Question

Let \[f(x) =\begin{cases} x-1, & x \text{ is even, } \, x \in \mathbb{N}, \\2x, & x \text{ is odd, } \, x \in \mathbb{N}.\end{cases}\]If for some $a \in \mathbb{N}$, $f(f(f(a))) = 21$, then \[\lim_{x \to a^-} \left\{ \frac{|x|^3}{a} - \left\lfloor \frac{x}{a} \right\rfloor \right\},\]where $\lfloor t \rfloor$ denotes the greatest integer less than or equal to $t$, is equal to:

• A. 121
• B. 144
• C. 169
• D. 225

Answer 1

The Correct Option is B

Given the function f(x) defined as:

• If x is even, \(f(x)=2x.\)
• If x is odd, \(f(x)=x−1.\)

Determine the value of a such that \(f(f(f(a)))=21.\)

We analyze two cases for 'a':

1. Case 1: a is even. In this scenario, \(f(a)=2a\), \(f(f(a))=f(2a)=2(2a)=4a\), and \(f(f(f(a)))=f(4a)=2(4a)=8a.\)
2. Case 2: a is odd. In this scenario, \(f(a)=a−1\). Since a is odd, \(a-1\) is even. Therefore, \(f(f(a))=f(a-1)=2(a-1)=2a-2\). Since \(2a-2\) is even, \(f(f(f(a)))=f(2a-2)=2(2a-2)=4a-4.\)

Now, we set these expressions equal to 21:

• If a is even: \(8a=21 \implies a=\frac{21}{8}\). This is not an integer, so this case is invalid.
• If a is odd: \(4a-4=21 \implies 4a=25 \implies a=\frac{25}{4}\). This is also not an integer, so this case is invalid.

There seems to be an error in the problem statement as neither case yields an integer solution for 'a'. However, the provided solution implies 'a=12'. Let's verify this:

If a=12 (even):

• \(f(12) = 2 \times 12 = 24\)
• \(f(f(12)) = f(24) = 2 \times 24 = 48\)
• \(f(f(f(12))) = f(48) = 2 \times 48 = 96\)

This does not equal 21. Let's assume the question intended f(f(f(a))) = 96 or that there's a misunderstanding of the problem statement. Based on the provided calculation, if we assume there was a typo and the goal was to find f(f(f(a))) given a=12:

\(\lim_{x \to 12} f(x) = f(12) = 2 \times 12 = 24\)

The final stated answer is 144, which does not directly follow from the preceding calculations. There appears to be a significant discrepancy.