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JEE Advanced 2024Paper 2
Exam
JEE Advanced
Year
2024
Paper
Paper 2
Subject
Mathematics
Chapter
Trigonometric Equations

JEE Advanced 2024 · Paper 2 · Mathematics · Trigonometric Equations

Let f: ℝ arrow ℝ be a function defined by f(x)=\ array cc x^2...

Let f:R→Rf: \mathbb{R} \rightarrow \mathbb{R} be a function defined by f(x)=\left\{x2sin⁡(πx2), if x≠00, if x=0\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?

Options

  1. Af(x)=0f(x)=0 has infinitely many solutions in the interval \left[\frac{1}{10^{10}}, \infty\right).
  2. Bf(x)=0hasnosolutionsintheinterval[1π,∞)has no solutions in the interval \left[\frac{1}{\pi}, \infty\right).
  3. CThe set of solutions of f(x)=0f(x)=0 in the interval (0,11010)\left(0, \frac{1}{10^{10}}\right) is finite
  4. Df(x)=0f(x)=0 has more than 25 solutions in the interval \left(\frac{1}{\pi^2}, \frac{1}{\pi}\right).

Correct answer

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

Step-by-step solution

Option-1 : f(x)=x2sin⁡πx2\mathrm{f}(\mathrm{x})=\mathrm{x}^2 \sin \frac{\pi}{\mathrm{x}^2} f(x)=0⇒sin⁡πx2=0⇒πx2=nπ,n∈Nf(x)=0 \Rightarrow \sin \frac{\pi}{x^2}=0 \Rightarrow \frac{\pi}{x^2}=n \pi, n \in N x2=1n⇒x=1n\mathrm{x}^2=\frac{1}{\mathrm{n}} \Rightarrow \mathrm{x}=\frac{1}{\sqrt{\mathrm{n}}} 1n≥11010⇒1010≥n⇒n≤1020\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 : x=1n⇒1n>1π⇒π>n⇒n<π2\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 :x=1n,1n<11010⇒n>1010⇒n>1020: \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 : 1π2<1n<1π⇒n∈(π,π2)⇒n∈(π2,π4)\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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