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10 Tips for IIT JEE Screening Test

  1. Do the easy questions first. Leave the difficult questions at first go.
  2. Reach the end of the question paper by following step 1.
  3. If you still have time, try to do the more difficult questions that you had left in the first go.
  4. Wild guessing should not be done. However, if two options out of four have been eliminated, then intelligent guessing may be done.
  5. Read the instructions given in the paper carefully. Nothing has officially been announced about the no. of quesions that will be asked or about the difficulty level.
  6. Reach the examination hall well in time. It may be a good idea to visit the place one day in advance, so that you do not have difficulty in locating it on the exam day in case you are unfamiliar with the location.
  7. If the paper is very tough, don't feel nervous. Remember that it is relative performance that counts. Others may find it tough also.
  8. Do not mark the bubbles in hurry. This may spoil all the work you have done. Remember that this is the only thing that is examined.
  9. Take care of your health. The most important thing in this respect in this season is drinking water.
  10. Do not study a new topic a few days before the exam. Strengthen the topics that you are good at.

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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$ $\Rightarrow n=7$