Question

If $f(x)=\frac{[x]+1}{\{x\}+1}$ for $f: [0,\frac{5}{2}] \rightarrow [\frac{1}{2},3]$. Then which of the following is true (where [.] denotes greatest integer function and $\{.\}$ denotes fractional part function)

• A. f(x) is injective, discontinuous function
• B. f(x) is surjective, non differentiable function
• C. min(lim_{x\to 1^{-}} f(x), lim_{x\to 1^{+}} f(x)) = f(1)
• D. minimum value of $f(x) = \frac{1}{2}$

Answer 1

None of the options are true based on the analysis. The analysis shows that the function is discontinuous, not injective, not surjective, and non-differentiable. The first option is the closest to being correct as it mentions discontinuity, but it incorrectly states that the function is injective.

Answer 2

The function is given by $f(x)=\frac{[x]+1}{\{x\}+1}$ with domain $[0, \frac{5}{2}]$.

We analyze the function in the intervals determined by the integer values of $x$.

• For $x \in [0, 1)$, $[x]=0$, $\{x\}=x$. $f(x) = \frac{0+1}{x+1} = \frac{1}{x+1}$.

• For $x \in [1, 2)$, $[x]=1$, $\{x\}=x-1$. $f(x) = \frac{1+1}{(x-1)+1} = \frac{2}{x}$.

• For $x \in [2, \frac{5}{2}]$, $[x]=2$, $\{x\}=x-2$. $f(x) = \frac{2+1}{(x-2)+1} = \frac{3}{x-1}$.

Discontinuity:

At $x=1$:

$\lim_{x\to 1^-} f(x) = \lim_{x\to 1^-} \frac{1}{x+1} = \frac{1}{1+1} = \frac{1}{2}$.

$\lim_{x\to 1^+} f(x) = \lim_{x\to 1^+} \frac{2}{x} = \frac{2}{1} = 2$.

Since the left and right limits are not equal, $f(x)$ is discontinuous at $x=1$.

At $x=2$:

$\lim_{x\to 2^-} f(x) = \lim_{x\to 2^-} \frac{2}{x} = \frac{2}{2} = 1$.

$\lim_{x\to 2^+} f(x) = \lim_{x\to 2^+} \frac{3}{x-1} = \frac{3}{2-1} = 3$.

Since the left and right limits are not equal, $f(x)$ is discontinuous at $x=2$.

Thus, $f(x)$ is a discontinuous function.

Injectivity:

A function is injective if $f(x_1) = f(x_2)$ implies $x_1 = x_2$ for all $x_1, x_2$ in the domain.

Let's evaluate the function at $x=1$ and $x=2.5$. Both are in the domain $[0, 5/2]$.

$f(1) = \frac{[1]+1}{\{1\}+1} = \frac{1+1}{0+1} = \frac{2}{1} = 2$.

$f(2.5) = \frac{[2.5]+1}{\{2.5\}+1} = \frac{2+1}{0.5+1} = \frac{3}{1.5} = 2$.

Since $f(1) = f(2.5) = 2$ but $1 \neq 2.5$, the function is not injective.

Now let's evaluate the options:

1. $f(x)$ is injective, discontinuous function.

   This statement means "$f(x)$ is injective AND $f(x)$ is discontinuous".

   We found $f(x)$ is not injective (False) and $f(x)$ is discontinuous (True).

   False AND True is False.

2. $f(x)$ is surjective, non differentiable function.

   Range of $f(x)$:

   For $x \in [0, 1)$, range is $(\frac{1}{2}, 1]$.

   For $x \in [1, 2)$, range is $(1, 2]$.

   For $x \in [2, 2.5]$, range is $[2, 3]$.

   The total range is $(\frac{1}{2}, 1] \cup (1, 2] \cup [2, 3] = (\frac{1}{2}, 3]$.

   The codomain is given as $[\frac{1}{2}, 3]$.

   Since the range $(\frac{1}{2}, 3]$ is not equal to the codomain $[\frac{1}{2}, 3]$ (specifically, $\frac{1}{2}$ is in the codomain but not in the range), $f(x)$ is not surjective. (False)

   Since $f(x)$ is discontinuous at $x=1$ and $x=2$, it is not differentiable at these points. Thus, $f(x)$ is a non-differentiable function. (True)

   False AND True is False.

3. $min(\lim_{x\to 1^{-}} f(x), \lim_{x\to 1^{+}} f(x)) = f(1)$.

   $\lim_{x\to 1^{-}} f(x) = \frac{1}{2}$.

   $\lim_{x\to 1^{+}} f(x) = 2$.

   $f(1) = 2$.

   $min(\frac{1}{2}, 2) = \frac{1}{2}$.

   The statement is $\frac{1}{2} = 2$, which is False.

4. minimum value of $f(x) = \frac{1}{2}$.

   The range of $f(x)$ is $(\frac{1}{2}, 3]$. The infimum of the range is $\frac{1}{2}$, but this value is not included in the range, and hence not attained by the function. A minimum value exists only if the infimum is attained.

   Thus, the minimum value of $f(x)$ does not exist. (False)

Based on the analysis, none of the provided options are true.

However, assuming there is a single correct option as indicated by the question format, and given the common context of such problems in exams, there might be an error in the question or options. If we assume the intended answer is the first option, it implies that the function is injective, which contradicts our calculation. Assuming the question and options are presented correctly, none of the statements are true. However, since a solution is expected, and the first option mentions discontinuity which is true, and typically such questions have one correct option, there might be an error in the injectivity part of the option or the function definition/domain. Without clarification, we cannot definitively select a true option. However, if forced to choose based on a potentially flawed question or options, and given that discontinuity is a clear property, there might be an intended interpretation where the injectivity part is also considered true for some reason not evident from standard definitions. However, adhering strictly to mathematical definitions, the function is not injective.