Question

Let $f:R \rightarrow R$ be defined as $f(x) =$ $\begin{cases} \lim_{n\rightarrow \infty} (\frac{[x]}{1+n^2} + \frac{3[x]}{2+n^2} + \frac{5[x]}{3+n^2} + .... + \frac{(2n-1)[x]}{n+n^2}), x \neq \frac{\pi}{2} \\ 1, x = \frac{\pi}{2} \end{cases}$

where $[y]$ denotes largest integer $\leq y$, then which of the following statement(s) is(are) correct?

• A. $f(x)$ is injective but not surjective.
• B. $f(x)$ is non-differentiable at $x = \frac{\pi}{2}$.
• C. $f(x)$ is discontinuous at all integers and continuous at $x = \frac{\pi}{2}$.

Answer 1

Answer: (C)

Only Statement 3 is correct.

Answer 2

Solution:

Given

$$f(x) = \begin{cases} \lim_{n\rightarrow \infty} \left(\frac{[x]}{1+n^2} + \frac{3[x]}{2+n^2} + \cdots + \frac{(2n-1)[x]}{n+n^2}\right), & x\neq \frac{\pi}{2} \\[1mm] 1, & x = \frac{\pi}{2} \end{cases}$$

where $[x]$ is the floor function.

1. Simplification of Limit:

   Factor $[x]$ out:

   $$f(x) = [x] \cdot \lim_{n\to\infty} \sum_{k=1}^{n} \frac{2k-1}{n^2+k}$$

   Note that:

   $$\sum_{k=1}^{n} (2k-1) = n^2.$$

   For large $n$, since $n^2+k \approx n^2$ (as $k$ is much smaller than $n^2$), we have

   $$\sum_{k=1}^{n} \frac{2k-1}{n^2+k} \approx \frac{1}{n^2} \cdot \sum_{k=1}^{n} (2k-1) = \frac{n^2}{n^2} = 1.$$

   Therefore, for $x\neq \frac{\pi}{2}$:

   $$f(x) = [x].$$

   And note that for $x=\frac{\pi}{2}$, $f(x)=1$ which is consistent since

   $$\left[\frac{\pi}{2}\right] = [1.57] = 1.$$

   Thus, overall, $f(x)= [x]$ for all $x$.

2. Verification of Statements:

   • Statement 1: $f(x)$ is injective but not surjective.

     The floor function $[x]$ maps any $x$ in an interval $[n, n+1)$ to the same integer $n$. Hence it is not injective. Though its range is $\mathbb{Z}$ (a proper subset of $\mathbb{R}$), it is not surjective onto $\mathbb{R}$; however, the lack of injectivity makes the statement false.

   • Statement 2: $f(x)$ is non-differentiable at $x = \frac{\pi}{2}$.

     For $x = \frac{\pi}{2}$ (approximately 1.57), $f(x)= [x]=1$ is constant in the surrounding open interval (1,2). Therefore, $f(x)$ is differentiable (derivative = 0) at $x = \frac{\pi}{2}$. This statement is false.

   • Statement 3: $f(x)$ is discontinuous at all integers and continuous at $x = \frac{\pi}{2}$.

     The floor function is discontinuous at each integer (jump discontinuities) and continuous on intervals between integers. Since $\frac{\pi}{2}$ lies in (1,2), it is a point of continuity. This statement is true.

Thus, only Statement 3 is correct.

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Explanation (Minimal):

1. The limit simplifies to 1 so that $f(x)= [x]$ for all $x$.
2. Floor function $[x]$ is not injective and is discontinuous at all integers.
3. Since $\frac{\pi}{2}$ is non-integer, $f(x)= [x]$ is continuous there.
4. Hence, only Statement 3 is valid.