(40) Find the bigger of the two numbers 100117 and 117100 (log table is not allowed).

Submission: Closed

Cracked by:

  1. Nachiket Vasant Vaidya, 12th, Sathaye College, Vile Parle, Mumbai.
  2. Vaibhav Tewari, 12th, Kendriya Vidyalaya Sector-30, Gandhinagar, Gujarat.
  3. Premraj Manohar Narkhede, 12, V G Vaze College, Mulund (E), Mumbai.
  4. Shashank Chorge, S.S. and L.S. Patkar College of Arts and Science, Goregaon, Mumbai.
  5. Swapneel Subhash Sonawane, 12th, D.G. Ruparel College, Mumbai.
  6. Sourabh Bhave

Solution Provided by Sourabh:

Take f(x) = x1/x.
Log f(x) = (1/x) log x.
f'(x)/f(x) = (1/x2) (1-log x)
f'(x) = [f(x)/x2] (1-log x)
f(x) and x2 are always positive and (1-log x) is negative for x > e.
Hence for x > e, f'(x) is negative.
Hence f(x) is monotonically decreasing for x > e.
Hence 1001/100 > 1171/117.
Hence 100117 > 117100.


(39)

A truncated cone is made of materials X and Y as shown. The radii of the two vertical surfaces are a and b (b > a). The resistivity of X is p and Y is a superconductor. Calculate the resistance between the two vertical surfaces. Assume, b - a << l.

Submission: Closed

Vinayak's solution edited by INDIANET (.pdf)

Cracked by:

  1. Vinayak Pathak, 12th Pass, Mumbai.
  2. Nachiket Vasant Vaidya, 12th, Sathaye College, Vile Parle, Mumbai.

(38) Resistances of certain volume of solution X and Y are 25 ohm and 100 ohm respectively. The two solutions are poured in a conductivity cell. What will be the resistance offered by the resulting solution assuming that the two solutions are miscible, non-reacting and there is no change is the degrees of dissociations of both X and Y after mixing.

Submission: Closed

Solution provided by Manish Sir


(37) Evaluate where [.] represents the greatest integer function and n is an integer.

Submission: Closed

Cracked by:

Rishabh Kothari, 12th, Columbas School, New Delhi.

Solution Provided by Rishabh:

We know that \([\frac {sinx}{x}]\) when x->0 is slightly less than unity because rate of change of sinx is less than that of x.

\(n.\frac {sinx}{x}\) when x->0 is slightly less than n. Therefore \([n.\frac {sinx}{x}]= n-1\).

Now, \([\frac {tanx}{x}]\) when x->0 is slightly greater than unity for similar reasons.

Therefore, \(n.\frac {tanx}{x}\) is slightly greater than n for x->0.

Hence, \([n.\frac {tanx}{x}]=n\).

Adding the two results, Answer= 2n-1.


(36)

Submission: Closed

Solution (.pdf)

Cracked by:

Royan John D'mello, St. Aloysius College, Mangalore.

Last modified: Sunday, 27 November 2011, 1:59 PM