JEE Advanced · Mathematics · 8. Trigonometric Equations
Let \(f: \mathbb{R} \rightarrow \mathbb{R}\) be a function defined by \(f(x)=\left\{\begin{array}{cc}x^2 \sin \left(\frac{\pi}{x^2}\right), & \text { if } x \neq 0 \\ 0, & \text { if } x=0\end{array}\right.\) Then which of the following statements is TRUE?
- A \(f(x)=0\) has infinitely many solutions in the interval \(\left[\frac{1}{10^{10}}, \infty\right)\).
- B \(f(x)=0\) has no solutions in the interval \(\left[\frac{1}{\pi}, \infty\right)\).
- C The set of solutions of \(f(x)=0\) in the interval \(\left(0, \frac{1}{10^{10}}\right)\) is finite
- D \(f(x)=0\) has more than 25 solutions in the interval \(\left(\frac{1}{\pi^2}, \frac{1}{\pi}\right)\).
Answer & Solution
Correct Answer
(D) \(f(x)=0\) has more than 25 solutions in the interval \(\left(\frac{1}{\pi^2}, \frac{1}{\pi}\right)\).
Step-by-step Solution
Detailed explanation
Option-1 : \(\mathrm{f}(\mathrm{x})=\mathrm{x}^2 \sin \frac{\pi}{\mathrm{x}^2}\)
\(f(x)=0 \Rightarrow \sin \frac{\pi}{x^2}=0 \Rightarrow \frac{\pi}{x^2}=n \pi, n \in N\)
\(\mathrm{x}^2=\frac{1}{\mathrm{n}} \Rightarrow \mathrm{x}=\frac{1}{\sqrt{\mathrm{n}}}\)
\(\frac{1}{\sqrt{\mathrm{n}}} \geq \frac{1}{10^{10}} \Rightarrow 10^{10} \geq \sqrt{\mathrm{n}} \Rightarrow \mathrm{n} \leq 10^{20}\), finite number of solutions
Option-2 : \(\mathrm{x}=\frac{1}{\sqrt{\mathrm{n}}} \Rightarrow \frac{1}{\sqrt{\mathrm{n}}}>\frac{1}{\pi} \Rightarrow \pi>\sqrt{\mathrm{n}} \Rightarrow \mathrm{n} < \pi^2\), Number of solutions is 9
Option-3 \(: \mathrm{x}=\frac{1}{\sqrt{\mathrm{n}}}, \frac{1}{\sqrt{\mathrm{n}}} < \frac{1}{10^{10}} \Rightarrow \sqrt{\mathrm{n}}>10^{10} \Rightarrow \mathrm{n}>10^{20}\), Infinite number of solutions
Option-4 : \(\frac{1}{\pi^2} < \frac{1}{\sqrt{\mathrm{n}}} < \frac{1}{\pi} \Rightarrow \sqrt{\mathrm{n}} \in\left(\pi, \pi^2\right) \Rightarrow \mathrm{n} \in\left(\pi^2, \pi^4\right)\), Definitely more than 25 solutions
\(f(x)=0 \Rightarrow \sin \frac{\pi}{x^2}=0 \Rightarrow \frac{\pi}{x^2}=n \pi, n \in N\)
\(\mathrm{x}^2=\frac{1}{\mathrm{n}} \Rightarrow \mathrm{x}=\frac{1}{\sqrt{\mathrm{n}}}\)
\(\frac{1}{\sqrt{\mathrm{n}}} \geq \frac{1}{10^{10}} \Rightarrow 10^{10} \geq \sqrt{\mathrm{n}} \Rightarrow \mathrm{n} \leq 10^{20}\), finite number of solutions
Option-2 : \(\mathrm{x}=\frac{1}{\sqrt{\mathrm{n}}} \Rightarrow \frac{1}{\sqrt{\mathrm{n}}}>\frac{1}{\pi} \Rightarrow \pi>\sqrt{\mathrm{n}} \Rightarrow \mathrm{n} < \pi^2\), Number of solutions is 9
Option-3 \(: \mathrm{x}=\frac{1}{\sqrt{\mathrm{n}}}, \frac{1}{\sqrt{\mathrm{n}}} < \frac{1}{10^{10}} \Rightarrow \sqrt{\mathrm{n}}>10^{10} \Rightarrow \mathrm{n}>10^{20}\), Infinite number of solutions
Option-4 : \(\frac{1}{\pi^2} < \frac{1}{\sqrt{\mathrm{n}}} < \frac{1}{\pi} \Rightarrow \sqrt{\mathrm{n}} \in\left(\pi, \pi^2\right) \Rightarrow \mathrm{n} \in\left(\pi^2, \pi^4\right)\), Definitely more than 25 solutions
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