Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function defined by $$ f(x)=\left\{\begin{array}{cl}

Question

Let $f: \mathbb{R} ightarrow \mathbb{R}$ be a function defined by $$ f(x)=\left\{\begin{array}{cl} x^{2} \sin \left(\frac{\pi}{x^{2}}ight), & 	ext { if } x 
eq 0 \ 0, & 	ext { if } x=0 \end{array}ight. $$ Then which of the following statements is TRUE?\

Professor PC Thomas Classes · Answer

(d): (a) $f(x)=x^{2} \sin \left(\frac{\pi}{x^{2}} ight)$\ $f(x)=0 \Rightarrow x^{2} \sin \left(\frac{\pi}{x^{2}} ight)=0 \quad($ As $x eq 0)$\ $\Rightarrow \quad \sin \left(\frac{\pi}{x^{2}} ight)=0$\ We have, $\frac{\pi}{x^{2}}=n \pi, n \in Z$ $$ \begin{gathered} x^{2}=\frac{1}{n} \Rightarrow x=\frac{1}{\sqrt{n}} \ herefore \frac{1}{10^{10}} \leq \frac{1}{\sqrt{n}}<\infty ext { i.e. } 10^{10} \geq \sqrt{n}>0 \ \sqrt{n} \leq 10^{10} \Rightarrow n \leq 10^{20} \end{gathered} $$ So, it has finite number of solutions.\ (b)We require $$ \frac{1}{\sqrt{n}} \geq \frac{1}{\pi} \Rightarrow \sqrt{n} \leq \pi \Rightarrow n \leq \pi^{2} $$ It has at least one solution.\ (c)$\frac{1}{\sqrt{n}}<\frac{1}{10^{10}} \Rightarrow \sqrt{n}>10^{10}$ is $n>10^{20}$ Thus, the number of solutions is infinite.\ (d)$\frac{1}{\pi^{2}}<\frac{1}{\sqrt{n}}<\frac{1}{\pi} \Rightarrow \pi^{2}>\sqrt{n}>\pi$\ i.e. $\pi^{2}