Sample Questions of ISI ME I (Mathematics) 2010 Discussion

[Amit Goyal]

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

[N Uday Kumar]

99

100(1-1/2+1/2-1/3+1/3-1/4+.....1/99-1/100)
100(1-1/100)
100*99/100
99

[duck]

please explain..

Thank you

..

[N Uday Kumar]

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

[N Uday]

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..

Thank you

..

Nidhi Jain

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[Amit Goyal]

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.

[duck]

Thank you.. :)

...

[duck]

Ans 2>>> the common difference of the AP series is  (r-1)

Ans1>> i think it is continously differentiable at x=0 ... i m not sure about dis answer..

....

[N Uday]

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.
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.

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[N Uday Kumar]

1) (d) not differentiable at x=0 because left hand limit does not exists so also not continuous

[Amit Goyal]

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 right-hand or left-hand derivative, respectively.
Now try this question again.
And Nidhi, answer you provided to the problem on AP-GP is correct

[duck]

the answer to that que is>> continuously differentiable at x=0.. :)

Thank you sir.. :)

..

[Amit Goyal]

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.

[duck]

[duck]

Q1>> its 12

binomial expansion for (1+x)^n = 1+nx+{n (n-1)/2}*x^2 + ........

=> first term = 1 ,
 second term = nx and
 third term = {n (n-1)/2}*x^2

As per given,
nx= -9                                    ..(I)  
{n (n-1)/2}*x^2= 297/8            ..(II)

Now, {n (n-1)/2}*x^2 can be written as nx (nx-x)/2     ....(III)

We know, nx= = -9 (from I)
substituting in (III) , we have>>

-9 (-9-x)
---------   = 297/8
    2

=> x= -3/4

=> n = 12    

                                                       

[N Uday Kumar]

2)Ans.(a)
 log(p/q)x = log(x)/log(p/q) = log(x)/[log(p)-log(q)] = 1/[log(x)p-log(x)q] = 1/[(1/a)-(1/b)] = ab/(b-a)

[Amit Goyal]

Perfect answers Uday and Nidhi. Keep it up.

[Amit Goyal]

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.

[Shweta Singhal]

for Q1 ans is (b) i.e. 7

[duck]

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