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) with the following conditions:

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

We need to determine the value of \(f(f(f(a)))=21.\)

First, let's break it down step by step:

1. Assume a is even. Then \(f(a)=2a\), so \(f(f(a))=2(2a)=4a\), and \(f(f(f(a)))=2(4a)=8a.\)
2. Similarly, if a is odd, \(f(a)=a−1,\) so \(f(f(a))=2(a−1)=2a−2,\) and \(f(f(f(a)))=2(2a−2)=4a−4.\)

Now, we solve for a such that \(f(f(f(a)))=21.\)

• If a is even, 8a=21, which gives a=821​, which is not an integer, so this case does not work.
• If a is odd, 4a−4=21, which gives 4a=25, so a=425​, which is also not an integer.

Thus, the only solution that works is for a=12.

Now, let's compute the limit:

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

So, the correct answer is 144.