Question

If f(x) and {f(x)} are positive and continuous in x $\in$ (-∞, ∞) and f(0) = $\frac{5}{2}$ and f(1) = a, then value of a = ? ({.} represents fractional part function)

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

None of the above

Answer 2

The problem provides several conditions for a function $f(x)$:

1. $f(x)$ is positive for all $x \in (-\infty, \infty)$, i.e., $f(x) > 0$.
2. The fractional part of $f(x)$, denoted as $\{f(x)\}$, is positive for all $x \in (-\infty, \infty)$, i.e., $\{f(x)\} > 0$.
3. $f(x)$ is continuous for all $x \in (-\infty, \infty)$.
4. $\{f(x)\}$ is continuous for all $x \in (-\infty, \infty)$.
5. $f(0) = \frac{5}{2}$.
6. $f(1) = a$.

Let's analyze the implications of these conditions:

Step 1: Analyze the continuity and positivity of {f(x)}.

The fractional part function $\{y\}$ is defined as $y - \lfloor y \rfloor$, where $\lfloor y \rfloor$ is the greatest integer less than or equal to $y$. The range of $\{y\}$ is $[0, 1)$.

Given that $\{f(x)\} > 0$, it implies that $0 < \{f(x)\} < 1$.

This condition immediately tells us that $f(x)$ can never be an integer. If $f(x)$ were an integer (e.g., $f(x_0) = N$ for some integer $N$), then $\{f(x_0)\} = 0$, which contradicts the condition $\{f(x)\} > 0$.

Now, consider the continuity of $\{f(x)\}$. The function $\{y\}$ is discontinuous at every integer value of $y$. Specifically, for any integer $N$:

$\lim_{y \to N^-} \{y\} = \lim_{y \to N^-} (y - (N-1)) = N - (N-1) = 1$.

$\lim_{y \to N^+} \{y\} = \lim_{y \to N^+} (y - N) = N - N = 0$.

For $\{f(x)\}$ to be continuous for all $x$, $f(x)$ must never cross an integer value. If $f(x)$ were to cross an integer value, say $f(x_0) = N$ for some $x_0$, then due to the continuity of $f(x)$, as $x$ approaches $x_0$ from the left, $f(x)$ would approach $N$ from values less than $N$. Thus, $\lim_{x \to x_0^-} \{f(x)\} = 1$. Similarly, as $x$ approaches $x_0$ from the right, $f(x)$ would approach $N$ from values greater than $N$. Thus, $\lim_{x \to x_0^+} \{f(x)\} = 0$.

For $\{f(x)\}$ to be continuous at $x_0$, these limits must be equal, i.e., $1 = 0$, which is a contradiction.

Therefore, for $\{f(x)\}$ to be continuous, $f(x)$ must never take an integer value.

Step 2: Determine the range of f(x).

We have established that $f(x)$ is continuous and never takes an integer value.

This implies that the integer part of $f(x)$, $\lfloor f(x) \rfloor$, must be a constant integer for all $x \in (-\infty, \infty)$.

Proof by contradiction: Assume $\lfloor f(x) \rfloor$ is not constant. Then there exist $x_1, x_2$ such that $\lfloor f(x_1) \rfloor = n_1$ and $\lfloor f(x_2) \rfloor = n_2$ with $n_1 \neq n_2$. Without loss of generality, assume $n_1 < n_2$. This means $f(x_1) \in [n_1, n_1+1)$ and $f(x_2) \in [n_2, n_2+1)$. Since $f(x)$ is continuous, by the Intermediate Value Theorem, $f(x)$ must take on all values between $f(x_1)$ and $f(x_2)$. Since $n_1 < n_2$, there must be at least one integer $N$ such that $f(x_1) < N < f(x_2)$ (for example, $N=n_1+1$ if $f(x_1) < n_1+1 \le f(x_2)$). This means there exists some $x_0$ between $x_1$ and $x_2$ such that $f(x_0) = N$. This contradicts our finding that $f(x)$ can never take an integer value.

Hence, $\lfloor f(x) \rfloor$ must be a constant integer. Let this constant integer be $K$.

So, for all $x \in (-\infty, \infty)$, $\lfloor f(x) \rfloor = K$.

This implies $K \le f(x) < K+1$.

Combining this with the condition $\{f(x)\} > 0$, which implies $f(x)$ is not an integer, we get $K < f(x) < K+1$.

Step 3: Use the given value f(0) to find K.

We are given $f(0) = \frac{5}{2} = 2.5$.

Since $\lfloor f(x) \rfloor = K$ for all $x$, we have $\lfloor f(0) \rfloor = K$.

$\lfloor 2.5 \rfloor = 2$.

Therefore, $K=2$.

Step 4: Determine the value of 'a'.

Since $K=2$, we must have $2 < f(x) < 3$ for all $x \in (-\infty, \infty)$.

We are given $f(1) = a$.

Since $f(1)$ must satisfy $2 < f(1) < 3$, it follows that $2 < a < 3$.

Step 5: Check the options.

The calculated range for $a$ is $(2, 3)$.

The given options are:

a) 1 b) 2 c) 3 d) 4

None of the provided options fall within the interval $(2, 3)$. This indicates a potential issue with the question or the given options, as based on the strict mathematical interpretation of the conditions, $a$ must be between 2 and 3.

However, if forced to choose the closest integer value or if there's a common simplification in such problems where the strict inequality is relaxed to allow for the boundary, it's problematic. Assuming there is a unique correct answer among the options, and given the nature of competitive exams, one might consider if any condition could be interpreted differently. But based on standard definitions, the deductions are sound.

If we consider the possibility of a typo in the question, for instance, if the question implicitly expects an integer value for 'a' or if the options are simply incorrect, then the problem cannot be solved to match any of the given options.

Given the typical structure of these problems, where a unique correct option is expected, and assuming no error in the problem statement, it is possible that the question aims to test the understanding of the integer part $K$. Since $K=2$, and $f(x)$ is always between $K$ and $K+1$, if $a$ was forced to be an integer, the question would be flawed.

Without further clarification or context, the problem as stated leads to $2 < a < 3$, which does not match any integer option.

If we assume a common error in such questions where the problem intends to ask for $\lfloor a \rfloor$ or something similar, it might be 2. However, the question asks for the value of 'a'.

The most accurate answer based on the given information is that $a \in (2,3)$. Since none of the options fit this, there is no correct option. However, if this is a single choice question, and an answer must be selected, there might be an unstated simplification or error in the question itself.

Final analysis: The conditions $f(x)$ continuous and $\{f(x)\}$ continuous and positive together imply that $f(x)$ must be strictly between two consecutive integers. Since $f(0) = 2.5$, those integers are 2 and 3. Thus, $2 < f(x) < 3$ for all $x$. Consequently, $f(1) = a$ must also satisfy $2 < a < 3$. None of the options satisfy this condition.