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(D) Given $f(x) = x^2 \sin \left(\frac{\pi}{x^2}ight)$ for $x 
eq 0$ and $f(0) = 0$.To find solutions of $f(x) = 0$,we set $x^2 \sin \left(\frac{\p

Vedclass · Accepted Answer

$f(x) = 0$ has more than $25$ solutions in the interval $\left(\frac{1}{\pi^2}, \frac{1}{\pi}\right)$.

Vedclass · Answer

(D) Given $f(x) = x^2 \sin \left(\frac{\pi}{x^2} ight)$ for $x eq 0$ and $f(0) = 0$.To find solutions of $f(x) = 0$,we set $x^2 \sin \left(\frac{\pi}{x^2} ight) = 0$.This implies $x = 0$ or $\sin \left(\frac{\pi}{x^2} ight) = 0$.Thus,$\frac{\pi}{x^2} = n\pi$ for $n \in \mathbb{N}$,which gives $x^2 = \frac{1}{n}$,or $x = \pm \frac{1}{\sqrt{n}}$.Checking Option-$A$: For $x \in \left[\frac{1}{10^{10}}, \infty ight)$,we have $\frac{1}{\sqrt{n}} \geq \frac{1}{10^{10}} \implies \sqrt{n} \leq 10^{10} \implies n \leq 10^{20}$. This is a finite number of solutions.Checking Option-$B$: For $x \in \left(\frac{1}{\pi}, \infty ight)$,we have $\frac{1}{\sqrt{n}} > \frac{1}{\pi} \implies \sqrt{n} Checking Option-$C$: For $x \in \left(0, \frac{1}{10^{10}} ight)$,we have $0 10^{10} \implies n > 10^{20}$. There are infinitely many such $n$,so the set of solutions is infinite.Checking Option-$D$: For $x \in \left(\frac{1}{\pi^2}, \frac{1}{\pi} ight)$,we have $\frac{1}{\pi^2} < \frac{1}{\sqrt{n}} < \frac{1}{\pi} \implies \pi < \sqrt{n} < \pi^2 \implies \pi^2 < n < \pi^4$. Since $\pi^2 \approx 9.86$ and $\pi^4 \approx 97.4$,$n$ can take values from $10$ to $97$. The number of solutions is $97 - 10 + 1 = 88$,which is greater than $25$. Thus,Option-$D$ is $TRUE$.

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

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