for choosing (c). The weightage system is also used internationally where an instructor can choose to give different positive and negative marks for various options. This scheme of marking makes things complicated and looks unlikely but is a possibility.)

In this context, there is a possibility that the questions having only one option correct have negative marking and questions having more than one option correct do not have negative marking. This of course may be the case if more than one option questions are asked.

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$f(x)=x^6+2x^4+x^3+2x+3 $

$\mathop {\lim }\limits_{x \to 1} \frac{{{x^n}f(1) - f(x)}}{{x - 1}} = 44$


Let $f(x)=x^6+2x^4+x^3+2x+3,x \in R $. Then the natural number n for which $\mathop {\lim }\limits_{x \to 1} \frac{{{x^n}f(1) - f(x)}}{{x - 1}} = 44$ is _ _ _ _ . Solution Since the limit has $\left[ {\frac{0}{0}} \right]$ form, L.H. Rule is applicable. Thus, $\mathop {\lim }\limits_{x \to 1} n{x^{n - 1}}f(1) - f'(x) = 44$ $\therefore nf(1) - f'(1) = 44$ $\therefore n.9 - ({6.1^5} + {8.1^3} + {3.1^2} + 2.1) = 44$ $ \Rightarrow 9n - 19 = 44$ $\Rightarrow n=7$