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Catalogs Discussion Forums -> Differential Calculus -> Odd function -> Go to message
This Post 5 points    (Olaaa!! Perrrfect answer.   in 1 votes )   [?]
1 replies   

Let g_i (x) represent any odd function,

Letf(x)=g_1(x).g_2(x)...g_{2n+1}(x)

For an odd function,

 f(-x)=-f(x)

f(-x)=g_1(-x).g_2(-x)...g_{2n+1}(-x)

                 =(-1)(2n+1)[g_1(x).g_2(x)...g_{2n+1}(x)]



\Rightarrowf(-x)=-f(x)

\Rightarrow\ it\ is\ an\ odd\ function

Catalogs Discussion Forums -> Coaching Institutes & Course Material -> please suggest how to prepare for eamcet to get below 1000 rank -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]
1 replies   
i have heard that intermediate marks weightage fr eamcet is being removed...is it true?
Catalogs Discussion Forums -> Physical Chemistry -> A Good question on vapour state -> Go to message
This Post 2 points    (Olaaa!! Perrrfect answer.   in 1 votes )   [?]
2 replies   

Vapour pressure does not depend on the surface area of the liquid.


No doubt increase in surface area increases the total rate of evaporation, but at the same time it increases the rate at which particles condense...and thus the vapour pressure remains unaffected




Catalogs Discussion Forums -> Algebra -> permuation & combination -> Go to message
This Post 17 points    (Olaaa!! Perrrfect answer.   in 4 votes )   [?]
1 replies   

The basic principle is that,

The no of ways of dividing n identical things into r ordered groups with 1,2,3,...k things in any group is same as the no of ways of expressing the natural number n as the sum of r integers using 1,2,3,...k and considering the order of addition.

This is same as the no of solutions of the equation



x_1+x_2+...x_r = n ,



where x_i can take any value 1,2,3,...k and this is equal to the coefficient of x^n in the expansion of



 {(x^1+x^2+...x^k)}^r



For example, consider a case where 3 identical balls are to be arranged into 2 groups, each group containing any number of things, r(r \ge 1)



the possible groups are group A with 1 ball, group B with 2 balls -(1,2)

                                    
group A with 2 balls, group A with 1 ball -(2,1)

                              
                                             



consider the expansion of  (x+x^2+x^3)^2

=(x+x^2+x^3)(x+x^2+x^3)

=x^{1+1}+\boxed{x^{2+1}}+x^{3+1}+\boxed{x^{2+1}}+x^{2+2}+x^{2+3}+x^{3+1}+x^{3+2}+x^{3+3}

the exponents which add to 3 are the second n 4th terms, x^{2+1}

which is same as the coefficient of x^3 in the expansion (x+x^2+x^3)^2 ...


once u have understood the basic idea you yourself will come to know when and where to apply it...

Catalogs Discussion Forums -> Integral Calculus -> Differential Equations -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]
1 replies   

Statement 1:The differential equation of all the circles on a plane must be of order 3

Statement 2:There is only one circle passing through three non collinear points



Does statement 2 explain statement 1? how?

Catalogs Discussion Forums -> Algebra -> permutation & combination -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]
16 replies   
hey u said the correct ans is 89...?
Catalogs Discussion Forums -> Algebra -> permutation & combination -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]
16 replies   

From among 4 oranges you can select either 0 or 1 or 2 or 3 or 4 oranges, so this can be done in (4+1)=5 ways

lly from among 5 apples..u can select 1 or more in (5+1) ways

lly from among 6 mangoes, u can select 1 or more in (6+1) ways.



but this includes the selection in which no fruit is selected...so no of selections which can be made = (4+1)(5+1)(6+1)-1

so total no of selections = 209

Catalogs Discussion Forums -> Electricity -> Electric potential ! -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]
3 replies   
yeah...sry...they are conducting shells
Catalogs Discussion Forums -> Electricity -> Electric potential ! -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]
3 replies   

Find the electric potentials at A,B,C,D,E,F & G, if the inner shell is given a charge +q and the outer shell is given a charge +q'


http://img230.imageshack.us/img230/7065/potentialnn6.png


(explain ur working)

Catalogs Discussion Forums -> Electricity -> Capacitors -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]
3 replies   

How to find the effective capacitance of a capacitor having many parallel plates as shown ?


Take surface area of each plate = A


AB=CD=EF=2d


BC=DE=d


(edited: P and Q are connected to the battery)

Catalogs Discussion Forums -> Algebra -> Permutation & combination -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]
9 replies   
u are welcome :)
Catalogs Discussion Forums -> Algebra -> Permutation & combination -> Go to message
This Post 10 points    (Olaaa!! Perrrfect answer.   in 2 votes )   [?]
9 replies   

@v89

ofc it can...infact to get the maximum number you SHOULD take that into consideration..



Let me explain my working..



suppose there are only two circles..they can intersect atmost at only 2 points..

Now take the case of three circles..the 1st can intersect the 2nd at points, 2nd can intersect the third at 2 points, and third can intersect the 1st at two

so total no of points of intersection = 6 = 3C2 * 2



now take the case of 4 circles...

 1-->2 at 2 pts

 1-->3 at 2 pts

 1-->4 at 2 pts

 2-->3 at 2 pts

 2-->4 at 2 pts

 3-->4 at 2 pts



so total no of points of intersection = 4C2 * 2



extending this to n circles...it will be nC2 * 2

Catalogs Discussion Forums -> Algebra -> Permutation & combination -> Go to message
This Post 2 points    (Olaaa!! Perrrfect answer.   in 1 votes )   [?]
9 replies   

Each circle can intersect another at atmost 2 points.


As we need the maximum no of points of intersection of 8 circles,


no of ways of selecting 2 circles out of 8 = 8C_2


they will intersect at  8C_2 \times 2 =56 points


(This qsn is quite similar to the famous qsn " n people at a party shook hands with each other.  How many 


handshakes were there altogether?")


 

Catalogs Discussion Forums -> Algebra -> permutation nd combination............ -> Go to message
This Post 5 points    (Olaaa!! Perrrfect answer.   in 1 votes )   [?]
2 replies   

Let us forget the zeros for a while


So u have 9 digits, 5 odd and 4 even with u


Let us take cases


five digit  numbers whose sum of digits is odd can be formed by 





2 even+3 odd (which can be arranged in 4C_2 \times 5C_3 \times 5! ways ) \longrightarrow \boxed{1}

4 even+1 odd (which can be arranged in 4C_4 \times 5C_1 \times 5!  ways ) \longrightarrow \boxed{2}



now let us take zero into consideration,

as 0 is also an even number , we need to select only



1 even + 3 odd 

3 even + 1 odd 



and arrange them

this can be done in

 

 (4C_1 \times 5C_3) \times 4! \times 4 \ \ \ +\ \ \ (4C_3 \times 5C_1)\times 4! \times 4 \longrightarrow \boxed{3}

 

 \boxed{1}+ \boxed{2}+\boxed{3} gives the no of five digit  numbers whose sum of digits is odd.


Catalogs Discussion Forums -> Algebra -> good questions from permutation and combination.........objective -> Go to message
This Post 5 points    (Olaaa!! Perrrfect answer.   in 1 votes )   [?]
3 replies   




 The possibilities are,


\underline {ABC} should be together in the same order ,this can be done in 3! ways

\underline {ABD} should be together in the same order, this can be done in 3! ways
\underline {AC}\ and\  \underline{BD} should be together in the same order, this can be done in 3! ways


so total no of ways = 6+6+6=18






 
 
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