Messages posted by titun

f(x) = -_{[ 0]}^{[ x]} log(cost)dt, so f ' (x) = -log(cosx)

Now, f ' (x) - 2f ' (

/4+x/2)/2 +2f ' (

/4 - x/2)(-1/2)

= -log(cosx) + log(cos(

/4+x/2)) + log(cos(

/4-x/2))

= -log(cosx) + log[(cosx)/2]

= -log(cosx) +log(cosx) - log2 = -log2

Now, integrate both sides to have,

f(x)-2f(

/4+x/2) +2f(

/4--x/2) = -xlog2 + c

c is an arbitary constant. Put x=0, you will get c=0 ( f(0) = 0 & f(

/4) gets cancelled).

Hence, you get,

f(x)-2f(

/4+x/2) +2f(

/4--x/2) = -xlog2

I have tried to explain all the steps as far as possible.

Hiiiii,

It's easy.

e^{(t^2}^-2)/2 (1/t) dt. Integrate this by parts taking e^{(t^2}^{-2)/2 as the first function and 1/t as the second function, you will get the following.}

e^{(t^2}^-2)/2 (1/t) dt = e^{(t^2}^-2)/2 lnt -

e^{(t^2}^-2)/2 tlnt dt

i.e

e^{(t^2}^-2)/2 (1/t) dt +

e^{(t^2}^-2)/2 tlnt dt = e^{(t^2}^-2)/2 lnt

Puneet sir, I think the method provided by Vikrant is correct. Take the following e.g

There are 2n+1 consecutive positive integers. We select 3 nos a,b,c. If a,b,c are in A.P.

2b=a+c. It is obvious that a+c = even integer. Now this implies that,

a & c are both odd integers or even integers.

(since, odd+odd = even & even + even = odd)

Now, that b is the A.M of the two nos a & c, it is obvious that b lies beteween a & c & b being an integer falls in that set of 2n+1 consecutive integers. In this way, all the criteria are satisfies.

e.g. From first 30 consecutive integers, if we select 2 odd nos like 1 & 9, then 5 being the AM of 1 & 9 lies within the first 30 integers.

Hiiii,

I think the answer is 36. The solution is given below:

275 = 25x11 The 8-digit no. will be divisible by 275 when it is divisible by 11 & 25. The no. will be divisible by 25 if the last two digits of the no is 75.

So, we fix the last 2 places of the no. by 75. The same no. will be divisible by 11 if the difference of the sum of the digits occupying the odd places & the sum of the digits occupying the even places is a multiple of 11. Let the 8-digit no. be abcdef75 where a,b,c,d,e,f belongs to the set {1,3,4,6,8,9}

abcdef75 is obviously divisible by 25. It will be also divisible by 11 if it satisfies the following criteria:

a+c+e+7 - (b+d+f+5) = 11k [k being an integer]

i.e a+c+e-(b+d+f) = 11k-2 = 11(k-1)+9 = 11m+9 [where (k-1) = m] ...............eqn.(1)

Again a,b,c,d,e,f are members of the set {1,3,4,6,8,9}

Therefore, a+b+c+d+e+f = 31 ...................eqn.(2)

(2) - (1), 2(b+d+f) = 22 - 11m

i.e b+d+f = 11 - 11(m/2) [ since, b+d+f is an integer, m/2 must be an integer, so let m=2n]

i.e b+d+f = 11 - 11n = 11z [z = (1-n)]

From, the above, we find that the sum of the nos occupying the 2nd, 4th, 6th places should be a multiple of 11.

The following can be the only cases:

Nos. occupying the 2nd,4th & 6th places(in any order) 1,6,4 Nos. occupying the 1st,3rd & 5th places(in any order) 3,8,9

Now, the nos in the odd places can arrange among themselves in 3! = 6 ways. For each such way, the nos in the even places can arrange themselves on 3! = 6 ways.

Therefore, total no. of 8-digit nos formed from 1,3,4,5,6,7,8,9 which is divisible by 275 = 6x6 = 36

Ans : 36

Note: I' ve not considered repetition of digits which will complicate the matters a little bit.

I've tried to give the solution in details. If there is any mistake or somewhere, I've failed to explain, plz let me know !!

IF A, B, C ARE THE CUBE ROOTS OF P (P<0) , THEN FOR ANY x, y, z the expression (xA + yB + zC)/(xB + yC + zA) IS EQUAL TO

a) 1 b) w c) w^2 d) None of these

1, w, w^2 are the cube roots of unity

I am just giving the outlines on how one should solve the problem.

Subtract the 2nd row from the 1st row, & the 3rd row from the second row in each of the two determinants to get (a-b)(b-c)(c-a) in each case.

Hence the two of the determinants are equal to (a-b)(b-c)(c-a)

Total probability is always equal to 1.

So p(1)+p(2)+p(3)+p(4)=1 i.e x=0.1

Whose time period of vibration are we talking about?

Is it that of the el. field?

I am thankful to Edison. The correct option is 3

The problem is very clear. I will first try to explain it & then give the solution.

6 nos are chosen from first consecutive 100 natural nos.

The first 3 of the chosen nos give the dimensions of the tiles(i.e its length, breadth & height). The last 3 chosen nos give the length, breadth & height of the rectangular box.

The tile will go into the box if the product of the first 3 nos (i.e the vol. of the tile) is less than the product of the last 3 nos (i.e. the vol. of the box)

There can be 3 cases:

1. The product of the first 3 nos i.e the vol of the tile < the product of the last 3 nos i.e. the vol of the box.

2. The product of the first 3 nos i.e the vol of the tile > the product of the last 3 nos i.e. the vol of the box.

3. The product of the first 3 nos i.e the vol of the tile = the product of the last 3 nos i.e. the vol of the box.

Hence, the reqd probability = 1/3

PLZ TELL ME IF I AM WRONG ANYWHERE

I do not agree with cvramana's method. I will give an example to contradict his method of approach.

e.g. What is the smallest natural no. with 3 divisors.

ANS: It is 4. The divisors are 1,2,4

It is not 6 as proposed by Mr. Cvramana (1x2x3).

Hence, in the first case, the sum of the divisors is 7 & in the 2nd case it is 6.

Hence the difference in answers.

I think the answer 744 is correct.

Asmita,

I don't agree with your solution.

Because, 54, 108 are divisible by 27 but the sum of the digits in either case is not divisible by 27.

So, can u provide another solution?

I think the question is wrong.

This is because z is purely immaginary and cosz is not defined. (since trigonometric function of an immaginary quantity is not defined)

The correct question should be:

Find out cos(mod.z) ?

Ans: In this case, mod.z = -log(2-

3) [negative sign is because mod.z is always a positive quantity]

So, cos(mod.z) = cos(log(2-

3))

A far as sumeet's solution is concerned, log_e(2-

3)ⁱ

^{is not defined. So, your solution is incorrect.}

Let a be the 81-didit number all digits of which are equal to 1. Then the number a is

1) divisible by 9 but not divisible by 27

2) divisible by 27 but not divisible by 81

3) divisible by 81 but not divisible by 243

4) divisible by 243

A circular wheel has a mass of m. It has 12 spokes each of which has a mass of m/4. The wheel is rotating with an angular speed of w. What is the amount of work done in stopping the wheel?

I WISH ALL OF YOU A VERY

HAPPY NEW YEAR.

Hey, ariyam66 I have found out your crafty move. The conjugate of 5+2

5 is

5-2

5, but the conjugate of 2(1+

5)+3 is 2(1-

5)+3 & not 2(1+

5)-3, as You have used.

I HAVE FOUND THE CATCH!!

The first one can easily be solved by defining a random variable X following binomial didtribution with parameters n=200 & p=0.5, q=0.5

The required probability is (200C1+200C3+........................+200C199)(1/2)^200 = 0.5

The answer to the 3rd problem is 14/200 = 7/100

1+n+n^2+.........+n^127 = (n^128-1)/(n-1) = (n^64+1)(n^32+1)(n^16+1)(n^8+1)(n^4+1)(n^2+1)(n+1)

(USING A^2-B^2=(A+B)(A-B))

Hence all the answers are correct.

Vinu, I want to point out something to you. You have written that,

Now, (n^x-1) is divisible by (n-1) for some x belonging to + I.

But for all x belonging to positive integers, (n^x-1) always have a factor n-1.

I want to ask Deepak, how did he know that a,b,c & d are all integers??

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