Sample Questions of ISI ME I (Mathematics) 2010 Discussion

[Viren]

nidhii...can u give an example..i m still not getting the point
thanks

[duck]

Hello.

Take f(x)= x
whenever x takes any irrational value , f(x) will also take the same.

Similarly, f(x) = x^2.
When x= 2, F(x) will be an irrational no,
and so on..

We can take many examples.

Now, suppose f(x)= 2.
but its given in the que tht f(1) = 1
therefore, any other constant value apart from 1 cannot be the function.

=> f(x)=1

hope it was helpful.. :)

[duck]

Sorry,
for f(x) = x^2
if x takes the values sqrt 22/7
=>f(x)= 22/7 which is an irrational no.

[duck]

Sorry for the mistake.
22/7 is a rational number.
Hence we cannot use this case to rule out the function ,f(x)= x^2.

:)

[reetika05]

Hey thanks so much..even my doubt gt cleared and a new concept added too.. :)

[ravian]

Solutions are as follows:

Let X be a Normally distributed random variable with mean 0 and variance 1. Let F(.) be the cumulative distribution function of the variable X. Then the expectation of F(X) is
(c) 1/2
(Hint: F(X) has uniform distribution on (0, 1))

i could not understand the solution.why do we have to consider limits only from 0 to 1?..please help.

[tim]

hi duck , i got his equation 1/2 +  2 * ( 1/2 ) ^2  +    3 * ( 1/2 ) ^3 +..
but i m nt able to solve it...so pls temme how did u get 2

[tim]

hi duck , i got his equation 1/2 +  2 * ( 1/2 ) ^2  +    3 * ( 1/2 ) ^3 +..
but i m nt able to solve it...so pls temme how did u get 2

[deepak]

Set S = 1/2 + 2* (1/2^2) + 3* (1/2^3) ..
Now Multiply both sides by 2 and subtract, you'll get S = 3/2 + (infinite GP with a = 1/4 and r = 1/2). Hence ans = 2

[Verma]

There is a question : The no. of possible permutations of the integer 1 to 7 such that numbers 1 and 2 always precede 3 and 6 and 7 always succeed the number 3 is?

[duck]

Hi.. :)

We need to find no. of permutations such that numbers 1 and 2 always precede 3 and 6 and 7 always succeed the number 3.

It means "3" can come only at "third", "fourth" , "fifth" place.
Now, fix "3" at "third" position.
Number of ways would be : 2!*4!

Now, fix "3" at "fourth" position.
Number of ways would be : 3!*3!

Now, fix "3" at "fifth" position.
Number of ways would be 4!*2!.

Just add all of them to get total number of ways.

 

[Verma]

Thanks for solving it.
 but i have a doubt !
 When we fix 3 at the fourth place such that the arrangement is  _ _ _ 3 _ _ _,
then for the first place we will have 4 choices i.e. 1,2,4,5
for the next place we will have 3 choices left,
and consequently 2 choices for the third place.

Moving to the right of 3, we will have 3 choices i.e 6,7 and 4 or 5(depending on which number was put before 3) for the fifth place,
2 choices for the sixth place and the last one will have 1 choice.

So wont it be 4*3*2*3*2, which makes the sum 240?
However the answer given by sir is however 168 contrary to both our answers!!

[duck]

Hey Sorry, there was a typo in the previous post and a correction..

Fix "3" at "third" position.
Number of ways would be : 3!*3!*2


Now, fix "3" at "fourth" position.
Number of ways would be : 4!*2!

Now, fix "3" at "fifth" position.
Number of ways would be 4!*2!.

Just add all of them to get total number of ways.

[gagan]

how is it continuously differentiable?

[duck]

Hi Gagan.. :)

First find the derivative of the function.
Now, that derivative is also a function.Check for its continuity.
You'll find it continuous.
Hence, Continuously differentiable.

[shreya lahiri]

how can u solve it??

[shreya lahiri]

how can u solve it??

[shreya lahiri]

pls if anyone can explain me the solutions!!

[XIPP]

SEE THIS. 
Is it sufficient?(Sorry, it is not sufficient. I forgot just taking 2, and in that case you choose none of the 3,4,5 and thus add 3C0=1 to the 7.)

Also, many many thanks for leading me to this thread.
Will you please link here such informative threads from this site?
I will be truly obliged (and it will surely help other aspirants, too.)

[duck]

Hi Shreya.. :)

Please post the questions.