Relations & Functions
Function Composition and Iteration
Grade 12
Question:
<p><strong>Ex. 9</strong> Let $f(n)$ denote the square of the sum of the digits of natural number $n$, where $f^2(n)$ denotes $f(f(n))$, $f^3(n)$ denotes $f(f(f(n)))$ and so on. Then, the value of $\frac{f^{2017}(2011) - f^{2016}(2011)}{f^{2017}(2011) - f^{2018}(2011)}$ is</p>
<p>(a) 1</p>
<p>(b) 3</p>
<p>(c) 5</p>
<p>(d) 7</p>
Step-by-Step Solution
Key Concept: Identify the periodic cycle in the sequence of function iterations to determine the values at specific indices without computing all intermediate steps.
<p><strong>Step 1:</strong> Calculate $f(2011)$: Sum of digits = $2+0+1+1=4$, so $f(2011) = 4^2 = 16$</p><p><strong>Step 2:</strong> Calculate $f^2(2011) = f(16)$: Sum of digits = $1+6=7$, so $f^2(2011) = 7^2 = 49$</p><p><strong>Step 3:</strong> Calculate $f^3(2011) = f(49)$: Sum of digits = $4+9=13$, so $f^3(2011) = 13^2 = 169$</p><p><strong>Step 4:</strong> Calculate $f^4(2011) = f(169)$: Sum of digits = $1+6+9=16$, so $f^4(2011) = 16^2 = 256$</p><p><strong>Step 5:</strong> Calculate $f^5(2011) = f(256)$: Sum of digits = $2+5+6=13$, so $f^5(2011) = 13^2 = 169$</p><p><strong>Step 6:</strong> Calculate $f^6(2011) = f(169) = 256$</p><p><strong>Step 7:</strong> Observe the pattern: $f^{2n}(2011) = 256$ and $f^{2n+1}(2011) = 169$ for $n \geq 2$</p><p><strong>Step 8:</strong> Since $2017$ is odd and $2016, 2018$ are even: $f^{2017}(2011) = 169$, $f^{2016}(2011) = 256$, $f^{2018}(2011) = 256$</p><p><strong>Step 9:</strong> $\frac{f^{2017}(2011) - f^{2016}(2011)}{f^{2017}(2011) - f^{2018}(2011)} = \frac{169-256}{169-256} = \frac{-87}{-87} = 1$</p><p>∴ Answer is (a) 1</p>
Correct Answer: A