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Do the following problem:
The value of 100[1/(1*2) + 1/(2*3) + ....... 1/(99*100)] is a) 99 b) 100 c) 101 d) (100)*(100)/99 
99
100(11/2+1/21/3+1/31/4+.....1/991/100) 100(11/100) 100*99/100 99 
please explain..
Thank you ..
:)

1/[n*(n+1)] can be written as [1/n  1/(n+1)]
i.e. 1/(1*2) = 1  1/2 1/(2*3) = 1/2  1/3 . . . 1/(99*100) = 1/99  1/100 
In reply to this post by duck
1/[n*(n+1)] can be written as [1/n  1/(n+1)]
i.e. 1/(1*2) = 1  1/2 1/(2*3) = 1/2  1/3 . . . 1/(99*100) = 1/99  1/100 On 5 April 2010 00:30, Nidhi Jain [via Discussion forum] <[hidden email]> wrote: please explain.. 
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Very well done Uday.
Lets do the next two then: The function f(x) = x(square_root(x) + square_root(x + 9)) is (a) continuously differentiable at x = 0, (b) continuous but not differentiable at x = 0, (c) differentiable but the derivative is not continuous at x = 0, (d) not differentiable at x = 0. Note: square_root(g(x)) stands for square root of g(x) Consider a GP series whose first term is 1 and the common ratio is a positive integer r(> 1). Consider an AP series whose first term is 1 and whose (r+2)th term coincides with the third term of the GP series. Then the common difference of the AP series is (a) r − 1, (b) r, (c) r + 1, (d) r + 2. 
Thank you.. :)
...
:)

Ans 2>>> the common difference of the AP series is (r1)
Ans1>> i think it is continously differentiable at x=0 ... i m not sure about dis answer.. ....
:)

In reply to this post by Amit Goyal
1) (d) not differentiable at x=0 because left hand limit does not exists so also not continuous
On 4 April 2010 20:41, Amit Goyal [via Discussion forum] <[hidden email]> wrote: Very well done Uday. 
1) (d) not differentiable at x=0 because left hand limit does not exists so also not continuous

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Definition: Let f be defined (and real valued) on [a, b]. For any x in [a, b] form the quotient g(t) = (f(t)  f(x))/(t  x) (a < t < b, t ≠ x) and define f'(x) = lim (as t → x) g(t) is the derivative of f at x. In particular at end ponits a and b, the derivative if it exists, is a righthand or lefthand derivative, respectively.
Now try this question again. And Nidhi, answer you provided to the problem on APGP is correct 
the answer to that que is>> continuously differentiable at x=0.. :)
Thank you sir.. :) ..
:)

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Good. Lets do the next two.
The first three terms of the binomial expansion ((1 + x) to the power n) are 1,−9, 297/8 respectively. What is the value of n? (a) 5 (b) 8 (c) 10 (d) 12 Given log(p)x = a and log(q)x = b, the value of log(p/q)x equals (a) ab/(b  a) (b) (b  a)/ab (c) (a  b)/ab (d) ab/(a  b) log(p)x stands for log of x with base p. And similarly read log(q)x & log(p/q)x. 
This post was updated on .
:)

Q1>> its 12
binomial expansion for (1+x)^n = 1+nx+{n (n1)/2}*x^2 + ........ => first term = 1 , second term = nx and third term = {n (n1)/2}*x^2 As per given, nx= 9 ..(I) {n (n1)/2}*x^2= 297/8 ..(II) Now, {n (n1)/2}*x^2 can be written as nx (nxx)/2 ....(III) We know, nx= = 9 (from I) substituting in (III) , we have>> 9 (9x)  = 297/8 2 => x= 3/4 => n = 12
:)

In reply to this post by Amit Goyal
2)Ans.(a)
log(p/q)x = log(x)/log(p/q) = log(x)/[log(p)log(q)] = 1/[log(x)plog(x)q] = 1/[(1/a)(1/b)] = ab/(ba) 
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Perfect answers Uday and Nidhi. Keep it up.

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Next three then:
Let P = {1, 2, 3, 4, 5} and Q = {1, 2}. The total number of subsets X of P such that X ∩ Q = {2} is (a) 6, (b) 7, (c) 8, (d) 9. An unbiased coin is tossed until a head appears. The expected number of tosses required is (a) 1, (b) 2, (c) 4, (d) ∞ Let X be a random variable with probability density function f(x) = c/sq(x) if x ≥ c, 0 if x < c. Then Expectation of X is (a) 0 (b) ∞ (c) 1/c (d) 1/sq(c) Note: sq(x) stands for square of x. 
for Q1 ans is (b) i.e. 7

In reply to this post by Amit Goyal
Q2 >>> its 2
Let "x" indicate the number of tosses of the coin The number of tosses of the coin would be * 1 if a head appears on the 1st throw * 2 if a tail appears on the 1st throw and a head appears on the 2nd throw * 3 if a tail appears on the 1st 2nd throws and a head appears on the 3rd throw * ... ... ... and so on.. We know, In a single throw with a coin, Probability of: P(H) = 1/2 = P(T) Now, probablity distribution would be x P(X=x) x*P 1 1/2 1/2 2 ( 1/2 ) ^2 2 * ( 1/2 ) ^2 3 ( 1/2 ) ^3 3 * ( 1/2 ) ^3 . . . . . . . . . . . . Therefore , expected number of tosses = summation x*P = 1/2 + 2 * ( 1/2 ) ^2 + 3 * ( 1/2 ) ^3 +... solving the above >> we get expected no = 2
:)

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