Answer: Option B
:
B
Method 1- Conventional Approach
f(x) = x${}^{15}$- 2013{x${}^{14}$ - x${}^{15}$ + x${}^{12}$ - ......... + x${}^2$ - x} or
f(x) = x${}^{15}$- 2013{-x + x${}^2$ - x${}^3$ + ........ - x${}^{13}$ + x${}^{14}$}
f(x) = x${}^{15}$ - $\left(\frac{2013x(x^{14}-1)}{(x+1)}\right)$
Now, Put x = 2012
f(2012) = (2012)${}^{15}$ - $\left(\frac{2013.2012({2012}^{14}-1)}{(2012+1)}\right)$
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.