Short Cuts and Tricks for IIT JEE!

Short Cuts and Tricks for IIT JEE!

Before reading further, short cuts and tricks can only help you to get those extra marks in IIT JEE. It will never help you get a selection. You can at best only expect to marginally save time and expect a few more marks. If you are reading this article for the heavenly potion that will get you through IIT JEE, read no more.

The whole analysis here is based on examples on how to solve a few problems! Your mind should be very alert to see things that are not ordinarily visible to a lot of other students! This specially helps in Mathematics.

Example 1:
One of the most common tricks in all summation questions is to put n=1.

IIT JEE 2000,equals:

A)

Now In this question, simply put n=r=2. We will get 1+2.2+1=6

The four options are respectively 3, 6, 12, and 6. So A and C are out of option! Now put n=3 and r=2 you will have solved this one. Isnt it?

Example 2:

Now lets do some differentiability questions! When you are given to find the differentiability of a function involving |x| or some term like |x-2| etc then simply check the derivative at the point of contention. If it is not zero, then the function is not differentiable! Lets see an example!

y=||x|-1| (IIT JEE 2005)

Here, the points of contentions are clearly -1,0,1. Remove the || sign. Just take y=x-1. Is the derivative at any of these points zero? No because derivative is 1! dy/dx=d(x-1)/dx=1. No need of checking anything else! This works in 99% of cases!

Another example: IIT JEE 1999: The function is not differentiable at..

Now see there are 3 points of contention, 1,2 and 0. Take 0 first .. it is to come from cos(|x|). Simply see the derivative of cos x at zero. It is sin 0=0 . Hence at zero it is differentiable!

Now look at its derivative at x=1 is zero! so the original function is differentiable at x=1!! Also derivative is not zero at x=2. Hence the original function is not differentiable at x=2! Verify it yourself!

Why this happens if very logical. dont think that there is any rocket science applicable here!

Example 3

There are questions where the function is split into many parts and you have to analyze the differentiability of the function at the point where the form is different . Example

f(x) = sinx; x>0
x; x<0

You have to check if the function is differentiable. What do you do? First of all check the continuity. Then no need to do any dirty limits! Just find the derivative of both sides. Derivative of x+ will be cos 0 = 1. Derivative of x- will be 1. Both are equal . Hence if the function is continuous, then it is differentiable!

IITJEE 2001 Question:
f:R->R is a function defined by f(x) = max {x, x^3}. The set of points where f is not differentiable is ?
Now the points of contention are x=x^3 the points given by x=1,-1 and 0. Check the derivative of x and x^3. 1 and 3x^2. Which are not equal for 1 or -1 or 0. Hence it is not differentiable at any of these three points!

Example 4 Find ¹²C₆
A. 924 B. 994
C. 1150 D.842
The Rank 5000 method: 12x11x10x9x8x7/(6x5x4x3x2x1) = 11x12x7=924
The Rank 500 method: It is a multiple of 11! which is the odd terms minus even terms should be a multiple of 11

Example 5. Find the value of
1/2!+1/4!+1/6!+....... (AIEEE 2004)

A. (e²-1) / 2 B. (e²-2) / e
C. (e²-1) / 2e D. (e-1)² / 2e

If you thought this was unsolvable if you did not know the expansion of e^x..... Think again! It is .. The first two terms are the only non negligible terms!

= 0.5 + 0.05 = 0.55

A .(e^2-1) / 2 > 1 easily
B .(e^2-2) / e > 1 again
C.(e^2-1) / 2e > 1 again!
D.(e-1)^2 / 2e was the only option with value less than 1!


Example 6The coefficient of xⁿ in the expansion of (1+x)(1-x)ⁿ is (AIEEE 2004)
A. (n-1) B. n (-1)^n-1
C. (n-1)²(-1)^n-1 D. (1-n)(-1)ⁿ

Easiest method: n=0
New Question: Co-efficient of x⁰ in the 1+x
the co-efficient of 1 in this expression is 1.

A. (n-1) = -1
B. n (-1)^n-1= 0
C. (n-1)²(-1)^n-1= -1
D. (1-n)(-1)ⁿ = 1