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Functions — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question
Let f(n)f(n) denote the square of the sum of the digits of natural number nn, where f2(n)=f(f(n))f^2(n) = f(f(n)), f3(n)=f(f(f(n)))f^3(n) = f(f(f(n))), and so on. Then the value of f2011(2011)−f2010(2011)f2013(2011)−f2012(2011)\dfrac{f^{2011}(2011)-f^{2010}(2011)}{f^{2013}(2011)-f^{2012}(2011)}.
Solution
Answer: 1
Step 1: Compute the iteration sequence
f(2011)=(2+0+1+1)2=16f(2011) = (2+0+1+1)^2 = 16
f2(2011)=f(16)=72=49f^2(2011) = f(16) = 7^2 = 49
f3(2011)=f(49)=132=169f^3(2011) = f(49) = 13^2 = 169
f4(2011)=f(169)=162=256f^4(2011) = f(169) = 16^2 = 256
f5(2011)=f(256)=132=169f^5(2011) = f(256) = 13^2 = 169
 Step 2: Identify the cycle For n≥3n \geq 3: fn(2011)=169f^n(2011) = 169 if nn is odd, 256256 if nn is even. Step 3: Evaluate the expression
f2011(2011)−f2010(2011)f2013(2011)−f2012(2011)=169−256169−256=−87−87=1\frac{f^{2011}(2011)-f^{2010}(2011)}{f^{2013}(2011)-f^{2012}(2011)} = \frac{169-256}{169-256} = \frac{-87}{-87} = 1
 Answer: 1
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