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September 7th, 2016, 12:31 PM
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Join Date: Mar 2012
Re: IIT JEE Short Tricks

As you will appear in IIT JEE exam and looking for short tricks for
Mathes problems , so here I am telling same :

1. Find the value of 36C3

A. 7282 B.7140


C. 6954 D.8326


Yeah,you got it right. The answer is 7140.

IF this is how you did it, you wasted approximately 45 seconds.

36C3= (36*35*34)/(3*2*1)
= ( 6*35*34)
on calculating
=7140
did you actually calculate the value of 6*35*34???
and then get the answer as 7140.

But was this necessary???Did you have to waste almost a minute when you could have got the correct answer in less than 15 seconds.

Yes,look at the options.Each option has the last digit different.
Hence if you find the last digit of the product it is sufficient.

The last digit of ( 6*35*34) is 0 which consumes only a few seconds.In the options only the option 'b' has the last digit as 0.Hence you can directly choose 7140 as the answer.

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

A.(e^2-1) / 2 B.(e^2-2) / e

C.(e^2-1) / 2e D.(e-1)^2 / 2e

This is also an irritating problem which requires a sound knowledge of the exponential series.This problem is tough to solve,and almost impossible, in less than a minute if you go by the normal method.


But if you examine the series carefully you may write it as,

1/2!+1/4!+1/6!+....... = 1/2 + 1/4.3.2.1 +1/6.5.4.3.2.1 +..........

= 0.5 + 1/24 +1/120 +..........

= 0.5 + 0.05 (approx) + 0.01 (approx)+ ....

This series can be approximated by taking only the first 2 terms because the succeeding terms become negligible.

thus

1/2!+1/4!+1/6!+....... = 0.5 + 0.05 (approximately)

Now lets check the options one by one

A .(e^2-1) / 2

=(2.71^2 -1) / 2 take 2.71^2 ~= 8 (ie:a little less than 3^2)

= 3.5
which cant be the answer.

Similarly

B .(e^2-2) / e =(8-2)/2.71 ~= 2 which too cant be the answer


C.(e^2-1) / 2e =(8-1)/(2.71*2) > 1


D.(e-1)^2 / 2e =(2.71-1)^2/(2*2.71) < 1


We note that the option D is closer to the answer.Hence the answer is D

3.The co efficient of x^n in the expansion of

(1+x)(1-x)^n is (AIEEE 2004)

A. (n-1) B. (-1)^(n-1) * n

C. (-1)^(n-1) * (n-1)^2 D. (-1)^n * (1-n)


Whenever you see terms like m,n etc it is highly advisable that you put particular values for m,n.
Here let us put n=0

Then the question becomes,find the co efficient of x^0 (which is 1) in the expansion
of (1+x)(1-x)^0=1+x

the co efficient of 1 in this expression is 1.Hence the correct option should lead us to the answer 1.

Let's check

A.(n-1)
= 0-1
= -1 therefore A cant be the answer

B.(-1)^(n-1) * n
=(-1)^(-1) * 0
= 0 Hence B cant be the answer.

C. (-1)^(n-1) * (n-1)^2
=(-1)^(-1) * (-1) ^2
= -1 Thus C is not the answer.


Since A,B,C dont qualify to be the answers,we can directly deduce that D is the correct answer.However lets check if this is true.

D. (-1)^n * (1-n)
= (-1)^0 * (1-0)
= 1
This is exactly what we should get if the option is right.Hence D is the correct answer.

THIS IS CALLED "THE METHOD OF SUBSTITUTION".

4. 3 particles,each of mass M are situated at the vertices of an
equilateral triangle of the side a.The only forces acting on
the particles are their mutual gravitational forces.It is
desired that each particle move in a circle,while maintaining
the original mutual separation a.Then the initial velocity
that should be given to each particle is

From the last sentence of the question it is clear that we are expected to find the value of initial velocity (the word velocity being more important here).Observing the options carefuly we notice that each of the options have different dimensions (you will know what i mean,if you have studied the chapter on 'Units and dimensions' in Physics).But since we need to find the value of initial velocity which should after all have the dimension [L1 T-1] ( irrespective of whether it is inital or final or whatever else ,so long as it is a velocity).Now check in the options to see which option has the dimension [L1 T-1].

Quantity Dimension

a (length) [M0 L1 T0]

G (Gravitational constant) [M-1 L3 T-2]

M (mass) [M1 L0 T0]

2 ∏ ,3 [M0 L0 T0]

a) [M0 L-1/2 T1 ] not= [M0 L1 T-1]



b) [M0 L-3/2 T1 ] not= [M0 L1 T-1]



c) [M0 L-1/2 T-1 ] not= [M0 L1 T-1]



d) [M0 L1 T-1 ] = [M0 L1 T-1]

Therefore option d is the correct answer.

Look,how an otherwise challenging and a tedious problem can be solved in less than half a minute.Such problems aren't very uncommon in the exams like IIT,AIEEE.CET s .You can always manage to find a couple of problems having such weak options in most of the exams.Hence this method is a valuable tool for the exams.But this should be taken with a pinch of salt.Dont try this method on questions with all or more than one options having the same dimensions. However if atleast one option has a different dimension then this method can be used to eliminate some options.


5.The sum of the first n terms of the series

1^2+2 . 2^2 + 3^2 +2 . 4^2 +5^2+2 . 6^2 + ............. is

n(n+1)^2/2 when n is even.

When n is odd the sum is (AIEEE 2004)


A.3n(n+1)/2 B.[n(n+1)/2]^2

C.n(n+1)^2/4 D.n^2*(n+1)/2


Without even bothering to see what the question actually is let us put a value for n,

ie; n=1 (note that this value should be odd)

Now the required sum = 1^2 (the first term of the series only)
= 1

Now lets check in the options

A.3n(n+1)/2
=3*1(1+1)/2
=3 CANT BE THE ANSWER

B.[n(n+1)/2]^2
=[1(2)/2]^2
= 1



C.n(n+1)^2/4

=1


D.n^2*(n+1)/2
=1

Now we see that the options B,C,D all give us the correct answer.But only one of them is correct.Hence we need to put another value for n.Lets put n=3 and check.

The required sum is 1+8+9=18

A.3n(n+1)/2
Already eliminated

B.[n(n+1)/2]^2
=[3(4)/2]^2
= 36 therefore not the answer


C.n(n+1)^2/4

=36 therefore not the answer


D.n^2*(n+1)/2
=18

Thus D is the correct answer.


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