Question

Let \(f:\R→\R\) be a function defined by
\(f(x) = \begin{cases}     x^2\sin(\frac{\pi}{x^2}), & \text{if } x \ne 0, \\     0, & \text{if } x = 0. \end{cases}\)
Then which of the following statements is TRUE ?

• A. f(x) = 0 has infinitely many solutions in the interval \([\frac{1}{10^{10}},\infin)\)
• B. f(x) = 0 has no solutions in the interval \([\frac{1}{\pi},\infin).\)
• C. The set of solutions of f(x) = 0 in the interval \((0,\frac{1}{10^{10}})\) is finite.
• D. f(x) = 0 has more than 25 solutions in the interval \((\frac{1}{\pi^2},\frac{1}{\pi})\)

Answer 1

The Correct Option is D

1. Given Function:
We are given the following function $ f(x) $, which is defined as:

$f(x) = \begin{cases} x^2 \sin\left(\frac{\pi}{x^2}\right), & \text{if } x \neq 0, \\ 0, & \text{if } x = 0. \end{cases}$

2. Analyzing $f(x) = 0$:
We are given that $ f(x) = 0 $, which implies:

$ \sin\left( \frac{\pi}{x^2} \right) = 0 $

3. Solving for $x$:
The sine function equals zero when its argument is a multiple of $\pi$, so we get:

$ \frac{\pi}{x^2} = n\pi $

4. Solving for $x^2$:
Simplifying the equation:

$ x^2 = \frac{1}{n} $

5. Solving for $x$:
Taking the square root of both sides:

$ x = \frac{1}{\sqrt{n}} $

6. Interpreting Values of $n$:
Now, we consider different intervals for $x$:

If $ x \in \left[ \frac{1}{10^{10}}, \infty \right) $:
We get:

$ \frac{1}{\sqrt{n}} \in \left[ 10^{10}, \infty \right) \Rightarrow n \in (0, (10^{10})^2) $

So, the finite values of $n$ are $ n = 1, 2, 3, \dots, 9 $.

If $ x \in \left[ 1, \pi \right) $:
We get:

$ \sqrt{n} \in \left( 0, \pi^2 \right) \Rightarrow n \in (0, \pi^2) $

So, $ n \in (0, \pi^2) $.

If $ x \in \left( 0, \frac{1}{10^{10}} \right) $:
We get:

$ \sqrt{n} \in (10^{10}, \infty) \Rightarrow n \text{ is infinite.} $

If $ x \in \left[ \frac{1}{\pi^2}, \frac{1}{\pi} \right) $:
We get:

$ \sqrt{n} \in (\pi, \pi^2) \Rightarrow n \in (\pi^2, \pi^4) $

There are more than 25 solutions for this case.

Final Answer:
The values of $n$ are finite in the first two cases, infinite in the third case, and there are more than 25 solutions in the fourth case.

Answer 2

To solve the problem, we analyze the solutions of the equation \(f(x) = 0\) for the function:

\[ f(x) = \begin{cases} x^2 \sin\left(\frac{\pi}{x^2}\right), & \text{if } x \neq 0, \\ 0, & \text{if } x = 0. \end{cases} \]

1. Condition for \(f(x) = 0\) when \(x \neq 0\):
\[ x^2 \sin\left(\frac{\pi}{x^2}\right) = 0 \implies \sin\left(\frac{\pi}{x^2}\right) = 0 \] \[ \Rightarrow \frac{\pi}{x^2} = n\pi, \quad n \in \mathbb{Z} \] \[ \Rightarrow \frac{1}{x^2} = n \implies x = \pm \frac{1}{\sqrt{n}}, \quad n \in \mathbb{N} \]

2. Solutions of \(f(x) = 0\) are at:
\[ x = \pm \frac{1}{\sqrt{n}}, \quad n=1,2,3,\dots \] Considering only positive \(x\), solutions are:
\[ x_n = \frac{1}{\sqrt{n}} \]

3. Analyze solutions in given intervals:

• Interval \([\frac{1}{10^{10}}, \infty)\):
  Since \(x_n = \frac{1}{\sqrt{n}}\), for \(x_n \geq \frac{1}{10^{10}}\), we have:
  \[ \frac{1}{\sqrt{n}} \geq \frac{1}{10^{10}} \implies \sqrt{n} \leq 10^{10} \implies n \leq 10^{20} \] There are infinitely many integers \(n \leq 10^{20}\), so infinitely many solutions in this interval.
  Statement 1 is FALSE.
• Interval \([\frac{1}{\pi}, \infty)\):
  Check solutions \(x_n = \frac{1}{\sqrt{n}} \geq \frac{1}{\pi}\), so:
  \[ \frac{1}{\sqrt{n}} \geq \frac{1}{\pi} \implies \sqrt{n} \leq \pi \implies n \leq \pi^2 \approx 9.87 \] So \(n = 1,2,...,9\) give solutions, which means solutions exist.
  Statement 2 is FALSE.
• Interval \((0, \frac{1}{10^{10}})\):
  Number of solutions \(x_n = \frac{1}{\sqrt{n}} < \frac{1}{10^{10}}\) means:
  \[ \frac{1}{\sqrt{n}} < \frac{1}{10^{10}} \implies \sqrt{n} > 10^{10} \implies n > 10^{20} \] Since \(n\) goes to infinity, infinitely many solutions in this interval.
  Statement 3 is FALSE (solutions are infinite, not finite).
• Interval \((\frac{1}{\pi^2}, \frac{1}{\pi})\):
  Number of solutions \(x_n = \frac{1}{\sqrt{n}}\) between \(\frac{1}{\pi^2}\) and \(\frac{1}{\pi}\) means:
  \[ \frac{1}{\pi^2} < \frac{1}{\sqrt{n}} < \frac{1}{\pi} \] Invert inequalities: \[ \pi^2 > \sqrt{n} > \pi \] Square all parts: \[ \pi^4 > n > \pi^2 \] \[ n \in ( \pi^2, \pi^4 ) \] Since \(\pi^2 \approx 9.87\), \(\pi^4 \approx 97.4\), the number of integers \(n\) in this range is about \(97 - 9 = 88\), which is more than 25.
  Statement 4 is TRUE.

Final Answer:
Statements 4 are TRUE.