Problems 36-40
(40) Find the bigger of the two numbers 100117 and 117100 (log table is not allowed).
Submission: Closed
Cracked by:
- Nachiket Vasant Vaidya, 12th, Sathaye College, Vile Parle, Mumbai.
- Vaibhav Tewari, 12th, Kendriya Vidyalaya Sector-30, Gandhinagar, Gujarat.
- Premraj Manohar Narkhede, 12, V G Vaze College, Mulund (E), Mumbai.
- Shashank Chorge, S.S. and L.S. Patkar College of Arts and Science, Goregaon, Mumbai.
- Swapneel Subhash Sonawane, 12th, D.G. Ruparel College, Mumbai.
- 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:
- Vinayak Pathak, 12th Pass, Mumbai.
- 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
Cracked by:
Royan John D'mello, St. Aloysius College, Mangalore.