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A function f(x) is defined as:
f(x) = x15- 2013x14 + 2013x13- 2013x12 + .... - 2013x2 + 2013x. Find the value of f(2012).


Options:
A .   2011
B .   2012
C .   2013
D .   2014
Answer: Option B
:
B

Method 1- Conventional Approach
f(x) = x15- 2013{x14 - x15 + x12 - ......... + x2 - x} or
f(x) = x15- 2013{-x + x2 - x3 + ........ - x13 + x14
f(x) = x15 - (2013x(x141)(x+1))
Now, Put x = 2012
f(2012) = (2012)15 - (2013.2012(2012141)(2012+1))
f(2012) = (2012)15 - 2012(2012^{14} - 1) 
f(2012) = 2012.
Method 2- Assumption
Assume 2012 = n and 2013 = (n+ 1). Put x=1. Now replace the options in terms of "n" as
a) (n-1)           b) n               c) (n + 1)                 d) (n + 2) 
f(1)=1 - 2 + 2 - 2 + 2 - 2 + 2 - ......................-2 + 2 
f(1) = 1. Options (a),(c) & (d) can be eliminated as only option (b) gives us 1. 



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